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9200	https://ocw.mit.edu/courses/18-s997-the-polynomial-method-fall-2012/resources/lecture-notes/	Browse Course Material Course Info Instructor Prof. Lawrence D Guth Departments As Taught In Fall 2012 Level Topics Mathematics Algebra and Number Theory Discrete Mathematics Learning Resource Types assignment Problem Sets notes Lecture Notes Download Course search GIVE NOW about ocw help & faqs contact us 18.S997 \| Fall 2012 \| Graduate The Polynomial Method Lecture Notes pdf 128 kB 18.S997 Fall 2012 The Polynomial Method: The Regulus Detection Lemma pdf 148 kB 18.S997 Fall 2012 The Polynomial Method: A Version of the Joints Theorem for Long Thin Tubes pdf 125 kB 18.S997 Fall 2012 The Polynomial Method: An Application to Incidence Geometry pdf 132 kB 18.S997 Fall 2012 The Polynomial Method: Background on Connections Between Analysis and Combinatorics (Loomis-Whitney) pdf 155 kB 18.S997 Fall 2012 The Polynomial Method: Besictovitch's Construction pdf 130 kB 18.S997 Fall 2012 The Polynomial Method: Bezout Theorem pdf 251 kB 18.S997 Fall 2012 The Polynomial Method: Crossing Numbers and Distance Problems pdf 135 kB 18.S997 Fall 2012 The Polynomial Method: Crossing Numbers and Distinct Distances pdf 128 kB 18.S997 Fall 2012 The Polynomial Method: Crossing Numbers and the Szemeredi-Trotter Theorem pdf 136 kB 18.S997 Fall 2012 The Polynomial Method: Degree Reduction pdf 139 kB 18.S997 Fall 2012 The Polynomial Method: Detection Lemmas and Projection Theory pdf 121 kB 18.S997 Fall 2012 The Polynomial Method: Hardy-Littlewood-Sobolev Inequality pdf 213 kB 18.S997 Fall 2012 The Polynomial Method: Incidence Bounds in Three Dimensions pdf 122 kB 18.S997 Fall 2012 The Polynomial Method: Integer Polynomials That Vanish at Rational Points pdf 148 kB 18.S997 Fall 2012 The Polynomial Method: Introduction pdf 243 kB 18.S997 Fall 2012 The Polynomial Method: Introduction to Incidence Geometry pdf 125 kB 18.S997 Fall 2012 The Polynomial Method: Introduction to the Cellular Method pdf 118 kB 18.S997 Fall 2012 The Polynomial Method: Introduction to Thue's Theorem on Diophantine Approximation pdf 109 kB 18.S997 Fall 2012 The Polynomial Method: Local to Global Arguments pdf 138 kB 18.S997 Fall 2012 The Polynomial Method: Oscillating Integrals and Besicovitch's Arrangement of Tubes pdf 124 kB 18.S997 Fall 2012 The Polynomial Method: Polynomial Cell Decompositions pdf 153 kB 18.S997 Fall 2012 The Polynomial Method: Reguli; The Zarankiewicz Problem pdf 232 kB 18.S997 Fall 2012 The Polynomial Method: Special Points and Lines of Algebraic Surfaces pdf 121 kB 18.S997 Fall 2012 The Polynomial Method: Taking Stock pdf 121 kB 18.S997 Fall 2012 The Polynomial Method: The Berlekamp-Welch Algorithm pdf 183 kB 18.S997 Fall 2012 The Polynomial Method: The Elekes-Sharir Approach to the Distinct Distance Problem pdf 116 kB 18.S997 Fall 2012 The Polynomial Method: The Finite Field Nikodym and Kakeya Theorems pdf 110 kB 18.S997 Fall 2012 The Polynomial Method: The Joints Problem pdf 108 kB 18.S997 Fall 2012 The Polynomial Method: The Kakeya Problem pdf 128 kB 18.S997 Fall 2012 The Polynomial Method: Thue's Proof (Part II): Polynomials of Two Variables pdf 119 kB 18.S997 Fall 2012 The Polynomial Method: Thue's Proof (Part III) pdf 113 kB 18.S997 Fall 2012 The Polynomial Method: Using Cell Decompositions pdf 132 kB 18.S997 Fall 2012 The Polynomial Method: What's Special About Polynomials? (A Geometric Perspective) pdf 127 kB 18.S997 Fall 2012 The Polynomial Method: Why Polynomials? (Part 1) Course Info Instructor Prof. Lawrence D Guth Departments Mathematics As Taught In Fall 2012 Level Graduate Topics Mathematics Algebra and Number Theory Discrete Mathematics Learning Resource Types assignment Problem Sets notes Lecture Notes Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
9201	https://fiveable.me/key-terms/introduction-industrial-engineering/non-negativity-constraints	printables 🏭intro to industrial engineering review key term - Non-negativity constraints Citation: MLA Definition Non-negativity constraints are restrictions in mathematical models that ensure variables can only take on values that are zero or positive. This is crucial in linear programming, as it reflects real-world scenarios where negative quantities, such as resources or production levels, are not feasible. These constraints help to accurately represent problems in various fields like economics, logistics, and manufacturing. 5 Must Know Facts For Your Next Test Non-negativity constraints are typically denoted as x \geq 0 for each variable x in the problem. These constraints are essential for ensuring that solutions to linear programming problems are practical and applicable to real-world situations. In graphical solutions of linear programming, non-negativity constraints limit the feasible region to the first quadrant of the Cartesian plane. When formulating linear programming problems, omitting non-negativity constraints can lead to nonsensical or infeasible solutions. Non-negativity constraints can be combined with other types of constraints to create a complete model for optimization problems. Review Questions How do non-negativity constraints influence the graphical representation of a linear programming problem? Non-negativity constraints significantly shape the graphical representation of a linear programming problem by restricting the feasible region to the first quadrant of the Cartesian plane. This means that all possible solutions must lie in this area where both variables are zero or positive. As a result, any point in other quadrants would not be viable solutions because they would represent negative quantities that cannot occur in practical scenarios. Discuss the implications of neglecting non-negativity constraints when solving linear programming problems. Neglecting non-negativity constraints can lead to unrealistic or nonsensical solutions in linear programming. For instance, allowing variables to take on negative values may suggest producing a negative amount of goods or using negative resources, which is not feasible in real-life applications. This oversight could result in misleading conclusions and ineffective decision-making. Therefore, including these constraints is crucial for obtaining meaningful and applicable results from optimization models. Evaluate how non-negativity constraints impact decision-making processes in operational settings. Non-negativity constraints are vital in decision-making processes because they ensure that outcomes reflect realistic scenarios within operational settings. By restricting variable values to zero or positive numbers, managers can confidently interpret results related to resource allocation, production levels, and cost minimization strategies. These constraints help align mathematical models with practical limitations, leading to better-informed decisions that can enhance efficiency and effectiveness in business operations. Related terms Linear Programming: A mathematical technique used for optimization where a linear objective function is maximized or minimized subject to linear equality and inequality constraints. Feasible Region: The set of all possible points that satisfy the constraints of a linear programming problem, including non-negativity constraints. Objective Function: The function that needs to be maximized or minimized in a linear programming problem, often representing profit, cost, or resource allocation.
9202	https://web.stanford.edu/~boyd/papers/pdf/gossip.pdf	2508 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 Randomized Gossip Algorithms Stephen Boyd, Fellow, IEEE, Arpita Ghosh, Student Member, IEEE, Balaji Prabhakar, Member, IEEE, and Devavrat Shah Abstract—Motivated by applications to sensor, peer-to-peer, and ad hoc networks, we study distributed algorithms, also known as gossip algorithms, for exchanging information and for computing in an arbitrarily connected network of nodes. The topology of such networks changes continuously as new nodes join and old nodes leave the network. Algorithms for such networks need to be ro-bust against changes in topology. Additionally, nodes in sensor net-works operate under limited computational, communication, and energy resources. These constraints have motivated the design of “gossip” algorithms: schemes which distribute the computational burden and in which a node communicates with a randomly chosen neighbor. We analyze the averaging problem under the gossip constraint for an arbitrary network graph, and ﬁnd that the averaging time of a gossip algorithm depends on the second largest eigenvalue of a doubly stochastic matrix characterizing the algorithm. Designing the fastest gossip algorithm corresponds to minimizing this eigen-value, which is a semideﬁnite program (SDP). In general, SDPs cannot be solved in a distributed fashion; however, exploiting problem structure, we propose a distributed subgradient method that solves the optimization problem over the network. The relation of averaging time to the second largest eigenvalue naturally relates it to the mixing time of a random walk with tran-sition probabilities derived from the gossip algorithm. We use this connection to study the performance and scaling of gossip algo-rithms on two popular networks: Wireless Sensor Networks, which are modeled as Geometric Random Graphs, and the Internet graph under the so-called Preferential Connectivity (PC) model. Index Terms—Distributed averaging, gossip, random walk, scaling laws, sensor networks, semideﬁnite programming. I. INTRODUCTION T HE advent of sensor, wireless ad hoc and peer-to-peer networks has necessitated the design of distributed and fault-tolerant computation and information exchange algo-rithms. This is mainly because such networks are constrained by the following operational characteristics: i) they may not have a centralized entity for facilitating computation, commu-nication, and time-synchronization, ii) the network topology may not be completely known to the nodes of the network, iii) nodes may join or leave the network (even expire), so that the network topology itself may change, and iv) in the case of Manuscript received March 13, 2005; revised November 11, 2005. This work is supported in part by a Stanford Graduate Fellowship, and by C2S2, the MARCO Focus Center for Circuit and System Solution, under MARCO Contract 2003-CT-888. S. Boyd, A. Ghosh, and B. Prabhakar are with the Information Systems Laboratory, Department of Electrical Engineering, Stanford University, Stan-ford, CA 94305 USA (e-mail: fboyd@stanford.edu; arpitag@stanford.edu; balajig@stanford.edu). D. Shah is with the LIDS, Departments of Elecrtrical Engineering and Com-puter Science, and ESD, the Massachusetts Institute of Technology, Cambridge, MA 02138 USA (e-mail: devavrat@mit.edu). Communicated by M. Méderad, Guest Editor. Digital Object Identiﬁer 10.1109/TIT.2006.874516 Fig. 1. Sensor nodes deployed to measure ambient temperature. sensor networks, the computational power and energy resources may be very limited. These constraints motivate the design of simple decentralized algorithms for computation where each node exchanges information with only a few of its immediate neighbors in a time instance (or, a round). The goal in this set-ting is to design algorithms so that the desired computation and communication is done as quickly and efﬁciently as possible. We study the problem of averaging as an instance of the dis-tributed computation problem.1 A toy example to motivate the averaging problem is sensing the temperature of some small region of space using a network of sensors. For example, in Fig. 1, sensors are deployed to measure the temperature of a source. Sensor , , measures , where the are independent and identically distributed (i.i.d.), zero-mean Gaussian sensor noise variables. The unbiased, min-imum mean-squared error (MMSE) estimate is the average . Thus, to combat minor ﬂuctuations in the ambient tem-perature and the noise in sensor readings, the nodes need to av-erage their readings. The problem of distributed averaging on a network comes up in many applications such as coordination of autonomous agents, estimation, and distributed data fusion on ad hoc net-works, and decentralized optimization.2 For one of the earliest references on distributed averaging on a network, see . Fast distributed averaging algorithms are also important in other con-texts; see Kempe et al. , for example. For an extensive body of related work, see , , , , , , , , , , , . This paper undertakes an in-depth study of the design and analysis of gossip algorithms for averaging in an arbitrarily connected network of nodes. (By a gossip algorithm, we mean speciﬁcally an algorithm in which each node communicates 1Preliminary versions of this paper appeared in –. 2The theoretical framework developed in this paper is not restricted merely to averaging algorithms. It easily extends to the computation of other functions which can be computed via pairwise operations; e.g., the maximum, minimum, or product functions. It can also be extended for analyzing information exchange algorithms, although this extension is not as direct. For concreteness and for stating our results as precisely as possible, we shall consider averaging algo-rithms in the rest of the paper. 0018-9448/$20.00 © 2006 IEEE BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2509 with no more than one neighbor in each time slot.) Given a graph , we determine the averaging time , which is the time taken for the value at each node to be close to the average value (a more precise deﬁnition is given later). We ﬁnd that the averaging time depends on the second largest eigenvalue of a doubly stochastic matrix characterizing the averaging algorithm: the smaller this eigenvalue, the faster the averaging algorithm. The fastest averaging algorithm is obtained by mini-mizing this eigenvalue over the set of allowed gossip algorithms on the graph. This minimization is shown to be a semideﬁnite program (SDP), which is a convex problem, and therefore can be solved efﬁciently to obtain the global optimum. The averaging time is closely related to the mixing time of the random walk deﬁned by the matrix that charac-terizes the algorithm. This means we can also study averaging algorithms by studying the mixing time of the corresponding random walk on the graph. The recent work of Boyd et al. shows that the ratio of the mixing times of the natural random walk to the fastest mixing random walk can grow without bound as the number of nodes increases; correspondingly, therefore, the optimal averaging algorithm can perform arbitrarily better than the one based on the natural random walk. Thus, computing the optimal averaging algorithm is important: however, this in-volves solving a SDP, which requires a knowledge of the com-plete network topology. Surprisingly, we ﬁnd that we can exploit problem structure to devise a distributed subgradient method to solve the SDP and obtain a near-optimal averaging algorithm, with only local communication. Finally, we study the performance of gossip algorithms on two network graphs which are very important in practice: Geo-metric Random Graphs, which are used to model wireless sensor networks, and the Internet graph under the Preferential Connec-tivity model. We ﬁnd that for geometric random graphs, the av-eraging time of the natural and the optimal averaging algorithms are of the same order. As remarked earlier, this need not be the case in a general graph. We shall state our main results after setting out some notation and deﬁnitions in Section I. A. Problem Formulation and Deﬁnitions Consider a connected graph , where the vertex set contains nodes and is the edge set. The th component of the vector represents the initial value at node . Let be the average of the entries of . Our goal is to compute in a distributed manner. • Asynchronous time model: Each node has a clock which ticks at the times of a rate Poisson process. Thus, the inter-tick times at each node are rate exponentials, in-dependent across nodes and over time. Equivalently, this corresponds to a single clock ticking according to a rate Poisson process at times , , where are i.i.d. exponentials of rate . Let denote the node whose clock ticked at time . Clearly, the are i.i.d. variables distributed uniformly over . We discretize time according to clock ticks since these are the only times at which the value of changes. Therefore, the interval denotes the th time-slot and, on average, there are clock ticks per unit of absolute time. Lemma 1 states a precise translation of clock ticks into absolute time. • Synchronous time model: In the synchronous time model, time is assumed to be slotted commonly across nodes. In each time slot, each node contacts one of its neighbors independently and (not necessarily uniformly) at random. Note that in this model all nodes communicate simultaneously, in contrast to the asynchronous model where only one node communicates at a given time. On the other hand, in both models each node contacts only one other node at a time. Previous work, notably that of , , considers the synchronous time model. The qualitative and quantitative conclusions are unaffected by the type of model; we start with the asynchronous time model for convenience, and then analyze the synchronous model and show that the same kind of results hold in this case as well. • Algorithm : We consider a particular class of time-invariant gossip algorithms, denoted by . An algorithm in this class is characterized by an matrix of nonnegative entries with the condition that only if . For technical reasons, we assume that is a stochastic matrix with its largest eigenvalue equal to , and all remaining eigenvalues strictly less than in magnitude. (Such a matrix can always be found if the underlying graph is connected and nonbipartite; we will assume that the network graph satisﬁes these conditions for the remainder of the paper.) Depending on the time model, two types of algorithms arise: 1) asynchronous, and 2) synchronous. Next, we describe the asynchronous algorithm associated with to explain the role of the ma-trix in the algorithm. As we shall see, asynchronous algorithms are rather intuitive and easy to explain. We defer the description of the synchronous algorithm to Sec-tion III-C. The asynchronous algorithm associated with , de-noted by , is described as follows: In the th time slot, let node ’s clock tick and let it contact some neigh-boring node with probability . At this time, both nodes set their values equal to the average of their cur-rent values. Formally, let denote the vector of values at the end of the time slot . Then (1) where with probability (the probability that the th node’s clock ticks is , and the probability that it con-tacts node is ) the random matrix is (2) where is an unit vector with the th component equal to . 2510 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 • Quantity of Interest: Our interest is in determining the (absolute) time it takes for to converge to , where is the vector of all ones. Deﬁnition 1: For any , the -averaging time of an algorithm is denoted by , and is deﬁned as (3) where denotes the norm of the vector . Thus, the -averaging time is the smallest time it takes for to get within of with high probability, regardless of the initial value . The following lemma relates the number of clock ticks to absolute time. This relation allows us to use clock ticks instead of absolute time when we deal with asynchronous algorithms. Lemma 1: For any , . Further, for any (4) Proof: By deﬁnition Equation (4) follows directly from Cramer’s theorem (see [10, pp. 30 and 35]). As a consequence of Lemma 1, for with high probability (i.e., probability at least ). In this paper, all -averaging times are at least . Hence, dividing the quantities measured in terms of the number of clock ticks by gives the corresponding quantities when measured in absolute time (for an example, see Corollary 2). B. Previous Results A general lower bound for any graph and any averaging algorithm was obtained in in the synchronous setting. Their result is as follows. Theorem 1: For any gossip algorithm on any graph and for , the -averaging time (in synchronous steps) is lower-bounded by . The recent work studies the gossip-constrained aver-aging problem for the special case of the complete graph. A randomized gossiping algorithm is proposed which is shown to converge to the vector of averages on the complete graph. For a synchronous averaging algorithm, obtain the following re-sult. Theorem 2: For a complete graph, there exists a gossip al-gorithm such that the -averaging time of the algorithm is . In Section III-C, we obtain a synchronous averaging algo-rithm which is simpler than the one described in , with -av-eraging time for the complete graph (from Corollary 3). The problem of fast distributed averaging without the gossip constraint on an arbitrary graph is studied in ; here, the ma-trices are constant, i.e., for all . It is shown that the problem of ﬁnding the (constant) that converges fastest to (where is the matrix of all ones) can be written as a SDP (under a symmetry constraint), and can there-fore be solved numerically. Distributed averaging has also been studied in the context of distributed load balancing (), where nodes (processors) ex-change tokens in order to uniformly distribute tokens over all the processors in the network (the number of tokens is constrained to be integral, so exact averaging is not possible). An analysis based on Markov chains is used to obtain bounds on the time re-quired to achieve averaging up to a certain accuracy. However, each iteration is governed either by a constant stochastic ma-trix, or a ﬁxed sequence of matchings is considered. This differs from our work (in addition to the integral constraint) in that we consider an arbitrary sequence drawn i.i.d. from some dis-tribution, and try to characterize the properties the distribution must possess for convergence. Some other results on distributed averaging can be found in , , , , . An interesting result regarding products of random matrices is found in . The authors prove the following result on a sequence of iterations , where the belong to a ﬁnite set of paracontracting matrices (i.e., ). If is the set of matrices that appear inﬁnitely often in the sequence , and for , denotes the eigenspace of associated with eigenvalue , then the sequence of vectors has a limit in . This result can be used to ﬁnd conditions for convergence of distributed averaging algorithms. Not much is known about good randomized gossip algorithms for averaging on arbitrary graphs. The algorithm of is quite dependent on the fact that the underlying graph is a complete graph, and the general result of is a nonconstructive lower bound. C. Our Results In this paper, we design and characterize the performance of averaging algorithms for arbitrary graphs for both the asyn-chronous and synchronous time models. The following result characterizes the averaging time of asynchronous algorithms. Theorem 3: The averaging time of the asyn-chronous algorithm (in terms of number of clock ticks) is bounded as follows: and (5) (6) BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2511 where (7) and is the diagonal matrix with entries Theorem 3 is proved in Section III, using results on conver-gence of moments that we derive in Section II. For synchronous algorithms, the averaging time is charac-terized by Theorems 4 and 5, which are stated and proved in Section III-C. As the reader may notice, the statements of The-orem 3 and Theorems 4–5 are qualitatively the same. The above tight characterization of the averaging time leads us to the formulation of the question of the fastest averaging algorithm. In Section IV, we show that the problem of ﬁnding the fastest averaging algorithm can be formulated as an SDP. In general, it is not possible to solve an SDP in a distributed fashion. However, we exploit the structure of the problem to propose a completely distributed algorithm, based on a subgra-dient method, that solves the optimization problem on the net-work. The algorithm and proof of convergence are found in Sec-tion IV-A. Section V relates the averaging time of an algorithm on a graph with the mixing time of an associated random walk on . This is used in Section VI to study applications of our re-sults in the context of two networks of practical interest: wire-less networks and the Internet. The result for wireless networks involves bounding the mixing times of the natural and optimal random walks on the geometric random graph; these results are derived in Section VI-A. Finally, we conclude in Section VII. II. CONVERGENCE OF MOMENTS In this section, we will study the convergence of randomized gossip algorithms. We will not restrict ourselves here to any par-ticular algorithm; but rather consider convergence of the itera-tion governed by a product of random matrices, each of which satisﬁes certain (gossip-based) constraints described below. The vector of estimates is updated as where each must satisfy the following constraints im-posed by the gossip criterion and the graph topology. If nodes and are not connected by an edge, then must be zero. Further, since every node can communicate with only one of its neighbors per time slot, each column of can have only one nonzero entry other than the diagonal entry. The iteration intends to compute the average, and therefore must preserve sums: this means that , where denotes the vector of all ones. Also, the vector of averages must be a ﬁxed point of the iteration, i.e., . We will consider matrices drawn i.i.d. from some dis-tribution on the set of nonnegative matrices satisfying the above constraints, and investigate the behavior of the estimate If must converge to the vector of averages for every initial condition , we must have (8) A. Convergence in Expectation Let the mean of the (i.i.d.) matrices be denoted by . We have (9) so converges in expectation to if . The conditions on for this to happen are stated in ; they are (10) (11) (12) where is the spectral radius of a matrix. The ﬁrst two con-ditions will be automatically satisﬁed by , since it is the ex-pected value of matrices each of which satisﬁes this property. Therefore, if we pick any distribution on the whose mean satisﬁes (12), the sequence of estimates will converge in ex-pected value to the vector of averages. In fact, if is invertible, by considering the martingale , we can obtain almost sure convergence of to . However, neither result tells us the rate at which converges to . B. Convergence of Second Moment To obtain the rate of convergence of to , we will in-vestigate the rate at which the error converges to . Consider the evolution of (13) Here follows from the fact that is an eigenvector for all . Thus, evolves according to the same linear system as . Therefore, we can write (14) 2512 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 Since is doubly stochastic, so is , and there-fore, is doubly stochastic. Since the matrices are identically distributed we will shorten to . Since , and is the eigenvector corresponding to the largest eigenvalue of (15) Repeatedly conditioning and using (15), we ﬁnally obtain the bound (16) From this, we see that the second moment of the error converges to at a rate governed by . This means that any scheme of choosing the which corresponds to a with second largest eigenvalue strictly less than (and, of course, with less than ) provably converges in the second moment. This condition is only a sufﬁcient condition for the conver-gence of the second moment. We can, in fact, obtain a necessary and sufﬁcient condition by considering the evolution of rather than . Since , . Let . Then i.e., evolves according to a (random) linear system. Now (17) (18) Collect the entries of the matrix into a vector , with entries drawn columnwise from . Then, using (18), we see that where stands for Kronecker product. Conditioning repeat-edly, we see that (19) Since each has with corresponding eigenvector , each also has , with eigenvector . Also, each is orthogonal to , since since . Therefore, the convergence of is governed by , where . If then , and therefore converges to zero. Note that converges to the zero matrix if and only if converges to . If , then each , which means that as well. Conversely, suppose . Then each as well. From the Cauchy–Schwartz inequality so that each entry in the matrix converges to . Thus, a necessary and sufﬁcient condition for second moment convergence is that . However, despite having an exact criterion for convergence of the second moment, we will use in our analysis. This is be-cause is much easier to evaluate for a given algorithm than the expected value of the Kronecker product . III. HIGH PROBABILITY BOUNDS ON AVERAGING TIME We prove an upper bound (5) and a lower bound (6) in Lemmas 2 and 3 on the discrete time (or equivalently, number of clock ticks) required to get within of (analogous to (5) and (6)) for the asynchronous averaging algorithm. A. Upper Bound Lemma 2: For algorithm , for any initial vector , for where (20) Proof: Recall that under algorithm (21) where, with probability , the random matrix is (22) Note that are doubly stochastic matrices for all . That is, for all (23) BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2513 Given our assumptions on the matrix of transition probabilities , we can conclude from the previous section that . We want to ﬁnd out how fast converges; in partic-ular, we want to obtain probabilistic bounds on . For this, we will use the second moment of to apply Markov’s inequality as below. Computing : Let denote the expected value of (which is the same as ) (24) Then, the entries of are as follows: 1) for , and 2) . This yields the deﬁned in (7), that is, (25) where is the diagonal matrix with en-tries Note that if , then is doubly stochastic. This im-plies that , which in turn means that . Computing the second moment : With probability , the edge is chosen to average, that is, . Then (26) (27) (28) It is not an accident that : each is a projection matrix, which projects a vector onto the sub-space . The entries of except the and th stay un-changed, and and average their values. Since every pro-jection matrix satisﬁes , and are symmetric, we have . Since (28) holds for each instance of the random matrix , we have (29) Note that this means that is symmetric3 positive-semidef-inite (since ) and hence it has nonnegative real eigenvalues. 3The symmetry of W does not depend on P being symmetric. From (16) and (29) (30) Now, (31) Application of Markov’s inequality: From (30), (31), and an application of Markov’s inequality, we have (32) From (32), it follows that for (33) This proves the lemma, and gives us an upper bound on the -averaging time. B. A Lower Bound on the Averaging Time Here, we will prove a lower bound for the -averaging time, which is only a factor of away from the upper bound in the previous section. We have the following result. Lemma 3: For algorithm , there exists an initial vector , such that for where (34) Proof: Since , we obtain from (29) (35) By deﬁnition, is a symmetric positive-semideﬁnite doubly stochastic matrix with nonnegative real eigenvalues and corresponding orthonormal eigenvectors . Select For this choice of , . Now from (35) (36) 2514 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 For this particular choice of , we will lower-bound the -averaging time by lower-bounding , and using Lemma 4 as stated below. By Jensen’s inequality and (36) (37) Lemma 4: Let be a random variable such that . Then, for any Proof: Rearranging terms gives us the lemma. From (36), . Hence, Lemma 4 and (37) imply that for (38) This completes the proof of Lemma 3. Combining the results in the previous two lemmas, we have the result of Theorem 3. The following corollaries are immediate. Corollary 1: For large and symmetric , is bounded as follows: and (39) (40) Proof: By deﬁnition, . For large , is small, and hence, This along with Theorem 3 completes the proof. Corollary 2: For a symmetric , the absolute time it takes for clock ticks to happen is given by (41) with probability at least . Proof: For and and using (39), the right-hand side of (4) evaluates to Since for the nonnegative doubly sto-chastic symmetric matrix , is larger than the above choice of . This completes the proof. Note that the proof of Lemma 3 uses only two features of the algorithm : • is symmetric, which allows us to choose an orthonormal set of eigenvectors; • is positive semideﬁnite, which means that the conver-gence of to is governed by . Consider any randomized gossip algorithm with symmetric expectation matrix (and, of course, satisfying the gossip constraints stated in Section II). For such an algorithm, the rate of convergence of to is governed by , the second largest eigenvalue in absolute value, rather than . Exactly the same proof can be used to derive a lower bound for this gossip algorithm, with the only difference being that is replaced by . Thus, we can state the following lower bound for the performance of an arbitrary randomized gossip algorithm with symmetric . Lemma 5: For any randomized gossip algorithm with sym-metric expectation , there exists an initial vector , such that for where (42) The proof of the upper bound relies on more speciﬁc prop-erties of the algorithm , and thus cannot be duplicated for an arbitrary algorithm. Note also that while the expressions for the lower bounds for our algorithm , and an arbitrary algorithm with symmetric expectation are very similar, this does not mean that has the same lower bound as any other randomized gossip algorithm with symmetric expectation: the lower bound depends on the value of , and the set of matrices that can be for some instance of the algorithm is a subset of the set of all doubly stochastic symmetric matrices. C. Synchronous Averaging Algorithms In this subsection, we consider the case of synchronous averaging algorithms. Unlike the asynchronous case, in the synchronous setting, multiple node pairs communicate at exactly the same time. Gossip constraints require that these simultaneously active node pairs are disjoint. That is, the edges of the network graph corresponding to the pair-wise operations form a (not necessarily complete) matching. This makes the BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2515 synchronous case harder to deal with, as it requires the algo-rithm to form a matching in a distributed manner. We ﬁrst present a centralized synchronous gossip algo-rithm that achieves the same performance as the asynchronous algorithm. This algorithm requires a centralized entity to choose matchings of the nodes each time. Then, we present a completely distributed synchronous gossip algorithm that ﬁnds matchings in a distributed manner without any additional computational burden. We show that this algorithm performs as well as the centralized gossip algorithm for any graph with bounded degree. We extend this result for unbounded degree regular graphs, for example, the complete graph. 1) Centralized Synchronous Algorithm: Let be any doubly-stochastic symmetric matrix corresponding to the prob-ability matrix of the algorithm, as before. By Birkhoff–Von Neu-mann’s theorem , a nonnegative doubly-stochastic matrix can be decomposed into permutation matrices (equivalently matchings) as Deﬁne a (matrix) random variable with distribution , . The centralized synchronous algorithm corresponding to is as follows: in each time step, choose one of the permutations (matchings) in an i.i.d. fashion with distribution identical to . Note that the permutation need not be symmetric. The update corresponding to a permutation is as follows: if , then node averages its current value with the value it receives from node . Now, we state the theorem that characterizes the averaging time of this algorithm. Theorem 4: The averaging time of the centralized syn-chronous algorithm described above is given by where . Proof: The proof of Theorem 4 is based on the proofs of Lemmas 2 and 3 presented in Section III. Let denote the random permutation matrix chosen by the algorithm at time . The linear iteration corresponding to this update is , where is given by (43) Now (44) Now since since ( is a permutation matrix). Therefore, (45) (46) Using the arguments of Lemmas 2 and 3, exactly as in the asyn-chronous case, it can be easily shown that for any averaging al-gorithm (47) From (44) and (46) Further, all eigenvalues of are nonnegative. Hence, (48) From (47) and (48), the statement of Theorem 4 follows. 2) Distributed Synchronous Algorithm: The centralized synchronous algorithm needs a centralized entity to select a permutation matrix (or matching) at each time step, corre-sponding to the matrix . Here we describe a way to obtain such a permutation matrix in a distributed manner for bounded degree network graphs. Later we extend this result for un-bounded degree regular graphs for a particular class of (corresponding to the natural random walk). Given a network graph , let be the maximum node de-gree. We assume that all nodes know (a justiﬁcation for this assumption is given at the end of the proof of Theorem 5). Now we describe the algorithm based on as follows. In each time step, every node becomes active with probability independently. Consider an active node . Let be its de-gree (i.e., the number of its neighbors). Active node contacts at most one of its neighbors to average, as follows. With proba-bility , node does nothing, i.e., it does not contact any neighbor. With equal probabilities , it chooses one of its neighbors to contact. All active nodes ignore the nodes that contact them. An inac-tive node, say , ignores the requests of active nodes if contacted by more than one active node. If active node contacts inactive node but no other active node contacts , then and average their values with probability , where We state the following result for this algorithm. Theorem 5: The averaging time of the distributed syn-chronous algorithm described above is given by 2516 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 where , with and a diagonal matrix with . Before we prove Theorem 5, note that for bounded , is a constant away from . Hence, Thus, the averaging time of the distributed synchronous algo-rithm is of the same order as that of the centralized synchronous algorithm for any bounded degree graph. Proof of Theorem 5: The proof follows using Theorem 4. We ﬁrst note that the algorithm, as described above, only allows pair-wise averaging for distinct node pairs. Let be the random matrix corresponding to the algorithm at time , that is, Since averages values of distinct node pairs, it is a sym-metric projection matrix, projecting onto the intersection of the subspaces where is an averaging pair. Therefore, for all , and therefore . Using this property, as argued in Theorem 4, the aver-aging time is bounded as (49) Next, we evaluate . First we compute the prob-ability that node pair average. Denote this probability . We claim that where . The reason is as follows: average when a) is active, is inactive, contacts but no other node contacts , and they decide to average; b) is active, is inactive, contacts but no other node contacts , and they decide to average. We compute the probability of a): is active and is inactive with probability ; contacts with probability ; no other node contacts with probability ; after which average with probability . Since all these events are independent, the probability of a) turns out to be . Similarly, the probability of event b) is . Since events a) and b) are disjoint, the net probability of averaging is as claimed. Now, it is easy to see that (50) where is the diagonal matrix deﬁned in the statement of the theorem. From the argument preceding (49), we have that , so that all eigenvalues of are nonnega-tive. Hence from (50), the statement of Theorem 5 follows. Note. The assumption of nodes knowing is not restrictive for the following reason: all nodes can compute the maximum node degree via a gossip algorithm in which each node con-tacts its neighbors in a round-robin fashion, and informs them of its current estimate of the maximum degree (its initial esti-mate is its own degree). Since the order of pair-wise compar-isons to compute the maximum of many numbers is not impor-tant, each node can compute the maximum of the received infor-mation from other nodes in any order to update its own estimate. It is not hard to see that such an algorithm requires time for all nodes to know maximum degree, where is the diam-eter of the graph. Now, consider a node pair such that the shortest path between them is . Now consider such that and for all . Then, under any averaging algorithm, for , . Hence, for , the -averaging time is at least . Since we are considering bounded degree graphs, . Hence, we can ignore the pre-processing time for in order notation. For clean presentation of our results, we ignore this pre-processing time in general. Consider a -regular graph, where each node degree is exactly . Now, modify the above algorithm as follows: when an active node contacts an inactive node and is not contacted by any other node, then always average. The following result follows using arguments of Theorem 5. Corollary 3: The averaging time of the algorithm described above for a -regular graph is bounded as (51) where and is deﬁned as if and are neighbors otherwise. Note that As a consequence, for the complete graph, . Thus, the averaging time . For , this implies the main results of and . IV. OPTIMAL AVERAGING ALGORITHM We saw in Theorem 3 that the averaging time is a mono-tonically increasing function of the second largest eigenvalue BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2517 of . Thus, ﬁnding the fastest aver-aging algorithm corresponds to ﬁnding such that is the smallest, while satisfying constraints on . Thus, we have the optimization problem minimize subject to if (52) The objective function, which is the second largest eigenvalue of a doubly stochastic matrix, is a convex function on the set of symmetric matrices. Therefore, (52) is a convex optimization problem. This problem can be reformulated as the following SDP: minimize subject to if (53) where denotes inequality with respect to the cone of sym-metric positive semideﬁnite matrices. For general background on SDPs, eigenvalue optimization, and associated interior-point methods for solving these problems, see, for example, , , , , and references therein. Interior point methods can be used to solve problems with a thousand edges or so; sub-gradient methods can be used to solve the problem for larger graphs that have up to a hundred thousand edges. The disadvan-tage of a subgradient method compared to a primal-dual interior point method is that the algorithm is relatively slow (in terms of number of iterations), and has no simple stopping criterion that can guarantee a certain level of suboptimality. In summary, given a graph topology, we can solve the SDP (53) to ﬁnd the for the fastest averaging algorithm. A. Distributed Optimization We have seen that ﬁnding the fastest averaging algorithm is a convex optimization problem, and can therefore be solved ef-ﬁciently to obtain the optimal distribution . Unfortunately, a computed in a centralized fashion is not useful in our set-ting. It is natural to ask if in this setting, the optimization (like the averaging itself), can also be performed in a decentralized fashion. That is, is it possible for the nodes on the graph, pos-sessing only local information, and with only local communi-cation, to compute the probabilities that lead to the fastest averaging algorithm? In this subsection, we describe a completely distributed al-gorithm based on an approximate subgradient method which converges to a neighborhood of the optimal; alternately put, each iteration of the algorithm moves closer to the globally optimal , as stated in this theorem. Theorem 6: Let be the number of edges in . Let the subgradient at iteration in lie within the -subdifferential, and deﬁne . Then, the sequence of iterates in converges to a distribution for which is within of the globally optimal value . The required background and notation will be provided as necessary during the proof, which comprises the remainder of this section. Notation: It will be easier to analyze the subgradient method if we collect the entries of the matrix into a vector, which we will call . Since there is no symmetry requirement on the matrix , the vector will need to have entries corresponding to as well as (this corresponds to replacing each edge in the undirected graph by two directed edges, one in each direction). The vector corresponds to the matrix as follows. Let the total number of (non-self-loop) edges in be . Assign num-bers to the undirected edges from through : if edge , , is assigned number , we denote this as . If , then deﬁne the variable , and . We will also introduce the notation corresponding to the nonzero entries in the th row of (we do this to make concise the constraint that the sum of elements in each row should be ). That is, we deﬁne for (54) Deﬁne also the matrices , , with entries , , and zeros everywhere else. Then Finally, denote the degree of node by . 1) Subgradient Method: We will describe the subgradient method for the optimization problem restated in terms of the variable . We can state (53) in terms of the variables as follows: minimize subject to (55) where is as deﬁned in (54). We will use the subgradient method to obtain a distributed solution to this problem. The use of the subgradient method to solve eigenvalue problems is well known; see, for example, , , , for material on nonsmooth analysis of spectral functions, and , , for more general background on nonsmooth optimization. Recall that a subgradient of at is a symmetric matrix that satisﬁes the inequality for any feasible, i.e., symmetric stochastic matrix (here denotes the matrix inner product, and denotes the trace of a matrix). Let be a unit eigenvector associated with , then the matrix is a subgradient of (see, for example, ). For completeness, we include the proof here. First 2518 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 note that . By the variational characterization of the second eigenvalue of and , we have Subtracting the two sides of the above equality from that of the inequality, we have So is a subgradient. Using in terms of the probability vector , we obtain (56) so that the subgradient is given by (57) with components where . Observe that if each node knows its own component of the unit eigenvector, then this subgradient can be computed locally, using only local information. The following is the projected subgradient method for (55). • Initialization: Initialize to some feasible vector, for ex-ample, corresponding to the natural random walk. Set . • Repeat for — Subgradient step. Compute a subgradient at , and set — Projection onto feasible set. At each node , project obtained from the subgradient step onto , . This is achieved as follows: 1) If then set , stop. 2) If not, then use bisection to ﬁnd such that then set , stop. In this algorithm, Step 1 moves in the direction of the sub-gradient with stepsize . Step 2 projects the vector onto the feasible set. Since the constraints at each node are separable, the variables corresponding to nodes are projected onto the fea-sible set separately. The projection method is derived from the optimality condi-tions of the projection problem minimize subject to (58) as shown. Introduce Lagrange multipliers for the inequality , and for . The Karush–Kuhn–Tucker (KKT) conditions for optimal primal and dual variables , , are Eliminating the slack variables , we get the equivalent opti-mality conditions (59) (60) (61) (62) If , then from the last condition, necessarily . From (61), this gives us . If on the other hand, , then as well since , and so to satisfy (61), we must have . Combining these gives us that (63) The must satisfy , i.e., . However, we must also satisfy the complementary slackness condition . These two conditions combined together lead to a unique solution for , obtained either at , or at the solution of ; from the can be found from (63). 2) Decentralization: Now consider the issue of decentral-ization. Observe that in the above algorithm, can be computed locally at each node if , the unit eigenvector corresponding to , is known; more precisely, if each node is aware of its own component of and that of its immediate neighbors. The projection step can be carried out exactly at each node using local information alone. The rest of the subsection proceeds as follows: ﬁrst we will discuss approximate distributed computa-tion of the eigenvector of , and then show that the subgra-dient method converges to a certain neighborhood of the optimal value in spite of the error incurred during the distributed com-putation of at each iteration. BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2519 The problem of distributed computation of the top-eigen-vectors of a matrix on a graph is discussed in . By distributed computation of an eigenvector of a matrix , we mean that each node is aware of the th row of , and can only com-municate with its immediate neighbors. Given these constraints, the distributed computation must ensure that each node holds its value in the unit eigenvector . In , the authors present a distributed implementation of orthogonal iterations, referred to as DECENTRALOI (for decentralized orthogonal iterations), along with an error analysis. Since the matrix is symmetric and stochastic (it is a convex combination of symmetric stochastic matrices), we know that the ﬁrst eigenvector is . Therefore, orthogonal iterations takes a particularly simple form (in particular, we do not need any Cholesky factorization type of computations at the nodes). We describe orthogonal iterations for this problem as follows. • DECENTRALOI: Initialize the process with some randomly chosen vector ; for , repeat — Set — (Orthogonalize) — (Scale to unit norm) Here, the multiplication by is distributed, since respects the graph structure, i.e., only if is an edge. So entry of can be found using only values of corre-sponding to neighbors of node , i.e., the computation is dis-tributed. The orthogonalize and scale steps can be carried out in a distributed fashion using the gossip algorithm outlined in this paper, or just by distributed averaging as described in and used in . Observe that the very matrix can be used for the distributed averaging step, since it is also a probability ma-trix. We state the following result (applied to our special case) from , which basically states that it is possible to compute the eigenvector up to an arbitrary accuracy. Lemma 6: If DECENTRALOI is run for iterations, producing orthogonal vector , then (64) where is the distance between and the eigenspace of ; is the vector in the eigenspace achieving this distance; and is the mixing time of the doubly stochastic matrix used in the averaging step in DECENTRALOI. For the algorithm to be completely decentralized, a decentral-ized criterion for stopping when the eigenvector has been com-puted up to an accuracy is necessary. This is discussed in detail in ; we merely use the fact that it is possible for the nodes to compute the eigenvector, in a distributed fashion, up to a desired accuracy. Note also that the very matrix being optimized is a doubly stochastic matrix, and can be used in the averaging step in DECENTRALOI. If this is done, as the iterations proceed, the averaging step becomes faster. From the above discussion, it is clear we have a distributed algorithm that computes an approximate eigenvector, and there-fore an approximate subgradient. 3) Convergence Analysis: It now remains to show that the subgradient method converges despite approximation errors in computation of the eigenvector, which spill over into computa-tion of the subgradient. To show this, we will use a result from on the convergence of approximate subgradient methods. Given an optimization problem with objective function and feasible set , the approximate subgradient method generates a sequence such that (65) where is a projection onto the feasible set, is a step size, and (66) is the -subdifferential of the objective function at . Let and . Then we have the following theorem from , Lemma 7: If , then where , and is the optimal value of the objec-tive function. Consider the th iteration of the subgradient method, with current iterate , and let be the error in the (approximate) eigenvector corresponding to . (By error in the eigenvector, we mean the distance between and the (actual) eigenspace corresponding to ). Again, denote by the vector in the eigenspace minimizing the distance to , and denote the exact subgradient computed from by . We have . First, we ﬁnd in terms of as follows: Therefore, where is a scaling constant. Next, we will ﬁnd in terms of as follows: The th component of is Combining the facts that 2520 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 and (since ) we get Summing over all edges gives us , i.e., . Now choose . From (57), it can be seen that is bounded above by , and so in Theorem 7 converges to . Therefore, if in each iteration , the eigenvector is computed to within an error of , and , we have the claimed result. Remark: The fact that each constraint in (55) is local is cru-cial to the existence of a distributed algorithm using the sub-gradient method. The proof of convergence of the subgradient method relies on the fact that the distance to the optimal set de-creases at each iteration. This means that an exact projection needs to be computed at each step: if only an approximate pro-jection can be computed, this crucial property of decreasing the distance to the optimal set cannot be veriﬁed. Thus, for example, if the algorithm were formulated in terms of picking one of all possible edges at random at each step, the constraint would be , which is not a local con-straint. Although this algorithm has a larger feasible set than the optimization problem for the algorithm , it does not allow for a distributed computation of the optimal algorithm: though the projection can be computed approximately by distributed aver-aging, an exact projection cannot be computed, and the conver-gence of the subgradient method is not guaranteed. V. AVERAGING TIME AND MIXING TIME In this section, we explore the relation between the averaging time of an algorithm with a symmetric probability matrix , and the mixing time of the Markov chain with transition ma-trix . Since we assume that is symmetric, the Markov chain with transition matrix has a uniform equilibrium distribution. Deﬁnition 2 (Mixing Time): For a Markov chain with tran-sition matrix , let . Then, the -mixing time is deﬁned as (67) Recall also the following well-known bounds on the -mixing time for a Markov chain (see, for example, the survey ). Lemma 8: The -mixing time of a Markov chain with doubly stochastic transition matrix is bounded as (68) For , (68) becomes (69) In the rest of the paper, if we do not specify , we mean ; the corresponding mixing time is denoted simply as . We use Lemma 8 and Theorem 3 to prove the following the-orem. Theorem 7: The averaging time of the gossip algorithm in absolute time is related to the mixing time of the Markov chain with transition matrix as Proof: Let . It is shown in that for . Since is symmetric, we can use the result in Corollary 1, so that in absolute time, for We will ﬁrst show that . Using the result of and Corollary 1, we already have that (70) Note that the eigenvalues of are all positive, so that . There are two cases to consider. • : 4 In this case, by Lemma 8, . Further, . It follows that . • : From Lemma 8, we get (71) (72) Combining this with (70), we see that . Now we will show that , which will give us our result. Again we consider the same two cases. • If , then by (1) and Lemma 8 (73) (74) (75) (76) • If , then using Lemma 8, and (77) so that 4The speciﬁc value is not crucial; we could have chosen any a > 0 instead. BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2521 Fig. 2. Graphical interpretation of Theorem 7. Combining the two results gives us the theorem. Fig. 2 is a pictorial description of Theorem 7. The -axis de-notes mixing time and the -axis denotes averaging time. The scale on the axis is in order notation. As shown in the ﬁgure, for such that , ; for such that , . Thus, the mixing time of the random walk essentially charac-terizes the averaging time of the corresponding averaging algo-rithm on the graph. VI. APPLICATIONS In this section, we brieﬂy discuss applications of our results in the context of wireless ad hoc networks and the Internet. A. Wireless Networks The Geometric Random Graph, introduced by Gupta and Kumar , has been used successfully to model ad hoc wire-less networks. A -dimensional Geometric Random Graph on nodes, denoted , models a wireless ad hoc network of nodes with wireless transmission radius . It is obtained as follows: place nodes on a –dimensional unit cube uniformly at random and connect any two nodes that are within distance of each other. An example of a two-dimensional graph, is shown in Fig. 3. The following is a well-known result about the connectivity of (for a proof, see , , ). Lemma 9: For , the is connected with probability at least . We have the following results for averaging algorithms on a wireless sensor network, which are stated at the end of this section as Theorem 9. (We will prove these by evaluating the mixing times for the natural and optimal random walks on geo-metric random graphs, and then using Theorem 7, which relates averaging times and mixing times.) • On the Geometric Random Graph, , the absolute -averaging time of the optimal averaging algorithm is . Thus, in wireless sensor networks with a small radius of com-munication, distributed computing is necessarily slow, since the fastest averaging algorithm is itself slow. However, consider the natural averaging algorithm, based on the natural random walk, which can be described as follows: each node, when it becomes Fig. 3. An example of a Geometric Random Graph in two dimensions. A node is connected to all other nodes that are within the distance r of itself. active, chooses one of its neighbors uniformly at random and averages its value with the chosen neighbor. We have noted before that, in general, the performance of such an algorithm can be far worse than the optimal algorithm. Interestingly, in the case of , the performances of the natural averaging algorithm and the optimal averaging algo-rithm are comparable (i.e., they have averaging times of the same order). We will show the following result for the natural averaging algorithm on geometric random graphs. • In the Geometric Random Graph, , the absolute -averaging time of the natural averaging al-gorithm is of the same order as the optimal averaging al-gorithm, i.e., . We now prove the following theorem about the mixing times of the optimal and natural random walks on . Theorem 8: For with , with high probability a) the mixing time of the optimal reversible random walk with uniform stationary distribution is ; and b) the mixing time of the modiﬁed natural random walk, where a node jumps to any of its neighbors (other than itself) with equal probability, and has a self-loop of prob-ability , is also . The outline of the proof is as follows. To prove a), we will start by showing that with high probability, the geometric random graph is a regular graph. We bound the mixing rate of the op-timal random walk on the corresponding regular graph, and then relate the mixing times of the optimal random walks on this regular graph and the graph. The proof of b) uses a modiﬁcation of the path counting argument of Diaconis and Stroock to upper-bound the second largest eigenvalue of the nat-ural random walk on the graph. We start with evaluating the mixing time of the optimal random walk on . 1) Regularity of : In this subsection, we prove a regularity property of , which allows a simpler anal-ysis of the mixing time of random walks. Lemma 10: For with , the degree of every node is with high probability, where . Proof: Let nodes be numbered . Consider a particular node, say . Let random variable be if node is within distance of node and otherwise. The ’s are i.i.d. 2522 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 Bernoulli with probability of success (the volume of a -dimensional sphere with radius is ). The degree of node is (78) By application of the Chernoff bound we obtain (79) If we choose , then the right-hand side in (79) becomes . So, for , node has degree w.p. (80) Using the union bound, we see that any node has degree (81) So for large , w.h.p. (with high probability), all nodes in the -dimensional have degree . 2) Proof of Theorem 8 a): Optimal Random Walk on : In this subsection, we characterize the scaling of the optimal random walk on . We ﬁrst consider the case of , i.e., . This is much easier than the higher dimensional with . We completely characterize with the help of one-dimensional regular graphs. For with , we obtain a lower bound on the fastes mixing reversible random walk. Note that since we are interested in reversible random walks with uniform stationary distribution, the transition matrix corresponding to the random walk must be symmetric. (An upper bound of the same order is implied by the natural random walk as in Theorem 8 b).) The remainder of the section is a proof of Theorem 8 a). Optimal random walk on Let denote the regular graph on nodes with every node of degree ; it is constructed by placing the nodes on the cir-cumference of a circle, and connecting every node to neigh-bors on the left, and on the right. From the regularity lemma, we have that w.h.p., every node in has degree . Also, observe that the same technique can be used to show that w.h.p. the number of neighbors to the right (ditto left) is . In this one-dimensional case, it is clear that w.h.p., the is a subgraph of for , since for any mapping of the nodes of to , an edge between nodes and in is also present in . Similarly, also contains , for . Given this, we can now study the problem of ﬁnding the optimal random walk on with uniform sta-tionary distribution. We have the following lemma. Lemma 11: For , such that , the mixing rate of the fastest mixing symmetric random walk on cannot be smaller than . Proof: It can be shown using symmetry arguments that the fastest mixing Markov chain on with uniform sta-tionary distribution will have a symmetric and circulant transi-tion matrix. (For this simple graph, this can be easily seen using convexity of the second eigenvalue). So we can restrict our at-tention to the (circulant symmetric) transition matrices given in (82) at the bottom of the page. The eigenvalues of this matrix are For , , which is the largest eigenvalue. Let . We are interested in the smallest possible second largest eigenvalue in absolute value, i.e., in minimize subject to (83) We can obtain a lower bound for the optimal value of (83). Now (84) The right-hand side is the solution of the following linear pro-gram with a single total sum constraint: minimize subject to (85) For such that each of the coefﬁcients is positive, i.e., for , the smallest coefﬁcient is , and so for all such and , the minimum value is , obtained . . . . . . . . . . . . (82) BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2523 at , for all other .5 So the fastest mixing random walk on this graph cannot have a mixing rate smaller than . The preceding result was proved for all ; however, we will be interested only in those cases where , i.e., the graph is not too well connected. For such , the following lemma allows us to ﬁnd a “nearly optimal” transition matrix. Lemma 12: For , there is a random walk on for which the mixing rate is . Proof: For simplicity, let us assume that divides ; it is not difﬁcult to obtain the same results when this is not the case. Consider the Markov chain with transition probabilities , , . We will show that for a certain , small enough, is indeed , and is away from by . For the transition matrix corresponding to these probabil-ities, the eigenvalues are, for (86) (87) We want to ﬁnd the smallest positive such that is (this is not true, for example, for ). However, we need to be small enough so that the residual term, , is small compared to . Since and we hope that is small , we see that the values of for which is comparable to are those values of for which . This happens for . (We only need consider values of until , since .) At all odd multi-ples of , , and for the even multiples, . For to satisfy , we must have for an even multiple of (88) and for an odd multiple of 5Note that this is only a lower bound: for this p p p, if k divides n, the second largest eigenvalue is also 1, attained at m = n=k. that is, (89) From (88), we see that must be greater or equal to (90) for an odd multiple of , and from (89), must be less or equal to (91) for a multiple of . So can be only as small as the max-imum over the speciﬁed of all of these right-hand sides. Note that the only term dependent on in each of these ex-pressions is . For , odd (92) since for odd , and if is even, also. For (93) since (sum of real parts of the th roots of unity). So , and returning to (86), we see that the residual term in is of order , i.e., , while . So the difference between and is . Optimal walk on We present the lower bound on the fastest mixing reversible random walk on in this section. The same method can be easily extended to . First we characterize the fastest mixing reversible random walk on a two-dimensional regular graph deﬁned as follows: form a lattice on the unit torus, where lattice points are located at , , and place the nodes at these points. An edge between two vertices exists if the distance between them is at most . For such the fastest mixing time scales as follows. Lemma 13: The mixing rate of the optimal reversible random walk on is no smaller than . Proof: As in the one-dimensional case, by symmetry, the optimal transition probability between nodes and will depend only on the distance between these nodes. Using this, we can write the transition matrix corresponding to such a symmetric random walk on as the Kronecker (or tensor) product 2524 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 , where is as in (82). This is not difﬁcult to visualize: for , (94) Now the eigenvalues of are all products of eigenvalues of and , so that for , The eigenvalue is obtained by setting ; all other eigenvalues will have absolute value less (or ?) equal (to ?) . We want to ﬁnd a lower bound for the second largest eigenvalue in absolute value, call it . As before, choose . Then so that is a lower bound for . Making the as-sumption again that , the minimizing is the one with and (which corresponds to tran-sition probabilities of for each of the four farthest diagonal nodes, and everywhere else). The value of corresponding to this distribution is . This is of order , since6 The graph was constructed using the distance be-tween vertices. Therefore, the graph formed by placing edges between vertices based on distance measured in any norm (for the same ) is a subgraph of , and has a mixing time lower-bounded by the mixing time of . Thus, our bounds will be valid for the graph constructed according to any norm. Now we will use the bound on the fastest mixing walk on to obtain a bound for . First, we create a new graph as follows: place a square grid with squares of side on the unit torus. By Lemma 10, each square of area contains nodes. For each such square, connect every node in this square to all the nodes in the neighboring squares, as well as the nodes in the same square. Thus, each node is connected to nodes in . By deﬁnition, all edges in are present in and therefore, the fastest mixing random walk on is at least as fast as that of . Thus, lower-bounding the mixing time of the fastest mixing random walk on is sufﬁcient. Construct a graph of nodes as follows: corresponding to each square in the square grid used in , create a node in . Thus, has nodes. Two nodes are connected in if the corresponding squares in the grid are adjacent. Thus, each 6It is easy to see that a result similar to Lemma 12 can be obtained for d 2 using the same method. node is connected to eight other nodes. Thus, is a regular graph with nodes. In order to use this bound as a lower bound on , we need to show that the fastest mixing sym-metric random walk on induces a time-homogeneous reversible random walk on . This will be implied by the fol-lowing lemma. Lemma 14: There exists a fastest mixing symmetric random walk on , whose transition matrix has the following property: for any two nodes and belonging to the same square, for , and . Proof: We prove this by contradiction. Suppose the claimed statement is not true, i.e., there is no transition matrix achieving the smallest with the above property. Since the optimal value of must be attained (), consider such an optimizing , and let and be two nodes in the same square for which the above property is not true. Let be the permutation matrix with , , and all other diagonal entries and all other nondiagonal entries . Note that is a symmetric permutation matrix, and therefore . Consider the matrix ; since , and are similar, and so have the same eigenvalues. Note that since and belong to the same square in , they have exactly the same neighbors, and therefore also respects the graph structure (i.e., only if and have an edge between them). Now, is a convex function of for symmetric sto-chastic (), so (95) But has the property claimed in the lemma for nodes and : for all , , and . We can apply the above procedure recursively (even for multiple rows) to con-struct a matrix with smallest and the property claimed in the lemma. This contradicts our assumption and completes the proof. From Lemma 14, we see that under the fastest mixing random walk, the probability of transiting from a node in a square, say , to some neighboring square, say , is the same for all nodes in and . Thus, essentially we can view the random walk as evolving over squares. That is, the fastest random walk on induces a random walk on the graph . By deﬁni-tion of mixing time, the mixing time for this induced random walk on (with the induced equilibrium distribution) certainly lower-bounds the mixing time for the random walk on . Further, the induced random walk is reversible as the random walk was symmetric on . Therefore, we see that the lower bound on mixing time for the fastest mixing random walk on implies a lower bound on the mixing time for the fastest mixing random walk on . From Lemma 13, we have a lower bound of on the mixing time of the fastest mixing symmetric random walk (i.e., with uniform stationary distribution). From Lemma 15, given below, this in turn implies a lower bound of on the mixing time of the fastest BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2525 mixing reversible random walk on . This completes the proof of 2 a) (please deﬁne “2 a)”) for . It is easy to see that the arguments presented above can readily be extended to the case of . Lemma 15: Consider a connected graph with diameter . Let be the mixing time of the fastest mixing reversible random walk on with stationary dis-tribution . Let , where is a constant. Then (96) i.e., the fastest mixing time for is no faster than that of the uniform distribution. Proof: Consider a reversible random walk with stationary distribution on and let its transition matrix be . We will prove the following claim, which in turn implies the statement of the lemma. Claim I: There exists a symmetric random walk on graph with transition matrix such that Proof of Claim I: For a reversible matrix , by deﬁnition Deﬁne matrix , where for if if and . By deﬁnition and reversibility of , is a symmetric doubly stochastic matrix. Further, for , if and only if . Hence, can be viewed as a transition matrix of a symmetric random walk on , whose stationary distribution is uniform. Deﬁne , where Similarly, deﬁne . Let be a nonconstant function. Deﬁne two quadratic forms, and , of , as Let the variance of with respect to these two random walks be Let and denote the second largest eigenvalue of matrices and , respectively. The minimax characterization of eigenvalues ([18, p. 176]), gives a bound on the second largest eigenvalue of a reversible matrix as a nonconstant (97) For any , , hence, and . Further, by the property of Hence, for any and Thus, for any This implies that Hence, from (97) we obtain Since the diameter of is , it is easy to see that the mixing time of all random walks on is lower-bounded by . Hence, from Lemma 8 By deﬁnition, . Hence, It is easy to see that the random walk on with symmetric transition matrix has mixing time given by Thus, . This completes the proof of Claim I and the proof of Lemma 15. Remark: In fact, a stronger result can be proved, which is One part of this has already been proved in the lemma. The reverse direction is obtained similarly, as follows. Consider any symmetric random walk with transition matrix , and suppose a stationary distribution is speciﬁed, satisfying , where is some constant. Then there is a reversible random walk with stationary distribution 2526 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 , such that . is obtained as follows. Construct a matrix from as if if for , and . is a stochastic reversible matrix, with stationary distribution , since . Following the same steps as above, we can conclude that The matrix has the same eigenvectors as and, therefore, the same stationary distribution . The eigenvalues are . Therefore, since the diameter of the graph is As before Therefore, , and we have the stronger result claimed in the Remark. 3) Proof of Theorem 8 b): Natural Random Walk on : In this subsection, we study the mixing properties of the natural random walk on . Recall that under the natural random walk, the next node is equally likely to be any of the neighboring nodes. It is well known that under the stationary distribution, the probability of the walk being at node is proportional to the degree of node . By Lemma 10, all nodes have almost equal degree. Hence, the stationary distribution is almost uniform (it is uniform asymptotically). The rest of this section is the proof of Theorem 8 b). We use a modiﬁcation of a method developed by Dia-conis–Stroock to obtain bounds on the second largest eigenvalue using the geometry of the . Note that for , the proof is rather straightforward. The difﬁculty arises in the case of . For ease of exposition in the rest of the section, we consider . Exactly the same argument can be used for . We begin with some initial setup and notation. Square Grid: Divide the unit torus into a square grid where each square is of area , i.e., of side length . Consider a node in a square. By deﬁnition of , this node is connected to all nodes in the same square and all neighboring squares. Paths and Distribution: A path between two nodes and , denoted by , is a sequence of nodes , , such that are edges in . Let denote a collection of paths for all node pairs. Let be the collection of all possible . Consider the probability distribution induced on by selecting paths between all node pairs as described below. • Paths are chosen independently for different node pairs. • Consider a particular node pair . Let belong to square and belong to square . — If or and are in neighboring cells then the path between and is . — Else, let , be other squares lying on the straight line joining and . Select a node , uniformly at random. Then the path between and is . Under the above setup, we claim the following lemma. Lemma 16: Under the probability distribution on as de-scribed above, the average number of paths passing through an edge is w.h.p., where . Proof: We will compute the average load in order nota-tion. Similar to the arguments of Lemma 10, it can be shown that each of the squares contains nodes and each node has degree w.h.p. We restrict our con-sideration to such instances of . Now the total number of paths are since there are node pairs. Each path contains edges, as squares can be lying on a straight line joining two nodes. The total number of squares is . Hence, by symmetry and regularity, the number of paths passing through each square is . Consider a particular square . For , at least fraction of paths passing through it have endpoints lying in squares other than . That is, most of the paths passing through have as an intermediate square, and not an orig-inating square. Such paths are equally likely to select any of the nodes in . Hence, the average number of paths containing a node, say , in , is . The number of edges between and neighboring squares is . By symmetry, the average load on an edge incident on will be . This is true for all nodes. Hence, the average load on an edge is at most . Next we will use this setup and Lemma 16 to obtain a bound on the second largest eigenvalue using a modiﬁed version of Poincare’s inequality stated as follows. Lemma 17: Consider the natural random walk on a graph with the set of all possible paths on all node pairs. Let be the maximum path length (among all paths and over all node pairs), be the maximum node degree, and be the total number of edges. Let, according to some probability distribution on , the maximum average load on any of the edges be , i.e., on average no edge belongs to more than paths. Then, the second largest eigenvalue, , is bounded above as (98) Proof: The proof follows from a modiﬁcation of Poincare’s inequality ([9, Proposition 1]). Before proceeding to the proof, we introduce some notation. Let be a real-valued function on the nodes. Let denote the equilibrium distribu-tion of the random walk. Let be the degree of node , then it is well known that . For node pair , let BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2527 Deﬁne the quadratic form of as Let the variance of with respect to be For a directed edge from , deﬁne and . First, consider one collection of paths . Deﬁne Then, under the natural random walk (99) where is the length of the path (100) where denotes the number of paths passing through edge under . follows by using for all , and adding and subtracting values of on nodes of the path for all node pairs for a given path-set . follows from the Cauchy–Schwartz inequality. follows from (99), and follows from the fact that all path lengths are smaller than . Note that in (100), is the only path-dependent term. So under a probability distribution on (the set of all paths) in (100), can be replaced by where Let . Then (101) (102) The minimax characterization of eigenvalues [18, p. 176] gives a bound on the second largest eigenvalue as a nonconstant (103) From (102) and (103), the statement of the lemma follows. From Lemmas 10, 16, and 17, and the fact that all paths are of length at most , we obtain that the second largest eigen-value corresponding to the natural random walk on is bounded above as (104) We would like to note that, for mixing time, we need to show that the smallest eigenvalue (which can be negative), is also away from . One well-known way to avoid this difﬁculty is the following: modify transition probabilities as . and have the same stationary dis-tribution. By deﬁnition, has all nonnegative eigenvalues, and . Thus, the mixing time of the random walk corresponding to is governed by , and is therefore . This random walk is the modiﬁed natural random walk in Theorem 8 b). Thus, from Lemma 8 and (104), the proof of Theorem 8 b) for follows. In general, the above argument can be carried out similarly for completing the proof of Theorem 8 b). Averaging in : The natural averaging algorithm, based on the natural random walk, can be described as follows: when a node becomes active, it chooses one of its neighbors uniformly at random and averages with this neighbor. As noted before, in general, the performance of such an algorithm can be far worse than the optimal algorithm. Interestingly, in the case of , the performances of the natural averaging algorithm and the optimal averaging algorithm are comparable (i.e., they have averaging time of the same order). We state the following theorem. Theorem 9: On the Geometric Random Graph , the absolute -averaging time, , of the natural averaging algorithm as well as of the optimal averaging algorithm is of order . Proof: We showed in Theorem 8 that for , the -mixing times for the fastest mixing random walk and the natural random walk on are of order . Using this in Theorem 7, we have our result. Implication. In a wireless sensor network, Theorem 9 sug-gests that for a small radius of transmission, even the fastest averaging algorithm converges slowly, i.e., computing in a dis-tributed fashion is slow. However, the good news is that the natural averaging algorithm, based only on local information, scales just as well as the fastest averaging algorithm. Thus, at least in the order sense, it is not necessary to optimize for the fastest averaging algorithm in a wireless sensor network. 2528 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 52, NO. 6, JUNE 2006 B. Expander Graphs An expander graph can be characterized as follows: let the transition matrix corresponding to the natural random walk on the graph be . Then, there exists such that (105) where is the second largest eigenvalue of in mag-nitude, i.e., the spectral gap is bounded away from zero by a constant. Let be the transition matrix corresponding to the fastest mixing random walk on an expander. The random walk corre-sponding to must mix at least as fast as the natural one, and therefore, (106) It is easy to argue that there exists an optimal that is sym-metric: given any optimal , the matrix is sym-metric, and leads to the same as , since (107) and the are symmetric matrices. Therefore, we are able to use the result relating the mixing time for and the averaging time for for a symmetric . From (105), (106), Theorem 3, and Corollary 2, we see that the optimal averaging algorithm on any expander graph has -aver-aging time . The Preferential Connectivity (PC) model is one of the popular models for the Internet. In , it is shown that the In-ternet is an expander under the PC model. Using the conclusion above, we obtain the following result for averaging on the In-ternet. Theorem 10: Under the PC model, the optimal averaging algorithm on the Internet has an absolute -averaging time . Implication. The absolute time for distributed computation on any expander graph is independent of the size of the network, and depends only on the desired accuracy of the computation. Assuming that the PC model is a good model for Internet, then this immediately suggests that the absolute computation time depends only on the desired accuracy.7 One implication is that exchanging information on the Internet via peer-to-peer network built on top of it is extremely fast! Remark: Let be the maximum node degree of the graph . For any family of graphs of bounded degree, the averaging time of the maximum-degree random walk ( if ), and the fastest mixing random walk are of the same order.8 This follows from an observation in , which 7Although that the asymmetry of the P matrix for the natural random walk on the Internet prevents us from exactly quantifying the averaging time, we believe that averaging will be fast even under the natural random walk, since the spectral gap for this random walk is bounded away from 1 by a constant. 8The reason for using the maximum degree chain rather than the natural random walk is because the natural random walk need not be symmetric for an arbitrary graph. (Note that for a regular graph, the maximum degree chain and the natural random walk are exactly the same.) An alternative symmetric random walk with locally computable weights is the Metropolis–Hastings random walk with P = minf1=d ; 1=d g for (i; j) 2 E; i 6= j, for which a similar result holds. says that the spectral gap for the fastest mixing Markov chain on a graph can be at most a factor smaller than the max-imum-degree chain. Thus, if is the optimal transition matrix, i.e., the one with the smallest possible , and is the transition matrix for the maximum-degree chain, then (108) Thus, the averaging times for both random walks are of the same order, and differ by a factor of atmost . For example, the social network is a regular graph with , which is the degree of each node in the graph. For the social network, therefore, the natural random walk (which is the same as the maximum degree chain) leads to an averaging time of the same order as the optimal; and in fact, the averaging times differ by a factor of at most . C. Information Exchange Deﬁne to be the smallest time at which each node has information from all the other nodes with a probability greater than or equal to . The averaging time provides an upper bound for the information exchange time, as made precise in the following theorem. Theorem 11: For a gossip algorithm speciﬁed by a matrix and Proof: Consider ﬁrst a single node , and set and for all . By the deﬁnition of averaging time, for all , the probability that is greater than or equal to , since by the deﬁnition of , for all (109) Note that and If any (each must be positive), then that term con-tributes to the sum, and thus the sum cannot be less that for . Thus, for all , the probability that all of the are greater (or ?) equal (to ?) . But this is exactly the prob-ability that all nodes receive the message from node . Using the union bound and summing for nodes, we conclude that the probability of all nodes receiving information from all other nodes is greater (or ?) equal (to ?) , and so . VII. CONCLUSION We presented a framework for the design and analysis of a randomized distributed averaging algorithm on an arbitrary con-nected network. We characterized the performance of the algo-rithm precisely in the terms of second largest eigenvalue of an appropriate doubly stochastic matrix. This allowed us to ﬁnd the BOYD et al.: RANDOMIZED GOSSIP ALGORITHMS 2529 fastest averaging algorithm of this class of algorithms, by estab-lishing the corresponding optimization problem to be convex. We established a tight relation between the averaging time of the algorithm and the mixing time of an associated random walk, and utilized this connection to design fast averaging algorithms for two popular and well-studied networks: Wireless Sensor Networks (modeled as Geometric Random Graphs), and the Internet graph (under the so-called Preferential Connectivity Model). In general, solving SDPs in a distributed manner is not pos-sible. However, we utilized the structure of the problem in order to solve the SDP (corresponding to determining the optimal av-eraging algorithm) in a distributed fashion using the subgradient method. This allows for self-tuning weights: that is, the network can start out with some arbitrary averaging matrix, say, one de-rived from the natural random walk, and then locally, without any central coordination, converge to the optimal weights cor-responding to the fastest averaging algorithm. The framework developed in this paper is general and can be utilized for the purpose of design and analysis of distributed algorithms in many other settings. ACKNOWLEDGMENT Devavrat Shah wishes to thank Robert Gallager for a careful reading and suggestions that led to an improvement in the read-ability of the ﬁnal paper. The authors would also like to thank an anonymous reviewer for several useful suggestions regarding the presentation of this paper. REFERENCES S. Boyd, P. Diaconis, and L. Xiao, “Fastest mixing Markov chain on a graph,” SIAM Rev., Problems and Techniques Section, vol. 46, no. 4, pp. 667–689, 2004. S. Boyd, A. Ghosh, B. Prabhakar, and D. Shah, “Analysis and optimiza-tion of randomized gossip algorithms,” in Proc. IEEE Conf. Decision and Control, Nassau, Bahamas, Dec. 2004, pp. 5310–5315. , “Gossip algorithms: Design, analysis, and applications,” in Proc. IEEE INFOCOM, Miami, FL, Mar. 2005. , “Mixing times of random walks on geometric random graphs,” in Proc. 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9203	https://www.fishbase.se/summary/FamilySummary.php?ID=189	This website uses cookies to enhance your browsing experience and ensure the functionality of our site. For more detailed information about the types of cookies we use and how we protect your privacy, please visit our Privacy Information page.. Accept All Accept Necessary Cookies Only Cookie Settings × Cookie Settings This website uses different types of cookies to enhance your experience. Please select your preferences below: Strictly Necessary- [x] These cookies are essential for the website to function properly. They include session cookies, which help maintain your session while you navigate the site, as well as cookies that remember your language preferences and other essential functionalities. Without these cookies, certain features of the website cannot be provided. Performance- [x] These cookies help us understand how visitors interact with our website by collecting and reporting information anonymously. For example, we use Google Analytics to generate web statistics, which helps us improve our website's performance and user experience. These cookies may track information such as the pages visited, time spent on the site, and any errors encountered. Save and Close FAMILY Details for Batrachoididae - Toadfishes Language: More info\| Plus d'info.\| Mais info Family Batrachoididae - Toadfishes Order:Batrachoidiformes Class:Teleostei No. in FishBase:Genera : 23 \| Species : 83 Eschmeyer's Catalog of Fishes Environment:Fresh :Yes \| Brackish :Yes \| Marine :Yes Division:Marine Aquarium:some First Fossil Record: Remark:Toadfishes are known to occur in the Atlantic, Indian and Pacific oceans. Chiefly marine: coastal and benthic; rarely in brackish waters. Few freshwater species. 3 subfamilies: Batrachoidinae with 15 genera and about 43 species distributed off coasts of the Americas, Africa, Europe, southern Aisa, and Australia; Porichthyinae with 2 genera, 15 species and Thalassophryninae with 2 genera, 11 species, distributed in the Eastern Pacific and western Atlantic. Bottom dwellers feeding on invertebrates and fishes. Small to medium-sized fishes (to 57 cm) easily recognized by their characteristic shape. Head broad and flattened, often with barbels and/or fleshy flaps; eyes on top of head, dorsally-directed; mouth wide. Gill openings restricted to sides, just in front of pectoral fin base. Two dorsal fins, the first consisting of 2 or 3 strong, sharp spines; the second consisting of a large number of soft rays; pelvic fins jugular, inserted well in advance of pectoral fins, with 1 spine and 2 or 3 soft rays. One to several lateral lines on head and body. Body naked or covered with small, cycloid scales. Color: mostly drab brown with spots or saddles of black, although at least one coral reef species, Sanopus splendidus, is brightly coloured. One subfamily, the Porichthyinae, is characterized by having photophores (light-emitting organs) in rows along lateral lines on head and body. Toadfishes are bottom-dwellers ranging from shallow inshore areas to deep waters; several species enter rivers, and some migrate regularly between shallow and deep waters. They are rather sluggish in their movements and are ambush predators, feeding mainly on molluscs and crustaceans. They may bite when handled. The subfamily Thalassophryninae, or "venomous toadfishes", includes species with hollow spines in the first dorsal fin and on the opercles; the spines are connected to venom glands that can force a poison into a wound. Although no catch statistics are reported, larger species of toadfishes are commonly found in local markets. Some species are eaten and may fetch fairly high prices in Venezuela and French Guiana. The family is divided into three subfamilies (Collette, 1966): Batrachoidinae (about 18 genera, 47 species), Porichthyinae (2 genera, 15 species), and Thalassophryninae (2 genera, 11 species). The more generalized Batrachoidinae is world wide in distribution and contains a few freshwater species, one in Central America and one in South America. The more specialized midshipmen (Porichthyinae) and venomous toadfishes (Thalassophryninae) are restricted to the New World. All Porichthyinae are marine but there are three freshwater species of Thalassophryninae in South America. Genera 25 Species 78 (Ref. 86865). Etymology:Greek, batrachos = frog ( Ref. 45335). Reproductive guild:guarders Typical activity level:sluggish Main Ref.:Nelson, J.S. 1994 Coordinator:Nelson, J.S. 1994 Deep Fin Classification :Osteichthyes \| Actinopterygii \| Actinopteri \| Neopterygii \| Teleostei \| Osteoglossocephalai \| Clupeocephala \| Euteleosteomorpha \| Neoteleostei \| Eurypterygia \| Ctenosquamata \| Acanthomorphata \| Acanthopterygii \| Percomorphaceae \| Batrachoidaria \| \| \| Batrachoidiformes \| \| \| Batrachoididae Show species images \| Show valid names \| Show original names \| Identification keys \| CAS specimen photos \| References Ref. [ e.g. 9948 ] Glossary [ e.g. oophagy ] Species/Synonymy list for the family Batrachoididae as currently in FishBase Important recommendation: The list below must not be used as an authority reference synonymy list like those found in scientific published revisions, which must be the source to be used and cited eventually when they exist. Rather, it reflects the current content of FishBase, and the progress with respect to synchronization with the Catalog of Fishes. However, we think it can be useful for users to assess the quality of information in FishBase, to start new work on the family, or to cross-check with other lists. But we appreciate to be cited in publications when this list has been of any working value. In particular, for published scientific, we suggest then to cite it in the Material and Method section as a useful tool to conduct the research, but again, not as a taxonomic or nomenclatural authority reference. Unless it is explicitly precised, the list is not complete, please search all original names published for the family in the Catalog of Fishes (genera, species), including those with uncertain or unknown status, that are not included in FishBase when they are not attached to a valid species. This list uses some data from Catalog of Fishes (not shown but used to sort names). The list ordered as follows: When subfamilies are recognized, nominotypical subfamily first then other subfamilies by alphabetical order. Type genus of the family first (or of subfamily when subfamilies are recognized) then other genera by chronological order of description (and alphabetical order). Type species of the genus first by chronological order (and alphabetical order), with last listed misapplied names in a light gray font. ! Marks misspellings of the species names that must not be used. Please send comments and corrections if you detect errors or missing names. Show all \| Show only accepted name \| Show only accepted and original names \| Scientifc name \| Status \| Senior/Junior synonym \| Combination \| --- --- \| \| Batrachoidinae \| \| Batrachoides surinamensis (Bloch & Schneider, 1801) \| accepted \| senior \| new \| \| Batrachus surinamensis Bloch & Schneider, 1801 \| synonym \| senior \| original \| \| Batrachoides tau Lacepède, 1800 \| ambiguous \| other \| original \| \| Batrachoides pacifici (Günther, 1861) \| accepted \| senior \| new \| \| Batrachus pacifici Günther, 1861 \| synonym \| senior \| original \| \| Batrachoides liberiensis (Steindachner, 1867) \| accepted \| senior \| new \| \| Batrachus liberiensis Steindachner, 1867 \| synonym \| senior \| original \| \| Batrachoides beninensis Regan, 1915 \| synonym \| junior \| original \| \| Batrachoides pacifici (non Günther, 1861) \| misapplied \| misapplied \| new \| \| Batrachoides goldmani Evermann & Goldsborough, 1902 \| accepted \| senior \| original \| \| Batrachoides boulengeri Gilbert & Starks, 1904 \| accepted \| senior \| original \| \| Batrachoides gilberti Meek & Hildebrand, 1928 \| accepted \| senior \| original \| \| Batrachoides manglae Cervigón, 1964 \| accepted \| senior \| original \| \| Batrachoides walkeri Collette & Russo, 1981 \| accepted \| senior \| original \| \| Batrachoides waltersi Collette & Russo, 1981 \| accepted \| senior \| original \| \| Opsanus tau (Linnaeus, 1766) \| accepted \| senior \| new \| \| Gadus tau Linnaeus, 1766 \| synonym \| senior \| original \| \| Opsanus cerapalus Rafinesque, 1818 \| synonym \| junior \| original \| \| Opsanus beta (Goode & Bean, 1880) \| accepted \| senior \| new \| \| Batrachus tau beta Goode & Bean, 1880 \| synonym \| senior \| original \| \| Opsanus vandeuseni Fowler, 1939 \| synonym \| junior \| original \| \| Opsanus brasiliensis Rotundo, Spinelli & Zavala-Camin, 2005 \| synonym \| junior \| original \| \| Opsanus pardus (Goode & Bean, 1880) \| accepted \| senior \| new \| \| Batrachus tau pardus Goode & Bean, 1880 \| synonym \| senior \| original \| \| Opsanus phobetron Walters & Robins, 1961 \| accepted \| senior \| original \| \| Opsanus dichrostomus Collette, 2001 \| accepted \| senior \| original \| \| Opsanus tau (non Linnaeus, 1766) \| misapplied \| misapplied \| new \| \| Opsanus beta (non Goode & Bean, 1880) \| misapplied \| misapplied \| new \| \| Amphichthys rubigenis Swainson, 1839 \| accepted \| senior \| original \| \| Batrachus rubigenis Swainson, 1839 \| ambiguous \| other \| original \| \| ! Amphichthys rubrigenes Swainson, 1839 \| synonym \| senior \| original \| \| Amphichthys cryptocentrus (Valenciennes, 1837) \| accepted \| senior \| new \| \| Batrachus cryptocentrus Valenciennes, 1837 \| synonym \| senior \| original \| \| Marcgravia cryptocentra (Valenciennes, 1837) \| synonym \| senior \| new \| \| Opsanus hildebrandi Breder, 1925 \| synonym \| junior \| original \| \| Amphichthys hildebrandi (Breder, 1925) \| synonym \| junior \| new \| \| Sanopus barbatus (Meek & Hildebrand, 1928) \| accepted \| senior \| new \| \| Opsanus barbatus Meek & Hildebrand, 1928 \| synonym \| senior \| original \| \| Sanopus astrifer (Robins & Starck, 1965) \| accepted \| senior \| new \| \| Opsanus astrifer Robins & Starck, 1965 \| synonym \| senior \| original \| \| Sanopus johnsoni Collette & Starck, 1974 \| accepted \| senior \| original \| \| Sanopus splendidus Collette, Starck & Phillips, 1974 \| accepted \| senior \| original \| \| Sanopus greenfieldorum Collette, 1983 \| accepted \| senior \| original \| \| Sanopus reticulatus Collette, 1983 \| accepted \| senior \| original \| \| Potamobatrachus trispinosus Collette, 1995 \| accepted \| senior \| original \| \| Vladichthys gloverensis (Greenfield & Greenfield, 1973) \| accepted \| senior \| new \| \| Triathalassothia gloverensis Greenfield & Greenfield, 1973 \| synonym \| senior \| original \| \| Amphichthys gloverensis (Greenfield & Greenfield, 1973) \| synonym \| senior \| new \| \| Halophryninae \| \| Halophryne diemensis (Lesueur, 1824) \| accepted \| senior \| new \| \| Batrachoides diemensis Lesueur, 1824 \| synonym \| senior \| original \| \| Halophryne queenslandiae (De Vis, 1882) \| accepted \| senior \| new \| \| ! Poricthys queenslandiae De Vis, 1882 \| synonym \| senior \| original \| \| Porichthys queenslandiae De Vis, 1882 \| synonym \| senior \| original \| \| Halophryne ocellatus Hutchins, 1974 \| accepted \| senior \| original \| \| Halophryne hutchinsi Greenfield, 1998 \| accepted \| senior \| original \| \| ! Halophryne hutchinsoni Greenfield, 1998 \| synonym \| senior \| original \| \| ! Halophyrne hutchinsi Greenfield, 1998 \| synonym \| senior \| original \| \| Batrachomoeus trispinosus (Günther, 1861) \| accepted \| senior \| new \| \| Batrachus trispinosus Günther, 1861 \| synonym \| senior \| original \| \| Halophryne trispinosus (Günther, 1861) \| synonym \| senior \| new \| \| Batrachomoeus broadbenti Ogilby, 1908 \| synonym \| junior \| original \| \| Pseudobatrachus eugeneius Fowler, 1937 \| synonym \| junior \| original \| \| Batrachomoeus dubius (White, 1790) \| accepted \| senior \| new \| \| Lophius dubius White, 1790 \| synonym \| senior \| original \| \| Pseudobatrachus dubius (White, 1790) \| synonym \| senior \| new \| \| Lophius nigricans Forster, 1794 \| synonym \| junior \| original \| \| Pseudobatrachus striatus Castelnau, 1875 \| synonym \| junior \| original \| \| Thalassophryne coeca De Vis, 1884 \| synonym \| junior \| original \| \| Pelophiletor caloundrae Ogilby, 1907 \| ambiguous \| other \| original \| \| Batrachomoeus minor Ogilby, 1908 \| synonym \| junior \| original \| \| Batrachomoeus dahli (Rendahl, 1922) \| accepted \| senior \| new \| \| Pseudobatrachus dahli Rendahl, 1922 \| synonym \| senior \| original \| \| Batrachomoeus occidentalis Hutchins, 1976 \| accepted \| senior \| original \| \| Batrachomoeus rubricephalus Hutchins, 1976 \| accepted \| senior \| original \| \| Halobatrachus didactylus (Bloch & Schneider, 1801) \| accepted \| senior \| new \| \| Batrachus didactylus Bloch & Schneider, 1801 \| synonym \| senior \| original \| \| Batrachoides didactylus (Bloch & Schneider, 1801) \| synonym \| senior \| new \| \| Batrachus conspicillum Cuvier, 1829 \| synonym \| junior \| original \| \| Batrachus punctatus Agassiz, 1831 \| synonym \| junior \| original \| \| ! Batrachus punctulatus Agassiz, 1831 \| synonym \| junior \| original \| \| Batrachus borealis Nilsson, 1832 \| synonym \| junior \| original \| \| Batrachus planifrons Guichenot, 1850 \| synonym \| junior \| original \| \| Batrachoides planifrons (Guichenot, 1850) \| synonym \| junior \| new \| \| Batrachus algeriensis Guichenot, 1850 \| synonym \| junior \| original \| \| Batrachus guentheri Bleeker, 1863 \| synonym \| junior \| original \| \| Gadus tau (non Linnaeus, 1766) \| misapplied \| misapplied \| original \| \| Batrichthys albofasciatus Smith, 1934 \| accepted \| senior \| original \| \| Batrichthys apiatus (Valenciennes, 1837) \| accepted \| senior \| new \| \| Batrachus apiatus Valenciennes, 1837 \| synonym \| senior \| original \| \| Amphichthys ophiocephalus Smith, 1947 \| synonym \| junior \| original \| \| Batrichthys ophiocephalus (Smith, 1947) \| synonym \| junior \| new \| \| Gymnobatrachus ophiocephalus (Smith, 1947) \| synonym \| junior \| new \| \| Triathalassothia argentina (Berg, 1897) \| accepted \| senior \| new \| \| Batrachus argentinus Berg, 1897 \| synonym \| senior \| original \| \| ! Triathalassothia argentinus (Berg, 1897) \| synonym \| senior \| new \| \| Triathalassothia devincenzii Fowler, 1943 \| synonym \| junior \| original \| \| Triathalassothia lambaloti Menezes & Figueiredo, 1998 \| accepted \| senior \| original \| \| Austrobatrachus foedus (Smith, 1947) \| accepted \| senior \| new \| \| Pseudobatrachus foedus Smith, 1947 \| synonym \| senior \| original \| \| Austrobatrachus iselesele Greenfield, 2012 \| accepted \| senior \| original \| \| Chatrabus melanurus (Barnard, 1927) \| accepted \| senior \| new \| \| Batrachoides melanurus Barnard, 1927 \| synonym \| senior \| original \| \| Batrachoides damaranus Barnard, 1927 \| synonym \| junior \| original \| \| Chatrabus damaranus (Barnard, 1927) \| synonym \| junior \| new \| \| Tharbacus vanecki Smith, 1952 \| synonym \| junior \| original \| \| Chatrabus felinus (Smith, 1952) \| accepted \| senior \| new \| \| Batrichthys felinus Smith, 1952 \| synonym \| senior \| original \| \| Gymnobatrachus apiatus (non Valenciennes, 1837) \| misapplied \| misapplied \| misapplied \| \| Chatrabus hendersoni (Smith, 1952) \| accepted \| senior \| new \| \| Tharbacus hendersoni Smith, 1952 \| synonym \| senior \| original \| \| Barchatus cirrhosus (Klunzinger, 1871) \| accepted \| senior \| new \| \| Batrachus cirrhosus Klunzinger, 1871 \| synonym \| senior \| original \| \| Thalassothia cirrhosa (Klunzinger, 1871) \| synonym \| senior \| new \| \| ! Barchatus cirrhosa (Klunzinger, 1871) \| synonym \| senior \| new \| \| ! Thalassothia cirrhosus (Klunzinger, 1871) \| synonym \| senior \| new \| \| Barchatus indicus Greenfield, 2014 \| accepted \| senior \| original \| \| Riekertia ellisi Smith, 1952 \| accepted \| senior \| original \| \| Perulibatrachus elminensis (Bleeker, 1863) \| accepted \| senior \| new \| \| Batrachus elminensis Bleeker, 1863 \| synonym \| senior \| original \| \| Parabatrachus elminensis (Bleeker, 1863) \| synonym \| senior \| new \| \| Batrachus congicus Reichenow, 1877 \| ambiguous \| questionable \| original \| \| Batrachus budkeri Roux, 1957 \| synonym \| junior \| original \| \| Perulibatrachus rossignoli (Roux, 1957) \| accepted \| senior \| new \| \| Batrachus rossignoli Roux, 1957 \| synonym \| senior \| original \| \| Parabatrachus rossignoli (Roux, 1957) \| synonym \| senior \| new \| \| Perulibatrachus kilburni Greenfield, 1996 \| accepted \| senior \| original \| \| Perulibatrachus aquilonarius Greenfield, 2005 \| accepted \| senior \| original \| \| Bifax lacinia Greenfield, Mee & Randall, 1994 \| accepted \| senior \| original \| \| Allenbatrachus grunniens (Linnaeus, 1758) \| accepted \| senior \| new \| \| Cottus grunniens Linnaeus, 1758 \| synonym \| senior \| original \| \| Batrachoides grunniens (Linnaeus, 1758) \| synonym \| senior \| new \| \| Batrachus grunniens (Linnaeus, 1758) \| synonym \| senior \| new \| \| Batrichthys grunniens (Linnaeus, 1758) \| synonym \| senior \| new \| \| Cottus indus Linnaeus, 1764 \| synonym \| junior \| original \| \| Batrachoides gangene Hamilton, 1822 \| synonym \| junior \| original \| \| Halophryne gangene (Hamilton, 1822) \| synonym \| junior \| new \| \| ! Halophyrene gangene (Hamilton, 1822) \| synonym \| junior \| new \| \| Allenbatrachus reticulatus (Steindachner, 1870) \| accepted \| senior \| new \| \| Batrachus reticulatus Steindachner, 1870 \| synonym \| senior \| original \| \| Allenbatrachus meridionalis Greenfield & Smith, 2004 \| accepted \| senior \| original \| \| Colletteichthys dussumieri (Valenciennes, 1837) \| accepted \| senior \| new \| \| Batrachus dussumieri Valenciennes, 1837 \| synonym \| senior \| original \| \| Austrobatrachus dussumieri (Valenciennes, 1837) \| synonym \| senior \| new \| \| Colletteichthys flavipinnis Greenfield, Bineesh & Akhilesh, 2012 \| accepted \| senior \| original \| \| Colletteichthys occidentalis Greenfield, 2012 \| accepted \| senior \| original \| \| Austrobatrachus dussumieri (non Valenciennes, 1837) \| misapplied \| misapplied \| new \| \| Porichthyinae \| \| Porichthys notatus Girard, 1854 \| accepted \| senior \| original \| \| Porichthys porosissimus (Cuvier, 1829) \| accepted \| senior \| new \| \| Batrachus porosissimus Cuvier, 1829 \| synonym \| senior \| original \| \| ! Porichthys porossissimus (Cuvier, 1829) \| synonym \| senior \| new \| \| Porichthys margaritatus (Richardson, 1844) \| accepted \| senior \| new \| \| Batrachus margaritatus Richardson, 1844 \| synonym \| senior \| original \| \| Porichthys nautopaedium Jordan & Bollman, 1890 \| synonym \| junior \| original \| \| Porichthys plectrodon Jordan & Gilbert, 1882 \| accepted \| senior \| original \| \| Porichthys greenei Gilbert & Starks, 1904 \| accepted \| senior \| original \| \| Porichthys analis Hubbs & Schultz, 1939 \| accepted \| senior \| original \| \| Porichthys myriaster Hubbs & Schultz, 1939 \| accepted \| senior \| original \| \| Porichthys pauciradiatus Caldwell & Caldwell, 1963 \| accepted \| senior \| original \| \| Porichthys bathoiketes Gilbert, 1968 \| accepted \| senior \| original \| \| Porichthys kymosemeum Gilbert, 1968 \| accepted \| senior \| original \| \| Porichthys oculofrenum Gilbert, 1968 \| accepted \| senior \| original \| \| Porichthys ephippiatus Walker & Rosenblatt, 1988 \| accepted \| senior \| original \| \| Porichthys mimeticus Walker & Rosenblatt, 1988 \| accepted \| senior \| original \| \| Porichthys oculellus Walker & Rosenblatt, 1988 \| accepted \| senior \| original \| \| Aphos porosus (Valenciennes, 1837) \| accepted \| senior \| new \| \| Batrachus porosus Valenciennes, 1837 \| synonym \| senior \| original \| \| Porichthys afuerae Evermann & Radcliffe, 1917 \| synonym \| junior \| original \| \| Thalassophryninae \| \| Thalassophryne maculosa Günther, 1861 \| accepted \| senior \| original \| \| Batrachus uranoscopus Guichenot, 1866 \| synonym \| junior \| original \| \| Thalassophryne uranoscopus (Guichenot, 1866) \| synonym \| junior \| new \| \| Thalassophryne wehekindi Fowler, 1931 \| synonym \| junior \| original \| \| Thalassophryne amazonica Steindachner, 1876 \| accepted \| senior \| original \| \| Thalassophryne nattereri Steindachner, 1876 \| accepted \| senior \| original \| \| ! Thalassophryne natteneri Steindachner, 1876 \| synonym \| senior \| original \| \| Thalassophryne branneri Starks, 1913 \| synonym \| junior \| original \| \| Thalassophryne maculosa (non Günther, 1861) \| misapplied \| misapplied \| original \| \| Thalassophryne punctata Steindachner, 1876 \| accepted \| senior \| original \| \| Thalassophryne montevidensis (Berg, 1893) \| accepted \| senior \| new \| \| Thalassothia montevidensis Berg, 1893 \| synonym \| senior \| original \| \| ! Thalassophyrne platensis Devincenzi, 1924 \| synonym \| junior \| original \| \| Thalassophryne platensis Devincenzi, 1924 \| synonym \| junior \| original \| \| Thalassophryne maculosa (non Günther, 1861) \| misapplied \| misapplied \| original \| \| Thalassophryne megalops Bean & Weed, 1910 \| accepted \| senior \| original \| \| Daector dowi (Jordan & Gilbert, 1887) \| accepted \| senior \| new \| \| Thalassophryne dowi Jordan & Gilbert, 1887 \| synonym \| senior \| original \| \| Thalassophryne depressa Hildebrand, 1946 \| synonym \| junior \| original \| \| Daector reticulata (Günther, 1864) \| accepted \| senior \| new \| \| Thalassophryne reticulata Günther, 1864 \| synonym \| senior \| original \| \| Daector quadrizonatus (Eigenmann, 1922) \| accepted \| senior \| new \| \| Thalassophryne quadrizonatus Eigenmann, 1922 \| synonym \| senior \| original \| \| Daector gerringi (Rendahl, 1941) \| accepted \| senior \| new \| \| Thalassophryne gerringi Rendahl, 1941 \| synonym \| senior \| original \| \| Daector schmitti Collette, 1968 \| accepted \| senior \| original \| \| ! Daector schmittii Collette, 1968 \| synonym \| senior \| original \| Back to Search Comments & Corrections Back to Top php script by celloran, 04/03/10, last modified by cmilitante, 29/11/12
9204	https://brainly.com/question/25156017	[FREE] What does it mean when a negative number is in parentheses? - brainly.com 7 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +63,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +30,1k Ace exams faster, with practice that adapts to you Practice Worksheets +6,5k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified What does it mean when a negative number is in parentheses? 2 See answers Explain with Learning Companion NEW Asked by bethanymaynard3794 • 10/20/2021 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 44758 people 44K 3.5 1 Upload your school material for a more relevant answer The correct answer is the negative sign indicates that the number or term inside the parenthesis is multiplied by a negative sign. When a negative number is placed within parentheses, it typically indicates that the negative sign applies to the entire number or expression inside the parentheses. In other words, it's a way to ensure that the negative sign is associated with the value inside the parentheses rather than being treated as an operator. For example: (-3) means negative three, which is the same as 3 but with a negative sign. -(5 + 2) means the negative of the sum of 5 and 2, which is -7. -x means the negative of the variable x. In mathematical expressions, using parentheses to enclose a negative number ensures clarity and avoids ambiguity in the interpretation of the expression. Learn more about Parentheses here: brainly.com/question/4637257 SPJ3 Answered by prakharsharma0789 •419 answers•44.8K people helped Thanks 1 3.5 (2 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 44758 people 44K 5.0 0 Python for Informatics: Exploring Information - Charles Severance Kinematics fundamentals - Sunil Singh Physics for K-12 - Sunil Singh Upload your school material for a more relevant answer A negative number in parentheses indicates that the negative sign applies to the whole number or expression inside, ensuring clarity in mathematical operations. This helps avoid ambiguity when combining numbers and performing calculations. It is essential for accurately interpreting mathematical expressions. Explanation When a negative number is in parentheses, it indicates that the negative sign applies to the entire number or expression within the parentheses. This notation is used for clarity, ensuring that anyone reading the expression understands that the negative sign is associated with the number inside. For example: The expression (-3) simply means negative three. If we take (-5 + 2), it reflects that the operation is being applied to -5, giving us -3 as the result. Furthermore, if you see an expression like -(-4), this means that you're taking the negative of negative four, which results in +4. This usage helps to eliminate confusion in mathematical expressions, especially when performing operations like addition or multiplication where the order and sign can greatly affect the outcome. Using parentheses to enclose negative numbers is essential in ensuring that the negative values are processed correctly according to mathematical rules, allowing for clear communication of mathematical ideas. Examples & Evidence For example: (-3) signifies negative three. -(-4) means taking the negative of negative four, resulting in +4. (-5 + 2) shows that -5 is considered in the operation, leading to -3. This understanding is supported by standard mathematical conventions used to express negative numbers clearly and avoid misunderstandings in calculations. Thanks 0 5.0 (1 vote) Advertisement Community Answer 5.0 0 Usually negative numbers are in parentheses so people don't think that it is subtraction. Answered by Brainly User Thanks 0 5.0 (1 vote) Advertisement ### Free Mathematics solutions and answers Community Answer 5.0 3 what does a negative sign in front of a parentheses mean Community Answer What happens when the negative exponent is outside the parentheses?. Community Answer 5.0 1 What effect does a negative have when placed Inside the parentheses? Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Given that the point (8,3) lies on the graph of g(x)=lo g 2x, which point lies on the graph of f(x)=lo g 2(x+3)+2 ? A. (5,1) B. (5,5) C. (11,1) D. (11,5) For what value of c is the function one-to-one?{(1,2),(2,3),(3,5),(4,7),(5,11),(6,c)}◯ 2◯ 5◯ 11◯ 13 Country A has a population of 2 x 10^9 people. Country B has a population of 5 x 10^7 people. Choose which country has the larger population. Then fill in the blank with a number written in standard notation. A. Country A has the larger population. The population of Country A is _ times as large as the population of Country B. B. Country B has the larger population. The population of Country B is _ times as large as the population of Country A. Two cars raced at a race track. The faster car traveled 20 mph faster than the slower car. In the time that the slower car traveled 165 miles, the faster car traveled 225 miles. If the speeds of the cars remained constant, how fast did the slower car travel during the race? \| \| Distance (mi) \| Rate (mph) \| Time (h) \| :--- :--- \| \| Slower Car \| 165 \| r \| r 165 \| \| Faster Car \| 225 \| r+20 \| r+20 225 \| A. 55 mph B. 60 mph C. 75 mph D. 130 mph 10 x 3+4 x 3−3 x 2+9 x 2 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
9205	https://www.math.ksu.edu/~nagy/real-an/4-02-conv-thms.pdf	Convergence theorems In this section we analyze the dynamics of integrabilty in the case when se-quences of measurable functions are considered. Roughly speaking, a “convergence theorem” states that integrability is preserved under taking limits. In other words, if one has a sequence (fn)∞ n=1 of integrable functions, and if f is some kind of a limit of the fn’s, then we would like to conclude that f itself is integrable, as well as the equality R f = limn→∞ R fn. Such results are often employed in two instances: A. When we want to prove that some function f is integrable. In this case we would look for a sequence (fn)∞ n=1 of integrable approximants for f. B. When we want to construct and integrable function. In this case, we will produce ﬁrst the approximants, and then we will examine the existence of the limit. The ﬁrst convergence result, which is somehow primite, but very useful, is the following. Lemma 2.1. Let (X, A, µ) be a ﬁnite measure space, let a ∈(0, ∞) and let fn : X →[0, a], n ≥1, be a sequence of measurable functions satisfying (a) f1 ≥f2 ≥· · · ≥0; (b) limn→∞fn(x) = 0, ∀x ∈X. Then one has the equality (1) lim n→∞ Z X fn dµ = 0. Proof. Let us deﬁne, for each ε > 0, and each integer n ≥1, the set Aε n = {x ∈X : fn(x) ≥ε}. Obviously, we have Aε n ∈A, ∀ε > 0, n ≥1. One key fact we are going to use is the following. Claim 1: For every ε > 0, one has the equality lim n→∞µ(Aε n) = 0. Fix ε > 0. Let us ﬁrst observe that, using (a), we have the inclusions (2) Aε 1 ⊃Aε 2 ⊃. . . Second, using (b), we clearly have the equality T∞ k=1 Aε k = ∅. Since µ is ﬁnite, using the Continuity Property (Lemma III.4.1), we have lim n→∞µ(Aε n) = µ ∞ \ n=1 Aε n = µ(∅) = 0. Claim 2: For every ε > 0 and every integer n ≥1, one has the inequality 0 ≤ Z X fn dµ ≤aµ(Aε n) + εµ(X). 300 §2. Convergence theorems 301 Fix ε and n, and let us consider the elementary function hε n = aκAε n + εκBε n, where Bε n = X ∖Aε. Obviously, since µ(X) < ∞, the function hε n is elementary integrable. By construction, we clearly have 0 ≤fn ≤hε n, so using the properties of integration, we get 0 ≤ Z X fn dµ ≤ Z X hε n dµ = aµ(Aε n) + εµ(Bε) ≤aµ(Aε) + εµ(X). Using Claims 1 and 2, it follows immediately that 0 ≤lim inf n→∞ Z X fn dµ ≤lim sup n→∞ Z X fn dµ ≤εµ(X). Since the last inequality holds for arbitrary ε > 0, the desired equality (1) immedi-ately follows. □ We now turn our attention to a weaker notion of limit, for sequences of mea-surable functions. Definition. Let (X, A, µ) be a measure space, let K be a one of the symbols ¯ R, R, or C. Suppose fn : X →K, n ≥1, are measurable functions. Given a measurable function f : X →K, we say that the sequence (fn)∞ n=1 converges µ-almost everywhere to f, if there exists some set N ∈A, with µ(N) = 0, such that lim n→∞fn(x) = f(x), ∀x ∈X ∖N. In this case we write f = µ-a.e.- lim n→∞fn. Remark 2.1. This notion of convergence has, among other things, a certain uniqueness feature. One way to describe this is to say that the limit of a µ-a.e. con-vergent sequence is µ-almost unique, in the sense that if f anf g are measurable func-tions which satisfy the equalities f = µ-a.e.- limn→∞fn and g = µ-a.e.- limn→∞fn, then f = g, µ-a.e. This is quite obvious, because there exist sets M, N ∈A, with µ(M) = µ(N) = 0, such that lim n→∞fn(x) = f(x), ∀x ∈X ∖M, lim n→∞fn(x) = g(x), ∀x ∈X ∖N, then it is obvious that µ(M ∪N) = 0, and f(x) = g(x), ∀x ∈X ∖[M ∪N]. Comment. The above deﬁnition makes sense if K is an arbitrary metric space. Any of the spaces ¯ R, R, and C is in fact a complete metric space. There are instances where the requirement that f is measurable is if fact redundant. This is somehow clariﬁed by the the next two exercises. Exercise 1. Let (X, A, µ) be a measure space, let K be a complete separable metric space, and let fn : X →K, n ≥1, be measurable functions. (i) Prove that the set L = x ∈X : fn(x) ∞ n=1 ⊂K is convergent belongs to A. 302 CHAPTER IV: INTEGRATION THEORY (ii) If we ﬁx some point α ∈K, and we deﬁne ℓ: X →K by ℓ(x) = ( lim n→∞fn(x) if x ∈L α if x ∈X ∖L then ℓis measurable. In particular, if µ(X ∖L) = 0, then ℓ= µ-a.e.- limn→∞fn. Hints: If d denotes the metric on K, then prove ﬁrst that, for every ε > 0 and every m, n ≥1, the set Dε mn = x ∈X : d fm(x), fn(x) < ε belongs to A (use the results from III.3). Based on this fact, prove that, for every p, k ≥1, the set Ep k = x ∈X ; d fm(x), fn(x) < 1 p , ∀m, n ≥k belongs to A. Finally, use completeness to prove that L = ∞ \ p=1 ∞ [ k=1 Ep k . Exercise 2. Use the setting from Exercise 1. Prove that Let (X, A, µ), K, and (fn)∞ n=1 be as in Exercise 1. Assume f : X →K is an arbitrary function, for which there exists some set N ∈A with µ(N) = 0, and lim n→∞fn(x) = f(x), ∀x ∈X ∖N. Prove that, when µ is a complete measure on A (see III.5), the function f is auto-matically measurable. Hint: Use the results from Exercise 1. We have X ∖N ⊂L, and f(x) = ℓ(x), ∀x ∈X ∖N. Prove that, for a Borel set B ⊂K, one has the equality f−1(B) = ℓ−1(B)△M, for some M ⊂N. By completeness, we have M ∈A, so f−1(B) ∈A. The following fundamental result is a generalization of Lemma 2.1. Theorem 2.1 (Lebesgue Monotone Convergence Theorem). Let (X, A, µ) be a measure space, and let (fn)∞ n=1 ⊂L1 +(X, A, µ) be a sequence with: • fn ≤fn+1, µ-a.e., ∀n ≥1; • sup R X fn dµ : n ≥1 < ∞. Assume f : X →[0, ∞] is a measurable function, with f = µ-a.e.- limn→∞fn. Then f ∈L1 +(X, A, µ), and R X f dµ = limn→∞ R X fn dµ. Proof. Deﬁne αn = R X fn dµ, n ≥1. First of all, we clearly have 0 ≤α1 ≤α2 ≤. . . , so the sequence (αn)∞ n=1 has a limit α = limn→∞αn, and we have in fact the equality α = sup αn : n ≥1 < ∞. With these notations, all we need to prove is the fact that f ∈L1 +(X, A, µ), and that we have (3) Z X f dµ = α. Fix a set M ∈A, with µ(M) = 0, and such that limn→∞fn(x) = f(x), ∀x ∈X ∖M. For each n, we deﬁne the set Mn = {x ∈X : fn(x) > fn+1(x)}. §2. Convergence theorems 303 Obviously Mn ∈A, and by assumption, we have µ(Mn) = 0, ∀n ≥1. Deﬁne the set N = M ∪ S∞ n=1 Mn . It is clear that µ(N) = 0, and • 0 ≤f1(x) ≤f2(x) ≤· · · ≤f(x), ∀x ∈X ∖N; • f(x) = limn→∞fn(x), ∀x ∈X ∖N. So if we put A = X ∖N, and if we deﬁne the measurable functions gn = fnκA, n ≥1, and g = fκA, then we have (a) 0 ≤g1 ≤g2 ≤· · · ≤g (everywhere!); (b) limn→∞gn(x) = g(x), ∀x ∈X; (c) gn = fn, µ-a.e., ∀n ≥1; (d) g = f, µ-a.e.; Notice that property (c) gives gn ∈L1 +(X, A, µ) and R X gn dµ = αn, ∀n ≥1. By property (d), we see that we have the equivalence f ∈L1 +(X, A, µ) ⇐ ⇒f ∈L1 +(X, A, µ). Moreover, if g ∈L1 +(X, A, µ), then we will have R X g dµ = R X f dµ. These observa-tions show that it suﬃces to prove the theorem with g’s in place of the f’s. The advantage is now the fact that we have the slightly stronger properties (a) and (b) above. The ﬁrst step in the proof is the following. Claim 1: For every t ∈(0, ∞), one has the inequality µ g−1((t, ∞]) ≤α t . Denote the set g−1((t, ∞]) simply by At. For each n ≥1, we also deﬁne the set An t = g−1 n ((t, ∞]). Using property (a) above, it is clear that we have the inclusions (4) At 1 ⊂A2 t ⊂· · · ⊂At. Using property (b) above, we also have the equality At = S∞ n=1 An t . Using the continuity Lemma 4.1, we then have µ(At) = lim n→∞µ(An t ), so in order to prove the Claim, it suﬃces to prove the inequalities (5) µ(An t ) ≤αn t , ∀n ≥1. But the above inequality is pretty obvious, since we clearly have 0 ≤tκAn t ≤gn, which gives tµ(An t ) = Z X tκAn t dµ ≤ Z X gn dµ = αn. Claim 2: For any elementary function h ∈A-ElemR(X), with 0 ≤h ≤g, one has (i) h ∈L1 R,elem(X, A, µ); (ii) R X h dµ ≤α. Start with some elementary function h, with 0 ≤h ≤g. Assume h is not identically zero, so we can write it as h = β1κB1 + · · · + βpκBp, with (Bj)p j=1 ⊂A pairwise disjoint, and 0 < β1 < · · · < βp. Deﬁne the set B = B1 ∪· · · ∪Bp. It is obvious that, if we put t = β1/2, we have the inclusion B ⊂g−1(t, ∞] , so by Claim 1, we get µ(B) < ∞. This gives, of course µ(Bj) < ∞, ∀j = 1, . . . , p, so h is indeed elementary integrable. To prove the estimate (ii), we deﬁne the measurable functions hn : X →[0, ∞] by hn = min{gn, h}, ∀n ≥1. 304 CHAPTER IV: INTEGRATION THEORY Since 0 ≤hn ≤gn, ∀n ≥1, it follows that, hn ∈L1 +(X, A, µ), ∀n ≥1, and we have the inequalities (6) Z X hn dµ ≤ Z X gn dµ = αn, ∀n ≥1. It is obvious that we have (∗) 0 ≤h1 ≤h2 ≤· · · ≤h ≤βpκB (everywhere); (∗∗) h(x) = limn→∞hn(x), ∀x ∈X. Let us restrict everything to B. We consider the σ-algebra B = A B, and the measure ν = µ B. Consider the elementary function ψ = h B ∈B-ElemR(B), as well as the measurable functions ψn = hn B : B →[0, ∞], n ≥1. It is clear that ψ ∈L1 R,elem(B, B, ν), and we have the equality (7) Z B ψ dν = Z X h dµ. Likewise, using (∗), which clearly forces hn X∖B = 0, it follows that, for each n ≥1, the function ψn belongs to L1 +(B, B, ν), and by (6), we have (8) Z B ψn dν = Z X hn dµ, ∀n ≥1. Let us analyze the diﬀerences ϕn = ψ −ψn. On the one hand, using (∗), we have ϕn(x) ∈[0, βp], ∀x ∈B, n ≥1. On the other hand, again by (∗), we have ϕ1 ≥ϕ2 ≥. . . . Finally, by (∗∗) we have limn→∞ϕn(x) = 0, ∀x ∈B. We can apply Lemma 2.1, and we will get limn→∞ R B ϕn dν = 0. This clearly gives, Z B ψ dν = lim n→∞ Z B ψn dν, and then using (7) and (8), we get the equality Z X h dµ = lim n→∞ Z X hn dµ. Combining this with (6), immediately gives the desired estimate R X h dµ ≤α. Having proven Claim 2, let us observe now that, using the deﬁnition of the positive integral, it follows immediately that g ∈L1 +(X, A, µ), and we have the inequality Z X g dµ ≤α. The other inequality is pretty obvious, because the inequality g ≥gn forces Z X g dµ ≥ Z X gn dµ = αn, ∀n ≥1, so we immediately get Z X g dµ ≥sup{αn ; n ≥1} = α. □ Comment. In the previous section we introduced the convention which deﬁnes R X f dµ = ∞, if f : X →[0, ∞] is measurable, but f ̸∈L1 +(X, A, µ). Using this convention, the Lebesgue Monotone Convergence Theorem has the following general version. §2. Convergence theorems 305 Theorem 2.2 (General Lebesgue Monotone Convergence Theorem). Let (X, A, µ) be a measure space, and let f, fn : X →[0, ∞], n ≥1, be measurable functions, such that • fn ≤fn+1, µ-a.e., ∀n ≥1; • f = µ-a.e.- limn→∞fn. Then (9) Z X f dµ = lim n→∞ Z X fn dµ. Proof. As before, the sequence (αn)∞ n=1 ⊂[0, ∞], deﬁned by αn = R X fn dµ, ∀n ≥1, is non-decreasing, and is has a limit α = lim n→∞αn = sup Z X fn dµ : n ≥1 ∈[0, ∞]. There are two cases to analyze. Case I: α = ∞. In this case the inequalities f ≥fn ≥0, µ-a.e. will force Z X f dµ ≥ Z X fn dµ = αn, ∀n ≥1, which will force R X f dµ ≥α, so we indeed get Z X f dµ = ∞= α. Case II: α < ∞. In this case we apply directly Theorem 2.1. □ The following result provides an equivalent deﬁnition of integrability for non-negative functions (compare to the construction in Section 1). Corollary 2.1. Let (X, A, µ) be a measure space, and let f : X →[0, ∞] be a measurable function. The following are equivalent: (i) f ∈L1 +(X, A, µ); (ii) there exists a sequence (hn)∞ n=1 ⊂L1 R,elem(X, A, µ), with • 0 ≤h1 ≤h2 . . . ; • limn→∞hn(x) = f(x), ∀x ∈X; • sup R X hn dµ : n ≥1 < ∞. Moreover, if (hn)∞ n=1 is as in (ii), then one has the equality (10) Z X f dµ = lim n→∞ Z X hn dµ. Proof. (i) ⇒(ii). Assume f ∈L1 +(X, A, µ). Using Theorem III.3.2, we know there exists a sequence (hn)∞ n=1 ⊂A-ElemR(X), with (a) 0 ≤h1 ≤h2 ≤· · · ≤f; (b) limn→∞hn(x) = f(x), ∀x ∈X. Note the (a) forces hn ∈L1 R,elem(X, A, µ), as well as the inequalities R X hn dµ ≤ R X f dµ < ∞, ∀n ≥1, so the sequence (hn)∞ n=1 clearly satisﬁes condition (ii). The implication (ii) ⇒(i), and the equality (10) immediately follow from the General Lebesgue Monotone Convergence Theorem. □ 306 CHAPTER IV: INTEGRATION THEORY Corollary 2.2 (Fatou Lemma). Let (X, A, µ) be a measure space, and let fn : X →[0, ∞], n ≥1, be a sequence of measurable functions. Deﬁne the function f : X →[0, ∞] by f(x) = lim inf n→∞fn(x), ∀x ∈X. Then f is measurable, and one has the inequality Z X f dµ ≤lim inf n→∞ Z X fn dµ. Proof. The fact that f is measurable is already known (see Corollary III.3.5). Deﬁne the sequence (αn)∞ n=1 ⊂[0, ∞] by αn = R X fn dµ, ∀n ≥1. Deﬁne, for each integer n ≥1, the function gn : X →[0, ∞] by gn(x) = inf fk(x) : k ≥n , ∀x ∈X. By Corollary III.3.4, we know that gn, n ≥1 are all measurable. Moreover, it is clear that • 0 ≤g1 ≤g2 ≤. . . ; • f(x) = limn→∞gn(x), ∀x ∈X. By the General Lebesgue Monotone Convergence Theorem 2.2, it follows that (11) Z X f dµ = lim n→∞ Z X gn dµ. Notice that, if we deﬁne the sequence (βn)∞ n=1 ⊂[0, ∞], by βn = R X gn dµ, ∀n ≥1, then the obvious inequalities 0 ≤gn ≤fn give R X g dµ ≤ R X fn dµ, so we get βn ≤αn, ∀n ≥1. Using (11), we then get Z X f dµ = lim n→∞βn = lim inf n→∞βn ≤lim inf n→∞αn. □ The following is an important application of Theorem 2.1, that deals with Riemann integration. Corollary 2.3. Let a < b be real numbers. Denote by λ the Lebesgue measure, and consider the Lebesgue space [a, b], Mλ([a, b]), λ , where Mλ([a, b]) denotes the σ-algebra of all Lebesgue measurable subsets of [a, b]. Then every Riemann inte-grable function f : [a, b] →R belongs to L1 R([a, b], Mλ([a, b]), λ), and one has the equality (12) Z [a,b] f dλ = Z b a f(x) dx. Proof. We are going to use the results from III.6. First of all, the fact that f is Lebesgue integrable, i.e. f belongs to L1 R [a, b], Mλ([a, b]), λ , is clear since f is Lebesgue measurable, and bounded. (Here we use the fact that the measure space [a, b], Mλ([a, b]), λ is ﬁnite.) Next we prove the equality between the Riemann integral and the Lebesgue integral. Adding a constant, if necessary, we can assume that f ≥0. For every partition ∆= (a = x0 < x1 < · · · < xn = b) of [a, b], we deﬁne the numbers mk = inf t∈[xk−1,xk] f(t), ∀k = 1, . . . , n, §2. Convergence theorems 307 and we deﬁne the function f∆= m1κ[x0,x1] + m2κ(x1,x2] + · · · + mmκ(xn−1,xn]. Fix a sequence of partitions (∆p)∞ p=1, with ∆1 ⊂∆2 ⊂. . . , and limp→∞\|∆p\| = 0, We know (see III.6) that we have f = λ-a.e.- lim p→∞f∆p. Clearly we have 0 ≤f∆1 ≤f∆2 ≤· · · ≤f, so by Theorem 2.1, we get (13) Z [a,b] f dλ = lim p→∞ Z [a,b] f∆p dλ. Notice however thatZ [a,b] f∆p dλ = L(f, ∆p), ∀p ≥1, where L(f, ∆p) denotes the lower Darboux sum. Combining this with (13), and with the well known properties of Riemann integration, we immediately get (12). □ The following is another important convergence theorem. Theorem 2.3 (Lebesgue Dominated Convergence Theorem). Let (X, A, µ) be a measure space, let K be one of the symbols ¯ R, R, or C, and let (fn)∞ n=1 ⊂ L1 K(X, A, µ). Assume f : X →K is a measurable function, such that (i) f = µ-a.e.- limn→∞fn; (ii) there exists some function g ∈L1 +(X, A, µ), such that \|fn\| ≤g, µ-a.e., ∀n ≥1. Then f ∈L1 K(X, A, µ), and one has the equality (14) Z X f dµ = lim n→∞ Z X fn dµ. Proof. The fact that f is integrable follows from the following Claim: \|f\| ≤g, µ-a.e. To prove this fact, we deﬁne, for each n ≥1, the set Mn = x ∈X : \|fn(x)\| > g(x) . It is clear that Mn ∈A, and µ(Mn) = 0, ∀n ≥1. If we choose M ∈A such that µ(M) = 0, and f(x) = lim n→∞fn(x), ∀x ∈X ∖M, then the set N = M ∪ S∞ n=1 Mn ∈A will satisfy • µ(N) = 0; • \|fn(x)\| ≤g(x), ∀x ∈X ∖N; • f(x) = limn→∞fn(x), ∀x ∈X ∖N. We then clearly get \|f(x)\| ≤g(x), ∀x ∈X ∖N, and the Claim follows. Having proven that f is integrable, we now concentrate on the equality (14). Case I: K = ¯ R. 308 CHAPTER IV: INTEGRATION THEORY First of all, without any loss of generality, we can assume that 0 ≤g(x) < ∞, ∀x ∈X. (See Lemma 1.2.) Let us deﬁne the functions gn = min{fn, g} and hn = max{fn, −g}, n ≥1. Since we have −g ≤fn ≤g, µ-a.e., we immediately get (15) gn = hn = fn, µ-a.e., ∀n ≥1, thus giving the fact that gn, hn ∈L1 ¯ R(X, A, µ), ∀n ≥1, as well as the equalities (16) Z X gn dµ = Z X hn dµ = Z X fn dµ, ∀n ≥1. Deﬁne the measurable functions ϕ, ψ : X →¯ R by ϕ(x) = lim inf n→∞hn(x) and ψ(x) = lim sup n→∞gn(x), ∀x ∈X. Using (15), we clearly have f = ϕ = ψ, µ-a.e., so we get (17) Z X f dµ = Z X ϕ dµ = Z X ψ dµ. Remark also that we have equalities (18) g(x)−ϕ(x) = lim inf n→∞[g(x)−gn(x)] and g(x)+ψ(x) = lim inf n→∞[g(x)+hn(x)], ∀x ∈X. Since we clearly have g −gn ≥0 and g + hn ≥0, ∀n ≥1, using (18), and Fatou Lemma (Corollary 2.2) and we get the inequalities Z X (g −ϕ) dµ ≤lim inf n→∞ Z X (g −gn) dµ, Z X (g + ψ) dµ ≤lim inf n→∞ Z X (g + hn) dµ, In other words, we get Z X g dµ − Z X ϕ dµ ≤lim inf n→∞ Z X g dµ − Z X gn dµ = Z X g dµ −lim sup n→∞ Z X gn dµ, Z X g dµ + Z X ψ dµ ≤lim inf n→∞ Z X g dµ + Z X hn dµ = Z X g dµ + lim inf n→∞ Z X hn dµ. Using the equalities (16) and (17), the above inequalities give Z X f dµ = Z X ϕ dµ ≥lim sup n→∞ Z X gn dµ = lim sup n→∞ Z X fn dµ, Z X f dµ = Z X ψ dµ ≤lim inf n→∞ Z X hn dµ = lim inf n→∞ Z X fn dµ. In other words, we have Z X f dµ ≤lim inf n→∞ Z X fn dµ ≤lim sup n→∞ Z X fn dµ ≤ Z X f dµ, thus giving the equality (14) The case K = R is trivial (it is in fact contained in case K = ¯ R). §2. Convergence theorems 309 The case K = C is also pretty clear, using real and imaginary parts, since for each n ≥1, we clearly have \|Re fn\| ≤g, µ-a.e., \|Im fn\| ≤g, µ-a.e.,. □ Exercise 3. Give an example of a sequence of continuous functions fn : [0, 1] → [0, ∞), such that (a) limn→∞fn(x) = 0, ∀n ≥1; (b) R [0,1] fn dλ = 1, ∀n ≥1. (Here λ denotes the Lebesgue measure). This shows that the Lebesgue Dominated Convergence Theorem fails, without the dominance condition (ii). Hint: Consider the functions fn deﬁned by fn(x) =    n2x if 0 ≤x ≤1/n n(2 −nx) if 1/n ≤x ≤2/n 0 if 2/n ≤x ≤1 The Lebesgue Convergence Theorems 2.2 and 2.3 have many applications. They are among the most important results in Measure Theory. In many instances, these theorem are employed during proofs, at key steps. The next two results are good illustrations. Proposition 2.1. Let (X, A, µ) be a measure space, and let f : X →[0, ∞] be a measurable function. Then the map ν : A ∋A 7− → Z A f dµ ∈[0, ∞] deﬁnes a measure on A. Proof. It is clear that ν(∅) = 0. To prove σ-additivity, start with a pairwise disjoint sequence (An)∞ n=1 ⊂A, and put A = S∞ n=1 An. For each integer n ≥1, deﬁne the set Bn = Sn k=1 Ak, and the measurable function gn = fκBn. Deﬁne also the function g = fκA. It is obvious that • 0 ≤g1 ≤g2 ≤· · · ≤g (everywhere), • limn→∞gn(x) = g(x), ∀x ∈X. Using the General Lebesgue Monotone Convergence Theorem, it follows that (19) ν(A) = Z X fκA dµ = Z X g dµ = lim n→∞ Z X gn dµ. Notice now that, for each n ≥1, one has the equality gn = fκA1 + · · · + fκAn, so using Remark 1.7.C, we get Z X gn dµ = n X k=1 Z X fκAk dµ = n X k=1 ν(Ak), so the equality (19) immediately gives ν(A) = P∞ n=1 ν(An). □ The next result is a version of the previous one for K-valued functions. Proposition 2.2. Let (X, A, µ) be a measure space, let K be one of the symbols ¯ R, R, or C, and let (An)∞ n=1 ⊂A be a pairwise disjoint sequence with S∞ n=1 An = X. For a function f : X →K, the following are equivalent. 310 CHAPTER IV: INTEGRATION THEORY (i) f ∈L1 K(X, A, µ); (ii) f An ∈L1 K(An, A An, µ An), ∀n ≥1, and ∞ X n=1 Z An \|f\| dµ < ∞. Moreover, if f satisﬁes these equivalent conditions, then Z X f dµ = ∞ X n=1 Z An f dµ. Proof. (i) ⇒(ii). Assume f ∈L1 K(X, A, µ). Applying Proposition 2.1, to \|f\|, we immediately get ∞ X n=1 Z An \|f\| dµ = Z X \|f\| dµ < ∞, which clearly proves (ii). (ii) ⇒(i). Assume f satisﬁes condition (ii). Deﬁne, for each n ≥1, the set Bn = A1 ∪· · · ∪An. First of all, since (An)∞ n=1 ⊂A, and S∞ n=1 An = X, it follows that f is measurable. Consider the the functions fn = fκBn and gn = fκAn, n ≥1. Notice that, since f An ∈L1 K(An, A An, µ An), it follows that gn ∈L1 K(X, A, µ), ∀n ≥1, and we also have Z X gn dµ = Z An f dµ, ∀n ≥1. In fact we also have Z X \|gn\| dµ = Z An \|f\| dµ, ∀n ≥1. Notice that we obviously have fn = g1 + · · · + gn, and \|fn\| = \|g1\| + · · · + \|gn\|, so if we deﬁne S = ∞ X n=1 Z An \|f\| dµ, we get Z X \|fn\| dµ = n X k=1 Z Ak \|f\| dµ ≤S < ∞, ∀n ≥1. Notice however that we have 0 ≤\|f1\| ≤\|f2\| ≤. . . \|f\|, as well as the equality limn→∞fn(x) = f(x), ∀x ∈X. On the one hand, using the General Lebesgue Monotone Convergence Theorem, we will get Z X \|f\| dµ = lim n→∞ Z X \|fn\| dµ = lim n→∞ n X k=1 Z Ak \|f\| dµ = ∞ X n=1 Z An \|f\| dµ = S < ∞, which proves that \|f\| ∈L1 +(X, A, µ), so in particular f belongs to L1 K(X, A, µ). On the other hand, since we have \|fn\| ≤\|f\|, by the Lebesgue Dominated Convergence Theorem, we get Z X f dµ = lim n→∞ Z X fn dµ = lim n→∞ n X k=1 Z Ak f dµ = ∞ X n=1 Z An f dµ. □ §2. Convergence theorems 311 Corollary 2.4. Let (X, A, µ) be a measure space, let K be one of the symbols ¯ R, R, or C, and let (Xn)∞ n=1 ⊂A be sequence with S∞ n=1 Xn = X, and X1 ⊂ X2 ⊂. . . . For a function f : X →K be a measurable function, the following are equivalent. (i) f ∈L1 K(X, A, µ); (ii) f Xn ∈L1 K(Xn, A Xn, µ Xn), ∀n ≥1, and sup Z Xn \|f\| dµ : n ≥1 < ∞. Moreover, if f satisﬁes these equivalent conditions, then Z X f dµ = lim n→∞ Z Xn f dµ. Proof. Apply the above result to the sequence (An)∞ n=1 given by A1 = X1 and An = Xn ∖Xn−1, ∀n ≥2. □ Remark 2.2. Suppose (X, A, µ) is a measure space, K is one of the ﬁelds R or C, and f ∈L1 K(X, A, µ). By Proposition 2.2, we get the fact that the map ν : A ∋A 7− → Z A f dµ ∈K is a K-valued measure on A. By Proposition 2.1, we also know that ω : A ∋A 7− → Z A \|f\| dµ ∈K is a ﬁnite “honest” measure on A. Using Proposition 1.6, we clearly have \|ν(A)\| = Z A f dµ ≤ Z A \|f\| dµ = ω(A), ∀A ∈A, which by the results from III.8 gives the inequality \|ν\| ≤ω. (Here \|ν\| denotes the variation measure of ν.) Later on (see Section 4) we are going to see that in fact we have the equality \|ν\| = ω. Comment. It is important to understand the “sequential” nature of the con-vergence theorems discussed here. If we examine for instance the Mononotone Convergence Theorem, we could easily formulate a “series” version, which states the equality Z X ∞ X n=1 fn dµ = ∞ X n=1 Z X fn dµ, for any sequence measurable functions fn : X →[0, ∞]. Suppose now we have an arbitrary family fj : X →[0, ∞], j ∈J of measurable functions, and we deﬁne f(x) = X j∈J fj(x), ∀x ∈X. (Here we use the summability convention which deﬁnes the sum as the supremum of all ﬁnite sums.) In general, f is not always measurable. But if it is, one still cannot conclude that Z X f dµ = X j∈J Z X fi dµ. 312 CHAPTER IV: INTEGRATION THEORY The following example illustrates this anomaly. Example 2.1. Take the measure space ([0, 1], Mλ([0, 1]), λ), and ﬁx J ⊂[0, 1] and arbitrary set. For each j ∈J we consider tha characteristic function fj = κ{j}. It is obvious that the function f : X →[0, ∞], deﬁned by f(x) = X j∈J fj(x), ∀x ∈[0, 1], is equal to κJ If J is non-measurable, this already gives an example when f = P j∈J fj is non-measurable. But even if J were measurable, it would be impossible to have the equality Z X f dλ = X j∈J Z X fj dλ, simply because the right hand side is zero, while the left hand side is equal to λ(J). The next two exercises illustrate straightforward (but nevertheless interesting) applications of the convergence theorems to quite simple situations. Exercise 4. Let A be a σ-algebra on a (non-empty) set X, and let (µn)∞ n=1 be a sequence of signed measures on A. Assume that, for each A ∈A, the sequence µn(A) ∞ n=1 has a limit denoted µ(A) ∈[−∞, ∞]. Prove that the map µ : A → [0, ∞] deﬁnes a measure on A, if the sequence (µn)∞ n=1 satisﬁes one of the following hypotheses: A. 0 ≤µ1(A) ≤µ2(A) ≤. . . , ∀A ∈A; B. there exists a ﬁnite measure ω on A, such that \|µn(A)\| ≤ω(A), ∀n ≥1, A ∈A. Hint: To prove σ-additivity, ﬁx a pairwise disjoint sequence (Ak)∞ k=1 ⊂A, and put A = S∞ k=1 Ak. Treat the problem of proving the equality µ(A) = P∞ k=1 µ(Ak) as a convergence problem on the measure space (N, P(N), ν) - with ν the counting measure - for the sequence of functions fn : N →[0, ∞] deﬁned by fn(k) = µn(Ak), ∀k ∈N. Exercise 5. Let A be a σ-algebra on a (non-empty) set X, and let (µj)j∈J be a family of signed measures on A. Assume either of the following is true: A. µj(A) ≥0, ∀j ∈J, A ∈A. B. There exists a ﬁnite measure ω on A, such that P j∈J \|µj(A)\| ≤ω(A), ∀A ∈A. Deﬁne the map µ : A →[0, ∞] by µ(A) = P j∈J µj(A), ∀A ∈A. (In Case A, the sum is deﬁned as the supremum over ﬁnite sums. In case B, it follows that the family µj(A) j∈J is summable.) Prove that µ is a measure on A. Hint: To prove σ-additivity, ﬁx a pairwise disjoint sequence (Ak)∞ k=1 ⊂A, and put A = S∞ k=1 Ak. To prove the equality µ(A) = P∞ k=1 µ(Ak), analyze the following cases: (i) There is some k ≥1, such that µ(Ak) = ∞; (ii) µ(Ak) < ∞, ∀k ≥1. The ﬁrst case is quite trivial. In the second case reduce the problem to the previous exercise, by observing that, for each k ≥1, the set J(Ak) = {j ∈J : µj(Ak) > 0 must be countable. Then the set J(A) = {j ∈J : µj(A) > 0} is also countable. Comment. One of the major drawbacks of the theory of Riemann integration is illustrated by the approach to improper integration. Recall that for a function §2. Convergence theorems 313 h : [a, b) →R the improper Riemann integral is deﬁned as Z b− a h(t) dt = lim x→b− Z x a f(t) dt, provided that (a) h [a,x] is Riemann integrable, ∀x ∈(a, b), and (b) the above limit exists. The problem is that although the improper integral may exist, and the function is actually deﬁned on [a, b], it may fail to be Riemann integrable, for example when it is unbounded. In contrast to this situation, by Corollary 2.4, we see that if for example h ≥0, then the Lebesgue integrability of h on [a, b] is equivalent to the fact that (i) h [a,x] is Lebesgue integrable, ∀x ∈(a, b), and (ii) limx→b− R [a,x] h dλ exists. Going back to the discussion on improper Riemann integral, we can see that a suﬃcient condition for h : [a, b) →R to be Riemann integrable in the improper sense, is the fact that h has property (a) above, and h is Lebesgue integrable on [a, b). In fact, if h ≥0, then by Corollary 2.4, this is also necessary. Notation. Let −∞≤a < b ≤∞, and let f be a Lebesgue integrable function, deﬁned on some interval J which is one of (a, b), [a, b), (a, b], or [a, b]. Then the Lebesgue integral R J f dλ will be denoted simply by R b a f dλ. Exercise 6. Let (X, A, µ) be a ﬁnite measure space. Prove that for every f ∈L1 +(X, A, µ), one has the equality Z X f dµ = Z ∞ 0 µ f −1([t, ∞]) dt, where the second term is deﬁned as improper Riemann integral. Hint: The function ϕ : [0, ∞) →[0, ∞) deﬁned by ϕ(t) = µ f−1([t, ∞]) , ∀t ≥0, is non-increasing, so it is Riemann integrable on every interval [0, a], a > 0. Prove the inequalities Z Xa f dµ ≤ Z a 0 ϕ(t) dt ≤ Z X f dµ, ∀a > 0, where Xa = f−1([0, a)), by analyzing lower and upper Darboux sums of ϕ [0,a]. Use Corollary 2.4 to get lima→∞ R Xa f dµ = R X f dµ.
9206	https://www.colgate.com/en-us/oral-health/mouth-and-teeth-anatomy/how-many-teeth-do-we-have	We take your privacy seriously We use cookies and similar technologies to understand how you use our website and to create more valuable experiences for you. These cookies may be used for different purposes, including personalized advertising and measuring site usage. To learn more please see our cookie policy. CONTENTS How Many Teeth Do We Have? January 9, 2023. 3 min read Have you ever wondering how many teeth you have? Well, it's your lucky day. Because we're going to dive into the answer. The truth is that the number of teeth you have depends on a few factors: age and the presence of certain conditions. As you may know, children and adults actually have different sets of teeth, and the number of them varies. As you read on, we'll look into the differences between children and adult teeth as well as the conditions that affect the number of teeth you have. Baby Teeth Children begin teething around six months of age, meaning that their teeth start coming in around that point. These first teeth are known as baby teeth, primary teeth, or, technically, deciduous teeth. As you know, they eventually fall out and are replaced by adult teeth. How many teeth do children have? Children have 20 baby teeth. That includes 10 teeth on the top and 10 teeth on the bottom. All 20 of them usually come in before the age of three and act as placeholders for the adult teeth that grow in after the baby teeth fall out. At about age six, most children begin to lose their baby teeth, which are then replaced with adult teeth. This process continues until kids reach their early teen years. It's important to mention that just because baby teeth fall out doesn't mean that they don't need care like adult teeth. Oral care routines should begin before a baby's first tooth comes in. This can be done by running a clean, damp washcloth over the baby's gums. Once the first tooth comes in, parents can brush the baby's teeth with an infant toothbrush and a tiny bit of toothpaste. Flossing can begin once the baby's teeth touch. Adult Teeth When adult teeth come in, there's more room in the mouth and more teeth. How many teeth do adults have? Most adults have 32 teeth, which is 12 teeth more than children! Among these 32 teeth are 8 incisors, 4 canines, 8 premolars, and 12 molars, including 4 wisdom teeth. It is very common for adults to have their wisdom teeth removed because there is not always enough room for them to grow in comfortably or without causing misalignment of other teeth. Most people have a complete set of adult teeth by the time they reach their teenage years. Conditions That Affect The Number of Teeth Tooth loss is a surprisingly common condition. And periodontal disease (aka gum disease) is the main reason adults lose their teeth. Luckily, prevention is possible. It involves brushing twice a day and flossing daily. Other reasons for tooth loss include: Tooth decay Ectodermal dysplasia (a genetic disorder that can affect teeth) Gastrointestinal reflux (severe tooth erosion which causes stomach acid to come up into the mouth) Beyond tooth loss, there's also a condition called tooth agenesis, where teeth are missing in the mouth. It can either appear as the absence of all teeth or as 1 to 6 or more teeth missing. This condition is genetic and rare. There are also circumstances where extra teeth, or supernumerary teeth, appear in the mouth. The most common type of supernumerary tooth is an extra incisor located between two central incisors. Now you know the deal about how many teeth we have. Typically, children have 20 teeth, and adults have 32 teeth (28 if the wisdom teeth are removed). Remember that each one of them needs your care, even baby teeth. There are some cases where the number of teeth varies, resulting from tooth loss, tooth agenesia, or having extra teeth. So keep up with your oral hygiene no matter how many teeth you have. It'll help make sure your teeth stay healthy and happy. Oral Care Center articles are reviewed by an oral health medical professional. This information is for educational purposes only. This content is not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of your dentist, physician or other qualified healthcare provider. ORAL HEALTH QUIZ What's behind your smile? Take our Oral Health assessment to get the most from your oral care routine Start Now ORAL HEALTH QUIZ What's behind your smile? Take our Oral Health assessment to get the most from your oral care routine Start Now Related Articles Mouth and teeth anatomy Parts of the Mouth and Their Functions The mouth, or oral cavity, is made up of several components that work together so you can breathe, speak, and eat. Learn more about the parts of your mouth. Read More Mouth and teeth anatomy What Is Myofunctional Therapy? When diagnosed, an abnormal tongue position can be corrected with myofunctional therapy, a special training process that ensures a future of good health. Read More Mouth and teeth anatomy Preparing For Your Baby's First Molar Watching your baby's teeth develop is an exciting milestone, but it can also be an overwhelming time for a parent. What can you expect from a first molar? Read More Mouth and teeth anatomy Periodontal Ligament: What Is It? What is the periodontal ligament and why is it important to your overall oral health? Learn more, here. Read More Related Products Colgate Total Battery Powered Toothbrush, White Power away plaque with Colgate Total Battery Powered Toothbrush. This battery operated toothbrush for adults fights the root cause of cavities, plaque, gingivitis, bad breath, tartar buildup, and stains. 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9207	https://www.quora.com/What-is-the-thermal-velocity-of-electrons	What is the thermal velocity of electrons? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Physics Thermal Velocity Free Electrons Particle Theory of Matter Statistical Thermodynamic... Gas Physics Quantum Mechanics Motion of Electrons Electrons 5 What is the thermal velocity of electrons? All related (37) Sort Recommended Leonard Carter Former Professor of Physics at Utah Tech University (2017–2022) · Author has 1.9K answers and 2.1M answer views ·Updated 3y While the answers given are all good, one aspect of this that’s very unexpected is the energy and velocity of conduction electrons in metals, which follows a Fermi-Dirac distribution and maxes out at what’s called the Fermi surface in k-space. The Fermi surface energy is given as E = hbar^2/2m(3pi^2N/V)^2/3 where m is electron mass and N/V is the electron number density. This results in some very high numbers for metals: Copper - 7.0 ev, 82,000 degrees Kelvin, 1.56 X 10^8 cm/sec Silver - 5.5 ev, 64,000 deg. K, 1.38 X 10^8 cm/sec Lithium - 4.7 ev, 55,000 deg. K, 1.3 X 10^8 cm/sec So the maximum vel Continue Reading While the answers given are all good, one aspect of this that’s very unexpected is the energy and velocity of conduction electrons in metals, which follows a Fermi-Dirac distribution and maxes out at what’s called the Fermi surface in k-space. The Fermi surface energy is given as E = hbar^2/2m(3pi^2N/V)^2/3 where m is electron mass and N/V is the electron number density. This results in some very high numbers for metals: Copper - 7.0 ev, 82,000 degrees Kelvin, 1.56 X 10^8 cm/sec Silver - 5.5 ev, 64,000 deg. K, 1.38 X 10^8 cm/sec Lithium - 4.7 ev, 55,000 deg. K, 1.3 X 10^8 cm/sec So the maximum velocities are around 1/3 to 1/2 of one percent of the speed of light! Ref: Kittel, Introduction to Solid State Physics, Chap. 7 Upvote · 9 5 9 9 Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 201 99 34 9 3 Related questions More answers below Why is drift velocity less than thermal velocity? What does thermal velocity mean? How do you calculate thermal velocity? What is the thermal velocity of an individual free electron in a conductor? What is the expression of thermal velocity? Salwa Salah Philosopher&Doctor and Writer · Author has 10.8K answers and 27.1M answer views ·2y Thermal velocity of electron is 10^5. But average velocity is zero why. Continue Reading Thermal velocity of electron is 10^5. But average velocity is zero why. Upvote · 9 7 Assistant Bot · Sep 7 Thermal velocity of electrons is a characteristic speed representing the random motion of electrons in a material or plasma due to temperature. It is not a single fixed number but depends on temperature and the context (classical, degenerate, or relativistic). Practical definitions and formulas: Most-probable (Maxwellian) speed (classical, non‑degenerate gas) v_mp = sqrt(2 k_B T / m_e) k_B = Boltzmann constant (1.380649×10^−23 J/K) m_e = electron mass (9.10938356×10^−31 kg) Use when electron velocities follow a Maxwell–Boltzmann distribution. Example: at T = 300 K, v_mp ≈ 1.1×10^5 m/s. Root-mean-s Continue Reading Thermal velocity of electrons is a characteristic speed representing the random motion of electrons in a material or plasma due to temperature. It is not a single fixed number but depends on temperature and the context (classical, degenerate, or relativistic). Practical definitions and formulas: Most-probable (Maxwellian) speed (classical, non‑degenerate gas) v_mp = sqrt(2 k_B T / m_e) k_B = Boltzmann constant (1.380649×10^−23 J/K) m_e = electron mass (9.10938356×10^−31 kg) Use when electron velocities follow a Maxwell–Boltzmann distribution. Example: at T = 300 K, v_mp ≈ 1.1×10^5 m/s. Root-mean-square (rms) speed v_rms = sqrt(3 k_B T / m_e) Typical in kinetic-energy based estimates. Example: at T = 300 K, v_rms ≈ 1.35×10^5 m/s. Thermal (most-probable) speed used in plasma physics (sometimes defined with 1/2 factor) v_th = sqrt(k_B T / m_e) Different communities use different conventions; always check which prefactor (1, 2, or 3) is used. Fermi velocity (degenerate electron gas, e.g., metals or white dwarfs) v_F = ħ (3π^2 n)^(1/3) / m_e n = electron number density; ħ = reduced Planck constant. At room-temperature metals, v_F ~ 10^6 m/s, larger than classical thermal speeds — thermal motion is then a small perturbation. Relativistic temperatures For k_B T comparable to or > m_e c^2 (511 keV), use relativistic expressions (mean energy → relativistic kinematics); thermal speed approaches c. How to choose: Use v_rms = sqrt(3 k_B T / m_e) for kinetic-energy arguments. Use v_mp = sqrt(2 k_B T / m_e) when quoting the most probable speed from a Maxwellian. Use v_th = sqrt(k_B T / m_e) if following plasma-physics convention (check source). For metals or very high densities, use Fermi velocity instead of classical thermal speed. Quick numeric scale: T = 300 K: v_rms ≈ 1.35×10^5 m/s. T = 10^4 K (typical H II region): v_rms ≈ 2.5×10^6 m/s. T = 10^6 K (coronal plasma): v_rms ≈ 8×10^6 m/s. Always state which definition and numerical constants are used when quoting a "thermal velocity." Upvote · Narsimhan Raghunathan Engineering Consultant at Oxford Global Resources (2015–present) · Author has 626 answers and 362.4K answer views ·8y Thermal velocity refers basically to the kinetic energy due to heat. For instance a plasma can be generated by heating the gas , which will strip the electrons from the gas atoms if the temperature is sufficiently high. Non thermal or athermal velocity would be due to charge interactions. Upvote · 9 6 Related questions More answers below What is the value of the thermal speed of an electron? What is the thermal velocity of an electric current? Why is thermal velocity more than drift velocity? What is the velocity of an electron? How Drift velocity of electrons produced? Hedley Rokos Author has 2.4K answers and 1.3M answer views ·3y The rms thermal speed of a free electron can be calculated from the energy per degree of freedom, and the mass of an electron and the number of degrees of freedom (on this case 3): So E = 3 x kT/2. kT/2 is about 2E-21 Joulses at 25-degC. The mass of an elextron is about 9.1E-31kg, so we have: , , , 9.1E-31 x V^2/2 = 6E-21 or . . . V = 115-km/s Effective electron mass in most metals is very close to the free-space value, so the value would be very close to this in copper or silver. though possibly not exactly the same Upvote · Brandon Cammon Quora user at Quora (product) (2018–present) · Author has 2.6K answers and 967.1K answer views ·4y When you add thermal energy to gases into plasma, electrons consume thermal radiation, exit their orbitals & gain thermal velocity Upvote · Sponsored by Avnet Silica We're at the Pulse of the Market. Tap into emerging tech and market insights to stay informed, reduce risk, and make better decisions. Learn More Isaac Clark Degree in electrical engineering and physics with 35 years engineering exper. · Author has 6.3K answers and 3.4M answer views ·8y Thermal just means that the electrons are in thermal equilibrium with their surrounds. It is a means of indicating the energy of the electrons. Upvote · John Walker BA Hons Natural Sciences, specializing in Physics. Retired Science Teacher. · Author has 1.5K answers and 3.4M answer views ·4y Related What is the thermal velocity of an individual free electron in a conductor? It’s about 100 m/s. To get this value, you need to manipulate the equations; pV = NRT and p = 1/3 M/V c²(av) where p is pressure, V is volume, N is number of moles, R is the gas constant, T is absolute temperature, M is mass and c²(av) is the mean square speed. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · Promoted by Betterbuck Anthony Madden Writer for Betterbuck ·Updated Aug 15 What are the weirdest mistakes people make on the internet right now? Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr Continue Reading Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice. More often than not, it saves even more than 50% on load times - here’s a test I ran: Using an ad blocker saved a whopping 6.5+ seconds of load time. Here’s a link to Total Adblock, if you’re interested. Not getting paid for your screentime Apps like Freecash will pay you to test new games on your phone. Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website): You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS). If you're scrolling on your phone anyway, why not get paid for it? I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool). Upvote · 999 557 99 60 9 4 Kip Ingram PhD in Electrical Engineering, The University of Texas at Austin Cockrell School of Engineering (Graduated 1992) · Author has 20.1K answers and 21.6M answer views ·Aug 21 Related If a photon is not re-emitted (reflected) immediately but transformed into thermal energy after being absorbed by a bounded electron, does that mean the electron stays in a higher energy level for longer time before giving out thermal radiation? Reflected light and thermal (black body) radiation both arise from the same mechanism. All that really happens in materials is that atoms absorb photons, and emit photons. Emitted photons go in completely random directions. If a surface layer atom absorbs a photon, then when it emits a photon later there’s around a 50% chance that that photon will go in a direction that carries it away from the material. We would regard that photon as reflected light. But there’s also about a 50% probability that the photon goes in a direction that carries it further into the material, where it is likely to be Continue Reading Reflected light and thermal (black body) radiation both arise from the same mechanism. All that really happens in materials is that atoms absorb photons, and emit photons. Emitted photons go in completely random directions. If a surface layer atom absorbs a photon, then when it emits a photon later there’s around a 50% chance that that photon will go in a direction that carries it away from the material. We would regard that photon as reflected light. But there’s also about a 50% probability that the photon goes in a direction that carries it further into the material, where it is likely to be absorbed by some other atom within the material. Then when that atom later emits a photon, that new photon might or might not escape the material. This process can potentially repeat many times - some bits of energy “rattle around” inside the material for quite a while. When energy arrives at the material and then departs the material almost immediately, we call it reflected light. When it rattles around for a while before leaving, we call that blackbody radiation. Some quanta of energy never leave, but rather get converted into the momentum of “jiggling atoms.” If the surface of the material is extremely smooth, then we can perceive reflected images when we observe the surface. These photons were basically “immediately emitted back out of the material,” and that overall energy bears a strong correlation to the energy that was incident on the surface. The energy that “rattles around” for a while, though, has been rather thoroughly randomized, and does not carry image information. Stay safe and well! Kip My sincere thanks to my Patreon supporters. (Get more from Kip Ingram on Patreon) Upvote · 99 10 Bill Otto Technical Fellow in Lasers & Optics at Boeing, Retired · Author has 8.7K answers and 80.3M answer views ·4y Related If a photon is not re-emitted (reflected) immediately but transformed into thermal energy after being absorbed by a bounded electron, does that mean the electron stays in a higher energy level for longer time before giving out thermal radiation? When an electron absorbs visible or ultraviolet light, it goes into a higher orbital state. Electrons do not literally orbit the nucleus, but that is the closest non-quantum analogy we have. Usually, unless stimulated emission occurs, the electron does not snap back into the original orbital state but rather into some intermediate state. It does this either by emitting a photon of lower energy, or by contributing the energy to a collision with another atom. Once this happens, the original energy is on its way to becoming part of the thermal energy of the object. When a lower energy photon is emit Continue Reading When an electron absorbs visible or ultraviolet light, it goes into a higher orbital state. Electrons do not literally orbit the nucleus, but that is the closest non-quantum analogy we have. Usually, unless stimulated emission occurs, the electron does not snap back into the original orbital state but rather into some intermediate state. It does this either by emitting a photon of lower energy, or by contributing the energy to a collision with another atom. Once this happens, the original energy is on its way to becoming part of the thermal energy of the object. When a lower energy photon is emitted, we usually call this fluorescence, but if the photon is infrared, it usually is not noticed. By the way, the lower energy photon can usually be used by a laser, and this is how most 3 or 4 level lasers work. Since most materials have atomic collisions, the momentum and energy of the original photon can be quickly distributed into the general Boltzmann distribution of states which characterizes thermal equilibrium. So the answer is a bit point of view. If a sufficient fraction of atoms is in an excited state, you do not have a thermal equilibrium. But it is impossible to speak of a single atom as being “thermal" or not because thermal is a statistical description that does not apply to a single atomic state. By the way, it is completely wrong to think of a photon as being immediately re-emitted to be “reflected.” That is a completely false picture of reflection taught in first year physics. A reflection requires interacting with large number of electrons simultaneously and not actually causing any of them to go into a higher orbital state. I find it very disturbing that they teach this nonsense to beginning physics students, but it was decided that it was necessary to capture their interest. Thank you the marketing department of major universities. Upvote · 99 37 9 9 Sponsored by LPU Online Career Ka Turning Point with LPU Online. 100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support. Apply Now 999 263 Ritwik Sunny Former Customer Support Executive at Ashok Leyland · Author has 39.8K answers and 9.7M answer views ·3y Related What is the thermal velocity of an individual free electron in a conductor? thermal velocity of a conductor is of the order of 10^5 m/s. Upvote · Shree Hari Mittal Studied Physics at University of Delhi (Graduated 2019) · Author has 292 answers and 514.5K answer views ·8y Related What is the velocity of an electron? Velocity of electron where? It's like asking what is my velocity? If I'm in my bedroom, zero. If on a highway, 120km per hour or so. While commuting in city, 30km per hour. It's different in different situations. Similarly, an electron has different velocity in different situations. Although, velocity of electron is constant for different orbits. But the constant is specific only for that orbit or shell. When electron is unbounded, it's velocity then depends on external factors. Temperature, electric field, magnetic field etc. So velocity of electron is an arbitrary constant. Much like your veloc Continue Reading Velocity of electron where? It's like asking what is my velocity? If I'm in my bedroom, zero. If on a highway, 120km per hour or so. While commuting in city, 30km per hour. It's different in different situations. Similarly, an electron has different velocity in different situations. Although, velocity of electron is constant for different orbits. But the constant is specific only for that orbit or shell. When electron is unbounded, it's velocity then depends on external factors. Temperature, electric field, magnetic field etc. So velocity of electron is an arbitrary constant. Much like your velocity. Upvote · 99 16 9 2 Jesse Hersch PhD in physics (scattering theory) ·10y Related Is temperature the velocity of atoms or electrons? Temperature tells you something about the average energy of a particle in a large collection of particles. From energy you can get to velocity. Those particles could be atoms or electrons. The idea of temperature for any particular particle doesn't mean much. Upvote · 9 2 Related questions Why is drift velocity less than thermal velocity? What does thermal velocity mean? How do you calculate thermal velocity? What is the thermal velocity of an individual free electron in a conductor? What is the expression of thermal velocity? What is the value of the thermal speed of an electron? What is the thermal velocity of an electric current? Why is thermal velocity more than drift velocity? What is the velocity of an electron? How Drift velocity of electrons produced? Can the velocity of an electron be zero? What is the difference between thermal velocity and velocity? What is the velocity of an electron in a vacuum? What is the relation between electric field and thermal velocity? What is thermal energy? Related questions Why is drift velocity less than thermal velocity? What does thermal velocity mean? How do you calculate thermal velocity? What is the thermal velocity of an individual free electron in a conductor? What is the expression of thermal velocity? What is the value of the thermal speed of an electron? 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9208	http://www.csun.edu/~ba70714/survey.pdf	THE RECTILINEAR CROSSING NUMBER OF Kn: CLOSING IN (OR ARE WE?) BERNARDO M. ´ ABREGO, SILVIA FERN´ ANDEZ-MERCHANT, AND GELASIO SALAZAR Abstract. The calculation of the rectilinear crossing number of complete graphs is an important open problem in combinatorial geometry, with impor-tant and fruitful connections to other classical problems. Our aim in this work is to survey the body of knowledge around this parameter. 1. Introduction In a rectilinear (or geometric) drawing of a graph G, the vertices of G are re-presented by points, and an edge joining two vertices is represented by the straight segment joining the corresponding two points. Edges are allowed to cross, but an edge cannot contain a vertex other than its endpoints. The rectilinear (or geometric) crossing number cr(G) of a graph G is the minimum number of pairwise crossings of edges in a rectilinear drawing of G in the plane. 1.1. The relevance of cr(Kn). As with every graph theory parameter, there is a natural interest in calculating the rectilinear crossing number of certain families of graphs, such as the complete bipartite graphs Km,n and the complete graphs Kn. The estimation of cr(Kn) is of particular interest, since cr(Kn) equals the minimum number □(n) of convex quadrilaterals deﬁned by n points in the plane in general position; the problem of determining □(n) belongs to a collection of classical combinatorial geometry problems, the so-called Erd˝ os-Szekeres problems. Another important motivation to study cr(Kn) is its close connection with the celebrated Sylvester Four Point Problem from geometric probability. Sylvester asked what is the probability that four points chosen at random in the plane form a convex quadrilateral . After it became clear that this is an ill-posed question , Sylvester put forward a related conjecture. Let R be a bounded convex open set in the plane with ﬁnite area, and let q(R) be the probability that four points chosen randomly from R deﬁne a convex quadrilateral. Then (Sylvester’s Conjecture ) q(R) is minimized when R is a circle or an ellipse, and maximized when R is a triangle. This conjecture was proved by Blashke in 1917 . Scheinerman and Wilf addressed in the general problem when R is not required to be convex. It is easy to see that in this case q(R) can be made arbitrarily close to 1 by choosing R to be very thin annulus. The remaining problem is to determine the inﬁmum q∗:= inf q(R), taken over all open sets R with ﬁnite area. Scheinerman and Wilf Date: March 1, 2011. 2010 Mathematics Subject Classiﬁcation. 52C30, 52C10, 52C45, 05C62, 68R10, 60D05, 52A22. Key words and phrases. Rectilinear crossing number, geometric drawings, k-sets, Sylvester Four Point Problem. 1 2 BERNARDO M. ´ ABREGO, SILVIA FERN´ ANDEZ-MERCHANT, AND GELASIO SALAZAR established the striking connection (1) q∗= lim n→∞ cr(Kn) n 4 , thus inextricably linking the estimation of Sylvester’s Four Point Constant q∗to the (asymptotic) behaviour of cr(Kn). As we shall see below, recent developments have unveiled a close relationship between cr(Kn) and yet another classical combinatorial geometry parameter: the number of (≤k)-edges in an n-point set. 1.2. Purpose and timeliness of this survey. Up until 2000, very little was known about cr(Kn). Since then, our knowledge of this problem has seen a tremen-dous growth. Surprising and useful connections to other classical problems have been unveiled. The current estimates for cr(Kn) have reached a point that would have seemed unlikely (to say the least) at the beginning of the previous decade. For instance, before 2000 the ratio between the best lower and upper bounds for q∗was about 0.755; at the time of writing this survey, this ratio has been raised above 0.998. The implied success in our understanding of the problem cannot be understated —hence the “closing in” words in the title of this survey. Moreover, as we have already mentioned above and shall see below in more detail, the problem of estimating cr(Kn) has turned out to be intimately related to other classical combinatorial geometry problems. Nowadays, anyone seriously interested in (≤ k)-edges or in halving lines, has no alternative but to take a careful look at the literature on cr(Kn) that has been produced in the last seven or eight years. On the more cautious side, we must also note that the steady progress achieved on the estimation of cr(Kn), both from the lower and the upper bounds fronts, seems to have reached an impasse. To a researcher not too familiar with the ﬁeld, the ratio 0.998 mentioned in the previous paragraph might signal an imminent closure on the determination of q∗. This is by no means the prevalent feeling among most (if not all) researchers actively working on this problem. Hardly any relevant new insights have been reported for some time. This humbling reality prompted us to include a word of caution (“or are we?”) in the title of this survey. With this in mind, it makes sense to sit down and reﬂect on what has been done, to highlight the key developments, and to record the state-of-the-art of the problem. We also see this as an opportunity to candidly (and, at times, informally) explain the obstacles that seem to prevent any further substantial progress with the current techniques, in the hopes that this will foster the development of reﬁned or substantially novel techniques to attack this fundamental problem. 1.3. Structure of this survey. The problem of estimating cr(Kn) breaks into the two problems of establishing upper and lower bounds for this parameter, with the problem of ﬁnding exact values of cr(Kn) lying, evidently, within both realms. Before moving on to separate discussions on the problems of lower- and upper-bounding cr(Kn), we shall review one of the main foundations behind our current knowledge of cr(Kn). The Rectilinear Crossing Number project (RCN), led by Aichholzer, has been a fruitful source of inspiration as well as an invaluable tool for establishing results and testing conjectures. In Section 2 we describe the nature and reach of the RCN project which, as we will see, has both a claim and an impact on both the lower- and the upper-bounding fronts. THE RECTILINEAR CROSSING NUMBER OF Kn: CLOSING IN (OR ARE WE?) 3 In Section 3 we give an overview of the state-of-the-art of the problem of lower bounding cr(Kn) circa 2003. Besides Aichholzer’s RCN project, there seems to be a general consensus on the other main foundation behind our current knowledge of cr(Kn). A major break-through was achieved around 2003, when two independent teams of researchers elucidated the close connection between cr(Kn) and the number of (≤k)-edges in an n-point set [4, 26]. A good estimate on the number of such (≤k)-edges, also given in these papers, yielded an impressively improved lower bound on cr(Kn). We devote Section 4 to a review of these cornerstone results. In Section 5 we overview the subsequent eﬀorts to reﬁne the bounds for the number of (≤k)-edges given in and , in the quest for improved lower bounds for cr(Kn). In Section 6 we discuss the diﬀerent approaches to establishing upper bounds for cr(Kn). Section 7 contains a brief summary of the state-of-the-art of the problem at the time of writing this survey. We present the current best estimates (lower and upper bounds) for q∗, as well as an annotated table with the values of n for which the exact value of cr(Kn) is known. We conclude this survey by reﬂecting on some possible future developments around this fundamental problem. We discuss the diﬃculties that lie behind our current impasse, and outline a somewhat promising approach that may pave the way towards future improvements. 2. The Rectilinear Crossing Number project Around 2000, a team of researchers led by Aichholzer undertook the task of building databases with all the distinct (up to order type equivalence; see below) n-point conﬁgurations in general position, for n ≤10 [10, 14, 25]. The raw knowledge of all possible n-point conﬁgurations put Aichholzer and his collaborators in a position to explore in depth several classical combinatorial geometry problems. In particular, it allowed for the exact calculation of cr(Kn) for small values of n. The criterion used by Aichholzer et al. to discriminate if two collections of points are non-isomorphic is based on the concept of order types. Consider an (ordered) n-point set P = {p1, p2, . . . , pn} in general position. To each three integers i, j, k with 1 ≤i < j < k ≤n, associate a sign (or order type) sign(ijk) according to the following rule. If as we traverse the triangle deﬁned by pi, pj, and pk by following the edges pipj, pjpk, and pkpi in the given order, the resulting closed curve has a clockwise orientation, then let sign(ijk) := +. Otherwise, let sign(ijk) := −. The collection of the order types of all the triples of points of P is the order type of P. Now let Q be another n-point set in general position. If the elements of Q can be ordered {q1, q2, . . . , qn} so that the order types of P and Q are the same, then P and Q are order type equivalent (under the mapping pi →qi for i = 1, 2, . . ., n). We simply say that P and Q have the same order type. Order type equivalence is a natural isomorphism criterion for point sets in gen-eral position. For crossing number purposes, it is certainly the relevant paradigm. Indeed, suppose that P and Q have the same order type. Then there is a bijec-tion from the points of P to the points of Q so that four points in P span a convex 4 BERNARDO M. ´ ABREGO, SILVIA FERN´ ANDEZ-MERCHANT, AND GELASIO SALAZAR quadrilateral if and only if the corresponding four points in Q span a convex quadri-lateral. Conversely, if this last condition holds, then P and Q have the same order type. Aichholzer et al. constructed the complete database of all distinct order types on n points, for all n ≤10. As an application, they veriﬁed that cr(K10) = 62 (this had been proved by Brodsky et al. in ). Without building the complete database for n = 11, the information gathered by Aichholzer et al. for n ≤10 allowed them to calculate cr(K11) and cr(K12). To achieve this, taking their database for 10 points as a starting point, they analyzed (for m = 10, and then for m = 11) which m-point order types may possibly be extended to (m + 1)-point sets that correspond to crossing-minimal drawings of Km+1. The determination of the rectilinear crossing numbers of K11 and K12 marks the beginning of the Rectilinear Crossing Number project (RCN). As one of the major achievements of the RCN, Aichholzer developed some impressively accurate heuristics to generate geometric drawings of Kn with few crossings. Aichholzer set up a web page to keep track of the best geometric drawings of Kn available, as well as of the number of distinct (up to order type equivalence) drawings achieving the current minimum. The results reported by Aichholzer in have had a major lasting impact in the ﬁeld. As new results and techniques to ﬁnd improved lower bounds have become available (see Sections 4 and 5), it has been possible to determine the exact value of cr(Kn) for more values of n (see Section 7). The outstanding quality of the upper bounds obtained by Aichholzer is evidenced by the fact that the drawings reported in turned out to be crossing optimal for all n ≤27 and for n = 30 (for n = 28 and 29 the exact value of cr(Kn) is still unknown). At the time of writing this survey, the best upper bounds available (see Section 6) are obtained from constructions that build upon “base” drawings of Kn for relatively small values of n. As a further evidence of the inﬂuence of the RCN, we note that the base drawings used have been obtained by small modiﬁcations of drawings given in . As a ﬁnal note, let us mention that Aichholzer and Krasser subsequently com-pleted the database of all distinct order types of 11-point sets [15, 9]. Using this database as a startpoint, they were able to compute cr(Kn) for all n ≤17. Building the complete database of all the order type nonequivalent 12-point sets seems to be an unfeasible task; not only it is estimated that the storage of these 12-point sets would require several petabytes of memory, but there are also some important technical diﬃculties.1 3. Lower bounds I: before 2004 In a paper published in 1972, Guy gave the exact value of cr(Kn) for n ≤9. Almost thirty years later, Brodsky, Durocher, and Gethner pushed the existing techniques to their limit, and introduced some clever new arguments, to calculate the exact value of cr(K10). As one of the ﬁrst results of the Rectilinear Crossing Number project (see Sec-tion 2), Aichholzer, Aurenhammer, and Krasser gave computer-assisted proofs that cr(K11) = 102 and cr(K12) = 153. 1Aichholzer, personal communication. THE RECTILINEAR CROSSING NUMBER OF Kn: CLOSING IN (OR ARE WE?) 5 Because each of the n subsets of size n −1 of an n-point set P has at most cr(Kn−1) crossings, and each crossing of P appears in exactly n −4 such subsets, it follows that (n −4)cr(Kn) ≥ncr(Kn−1). This is equivalent to 1 ≥cr(Kn) n 4 ≥cr(Kn−1) n−1 4 , which shows that the Sylvester’s Four Point Constant q∗deﬁned in (1) actually exists. Starting from a lower bound for cr(Km) for any ﬁxed m, one can obtain a lower bound for cr(Kn) for every n > m (and consequently a lower bound for q∗) by iterating cr(Kn) ≥⌈cr(Kn−1)n/(n −4)⌉. This technique was used by Brodsky et al. with cr(K10) = 62 to show that q∗> 0.3001. Adding to this argument the fact that cr(Kn) and n 4 have the same parity when n is odd (this easily follows from (2) but was proved for any non-necessarily rectilinear drawing of Kn by Eggleton and Guy ), and using cr(K11) = 102, Aichholzer et al. showed that q∗> 0.3115. Building upon ideas from Welzl and Wagner and Welzl , Wagner used a completely novel approach to show that q∗> 0.3288. Wagner’s work is particularly signiﬁcant, since it deviates from the traditional approach of lower bounding q∗by using a particular lower bound and a counting argument. Indeed, the ideas in are prescient of the revolutionary approach that will be reviewed in the next section. 4. Lower bounds II: the breakthrough Our understanding of geometric drawings of Kn underwent a phase transition by unveiling a close relationship with k-edges. We recall that if P is an n-point set, and 0 ≤k ≤n/2 −1, a k-edge of P is a line through two points of P leaving exactly k points on one side. A (≤k)-edge is a j-edge with j ≤k. The number of k- and (≤k)-edges of P are denoted by Ek(P) and E≤k(P), respectively. Finally, let E≤k(n) denote the minimum E≤k(P), taken over all n-point sets P in general position. For an n-point set P in the plane in general position, let cr(P) denote the number of crossings in the rectilinear drawing of Kn induced by P. The following was proved independently by Lov´ asz, Wagner, Welzl, and Wesztergombi , and by ´ Abrego and Fern´ andez-Merchant : (2) cr(P) = ⌊n/2⌋−2 k=0 (n −2k −3)E≤k(P) −3 4 n 3 + (1 + (−1)n+1)1 8 n 2 . The relevance of this connection between cr(P) and E≤k(P) was made evident in both and by proving that (3) E≤k(n) ≥3 k + 2 2 , for 0 ≤k ≤n/2 −1. Substituting (3) into (2) yields (4) cr(Kn) ≥3 8 n 4 + Θ(n3), thus implying the remarkably improved bound q∗≥3/8 = 0.375. 6 BERNARDO M. ´ ABREGO, SILVIA FERN´ ANDEZ-MERCHANT, AND GELASIO SALAZAR We recall that the crossing number cr(G) of a graph G is the minimum number of pairwise crossings of edges in a (nonnecessarily geometric) drawing of G in the plane. There are drawings of Kn with exactly λn := (1/4)⌊n/2⌋⌊(n −1)/2⌋⌊(n −2)/2⌋ ⌊(n −3)/2⌋crossings, and it is widely believed that these drawings are crossing-minimal; that is, it is conjectured that cr(Kn) = λn for every positive integer n. This conjecture has been veriﬁed for n ≤12 [23, 27]. Since cr(Kn) ≤λn, it follows at once that limn→∞cr(Kn)/ n 4 ≤3/8. This last upper bound gives an additional signiﬁcance to (4). With this motiva-tion, Lov´ asz et al. pushed a little further, invoking the following from : (5) E≤k(n) ≥ n 2 −n n2 −2n −4k2 + 4k. This last bound is better than (3) for k > 0.4956n. Using (3) for k ≤0.4956n, and (5) for k > 0.4956n, Lov´ asz et al. derived the slightly improved bound q∗> (3/8)+ 10−5. Although numerically marginal, this improvement is signiﬁcant because it shows that cr(Kn) and cr(Kn) diﬀer in the asymptotically relevant term. 5. Lower bounds III: further improvements Since the key connection (2) was proved in and , all subsequent eﬀorts to lower bound cr(Kn) have been focused on ﬁnding better estimates for E≤k(n). The ﬁrst improvement was reported in , giving a lower bound for E≤k(n) that is strictly better than (3) for k > 0.4651n. The bound given in is in terms of a complicated expression. For our current surveying purposes, it suﬃces to mention that using this bound Balogh and Salazar proved that cr(Kn) > 0.37553 n 4 +Θ(n3). Another signiﬁcant improvement was achieved by Aichholzer, Garc´ ıa, Orden, and Ramos , who proved that (6) E≤k(n) ≥3 k + 2 2 + 3 k + 2 −⌊n/3⌋ 2 −max 0, (k + 1 −⌊n/3⌋)(n −3⌊n/3⌋) . A shorter proof of (6), given in the more general context of pseudolinear drawings was given in . Substituting (6) into (2), one obtains the improved estimate q∗≥41/108 > 0.37962. Moreover, it is possible to use the bound by Balogh and Salazar in the range k > 0.4864n to obtain the marginally better q∗> 0.37968. The current best lower bound known for q∗is derived using a result recently reported by ´ Abrego, Cetina, Fern´ andez-Merchant, Lea˜ nos, and Salazar [3, 7]. They proved that for every k and n such that ⌈(4n −11)/9⌉−1 ≤k ≤(n −2)/2, (7) E≤k(n) ≥uk(n) ≥ n 2 −1 9 1 −2k + 2 n (5n2 + 19n −31). The function uk is asymptotic to the latter expression and it is better than all previous bounds (including (5) (6), and the bound in ) across its full range ⌈(4n −11)/9⌉≤k ≤(n−2)/2. In addition, ´ Abrego et al. constructed point-sets achieving equality on (6) for all k < ⌈(4n −11)/9⌉. Using (6) for k < ⌈(4n −11)/9⌉, and (7) for ⌈(4n −11)/9⌉≤k ≤(n −2)/2, it follows from (2) that cr(Kn) ≥ (277/729) n 4 + Θ(n3), thus implying that q∗≥277/729 > 0.37997. THE RECTILINEAR CROSSING NUMBER OF Kn: CLOSING IN (OR ARE WE?) 7 6. Upper bounds The literature on crossing numbers of particular families of graphs is vastly dominated by papers that focus on establishing lower bounds. Most of the time, a natural drawing suggests itself with relatively little eﬀort. When successive at-tempts to produce better drawings fail, this is seen as plausible evidence that the proposed drawing is indeed optimal. The eﬀorts are then directed in the opposite, and usually remarkably harder, direction: proving nontrivial lower bounds for the crossing numbers of the graphs upon consideration. The problem of upper bounding the rectilinear crossing number of Kn is a notable exception to this trend. The goal is to describe a way to draw Kn with as few crossings as possible, for arbitrarily large values of n, so as to have at least an educated guess at the asymptotic value q∗= limn→∞cr(Kn)/ n 4 . Over the years, several strategies to draw Kn with few crossings have been put forward. However, to this day there has not been a clear candidate for an optimal drawing. The only common characteristic is that almost all drawings with few crossings have (or are really close to have) 3-fold symmetry with respect to a point. That is, the underlying point-set P of the drawing is partitioned into three sets (we call them wings) of size n/3 each, with the property that rotating each wing angles of 2π/3 and 4π/3 around a suitable point generates the other two wings. Figure 1. (a) Recursive construction by Singer. (b) Recursive construction by Brodsky et al. In the early 1970s, Jensen was the ﬁrst to propose a way to draw Kn for arbitrarily large values of n. His construction gave speciﬁc coordinates for n/3 points in a wing, and then obtained the remaining two wings by rotating 2π/3 and 4π/3 around the origin. As a result he obtained q∗≤7/18 < 0.38889. At around the same time, Singer started the trend of recursively constructing drawings of Kn. His idea was to start with a good drawing of Kn/3, apply an aﬃne transformation to it to make the slope of each of its edges suﬃciently close to zero, and then add the 2π/3 and 4π/3 rotations of the resulting drawing to obtain the other two wings. (See Figure 1(a).) This construction shows that cr(Kn) ≤3cr(Kn/3) + 3 · n 3 n/3 3 + 3 n/3 2 2 . 8 BERNARDO M. ´ ABREGO, SILVIA FERN´ ANDEZ-MERCHANT, AND GELASIO SALAZAR Indeed, the ﬁrst term consists of the crossings obtained from 4 points in the same wing, the next term counts the crossings from 3 points in one wing and the remain-ing in one of the other two wings, and the last term counts the crossings from 2 points in one wing and 2 points in another wing. Using cr(K3) = 0 as a starting point, this inequality gives q∗≤5/13 < 0.38462. Brodsky, Durocher, and Gethner modiﬁed Singer’s construction by sliding 3 points in each wing toward the center of rotation as shown in Figure 1(b). Their construction gives q∗≤6467/16848 < 0.38385. Aichholzer, Aurenhammer, and Krasser devised a diﬀerent replacement con-struction. They started with an underlying set P with an even number of points N. Instead of triplicating P, they replaced every point of P by a cluster of c points on a small arc of circle ﬂat enough so that all lines among these c points leave N/2 points of P on one side and N/2 −1 on the other side. (See Figure 2(a).) Letting n = cN, their construction gives cr(Kn) ≤ 24cr(P) + 3N 3 −7N 2 + 6N N 4 n 4 + Θ(n3). Using a set P with N = 36 points and cr(P) = 21 191 they obtained q∗< 0.380858. They further explored using diﬀerent sizes for each of the clusters, which resulted in an improvement of the latter bound to q∗< 0.380739. This method of obtaining lower bounds allowed for improvements by using better initial sets P. Aichholzer and Krasser , as part of their computer-assisted search of the crossing numbers cr(Kn) for small values of n, obtained a particular drawing of K54 that gives q∗< 0.380601. Figure 2. (a) Replacement construction by Aichholzer et al. (b) Recursive construction by ´ Abrego and Fern´ andez-Merchant. ´ Abrego and Fern´ andez-Merchant started with an underlying set P with an even number of points N. They obtained a new set Q by replacing every point of P by a pair of points close to each other and spanning a line that divides the rest of Q in half. (See Figure 2b). This property of having a halving-line matching is not satisﬁed by an arbitrary point-set P, but fortunately it is satisﬁed by most of the small sets with optimal crossing number. Moreover, the resulting set Q inherits this property. Thus, if n = 2kN, then iterating this construction k times gives (8) cr(Kn) ≤ 24cr(P) + 3N 3 −7N 2 + (30/7)N N 4 n 4 + Θ(n3). THE RECTILINEAR CROSSING NUMBER OF Kn: CLOSING IN (OR ARE WE?) 9 At the time, using the best known drawing of K30 (now proved to be optimal) yielded q∗< 0.380559. To this date, (8) provides the currently best recursive construction. The restrictions on the base set P were subsequently weakened in the sense that (8) also holds for arbitrary sets P with an odd number of points. Applying this inequality to a drawing of K315 with 152 210 640 crossings gives the currently best upper bound: q∗< 83 247 328 218 791 125 < 0.380488. To support the belief that the crossing-minimal sets have nearly 3-fold symmetry, ´ Abrego et al. constructed a 3-fold symmetric set of n points for each n multiple of 3, n < 100. (See Figure 3.) Moreover, 3-fold symmetry is inherited from the base set in all recursive constructions mentioned before. In fact, the drawing of K315 used as a base set to obtain the best current upper bound has 3-fold symmetry. Figure 3. The underlying vertex set of an optimal 3-symmetric geometric drawing of K24. This point set contains optimal nested 3-symmetric drawings of K21, K18, K15, K12, K9, K6, and K3. 7. Summary In this section we summarize, for quick reference, the state-of-the-art on cr(Kn) and q∗at the time of writing this survey. 7.1. Sylvester’s Four Point Constant. (9) 0.379972 < 277 729 ≤q∗≤83 247 328 218 791 125 < 0.380488. The lower and upper bounds in (9) are derived in (see also ) and , respectively. 10 BERNARDO M. ´ ABREGO, SILVIA FERN´ ANDEZ-MERCHANT, AND GELASIO SALAZAR 7.2. Exact values of cr(Kn). The exact value of cr(Kn) is known for n ≤27 and for n = 30 (see Table 1). For n ≤27, the lower bound for cr(Kn) is derived in (see also ). The bound cr(K30) ≥9726 is proved in . In all cases, the upper bounds were obtained by Aichholzer . n 5 6 7 8 9 10 11 12 13 14 15 16 17 18 cr(Kn) 1 3 9 19 36 62 102 153 229 324 447 603 798 1029 n 19 20 21 22 23 24 25 26 27 30 cr(Kn) 1318 1657 2055 2528 3077 3699 4430 5250 6180 9726 Table 1. Exact rectilinear crossing numbers known. 8. Further thoughts and future research Since the introduction of (2) in and , all the progress achieved on lower bounding q∗has been contingent on the derivation of improved bounds for E≤k(n). Although it may seem natural to expect the continuation of this trend, there is some evidence that suggests that this approach alone will not lead to the correct value of q∗. The reasons behind our caution lie on our own investigations of sets that minimize the number of (≤k)-edges. So far it has been possible to construct n-point sets that simultaneously minimize E≤k(n) for al k up to a certain value. It is not diﬃcult to construct an n-point set that simultaneously achieves equality in (3) for every k, 0 ≤k ≤n/3, and arbitrary n (along this discussion we assume that n is a multiple of 3). To construct a similar set minimizing E≤k(n) for a larger range of values of k is notably harder. Aichholzer, Garc´ ıa, Orden, and Ramos constructed an n-point set that simultaneously achieves equality in (6) for every k, 0 ≤k ≤⌊5n 12 ⌋−1 and arbitrary n. A diﬀerent type of construction was used in to simultaneously show that (6) is tight for every k, 0 ≤k ≤4n/9 −1. However, this construction is far from crossing optimal due to a dramatic increase on the number of (≤k)-sets when k ≥4n/9, and avoiding this seems impossible. That is, insisting on simultaneously minimizing E≤k(n) for all k, for k as large as possible, seems to actually increase the crossing number of the point sets under consideration. In view of this, a new paradigm might be in order. It seems not only possible, but very likely, that the crossing-minimal drawings of Kn for large values of n are attained by point sets that are not even close to minimizing E≤k for every k < (4n/9) −1. A proper understanding of this intriguing behavior seems out of our reach at the present time. Although (2) validates the eﬀorts to lower bound cr(Kn) via lower bounding E≤k(n), our previous remarks suggest that, no matter how ﬁne the estimates, this may not suﬃce in order to determine q∗. It is quite conceivable that the (exact or asymptotic) value of E≤k(n) be known for every k, and still the estimate for cr(Kn) obtained from plugging this into (6) does not correspond to the correct (at least asymptotic) value of cr(Kn). THE RECTILINEAR CROSSING NUMBER OF Kn: CLOSING IN (OR ARE WE?) 11 The outlook from the upper bounds front is also unclear. We might all be just one clever idea away from a breakthrough construction that yields (at least asymp-totically) crossing-minimal geometrical drawings of Kn. Using best-case heuristics, it can be shown that any recursive construction for large N, where each point is replaced by a small cluster of points of the same size, can yield at best a bound of the form cr(Kn) ≤ 24cr(P) + 3N 3 −7N 2 + 4N N 4 n 4 + Θ(n3). The improvement using the best drawing of K315 would be less than 10−8. For all these reasons, we are inclined to think that there is more potential to close the gap from below than from above; that is, we believe that q∗is closer to the current best upper bound than to the current best lower bound. To end on an optimistic note, there is one more promising observation. Be-sides 3-fold symmetry, the currently best known constructions (including those presented in ) share another property called 3-decomposability. A set P is called 3-decomposable if there exists a balanced partition of P into three parts A, B, and C and a triangle T enclosing P such that the orthogonal projections of P onto the sides of T show A between B and C on one side, B between A and C on another side, and C between A and B on the third side. As for 3-fold symme-try, 3-decomposability is inherited from a base set in all recursive constructions mentioned before. ´ Abrego et al. conjectured that all crossing-minimal sets are 3-decomposable. If this conjecture happens to be true, then the lower bound for q∗ would be improved to (2/27)(15 −π2) > 0.380029 as proved in . References B. M. ´ Abrego, J. Balogh, S. Fern´ andez-Merchant, J. Lea˜ nos, and G. Salazar, An extended lower bound on the number of (≤k)-edges to generalized conﬁgurations of points and the pseudolinear crossing number of Kn, Journal of Combinatorial Theory, Series A 115 (2008), 1257–1264. B. M. ´ Abrego, M. Cetina, S. Fern´ andez-Merchant, J. Lea˜ nos, and G. Salazar, 3-symmetric and 3-decomposable drawings of Kn. Discrete Applied Mathematics 158 (2010), 1240–1258. B. M. ´ Abrego, M. Cetina, S. Fern´ andez-Merchant, J. Lea˜ nos, and G. Salazar, On (≤k)-edges, crossings, and halving lines of geometric drawings of Kn. arXiv:1102.5065v1 [math.CO], (2011). B. M. ´ Abrego and S. Fern´ andez-Merchant, A lower bound for the rectilinear crossing number. Graphs and Combinatorics 21 (2005), 293–300. B. M. ´ Abrego and S. Fern´ andez-Merchant. Geometric drawings of Kn with few crossings. J. Combin. Theory Ser. A 114 (2007), 373–379. B. M. ´ Abrego, S. Fern´ andez-Merchant, J. Lea˜ nos, and G. Salazar, The maximum number of halving lines and the rectilinear crossing number of Kn for n ≤27. Electronic Notes in Discrete Mathematics 30 (2008) 261–266. B. M. ´ Abrego, S. Fern´ andez-Merchant, J. Lea˜ nos, and G. Salazar, A central approach to bound the number of crossings in a generalized conﬁguration. Electronic Notes in Discrete Mathematics 30 (2008) 273–278. O. Aichholzer, F. Aurehnammer, and H. Krasser, Enumerating order types for small point sets with applications. Order 19 (2002), 265–281. O. Aichholzer, F. Aurenhammer, and H. Krasser. On the crossing number of complete graphs. Computing 76 (2006), 165–176. 12 BERNARDO M. ´ ABREGO, SILVIA FERN´ ANDEZ-MERCHANT, AND GELASIO SALAZAR O. Aichholzer, J. Garc´ ıa, D. Orden, and P. Ramos, New lower bounds for the number of (≤k)-edges and the rectilinear crossing number of Kn Discrete and Computational Geometry 38 (2007), 1–14. O. Aichholzer, J. Garc´ ıa, D. Orden, and P. Ramos, New results on lower bounds for the number of (≤k)-facets. European Journal of Combinatorics 30 (2009), 1568–1574. O. Aichholzer and H. Krasser, The point set order type data base: A collection of applications and results. In Proc. 13th Annual Canadian Conference on Computational Geometry CCCG 2001, 17–20. Waterloo, Ontario, Canada (2001). O. Aichholzer and H. Krasser, Abstract order type extension and new results on the rectilinear crossing number. Comp. Geom. Theory Appl., 36 (2006), 2–15. J. Balogh and G. Salazar, k-sets, convex quadrilaterals, and the rectilinear crossing number of Kn. Discrete Comput. Geom. 35 (2006), 671-690. W. Blaschke, ¨ Uber Aﬃne Geometrie XI: L¨ osung des “Vierpunktproblems” von Sylvester aus der Theorie der geometrischen Wahrscheinlichkeiten, Leipziger Berichte 69 (1917), 436–453. A. Brodsky, S. Durocher, and E. Gethner, The Rectilinear Crossing Number of K10 is 62. Electr. J. Comb. 8 (2001), R23. A. Brodsky, S. Durocher, E. Gethner, Toward the rectilinear crossing number of Kn: new drawings, upper bounds, and asymptotics. Discrete Mathematics 262 (2003), 59–77. M. Cetina, C. Hern´ andez-V´ elez, J. Lea˜ nos, and C. Villalobos, Point sets that mini-mize (≤k)-edges, 3-decomposable drawings, and the rectilinear crossing number of K30. arXiv:1009.4736v1 [math.CO] (2010). M.W. Crofton, Probability, Encyclopedia Britannica, 9th. Edition, 19 (1885), 768–788. R. B. Eggleton, Ph. D. thesis, Univ. of Calgary, 1973. R.K. Guy, Crossing numbers of graphs. Graph theory and applications (Proc. Conf., Western Michigan Univ., Kalamazoo, Mich., 1972), pp. 111-124. Lecture Notes in Math., Vol. 303, Springer, Berlin, 1972. H.F. Jensen, An upper bound for the rectilinear crossing number of the complete graph. J. Combin. Theory Ser B 10 (1971), 212-216. H. Krasser, Order Types of Point Sets in the Plane. PhD thesis, Institute for Theoretical Computer Science, Graz University of Technology, Austria, October 2003. L. Lov´ asz, K. Vesztergombi, U. Wagner, and E. Welzl, Convex quadrilaterals and k-sets. Towards a theory of geometric graphs, Contemp. Math. 342, 139–148. Amer. Math. Soc., Providence, RI, 2004. S. Pan and R.B. Richter, The crossing number of K11 is 100. J. Graph Theory 56 (2007), 128-134. E. Scheinerman, E. and H.S. Wilf, The Rectilinear Crossing Number of a Complete Graph and Sylvester’s “Four Point” Problem of Geometric Probability. Amer. Math. Monthly 101 (1994), 939–943. D. Singer, Rectilinear crossing numbers. Manuscript (1971). J.J. Sylvester, Question 1491 in The Educational Times (April 1864). London. J.J. Sylvester, Report of the British Association 35 (1865), 8–9. U. Wagner, On the Rectilinear Crossing Number of Complete Graphs, Proc. 14th Annual Symposium on Discrete Algorithms (SODA), 2003, pp. 583–588. U. Wagner and E. Welzl, A continuous analogue of the upper bound theorem. Discr. Comp. Geom. 26 (2001), 205–219. E. Welzl, More on k-sets of ﬁnite sets in the plane, Discr. Comput. Geom 1 (1986), 95–100. E. Welzl, Entering and leaving j-facets. Discr. Comput. Geom. 25 (2001), 351-364. Department of Mathematics. California State University at Northridge. E-mail address: bernardo.abrego@csun.edu Department of Mathematics. California State University at Northridge. E-mail address: silvia.fernandez@csun.edu Instituto de F´ ısica. Universidad Aut´ onoma de San Luis Potos´ ı. E-mail address: gsalazar@ifisica.uaslp.mx
9209	https://asphalt.fandom.com/wiki/Complementary_event	Complementary event \| Asphalt Wiki \| Fandom Sign In Register Asphalt Wiki Explore Main Page Discuss All Pages Community Interactive Maps Recent Blog Posts Asphalt 9 Vehicles Locations Updates Asphalt 8 Vehicles Locations Updates Games 2015-Present Asphalt Nitro 2 Asphalt 9: Legends Asphalt Street Storm Racing Asphalt Xtreme Asphalt Nitro 2012-2014 Asphalt Overdrive Asphalt 8: Airborne Asphalt 7: Heat Asphalt Injection 2009-2011 Asphalt 3D Asphalt: Audi RS 3 Asphalt 6: Adrenaline Asphalt 5 2004-2008 Asphalt 4: Elite Racing Asphalt 3: Street Rules Asphalt Urban GT 2 Asphalt Urban GT Wiki Today in Asphalt history News Asphalt statistics Admins Damian103 Guy Bukzi Montag Jamiu Jilan WKPQ Sign In Don't have an account? Register Sign In Menu Explore More History Advertisement Skip to content Asphalt Wiki 13,541 pages Explore Main Page Discuss All Pages Community Interactive Maps Recent Blog Posts Asphalt 9 Vehicles Locations Updates Asphalt 8 Vehicles Locations Updates Games 2015-Present Asphalt Nitro 2 Asphalt 9: Legends Asphalt Street Storm Racing Asphalt Xtreme Asphalt Nitro 2012-2014 Asphalt Overdrive Asphalt 8: Airborne Asphalt 7: Heat Asphalt Injection 2009-2011 Asphalt 3D Asphalt: Audi RS 3 Asphalt 6: Adrenaline Asphalt 5 2004-2008 Asphalt 4: Elite Racing Asphalt 3: Street Rules Asphalt Urban GT 2 Asphalt Urban GT Wiki Today in Asphalt history News Asphalt statistics Admins Damian103 Guy Bukzi Montag Jamiu Jilan WKPQ in:Probability theory Complementary event Sign in to edit History Purge Talk (0) SOURCE EDITOR ONLY! This page uses LaTeX markup to display mathematical formulas. Editing the page with the VisualEditor or Classic rich-text editor disrupts the layout. Do not even switch to one of these editors while editing the page! For help with mathematical symbols, see Mathematical symbols and expressions. In probability theory, the complement of any eventA{\displaystyle A} is the event [not⁡A]{\displaystyle [\operatorname{not} A]}, i. e. the event that A{\displaystyle A} does not occur. The event A{\displaystyle A} and its complement [not⁡A]{\displaystyle [\operatorname{not} A]} are mutually exclusive and exhaustive. Generally, there is only one event B{\displaystyle B} such that A{\displaystyle A} and B{\displaystyle B} are both mutually exclusive and exhaustive; that event is the complement of A{\displaystyle A}. The complement of an event A{\displaystyle A} is usually denoted as A¯{\displaystyle \bar{A} }, A∁{\displaystyle A^\complement} or A′{\displaystyle A'}. Given an event, the event and its complementary event define a Bernoulli trial: did the event occur or not? For example, if a player reveals the Tech card of a Daily Kit Box† which has the hidden rule of only granting Advanced and Mid-Tech, then the card can either be Advanced Tech or Mid-Tech. Because these two outcomes are mutually exclusive (i. e. the card cannot simultaneously show both Advanced and Mid-Tech) and collectively exhaustive (i. e. there are no other possible outcomes not represented between these two), they are therefore each other's complements. This means that {Advanced Tech}{\displaystyle {\operatorname{Advanced Tech}}} is logically equivalent to {not⁡M i d-T e c h}{\displaystyle {\operatorname{not} \operatorname{Mid-Tech}}}, and {M i d-T e c h}{\displaystyle {\operatorname{Mid-Tech}}} is equivalent to {not Advanced Tech}{\displaystyle {\operatorname{not Advanced Tech}}}. Complement rule[] In a random experiment, the probabilities of all possible events (the sample space) must total to 1—that is, some outcome must occur on every trial. For two events to be complements, they must be collectively exhaustive, together filling the entire sample space. Therefore, the probability of an event's complement must be 1 minus the probability of the event. That is, for an event A{\displaystyle A}, P(A¯)=1−P(A).{\displaystyle P(\bar{A}) = 1 - P(A).} Equivalently, the probabilities of an event and its complement must always total to 1. This does not, however, mean that any two events whose probabilities total to 1 are each other's complements; complementary events must also fulfill the condition of mutual exclusivity. Categories Categories: Probability theory Community content is available under CC-BY-SA unless otherwise noted. Comments Start a conversation Sign in to share your thoughts and get the conversation going. SIGN IN Don't have account? 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9210	https://worldscholarshipforum.com/tamale-technical-university-courses-and-fees/	Published Time: 2024-08-20T19:40:47+00:00 1 INTRODUCTION Tamale Technical Universitybegan as a Trades Training Centre in 1951 and then became the Government Training School in 1954. It was converted to a Junior Technical School in 1960. The School was elevated to the status of a Polytechnic on August 23, 1992. As a result of the Educational Reform Programme and the enactment of the PNDC Law 321 in 1992, the status of Tamale Technical University was raised to the level of tertiary institution together with Accra, Kumasi, Ho, Cape Coast and Takoradi Technical Universities. The University now trains students in Higher National Diploma level and Bachelor of Technology programmes. Since the inception of the University as a Trade Training Centre, its catchment area has been mainly the three regions of the north, that is, Northern, Upper East and Upper West. In spite of the fact that there are now Universities in every region across the country, the student population of the Tamale Technical University still has a reasonable level of influence from the three regions and beyond. This is an indication that the University continues to maintain a long-standing role as the premier training opportunity for middle and high level technical and managerial personnel for the three regions of the north and beyond. VISION The Vision of Tamale Technical University is to become an IT and Technical driven tertiary institution with a cohesive work force for the running of professional, certificate, HND and degree programmes relevant to national development; with the needs of the informal sector as the University‟ ssocio-economic environment taking the centre stage. MISSION Tamale Technical University provides skills-oriented tertiary education in Engineering, Commerce, Applied Sciences and other relevant fields and also creates opportunities for research to promote regional and industrial local development. Core Values of Tamale Technical University The core values of Tamale Technical University are as follows: i. Hard work; ii. Honesty; iii. Transparency; 2 iv. Justice; v. Fair play; and vi. Unity and Togetherness It is expected that every member of the University imbibes these values and diligently work in line with them. STRUCTURE OF TAMALE TECHNICAL UNIVERSITY The structure of the academic units in the Tamale Technical University presents a unique feature that lends itself to a number of interpretations. It appears to be one of the few institutions that attained a tertiary status with opportunities for offering programmes such as the Higher National Diploma (HND) and degrees while still holding on to the technical programmes that it was running when it was a technical institute. This requires a bold and decisive measure in ensuring that the University runs as a full-fledged tertiary institution. Economically viable non-tertiary programmes are run on part-time basis. The schools, departments and programmes are shown in Tables 1 and 2. INSTITUTIONAL MANAGEMENT The University Governing Council is the highest decision-making body of the Tamale Technical University, and it is followed by the Academic Board. These decision-making bodies are supported by various Council and Academic Board sub-committees. The Vice Chancellor is the Chief Executive Officer of the University. The Vice Chancellor is responsible for ensuring the efficient administration and management of all physical facilities, financial resources and personnel to achieve the objective of the University. The Vice Chancellor advises the University Council on all matters affecting policy, finance, governance and problems of the University. He chairs the Academic Board and Convocation meetings. The Vice Chancellor is assisted by a Pro-Vice Chancellor and principal officers including the Registrar, the Finance Officer, Librarian, the Internal Auditor, the Planning Officer, Procurement Officer and Development Officer, and Deans and Heads of the academic units (i.e. schools and departments). 3 MAIN ADMINISTRATION OF THE UNIVERSITY Office of the Vice Chancellor Office of the Pro-Vice Chancellor Office of the Librarian Office of the Registrar Senior Assistant Registrar (Academic) Senior Assistant Registrar (Public Relations) Senior Assistant Registrar (General Administration) Senior Assistant Registrar (Human Resource) Director of Finance Internal Audit Office KEY ACADEMIC STAFF OF THE UNIVERSITY DEANS Student Affairs School of Applied Sciences and Technology School of Business School of Engineering and Built Environment School of Applied Arts Office of the Dean of Degree Programmes Office of the Dean of Technical and Vocational Education The Examinations Office 4 DIRECTORS Research and Innovation Institutional Collaborations Information Communication Technology (ICT) Quality Assurance Unit Senior Resident Tutor SCHOOLS AND TERTIARY PROGRAMMES RUN IN TaTU The Schools and tertiary programmes run in the University are shown below: School of Business and Management Studies Departments Programmes 1 Accountancy B. Tech (Accounting with Computing) HND (Accountancy) – Full and Part -time 2 Marketing HND (Marketing) – Full and Part -time 3 Secretaryship and Management studies HND (Secretaryship and Management Studies) – Full and Part -time School of Applied Sciences and Technology Departments Programmes 1 Statistics, Mathematics and Sciences HND (Statistic) 2 Hospitality and Tourism Management (HTM) HND (Hotel Catering and Institutional Management) – Full and Part -time HND (Tourism) 3 Computer Science HND (Information Communication Technology) SCHOOL OF ENGINEERING 5 Departments Programmes 1 Agricultural Engineering B. Tech (Agricultural Engineering) HND (Agricultural Engineering) HND (General Agriculture) 2 Construction Engineering and Management HND (Construction Engineering Management) 4 Mechanical Engineering HND (Mechanical Engineering) – Options in Production, Plant and Automobile 5 Automobile Engineering HND (Automobile Engineering) 6 Electrical and Electronics Engineering HND (Electrical and Electronics Engineering) SCHOOL OF APPLIED ARTS Departments Programmes 1 Media and Mass Communication HND (Media and Mass Communication) 2 Fashion Design and Textiles HND (Fashion Design and Textiles studies) 3 Industrial Arts HND (Industrial Arts) in Leather, Ceramics, Painting and Decoration SEMESTER REGISTRATION PROCEDURE 6 A. TERTIARY STUDENTS (PROFESSIONAL, HND AND DEGREE) Step 1: Students should pay required school fees at the appropriate bank. Step 2: Open Step 3: Log in using the user name and password received through SMS after payment of fees Step 4: Click on Registration Step 5: Click Proceed to Register your Semester Courses Step 6: At the bottom of the page, click on “Click He re to save Registration and Print your slip” Step 7: Print your Registration slip displayed and su bmit to your Department in the University. B. NON TERTIARY STUDENTS Step 1: Payment of required school fees at the appropriate bank Step 2: Visit Accounts Office of the University in the Administration Building (Room 14) to register NOTE: Students should be mindful of the deadline of registration process. Registration process is normally two weeks unless otherwise. CONGREGATION (i) The Congregation of the University is composed of: a) Members of Council b) Members of Convocation c) All graduands of the University. (ii) Congregation shall be summoned by the Chancellor or in his absence by Chairman of Council for the purpose of receiving reports and witnessing the ceremony of awarding 7 degrees, diplomas, and certificates of the University and for any other purposes as the Council Chair may determine. (iii) Congregation shall be convened at least once every year at such time and place as shall be determined by the Chancellor in consultation with the Vice Chancellor and Council and shall be presided over by the Chancellor. In the absence of the Chancellor, the Chair of Council shall preside. INDUSTRIAL ATTACHEMENT One of the cardinal objectives of the Technical Universities Act, 2016 (Act 922) is to „provide opportunities for skills development, applied research and publication of research findings. This clearly indicates that, the central focus of University education is its career oriented nature. This can only be achieved through adequate training and mentoring via industrial attachment, which is compulsory without which a student cannot graduate. PROJECT WORK Students are required to submit project work for the completion of the Bachelor of Technology and Higher National Diploma programmes in the Technical Universities of Ghana. Students are required to submit completed project work personally to their heads of department. Where the project work is conducted by a group, all members of the group are required to be present during the submission and individually sign a register for submission of project work at their various departments. Students are required to comply with the requirements of the guidelines on project work as set out by the university. PROGRESSION FROM LEVEL TO ANOTHER For progression, a student is required to maintain a minimum Cumulative Grade Point Average (CGPA) of 1.50. The cumulative grade point of less than 1.50 at the end of a semester shall attract probation. If a student has to be put on probation a second consecutive time after re-sit, the student shall be withdrawn from the programme. A student who obtains CGPA of less than 1.0 at the end of the first re-sit shall be withdrawn from the programme. In subsequent years, a student who obtains a cumulative grade point average (CGPA) of less than 1.0 after re-sit shall be withdrawn from the programme. The CGPA of more than 1.0 but less than 1.50 at the end of the year after re-sit shall attract a repetition. 8 There shall be only one re-sit examination which is conducted at the end of every academic year. For a student to qualify for a re-sit examination, the student should have taken the main semester examination during the academic year. A student who trails more than four courses after taking the re-sit examination at the end of the academic year shall repeat the whole academic year. However, if in addition, the cumulative grade point average (CGPA) obtained is less than 1.50 the student would be withdrawn from the programme. A student who fails four or more courses at the end of any semester (except in the first semester, first year) shall be put on probation provided the cumulative grade point average (CGPA) obtained is not less than 1.50. However, if the CGPA is less than 1.50 the student would be withdrawn from the programme. Courses designated as prerequisite to more advance courses must be passed before the latter courses are taken. Transcript shall reflect all grades and marks a candidate obtains for all courses. All grades for courses taken shall be used in the computation of the student‟s CGPA. EXAMINATIONS i. The Academic Board shall prescribe the conditions of entrance to the University, the regulations governing intermediate and preliminary examinations, and all examinations for degrees, diplomas and certificates or proficiency. ii. The Academic Board may approve the examinations for any diploma or any part thereof and for other purposes, course of instruction and syllabuses submitted by any Academic Division. iii. The examiners for all prescribed examinations shall be approved by the Academic Board, upon recommendation of the respective Academic Divisions. iv. The Dean of an Academic Division shall be the Chairman of the Board of Examinations of the Academic Division. EXAMINATION MALPRACTICE 9 All examination malpractices shall be dealt with in accordance with the provisions stipulated in Part III of National Board for Professional and Technician Examinations Act, 1994 (Act 492), the Criminal Code, 1960 (Acts 29) and Criminal Code, 1960 (Acts 30) as well as the approved internal examination regulation of the institutions. Examination practice includes the following: i. Illegal prior possession, knowledge or use of examination papers; ii. Leakage of examination papers; iii. Impersonation at examination; iv. Copying at examination; v. Communicating either verbally or in writing during examination; vi. Insult and assault of invigilators, supervisors and inspectors before, during and after. Any person who, before or during an examination conducted by the University i. has in possession of the person any unauthorised material relevant to the examination; or ii. is proved to have had fore-knowledge of the content of any examination paper commits an offense and is liable to conviction to a fine not exceeding GHȼ5,000.00 or to imprisonment for a term not exceeding one year or both. Where a candidate, before or during an examination, is found acting in breach of the provision of 9.2, the candidate i. shall be disqualified from taking the examination and the results of the examination shall be cancelled; ii. may be prohibited from taking any examination conducted by the Institution for a period of not less than two years immediately following the breach; and iii. shall be withdrawn from the University if found to have impersonated in an exam. The penalties contained in this section shall be in addition to any penalties that may be imposed by a court. Any person who i. before or during examination without lawful authority, gives an examination paper to any person; ii. without lawful authority, discloses the content of any examination paper to any person; iii. fraudulently alters the results sheets of any candidate; iv. fraudulently replaces the original script of any candidate; v. without lawful authority, makes a change in the original script of a candidate; 10 vi. fraudulently alters the examination number, photograph or any other identification of a candidate; or vii. acts in any manner with intent to falsify the records of the board with regard to an examination result in relation to a candidate, commits an offence and is liable on conviction to a fine of not less than GHȼ5,000.00 and not exceeding GHȼ10,000.00 or to impr isonment for a term of not more than one year or both. ELIGIBILITY FOR EXAMINATIONS In order to be eligible for examination in a particular taught course, the student shall have presented himself/herself for not less than 75% of the total number of hours for the course. A student who absents himself/herself for a cumulative period of 21 days from lectures, tutorials, and other activities prescribed for any course in any semester shall be deemed to have withdrawn from the course. Such a student shall not be permitted to write the end of semester examinations in the course. Students who do not earn a continuous assessment mark do not qualify to take part in the End-of-semester examinations. REGISTRATION FOR EXAMINATIONS Registration for any examination shall require the endorsement of the registration by the Head of Department to effect that the candidate has pursued satisfactorily the approved course(s) of study in each subject being offered over the prescribed period. A candidate‟s registration shall not be valid unless it is so endorsed. A list of students registered for the semester examinations shall be published prior to the examination. Students whose names do not appear in any course list shall not be allowed into the end-of-semester examination for that course. Students who are duly registered for a course but fail to take the end-of-semester examination shall be deemed to have absented themselves from the examination of that particular course, for which a grade X or I shall be awarded. SEMESTER EXAMINATIONS An end-of-semester examination shall normally be required for every course. An examination schedule (time table) showing time and place of examination shall be published each semester, along with the rules and regulations that will guide the conduct of the examinations. 11 All University examinations shall generally be conducted under the rules and regulations spelt in Appendix … of this handbook. It shall be the responsibility of the student to be well-informed of the semester examination schedule, rules and regulations. The University assumes no responsibility for incorrect information concerning any aspect of the examinations that has been furnished by persons who are not authorized to give such information. All enquiries should be made at the Examinations Office. DEFEREMENT OF PROGRAMME A student is allowed a maximum period of two years to defer a programme. A student can defer a programme for one year within the course period and one year to redeem him/her, when referred after the final year exams. In general, a grace period of two years is allowed for students to redeem themselves, after the normal three-year period for HND programme. DEFERMENT OF EXAMINATION A student, after successful completion of a semester can, upon a written application to the Assistant Registrar for Academic Affairs, giving reasons, and after approval has been granted, defer his/her examinations. A student who wishes to defer his/her examinations shall first discuss the matter with his/her Head of Department. In all cases of deferment of examination(s), the applicant shall obtain written permission from the Assistant Registrar for Academic Affairs before leaving the University. A student who will be unable to take the end-of-semester examinations on grounds of ill-health shall, on application to the Assistant Registrar for Academic Affairs and copied to the Examinations Officer, Head(s) of Department and Dean of Students, and on provision of a medical certificate issued or endorsed by a recognised medical officer, be allowed to take the supplementary (re-sit) examination as his/her main examination. ILLNESS DURING THE EXAMINATION PERIOD If a student is prevented by illness from taking the entire or part of an examination, he/she shall immediately report to the University Hospital for a medical report on the state of his/her health. The report will state whether the student can take the examination or not. The Medical Officer shall state, in the report to the Examinations Officer , the name and the examination number of the student, the nature of the illness and whether, in his/her 12 opinion, if the student is capable of taking the examination, he/she should do so at the hospital or at the examination hall. Where it is recommended that the student should take the examination at the hospital, the Examinations Officer will arrange for the student to take the examination at the hospital under the supervision of an invigilator to be appointed by the Examinations Officer . SUPPLEMENTARY EXAMINATIONS A supplementary (re-sit) examination shall be conducted at the end of each academic year. A student who fails the main examination in a course qualifies for a supplementary examination. If the conditions for the award of a degree have not been met a student may re-sit a failed course up to the permitted number of attempts, as long as the registration has not expired. DISCIPLINE OF STUDENTS Any individual or collective action which disrupts or threatens to disrupt the academic and or normal life on campus or which brings the name of the University into disrepute shall be a major offence, subject to instant dismissal of the perpetrators. Except with respect to matters falling within the competence of the Residence Committee, it shall be the responsibility of the Vice Chancellor on the advice of and in consultation with the relevant Head of Academic Division and Head of Department, and subject to any regulations made by the Academic Board to provide for the discipline of students of the University. The Vice Chancellor may make such delegation of his authority subject to such review procedures as considered appropriate. A student who is deemed to be in breach of discipline may be liable to one or more of the following penalties: i. a warning; ii. a reprimand; iii. a fine; iv. suspension from the use of the University service or facilities for a stipulated period; v. requirement to make good to the satisfaction of the University any damage or injury caused to the property of the University or an Institution attended as a part of a University programme; vi. rustication from attendance at the University for a stated period; vii. withdrawal from the University for cheating in University Examinations; viii. expulsion from the University; and 13 ix. penalties iv – vii are major penalties. It shall be misconduct for a Junior Member (student) of the University i. to be absent from the campus, without permission or reasonable excuse; ii. to be absent from lectures and other prescribed assignment without permission or reasonable excuse; iii. to be insubordinate; iv. to indulge in any anti-social activities while in residence or outside the campus which tend to bring the name of the University into disrepute; and v. to cheat in examination. All major penalties imposed on a Junior Member shall be subject to satisfaction of the Vice Chancellor. The Vice Chancellor shall appoint an ad hoc committee with the following composition to make findings in a disciplinary case likely to attract the imposition of a major penalty against a Junior Member i. Dean of Student Affairs - Chairman ii. Senior Resident Tutor - Member iii. one person representing the Hall Master - Member iv. one person representing the SRC - Member v. an Assistant Registrar - Member/ Secretary APPEAL Any student who is aggrieved by any disciplinary action may appeal to the Vice Chancellor of the University for review. The Vice Chancellor will then refer the matter to the disciplinary committee for review. RULES AND REGULATIONS FOR NON-TERTIARY In respect of non-tertiary students, the rules and regulations of the Ghana Education Service (GES) shall apply. STUDENT GATHERING/DEMONSTRATION/PROCESSION i. Any student wishing to organise a procession/demonstration in the University shall notify the Vice Chancellor in writing. ii. Such notice shall be given to the Vice Chancellor at least 48 hours before the procession/demonstration is due to begin. iii. The notification shall state the purpose of the procession/demonstration and the name(s) of the organiser(s). 14 iv. No procession/demonstration shall be held between the hours of 6:00 p.m. and 6:00 a.m. v. All processions/demonstrations shall follow a route approved by the Vice Chancellor and keep as close to the left side of the road as to facilitate free passage of vehicular and pedestrian movement; vi. Students in a procession/demonstration shall do nothing to obstruct traffic; vii. During the procession/demonstration, nothing shall be done and said that might occasion violence or cause a breach of the peace. viii. The organiser(s) of any acts of violence and/or breach of University, hostel or other regulations that may occur during the procession/demonstration shall be responsible for their actions. ix. Notwithstanding the provision in sub-paragraph (vii), any individual who commits an act of violence or breach University, hostel or residence regulations during any procession/demonstration shall be held responsible for their actions. x. The Vice Chancellor may prescribe any special conditions, limitations or restrictions as may be considered appropriate in the circumstances. xi. If in the Vice Chancellor‟s opinion, the procession/demonstration shall be likely t o lead to a breach of the peace or cause serious interference with the work of the University, the Vice Chancellor may refuse to approve the procession/demonstration. xii. The fact that a procession/demonstration is not prohibited in no way implies that the University has either approved of or is in sympathy with its objectives. xiii. The Police/Army shall be alerted by the Vice Chancellor as soon as emergency occurs. If the situation gets out of hand or threatens to get out of hand and/or violence is resorted to, the Vice Chancellor shall ask the Police/Army to assist to restore order. xiv. The organisers of an act of violence and/or breach of University, hostel or any other regulation that may occur during the procession/demonstration shall be held responsible for their actions. Notwithstanding the provision in sub-section viii, any individual who commits an act of violence or breach of University, hostel or residential regulations during any procession/demonstrations shall be held responsible for their actions. 15 xv. GRADING OFASSESSMENTS Students in a course shall be graded as follows: LETTER GRADE (LG) MARK RANK NUMERICAL EQUIVALENT (NE) A+ A B+ B C+ C D+ D F 85-100 80-84 75-79 70-74 65-69 60-64 55-59 50-54 0 - 49 9 8 7 6 5 4 3 1 0 Where there is collaboration with another institution the grading system of awarding institution shall apply. FINAL AWARDS Final award shall be classified as follows: CLASS OF AWARDS C.G.P.A 1st Class 2nd Class Upper 2nd Class Lower Pass Fail 4.00 and above 3.00 – 3.99 2.00 – 2.99 1.50 – 1.99 Below 1.50 These criteria may be subjected to periodic review. 16 ADMISSION REQUIREMENTS A. BACHELOR OF TECHNOLOGY PROGRAMMES Accounting with Computing Entry Requirements The applicant must have two (2) years‟ work experience and any of the following qualifications: i. HND (Accountancy) certificate with a minimum of second class lower. ii. Passes in seven (7) or more subjects in ACCA, ICA (Ghana) or its equivalents. iii. Post-graduate certificate in Computerized Accounting. iv. Professional Diploma in Computerized Accounting certificate with credit passes (A-D for SSSCE or A1-C6 for WASSCE) in three (3) core subjects including English Language, Mathematics and Integrated Science or Social Studies. Agricultural Engineering Entry Requirements HND graduates must be competent or minimum of second class lower in Agricultural Engineering from recognized tertiary institutions and at least two (2) years of post-qualification industrial/work experience. Candidates with equivalent qualification/certificates can also apply. B. HIGHER NATIONAL DIPLOMA (HND) PROGRAMMES Arts, Humanities and Business Programmes The following programmes are offered: i. HND Accountancy – Morning and Evening Sessions; ii. HND Secretaryship and Management Studies – Morning and Evening Sessions iii. HND Marketing – Morning and Evening Sessions; iv. HND Fashion Design and Textiles Studies; v. HND Industrial Arts (Leather Works, Ceramics, Painting & Decorating and Textiles options); vi. HND Tourism; and vii. HND Media and Mass Communication. Entry Requirements for Arts, Humanities and Business (Direct Entry) 17 WASSCE Candidates: Six (6) credit passes (A1 –D7) comprising two (2) core subjects; English Language, Mathematics, and must possess a minimum of C6 in any three (3) of the passes relevant to the area of specialization. SSSCE Candidates: Six (6) passes (A – D) in all subjects; including English and Mathematics and at least three (3) of the passes must be relevant to the area of specialization. GBCE/RSAII Candidates: Credit passes (A – D) in five (5) subjects comprising three (3) core subjects; English language, Mathematics and Science or Social Studies, plus two (2) relevant elective subjects. DBS/ABC/RSA III /Senior Stenographer Candidates: Applicants must possess the following certificates; DBS Certificate or Diploma Certificate in ABCE or Diploma Certificate in RSA III, in addition to four WASSCE credit passes (A1 – C6) or four SSSCE passes (A – D) or four GBCE passes or four RSA II passes including English Language and Mathematics. GCE Advanced Level Candidates: Passes in Three (3) relevant subjects and five (5) credit passes in GCE „O‟ Level subjects including English Language and Mathematics. Science and Technology Programmes The following programmes are offered: i. HND Mechanical Engineering (Plant, Production & Automobile options) ii. HND Agricultural Engineering iii. HND Building Technology iv. HND Statistics v. HND Hotel, Catering and Institutional Management (HCIM) – Morning, Evening and weekends Session vi. HND Electricals/ Electronics Engineering vii. HND Information and Communication Technology – Morning and Evening Sessions Entry Requirements for Science and Technology Programmes (Direct Entry) WASSCE Candidates: Six (6) credit passes (A1 –D7) comprising two (2) core subjects; English Language, Mathematics, and must possess a minimum of C6 in any three (3) of the passes relevant to the area of specialization. 18 SSSCE Candidates: Six (6) passes (A – D) in all subjects; including English and Mathematics and at least three (3) of the passes must be relevant to the area of specialization. GCE Advanced Level Candidates: Passes in three (3) relevant subjects and six (6) credit passes in GCE „O‟ Level subjects including En glish Language and Mathematics. Technical School (Certificate II) Applicants (NABPTEX &TEU): Applicants must have passes (A-E) in English Language, Mathematics, Integrated Science or Social studies plus three (3) passes (A-E) in relevant elective subjects. Technician Applicants (MET 1 / MVT 1 / AET 1 / AMW): Applicants with Technician certificate (Part I and Part II) must have four (4) SSSCE/WASSCE Credits including English Language and Mathematics. Access Applicants: Passes in English language, Mathematics and Science organized by NABPTEX, plus Technician certificate or SSSCE/WASSCE passes in one (1) core subject, one (1) science elective subject and any other science subject. Other entry Requirements Certificate or Diploma holders from recognized higher institutions are eligible to apply for any related HND programme. However, such diplomas/certificates must be included in the database of National Accreditation Board (NAB) of recognized credentials. General Entry Requirements for Mature HND Applicants Applicants who have evidence of completing school at Advanced or Technical programmes, GCE „O‟ Level, GCE “A” Level, ABC, RSA, Stenographer, WASSCE/SSSCE, Teachers certificate or any recognized education and obtained the age of 25 years and above but have not met the above stated requirements are eligible to apply as mature candidates. In addition, such applicants: i. Must pass a written examination and an interview to be conducted by the University. ii. Must provide evidence of at least two (2) years‟ work experience. iii. Must provide a Birth Certificate at the interview. C. DIPLOMA IN BUSINESS STUDIES (DBS) PROGRAMMES The University offers the following DBS options: 19 i. Accounting ii. Secretarial iii. Management iv. Marketing v. Statistics vi. Purchasing and Supply vii. Information Technology viii. Entrepreneurship ix. Banking x. Finance General Entry Requirements for DBS Programmes i. Four (4) SSSCE passes (A – D) or WASSCE Credits passes (A1 – C6) including English Language and Mathematics); ii. Four (4) Ghana Commercial Certificate Examination passes (including English Language and Mathematics); iii. NACVET/NVTI, Typist Certificate (for Secretarial Option); iv. Four (4) ABC passes (A – D) including English Language and Mathematics; and v. Four (4) RSA II passes including English Language and Mathematics. D. THREE-YEAR CERTIFICATE TWO COURSES IN BUSINESS STUDIES Holders of BECE results with six (6) passes including English language and Core Mathematics are eligible to apply for three-year business studies programme. The programme offers opportunity for the business certificate two holders to progress to the University to pursue HND programme in the business areas. The options available are: (1) Accounting option, (2) Secretarial option and (3) Information Technology option. E. ASSOCIATION OF BUSINESS MANAGERS AND ADMINISTRATORS (ABMA) PROGRAMMES – EVENING SESSIONS ABMA Education is an awarding organization based in the UK and recognized by National Accreditation Board (NAB), Ghana. Over 60,000 ABMA examinations are conducted every year worldwide. Applicants have the opportunity now to pursue ABMA programmes in Tamale Technical University in the following areas: i. Computing and information systems 20 ii. HIV/AIDS Management iii. Human Resource Management iv. Business Management v. Marketing Management The ABMA programmes are available at level 4 Diploma, level 5 Diploma and level 6 Diploma. Entry Requirements for ABMA programme Level 4 Diploma SSSCE/WASSCE - Passes in English Language and Mathematics GBCE - Passes in English Language and Mathematics GCE “O” Level - Passes in English Language and Mathematics ABCE/A‟ Level/RSAIII - 1 Pass Plus passes in English Language and Mathematics RSA II - 3 Passes Plus passes in English Language and Mathematics DBS - 2 passes Plus passes in English Language and Mathematics Level 5 Diploma ABMA level 4 Diploma 2 passes at A‟ level/ABCE at Grade A -D and GCSE including English Language and Mathematics at grade A-D 4 GBCE passes including English Language and Mathematics, 4 SSSCE/WASSCE passes including English and Mathematics at grade (A1-C6) or (A-D) 3 DBS passes Plus SSSCE/WASSCE passes in English and Mathematics F. ADVANCEDAND TECHNICIAN PROGRAMMES i. Advanced Craft - Cookery For The Catering Industry ii. Advanced Craft - Fashion And Design i. Advanced Craft - Painting And Decorating ii. Advanced Craft - Block-Laying And Concreting iii. Advanced Craft - Carpentry And Joinery iv. Advanced Craft - Welding And Fabrication v. Mechanical Engineering Technician Part I & II vi. Motor Vehicle Technician Part I & II vii. Electrical Engineering Technician Part I &II viii. Radio, TV& Electronics Part I &II ix. Construction Technician Course Part I &II x. Agricultural Engineering Technician Part I & II 21 General Entry Requirements for Advanced and Technician Programmes (1) CTC - 1 i. Four (4) SSSCE/WASSCE passes including English Language, Mathematics and Technical Drawing. ii. Technicians Construction Option Subjects; Craft (B/C, C/J, P/D Plumbing) or Advanced (B/C and C/J) (2) MET - 1 i. Four (4) SSSCE/WASSCE passes including English language, Mathematics and Integrated Science. ii. Technical Option Subjects; Engineering Intermediate Craft, General Engineering (GE), Mechanical Engineering Craft Practice (MECP). (3) MVT – 1 i. Four (4) SSSCE passes including English Language, Mathematics and Integrated Science. ii. Technical Option Subjects, Motor Vehicle Mechanics Works (MVM), Intermediate General Engineering (GE) (4) EET – 1 and Radio TV &Electronics Part I i. Four (4) SSSCE/WASSCE passes including English Language, Mathematics and Integrated Science and any two of the following – Applied Electricity/Electronics, Physics, Mathematics. ii. Electrical Installation work course C. iii. Electrical Installation Work course B or Electrical Installation Craft practice. iv. Radio, Television and Electronics Mechanics (5) CTC II/MET II/ MVT II/ EET II/ RADIO TV AND ELECTRONICS PART II Applicants should be holders of CTC I/ MET I/ MVT I/ Radio TV Part I Advanced Catering – 812/2: Holders of Cookery for the Catering Industry Craft Advanced Fashion: Holders of Intermediate Fashion Advanced Welding: Holders of Intermediate Welding 22 G. ACCESS OR PRE-HND COURSE The Pre- HND or Access Course (English Language, Mathematics and Science) is organized by National Board for Professional and Technician Examinations (NABPTEX) to foster enrolment into the following programmes: i. HND (Agricultural Engineering) ii. HND (Mechanical Engineering) iii. HND (Automotive Engineering) iv. HND (Building Technology) v. HND (Hotel, Catering and Institutional Management) vi. HND (Industrial Arts) vii. HND (Fashion Design and Textiles Studies) viii. HND (Electrical and Electronics Engineering) ix. HND (Information and Communication Technology) Note : Duration for the course is nine (9) months. Entry Requirements Applicants should possess the following: i. Intermediate Craft Certificate or its equivalent in the relevant programme stated above. ii. SSSCE and WASSCE passes in three (3) subjects comprising at least one (1) core subject (English Language, Mathematics and Integrated Science or Social Studies) plus one (1) Science elective subject (Physics, Chemistry, Biology, Elective Mathematics, Agriculture Science, Metal Works) and any other science subject. H. CHARTERED INSTITUTE OF LOGISTICS AND TRANSPORT (CILT) PROGRAMMES i. CILT INTERNATIONAL CERTFICATE ii. CILT INTERNATIONAL DIPLOMA Time of Delivery : Evening or Week-ends ENTRY REQUIREMENTS AND DURATION FOR CILT PROGRAMMES CILT INTERNATIONAL CERTIFICATE: Entry requirements: 23 Applicants who have evidence of completing school at Advance or Technician Part I, ii, and iii programmes, GCE „O‟ Level, GCE “A” Level and WASSCE/SSSCE, or equivalent professional qualifications. Applicants should have worked in logistics and transport environment. Duration : Six months Optional subjects: i. Warehousing and stores operations ii. Freight operations iii. Fleet management iv. Passenger Transport operations v. Port Operations CILT INTERNATIONAL DIPLOMA: Entry requirements:  International Certificate in Logistics and Transport  University degree/ HND or equivalence with relevant area of study. Plus exemption at the certificate level, and evidence of work experience in a logistics and transport environment Duration : One year (two semesters) I. INTERMEDIATE CRAFT CERTIFICATE PROGRAMMES The following intermediate craft certificate programmes are offered: i. Radio, Television and Electronics Mechanics ii. Cookery for the Catering industry Part 1 iii. Block-laying and Concreting, Craft iv. Carpentry and Joinery, Craft v. Painting and Decorating, Craft vi. Welding Craft Practice vii. Mechanical Engineering Craft Practice viii. Motor Vehicle Mechanics work ix. Agricultural Mechanics work x. Electrical Installation work xi. Intermediate Fashion Entry requirements: Applicants with BECE results must have passes in English Language, Mathematics and Integrated Science in addition to passes in three (3) other subjects with an aggregate score of not more than 30. Applicants with SSSCE/WASSCE results must have passes in at least three (3) subjects. 24 J. CERTIFICATE COURSES The Mechanical Engineering Department runs the following certificate courses: i. Repairs of Small Engines. ii. Industrial Mechanics. Entry requirements: Applicants with BECE results must have passes in English Language, Mathematics and Integrated Science in addition to passes in three (3) other subjects with an aggregate score of not more than 30. K. MODE OF PAYMENT AND DELIVERY OF APPLICATION FORM Application forms for admission into Tamale Technical University are obtained online. Applicants should pay for e-vouchers from any branch of the HFC Bank or GN Bank. Applicants should log in; www.tatu.edu.gh or tamalepoly.edu.gh with e-voucher pin to complete the application form online. Two (2) copies of the completed application forms should be printed and submitted to: The Registrar Tamale Technical University (TaTU) P.O. Box 3 E/R Tamale
9211	https://bjc.edc.org/bjc-r/cur/programming/5-algorithms/4-unsolvable-undecidable/1-logical-inconsistency.html?topic=nyc_bjc%2F5-algorithms.topic&course=bjc4nyc.html	Proof by Contradiction Averaging just under 50 minutes, Lab 5: Unsolvable and Undecidable Problems, was the lab to which teachers devoted the least amount of time. It was also found to be the least successful, with only 20% reporting very successful and the majority (71%) noting the lab was neither unsuccessful or very successful. Teachers reported supplementing with videos from YouTube PG: Language inconsistency! Edit to clean up typos. I /really/ like the ideas here but the feedback has not been good. BH: But change the "at least four" problem to make it clear that you're only allowed to ask one question. MF: I want review this page and cut down on the switching back and forth among colored boxes. "Zoey and I are from the same family." is a poor choice for motivating proof by contradiction because it doesn't require it. --MF, 9/1/19 Also, this lab overdoes decidability and solvability which has been simplified in the standards now (and is fully covered in 5.1.6 Heuristic Solutions. When we revise, we should distill down to the core of what we want to teach. --MF, 9/1/19 In this lab, you will learn that some problems can't be solved at all. On this page, you will solve logic puzzles by finding a contradiction, that is, by showing that one possibility has to be true because the other possibility doesn't make sense. Imagine an island somewhere with two large families. One family (unlike normal people) can tell only the truth, even when they'd rather lie. This Truth-teller family can't ever make false statements, even by mistake. The other family, the False-Teller family, is just as reliable but in the opposite way: they can't make true statements ever. You are visiting the island and meet two of its people, Diego and Zoey. Diego says, "Zoey and I are from the same family." Can you say for sure which family Zoey is from? If so, which family? Can you say for sure which family Diego is from? If so, which family? Jasmine, Omar, and Morgan are considering the problem above. Jasmine: I'm pretty sure Zoey is a Truth-teller, but I don't know how to prove it. : Proof by Contradiction A proof by contradiction is a two-step proof that a statement is false, which is done by assuming the statement is true based on that assumption, proving something known to be false (that is, showing the assumption creates a contradiction) Omar: Sometimes it's easier to prove that something is false than to prove that something is true. So let's assume the opposite of what you want to prove, and see where that leads us. Let's assume that Zoey is a False-teller. Jasmine: Okay. So if Diego is a Truth-teller, then what he said is true, and they are from the same family, the Truth-tellers. But we assumed that Zoey is a False-teller, so they're actually from different families, and so Diego can't be a Truth-teller. Morgan: So, Diego has to be a False-teller. Omar: But that won't work either! If Diego is a False-teller, then what he said is false, and they are from different families. But they are both False-tellers, so they're actually in the same family, and so Diego can't be a False-teller either. Jasmine: No matter what family Diego is from, our assumption that Zoey is a False-teller led us to a contradiction. Zoey can't be a False-teller, so has to be a Truth-teller. We proved it. Imagine you meet someone named Derek on the island and you ask him if he's from the Truth-teller family. What does he answer? What if you ask Derek if he's from the False-Teller family? Jasmine and Omar are exploring logical statements of their own. Jasmine: The statement I'm making right now is false. Omar: (Thinks for a moment) Wait! What?!? Jasmine claims her statement is false. What do you think? Explain your thinking clearly. Omar: That's very clever, Jasmine. Your statement can't be true, and it can't be false. So neither a Truth-teller nor a False-teller could say that. There are four kinds of true/false statements: : Undecidable Statement vs Self-Contradictory Statement An undecidable statement might be true or might be false; we don't know which. A self-contradictory statement can be neither true nor false. Provably True: For example in this problem, "Zoey is a Truth-teller." Provably False: For example, "Zoey is a False-teller." Undecidable: For example, "Diego is a Truth-teller." Self-Contradictory: Such as "This statement is false." What questions can you ask in order to determine whether a person is a Truth-teller or a False-Teller? Talking with others, find at least four questions that will work reliably. If Diego were a Truth-teller, how would he answer your questions? Check to make sure that if he were a False-Teller, he'd answer differently. Theorem: All positive integers are interesting. Proof: In order to prove this by contradiction, assume that not all positive integers are interesting. Since positive integers don't go down forever, there must be a smallest non-interesting positive integer. But isn't that an interesting thing about that number? ;) On that island of Truth-tellers and False-Tellers, you meet Max and Min. Max says "Min and I are both liars!" Which kind of statement is this? Is it self-contradictory? Is it undecidable (it could be either, but there's no way to tell)? Or is it definitely resolvable? If resolvable, who's in what family?
9212	https://fiveable.me/discrete-mathematics/unit-7/binomial-coefficients-identities/study-guide/os6tgHwnGojtQ2Uj	printables 🧩Discrete Mathematics Unit 7 Review 7.3 Binomial Coefficients and Identities 🧩Discrete Mathematics Unit 7 Review 7.3 Binomial Coefficients and Identities Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 APA 🧩Discrete Mathematics Unit & Topic Study Guides 7.1 Basic Counting Principles 7.2 Permutations and Combinations 7.3 Binomial Coefficients and Identities 7.4 Introduction to Probability Theory Binomial coefficients are the building blocks of combinatorics. They show up everywhere, from Pascal's Triangle to probability calculations. These powerful tools help us count combinations and expand algebraic expressions, making them essential for tackling complex counting problems. Mastering binomial coefficients opens doors to understanding advanced topics in discrete math. We'll explore their properties, identities, and applications, seeing how they connect to other concepts in counting and probability. Get ready to unlock the secrets of these mathematical powerhouses! Binomial Coefficients and Pascal's Triangle Understanding Binomial Coefficients Binomial coefficient represents the number of ways to choose k items from n items without replacement and without order Denoted as (kn) or C(n,k) Calculated using the formula (kn)=k!(n−k)!n! Useful in probability, combinatorics, and algebra Symmetric property: (kn)=(n−kn) Appears in the expansion of binomial expressions (x+y)n Can be computed recursively using the formula (kn)=(k−1n−1)+(kn−1) Exploring Pascal's Triangle Triangular array of binomial coefficients arranged in rows Each number equals the sum of the two numbers directly above it First few rows of Pascal's Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Nth row contains the coefficients of (x+y)n when expanded Exhibits numerous patterns and properties (prime number patterns, Fibonacci sequence) Used to calculate combinations and probabilities efficiently Provides a visual representation of binomial coefficients Applying Combinatorial Proofs Method of proving identities involving binomial coefficients Involves interpreting both sides of an equation as counting the same set in different ways Steps for combinatorial proof: Interpret left side of equation as counting a specific set Interpret right side as counting the same set in a different way Conclude that both sides must be equal Useful for proving binomial identities and Pascal's Triangle properties Provides intuitive understanding of complex algebraic relationships Can be applied to prove identities like (kn)=(k−1n−1)+(kn−1) Binomial Theorem and Expansion Exploring the Binomial Theorem Fundamental theorem in algebra and combinatorics Describes the algebraic expansion of powers of a binomial General form: (x+y)n=∑k=0n(kn)xn−kyk Allows quick expansion of binomial expressions without manual multiplication Coefficients in the expansion are binomial coefficients Applies to any positive integer exponent n Generalizes to multinomial theorem for expressions with more than two terms Applying Binomial Expansion Process of using the Binomial Theorem to expand $(x+y)^n$ Steps for binomial expansion: Identify the terms x and y, and the exponent n Write out each term using the formula $\binom{n}{k} x^{n-k} y^k$ Simplify and combine like terms if necessary Useful in various mathematical fields (calculus, probability, statistics) Can be used to approximate functions using Taylor series Allows for quick calculation of specific terms in the expansion Expanded form reveals patterns and properties of the original expression Understanding Binomial Identities Equations involving binomial coefficients that hold true for all values Common binomial identities: ∑k=0n(kn)=2n (kn)=(k−1n−1)+(kn−1) ∑k=0n(kn)(−1)k=0 Used to simplify complex expressions and solve combinatorial problems Can be proved using algebraic manipulation, induction, or combinatorial arguments Provide insights into the relationships between different combinations Applicable in probability calculations and statistical analysis Advanced Binomial Identities Exploring Vandermonde's Identity Important combinatorial identity named after Alexandre-Théophile Vandermonde States that ∑k=0r(km)(r−kn)=(rm+n) Interprets as the number of ways to choose r items from two separate groups Generalizes to multinomial coefficients for more than two groups Useful in probability theory and combinatorial mathematics Can be proved using generating functions or combinatorial arguments Applies in various fields (coding theory, number theory, statistical mechanics) Applying Combinatorial Proofs to Advanced Identities Extends the concept of combinatorial proofs to more complex binomial identities Steps for proving advanced identities: Interpret each side as counting a specific set of objects Demonstrate that both sides count the same set in different ways Conclude the equality of both expressions Requires creative thinking to find appropriate counting interpretations Often involves breaking down complex scenarios into simpler subproblems Can be used to prove identities like ∑k=0n(kn)2=(n2n) Provides intuitive understanding of relationships between different combinations Exploring Additional Binomial Identities Numerous other binomial identities exist beyond the basic and Vandermonde's identity Examples include: ∑k=0nk(kn)=n2n−1 (kn)=kn(k−1n−1) ∑k=0n(kn)xk=(1+x)n These identities reveal deeper relationships between combinations and series Often used in solving complex combinatorial problems and simplifying expressions Can be proved using various methods (algebraic manipulation, induction, generating functions) Understanding these identities enhances problem-solving skills in discrete mathematics
9213	https://mathbitsnotebook.com/Algebra1/LinearEquations/LELiteralEquationPractice.html	Literal Equations Practice - MathBitsNotebook(A1) Literal Equations - Practice MathBitsNotebook.com Topical Outline \| Algebra 1 Outline \| MathBits' Teacher Resources Terms of UseContact Person:Donna Roberts Directions: Read carefully to determine the variable for which you are to solve. 1.If k = am + 3 mx, the value of m in terms of a, k and x can be expressed as: Choose: 2.If ba - cd = b, which expression represents a? Choose: 3.If , then t equals: Choose: sr+ 2 x sr- 2 x 4.The volume of a pyramid is given by . What is h expressed in terms of B and V? Choose: 5.For the equation , express r in terms of t and w. Choose: r = 3 w - t r = 3 tw r = 3(w - t) 6.If P = Q 2•O, express O in terms of P and Q. Choose: P - Q 2 P + Q 2 7.Given: . Solve for C. Choose: 8.A formula is expressed as D = a(2 + kt). Express k in terms of D, a and t. Choose: D- 2 at 9.If A = C(1 - S 2), express S in terms of A and C. Choose: NOTE:There-posting of materials(in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". Topical Outline \| Algebra 1 Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of UseContact Person:Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved.
9214	https://www.mytutor.co.uk/answers/20355/A-Level/Maths/Prove-using-the-product-rule-that-the-derivative-of-x-n-is-nx-n-1-where-n-is-a-natural-number-What-if-n-is-an-integer-or-n-is-rational/	Prove, using the product rule that, the derivative of x^{n} is nx^{n-1} where n is a natural number. What if n is an integer or n is rational? \| MyTutor Find a tutorHow it worksPricesResourcesFor schoolsBecome a tutor +44 (0) 203 773 6024 Log inSign up Answers>Maths>A Level>Article Prove, using the product rule that, the derivative of x^{n} is nx^{n-1} where n is a natural number. What if n is an integer or n is rational? We know that d / dx (x) = 1. Looking at x^2 as a specific example, and using the product rule, we see that d/dx(x^2) = xd/dx(x) + d/dx(x)x = x + x = 2x. Similarly for x^3, d/dx(x^3) = 3 x^2 d/dx(x) = 3 x^2 Following this logic for general n, we have that d/dx(x^n) = nx^(n-1).If n is negative then we need to use that the derivative of 1 / x = - 1 / x^2. To show this, let y = 1/x so xy = 1. Then differentiate with respect to x to get that xdy/dx + y = 0, and so xdy/dx = -1 / x which is what we need.We can now proceed in the same way. d/dx(1 / x^-n) = -n 1 / x^(-n - 1) -1 / x^2 = nx^(n-1).For rational n we need a slightly different approach. Let y = x^n = x^(a/b) where a and b are integers, so we have that y^b = x^a. Differentiating this with respect to x gives, using the previous parts, that by^(b-1)dy/dx = ax^(a-1) and so dy/dx = (a/b)(x^(a - 1) / y^(b-1)) = (a / b)(x^(a-1) / x^(a(b-1) / b)) = (a/b)x^((a - 1) - a(b-1)/b) = (a/b)(x^(ba - b - ab + a) / b) = (a/b)x^(a / b - 1). AH Answered by Alfred H. •Maths tutor 10446 Views See similar Maths A Level tutors 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 Find a tutor 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. Download handbook Free weekly group tutorials Join MyTutor Squads for free (and fun) help with Maths, Coding & Study Skills. Sign up Related Maths A Level answers All answers ▸ #### Express 6sin(2x)+5cos(x) in the form Rsin(x+a) (0degrees<x<90degrees) Answered by Jacob R. #### A line L is parallel to y=4x+5 and passes through the point (-1, 6). Find the equation of the line L in the form y=ax+b . Find also the coordinates of its intersections with the axes. Answered by Matt M. #### How do I integrate cos^2(x)? Answered by Daniel F. #### What is Mathematical Induction? Answered by Ayesha H. We're here to help Contact us+44 (0) 203 773 6020 Company Information CareersBlogSubject answersBecome a tutorSchoolsSafeguarding policyFAQsUsing the Online Lesson SpaceTestimonials & pressSitemapTerms & ConditionsPrivacy Policy Popular Requests Maths tutorChemistry tutorPhysics tutorBiology tutorEnglish tutorGCSE tutorsA level tutorsIB tutorsPhysics & Maths tutorsChemistry & Maths tutorsGCSE Maths tutors ##### 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
9215	https://brainly.com/question/30681537	[FREE] 3. Find the height of the tree below to the nearest foot. - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +66,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +20,6k Ace exams faster, with practice that adapts to you Practice Worksheets +7,2k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified 1. Find the height of the tree below to the nearest foot. 1 See answer Explain with Learning Companion NEW Asked by brownskinjada • 02/11/2023 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 25432329 people 25M 5.0 0 Upload your school material for a more relevant answer The height of the tree to the nearest foot is 27 feet. How to find the height of a tree? The height of the tree can be found as follows: The height of the tree can be found using trigonometric ratios. Therefore, let tan ∅ = opposite / adjacent Therefore, opposite side = adjacent tan ∅ Hence, height of the tree = 20 + x tan 45° = x tan 75° 20 + x tan 45° = x tan 75° 20 tan 45 + x tan 45° = x tan 75° 20 tan 45 = x tan 75° - x tan 45° 20 tan 45 = x(tan 75 - tan 45) 20 tan 45 = x(3.73205080757 - 1) 2.73205x = 20 x = 20 / 2.73 x = 7.32600732601 x = 7.34 Therefore, height of the tree = x tan 75° height of the tree = 7.34 × tan 75° height of the tree = 27.3186119114 height of the tree = 27 ft learn more on right triangle here: brainly.com/question/8421388 SPJ1 Answered by vintechnology •7.1K answers•25.4M people helped Thanks 0 5.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 25432329 people 25M 5.0 0 Forest Measurements - An Applied Approach - Joan DeYoung Upload your school material for a more relevant answer To find the height of the tree, use the tangent function relating the angle of elevation and the horizontal distance from the tree. Measure the angle from your position to the top of the tree, then apply the tangent ratio to calculate the height, rounding to the nearest foot. For example, if the angle is 75 degrees and you're 20 feet away, the height would be approximately 19 feet. Explanation To find the height of the tree, we can use the principles of trigonometry, specifically focusing on a right triangle formed between the tree, the ground, and the line of sight to the top of the tree. Here’s a step-by-step method: Establish Your Position: Stand a known horizontal distance away from the base of the tree. This distance will be the 'adjacent' side of our right triangle. Measure the Angle: Using a clinometer, measure the angle of elevation (let's call this angle θ) from your standing position to the top of the tree. Apply the Tangent Function: We can use the tangent ratio from trigonometry which states that: tan(θ)=horizontal distance height of the tree Rearranging the Formula: We can rearrange this to calculate the height of the tree: height of the tree=tan(θ)×horizontal distance Calculate the Height: Substitute your measured angle and the horizontal distance into the formula to find the height. For example, if the angle measured was 75 degrees and you were standing 20 feet away: height=tan(75°)×20≈19.2 feet Combine Measurements if Needed: If your measurement position is elevated (e.g., you are standing on a hill), add that height to your calculation to get the total height of the tree. To round your final answer, use the nearest foot for increased clarity. If your calculations produce a height of 19.2 feet, round it to 19 feet. Following these steps will give you a correct estimate of the tree's height based on the measurements you take. Examples & Evidence For instance, standing 30 feet away and measuring a 60-degree angle to the top of the tree, you would calculate the height as tan(60°)×30≈51.96 feet, rounding to 52 feet. This procedure is based on the properties of right triangles and trigonometric functions, widely used in fields such as surveying and physics. Thanks 0 5.0 (1 vote) Advertisement brownskinjada has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Simplify. (v 3)−2 Write your answer without using negative exponents. Suppose X is a binomial random variable with 26 trials and a probability of success of 0.54. Find P(X>13). Round your answer to four decimal places. Solve for w. 8.4=w+1 Suppose X is a binomial random variable with 37 trials and a probability of success of 0.43. Find P(X=14). Round your answer to four decimal places. Solve for x: 3(x−1)(x+3)−7 x=9+x Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
9216	https://pmc.ncbi.nlm.nih.gov/articles/PMC10332772/	Anticholinergic Toxicity in the Emergency Department - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Educ Teach Emerg Med . 2023 Jan 31;8(1):S25–S47. doi: 10.21980/J8D07Z Search in PMC Search in PubMed View in NLM Catalog Add to search Anticholinergic Toxicity in the Emergency Department C Eric McCoy C Eric McCoy, MD, MPH University of California, Irvine, Department of Emergency Medicine, Orange, CA Find articles by C Eric McCoy ,✉, Reid Honda Reid Honda, MD ^Queens Medical Center, Department of Emergency Medicine, Honolulu, HI Find articles by Reid Honda ^ Author information Article notes Copyright and License information University of California, Irvine, Department of Emergency Medicine, Orange, CA ^Queens Medical Center, Department of Emergency Medicine, Honolulu, HI ✉ Correspondence should be addressed to C Eric McCoy, MD, MPH at cmccoy@hs.uci.edu ✉ Corresponding author. Received 2022 Apr 26; Accepted 2022 Aug 4; Collection date 2023 Jan. © 2023 McCoy, et al. This is an open access article distributed in accordance with the terms of the Creative Commons Attribution (CC BY 4.0) License. See: PMC Copyright notice PMCID: PMC10332772 PMID: 37465041 Abstract Audience Emergency medicine residents, internal medicine residents, family medicine residents, community physicians, pediatricians, toxicology fellows Introduction There are over 600 compounds which contain anticholinergic properties.1 Medications with anticholinergic properties include antihistamines, atropine, tricyclic antidepressants, antipsychotics, topical mydriatics, antispasmodics, sleep aids, and cold preparations. 1–4 Plants that possess anticholinergic properties such as jimson weed, and street drugs cut with anticholinergics such as scopolamine are sources of accidental or intentional ingestion.1,2,4 Anticholinergic toxicity can cause a myriad of signs and symptoms, including agitation, seizures, hyperthermia, cardiac dysrhythmias, and death. Since poisoning from anticholinergic medications is frequently encountered in the emergency department, is it essential that emergency physicians be familiar with how to manage this toxidrome. This simulation case will allow the learner to evaluate and manage a patient presenting with anticholinergic toxicity. Educational Objectives By the end of this simulation case, learners will be able to: 1) describe the classic clinical presentation of anticholinergic toxicity, 2) discuss common medications and substances that may lead to anticholinergic toxicity, 3) recognize the electrocardiogram (ECG) findings in anticholinergic toxicity that require specific therapy, and 4) review the management of anticholinergic toxicity. Educational Methods This simulation is taught using a high- or moderate-fidelity manikin. Research Methods The educational content was evaluated by the learners immediately after completion and debriefing of the scenario. This case was initially piloted with approximately twenty emergency medicine residents. The group was comprised of first, second-, and third-year residents from a three-year emergency medicine residency. The efficacy of the content was assessed by oral feedback. Results Overall, the case was well received by learners, who felt it was useful and were engaged throughout the session. The overall feedback was positive and the case was well-received by learners. Discussion This scenario was eventually tested on over 100 learners over the course of several years, and the overall feedback was positive. It was found to be effective when debriefing sessions using various debriefing techniques (such as advocacy/inquiry) were utilized to discuss both the learners’ performance in the case, as well as the debriefing pearls (located at the end of this manuscript). Topics Anticholinergic toxicity, altered mental status, toxicology. USER GUIDE \| List of Resources: \| \| Abstract \| 25 \| \| User Guide \| 27 \| \| Instructor Materials \| 28 \| \| Operator Materials \| 37 \| \| Debriefing and Evaluation Pearls \| 40 \| \| Simulation Assessment \| 43 \| Open in a new tab Learner Audience: Interns, junior residents, senior residents, community physicians, pediatricians, toxicology fellows Time Required for Implementation: Instructor Preparation: 20–30 minutes Time for case: 10–15 minutes Time for debriefing: 15–30 minutes Recommended Number of Learners per Instructor: 3–5 Topics: Anticholinergic toxicity, altered mental status, toxicology. Objectives: By the end of this simulation case, learners will be able to: Describe the classic clinical presentation of anticholinergic toxicity Discuss common medications and substances that may lead to anticholinergic toxicity Recognize the ECG findings in anticholinergic toxicity that require specific therapy Review the management of anticholinergic toxicity Linked objectives and methods During this simulation, the participants will need to make the diagnosis of anticholinergic toxicity in a critically ill patient presenting with undifferentiated altered mental status (objective 1). If a thorough history and collateral information is obtained, the participants will realize that the patient overdosed on diphenhydramine and jimson weed (objective 2). During the case, the patient will be tachycardic with a widened QRS interval, requiring treatment with sodium bicarbonate (objective 3). In addition, the patient will require treatment with benzodiazepines, supportive care, and potential toxicology/poison center consultation (objective 4). Recommended pre-reading for instructor Fulton J, Nelson L. Anticholinergics. In: Adams J, ed. Emergency Medicine: Clinical Essentials. 2 nd Edition. Philadelphia, PA: Elsevier; 2012:1240–1245. Wax PM. Anticholinergic toxicity. In: Tintinalli JE, Stapczynski J, Ma O, Cline DM,, Meckler GD, eds. Tintinalli’s Emergency Medicine: A Comprehensive Study Guide. 8 th ed. New York, NY: McGraw-Hill; 2016:1143–1146. Results and tips for successful implementation This simulation scenario is best implemented in an educational environment that includes high- or moderate-fidelity simulation resources along with monitors that will allow the learner(s) to independently interpret vital signs and visual stimuli (ie, a simulation center or equivalent educational space). This scenario was tested on more than 100 learners over the course of several years. Performance is best measured with a critical actions checklist that can be used during the real-time assessment of the simulation scenario by proctors with the primary purpose of participant evaluation. The educational content was evaluated by the learners immediately after completion and debriefing of the scenario. This case was initially piloted with approximately twenty emergency medicine residents. The group was comprised of first, second-, and third-year residents from a three-year emergency medicine residency. The efficacy of the content was assessed by oral feedback. Overall, the case was well received by learners, who felt it was useful and were engaged throughout the session. The overall feedback was positive, and the case was well-received by learners. Supplementary Information jetem-8-1-S25-supp1.pptx (280.9KB, pptx) jetem-8-1-S25-supp2.jpg (70.7KB, jpg) jetem-8-1-S25-supp3.jpg (132KB, jpg) jetem-8-1-S25-supp4.jpg (85.5KB, jpg) jetem-8-1-S25-supp5.jpg (235.5KB, jpg) jetem-8-1-S25-supp6.jpg (113.4KB, jpg) jetem-8-1-S25-supp7.jpg (95.5KB, jpg) INSTRUCTOR MATERIALS Case Title: Anticholinergic Toxicity in the Emergency Department Case Description & Diagnosis (short synopsis): An 18-year-old male is brought in by ambulance after being found altered and wandering around a mobile home park. Upon arrival to the emergency department, he is confused, combative, febrile, and tachycardic. Upon further questioning of his friend at bedside, it is discovered that the patient and his friends were drinking tea made out of jimson weed and liquid diphenhydramine, as part of a viral online social media challenge, prior to his altered level of consciousness. The patient is ultimately found to be presenting with an anticholinergic toxidrome secondary to consumption of jimson weed and diphenhydramine. Equipment or Props Needed: High- (or moderate-) fidelity simulation mannequin Cardiac monitor, pulse oximeter Blood pressure cuff Intravenous (IV) catheter and lines Crash cart with defibrillator Bag valve mask Intubation equipment (endotracheal tube, laryngoscope) Syringes to simulate medications Normal saline Confederates needed: Nurse to assist with management of patient Paramedics who bring patient into emergency department (this can be played by simulation operator overhead) Friend who arrives with patient and paramedics (this can be played by simulation operator overhead) Stimulus Inventory: 1 Electrocardiogram (ECG) showing sinus tachycardia with widened QRS interval (>100ms) 2 Chest Radiograph (CXR) 3 Complete blood count (CBC) 4 Comprehensive metabolic panel (CMP) 5 Urinalysis (UA) and Urine Drug Screen Open in a new tab Background and brief information: The scenario takes place in an emergency department at a community hospital. The patient is an 18-year-old male brought in by paramedics with a chief complaint of altered mental status. Initial presentation: The patient is confused and intermittently agitated upon arrival. His vital signs are significant for a fever and tachycardia. His physical exam is significant for dry and flushed skin, dilated pupils, and altered mental status. He is unable to provide further history. How the scene unfolds: The patient arrives to the emergency department confused and unable to provide further history. If asked, the paramedics will report that the patient was found wandering about his mobile home park confused. They will report that the patient was tachycardic on scene, and that his finger-stick blood glucose was normal. They will state that the patient was intermittently combative en route, and would not cooperate with the non-rebreather face mask that they attempted to place on him. If the participants ask the patient’s friend at bedside for more information, he will state that the patient and some friends were taking part in a viral online social media challenge, and were drinking a tea made out of jimson weed and liquid diphenhydramine, which the patient consumed shortly before becoming altered. On arrival, the patient is tachycardic and hyperthermic. His physical exam is significant for dry, flushed skin, dilated pupils, and altered mental status. The patient’s ECG will be significant for tachycardia with a QRS >100ms. The participants should make the diagnosis of anticholinergic toxicity. The participants may initiate treatment with activated charcoal. The participants should also contact the poison control center. They may control the patient’s agitation using benzodiazepines. If the diagnosis of anticholinergic toxidrome is delayed and benzodiazepines are not administered, the patient will become increasingly agitated and will have a seizure. The participants should recognize the concerning finding of the QRS >100msec on ECG, and should administer sodium bicarbonate. If the participants do not recognize this, but consult the poison control center, the toxicologist will prompt them to review the ECG and initiate sodium bicarbonate. If sodium bicarbonate is not given, the patient will go into ventricular fibrillation. If the participants administer physostigmine, the patient will go into cardiac arrest. The patient should ultimately be admitted to the intensive care unit for further observation and management. Critical actions: Assess airway, breathing, and circulation (ABCs) Place patient on cardiac monitors and pulse oximeter, and obtain initial set of vital signs Establish two large bore IV lines Administer IV fluids (at least 1 liter) Make the diagnosis of anticholinergic toxicity Contact the poison control center/obtain toxicologist consult Give sodium bicarbonate for QRS >100msec on ECG Give benzodiazepines to control agitation Admit patient to the intensive care unit Case Title: Anticholinergic Toxicity in the Emergency Department Chief Complaint: The patient is an 18-year-old male presenting with a chief complaint of altered mental status Vitals:Heart Rate (HR) 137 Blood Pressure (BP) 167/95 Respiratory Rate (RR) 22 Temperature (T) 38.7°C Oxygen Saturation (O 2 Sat) 98% on room air (RA) Open in a new tab General Appearance: Appears stated age, in moderate distress Primary Survey: Airway: Patent Breathing: Clear to auscultation bilaterally, slightly tachypneic Circulation: 2+ femoral pulses bilaterally History: History of present illness: The patient is an 18-year-old male who presents brought in by paramedics with altered mental status. The patient will be too altered to provide any history. If asked, the paramedics will report that the patient was found wandering about a mobile home park altered. They will report that he was tachycardic in the field, and that his finger-stick blood glucose was normal. They will state that he was intermittently agitated en route. If the friend at bedside is questioned, he will state that the patient and his friends were drinking tea made out of jimson weed and liquid diphenhydramine earlier in the day. Past medical history: Unable to obtain from patient due to altered mental status Past surgical history: Unable to obtain from patient due to altered mental status Patient’s medications: Unable to obtain from patient due to altered mental status Allergies: Unable to obtain from patient due to altered mental status Social history: Unable to obtain from patient due to altered mental status Family history: Unable to obtain from patient due to altered mental status Secondary Survey/Physical Examination: General appearance: Awake, confused, dazed look on face, appears to be responding to internal stimuli HEENT): ○ Head: No external signs of trauma ○ Eyes: Pupils dilated 8mm bilaterally, extra-ocular movements intact ○ Ears: within normal limits (WNL) ○ Nose: WNL ○ Throat: No tonsillar exudates, no erythema Neck: No midline C-spine tenderness, full range of motion, no swelling, trachea midline Heart: Tachycardic, no audible murmurs Lungs: Clear to auscultation bilaterally, slightly tachypneic Abdominal/gastrointestinal: Soft, non-distended, non-tender to palpation, no guarding, no rigidity. Slight fullness noted to suprapubic region. Decreased bowel sounds. Genitourinary: Normal appearing external male genitalia Rectal: Normal tone, no melena, no gross blood Extremities: No gross deformities. No lower extremity edema. Back: No thoracic or lumbar spine tenderness, no step-offs Neuro: Awake. Not answering questions appropriately. No facial droop. Intermittently following commands. Intermittently agitated. Appears to be responding to internal stimuli, reaching for things in the air. Moves all four extremities, no significant weakness noted on allowable exam. Skin: Dry, flushed skin, warm to touch, no diaphoresis Lymph: WNL Psych: Intermittently agitated, appears to be responding to internal stimuli Results: Open in a new tab Electrocardiogram (ECG) – sinus tachycardia with widened QRS interval Image source: (author’s own image) Open in a new tab Chest Radiograph (CXR) Image source: Author’s own image Complete blood count (CBC) White blood count (WBC)15.0 ×1000/mm 3 Hemoglobin (Hgb)14.0 g/dL Hematocrit (HCT)40.0% Platelet (Plt)250 ×1000/mm 3 Open in a new tab Comprehensive metabolic panel (BMP) Sodium 140 mEq/L Chloride 100 mEq/L Potassium 4.7 mEq/L Bicarbonate (HCO 3)19 mEq/L Anion Gab 21 mEq/L Blood Urea Nitrogen (BUN)30 mg/dL Creatine (Cr)1.1 mg/dL Glucose 100 mg/dL Calcium 9.5 mg/dL Total Protein 7.0 g/dL Albumin 40 units/L Alkaline Phoshatase 100 units/L Aspartate aminotransferase (AST)90 units/L Alanine aminotransferase (ALT)60 units/L Open in a new tab Urinalysis (UA) Color yellow Appearance clear Specific gravity 1.010 pH 7.0 Glucose negative Ketones negative Hemoglobin negative Leukocyte esterase negative Nitrites negative White blood cells (WBC)0 WBCs/high powered field (HPF) Red blood cells (RBC)0 RBCs/HPF Squamous epithelial cells 1 cells/HPF Bacteria negative Open in a new tab Urine Toxicology Screen: Tetrahydrocannabinol (THC)None Detected Phencyclidine (PCP)None Detected Benzodiazepines None Detected Barbiturates None Detected Methadone None Detected Cocaine None Detected Amphetamines None Detected MDMA None Detected Open in a new tab OPERATOR MATERIALS SIMULATION EVENTS TABLE: \| Minute (state) \| Participant action/ trigger \| Patient status (simulator response) & operator prompts \| Monitor display (vital signs) \| :---: :--- \| \| 0:00 (Baseline) \| Assess ABCs Attach patient to cardiac monitor and pulse ox Obtain initial set of vitals \| Patient arrives confused and intermittently agitated. \| T 38.7°C HR 137 BP 167/95 RR 22 O2 98% on RA \| \| 02:00 \| Obtain history from paramedics Obtain history from patient’s friend Perform physical exam Obtain finger-stick glucose \| If asked, paramedics will state that patient was found wandering around mobile home park confused. If asked, patient’s friend will state that patient was making tea with jimson weed and liquid diphenhydramine, which he consumed prior to becoming altered. Physical exam will be significant for dry, warm, flushed skin, mydriasis, tachycardia, and confusion. Finger-stick glucose will be normal. \| T 38.7°C HR 137 BP 167/95 RR 22 O2 98% on RA \| \| 04:00 \| Make diagnosis of anticholinergic toxicity Obtain IV access Order labs Order ECG \| Patient’s agitation may be treated with benzodiazepines. If benzodiazepine administered: Patient’s agitation and tachycardia will improve. If diagnosis of anticholinergic toxicity not made by this time and patient not given benzodiazepine, patient will become increasingly agitated and have a seizure. If patient has seizure, it will resolve with benzodiazepine administration. If benzodiazepine is given, vitals will stabilize, and move on to time 6:00. If benzodiazepine is not given, patient will go into cardiac arrest and expire. \| If benzo given: HR 115 BP 140/70 RR 18 O2 98% on RA If benzo not given, seizure: HR 150 BP 170/90 RR 28 O2 88% on RA If benzo given, seizure resolves: HR 120 BP 140/90 RR 18 O2 98% on RA \| \| 06:00 \| Poison control/ toxicology consult should be obtained Lab results will return \| ECG is available demonstrating sinus tachycardia with a QRS >100msec. Participants should recognize wide QRS complex and initiate treatment with sodium bicarbonate. If poison control/toxicology consult obtained, they will prompt participants to review ECG and give sodium bicarbonate. They will also recommend that activated charcoal be given. They will recommend against physostigmine, because patient’s QRS is >100ms. If sodium bicarb is given: Patient’s vital signs will remain stable, proceed to time 10:00 If sodium bicarb not given: Patient will go into Ventricular fibrillation. If ACLS started, patient will not obtain return of spontaneous circulation (ROSC) until sodium bicarb is given. If sodium bicarb given, patient will obtain ROSC, proceed to time 10:00. If sodium bicarb still not administered, patient will expire and case will end. If physostigmine is given, patient will decompensate into cardiac arrest and case will end. \| If sodium bicarb given: HR 110 BP 130/60 RR 18 O2 98% on RA If sodium bicarb not given: HR v Fib BP 0 RR 0 O2 undetectable \| \| 10:00 \| Admit the patient to the ICU \| If fluids, benzodiazepines, and sodium bicarb given, patient’s agitation and vital signs will improve, and patient should be admitted to the ICU. If no treatment was started, and/or if the diagnosis of anticholinergic toxicity was not made, the patient will have further seizures requiring intubation, and will ultimately go into cardiac arrest and expire. \| HR 90 BP 140/160 RR 18 O2 98% on RA \| Open in a new tab Diagnosis: Anticholinergic toxicity secondary to jimson weed ingestion Disposition: Intensive care unit DEBRIEFING AND EVALUATION PEARLS Anticholinergic Toxicity Introduction: ○ There are over 600 compounds which contain anticholinergic properties.1 Medications with anticholinergic properties include antihistamines, atropine, tricyclic antidepressants, antipsychotics, topical mydriatics, antispasmodics, sleep aids, and cold preparations.1,2,3,4 Plants such as jimson weed (datura stramonium) and deadly nightshade (atropa belladonna) also have anticholinergic properties and are a common source of accidental or intentional ingestion.1,2,4 Street drugs such as heroin and cocaine have been known to be cut with anticholinergics such as scopolamine. 1,2 Thus, it is not surprising that anticholinergic poisonings are frequently seen in the emergency department, and recognition of the anticholinergic toxidrome is a necessary clinical skill.1,4 Pathophysiology and Clinical Manifestation: ○ Muscarinic acetylcholine receptors are located in the central nervous system, heart, smooth muscle, secretory glands, and ciliary body of the eye.1 ○ Normally, the neurotransmitter acetylcholine binds to these muscarinic receptors1 ○ Anticholinergic drugs competitively inhibit this binding, producing the following clinical effects: “Mad as a hatter”Blockage of CNS muscarinic receptors Delirium, hallucinations, anxiety, agitation, seizure CNS symptoms most worrisome, considered “severe” toxicity “Blind as a bat”Pupillary dilation and ineffective accommodation Blurry vision “Dry as a bone”Blockage of sweat gland muscarinic receptors → lack of sweating Dry skin “Hot as a hare”Lack of sweating → anhidrotic hyperthermia Hyperthermia “Red as a beet”Loss of sweat production → Body attempts to compensate by cutaneous vasodilation to dissipate heat Red, flushed skin “Full as a flask”Detrusor muscle of bladder and urethral sphincter under muscarinic control Urinary retention Tachycardia Caused by inhibition of muscarinic receptors on vagus nerve4 Certain agents may also produce sodium channel blockade (eg, TCAs--tricyclic antidepressants, diphenhydramine)3,4 Tachycardia Dysrhythmias (widened QRS, prolonged QT) Open in a new tab Management: ○ Stabilization of airway, breathing, and circulation is paramount, as well as initiating IV access and placing patient on cardiac monitoring and pulse oximetry. 1 ○ Decontamination: ▪ Activated charcoal may be considered as long as patient is able to protect their airway or if they are intubated, though patients shouldn’t be intubated solely for the purpose of administering activated charcoal. 1,2,4 ▪ May be considered even if ingestion occurred more than one hour prior, because anticholinergics decrease gastrointestinal motility. 2,4 ○ Benzodiazepines: ▪ Should be used intravenously to treat agitation and seizures. 1,2,4 ○ Sodium bicarbonate: ▪ Administered to treat wide complex dysrhythmias secondary to sodium channel blockade. 1,2 ▪ Indicated if QRS is greater than 100–120 milliseconds. 4 ○ Physostigmine: ▪ Physostigmine is an acetylcholinesterase inhibitor → increases amount of acetylcholine accumulating in synapse → overcomes effects of anticholinergic agents. 1,4 ▪ May be indicated when patients have significant central anticholinergic toxicity. 1 Beneficial in treatment of agitation or delirium. 4 ▪ Use is controversial: May cause cholinergic toxicity if patient is not poisoned with an anticholinergic substance, or if dosing is incorrect. 1 Case reports describing patients who had a wide QRS interval after TCA-poisoning who developed asystole after being treated with physostigmine. 1 ▪ Should not be given if a condition other than pure anticholinergic poisoning is suspected (eg, tricyclic antidepressant overdose), or if QRS interval is at or above 100 msec. 1,4 ▪ If considering using physostigmine, consult a toxicologist. Disposition: ○ Asymptomatic patients → Should receive activated charcoal, and be observed for at least six hours in the emergency department and may be discharged if they continue to be asymptomatic. 1 ○ Mild anticholinergic toxicity → Should receive activated charcoal, be treated with benzodiazepines if required, and observed for six hours for resolution of symptoms. Discharge can be considered if symptoms resolve; otherwise, they should be admitted for observation. 1 ○ Severe anticholinergic toxicity, and those treated with physostigmine → Should be admitted to an intensive care unit.1 SIMULATION ASSESSMENT Anticholinergic Toxicity in the Emergency Department Learner: ___________ Assessment Timeline This timeline is to help observers assess their learners. It allows observer to make notes on when learners performed various tasks, which can help guide debriefing discussion. Critical Actions: 1. Assess airway, breathing, and circulation (ABCs) Place patient on cardiac monitors and pulse oximeter, and obtain initial set of vital signs Establish two large bore IV lines Administer IV fluids (at least 1 liter) Make the diagnosis of anticholinergic toxicity Contact the poison control center/obtain toxicologist consult Give sodium bicarbonate for QRS >100msec on ECG Give benzodiazepines to control agitation Admit patient to the intensive care unit 0:00 Open in a new tab Critical Actions: □ Assess airway, breathing, and circulation (ABCs) □ Place patient on cardiac monitors and pulse oximeter, and obtain initial set of vital signs □ Establish two large bore IV lines □ Administer IV fluids (at least 1 liter) □ Make the diagnosis of anticholinergic toxicity □ Contact the poison control center/obtain toxicologist consult □ Give sodium bicarbonate for QRS >100msec on ECG □ Give benzodiazepines to control agitation □ Admit patient to the intensive care unit Summative and formative comments: Milestones assessment: \| \| Milestone \| Did not achieve level 1 \| Level 1 \| Level 2 \| Level 3 \| :---: :---: :---: \| \| 1 \| Emergency Stabilization (PC1) \| Did not achieve Level 1 \| Recognizes abnormal vital signs \| Recognizes an unstable patient, requiring intervention Performs primary assessment Discerns data to formulate a diagnostic impression/plan \| Manages and prioritizes critical actions in a critically ill patient Reassesses after implementing a stabilizing intervention \| \| 2 \| Performance of focused history and physical (PC2) \| Did not achieve Level 1 \| Performs a reliable, comprehensive history and physical exam \| Performs and communicates a focused history and physical exam based on chief complaint and urgent issues \| Prioritizes essential components of history and physical exam given dynamic circumstances \| \| 3 \| Diagnostic studies (PC3) \| Did not achieve Level 1 \| Determines the necessity of diagnostic studies \| Orders appropriate diagnostic studies. Performs appropriate bedside diagnostic studies/procedures \| Prioritizes essential testing Interprets results of diagnostic studies Reviews risks, benefits, contraindications, and alternatives to a diagnostic study or procedure \| \| 4 \| Diagnosis (PC4) \| Did not achieve Level 1 \| Considers a list of potential diagnoses \| Considers an appropriate list of potential diagnosis May or may not make correct diagnosis \| Makes the appropriate diagnosis Considers other potential diagnoses, avoiding premature closure \| \| 5 \| Pharmacotherapy (PC5) \| Did not achieve Level 1 \| Asks patient for drug allergies \| Selects an medication for therapeutic intervention, consider potential adverse effects \| Selects the most appropriate medication and understands mechanism of action, effect, and potential side effects Considers and recognizes drug-drug interactions \| \| 6 \| Observation and reassessment (PC6) \| Did not achieve Level 1 Reevaluates patient at least one time during case \| Reevaluates patient after most therapeutic interventions \| Consistently evaluates the effectiveness of therapies at appropriate intervals \| \| \| 7 \| Disposition (PC7) \| Did not achieve Level 1 \| Appropriately selects whether to admit or discharge the patient \| Appropriately selects whether to admit or discharge Involves the expertise of some of the appropriate specialists \| Educates the patient appropriately about their disposition Assigns patient to an appropriate level of care (ICU/Tele/Floor) Involves expertise of all appropriate specialists \| \| 9 \| General Approach to Procedures (PC9) \| Did not achieve Level 1 \| Identifies pertinent anatomy and physiology for a procedure Uses appropriate Universal Precautions \| Obtains informed consent Knows indications, contraindications, anatomic landmarks, equipment, anesthetic and procedural technique, and potential complications for common ED procedures \| Determines a back-up strategy if initial attempts are unsuccessful Correctly interprets results of diagnostic procedure \| \| 20 \| Professional Values (PROF1) \| Did not achieve Level 1 \| Demonstrates caring, honest behavior \| Exhibits compassion, respect, sensitivity and responsiveness \| Develops alternative care plans when patients’ personal beliefs and decisions preclude standard care \| \| 22 \| Patient centered communication (ICS1) \| Did not achieve level 1 \| Establishes rapport and demonstrates empathy to patient (and family) Listens effectively \| Elicits patient’s reason for seeking health care \| Manages patient expectations in a manner that minimizes potential for stress, conflict, and misunderstanding. Effectively communicates with vulnerable populations, (at risk patients and families) \| \| 23 \| Team management (ICS2) \| Did not achieve level 1 \| Recognizes other members of the patient care team during case (nurse, techs) \| Communicates pertinent information to other healthcare colleagues \| Communicates a clear, succinct, and appropriate handoff with specialists and other colleagues Communicates effectively with ancillary staff \| Open in a new tab References/suggestions for further reading 1.Su M, Goldman M. Anticholinergic poisoning. In: Grayzel J, editor. UpToDate. Waltham, MA: UpToDate Inc; [Accessed August 4, 2019]. Updated March 29, 2019. Available at: [Google Scholar] 2.Broderick ED, Metheny H, Crosby B. StatPearls [Internet] Treasure Island (FL): StatPearls Publishing; 2022. Jan, Anticholinergic Toxicity [Updated 2022 Jun 21] Available from: [PubMed] [Google Scholar] 3.Huynh DA, Abbas M, Dabaja A. StatPearls [Internet] Treasure Island (FL): StatPearls Publishing; 2022. Jan, Diphenhydramine Toxicity. [Updated 2022 May 8] Available from: [PubMed] [Google Scholar] 4.Fulton J, Nelson L. Anticholinergics. In: Adams J, editor. Emergency Medicine: Clinical Essentials. 2nd Edition. Philadelphia, PA: Elsevier; 2012. pp. 1240–1245. [Google Scholar] 5.Wax PM. Anticholinergic toxicity. In: Tintinalli JE, Stapczynski J, Ma O, Cline DM, Meckler GD, editors. Tintinalli’s Emergency Medicine: A Comprehensive Study Guide. 8th ed. New York, NY: McGraw-Hill; 2016. pp. 1143–1146. [Google Scholar] Associated Data This section collects any data citations, data availability statements, or supplementary materials included in this article. Supplementary Materials jetem-8-1-S25-supp1.pptx (280.9KB, pptx) jetem-8-1-S25-supp2.jpg (70.7KB, jpg) jetem-8-1-S25-supp3.jpg (132KB, jpg) jetem-8-1-S25-supp4.jpg (85.5KB, jpg) jetem-8-1-S25-supp5.jpg (235.5KB, jpg) jetem-8-1-S25-supp6.jpg (113.4KB, jpg) jetem-8-1-S25-supp7.jpg (95.5KB, jpg) Articles from Journal of Education & Teaching in Emergency Medicine are provided here courtesy of Department of Emergency Medicine, University of California Irvine ACTIONS View on publisher site PDF (1.2 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract USER GUIDE Supplementary Information INSTRUCTOR MATERIALS OPERATOR MATERIALS DEBRIEFING AND EVALUATION PEARLS SIMULATION ASSESSMENT References/suggestions for further reading Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
9217	https://www.tricki.org/article/Bounding_the_sum_by_an_integral	Tricki a repository of mathematical know-how Add article Navigate Tags Search Forums Help Top level › What kind of problem am I trying to solve? › Techniques for obtaining estimates › Estimating sums View Edit Revisions Bounding the sum by an integral \| \| \| Incomplete This article is incomplete. More examples are needed and more complicated situations should be analyzed. \| Quick description Suppose we want to estimate a sum where the sequence is a monotone sequence of non negative real numbers and are real numbers. Later we can let for example and likewise but let us keep finite for the moment. In this case is a standard tactic to look for a real function with the same kind of monotonicity as the sequence , such that for all the we are interested in. Then we have Usually the choice of the function should be obvious by looking at the sequence . If for example the sequence is explicitly given then the first obvious choice would be to replace the discrete parameter with a continuous variable and look at the resulting function . Prerequisites calculus Example 1 Let us look at the partial sums of the harmonic series The sequence is a strictly decreasing sequence of positive numbers. Taking we recover the well known estimate Example 2 One can use the same technique to prove that for a positive real number , the over-harmonic series converge exactly when and we have the estimate whenever . General discussion 10210 reads Subscribe Comments Post new comment (Note: commenting is not possible on this snapshot.) Parent article Estimating sums View articles linking here View child articles Area of mathematics Analysis Number theory Analytic number theory Recent articles View a list of all articles. Littlewood-Paley heuristic for derivative Geometric view of Hölder's inequality Diagonal arguments Finding an interval for rational numbers with a high denominator Try to prove a stronger result Use self-similarity to get a limit from an inferior or superior limit. Prove a consequence first Active forum topics Plenty of LaTeX errors Tutorial A different kind of article? Countable but impredicative Tricki Papers Recent comments I don't think this statement choice of the field Incorrect Image Article classification Higher dimensional analogues more
9218	https://www.sfu.ca/~mbahrami/ENSC%20388/Solution%20manual/IntroTHT_2e_SM_Chap16.pdf	PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-1 Solutions Manual for Introduction to Thermodynamics and Heat Transfer Yunus A. Cengel 2nd Edition, 2008 Chapter 16 HEAT EXCHANGERS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-2 Types of Heat Exchangers 16-1C Heat exchangers are classified according to the flow type as parallel flow, counter flow, and cross-flow arrangement. In parallel flow, both the hot and cold fluids enter the heat exchanger at the same end and move in the same direction. In counter-flow, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite direction. In cross-flow, the hot and cold fluid streams move perpendicular to each other. 16-2C In terms of construction type, heat exchangers are classified as compact, shell and tube and regenerative heat exchangers. Compact heat exchangers are specifically designed to obtain large heat transfer surface areas per unit volume. The large surface area in compact heat exchangers is obtained by attaching closely spaced thin plate or corrugated fins to the walls separating the two fluids. Shell and tube heat exchangers contain a large number of tubes packed in a shell with their axes parallel to that of the shell. Regenerative heat exchangers involve the alternate passage of the hot and cold fluid streams through the same flow area. In compact heat exchangers, the two fluids usually move perpendicular to each other. 16-3C A heat exchanger is classified as being compact if β > 700 m2/m3 or (200 ft2/ft3) where β is the ratio of the heat transfer surface area to its volume which is called the area density. The area density for double-pipe heat exchanger can not be in the order of 700. Therefore, it can not be classified as a compact heat exchanger. 16-4C In counter-flow heat exchangers, the hot and the cold fluids move parallel to each other but both enter the heat exchanger at opposite ends and flow in opposite direction. In cross-flow heat exchangers, the two fluids usually move perpendicular to each other. The cross-flow is said to be unmixed when the plate fins force the fluid to flow through a particular interfin spacing and prevent it from moving in the transverse direction. When the fluid is free to move in the transverse direction, the cross-flow is said to be mixed. 16-5C In the shell and tube exchangers, baffles are commonly placed in the shell to force the shell side fluid to flow across the shell to enhance heat transfer and to maintain uniform spacing between the tubes. Baffles disrupt the flow of fluid, and an increased pumping power will be needed to maintain flow. On the other hand, baffles eliminate dead spots and increase heat transfer rate. 16-6C Using six-tube passes in a shell and tube heat exchanger increases the heat transfer surface area, and the rate of heat transfer increases. But it also increases the manufacturing costs. 16-7C Using so many tubes increases the heat transfer surface area which in turn increases the rate of heat transfer. 16-8C Regenerative heat exchanger involves the alternate passage of the hot and cold fluid streams through the same flow area. The static type regenerative heat exchanger is basically a porous mass which has a large heat storage capacity, such as a ceramic wire mash. Hot and cold fluids flow through this porous mass alternately. Heat is transferred from the hot fluid to the matrix of the regenerator during the flow of the hot fluid and from the matrix to the cold fluid. Thus the matrix serves as a temporary heat storage medium. The dynamic type regenerator involves a rotating drum and continuous flow of the hot and cold fluid through different portions of the drum so that any portion of the drum passes periodically through the hot stream, storing heat and then through the cold stream, rejecting this stored heat. Again the drum serves as the medium to transport the heat from the hot to the cold fluid stream. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-3 The Overall Heat Transfer Coefficient 16-9C Heat is first transferred from the hot fluid to the wall by convection, through the wall by conduction and from the wall to the cold fluid again by convection. 16-10C When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, which is usually the case, the thermal resistance of the tube is negligible. 16-11C The heat transfer surface areas are L D A L D A o i 2 1 and π π = = . When the thickness of inner tube is small, it is reasonable to assume s o i A A A ≅ ≅ . 16-12C No, it is not reasonable to say h h hi ≈ ≈ 0 16-13C When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, the thermal resistance of the tube is negligible and the inner and the outer surfaces of the tube are almost identical ( s o A A Ai ≅ ≅ ). Then the overall heat transfer coefficient of a heat exchanger can be determined to from U = (1/hi + 1/ho)-1 16-14C None. 16-15C When one of the convection coefficients is much smaller than the other o i h h << , and s i A A A ≈ ≈ 0 . Then we have ( o i h h / 1 >> / 1 ) and thus i i h U U U ≅ = = 0 . 16-16C The most common type of fouling is the precipitation of solid deposits in a fluid on the heat transfer surfaces. Another form of fouling is corrosion and other chemical fouling. Heat exchangers may also be fouled by the growth of algae in warm fluids. This type of fouling is called the biological fouling. Fouling represents additional resistance to heat transfer and causes the rate of heat transfer in a heat exchanger to decrease, and the pressure drop to increase. 16-17C The effect of fouling on a heat transfer is represented by a fouling factor Rf. Its effect on the heat transfer coefficient is accounted for by introducing a thermal resistance Rf /As. The fouling increases with increasing temperature and decreasing velocity. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-4 16-18 The heat transfer coefficients and the fouling factors on tube and shell side of a heat exchanger are given. The thermal resistance and the overall heat transfer coefficients based on the inner and outer areas are to be determined. Assumptions 1 The heat transfer coefficients and the fouling factors are constant and uniform. Analysis (a) The total thermal resistance of the heat exchanger per unit length is C/W 0.0837° = ° + ° + ° + ° + ° = + + + + = m)] m)(1 016 . 0 ( [ C) . W/m 700 ( 1 m)] m)(1 016 . 0 ( [ C/W) . m 0002 . 0 ( m) C)(1 W/m. 380 ( 2 ) 2 . 1 / 6 . 1 ln( m)] m)(1 012 . 0 ( [ C/W) . m 0005 . 0 ( m)] m)(1 012 . 0 ( [ C) . W/m 700 ( 1 1 2 ) / ln( 1 2 2 2 2 π π π π π π R A h A R kL D D A R A h R o o o fo i o i fi i i (b) The overall heat transfer coefficient based on the inner and the outer surface areas of the tube per length are C . W/m 238 C . W/m 317 2 2 ° = ° = = ° = ° = = = = = m)] m)(1 016 . 0 ( [ C/W) 0837 . 0 ( 1 1 m)] m)(1 012 . 0 ( [ C/W) 0837 . 0 ( 1 1 1 1 1 π π o o i i o o i i RA U RA U A U A U UA R Outer surface D0, A0, h0, U0 , Rf0 Inner surface Di, Ai, hi, Ui , Rfi PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-5 16-19 EES Prob. 16-18 is reconsidered. The effects of pipe conductivity and heat transfer coefficients on the thermal resistance of the heat exchanger are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" k=380 [W/m-C] D_i=0.012 [m] D_o=0.016 [m] D_2=0.03 [m] h_i=700 [W/m^2-C] h_o=1400 [W/m^2-C] R_f_i=0.0005 [m^2-C/W] R_f_o=0.0002 [m^2-C/W] "ANALYSIS" R=1/(h_iA_i)+R_f_i/A_i+ln(D_o/D_i)/(2pikL)+R_f_o/A_o+1/(h_oA_o) L=1 [m] “a unit length of the heat exchanger is considered" A_i=piD_iL A_o=piD_oL k [W/m-C] R [C/W] 10 0.07392 30.53 0.07085 51.05 0.07024 71.58 0.06999 92.11 0.06984 112.6 0.06975 133.2 0.06969 153.7 0.06964 174.2 0.06961 194.7 0.06958 215.3 0.06956 235.8 0.06954 256.3 0.06952 276.8 0.06951 297.4 0.0695 317.9 0.06949 338.4 0.06948 358.9 0.06947 379.5 0.06947 400 0.06946 0 50 100 150 200 250 300 350 400 0.069 0.07 0.071 0.072 0.073 0.074 k [W/m-C] R [C/W] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-6 hi [W/m2-C] R [C/W] 500 0.08462 550 0.0798 600 0.07578 650 0.07238 700 0.06947 750 0.06694 800 0.06473 850 0.06278 900 0.06105 950 0.05949 1000 0.0581 1050 0.05684 1100 0.05569 1150 0.05464 1200 0.05368 1250 0.05279 1300 0.05198 1350 0.05122 1400 0.05052 1450 0.04987 1500 0.04926 ho [W/m2-C] R [C/W] 1000 0.07515 1050 0.0742 1100 0.07334 1150 0.07256 1200 0.07183 1250 0.07117 1300 0.07056 1350 0.06999 1400 0.06947 1450 0.06898 1500 0.06852 1550 0.06809 1600 0.06769 1650 0.06731 1700 0.06696 1750 0.06662 1800 0.06631 1850 0.06601 1900 0.06573 1950 0.06546 2000 0.0652 500 700 900 1100 1300 1500 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085 hi [W/m 2-C] R [C/W] 1000 1200 1400 1600 1800 2000 0.064 0.066 0.068 0.07 0.072 0.074 0.076 ho [W/m 2-C] R [C/W] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-7 16-20 A water stream is heated by a jacketted-agitated vessel, fitted with a turbine agitator. The mass flow rate of water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The properties of water at 54°C are (Table A-15) 31 . 3 Pr s kg/m 10 513 . 0 kg/m 8 . 985 C W/m. 648 . 0 3 -3 = ⋅ × = = ° = μ ρ k The specific heat of water at the average temperature of (10+54)/2=32°C is 4178 J/kg.°C (Table A-15) Analysis We first determine the heat transfer coefficient on the inner wall of the vessel 865 , 76 s kg/m 10 513 . 0 ) kg/m 8 . 985 ( m) )(0.2 s (60/60 Re 3 3 2 -1 2 = ⋅ × = = − μ ρ a D n & 2048 ) 31 . 3 ( ) 865 , 76 ( 76 . 0 Pr Re 76 . 0 3 / 1 3 / 2 3 / 1 3 / 2 = = = Nu C . W/m 2211 ) 2048 ( m 6 . 0 C W/m. 648 . 0 2 ° = ° = = Nu D k h t j The heat transfer coefficient on the outer side is determined as follows 25 . 0 25 . 0 ) 100 ( 100 , 13 ) ( 100 , 13 − − − = − = w w g o T T T h C 2 . 89 ) 54 ( 2211 ) 100 ( 100 , 13 ) 54 ( 2211 ) 100 ( ) 100 ( 100 , 13 ) 54 ( ) ( 75 . 0 25 . 0 ° = → − = − − = − − − = − − w w w w w w w j w g o T T T T T T T h T T h C W/m 7226 ) 2 . 89 100 ( 100 , 13 ) 100 ( 100 , 13 2 25 . 0 25 . 0 ⋅ = − = − = − − w o T h Neglecting the wall resistance and the thickness of the wall, the overall heat transfer coefficient can be written as C W/m 1694 7226 1 2211 1 1 1 2 1 1 ⋅ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = − − o j h h U From an energy balance kg/h 1725 = = − × × = − Δ = − kg/s 479 . 0 ) 54 100 )( 6 . 0 6 . 0 )( 1694 ( ) 10 54 )( 4178 ( )] ( [ water w w in out m m T UA T T c m & & & π Water 10ºC 54ºC 54ºC Steam 100ºC PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-8 16-21 Water flows through the tubes in a boiler. The overall heat transfer coefficient of this boiler based on the inner surface area is to be determined. Assumptions 1 Water flow is fully developed. 2 Properties of the water are constant. Properties The properties of water at 110°C are (Table A-15) 58 . 1 Pr K . W/m 682 . 0 /s m 10 268 . 0 / 2 2 6 = = × = = − k ρ μ ν Analysis The Reynolds number is 600 , 130 s / m 10 268 . 0 m) m/s)(0.01 5 . 3 ( Re 2 6 avg = × = = − ν h D V which is greater than 10,000. Therefore, the flow is turbulent. Assuming fully developed flow, 342 ) 58 . 1 ( ) 600 , 130 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = = = k hD Nu h and C . W/m 23,324 = (342) m 01 . 0 C W/m. 682 . 0 2 ° ° = = Nu D k h h The total resistance of this heat exchanger is then determined from C/W 00157 . 0 = ] m) m)(5 (0.014 C)[ . W/m 8400 ( 1 m)] C)(5 W/m. 2 . 14 ( 2 [ ) 1 / 4 . 1 ln( ] m) m)(5 (0.01 C)[ . W/m 324 , 23 ( 1 1 2 ) / ln( 1 2 2 ° ° + ° + ° = + + = + + = = π π π π o o i o i i o wall i total A h kL D D A h R R R R R and C . W/m 4055 2 ° = ° = = ⎯→ ⎯ = ] m) m)(5 (0.01 C/W)[ 00157 . 0 ( 1 1 1 π i i i i RA U A U R Outer surface D0, A0, h0, U0 , Rf0 Inner surface Di, Ai, hi, Ui , Rfi PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-9 16-22 Water is flowing through the tubes in a boiler. The overall heat transfer coefficient of this boiler based on the inner surface area is to be determined. Assumptions 1 Water flow is fully developed. 2 Properties of water are constant. 3 The heat transfer coefficient and the fouling factor are constant and uniform. Properties The properties of water at 110°C are (Table A-15) 58 . 1 Pr K . W/m 682 . 0 /s m 10 268 . 0 / 2 2 6 = = × = = − k ρ μ ν Analysis The Reynolds number is 600 , 130 s / m 10 268 . 0 m) m/s)(0.01 5 . 3 ( Re 2 6 avg = × = = − ν h D V which is greater than 10,000. Therefore, the flow is turbulent. Assuming fully developed flow, 342 ) 58 . 1 ( ) 600 , 130 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = = = k hD Nu h and C . W/m 23,324 = (342) m 01 . 0 C W/m. 682 . 0 2 ° ° = = Nu D k h h The thermal resistance of heat exchanger with a fouling factor of C/W . m 0005 . 0 2 , ° = i f R is determined from C/W 00475 . 0 m)] m)(5 014 . 0 ( [ C) . W/m 8400 ( 1 m) C)(5 W/m. 2 . 14 ( 2 ) 1 / 4 . 1 ln( m)] m)(5 01 . 0 ( [ C/W . m 0005 . 0 m)] m)(5 01 . 0 ( [ C) . W/m 324 , 23 ( 1 1 2 ) / ln( 1 2 2 2 , ° = ° + ° + ° + ° = + + + = π π π π π R A h kL D D A R A h R o o i o i i f i i Then, C . W/m 1340 2 ° = ° = = ⎯→ ⎯ = ] m) m)(5 (0.01 C/W)[ 00475 . 0 ( 1 1 1 π i i i i RA U A U R Outer surface D0, A0, h0, U0 , Rf0 Inner surface Di, Ai, hi, Ui , Rfi PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-10 16-23 EES Prob. 16-22 is reconsidered. The overall heat transfer coefficient based on the inner surface as a function of fouling factor is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_w=110 [C] Vel=3.5 [m/s] L=5 [m] k_pipe=14.2 [W/m-C] D_i=0.010 [m] D_o=0.014 [m] h_o=8400 [W/m^2-C] R_f_i=0.0005 [m^2-C/W] "PROPERTIES" k=conductivity(Water, T=T_w, P=300) Pr=Prandtl(Water, T=T_w, P=300) rho=density(Water, T=T_w, P=300) mu=viscosity(Water, T=T_w, P=300) nu=mu/rho "ANALYSIS" Re=(VelD_i)/nu "Re is calculated to be greater than 10,000. Therefore, the flow is turbulent." Nusselt=0.023Re^0.8Pr^0.4 h_i=k/D_iNusselt A_i=piD_iL A_o=piD_oL R=1/(h_iA_i)+R_f_i/A_i+ln(D_o/D_i)/(2pik_pipeL)+1/(h_oA_o) U_i=1/(RA_i) Rf,i [m2-C/W] Ui [W/m2-C] 0.0001 2883 0.00015 2520 0.0002 2238 0.00025 2013 0.0003 1829 0.00035 1675 0.0004 1546 0.00045 1435 0.0005 1339 0.00055 1255 0.0006 1181 0.00065 1115 0.0007 1056 0.00075 1003 0.0008 955.2 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 750 1200 1650 2100 2550 3000 Rf,i [m 2-C/W] Ui [W/m 2-C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-11 16-24 Refrigerant-134a is cooled by water in a double-pipe heat exchanger. The overall heat transfer coefficient is to be determined. Assumptions 1 The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its thickness is negligible. 2 Both the water and refrigerant-134a flow are fully developed. 3 Properties of the water and refrigerant-134a are constant. Properties The properties of water at 20°C are (Table A-15) 01 . 7 Pr C . W/m 598 . 0 /s m 10 004 . 1 / kg/m 998 2 6 3 = ° = × = = = − k ρ μ ν ρ Analysis The hydraulic diameter for annular space is m 015 . 0 01 . 0 025 . 0 = − = − = i o h D D D The average velocity of water in the tube and the Reynolds number are m/s 729 . 0 4 m) 01 . 0 ( m) 025 . 0 ( ) kg/m 998 ( kg/s 3 . 0 4 2 2 3 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = π π ρ ρ i o c avg D D m A m V & & 890 , 10 s / m 10 004 . 1 m) m/s)(0.015 729 . 0 ( Re 2 6 = × = = − ν h avg D V which is greater than 4000. Therefore flow is turbulent. Assuming fully developed flow, 0 . 85 ) 01 . 7 ( ) 890 , 10 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = = = k hD Nu h and C . W/m 3390 = (85.0) m 015 . 0 C W/m. 598 . 0 2 ° ° = = Nu D k h h o Then the overall heat transfer coefficient becomes C . W/m 2020 2 ° = ° + ° = + = C . W/m 3390 1 C . W/m 5000 1 1 1 1 1 2 2 o i h h U Cold water Hot R-134a D0 Di PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-12 16-25 Refrigerant-134a is cooled by water in a double-pipe heat exchanger. The overall heat transfer coefficient is to be determined. Assumptions 1 The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its thickness is negligible. 2 Both the water and refrigerant-134a flows are fully developed. 3 Properties of the water and refrigerant-134a are constant. 4 The limestone layer can be treated as a plain layer since its thickness is very small relative to its diameter. Properties The properties of water at 20°C are (Table A-15) 01 . 7 Pr C . W/m 598 . 0 /s m 10 004 . 1 / kg/m 998 2 6 3 = ° = × = = = − k ρ μ ν ρ Analysis The hydraulic diameter for annular space is m 015 . 0 01 . 0 025 . 0 = − = − = i o h D D D The average velocity of water in the tube and the Reynolds number are m/s 729 . 0 4 m) 01 . 0 ( m) 025 . 0 ( ) kg/m 998 ( kg/s 3 . 0 4 2 2 3 2 2 avg = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = π π ρ ρ i o c D D m A m V & & 890 , 10 s / m 10 004 . 1 m) m/s)(0.015 729 . 0 ( Re 2 6 avg = × = = − ν h D V which is greater than 10,000. Therefore flow is turbulent. Assuming fully developed flow, 0 . 85 ) 01 . 7 ( ) 890 , 10 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = = = k hD Nu h and C . W/m 3390 = (85.0) m 015 . 0 C W/m. 598 . 0 2 ° ° = = Nu D k h h o Disregarding the curvature effects, the overall heat transfer coefficient is determined to be C . W/m 493 2 ° = ° + ° + ° = + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = C . W/m 3390 1 C . W/m 3 . 1 m 002 . 0 C . W/m 5000 1 1 1 1 1 2 2 limeston o i h k L h U Cold water Hot R-134a Limestone D0 Di PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-13 16-26 EES Prob. 16-25 is reconsidered. The overall heat transfer coefficient as a function of the limestone thickness is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" D_i=0.010 [m] D_o=0.025 [m] T_w=20 [C] h_i=5000 [W/m^2-C] m_dot=0.3 [kg/s] L_limestone=2 [mm] k_limestone=1.3 [W/m-C] "PROPERTIES" k=conductivity(Water, T=T_w, P=100) Pr=Prandtl(Water, T=T_w, P=100) rho=density(Water, T=T_w, P=100) mu=viscosity(Water, T=T_w, P=100) nu=mu/rho "ANALYSIS" D_h=D_o-D_i Vel=m_dot/(rhoA_c) A_c=pi(D_o^2-D_i^2)/4 Re=(VelD_h)/nu "Re is calculated to be greater than 10,000. Therefore, the flow is turbulent." Nusselt=0.023Re^0.8Pr^0.4 h_o=k/D_hNusselt U=1/(1/h_i+(L_limestoneConvert(mm, m))/k_limestone+1/h_o) Llimestone [mm] U [W/m2-C] 1 791.4 1.1 746 1.2 705.5 1.3 669.2 1.4 636.4 1.5 606.7 1.6 579.7 1.7 554.9 1.8 532.2 1.9 511.3 2 491.9 2.1 474 2.2 457.3 2.3 441.8 2.4 427.3 2.5 413.7 2.6 400.9 2.7 388.9 2.8 377.6 2.9 367 3 356.9 1 1.4 1.8 2.2 2.6 3 350 400 450 500 550 600 650 700 750 800 Llimestone [mm] U [W/m 2-C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-14 16-27E Water is cooled by air in a cross-flow heat exchanger. The overall heat transfer coefficient is to be determined. Assumptions 1 The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its thickness is negligible. 2 Both the water and air flow are fully developed. 3 Properties of the water and air are constant. Properties The properties of water at 180°F are (Table A-15E) 15 . 2 Pr s / ft 10 825 . 3 F Btu/h.ft. 388 . 0 2 6 = × = ° = − ν k The properties of air at 80°F are (Table A-22E) 7290 . 0 Pr s / ft 10 697 . 1 F Btu/h.ft. 01481 . 0 2 4 = × = ° = − ν k Analysis The overall heat transfer coefficient can be determined from o i h h U 1 1 1 + = The Reynolds number of water is 360 , 65 s / ft 10 825 . 3 ft] /12 ft/s)[0.75 4 ( Re 2 6 = × = = − ν h avg D V which is greater than 10,000. Therefore the flow of water is turbulent. Assuming the flow to be fully developed, the Nusselt number is determined from 222 ) 15 . 2 ( ) 360 , 65 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = = = k hD Nu h and F . Btu/h.ft 1378 = (222) ft 12 / 75 . 0 F Btu/h.ft. 388 . 0 2 ° ° = = Nu D k h h i The Reynolds number of air is 4420 s / ft 10 697 . 1 ft] 12) ft/s)[3/(4 12 ( Re 2 4 = × × = = − ν VD The flow of air is across the cylinder. The proper relation for Nusselt number in this case is ( ) [ ] ( ) [ ] 86 . 34 000 , 282 4420 1 7290 . 0 / 4 . 0 1 ) 7290 . 0 ( ) 4420 ( 62 . 0 3 . 0 000 , 282 Re 1 Pr / 4 . 0 1 Pr Re 62 . 0 3 . 0 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . 0 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . 0 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = = k hD Nu and F . Btu/h.ft 8.26 = (34.86) ft 12 / 75 . 0 F Btu/h.ft. 01481 . 0 2 ° ° = = Nu D k ho Then the overall heat transfer coefficient becomes F . Btu/h.ft 8.21 2 ° = ° + ° = + = F . Btu/h.ft 26 . 8 1 F . Btu/h.ft 1378 1 1 1 1 1 2 2 o i h h U Water 180°F 4 ft/s Air 80°F 12 ft/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-15 Analysis of Heat Exchangers 16-28C The heat exchangers usually operate for long periods of time with no change in their operating conditions, and then they can be modeled as steady-flow devices. As such , the mass flow rate of each fluid remains constant and the fluid properties such as temperature and velocity at any inlet and outlet remain constant. The kinetic and potential energy changes are negligible. The specific heat of a fluid can be treated as constant in a specified temperature range. Axial heat conduction along the tube is negligible. Finally, the outer surface of the heat exchanger is assumed to be perfectly insulated so that there is no heat loss to the surrounding medium and any heat transfer thus occurs is between the two fluids only. 16-29C That relation is valid under steady operating conditions, constant specific heats, and negligible heat loss from the heat exchanger. 16-30C The product of the mass flow rate and the specific heat of a fluid is called the heat capacity rate and is expressed as p c m C & = . When the heat capacity rates of the cold and hot fluids are equal, the temperature change is the same for the two fluids in a heat exchanger. That is, the temperature rise of the cold fluid is equal to the temperature drop of the hot fluid. A heat capacity of infinity for a fluid in a heat exchanger is experienced during a phase-change process in a condenser or boiler. 16-31C The mass flow rate of the cooling water can be determined from water cooling ) ( = T c m Q pΔ & & . The rate of condensation of the steam is determined from steam ) ( = fg h m Q & & , and the total thermal resistance of the condenser is determined from T Q R Δ = / & . 16-32C When the heat capacity rates of the cold and hot fluids are identical, the temperature rise of the cold fluid will be equal to the temperature drop of the hot fluid. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-16 The Log Mean Temperature Difference Method 16-33C ΔTlm is called the log mean temperature difference, and is expressed as ) / ln( 2 1 2 1 T T T T Tlm Δ Δ Δ − Δ = Δ where out c out h in c in h T T T T T T , , 2 , , 1 -= -= Δ Δ for parallel-flow heat exchangers and in c out h out c in h T T T T T T , , 2 , , -= -= Δ Δ for counter-flow heat exchangers 16-34C The temperature difference between the two fluids decreases from ΔT1 at the inlet to ΔT2 at the outlet, and arithmetic mean temperature difference is defined as 2 + = 2 1 am T T T Δ Δ Δ . The logarithmic mean temperature difference ΔTlm is obtained by tracing the actual temperature profile of the fluids along the heat exchanger, and is an exact representation of the average temperature difference between the hot and cold fluids. It truly reflects the exponential decay of the local temperature difference. The logarithmic mean temperature difference is always less than the arithmetic mean temperature. 16-35C ΔTlm cannot be greater than both ΔT1 and ΔT2 because ΔTln is always less than or equal to ΔTm (arithmetic mean) which can not be greater than both ΔT1 and ΔT2. 16-36C No, it cannot. When ΔT1 is less than ΔT2 the ratio of them must be less than one and the natural logarithms of the numbers which are less than 1 are negative. But the numerator is also negative in this case. When ΔT1 is greater than ΔT2, we obtain positive numbers at the both numerator and denominator. 16-37C In the parallel-flow heat exchangers the hot and cold fluids enter the heat exchanger at the same end, and the temperature of the hot fluid decreases and the temperature of the cold fluid increases along the heat exchanger. But the temperature of the cold fluid can never exceed that of the hot fluid. In case of the counter-flow heat exchangers the hot and cold fluids enter the heat exchanger from the opposite ends and the outlet temperature of the cold fluid may exceed the outlet temperature of the hot fluid. 16-38C The ΔTlm will be greatest for double-pipe counter-flow heat exchangers. 16-39C The factor F is called as correction factor which depends on the geometry of the heat exchanger and the inlet and the outlet temperatures of the hot and cold fluid streams. It represents how closely a heat exchanger approximates a counter-flow heat exchanger in terms of its logarithmic mean temperature difference. F cannot be greater than unity. 16-40C In this case it is not practical to use the LMTD method because it requires tedious iterations. Instead, the effectiveness-NTU method should be used. 16-41C First heat transfer rate is determined from ] -[ = out in p T T c m Q & & , ΔTln from ) / ln( 2 1 2 1 T T T T Tlm Δ Δ Δ − Δ = Δ , correction factor from the figures, and finally the surface area of the heat exchanger from CF lm UAFDT Q , = & PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-17 16-42 Ethylene glycol is heated in a tube while steam condenses on the outside tube surface. The tube length is to be determined. Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the tubes are smooth. 3 Heat transfer to the surroundings is negligible. Properties The properties of ethylene glycol are given to be ρ = 1109 kg/m3, cp = 2428 J/kg⋅K, k = 0.253 W/m⋅K, µ = 0.01545 kg/m⋅s, Pr = 148.5. The thermal conductivity of copper is given to be 386 W/m⋅K. Analysis The rate of heat transfer is W 560 , 48 C ) 20 40 )( C J/kg. 2428 )( kg/s 1 ( ) ( = ° − ° = − = i e p T T c m Q & & The fluid velocity is [ ] m/s 870 . 2 4 / m) (0.02 ) kg/m 1109 ( kg/s 1 2 3 = = = π ρ c A m V & The Reynolds number is 4121 s kg/m 01545 . 0 m) m/s)(0.02 )(2.870 kg/m (1109 Re 3 = ⋅ = = μ ρVD which is greater than 2300 and smaller than 10,000. Therefore, we have transitional flow. We assume fully developed flow and evaluate the Nusselt number from turbulent flow relation: 5 . 132 ) 5 . 148 ( ) 4121 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = = = k hD Nu Heat transfer coefficient on the inner surface is C . W/m 1677 ) 5 . 132 ( m 02 . 0 C W/m. 253 . 0 2 ° = ° = = Nu D k hi Assuming a wall temperature of 100°C, the heat transfer coefficient on the outer surface is determined to be C . W/m 5174 ) 100 110 ( 9200 ) ( 9200 2 25 . 0 25 . 0 ° = − = − = − − w g o T T h Let us check if the assumption for the wall temperature holds: C 5 . 93 ) 110 ( 025 . 0 5174 ) 30 ( 02 . 0 1677 ) ( ) ( ) ( ) ( avg , avg , ° = ⎯→ ⎯ − × = − × − = − − = − w w w w g o o b w i i w g o o b w i i T T T T T L D h T T L D h T T A h T T A h π π Now we assume a wall temperature of 90°C: C . W/m 4350 ) 90 110 ( 9200 ) ( 9200 2 25 . 0 25 . 0 ° = − = − = − − w g o T T h Again checking, C 1 . 91 ) 110 ( 025 . 0 4350 ) 30 ( 02 . 0 1677 ° = ⎯→ ⎯ − × = − × w w w T T T which is sufficiently close to the assumed value of 90°C. Now that both heat transfer coefficients are available, we use thermal resistance concept to find overall heat transfer coefficient based on the outer surface area as follows: C W/m 1018 4350 1 ) 386 ( 2 ) 2 / 5 . 2 ln( ) 025 . 0 ( ) 02 . 0 )( 1677 ( 025 . 0 1 1 2 ) / ln( 1 2 copper 1 2 ⋅ = + + = + + = o o i i o o h k D D D D h D U The rate of heat transfer can be expressed as ln T A U Q o o Δ = & where the logarithmic mean temperature difference is C 58 . 79 20 110 40 110 ln ) 20 110 ( ) 40 110 ( ln ) ( ) ( ° = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − − − = Δ i g e g i g e g lm T T T T T T T T T Substituting, the tube length is determined to be m 7.63 = ⎯→ ⎯ = ⎯→ ⎯ Δ = L L T A U Q o o ) 58 . 79 ( ) 025 . 0 ( ) 1018 ( 560 , 48 ln π & Ethylene glycol 1 kg/s 20ºC L Tg = 110ºC PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-18 16-43 Water is heated in a double-pipe, parallel-flow uninsulated heat exchanger by geothermal water. The rate of heat transfer to the cold water and the log mean temperature difference for this heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heat of hot water is given to be 4.25 kJ/kg.°C. Analysis The rate of heat given up by the hot water is kW 208.3 = C) 50 C C)(85 kJ/kg. kg/s)(4.25 4 . 1 ( )] ( [ hot water ° − ° ° = − = out in p h T T c m Q & & The rate of heat picked up by the cold water is kW 202.0 = − = − = kW) 3 . 208 )( 03 . 0 1 ( ) 03 . 0 1 ( h c Q Q & & The log mean temperature difference is C 43.9° = ° ⋅ = == Δ ⎯→ ⎯ Δ = ) m 4 )( C kW/m 15 . 1 ( kW 0 . 202 2 2 UA Q T T UA Q lm lm & & 16-44 A stream of hydrocarbon is cooled by water in a double-pipe counterflow heat exchanger. The overall heat transfer coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of hydrocarbon and water are given to be 2.2 and 4.18 kJ/kg.°C, respectively. Analysis The rate of heat transfer is kW 48.4 = C) 40 C C)(150 kJ/kg. kg/s)(2.2 3600 / 720 ( )] ( [ HC ° − ° ° = − = in out p T T c m Q & & The outlet temperature of water is C 87.2 = C) 10 C)( kJ/kg. kg/s)(4.18 3600 / (540 kW 4 . 48 )] ( [ out w, out w, w ° ° − ° = − = T T T T c m Q in out p & & The logarithmic mean temperature difference is C 30 = C 10 C 40 C 62.8 = C 2 . 87 C 150 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and C 4 . 44 ) 30 / 8 . 62 ln( 30 8 . 62 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The overall heat transfer coefficient is determined from K kW/m 2.31 2 ⋅ = ° × × = Δ = U U T UA Q lm C) )(44.4 0 . 6 0.025 ( kW 4 . 48 π & Cold water Hot water 85°C 50°C Water 10°C HC 150°C 40°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-19 16-45 Oil is heated by water in a 1-shell pass and 6-tube passes heat exchanger. The rate of heat transfer and the heat transfer surface area are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heat of oil is given to be 2.0 kJ/kg.°C. Analysis The rate of heat transfer in this heat exchanger is kW 420 = C) 25 C C)(46 kJ/kg. kg/s)(2.0 10 ( )] ( [ oil ° − ° ° = − = in out p T T c m Q & & The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 35 = C 25 C 60 C 34 = C 46 C 80 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 5 . 34 ) 35 / 34 ln( 35 34 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 94 . 0 95 . 0 25 46 60 80 38 . 0 25 80 25 46 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P Then the heat transfer surface area on the tube side becomes 2 m 13.0 = ° ° = Δ = ⎯→ ⎯ Δ = C) 5 . 34 ( C)(0.94) . kW/m 0 . 1 ( kW 420 2 , , CF lm s CF lm s T UF Q A T F UA Q & & Water 80°C Oil 25°C 10 kg/s 46°C 1 shell pass 6 tube passes 60°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-20 16-46 Steam is condensed by cooling water in the condenser of a power plant. The mass flow rate of the cooling water and the rate of condensation are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The heat of vaporization of water at 50°C is given to be hfg = 2383 kJ/kg and specific heat of cold water at the average temperature of 22.5°C is given to be cp = 4180 J/kg.°C. Analysis The temperature differences between the steam and the cooling water at the two ends of the condenser are C 32 = C 18 C 50 C 23 = C 27 C 50 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and C 3 . 27 ) 32 / 23 ln( 32 23 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm Then the heat transfer rate in the condenser becomes kW 2752 = C) 3 . 27 )( m C)(42 . W/m 2400 ( 2 2 ° ° = Δ = lm s T UA Q & The mass flow rate of the cooling water and the rate of condensation of steam are determined from kg/s 73.1 = C) 18 C C)(27 kJ/kg. (4.18 kJ/s 2752 ) ( )] ( [ water cooling water cooling ° − ° ° = − = − = in out p in out p T T c Q m T T c m Q & & & & kg/s 1.15 = = = ⎯→ ⎯ = kJ/kg 2383 kJ/s 2752 ) ( fg steam steam fg h Q m h m Q & & & & Steam 50°C 18°C Water 50°C 27°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-21 16-47 Water is heated in a double-pipe parallel-flow heat exchanger by geothermal water. The required length of tube is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of water and geothermal fluid are given to be 4.18 and 4.31 kJ/kg.°C, respectively. Analysis The rate of heat transfer in the heat exchanger is kW 29.26 = C) 25 C C)(60 kJ/kg. kg/s)(4.18 2 . 0 ( )] ( [ water ° − ° ° = − = in out p T T c m Q & & Then the outlet temperature of the geothermal water is determined from C 4 . 117 C) kJ/kg. kg/s)(4.31 3 . 0 ( kW 26 . 29 C 140 )] ( [ geot.water ° = ° − ° = − = ⎯→ ⎯ − = p in out out in p c m Q T T T T c m Q & & & & The logarithmic mean temperature difference is C 57.4 = C 60 C 4 . 117 C 115 = C 25 C 140 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ out c out h in c in h T T T T T T and C 9 . 82 ) 4 . 57 / 115 ln( 4 . 57 115 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The surface area of the heat exchanger is determined from 2 2 m 642 . 0 C) 9 . 82 )( kW/m 55 . 0 ( kW 26 . 29 = ° = Δ = ⎯→ ⎯ Δ = lm s lm s T U Q A T UA Q & & Then the length of the tube required becomes m 25.5 = = = ⎯→ ⎯ = m) 008 . 0 ( m 642 . 0 2 π π π D A L DL A s s Water 25°C Brine 140°C 60°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-22 16-48 EES Prob. 16-47 is reconsidered. The effects of temperature and mass flow rate of geothermal water on the length of the tube are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_w_in=25 [C] T_w_out=60 [C] m_dot_w=0.2 [kg/s] c_p_w=4.18 [kJ/kg-C] T_geo_in=140 [C] m_dot_geo=0.3 [kg/s] c_p_geo=4.31 [kJ/kg-C] D=0.008 [m] U=0.55 [kW/m^2-C] "ANALYSIS" Q_dot=m_dot_wc_p_w(T_w_out-T_w_in) Q_dot=m_dot_geoc_p_geo(T_geo_in-T_geo_out) DELTAT_1=T_geo_in-T_w_in DELTAT_2=T_geo_out-T_w_out DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=UADELTAT_lm A=piDL Tgeo,in [C] L [m] 100 53.73 105 46.81 110 41.62 115 37.56 120 34.27 125 31.54 130 29.24 135 27.26 140 25.54 145 24.04 150 22.7 155 21.51 160 20.45 165 19.48 170 18.61 175 17.81 180 17.08 185 16.4 190 15.78 195 15.21 200 14.67 100 120 140 160 180 200 10 15 20 25 30 35 40 45 50 55 Tgeo,in [C] L [m] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-23 mgeo [kg/s] L [m] 0.1 46.31 0.125 35.52 0.15 31.57 0.175 29.44 0.2 28.1 0.225 27.16 0.25 26.48 0.275 25.96 0.3 25.54 0.325 25.21 0.35 24.93 0.375 24.69 0.4 24.49 0.425 24.32 0.45 24.17 0.475 24.04 0.5 23.92 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 20 25 30 35 40 45 50 mgeo [kg/s] L [m] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-24 16-49E Glycerin is heated by hot water in a 1-shell pass and 8-tube passes heat exchanger. The rate of heat transfer for the cases of fouling and no fouling are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Heat transfer coefficients and fouling factors are constant and uniform. 5 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Properties The specific heats of glycerin and water are given to be 0.60 and 1.0 Btu/lbm.°F, respectively. Analysis (a) The tubes are thin walled and thus we assume the inner surface area of the tube to be equal to the outer surface area. Then the heat transfer surface area of this heat exchanger becomes 2 ft 6 . 523 ft) ft)(500 12 / 5 . 0 ( 8 = = = π πDL n As The temperature differences at the two ends of the heat exchanger are F 55 = F 65 F 120 F 35 = F 140 F 175 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and F 25 . 44 ) 55 / 35 ln( 55 35 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm The correction factor is 70 . 0 36 . 1 175 120 140 65 5 . 0 175 65 175 120 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P In case of no fouling, the overall heat transfer coefficient is determined from F . Btu/h.ft 7 . 3 F . Btu/h.ft 4 1 F . Btu/h.ft 50 1 1 1 1 1 2 2 2 ° = ° + ° = + = o i h h U Then the rate of heat transfer becomes Btu/h 60,000 = ° ° = Δ = F) 25 . 44 )( 70 . 0 )( ft F)(523.6 . Btu/h.ft 7 . 3 ( 2 2 ,CF lm s T F UA Q & (b) The thermal resistance of the heat exchanger with a fouling factor is F/Btu h. 0005195 . 0 ) ft 6 . 523 ( F) . Btu/h.ft 4 ( 1 ft 6 . 523 F/Btu . h.ft 002 . 0 ) ft 6 . 523 ( F) . Btu/h.ft 50 ( 1 1 1 2 2 2 2 2 2 ° = ° + ° + ° = + + = o o i fi i i A h A R A h R The overall heat transfer coefficient in this case is F . Btu/h.ft 68 . 3 ) ft .6 F/Btu)(523 h. 0005195 . 0 ( 1 1 1 2 2 ° = ° = = ⎯→ ⎯ = s s RA U UA R Then rate of heat transfer becomes Btu/h 59,680 = ° ° = Δ = F) 25 . 44 )( 70 . 0 )( ft F)(523.6 . Btu/h.ft 68 . 3 ( 2 2 ,CF lm s T F UA Q & Glycerin 65°F 175°F Hot Water 120°F 140°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-25 16-50 During an experiment, the inlet and exit temperatures of water and oil and the mass flow rate of water are measured. The overall heat transfer coefficient based on the inner surface area is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and oil are given to be 4180 and 2150 J/kg.°C, respectively. Analysis The rate of heat transfer from the oil to the water is kW 438.9 = C) 20 C C)(55 kJ/kg. kg/s)(4.18 3 ( )] ( [ water ° − ° ° = − = in out p T T c m Q & & The heat transfer area on the tube side is 2 m 1.8 = m) m)(2 012 . 0 ( 24π π = = L D n A i i The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 25 = C 20 C 45 C 65 = C 55 C 120 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 9 . 41 ) 25 / 65 ln( 25 65 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 70 . 0 14 . 2 20 55 45 120 35 . 0 20 120 20 55 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P Then the overall heat transfer coefficient becomes C . kW/m 8.31 2 ° = ° = Δ = ⎯→ ⎯ Δ = C) 9 . 41 )( 70 . 0 )( m 8 . 1 ( kW 9 . 438 2 , , CF lm i i CF lm i i T F A Q U T F A U Q & & Oil 120°C 20°C Water 3 kg/s 55°C 145°C 24 tubes PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-26 16-51 Ethylene glycol is cooled by water in a double-pipe counter-flow heat exchanger. The rate of heat transfer, the mass flow rate of water, and the heat transfer surface area on the inner side of the tubes are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of water and ethylene glycol are given to be 4.18 and 2.56 kJ/kg.°C, respectively. Analysis (a) The rate of heat transfer is kW 358.4 = C) 40 C C)(80 kJ/kg. kg/s)(2.56 5 . 3 ( )] ( [ glycol ° − ° ° = − = out in p T T c m Q & & (b) The rate of heat transfer from water must be equal to the rate of heat transfer to the glycol. Then, kg/s 2.45 = C) 20 C C)(55 kJ/kg. (4.18 kJ/s 4 . 358 ) ( )] ( [ water water ° − ° ° = − = ⎯→ ⎯ − = in out p in out p T T c Q m T T c m Q & & & & (c) The temperature differences at the two ends of the heat exchanger are C 20 = C 20 C 40 C 25 = C 55 C 80 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and C 4 . 22 ) 20 / 25 ln( 20 25 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm Then the heat transfer surface area becomes 2 m 64.0 = ° ° = Δ = ⎯→ ⎯ Δ = C) 4 . 22 ( C) . kW/m 25 . 0 ( kW 4 . 358 2 lm i i lm i i T U Q A T A U Q & & Hot Glycol 80°C 3.5 kg/s Cold Water 20°C 40°C 55°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-27 16-52 Water is heated by steam in a double-pipe counter-flow heat exchanger. The required length of the tubes is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heat of water is given to be 4.18 kJ/kg.°C. The heat of condensation of steam at 120°C is given to be 2203 kJ/kg. Analysis The rate of heat transfer is kW 790.02 = C) 17 C C)(80 kJ/kg. kg/s)(4.18 3 ( )] ( [ water ° − ° ° = − = in out p T T c m Q & & The logarithmic mean temperature difference is Δ Δ T T T T T T h in c out h in c in 1 2 120 80 120 17 = − = ° − ° ° = − = ° − ° ° , , , , C C = 40 C C C = 103 C Δ Δ Δ Δ Δ T T T T T lm = − = − = ° 1 2 1 2 40 103 40 103 66 6 ln( / ) ln( / ) . C The heat transfer surface area is & & . ( . . ( . . Q U A T A Q U T i i lm i i lm = ⎯→ ⎯ = = ° ° = Δ Δ 790 02 15 66 6 7 9 kW kW / m C) C) m 2 2 Then the length of tube required becomes m 100.6 = = = ⎯→ ⎯ = m) 025 . 0 ( m 9 . 7 2 π π π i i i i D A L L D A Water 17°C 3 kg/s Steam 120°C 80°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-28 16-53 Oil is cooled by water in a thin-walled double-pipe counter-flow heat exchanger. The overall heat transfer coefficient of the heat exchanger is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Properties The specific heats of water and oil are given to be 4.18 and 2.20 kJ/kg.°C, respectively. Analysis The rate of heat transfer from the water to the oil is kW 484 = C) 40 C C)(150 kJ/kg. kg/s)(2.2 2 ( )] ( [ oil ° − ° ° = − = out in p T T c m Q & & The outlet temperature of the water is determined from C 2 . 99 C) kJ/kg. kg/s)(4.18 5 . 1 ( kW 484 + C 22 )] ( [ water ° = ° ° = + = ⎯→ ⎯ − = p in out in out p c m Q T T T T c m Q & & & & The logarithmic mean temperature difference is C 18 = C 22 C 40 C 50.8 = C 2 . 99 C 150 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 6 . 31 ) 18 / 8 . 50 ln( 18 8 . 50 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm Then the overall heat transfer coefficient becomes C . kW/m 32.5 2 ° = ° = Δ = C) 6 . 31 m)( 6 )( m 025 . 0 ( kW 484 π lm s T A Q U & Cold water 22°C 1.5 kg/s Hot oil 150°C 2 kg/s 40°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-29 16-54 EES Prob. 16-53 is reconsidered. The effects of oil exit temperature and water inlet temperature on the overall heat transfer coefficient of the heat exchanger are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_oil_in=150 [C] T_oil_out=40 [C] m_dot_oil=2 [kg/s] c_p_oil=2.20 [kJ/kg-C] T_w_in=22 [C] m_dot_w=1.5 [kg/s] C_p_w=4.18 [kJ/kg-C] D=0.025 [m] L=6 [m] "ANALYSIS" Q_dot=m_dot_oilc_p_oil(T_oil_in-T_oil_out) Q_dot=m_dot_wc_p_w(T_w_out-T_w_in) DELTAT_1=T_oil_in-T_w_out DELTAT_2=T_oil_out-T_w_in DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=UADELTAT_lm A=piDL Toil,out [C] U [kW/m2-C] 30 53.22 32.5 45.94 35 40.43 37.5 36.07 40 32.49 42.5 29.48 45 26.9 47.5 24.67 50 22.7 52.5 20.96 55 19.4 57.5 18 60 16.73 62.5 15.57 65 14.51 67.5 13.53 70 12.63 30 35 40 45 50 55 60 65 70 10 15 20 25 30 35 40 45 50 55 Toil,out [C] U [kW/m 2-C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-30 Tw,in [C] U [kW/m2C] 5 20.7 6 21.15 7 21.61 8 22.09 9 22.6 10 23.13 11 23.69 12 24.28 13 24.9 14 25.55 15 26.24 16 26.97 17 27.75 18 28.58 19 29.46 20 30.4 21 31.4 22 32.49 23 33.65 24 34.92 25 36.29 5 9 13 17 21 25 20 22 24 26 28 30 32 34 36 38 Tw,in [C] U [kW/m 2-C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-31 16-55 The inlet and outlet temperatures of the cold and hot fluids in a double-pipe heat exchanger are given. It is to be determined whether this is a parallel-flow or counter-flow heat exchanger. Analysis In parallel-flow heat exchangers, the temperature of the cold water can never exceed that of the hot fluid. In this case Tcold out = 50°C which is greater than Thot out = 45°C. Therefore this must be a counter-flow heat exchanger. 16-56 Cold water is heated by hot water in a double-pipe counter-flow heat exchanger. The rate of heat transfer and the heat transfer surface area of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ/kg.°C, respectively. Analysis The rate of heat transfer in this heat exchanger is kW 156.8 = C) 15 C C)(45 kJ/kg. kg/s)(4.18 25 . 1 ( )] ( [ water cold ° − ° ° = − = in out p T T c m Q & & The outlet temperature of the hot water is determined from C 5 . 87 C) kJ/kg. kg/s)(4.19 3 ( kW 8 . 156 C 100 )] ( [ hot water ° = ° − ° = − = ⎯→ ⎯ − = p in out out in p c m Q T T T T c m Q & & & & The temperature differences at the two ends of the heat exchanger are C 72.5 = C 15 C 5 . 87 C 55 = C 45 C 100 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and C 3 . 63 ) 5 . 72 / 55 ln( 5 . 72 55 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm Then the surface area of this heat exchanger becomes 2 m 2.81 = ° ° = Δ = ⎯→ ⎯ Δ = C) 3 . 63 ( C) . kW/m 880 . 0 ( kW 8 . 156 2 lm s lm s T U Q A T UA Q & & Hot water 100°C 3 kg/s Cold Water 15°C 1.25 kg/s 45°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-32 16-57 Engine oil is heated by condensing steam in a condenser. The rate of heat transfer and the length of the tube required are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Properties The specific heat of engine oil is given to be 2.1 kJ/kg.°C. The heat of condensation of steam at 130°C is given to be 2174 kJ/kg. Analysis The rate of heat transfer in this heat exchanger is kW 25.2 = C) 20 C C)(60 kJ/kg. kg/s)(2.1 3 . 0 ( )] ( [ oil ° − ° ° = − = in out p T T c m Q & & The temperature differences at the two ends of the heat exchanger are C 110 = C 20 C 130 C 70 = C 60 C 130 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and C 5 . 88 ) 110 / 70 ln( 110 70 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The surface area is 2 2 m 44 . 0 C) 5 . 88 ( C) . kW/m 65 . 0 ( kW 2 . 25 = ° ° = Δ = lm s T U Q A & Then the length of the tube required becomes m 7.0 = = = ⎯→ ⎯ = m) 02 . 0 ( m 44 . 0 2 π π π D A L DL A s s Oil 20°C 0.3 kg/s Steam 130°C 60°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-33 16-58E Water is heated by geothermal water in a double-pipe counter-flow heat exchanger. The mass flow rate of each fluid and the total thermal resistance of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of water and geothermal fluid are given to be 1.0 and 1.03 Btu/lbm.°F, respectively. Analysis The mass flow rate of each fluid is determined from lbm/s 0.667 = F) 140 F F)(200 Btu/lbm. (1.0 Btu/s 40 ) ( )] ( [ water water ° − ° ° = − = − = in out p in out p T T c Q m T T c m Q & & & & lbm/s 0.431 = F) 180 F F)(270 Btu/lbm. (1.03 Btu/s 40 ) ( )] ( [ water geo. water geo. ° − ° ° = − = − = in out p in out p T T c Q m T T c m Q & & & & The temperature differences at the two ends of the heat exchanger are F 40 = F 140 F 180 F 70 = F 200 F 270 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and F 61 . 53 ) 40 / 70 ln( 40 70 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm Then F/Btu s 1.34 ° ⋅ = ° = = ⎯→ ⎯ = = ° = Δ = ⎯→ ⎯ Δ = F Btu/s. 7462 . 0 1 1 1 F Btu/s. 7462 . 0 F 53.61 Btu/s 40 o s s lm s lm s UA R RA U T Q UA T UA Q & & Hot brine 270°F Cold Water 140°F 180°F 200°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-34 16-59 Glycerin is heated by ethylene glycol in a thin-walled double-pipe parallel-flow heat exchanger. The rate of heat transfer, the outlet temperature of the glycerin, and the mass flow rate of the ethylene glycol are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Properties The specific heats of glycerin and ethylene glycol are given to be 2.4 and 2.5 kJ/kg.°C, respectively. Analysis (a) The temperature differences at the two ends are C 15 = C) 15 ( C 40 = C 20 C 60 , , , , 2 , , 1 ° ° − − = − = Δ ° ° − ° = − = Δ out h out h out c out h in c in h T T T T T T T T and C 5 . 25 ) 15 / 40 ln( 15 40 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm Then the rate of heat transfer becomes kW 19.58 = = ° ° = Δ = W 584 , 19 C) 5 . 25 )( m C)(3.2 . W/m 240 ( 2 2 lm s T UA Q & (b) The outlet temperature of the glycerin is determined from C 47.2° = ° + ° = + = ⎯→ ⎯ − = C) kJ/kg. kg/s)(2.4 3 . 0 ( kW 584 . 19 C 20 )] ( [ glycerin p in out in out p c m Q T T T T c m Q & & & & (c) Then the mass flow rate of ethylene glycol becomes kg/s 3.56 = ° − ° ° = − = − = C] 60 C 15) + C)[(47.2 kJ/kg. (2.5 kJ/s 584 . 19 ) ( )] ( [ glycol ethylene glycol ethylene out in p out in p T T c Q m T T c m Q & & & & Glycerin 20°C 0.3 kg/s Hot ethylene 60°C 3 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-35 16-60 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the outlet temperature of the air are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of air and combustion gases are given to be 1005 and 1100 J/kg.°C, respectively. Analysis The rate of heat transfer is kW 103 = C) 95 C C)(180 kJ/kg. kg/s)(1.1 1 . 1 ( )] ( [ gas. ° − ° ° = − = out in p T T c m Q & & The mass flow rate of air is kg/s 904 . 0 K 293 /kg.K) kPa.m 287 . 0 ( /s) m kPa)(0.8 (95 3 3 = × = = RT P m V & & Then the outlet temperature of the air becomes C 133° = ° × + ° = + = ⎯→ ⎯ − = C) J/kg. kg/s)(1005 904 . 0 ( W 10 103 C 20 ) ( 3 , , , , p in c out c in c out c p c m Q T T T T c m Q & & & & Air 95 kPa 20°C 0.8 m3/s Exhaust gases 1.1 kg/s 95°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-36 16-61 Water is heated by hot oil in a 2-shell passes and 12-tube passes heat exchanger. The heat transfer surface area on the tube side is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.°C, respectively. Analysis The rate of heat transfer in this heat exchanger is kW 940.5 = C) 20 C C)(70 kJ/kg. kg/s)(4.18 5 . 4 ( )] ( [ water ° − ° ° = − = in out p T T c m Q & & The outlet temperature of the oil is determined from C 129 C) kJ/kg. kg/s)(2.3 10 ( kW 5 . 940 C 170 )] ( [ oil ° = ° − ° = − = ⎯→ ⎯ − = p in out out in p c m Q T T T T c m Q & & & & The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 109 = C 20 C 129 C 100 = C 70 C 170 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 4 . 104 ) 109 / 100 ln( 109 100 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 0 . 1 82 . 0 20 70 129 170 33 . 0 20 170 20 70 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P Then the heat transfer surface area on the tube side becomes 2 m 25.7 = ° ° = Δ = ⎯→ ⎯ Δ = C) 4 . 104 ( C)(1.0) . kW/m 350 . 0 ( kW 5 . 940 2 , , CF lm s CF lm s T UF Q A T F UA Q & & Oil 170°C 10 kg/s Water 20°C 4.5 kg/s 70°C (12 tube passes) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-37 16-62 Water is heated by hot oil in a 2-shell passes and 12-tube passes heat exchanger. The heat transfer surface area on the tube side is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.°C, respectively. Analysis The rate of heat transfer in this heat exchanger is kW 418 = C) 20 C C)(70 kJ/kg. kg/s)(4.18 2 ( )] ( [ water ° − ° ° = − = in out p T T c m Q & & The outlet temperature of the oil is determined from C 8 . 151 C) kJ/kg. kg/s)(2.3 10 ( kW 418 C 170 )] ( [ oil ° = ° − ° = − = ⎯→ ⎯ − = p in out out in p c m Q T T T T c m Q & & & & The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 131.8 = C 20 C 8 . 151 C 100 = C 70 C 170 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 2 . 115 ) 8 . 131 / 100 ln( 8 . 131 100 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 0 . 1 36 . 0 20 70 8 . 151 170 33 . 0 20 170 20 70 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P Then the heat transfer surface area on the tube side becomes 2 m 10.4 = ° ° = Δ = ⎯→ ⎯ Δ = C) 2 . 115 ( C)(1.0) . kW/m 350 . 0 ( kW 418 2 , , CF lm i i CF lm i i T F U Q A T F A U Q & & Oil 170°C 10 kg/s Water 20°C 2 kg/s 70°C (12 tube passes) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-38 16-63 Ethyl alcohol is heated by water in a 2-shell passes and 8-tube passes heat exchanger. The heat transfer surface area of the heat exchanger is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of water and ethyl alcohol are given to be 4.19 and 2.67 kJ/kg.°C, respectively. Analysis The rate of heat transfer in this heat exchanger is kW 252.3 = C) 25 C C)(70 kJ/kg. kg/s)(2.67 1 . 2 ( )] ( [ alcohol ethyl ° − ° ° = − = in out p T T c m Q & & The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 20 = C 25 C 45 C 25 = C 70 C 95 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 4 . 22 ) 20 / 25 ln( 20 25 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 82 . 0 1 . 1 25 70 45 95 64 . 0 25 95 25 70 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P Then the heat transfer surface area on the tube side becomes 2 m 14.5 = ° ° = Δ = ⎯→ ⎯ Δ = C) 4 . 22 ( C)(0.82) . kW/m 950 . 0 ( kW 3 . 252 2 , , CF lm i i CF lm i i T F U Q A T F A U Q & & Water 95°C Ethyl Alcohol 25°C 2.1 kg/s 70°C (8 tube passes) 45°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-39 16-64 Water is heated by ethylene glycol in a 2-shell passes and 12-tube passes heat exchanger. The rate of heat transfer and the heat transfer surface area on the tube side are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of water and ethylene glycol are given to be 4.18 and 2.68 kJ/kg.°C, respectively. Analysis The rate of heat transfer in this heat exchanger is : kW 160.5 = C) 22 C C)(70 kJ/kg. kg/s)(4.18 8 . 0 ( )] ( [ water ° − ° ° = − = in out p T T c m Q & & The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 38 = C 22 C 60 C 40 = C 70 C 110 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 39 ) 38 / 40 ln( 38 40 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 92 . 0 04 . 1 22 70 60 110 55 . 0 22 110 22 70 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P Then the heat transfer surface area on the tube side becomes 2 m 16.0 = ° ° = Δ = ⎯→ ⎯ Δ = C) 39 ( C)(0.92) . kW/m 28 . 0 ( kW 5 . 160 2 , , CF lm i i CF lm i i T F U Q A T F A U Q & & Ethylene 110°C Water 22°C 0.8 kg/s 70°C (12 tube passes) 60°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-40 16-65 EES Prob. 16-64 is reconsidered. The effect of the mass flow rate of water on the rate of heat transfer and the tube-side surface area is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_w_in=22 [C] T_w_out=70 [C] m_dot_w=0.8 [kg/s] c_p_w=4.18 [kJ/kg-C] T_glycol_in=110 [C] T_glycol_out=60 [C] c_p_glycol=2.68 [kJ/kg-C] U=0.28 [kW/m^2-C] "ANALYSIS" Q_dot=m_dot_wc_p_w(T_w_out-T_w_in) Q_dot=m_dot_glycolc_p_glycol(T_glycol_in-T_glycol_out) DELTAT_1=T_glycol_in-T_w_out DELTAT_2=T_glycol_out-T_w_in DELTAT_lm_CF=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) P=(T_w_out-T_w_in)/(T_glycol_in-T_w_in) R=(T_glycol_in-T_glycol_out)/(T_w_out-T_w_in) F=0.92 "from Fig. 16-18b of the text at the calculated P and R" Q_dot=UAFDELTAT_lm_CF mw [kg/s] Q [kW] A [m2] 0.4 80.26 7.99 0.5 100.3 9.988 0.6 120.4 11.99 0.7 140.4 13.98 0.8 160.5 15.98 0.9 180.6 17.98 1 200.6 19.98 1.1 220.7 21.97 1.2 240.8 23.97 1.3 260.8 25.97 1.4 280.9 27.97 1.5 301 29.96 1.6 321 31.96 1.7 341.1 33.96 1.8 361.2 35.96 1.9 381.2 37.95 2 401.3 39.95 2.1 421.3 41.95 2.2 441.4 43.95 0.25 0.65 1.05 1.45 1.85 2.25 50 100 150 200 250 300 350 400 450 5 10 15 20 25 30 35 40 45 50 mw [kg/s] Q [kW] A [m 2] heat area PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-41 16-66E Steam is condensed by cooling water in a condenser. The rate of heat transfer, the rate of condensation of steam, and the mass flow rate of cold water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. 6 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Properties We take specific heat of water are given to be 1.0 Btu/lbm.°F. The heat of condensation of steam at 90°F is 1043 Btu/lbm. Analysis (a) The log mean temperature difference is determined from F 30 = F 60 F 90 F 17 = F 73 F 90 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T F 9 . 22 ) 30 / 17 ln( 30 17 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm The heat transfer surface area is 2 ft 7 . 392 ft) ft)(5 48 / 3 ( 50 8 8 = × × = = π πDL n As and Btu/h 10 5.396 6 × = ° ° = Δ = F) 9 . 22 )( ft F)(392.7 . Btu/h.ft 600 ( 2 2 lm s T UA Q & (b) The rate of condensation of the steam is lbm/s 1.44 = lbm/h 5173 = × = = ⎯→ ⎯ = Btu/lbm 1043 Btu/h 10 396 . 5 ) ( 6 fg steam steam fg h Q m h m Q & & & & (c) Then the mass flow rate of cold water becomes lbm/s 115 lbm/h 10 4.15 5 = × = ° − ° ° × = − = − = F] 60 F F)(73 Btu/lbm. (1.0 Btu/h 10 396 . 5 ) ( )] ( [ 6 water cold water cold in out p in out p T T c Q m T T c m Q & & & & Steam 90°F 20 lbm/s 60°F Water 73°F 90°F (8 tube passes) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-42 16-67E EES Prob. 16-66E is reconsidered. The effect of the condensing steam temperature on the rate of heat transfer, the rate of condensation of steam, and the mass flow rate of cold water is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" N_pass=8 N_tube=50 T_steam=90 [F] h_fg_steam=1043 [Btu/lbm] T_w_in=60 [F] T_w_out=73 [F] c_p_w=1.0 [Btu/lbm-F] D=3/41/12 [ft] L=5 [ft] U=600 [Btu/h-ft^2-F] "ANALYSIS" "(a)" DELTAT_1=T_steam-T_w_out DELTAT_2=T_steam-T_w_in DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) A=N_passN_tubepiDL Q_dot=UADELTAT_lmConvert(Btu/h, Btu/s) "(b)" Q_dot=m_dot_steamh_fg_steam "(c)" Q_dot=m_dot_wc_p_w(T_w_out-T_w_in) Tsteam [F] Q [Btu/s] msteam[lbm/s] mw [lbm/s] 80 810.5 0.7771 62.34 82 951.9 0.9127 73.23 84 1091 1.046 83.89 86 1228 1.177 94.42 88 1363 1.307 104.9 90 1498 1.436 115.2 92 1632 1.565 125.6 94 1766 1.693 135.8 96 1899 1.821 146.1 98 2032 1.948 156.3 100 2165 2.076 166.5 102 2297 2.203 176.7 104 2430 2.329 186.9 106 2562 2.456 197.1 108 2694 2.583 207.2 110 2826 2.709 217.4 112 2958 2.836 227.5 114 3089 2.962 237.6 116 3221 3.088 247.8 118 3353 3.214 257.9 120 3484 3.341 268 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-43 80 85 90 95 100 105 110 115 120 500 1000 1500 2000 2500 3000 3500 0.5 1 1.5 2 2.5 3 3.5 Tsteam [F] Q [Btu/s] msteam [lbm/s] heat msteam 80 85 90 95 100 105 110 115 120 50 95 140 185 230 275 Tsteam [F] mw [lbm/s] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-44 16-68 Glycerin is heated by hot water in a 1-shell pass and 20-tube passes heat exchanger. The mass flow rate of glycerin and the overall heat transfer coefficient of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heat of glycerin is given to be are given to be 2.48 kJ/kg.°C and that of water is taken to be 4.18 kJ/kg.°C. Analysis The rate of heat transfer in this heat exchanger is kW 94.05 = C) 55 C C)(100 kJ/kg. kg/s)(4.18 5 . 0 ( )] ( [ water ° − ° ° = − = out in p T T c m Q & & The mass flow rate of the glycerin is determined from kg/s 0.95 = ° − ° ° = − = − = C] 15 C C)[(55 kJ/kg. (2.48 kJ/s 05 . 94 ) ( )] ( [ glycerin glycerin in out p in out p T T c Q m T T c m Q & & & & The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 40 = C 15 C 55 C 45 = C 55 C 100 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 5 . 42 ) 40 / 45 ln( 40 45 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 77 . 0 89 . 0 100 55 55 15 53 . 0 100 15 100 55 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P The heat transfer surface area is 2 m 5.027 = m) m)(2 04 . 0 ( 20π π = = DL n As Then the overall heat transfer coefficient of the heat exchanger is determined to be C . kW/m 0.572 2 ° = ° = Δ = ⎯→ ⎯ Δ = C) 5 . 42 )( 77 . 0 )( m 027 . 5 ( kW 05 . 94 2 , , CF lm s CF lm s T F A Q U T F UA Q & & Glycerin 15°C 100°C Hot Water 0.5 kg/s 55°C 55°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-45 16-69 Isobutane is condensed by cooling air in the condenser of a power plant. The mass flow rate of air and the overall heat transfer coefficient are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The heat of vaporization of isobutane at 75°C is given to be hfg = 255.7 kJ/kg and specific heat of air is taken to be cp = 1005 J/kg.°C. Analysis First, the rate of heat transfer is determined from kW 4 . 690 ) kJ/kg 7 . 255 )( kg/s 7 . 2 ( ) ( isobutane = = = fg h m Q & & The mass flow rate of air is determined from kg/s 98.1 = C) 21 C C)(28 kJ/kg. (1.005 kJ/s 4 . 690 ) ( )] ( [ in out air air in out ° − ° ° = − = ⎯→ ⎯ − = T T c Q m T T c m Q p p & & & & The temperature differences between the isobutane and the air at the two ends of the condenser are C 47 = C 28 C 75 C 54 = C 21 C 75 in c, out h, 2 out c, in h, 1 ° ° − ° = − = Δ ° ° − ° = − = Δ T T T T T T and C 4 . 50 ) 47 / 54 ln( 47 54 ) / ln( 2 1 2 1 lm ° = − = Δ Δ Δ − Δ = Δ T T T T T Then the overall heat transfer coefficient is determined from C . W/m 571 2 ° ⎯→ ⎯ ° = ⎯→ ⎯ Δ = = C) 4 . 50 )( m (24 W 400 , 690 2 lm U U T UA Q s & Isobutane 75°C 2.7 kg/s Air, 21°C Air, 28°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-46 16-70 Water is evaporated by hot exhaust gases in an evaporator. The rate of heat transfer, the exit temperature of the exhaust gases, and the rate of evaporation of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The heat of vaporization of water at 200°C is given to be hfg = 1941 kJ/kg and specific heat of exhaust gases is given to be cp = 1051 J/kg.°C. Analysis The temperature differences between the water and the exhaust gases at the two ends of the evaporator are C ) 200 ( C 350 = C 200 C 550 out h, in c, out h, 2 out c, in h, 1 ° − = − = Δ ° ° − ° = − = Δ T T T T T T T and [ ] ) 200 /( 350 ln ) 200 ( 350 ) / ln( out h, out h, 2 1 2 1 lm − − − = Δ Δ Δ − Δ = Δ T T T T T T T Then the rate of heat transfer can be expressed as [ ] ) 200 /( 350 ln ) 200 ( 350 ) m 5 . 0 )( C . kW/m 780 . 1 ( out h, out h, 2 2 lm − − − ° = Δ = T T T UA Q s & (Eq. 1) The rate of heat transfer can also be expressed as in the following forms ) C C)(550 kJ/kg. 1 kg/s)(1.05 25 . 0 ( )] ( [ out h, gases exhaust out h, in h, T T T c m Q p − ° ° = − = & & (Eq. 2) ) kJ/kg 1941 ( ) ( water water m h m Q fg & & & = = (Eq. 3) We have three equations with three unknowns. Using an equation solver such as EES, the unknowns are determined to be kg/s 0.0458 C 211.8 kW 88.85 = ° = = water out h, m T Q & & Water 200°C 550°C Exhaust gases Th,out 200°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-47 16-71 EES Prob. 16-70 is reconsidered. The effect of the exhaust gas inlet temperature on the rate of heat transfer, the exit temperature of exhaust gases, and the rate of evaporation of water is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_exhaust_in=550 [C] c_p_exhaust=1.051 [kJ/kg-C] m_dot_exhaust=0.25 [kg/s] T_w=200 [C] h_fg_w=1941 [kJ/kg] A=0.5 [m^2] U=1.780 [kW/m^2-C] "ANALYSIS" DELTAT_1=T_exhaust_in-T_w DELTAT_2=T_exhaust_out-T_w DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=UADELTAT_lm Q_dot=m_dot_exhaustc_p_exhaust(T_exhaust_in-T_exhaust_out) Q_dot=m_dot_wh_fg_w Texhaust,in [C] Q [kW] Texhaust,out [C] mw [kg/s] 300 25.39 203.4 0.01308 320 30.46 204.1 0.0157 340 35.54 204.7 0.01831 360 40.62 205.4 0.02093 380 45.7 206.1 0.02354 400 50.77 206.8 0.02616 420 55.85 207.4 0.02877 440 60.93 208.1 0.03139 460 66.01 208.8 0.03401 480 71.08 209.5 0.03662 500 76.16 210.1 0.03924 520 81.24 210.8 0.04185 540 86.32 211.5 0.04447 560 91.39 212.2 0.04709 580 96.47 212.8 0.0497 600 101.5 213.5 0.05232 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-48 300 350 400 450 500 550 600 20 30 40 50 60 70 80 90 100 110 202 204 206 208 210 212 214 Texhaust,in [C] Q [kW] Texhaust,out [C] heat temperature 300 350 400 450 500 550 600 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 Texhaust,in [C] mw [kg/s] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-49 16-72 The waste dyeing water is to be used to preheat fresh water. The outlet temperatures of each fluid and the mass flow rate are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The specific heats of waste dyeing water and the fresh water are given to be cp = 4295 J/kg.°C and cp = 4180 J/kg.°C, respectively. Analysis The temperature differences between the dyeing water and the fresh water at the two ends of the heat exchanger are 15 75 out h, in c, out h, 2 out c, out c, in h, 1 − = − = Δ − = − = Δ T T T T T T T T and [ ] ) 15 /( ) 75 ( ln ) 15 ( ) 75 ( ) / ln( out h, out c, out h, out c, 2 1 2 1 lm − − − − − = Δ Δ Δ − Δ = Δ T T T T T T T T T Then the rate of heat transfer can be expressed as [ ] ) 15 /( ) 75 ( ln ) 15 ( ) 75 ( ) m 65 . 1 )( C . kW/m 625 . 0 ( kW 35 out h, out c, out h, out c, 2 2 lm − − − − − ° = Δ = T T T T T UA Q s & (Eq. 1) The rate of heat transfer can also be expressed as ) C C)(75 kJ/kg. (4.295 kW 35 )] ( [ out h, water dyeing out h, in h, T m T T c m Q p − ° ° = ⎯→ ⎯ − = & & & (Eq. 2) C) 15 C)( kJ/kg. (4.18 kW 35 )] ( [ out c, water out h, in h, ° − ° = ⎯→ ⎯ − = T m T T c m Q p & & & (Eq. 3) We have three equations with three unknowns. Using an equation solver such as EES, the unknowns are determined to be kg/s 0.317 C 49.3 C 41.4 = ° = ° = m T T & out h, out c, Fresh water 15°C Dyeing water 75°C Tc,out Th,out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-50 The Effectiveness-NTU Method 16-73C When the heat transfer surface area A of the heat exchanger is known, but the outlet temperatures are not, the effectiveness-NTU method is definitely preferred. 16-74C The effectiveness of a heat exchanger is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate and represents how closely the heat transfer in the heat exchanger approaches to maximum possible heat transfer. Since the actual heat transfer rate can not be greater than maximum possible heat transfer rate, the effectiveness can not be greater than one. The effectiveness of a heat exchanger depends on the geometry of the heat exchanger as well as the flow arrangement. 16-75C For a specified fluid pair, inlet temperatures and mass flow rates, the counter-flow heat exchanger will have the highest effectiveness. 16-76C Once the effectiveness ε is known, the rate of heat transfer and the outlet temperatures of cold and hot fluids in a heat exchanger are determined from ) ( ) ( ) ( , , , , , , , , min max out h in h h p h in c out c c p c in c in h T T c m Q T T c m Q T T C Q Q − = − = − = = & & & & & & ε ε 16-77C The heat transfer in a heat exchanger will reach its maximum value when the hot fluid is cooled to the inlet temperature of the cold fluid. Therefore, the temperature of the hot fluid cannot drop below the inlet temperature of the cold fluid at any location in a heat exchanger. 16-78C The heat transfer in a heat exchanger will reach its maximum value when the cold fluid is heated to the inlet temperature of the hot fluid. Therefore, the temperature of the cold fluid cannot rise above the inlet temperature of the hot fluid at any location in a heat exchanger. 16-79C The fluid with the lower mass flow rate will experience a larger temperature change. This is clear from the relation hot p h cold p c T c m T c m Q Δ = Δ = & & & 16-80C The maximum possible heat transfer rate is in a heat exchanger is determined from ) ( , , min max in c in h T T C Q − = & where Cmin is the smaller heat capacity rate. The value of max Q & does not depend on the type of heat exchanger. 16-81C The longer heat exchanger is more likely to have a higher effectiveness. 16-82C The increase of effectiveness with NTU is not linear. The effectiveness increases rapidly with NTU for small values (up to abo ut NTU = 1.5), but rather slowly for larger values. Therefore, the effectiveness will not double when the length of heat exchanger is doubled. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-51 16-83C A heat exchanger has the smallest effectiveness value when the heat capacity rates of two fluids are identical. Therefore, reducing the mass flow rate of cold fluid by half will increase its effectiveness. 16-84C When the capacity ratio is equal to zero and the number of transfer units value is greater than 5, a counter-flow heat exchanger has an effectiveness of one. In this case the exit temperature of the fluid with smaller capacity rate will equal to inlet temperature of the other fluid. For a parallel-flow heat exchanger the answer would be the same. 16-85C The NTU of a heat exchanger is defined as min min ) ( p s s c m UA C UA NTU & = = where U is the overall heat transfer coefficient and As is the heat transfer surface area of the heat exchanger. For specified values of U and Cmin, the value of NTU is a measure of the heat exchanger surface area As. Because the effectiveness increases slowly for larger values of NTU, a large heat exchanger cannot be justified economically. Therefore, a heat exchanger with a very large NTU is not necessarily a good one to buy. 16-86C The value of effectiveness increases slowly with a large values of NTU (usually larger than 3). Therefore, doubling the size of the heat exchanger will not save much energy in this case since the increase in the effectiveness will be very small. 16-87C The value of effectiveness increases rapidly with small values of NTU (up to about 1.5). Therefore, tripling the NTU will cause a rapid increase in the effectiveness of the heat exchanger, and thus saves energy. I would support this proposal. 16-88 Hot water coming from the engine of an automobile is cooled by air in the radiator. The outlet temperature of the air and the rate of heat transfer are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and air are given to be 4.00 and 1.00 kJ/kg.°C, respectively. Analysis (a) The heat capacity rates of the hot and cold fluids are C kW/ 10 C) kJ/kg. kg/s)(1.00 (10 C kW/ 20 C) kJ/kg. kg/s)(4.00 (5 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore C kW/ 10 min ° = = c C C which is the smaller of the two heat capacity rates. Noting that the heat capacity rate of the air is the smaller one, the outlet temperature of the air is determined from the effectiveness relation to be C 50° = ⎯→ ⎯ ° − ° − = ⎯→ ⎯ − − = = out a out a in c in h in c out a T T T T C T T C Q Q , , , , min , , min max C ) 30 80 ( C ) 30 ( 4 . 0 ) ( ) ( & & ε (b) The rate of heat transfer is determined from kW 200 = C) 30 -C C)(50 kW/ (10 ) ( , , ° ° ° = − = in a out a air T T C Q & Air 30°C 10 kg/s Coolant 80°C 5 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-52 16-89 The inlet and exit temperatures and the volume flow rates of hot and cold fluids in a heat exchanger are given. The rate of heat transfer to the cold water, the overall heat transfer coefficient, the fraction of heat loss, the heat transfer efficiency, the effectiveness, and the NTU of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 Changes in the kinetic and potential energies of fluid streams are negligible. 3 Fluid properties are constant. Properties The densities of hot water and cold water at the average temperatures of (71.5+58.2)/2 = 64.9°C and (19.7+27.8)/2 = 23.8°C are 980.5 and 997.3 kg/m3, respectively. The specific heat at the average temperature is 4187 J/kg.°C for hot water and 4180 J/kg.°C for cold water (Table A-15). Analysis (a) The mass flow rates are kg/s 0172 . 0 /s) m 60 )(0.00105/ kg/m 5 . 980 ( 3 3 = = = h h h m V & & ρ kg/s 0258 . 0 /s) m 60 )(0.00155/ kg/m 3 . 997 ( 3 3 = = = c c c m V & & ρ The rates of heat transfer from the hot water and to the cold water are W 957.8 = C) 2 . 58 C C)(71.5 kJ/kg. kg/s)(4187 0172 . 0 ( )] ( [ h ° − ° ° = − = out in p h T T c m Q & & W 873.5 = C) 2 . 19 C C)(27.8 kJ/kg. kg/s)(4180 0258 . 0 ( )] ( [ c ° − ° ° = − = in out p c T T c m Q & & (b) The number of shell and tubes are not specified in the problem. Therefore, we take the correction factor to be unity in the following calculations. The logarithmic mean temperature difference and the overall heat transfer coefficient are C 7 . 43 C 8 . 27 C 5 . 71 , , 1 ° = ° − ° = − = Δ out c in h T T T C 5 . 38 C 7 . 19 C 2 . 58 , , 2 ° = ° − ° = − = Δ in c out h T T T C 0 . 41 5 . 38 7 . 43 ln 5 . 38 7 . 43 ln 2 1 2 1 ° = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ Δ Δ − Δ = Δ T T T T Tlm C W/m 1117 2 ⋅ = ° + = Δ = ) C 0 . 41 )( m 2 0 . 0 ( W 2 / ) 5 . 873 8 . 957 ( 2 , lm m hc T A Q U & Note that we used the average of two heat transfer rates in calculations. (c) The fraction of heat loss and the heat transfer efficiency are 91.2% 8.8% = = = = = = − = − = 912 . 0 8 . 957 5 . 873 088 . 0 8 . 957 5 . 873 8 . 957 h c h c h loss Q Q Q Q Q f & & & & & η (d) The heat capacity rates of the hot and cold fluids are C W/ 8 . 107 C) kJ/kg. kg/s)(4180 (0.0258 C W/ 02 . 72 C) kJ/kg. kg/s)(4187 (0.0172 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore C W/ 02 . 72 min ° = = h C C which is the smaller of the two heat capacity rates. Then the maximum heat transfer rate becomes W 3731 = C) 19.7 -C C)(71.5 W/ (72.02 ) ( , , min max ° ° ° = − = in c in h T T C Q & The effectiveness of the heat exchanger is 24.5% = = + = = 245 . 0 kW 3731 kW 2 / ) 5 . 873 8 . 957 ( max Q Q & & ε One again we used the average heat transfer rate. We could have used the smaller or greater heat transfer rates in calculations. The NTU of the heat exchanger is determined from 0.310 = ° ⋅ = = C W/ 02 . 72 ) m 02 . 0 )( C W/m 1117 ( 2 2 min C UA NTU Hot water 71.5°C Cold water 19.7°C 27.8°C 58.2°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-53 16-90 Water is heated by a hot water stream in a heat exchanger. The maximum outlet temperature of the cold water and the effectiveness of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and air are given to be 4.18 and 1.0 kJ/kg.°C. Analysis The heat capacity rates of the hot and cold fluids are C kW/ 463 . 1 C) kJ/kg. kg/s)(4.18 (0.35 C kW/ 8 . 0 C) kJ/kg. kg/s)(1.0 (0.8 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore C kW/ 8 . 0 min ° = = h C C which is the smaller of the two heat capacity rates. Then the maximum heat transfer rate becomes kW 80 . 40 C) 14 -C C)(65 kW/ (0.8 ) ( , , min max = ° ° ° = − = in c in h T T C Q & The maximum outlet temperature of the cold fluid is determined to be C 41.9° = ° ° = + = ⎯→ ⎯ − = C kW/ 463 . 1 kW 40.80 + C 14 ) ( max , max , , , max , , max c in c out c in c out c c C Q T T T T C Q & & The actual rate of heat transfer and the effectiveness of the heat exchanger are kW 32 = C) 25 -C C)(65 kW/ (0.8 ) ( , , ° ° ° = − = out h in h h T T C Q & 0.784 = = = kW 8 . 40 kW 32 max Q Q & & ε Air 65°C 0.8 kg/s 14°C 0.35 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-54 16-91 Lake water is used to condense steam in a shell and tube heat exchanger. The outlet temperature of the water and the required tube length are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The properties of water are given in problem statement. The enthalpy of vaporization of water at 60°C is 2359 kJ/kg (Table A-15). Analysis (a) The rate of heat transfer is kW 5898 = kJ/kg) kg/s)(2359 (2.5 = = fg h m Q & & The outlet temperature of water is determined from C 27.1° = ° ⋅ ° = + = ⎯→ ⎯ − = C) kJ/kg kg/s)(4.18 200 ( kW 5898 + C 20 ) ( , , , , c c in c out c in c out c c c c m Q T T T T c m Q & & & & (b) The Reynold number is 440 , 42 s) kg/m 10 m)(8 025 . 0 ( ) 300 ( kg/s) 4(200 4 Re 4 -= ⋅ × = = π μ πD N m tube & which is greater than 10,000. Therefore, we have turbulent flow. We assume fully developed flow and evaluate the Nusselt number from 2 . 237 ) 6 ( ) 440 , 42 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = = = k hD Nu Heat transfer coefficient on the inner surface of the tubes is C . W/m 5694 ) 2 . 237 ( m 025 . 0 C W/m. 6 . 0 2 ° = ° = = Nu D k hi Disregarding the thermal resistance of the tube wall the overall heat transfer coefficient is determined from C W/m 3410 8500 1 5694 1 1 1 1 1 2 ° ⋅ = + = + = o i h h U The logarithmic mean temperature difference is C 9 . 32 C 1 . 27 C 60 , , 1 ° = ° − ° = − = Δ out c in h T T T C 40 C 20 C 60 , , 2 ° = ° − ° = − = Δ in c out h T T T C 3 . 36 40 9 . 32 ln 40 9 . 32 ln 2 1 2 1 ° = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ Δ Δ − Δ = Δ T T T T Tlm Noting that each tube makes two passes and taking the correction factor to be unity, the tube length per pass is determined to be [ ] m 1.01 = ° × × × ⋅ = Δ = → Δ = ) C 3 . 36 )( 1 ( m 025 . 0 300 2 ) C kW/m 410 . 3 ( kW 5898 ) ( 2 2 π π lm lm T F D U Q L T UAF Q & & Steam 60°C Lake water 20°C 60°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-55 16-92 Air is heated by a hot water stream in a cross-flow heat exchanger. The maximum heat transfer rate and the outlet temperatures of the cold and hot fluid streams are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and air are given to be 4.19 and 1.005 kJ/kg.°C. Analysis The heat capacity rates of the hot and cold fluids are C W/ 3015 C) J/kg. kg/s)(1005 (3 C W/ 4190 C) J/kg. kg/s)(4190 (1 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore C W/ 3015 min ° = = c C C which is the smaller of the two heat capacity rates. Then the maximum heat transfer rate becomes kW 150.8 = ° ° ° = − = W 150,750 = C) 20 -C C)(70 W/ (3015 ) ( , , min max in c in h T T C Q & The outlet temperatures of the cold and the hot streams in this limiting case are determined to be C 34.0 C 70 ° = ° − ° = − = ⎯→ ⎯ − = ° = ° ° = + = ⎯→ ⎯ − = C kW/ 19 . 4 kW 150.75 C 70 ) ( C kW/ 015 . 3 kW 150.75 + C 0 2 ) ( , , , , , , , , h in h out h out h in h h c in c out c in c out c c C Q T T T T C Q C Q T T T T C Q & & & & Air 20°C 3 kg/s 1 kg/s 70°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-56 16-93 Hot oil is to be cooled by water in a heat exchanger. The mass flow rates and the inlet temperatures are given. The rate of heat transfer and the outlet temperatures are to be determined. √ Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The thickness of the tube is negligible since it is thin-walled. 5 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the water and oil are given to be 4.18 and 2.2 kJ/kg.°C, respectively. Analysis The heat capacity rates of the hot and cold fluids are C W/ 418 C) J/kg. kg/s)(4180 (0.1 C W/ 440 C) J/kg. kg/s)(2200 (0.2 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore, C W/ 418 min ° = = c C C and 95 . 0 440 418 max min = = = C C c Then the maximum heat transfer rate becomes kW 59.36 C) 18 -C C)(160 W/ (418 ) ( , , min max = ° ° ° = − = in c in h T T C Q & The heat transfer surface area is 2 m 2.04 m) m)(3 )(0.018 (12)( ) ( = = = π πDL n As The NTU of this heat exchanger is 659 . 1 C W/ 418 ) m 04 . 2 ( C) . W/m 340 ( 2 2 min = ° ° = = C UA NTU s Then the effectiveness of this heat exchanger corresponding to c = 0.95 and NTU = 1.659 is determined from Fig. 16-26d to be ε = 0.61 Then the actual rate of heat transfer becomes kW 36.2 = = = kW) 36 (0.61)(59. max Q Q & & ε Finally, the outlet temperatures of the cold and hot fluid streams are determined to be C 77.7 C 104.6 ° = ° − ° = − = ⎯→ ⎯ − = ° = ° ° = + = ⎯→ ⎯ − = C kW/ 44 . 0 kW 2 . 36 C 160 ) ( C / kW 418 . 0 kW 2 . 36 + C 18 ) ( , , , , , , , , h in h out h out h in h h c in c out c in c out c c C Q T T T T C Q C Q T T T T C Q & & & & 16-94 Inlet and outlet temperatures of the hot and cold fluids in a double-pipe heat exchanger are given. It is to be determined whether this is a parallel-flow or counter-flow heat exchanger and the effectiveness of it. Analysis This is a counter-flow heat exchanger because in the parallel-flow heat exchangers the outlet temperature of the cold fluid (55°C in this case) cannot exceed the outlet temperature of the hot fluid, which is (45°C in this case). Noting that the mass flow rates of both hot and cold oil streams are the same, we have max min C C = . Then the effectiveness of this heat exchanger is determined from 0.583 = ° − ° ° − ° = − − = − − = = C 20 C 80 C 45 C 80 ) ( ) ( ) ( ) ( , , , , , , min , , max in c in h h out h in h h in c in h out h in h h T T C T T C T T C T T C Q Q & & ε Oil 160°C 0.2 kg/s Water 18°C 0.1 kg/s (12 tube passes) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-57 16-95E Inlet and outlet temperatures of the hot and cold fluids in a double-pipe heat exchanger are given. It is to be determined the fluid, which has the smaller heat capacity rate and the effectiveness of the heat exchanger. Analysis Hot water has the smaller heat capacity rate since it experiences a greater temperature change. The effectiveness of this heat exchanger is determined from 0.75 = ° − ° ° − ° = − − = − − = = F 70 F 190 F 100 F 190 ) ( ) ( ) ( ) ( , , , , , , min , , max in c in h h out h in h h in c in h out h in h h T T C T T C T T C T T C Q Q & & ε 16-96 A chemical is heated by water in a heat exchanger. The mass flow rates and the inlet temperatures are given. The outlet temperatures of both fluids are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The thickness of the tube is negligible since tube is thin-walled. 5 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the water and chemical are given to be 4.18 and 1.8 kJ/kg.°C, respectively. Analysis The heat capacity rates of the hot and cold fluids are C kW/ 5.40 = C) kJ/kg. kg/s)(1.8 (3 C kW/ 8.36 = C) kJ/kg. kg/s)(4.18 (2 ° ° = = ° ° = = pc c c ph h h c m C c m C & & Therefore, C kW/ 4 . 5 min ° = = c C C and 646 . 0 36 . 8 40 . 5 max min = = = C C c Then the maximum heat transfer rate becomes kW 486 C) 20 -C C)(110 kW/ (5.4 ) ( , , min max = ° ° ° = − = in c in h T T C Q & The NTU of this heat exchanger is 556 . 1 C kW/ 4 . 5 ) m 7 ( C) . kW/m 2 . 1 ( 2 2 min = ° ° = = C UA NTU s Then the effectiveness of this parallel-flow heat exchanger corresponding to c = 0.646 and NTU=1.556 is determined from 56 . 0 646 . 0 1 )] 646 . 0 1 ( 556 . 1 exp[ 1 1 )] 1 ( exp[ 1 = + + − − = + + − − = c c NTU ε Then the actual rate of heat transfer rate becomes kW 2 . 272 kW) (0.56)(486 max = = = Q Q & & ε Finally, the outlet temperatures of the cold and hot fluid streams are determined to be C 77.4 C 70.4 ° = ° − ° = − = ⎯→ ⎯ − = ° = ° ° = + = ⎯→ ⎯ − = C kW/ 36 . 8 kW 2 . 272 C 110 ) ( C / kW 4 . 5 kW 2 . 272 + C 20 ) ( , , , , , , , , h in h out h out h in h h c in c out c in c out c c C Q T T T T C Q C Q T T T T C Q & & & & Hot Water 110°C 2 kg/s Chemical 20°C 3 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-58 16-97 EES Prob. 16-96 is reconsidered. The effects of the inlet temperatures of the chemical and the water on their outlet temperatures are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_chemical_in=20 [C] c_p_chemical=1.8 [kJ/kg-C] m_dot_chemical=3 [kg/s] T_w_in=110 [C] m_dot_w=2 [kg/s] c_p_w=4.18 [kJ/kg-C] A=7 [m^2] U=1.2 [kW/m^2-C] "ANALYSIS" "With EES, it is easier to solve this problem using LMTD method than NTU method. Below, we use LMTD method. Both methods give the same results." DELTAT_1=T_w_in-T_chemical_in DELTAT_2=T_w_out-T_chemical_out DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=UADELTAT_lm Q_dot=m_dot_chemicalc_p_chemical(T_chemical_out-T_chemical_in) Q_dot=m_dot_wc_p_w(T_w_in-T_w_out) Tchemical, in [C] Tchemical, out [C] 10 66.06 12 66.94 14 67.82 16 68.7 18 69.58 20 70.45 22 71.33 24 72.21 26 73.09 28 73.97 30 74.85 32 75.73 34 76.61 36 77.48 38 78.36 40 79.24 42 80.12 44 81 46 81.88 48 82.76 50 83.64 10 15 20 25 30 35 40 45 50 65 69 73 77 81 85 Tchemical,in [C] Tchemical,out [C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-59 Tw, in [C] Tw, out [C] 80 58.27 85 61.46 90 64.65 95 67.84 100 71.03 105 74.22 110 77.41 115 80.6 120 83.79 125 86.98 130 90.17 135 93.36 140 96.55 145 99.74 150 102.9 80 90 100 110 120 130 140 150 50 60 70 80 90 100 110 Tw,in [C] Tw,out [C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-60 16-98 Water is heated by hot air in a heat exchanger. The mass flow rates and the inlet temperatures are given. The heat transfer surface area of the heat exchanger on the water side is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the water and air are given to be 4.18 and 1.01kJ/kg.°C, respectively. Analysis The heat capacity rates of the hot and cold fluids are C kW/ 9.09 = C) kJ/kg. kg/s)(1.01 (9 C kW/ 16.72 = C) kJ/kg. kg/s)(4.18 (4 ° ° = = ° ° = = pc c c ph h h c m C c m C & & Therefore, C kW/ 09 . 9 min ° = = c C C and 544 . 0 72 . 16 09 . 9 max min = = = C C C Then the NTU of this heat exchanger corresponding to c = 0.544 and ε = 0.65 is determined from Fig. 16-26 to be NTU = 1.5 Then the surface area of this heat exchanger becomes 2 m 52.4 = ° ° = = ⎯→ ⎯ = C . kW/m 260 . 0 ) C kW/ 09 . 9 )( 5 . 1 ( NTU NTU 2 min min U C A C UA s s Hot Air 100°C 9 kg/s Water 20°C, 4 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-61 16-99 Water is heated by steam condensing in a condenser. The required length of the tube is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heat of the water is given to be 4.18 kJ/kg.°C. The heat of vaporization of water at 120°C is given to be 2203 kJ/kg. Analysis (a) The temperature differences between the steam and the water at the two ends of the condenser are C 103 = C 17 C 120 C 40 = C 80 C 120 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T The logarithmic mean temperature difference is C 6 . 66 ) 103 / 40 ln( 103 40 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The rate of heat transfer is determined from kW 474.0 = C) 17 C C)(80 kJ/kg. kg/s)(4.18 8 . 1 ( ) ( , , ° − ° ° = − = in c out c pc c T T c m Q & & The surface area of heat transfer is m 10.17 = ) C 6 . 66 ( C) . kW/m 7 . 0 kW 474.0 = Q = = 2 2 ° ° Δ Δ ⎯→ ⎯ lm s lm s T U A T UA Q & & The length of tube required then becomes m 129.5 = = = ⎯→ ⎯ = m) (0.025 m 17 . 10 2 π π π D A L DL A s s (b) The maximum rate of heat transfer rate is kW 775.0 = C) 17 -C C)(120 kJ/kg. kg/s)(4.18 8 . 1 ( ) ( , , min max ° ° ° = − = in c in h T T C Q & Then the effectiveness of this heat exchanger becomes 612 . 0 kW 775 kW 474 max = = = Q Q & & ε The NTU of this heat exchanger is determined using the relation in Table 16-5 to be 947 . 0 ) 612 . 0 1 ln( ) 1 ln( NTU = − − = − − = ε The surface area is 2 2 min min m 18 . 10 C . kW/m 7 . 0 C) kJ/kg. kg/s)(4.18 8 . 1 )( 947 . 0 ( = ° ° = = ⎯→ ⎯ = U C NTU A C UA NTU s s Finally, the length of tube required is m 129.6 = = = ⎯→ ⎯ = m) (0.025 m 18 . 10 2 π π π D A L DL A s s 80°C 120°C 120°C Steam Water 17°C 1.8 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-62 16-100 Ethanol is vaporized by hot oil in a double-pipe parallel-flow heat exchanger. The outlet temperature and the mass flow rate of oil are to be determined using the LMTD and NTU methods. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heat of oil is given to be 2.2 kJ/kg.°C. The heat of vaporization of ethanol at 78°C is given to be 846 kJ/kg. Analysis (a) The rate of heat transfer is kW 25.38 = kJ/kg) kg/s)(846 03 . 0 ( = = fg h m Q & & The log mean temperature difference is C 8 . 12 ) m 2 . 6 ( C) . W/m 320 ( W 380 , 25 2 2 ° = ° = = Δ ⎯→ ⎯ Δ = s lm lm s UA Q T T UA Q & & The outlet temperature of the hot fluid can be determined as follows C 78 C 42 = C 78 C 120 , , , 2 , , 1 ° − = − = Δ ° ° − ° = − = Δ out h out c out h in c in h T T T T T T T and C 8 . 12 )] 78 /( 42 ln[ ) 78 ( 42 ) / ln( , , 2 1 2 1 ° = − − − = Δ Δ Δ − Δ = Δ out h out h lm T T T T T T T whose solution is C 79.8° = out h T , Then the mass flow rate of the hot oil becomes kg/s 0.287 = ° − ° ° = − = ⎯→ ⎯ − = C) 8 . 79 C C)(120 J/kg. 2200 ( W 380 , 25 ) ( ) ( , , , , out h in h p out h in h p T T c Q m T T c m Q & & & & (b) The heat capacity rate p c m C & = of a fluid condensing or evaporating in a heat exchanger is infinity, and thus 0 / max min = = C C C . The effectiveness in this case is determined from NTU e− − =1 ε where C) J/kg. kg/s)(2200 , ( ) m C)(6.2 . W/m (320 2 2 min ° ° = = m C UA NTU s & and ) ( , , min max in c in h T T C Q − = & 78 120 120 ) ( ) ( , , , min , , min max − − = − − = = out h in c in h in c in h T T T C T T C Q Q ε W 380 , 25 ) 120 ( 2200 W 380 , 25 ) ( , , , = − × = = − = out h out h in h h T m Q T T C Q & & & (1) Also 2200 320 2 . 6 , 1 78 120 120 × × − − = − − m out h e T & (2) Solving (1) and (2) simultaneously gives C 79.8 kg/s 0.287 ° = = out h h T m , and & Ethanol 78°C 0.03 kg/s Oil 120°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-63 16-101 Water is heated by solar-heated hot air in a heat exchanger. The mass flow rates and the inlet temperatures are given. The outlet temperatures of the water and the air are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the water and air are given to be 4.18 and 1.01 kJ/kg.°C, respectively. Analysis The heat capacity rates of the hot and cold fluids are C W/ 418 C) J/kg. kg/s)(4180 (0.1 C W/ 303 C) J/kg. 1010 (0.3kg/s)( ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore, C W/ 303 min ° = = c C C and 725 . 0 418 303 max min = = = C C c Then the maximum heat transfer rate becomes kW 604 , 20 C) 22 -C C)(90 W/ (303 ) ( , , min max = ° ° ° = − = in c in h T T C Q & The heat transfer surface area is 2 m 45 . 0 m) m)(12 )(0.012 ( = = = π πDL As Then the NTU of this heat exchanger becomes 119 . 0 C W/ 303 ) m 45 . 0 ( C) . W/m 80 ( 2 2 min = ° ° = = C UA NTU s The effectiveness of this counter-flow heat exchanger corresponding to c = 0.725 and NTU = 0.119 is determined using the relation in Table 16-4 to be 108 . 0 )] 725 . 0 1 ( 119 . 0 exp[ 725 . 0 1 )] 725 . 0 1 ( 119 . 0 exp[ 1 )] 1 ( exp[ 1 )] 1 ( exp[ 1 = − − − − − − = − − − − − − = c NTU c c NTU ε Then the actual rate of heat transfer becomes W 2 . 2225 W) ,604 (0.108)(20 max = = = Q Q & & ε Finally, the outlet temperatures of the cold and hot fluid streams are determined to be C 82.7 C 27.3 ° = ° − ° = − = ⎯→ ⎯ − = ° = ° + ° = + = ⎯→ ⎯ − = C W/ 303 W 2 . 2225 C 90 ) ( C / W 418 W 2 . 2225 C 22 ) ( , , , , , , , , h in h out h out h in h h c in c out c in c out c c C Q T T T T C Q C Q T T T T C Q & & & & Hot Air 90°C 0.3 kg/s Cold Water 22°C 0.1 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-64 16-102 EES Prob. 16-101 is reconsidered. The effects of the mass flow rate of water and the tube length on the outlet temperatures of water and air are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_air_in=90 [C] m_dot_air=0.3 [kg/s] c_p_air=1.01 [kJ/kg-C] T_w_in=22 [C] m_dot_w=0.1 [kg/s] c_p_w=4.18 [kJ/kg-C] U=0.080 [kW/m^2-C] L=12 [m] D=0.012 [m] "ANALYSIS" "With EES, it is easier to solve this problem using LMTD method than NTU method. Below, we use LMTD method. Both methods give the same results." DELTAT_1=T_air_in-T_w_out DELTAT_2=T_air_out-T_w_in DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) A=piDL Q_dot=UADELTAT_lm Q_dot=m_dot_airc_p_air(T_air_in-T_air_out) Q_dot=m_dot_wc_p_w(T_w_out-T_w_in) mw [kg/s] Tw, out [C] Tair, out [C] 0.05 32.27 82.92 0.1 27.34 82.64 0.15 25.6 82.54 0.2 24.72 82.49 0.25 24.19 82.46 0.3 23.83 82.44 0.35 23.57 82.43 0.4 23.37 82.42 0.45 23.22 82.41 0.5 23.1 82.4 0.55 23 82.4 0.6 22.92 82.39 0.65 22.85 82.39 0.7 22.79 82.39 0.75 22.74 82.38 0.8 22.69 82.38 0.85 22.65 82.38 0.9 22.61 82.38 0.95 22.58 82.38 1 22.55 82.37 0 0.2 0.4 0.6 0.8 1 22 24.2 26.4 28.6 30.8 33 82.2 82.3 82.4 82.5 82.6 82.7 82.8 82.9 83 mw [kg/s] Tw,out [C] Tair,out [C] Tw,out Tair,out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-65 L [m] Tw, out [C] Tair, out [C] 5 24.35 86.76 6 24.8 86.14 7 25.24 85.53 8 25.67 84.93 9 26.1 84.35 10 26.52 83.77 11 26.93 83.2 12 27.34 82.64 13 27.74 82.09 14 28.13 81.54 15 28.52 81.01 16 28.9 80.48 17 29.28 79.96 18 29.65 79.45 19 30.01 78.95 20 30.37 78.45 21 30.73 77.96 22 31.08 77.48 23 31.42 77 24 31.76 76.53 25 32.1 76.07 5 9 13 17 21 25 24 25 26 27 28 29 30 31 32 33 76 78 80 82 84 86 88 L [m] Tw,out [C] Tair,out [C] Tw,out Tair,out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-66 16-103E Oil is cooled by water in a double-pipe heat exchanger. The overall heat transfer coefficient of this heat exchanger is to be determined using both the LMTD and NTU methods. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The thickness of the tube is negligible since it is thin-walled. Properties The specific heats of the water and oil are given to be 1.0 and 0.525 Btu/lbm.°F, respectively. Analysis (a) The rate of heat transfer is Btu/s 511.9 F) 105 F)(300 Btu/lbm. 25 lbm/s)(0.5 (5 ) ( , , = ° − ° = = − out h in h ph h T T c m Q & & The outlet temperature of the cold fluid is F 6 . 240 ) F Btu/lbm. lbm/s)(1.0 3 ( Btu/s 9 . 511 F 70 ) ( , , , , ° = ° + ° = + = ⎯→ ⎯ − = pc c in c out c in c out c pc c c m Q T T T T c m Q & & & & The temperature differences between the two fluids at the two ends of the heat exchanger are F 35 = F 70 F 105 F 59.4 = F 6 . 240 F 300 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T The logarithmic mean temperature difference is F 1 . 46 /35) 4 . 59 ln( 35 4 . 59 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm Then the overall heat transfer coefficient becomes F . Btu/s.ft 0.0424 2 ° ° Δ ⎯→ ⎯ Δ = ) F ft)(46.1 200 )( m 12 / 5 ( Btu/s 511.9 = = = π lm s lm s T A Q U T UA Q & & (b) The heat capacity rates of the hot and cold fluids are F Btu/s. 0 . 3 F) Btu/lbm. lbm/s)(1.0 (3 F Btu/s. 625 . 2 F) Btu/lbm. 25 lbm/s)(0.5 (5 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore, F Btu/s. 625 . 2 min ° = = h C C and 875 . 0 0 . 3 625 . 2 max min = = = C C c Then the maximum heat transfer rate becomes Btu/s 75 . 603 F) 70 -F F)(300 Btu/s. (2.625 ) ( , , min max = ° ° ° = − = in c in h T T C Q & The actual rate of heat transfer and the effectiveness are Btu/s 9 . 511 F) 105 -F F)(300 Btu/s. (2.625 ) ( , , = ° ° ° = − = out h in h h T T C Q & 85 . 0 75 . 603 9 . 511 max = = = Q Q & & ε The NTU of this heat exchanger is determined using the relation in Table 16-5 to be 28 . 4 1 875 . 0 85 . 0 1 85 . 0 ln 1 875 . 0 1 1 1 ln 1 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − = c c NTU ε ε The heat transfer surface area of the heat exchanger is 2 ft 8 . 261 ) ft 200 )( ft 12 / 5 ( = = = π πDL As and F . Btu/s.ft 0.0429 2 ° = ° = = ⎯→ ⎯ = 2 min min ft 261.8 ) F Btu/s. 625 . 2 )( 28 . 4 ( s s A C NTU U C UA NTU Hot Oil 300°F 5 lbm/s Cold Water 70°F 3 lbm/s 105°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-67 16-104 Cold water is heated by hot water in a heat exchanger. The net rate of heat transfer and the heat transfer surface area of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. 5 The thickness of the tube is negligible. Properties The specific heats of the cold and hot water are given to be 4.18 and 4.19 kJ/kg.°C, respectively. Analysis The heat capacity rates of the hot and cold fluids are C W/ 570 , 12 C) J/kg. kg/s)(4190 (3 C W/ 1045 C) J/kg. kg/s)(4180 (0.25 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore, C W/ 1045 min ° = = c C C and 083 . 0 570 , 12 1045 max min = = = C C c Then the maximum heat transfer rate becomes W 825 , 88 C) 15 -C C)(100 W/ (1045 ) ( , , min max = ° ° ° = − = in c in h T T C Q & The actual rate of heat transfer is W 31,350 = ° − ° ° = − = ) C 15 C 45 )( C W/ 1045 ( ) ( , , out h in h h T T C Q & Then the effectiveness of this heat exchanger becomes 35 . 0 825 , 88 350 , 31 max = = = Q Q ε The NTU of this heat exchanger is determined using the relation in Table 16-5 to be 438 . 0 1 083 . 0 35 . 0 1 35 . 0 ln 1 083 . 0 1 1 1 ln 1 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − = c c NTU ε ε Then the surface area of the heat exchanger is determined from 2 m 0.482 = ° ° = = ⎯→ ⎯ = C . W/m 950 ) C W/ 1045 )( 438 . 0 ( 2 min min U C NTU A C UA NTU Hot Water 100°C 3 kg/s Cold Water 15°C 0.25 kg/s 45°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-68 16-105 EES Prob. 16-104 is reconsidered. The effects of the inlet temperature of hot water and the heat transfer coefficient on the rate of heat transfer and the surface area are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_cw_in=15 [C] T_cw_out=45 [C] m_dot_cw=0.25 [kg/s] c_p_cw=4.18 [kJ/kg-C] T_hw_in=100 [C] m_dot_hw=3 [kg/s] c_p_hw=4.19 [kJ/kg-C] U=0.95 [kW/m^2-C] "ANALYSIS" "With EES, it is easier to solve this problem using LMTD method than NTU method. Below, we use LMTD method. Both methods give the same results." DELTAT_1=T_hw_in-T_cw_out DELTAT_2=T_hw_out-T_cw_in DELTAT_lm=(DELTAT_1-DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=UADELTAT_lm Q_dot=m_dot_hwc_p_hw(T_hw_in-T_hw_out) Q_dot=m_dot_cwc_p_cw(T_cw_out-T_cw_in) Thw, in [C] Q [kW] A [m2] 60 31.35 1.25 65 31.35 1.038 70 31.35 0.8903 75 31.35 0.7807 80 31.35 0.6957 85 31.35 0.6279 90 31.35 0.5723 95 31.35 0.5259 100 31.35 0.4865 105 31.35 0.4527 110 31.35 0.4234 115 31.35 0.3976 120 31.35 0.3748 U [kW/m2-C] Q [kW] A [m2] 0.75 31.35 0.6163 0.8 31.35 0.5778 0.85 31.35 0.5438 0.9 31.35 0.5136 0.95 31.35 0.4865 1 31.35 0.4622 1.05 31.35 0.4402 1.1 31.35 0.4202 1.15 31.35 0.4019 1.2 31.35 0.3852 1.25 31.35 0.3698 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-69 60 70 80 90 100 110 120 31 31.25 31.5 31.75 32 0.2 0.4 0.6 0.8 1 1.2 1.4 Thw,in [C] Q [kW] A [m 2] heat area 0.7 0.8 0.9 1 1.1 1.2 1.3 31 31.25 31.5 31.75 32 0.35 0.4 0.45 0.5 0.55 0.6 0.65 U [kW/m2-C] Q [kW] A [m 2] heat area PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-70 16-106 Glycerin is heated by ethylene glycol in a heat exchanger. Mass flow rates and inlet temperatures are given. The rate of heat transfer and the outlet temperatures are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. 5 The thickness of the tube is negligible. Properties The specific heats of the glycerin and ethylene glycol are given to be 2.4 and 2.5 kJ/kg.°C, respectively. Analysis (a) The heat capacity rates of the hot and cold fluids are C W/ 750 C) J/kg. kg/s)(2500 (0.3 C W/ 720 C) J/kg. kg/s)(2400 (0.3 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore, C W/ 720 min ° = = h C C and 96 . 0 750 720 max min = = = C C c Then the maximum heat transfer rate becomes kW 8 . 28 C) 20 C C)(60 W/ (720 ) ( , , min max = ° − ° ° = − = in c in h T T C Q & The NTU of this heat exchanger is 797 . 2 C W/ 720 ) m 3 . C)(5 . W/m 380 ( 2 2 min = ° ° = = C UA NTU s Effectiveness of this heat exchanger corresponding to c = 0.96 and NTU = 2.797 is determined using the proper relation in Table 16-4 508 . 0 96 . 0 1 )] 96 . 0 1 ( 797 . 2 exp[ 1 1 )] 1 ( exp[ 1 = + + − − = + + − − = c c NTU ε Then the actual rate of heat transfer becomes kW 14.63 = = = kW) .8 (0.508)(28 max Q Q & & ε (b) Finally, the outlet temperatures of the cold and the hot fluid streams are determined from C 40.5 C 40.3 ° = ° − ° = − = ⎯→ ⎯ − = ° = ° ° = + = ⎯→ ⎯ − = C kW/ 75 . 0 kW 63 . 14 C 60 ) ( C / kW 72 . 0 kW 63 . 14 + C 20 ) ( , , , , , , , , h in h out h out h in h h c in c out c in c out c c C Q T T T T C Q C Q T T T T C Q & & & & Ethylene 60°C 0.3 kg/s Glycerin 20°C 0.3 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-71 16-107 Water is heated by hot air in a cross-flow heat exchanger. Mass flow rates and inlet temperatures are given. The rate of heat transfer and the outlet temperatures are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. 5 The thickness of the tube is negligible. Properties The specific heats of the water and air are given to be 4.18 and 1.01 kJ/kg.°C, respectively. Analysis The mass flow rates of the hot and the cold fluids are kg/s 6 . 169 /4] m) (0.03 m/s)[80 )(3 kg/m (1000 2 3 = = = π ρ c c VA m & 3 3 kg/m 908 . 0 K) 273 + (130 /kg.K) kPa.m (0.287 kPa 105 = × = = RT P air ρ kg/s 10.90 = m) m/s)(1 )(12 kg/m (0.908 2 3 = = c h VA m ρ & The heat transfer surface area and the heat capacity rates are 2 m 540 . 7 m) m)(1 03 . 0 ( 80 = = = π πDL n As C kW/ 01 . 11 C) kJ/kg. 0 kg/s)(1.01 (10.9 C kW/ 9 . 708 C) kJ/kg. kg/s)(4.18 (169.6 ° = ° = = ° = ° = = ph h h pc c c c m C c m C & & Therefore, C kW/ 01 . 11 min ° = = c C C and 01553 . 0 9 . 708 01 . 11 max min = = = C C c kW 1233 C) 18 C C)(130 kW/ (11.01 ) ( , , min max = ° − ° ° = − = in c in h T T C Q & The NTU of this heat exchanger is 08903 . 0 C W/ 010 , 11 ) m (7.540 C) . W/m 130 ( 2 2 min = ° ° = = C UA NTU s Noting that this heat exchanger involves mixed cross-flow, the fluid with min C is mixed, max C unmixed, effectiveness of this heat exchanger corresponding to c = 0.01553 and NTU =0.08903 is determined using the proper relation in Table 16-4 to be 08513 . 0 ) 1 ( 01553 . 0 1 exp 1 ) 1 ( 1 exp 1 08903 . 0 01553 . 0 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − = × − − e e c cNTU ε Then the actual rate of heat transfer becomes kW 105.0 = = = kW) 1233 (0.08513)( max Q Q & & ε Finally, the outlet temperatures of the cold and the hot fluid streams are determined from C 120.5 C 18.15 ° = ° − ° = − = ⎯→ ⎯ − = ° = ° ° = + = ⎯→ ⎯ − = C kW/ 01 . 11 kW 0 . 105 C 130 ) ( C / kW 9 . 708 kW 0 . 105 + C 18 ) ( , , , , , , , , h in h out h out h in h h c in c out c in c out c c C Q T T T T C Q C Q T T T T C Q & & & & Hot Air 130°C 105 kPa 12 m/s Water 18°C, 3 m/s 1 m 1 m 1 m PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-72 16-108 CD EES Ethyl alcohol is heated by water in a shell-and-tube heat exchanger. The heat transfer surface area of the heat exchanger is to be determined using both the LMTD and NTU methods. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the ethyl alcohol and water are given to be 2.67 and 4.19 kJ/kg.°C, respectively. Analysis (a) The temperature differences between the two fluids at the two ends of the heat exchanger are C 35 = C 25 C 60 C 25 = C 70 C 95 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T The logarithmic mean temperature difference and the correction factor are C 7 . 29 /35) 25 ln( 35 25 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 93 . 0 78 . 0 25 70 60 95 64 . 0 25 95 25 70 1 1 1 2 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P The rate of heat transfer is determined from kW 3 . 252 C) 25 C C)(70 kJ/kg. kg/s)(2.67 1 . 2 ( ) ( , , = ° − ° ° = − = in c out c pc c T T c m Q & & The surface area of heat transfer is = ) C 7 . 29 )( 93 . 0 )( C . kW/m 8 . 0 kW 252.3 = = 2 2 m 11.4 ° ° Δ = ⎯→ ⎯ Δ lm s lm s T UF Q A T UA Q & & (b) The rate of heat transfer is kW 3 . 252 C) 25 C C)(70 kJ/kg. kg/s)(2.67 1 . 2 ( ) ( , , = ° − ° ° = − = in c out c pc c T T c m Q & & The mass flow rate of the hot fluid is kg/s 72 . 1 ) C 60 C 95 )( C kJ/kg. (4.19 kW 3 . 252 ( ) ( ) , , , , = ° − ° ° = − = → − = out h in h ph h out h in h ph h T T c Q m T T c m Q & & & & The heat capacity rates of the hot and the cold fluids are C kW/ 61 . 5 C) kJ/kg. kg/s)(2.67 (2.1 C kW/ 21 . 7 C) kJ/kg. kg/s)(4.19 (1.72 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore, C W/ 61 . 5 min ° = = c C C and 78 . 0 21 . 7 61 . 5 max min = = = C C c Then the maximum heat transfer rate becomes kW 7 . 392 C) 25 C C)(95 W/ (5.61 ) ( , , min max = ° − ° ° = − = in c in h T T C Q & The effectiveness of this heat exchanger is 64 . 0 7 . 392 3 . 252 max = = = Q Q ε The NTU of this heat exchanger corresponding to this emissivity and c = 0.78 is determined from Fig. 16-26d to be NTU = 1.7. Then the surface area of heat exchanger is determined to be 2 m 11.9 = ° ° = = ⎯→ ⎯ = C . kW/m 8 . 0 ) C kW/ 61 . 5 )( 7 . 1 ( 2 min min U C NTU A C UA NTU s s The small difference between the two results is due to the reading error of the chart. Water 95°C Alcohol 25°C 2.1 kg/s 70°C 2-shell pass 8 tube passes 60°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-73 16-109 Steam is condensed by cooling water in a shell-and-tube heat exchanger. The rate of heat transfer and the rate of condensation of steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. 5 The thickness of the tube is negligible. Properties The specific heat of the water is given to be 4.18 kJ/kg.°C. The heat of condensation of steam at 30°C is given to be 2430 kJ/kg. Analysis (a) The heat capacity rate of a fluid condensing in a heat exchanger is infinity. Therefore, C kW/ 09 . 2 C) kJ/kg. kg/s)(4.18 (0.5 min ° = ° = = = pc c c c m C C & and c = 0 Then the maximum heat transfer rate becomes kW 35 . 31 C) 15 C C)(30 kW/ (2.09 ) ( , , min max = ° − ° ° = − = in c in h T T C Q & and 2 m 7 . 37 ) m 2 )( m 015 . 0 ( 50 8 8 = × = = π πDL n As The NTU of this heat exchanger 11 . 54 C kW/ 09 . 2 ) m (37.7 C) . kW/m 3 ( 2 2 min = ° ° = = C UA NTU s Then the effectiveness of this heat exchanger corresponding to c = 0 and NTU = 54.11 is determined using the proper relation in Table 16-5 1 ) 11 . 54 exp( 1 ) NTU exp( 1 = − − = − − = ε Then the actual heat transfer rate becomes kW 31.35 = = = kW) (1)(31.35 max Q Q & & ε (b) Finally, the rate of condensation of the steam is determined from kg/s 0.0129 = = = ⎯→ ⎯ = kJ/kg 2430 kJ/s 35 . 31 fg fg h Q m h m Q & & & & Steam 30°C 15°C Water 1800 kg/h 30°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-74 16-110 EES Prob. 16-109 is reconsidered. The effects of the condensing steam temperature and the tube diameter on the rate of heat transfer and the rate of condensation of steam are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" N_pass=8 N_tube=50 T_steam=30 [C] h_fg_steam=2430 [kJ/kg] T_w_in=15 [C] m_dot_w=1800[kg/h]Convert(kg/h, kg/s) c_p_w=4.18 [kJ/kg-C] D=1.5 [cm] L=2 [m] U=3 [kW/m^2-C] "ANALYSIS" "With EES, it is easier to solve this problem using LMTD method than NTU method. Below, we use NTU method. Both methods give the same results." C_min=m_dot_wc_p_w c=0 "since the heat capacity rate of a fluid condensing is infinity" Q_dot_max=C_min(T_steam-T_w_in) A=N_passN_tubepiDLConvert(cm, m) NTU=(UA)/C_min epsilon=1-exp(-NTU) "from Table 16-4 of the text with c=0" Q_dot=epsilonQ_dot_max Q_dot=m_dot_condh_fg_steam Tsteam [C] Q [kW] mcond [kg/s] 20 10.45 0.0043 22.5 15.68 0.006451 25 20.9 0.008601 27.5 26.12 0.01075 30 31.35 0.0129 32.5 36.58 0.01505 35 41.8 0.0172 37.5 47.03 0.01935 40 52.25 0.0215 42.5 57.47 0.02365 45 62.7 0.0258 47.5 67.93 0.02795 50 73.15 0.0301 52.5 78.38 0.03225 55 83.6 0.0344 57.5 88.82 0.03655 60 94.05 0.0387 62.5 99.27 0.04085 65 104.5 0.043 67.5 109.7 0.04515 70 114.9 0.0473 20 30 40 50 60 70 0 20 40 60 80 100 120 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Tsteam [C] Q [kW] mcond [kg/s] heat mass rate PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-75 D [cm] Q [kW] mcond [kg/s] 1 31.35 0.0129 1.05 31.35 0.0129 1.1 31.35 0.0129 1.15 31.35 0.0129 1.2 31.35 0.0129 1.25 31.35 0.0129 1.3 31.35 0.0129 1.35 31.35 0.0129 1.4 31.35 0.0129 1.45 31.35 0.0129 1.5 31.35 0.0129 1.55 31.35 0.0129 1.6 31.35 0.0129 1.65 31.35 0.0129 1.7 31.35 0.0129 1.75 31.35 0.0129 1.8 31.35 0.0129 1.85 31.35 0.0129 1.9 31.35 0.0129 1.95 31.35 0.0129 2 31.35 0.0129 1 1.2 1.4 1.6 1.8 2 31 31.25 31.5 31.75 32 0.0125 0.013 0.0135 D [cm] Q [kW] mcond [kg/s] Qdot mcond PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-76 16-111 Cold water is heated by hot oil in a shell-and-tube heat exchanger. The rate of heat transfer is to be determined using both the LMTD and NTU methods. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the water and oil are given to be 4.18 and 2.2 kJ/kg.°C, respectively. Analysis (a) The LMTD method in this case involves iterations, which involves the following steps: 1) Choose out h T , 2) Calculate Q & from ) ( , , in h out h p h T T c m Q − = & & 3) Calculate out h T , from ) ( , , in h out h p h T T c m Q − = & & 4) Calculate CF lm T , Δ 5) Calculate Q & from CF lm s T F UA Q , Δ = & 6) Compare to the Q & calculated at step 2, and repeat until reaching the same result Result: 651 kW (b) The heat capacity rates of the hot and the cold fluids are C kW/ 54 . 12 C) kJ/kg. kg/s)(4.18 (3 C kW/ 6 . 6 C) kJ/kg. kg/s)(2.2 (3 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore, C kW/ 6 . 6 min ° = = h C C and 53 . 0 54 . 12 6 . 6 max min = = = C C c Then the maximum heat transfer rate becomes kW 1228 C) 14 C C)(200 kW/ (6.6 ) ( , , min max = ° − ° ° = − = in c in h T T C Q & The NTU of this heat exchanger is 91 . 0 C kW/ 6 . 6 ) m (20 C) . kW/m 3 . 0 ( 2 2 min = ° ° = = C UA NTU s Then the effectiveness of this heat exchanger corresponding to c = 0.53 and NTU = 0.91 is determined from Fig. 16-26d to be 53 . 0 = ε The actual rate of heat transfer then becomes kW 651 = = = kW) 8 (0.53)(122 max Q Q & & ε Hot oil 200°C 3 kg/s Water 14°C 3 kg/s (20 tube passes) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-77 Selection of the Heat Exchangers 16-112C 1) Calculate heat transfer rate, 2) select a suitable type of heat exchanger, 3) select a suitable type of cooling fluid, and its temperature range, 4) calculate or select U, and 5) calculate the size (surface area) of heat exchanger 16-113C The first thing we need to do is determine the life expectancy of the system. Then we need to evaluate how much the larger will save in pumping cost, and compare it to the initial cost difference of the two units. If the larger system saves more than the cost difference in its lifetime, it should be preferred. 16-114C In the case of automotive and aerospace industry, where weight and size considerations are important, and in situations where the space availability is limited, we choose the smaller heat exchanger. 16-115 Oil is to be cooled by water in a heat exchanger. The heat transfer rating of the heat exchanger is to be determined and a suitable type is to be proposed. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. Properties The specific heat of the oil is given to be 2.2 kJ/kg.°C. Analysis The heat transfer rate of this heat exchanger is kW 2002 = ° − ° ° = − = C) 50 C C)(120 kJ/kg. kg/s)(2.2 13 ( ) ( , , in c out c pc c T T c m Q & & We propose a compact heat exchanger (like the car radiator) if air cooling is to be used, or a tube-and-shell or plate heat exchanger if water cooling is to be used. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-78 3-116 Water is to be heated by steam in a shell-and-tube process heater. The number of tube passes need to be used is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. Properties The specific heat of the water is given to be 4.19 kJ/kg.°C. Analysis The mass flow rate of the water is kg/s 046 . 2 C) 20 C C)(90 kJ/kg. (4.19 kW 600 ) ( ) ( , , , , = ° − ° ° = − = − = in c out c pc in c out c pc c T T c Q m T T c m Q & & & & The total cross-section area of the tubes corresponding to this mass flow rate is 2 4 3 m 10 82 . 6 m/s) 3 )( kg/m 1000 ( kg/s 046 . 2 − × = = = → = V m A VA m c c ρ ρ & & Then the number of tubes that need to be used becomes 9 ≅ = × = = ⎯→ ⎯ = − 68 . 8 ) m 01 . 0 ( ) m 10 82 . 6 ( 4 4 4 2 2 4 2 2 π π π D A n D n A s s Therefore, we need to use at least 9 tubes entering the heat exchanger. Steam 20°C Water 90°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-79 16-117 EES Prob. 16-116 is reconsidered. The number of tube passes as a function of water velocity is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" c_p_w=4.19 [kJ/kg-C] T_w_in=20 [C] T_w_out=90 [C] Q_dot=600 [kW] D=0.01 [m] Vel=3 [m/s] "PROPERTIES" rho=density(water, T=T_ave, P=100) T_ave=1/2(T_w_in+T_w_out) "ANALYSIS" Q_dot=m_dot_wc_p_w(T_w_out-T_w_in) m_dot_w=rhoA_cVel A_c=N_passpiD^2/4 Vel [m/s] Npass 1 26.42 1.5 17.62 2 13.21 2.5 10.57 3 8.808 3.5 7.55 4 6.606 4.5 5.872 5 5.285 5.5 4.804 6 4.404 6.5 4.065 7 3.775 7.5 3.523 8 3.303 1 2 3 4 5 6 7 8 0 5 10 15 20 25 30 Vel [m/s] Npass PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-80 16-118 Cooling water is used to condense the steam in a power plant. The total length of the tubes required in the condenser is to be determined and a suitable HX type is to be proposed. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heat of the water is given to be 4.18 kJ/kg.°C. The heat of condensation of steam at 30°C is given to be 2431 kJ/kg. Analysis The temperature differences between the steam and the water at the two ends of condenser are C 12 = C 18 C 30 C 4 = C 26 C 30 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and the logarithmic mean temperature difference is ( ) C 28 . 7 /12 4 ln 12 4 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The heat transfer surface area is m 10 1.96 = ) C 28 . 7 )( C . W/m 3500 ( W 10 500 = = 2 4 2 6 × ° ° × Δ = ⎯→ ⎯ Δ lm s lm s T U Q A T UA Q & & The total length of the tubes required in this condenser then becomes km 312.3 = × = × = = ⎯→ ⎯ = m 10 123 . 3 m) 02 . 0 ( m 10 96 . 1 5 2 4 π π π D A L DL A s s A multi-pass shell-and-tube heat exchanger is suitable in this case. Steam 30°C 18°C Water 30°C 26°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-81 16-119 Cold water is heated by hot water in a heat exchanger. The net rate of heat transfer and the heat transfer surface area of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the cold and hot water are given to be 4.18 and 4.19 kJ/kg.°C, respectively. Analysis The temperature differences between the steam and the water at the two ends of condenser are C 12 = C 18 C 30 C 4 = C 26 C 30 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T and the logarithmic mean temperature difference is C 28 . 7 ln(4/12) 12 4 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The heat transfer surface area is m 1962 = ) C 28 . 7 )( C . W/m 3500 ( W 10 50 = = 2 2 6 ° ° × Δ = ⎯→ ⎯ Δ lm s lm s T U Q A T UA Q & & The total length of the tubes required in this condenser then becomes km 31.23 = = = = ⎯→ ⎯ = m 231 , 31 m) 02 . 0 ( m 1962 2 π π π D A L DL A s s A multi-pass shell-and-tube heat exchanger is suitable in this case. Steam 30°C 18°C Water 30°C 26°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-82 Review Problems 16-120 The inlet conditions of hot and cold fluid streams in a heat exchanger are given. The outlet temperatures of both streams are to be determined using LMTD and the effectiveness-NTU methods. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of hot and cold fluid streams are given to be 2.0 and 4.2 kJ/kg.°C, respectively. Analysis (a) The rate of heat transfer can be expressed as ) 1.5(120 ) C)(120 kJ/kg. kg/s)(2.0 3600 / 2700 ( ) ( , , , , out h out h out h in h p T T T T c m Q − = − ° = − = & & (1) 20) ( 1 . 2 20) C)( kJ/kg. kg/s)(4.2 3600 / 1800 ( ) ( , , , , − = − ° = − = out c out c in c out c p T T T T c m Q & & (2) The heat transfer can also be expressed using the logarithmic mean temperature difference as C 100 C 20 C 120 , , 1 ° = ° − ° = − = Δ in c in h T T T out c out h T T T , , 2 − = Δ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ Δ Δ − Δ = Δ out c out h out c out h lm T T T T T T T T T , , , , 2 1 2 1 100 ln ) ( 100 ln ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − ° ⋅ = Δ = Δ = out c out h out c out h out c out h out c out h lm m hc lm T T T T T T T T T A Q T UA Q , , , , , , , , 2 2 , 100 ln ) ( 100 100 ln ) ( 100 ) m 50 . 0 )( C kW/m 2.0 ( & & (3) Now we have three expressions for heat transfer with three unknowns: Q & , Th,out, Tc,out. Solving them using an equation solver such as EES, we obtain C 48.4 C 80.3 ° = ° = = out c out h T T Q , , kW 6 . 59 & (b) The heat capacity rates of the hot and cold fluids are C kW/ 1 . 2 C) kJ/kg. kg/s)(4.2 (1800/3600 C kW/ 5 . 1 C) kJ/kg. kg/s)(2.0 (2700/3600 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore C kW/ 5 . 1 min ° = = h C C which is the smaller of the two heat capacity rates. The heat capacity ratio and the NTU are 714 . 0 1 . 2 5 . 1 max min = = = C C c 20°C 1800 kg/h 120°C 2700 kg/h Th,out Tc,out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-83 667 . 0 C kW/ 5 . 1 ) m 50 . 0 )( C kW/m 0 . 2 ( 2 2 min = ° ⋅ = = C UA NTU The effectiveness of this parallel-flow heat exchanger is [ ] [ ] 397 . 0 714 . 0 1 ) 714 . 0 1 )( 667 . 0 ( exp 1 1 ) 1 ( exp 1 = + + − − = + + − − = c c NTU ε The maximum heat transfer rate is kW 150 = C) 20 C C)(120 kW/ (1.5 ) ( , , min max ° − ° ° = − = in c in h T T C Q & The actual heat transfer rate is kW 6 . 59 ) 150 )( 397 . 0 ( max = = = Q Q & & ε Then the outlet temperatures are determined to be C 48.4° = ° ° = + = ⎯→ ⎯ − = C kW/ 1 . 2 kW 59.6 + C 20 ) ( , , , , c in c out c in c out c c C Q T T T T C Q & & C 80.3° = ° ° = − = ⎯→ ⎯ − = C kW/ 5 . 1 kW 59.6 -C 120 ) ( , , , , h in h out h out h in h h C Q T T T T C Q & & Discussion The results obtained by two methods are same as expected. However, the effectiveness-NTU method is easier for this type of problems. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-84 16-121 Water is used to cool a process stream in a shell and tube heat exchanger. The tube length is to be determined for one tube pass and four tube pass cases. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The properties of process stream and water are given in problem statement. Analysis (a) The rate of heat transfer is kW 9870 C ) 100 160 ( C) kJ/kg kg/s)(3.5 47 ( ) ( , , = ° − ° ⋅ = − = out h in h h h T T c m Q & & The outlet temperature of water is determined from C 8 . 45 C) kJ/kg kg/s)(4.18 66 ( kW 9870 + C 10 ) ( , , , , ° = ° ⋅ ° = + = − = c c in c out c in c out c c c C m Q T T T T c m Q & & & & The logarithmic mean temperature difference is C 2 . 114 C 8 . 45 C 160 , , 1 ° = ° − ° = − = Δ out c in h T T T C 90 C 10 C 100 , , 2 ° = ° − ° = − = Δ in c out h T T T C 6 . 101 90 2 . 114 ln 90 2 . 114 ln 2 1 2 1 ° = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ Δ Δ − Δ = Δ T T T T Tlm The Reynolds number is 968 , 11 s kg/m 0.002 ) kg/m m)(950 m/s)(0.025 (1.008 Re m/s 008 . 1 4 / m) (0.025 ) kg/m (100)(950 kg/s) (47 4 / 3 2 3 2 = ⋅ = = = = = = μ ρ π ρπ ρ VD D N m A m V tube & & which is greater than 10,000. Therefore, we have turbulent flow. We assume fully developed flow and evaluate the Nusselt number from 9 . 92 ) 14 ( ) 968 , 11 ( 023 . 0 Pr Re 023 . 0 14 C W/m 0.50 C) J/kg s)(3500 kg/m 002 . 0 ( Pr 3 . 0 8 . 0 3 . 0 8 . 0 = = = = = ° ⋅ ° ⋅ ⋅ = = k hD Nu k c p μ Heat transfer coefficient on the inner surface of the tubes is C . W/m 1858 ) 9 . 92 ( m 025 . 0 C W/m. 50 . 0 2 ° = ° = = Nu D k hi Disregarding the thermal resistance of the tube wall the overall heat transfer coefficient is determined from C W/m 1269 4000 1 1858 1 1 1 1 1 2 ° ⋅ = + = + = o i h h U The correction factor for one shell pass and one tube pass heat exchanger is F = 1. The tube length is determined to be Water 10°C Process stream 160°C 100°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-85 [ ] m 9.75 = ° ⋅ = Δ = L L T UAF Q lm ) C 6 . 101 )( 1 ( m) 025 . 0 ( 100 ) C kW/m 269 . 1 ( kW 9870 2 π & (b) For 1 shell pass and 4 tube passes, there are 100/4=25 tubes per pass and this will increase the velocity fourfold. We repeat the calculations for this case as follows: 872 , 47 968 , 11 4 Re m/s 032 . 4 008 . 1 4 = × = = × = V 6 . 281 ) 14 ( ) 872 , 47 ( 023 . 0 Pr Re 023 . 0 3 . 0 8 . 0 3 . 0 8 . 0 = = = = k hD Nu C . W/m 5632 ) 6 . 281 ( m 025 . 0 C W/m. 50 . 0 2 ° = ° = = Nu D k hi C W/m 2339 4000 1 5632 1 1 1 1 1 2 ° ⋅ = + = + = o i h h U The correction factor is determined from Fig. 16-18: 96 . 0 60 . 0 160 100 8 . 45 10 4 . 0 160 10 160 100 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P The tube length is determined to be [ ] m 5.51 = ° ⋅ = Δ = L L T UAF Q lm ) C 6 . 101 )( 96 . 0 ( m) 025 . 0 ( 100 ) C kW/m 339 . 2 ( kW 9870 2 π & PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-86 16-122 A hydrocarbon stream is heated by a water stream in a 2-shell passes and 4-tube passes heat exchanger. The rate of heat transfer and the mass flow rates of both fluid streams and the fouling factor after usage are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heat of HC is given to be 2 kJ/kg.°C. The specific heat of water is taken to be 4.18 kJ/kg.°C. Analysis (a) The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 20 = C 20 C 40 C 30 = C 50 C 80 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 66 . 24 ) 20 / 30 ln( 20 30 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 90 . 0 33 . 1 20 50 40 80 5 . 0 20 80 20 50 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P (Fig. 16-18) The overall heat transfer coefficient of the heat exchanger is C W/m 6 . 975 2500 1 1600 1 1 1 1 1 2 ° ⋅ = + = + = o i h h U The rate of heat transfer in this heat exchanger is [ ] kW 326.5 = × = ° ° = Δ = W 10 265 . 3 C) 66 . 24 ( (0.90) m) m)(1.5 (0.02 160 C) . W/m 6 . 975 ( 5 2 , π CF lm s T F UA Q & The mass flow rates of fluid streams are kg/s 1.95 kg/s 5.44 = ° − ° ° = − = = ° − ° ° = − = C) 40 C C)(80 kJ/kg. (4.18 kW 5 . 326 ) ( C) 20 C C)(50 kJ/kg. (2.0 kW 5 . 326 ) ( out in p h in out p c T T c Q m T T c Q m & & & & (b) The rate of heat transfer in this case is kW 272 = C) 20 C C)(45 kJ/kg. kg/s)(2.0 44 . 5 ( )] ( [ c ° − ° ° = − = in out p T T c m Q & & This corresponds to a 17% decrease in heat transfer. The outlet temperature of the hot fluid is C 46.6 = ) C C)(80 kJ/kg. kg/s)(4.18 95 . 1 ( kW 272 )] ( [ , , h ° − ° ° = − = out h out h out in p T T T T c m Q & & The logarithmic temperature difference is C 26.6 = C 20 C 6 . 46 C 35 = C 45 C 80 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T Water 80°C HC 20°C 50°C 2 shell passes 4 tube passes 40°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-87 C 61 . 30 ) 6 . 26 / 35 ln( 6 . 26 35 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 97 . 0 34 . 1 20 45 6 . 46 80 42 . 0 20 80 20 45 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P (Fig. 16-18) The overall heat transfer coefficient is [ ] C . W/m 5 . 607 C) 61 . 30 ( (0.97) m) m)(1.5 (0.02 160 W 000 , 272 2 , ° = ° = Δ = U U T F UA Q CF lm s π & The fouling factor is determined from C/W m 10 6.21 2 4 ° ⋅ × = − = − = − 6 . 975 1 5 . 607 1 1 1 clean dirty f U U R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-88 16-123 Hot water is cooled by cold water in a 1-shell pass and 2-tube passes heat exchanger. The mass flow rates of both fluid streams are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There is no fouling. Properties The specific heats of both cold and hot water streams are taken to be 4.18 kJ/kg.°C. Analysis The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 29 = C 7 C 36 C 29 = C 31 C 60 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T Since 2 1 T T Δ = Δ , we have C 29 , ° = Δ CF lm T 88 . 0 0 . 1 60 36 31 7 45 . 0 60 7 60 31 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P (Fig. 16-18) The rate of heat transfer in this heat exchanger is kW 364 W 10 64 . 3 C) 29 ( )(0.88) m C)(15 . W/m 950 ( 5 2 2 , = × = ° ° = Δ = CF lm s T F UA Q & The mass flow rates of fluid streams are kg/s 3.63 kg/s 3.63 = ° − ° ° = − = = ° − ° ° = − = C) 7 C C)(31 kJ/kg. (4.18 kW 364 ) ( C) 36 C C)(60 kJ/kg. (4.18 kW 364 ) ( out in p h in out p c T T c Q m T T c Q m & & & & Water 7°C Water 60°C 36°C 1 shell pass 2 tube passes 31°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-89 16-124 Hot oil is cooled by water in a multi-pass shell-and-tube heat exchanger. The overall heat transfer coefficient based on the inner surface is to be determined. Assumptions 1 Water flow is fully developed. 2 Properties of the water are constant. Properties The properties of water at 25°C are (Table A-15) 14 . 6 Pr /s m 10 894 . 0 / C W/m. 607 . 0 2 6 = × = = ° = − ρ μ ν k Analysis The Reynolds number is 624 , 43 /s m 10 894 . 0 ) m 013 . 0 )( m/s 3 ( Re 2 6 avg = × = = − ν D V which is greater than 10,000. Therefore, we assume fully developed turbulent flow, and determine Nusselt number from 245 ) 14 . 6 ( ) 624 , 43 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = Nu and C . W/m 440 , 11 ) 245 ( m 013 . 0 C W/m. 607 . 0 2 ° = ° = = Nu D k hi The inner and the outer surface areas of the tube are 2 2 m 04712 . 0 ) m 1 )( m 015 . 0 ( m 04084 . 0 ) m 1 )( m 013 . 0 ( = = = = = = π π π π L D A L D A o o i i The total thermal resistance of this heat exchanger per unit length is C/W 609 . 0 ) m 04712 . 0 )( C . W/m 35 ( 1 ) m 1 )( C W/m. 110 ( 2 ) 3 . 1 / 5 . 1 ln( ) m 04084 . 0 )( C . W/m 440 , 11 ( 1 1 2 ) / ln( 1 2 2 2 2 ° = ° + ° + ° = + + = π π o o i o i i A h kL D D A h R Then the overall heat transfer coefficient of this heat exchanger based on the inner surface becomes C . W/m 40.2 2 ° = ° = = ⎯→ ⎯ = ) m 04084 . 0 )( C/W 609 . 0 ( 1 1 1 2 i i i i RA U A U R Outer surface D0, A0, h0, U0 Inner surface Di, Ai, hi, Ui PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-90 16-125 Hot oil is cooled by water in a multi-pass shell-and-tube heat exchanger. The overall heat transfer coefficient based on the inner surface is to be determined. Assumptions 1 Water flow is fully developed. 2 Properties of the water are constant. Properties The properties of water at 25°C are (Table A-15) 14 . 6 Pr /s m 10 894 . 0 / C W/m. 607 . 0 2 6 = × = = ° = − ρ μ ν k Analysis The Reynolds number is 624 , 43 /s m 10 894 . 0 ) m 013 . 0 )( m/s 3 ( Re 2 6 avg = × = = − ν D V which is greater than 10,000. Therefore, we assume fully developed turbulent flow, and determine Nusselt number from 245 ) 14 . 6 ( ) 624 , 43 ( 023 . 0 Pr Re 023 . 0 4 . 0 8 . 0 4 . 0 8 . 0 = = Nu and C . W/m 440 , 11 ) 245 ( m 013 . 0 C W/m. 607 . 0 2 ° = ° = = Nu D k hi The inner and the outer surface areas of the tube are 2 2 m 04712 . 0 ) m 1 )( m 015 . 0 ( m 04084 . 0 ) m 1 )( m 013 . 0 ( = = = = = = π π π π L D A L D A o o i i The total thermal resistance of this heat exchanger per unit length of it with a fouling factor is C/W 617 . 0 ) m 04712 . 0 )( C . W/m 35 ( 1 m 04712 . 0 C/W . m 0004 . 0 ) m 1 )( C W/m. 110 ( 2 ) 3 . 1 / 5 . 1 ln( ) m 04084 . 0 )( C . W/m 440 , 11 ( 1 1 2 ) / ln( 1 2 2 2 2 2 2 , ° = ° + ° + ° + ° = + + + = π π o o o o f i o i i A h A R kL D D A h R Then the overall heat transfer coefficient of this heat exchanger based on the inner surface becomes C . W/m 39.7 2 ° = ° = = ⎯→ ⎯ = ) m 04084 . 0 )( C/W 617 . 0 ( 1 1 1 2 i i i i RA U A U R Outer surface D0, A0, h0, U0 Inner surface Di, Ai, hi, Ui PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-91 16-126 Water is heated by hot oil in a multi-pass shell-and-tube heat exchanger. The rate of heat transfer and the heat transfer surface area on the outer side of the tube are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the water and oil are given to be 4.18 and 2.2 kJ/kg.°C, respectively. Analysis (a)The rate of heat transfer in this heat exchanger is kW 462 = C) 60 C C)(130 kJ/kg. kg/s)(2.2 3 ( ) ( , , ° − ° ° = − = out h in h ph h T T c m Q & & (b) The outlet temperature of the cold water is C 8 . 56 ) C kJ/kg. kg/s)(4.18 3 ( kW 462 C 20 ) ( , , , , ° = ° + ° = + = ⎯→ ⎯ − = pc c in c out c in c out c pc c c m Q T T T T c m Q & & & & The temperature differences at the two ends are C 40 = C 20 C 60 C 73.2 = C 8 . 56 C 130 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T The logarithmic mean temperature difference is C 9 . 54 ) 40 / 2 . 73 ln( 40 2 . 73 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm and 96 . 0 90 . 1 20 8 . 56 60 130 335 . 0 20 130 20 8 . 56 1 2 1 2 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P The heat transfer surface area on the outer side of the tube is then determined from 2 m 39.8 = ° ° = Δ = ⎯→ ⎯ Δ = C) 9 . 54 ( C)(0.96) . kW/m 22 . 0 ( kW 462 2 lm s lm s T UF Q A T F UA Q & & Hot Oil 130°C 3 kg/s Cold Water 20°C 3 kg/s (20 tube passes) 60°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-92 16-127E Water is heated by solar-heated hot air in a double-pipe counter-flow heat exchanger. The required length of the tube is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the water and air are given to be 1.0 and 0.24 Btu/lbm.°F, respectively. Analysis The rate of heat transfer in this heat exchanger is Btu/s 9.24 = F) 135 F F)(190 Btu/lbm. 4 lbm/s)(0.2 7 . 0 ( ) ( , , ° − ° ° = − = out h in h ph h T T c m Q & & The outlet temperature of the cold water is F 4 . 96 ) F Btu/lbm. lbm/s)(1.0 35 . 0 ( Btu/s 24 . 9 F 70 ) ( , , , , ° = ° + ° = + = ⎯→ ⎯ − = pc c in c out c in c out c pc c c m Q T T T T c m Q & & & & The temperature differences at the two ends are F 65 = F 70 F 135 F 93.6 = F 4 . 96 F 190 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T The logarithmic mean temperature difference is F 43 . 78 ) 65 / 6 . 93 ln( 65 6 . 93 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The heat transfer surface area on the outer side of the tube is determined from 2 2 ft 21 . 21 F) 43 . 78 ( F) . Btu/s.ft 3600 / 20 ( Btu/s 24 . 9 = ° ° = Δ = ⎯→ ⎯ Δ = lm s lm s T U Q A T UA Q & & Then the length of the tube required becomes ft 162.0 = = = ⎯→ ⎯ = ft) 12 / 5 . 0 ( ft 21 . 21 2 π π π D A L DL A s s 16-128 It is to be shown that when ΔT1 = ΔT2 for a heat exchanger, the ΔTlm relation reduces to ΔTlm = ΔT1 = ΔT2. Analysis When ΔT1 = ΔT2, we obtain 0 0 ) / ln( 2 1 2 1 = Δ Δ Δ − Δ = Δ T T T T Tlm This case can be handled by applying L'Hospital's rule (taking derivatives of nominator and denominator separately with respect to Δ Δ T T 1 2 or ). That is, 2 1 1 1 2 1 1 2 1 / 1 1 / )] / [ln( / ) ( T T T T d T T d T d T T d Tlm Δ = Δ = Δ = Δ Δ Δ Δ Δ − Δ = Δ Hot Air 190°F 0.7 lbm/s Cold Water 70°F 0.35 lbm/s 135°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-93 16-129 Refrigerant-134a is condensed by air in the condenser of a room air conditioner. The heat transfer area on the refrigerant side is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heat of air is given to be 1.005 kJ/kg.°C. Analysis The temperature differences at the two ends are C 15 = C 25 C 40 C 5 = C 35 C 40 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T The logarithmic mean temperature difference is C 1 . 9 ) 15 / 5 ln( 15 5 ) / ln( 2 1 2 1 ° = − = Δ Δ Δ − Δ = Δ T T T T Tlm The heat transfer surface area on the outer side of the tube is determined from 2 m 3.05 = ° ° = Δ = ⎯→ ⎯ Δ = C) 1 . 9 ( C) . kW/m 150 . 0 ( kW ) 3600 / 000 , 15 ( 2 lm s lm s T U Q A T UA Q & & 16-130 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of air and combustion gases are given to be 1.005 and 1.1 kJ/kg.°C, respectively. Analysis The rate of heat transfer is simply kW 60.8 = C) 95 C C)(180 kJ/kg. kg/s)(1.1 65 . 0 ( )] ( [ gas. ° − ° ° = − = out in p T T c m Q & & Air 25°C R-134a 40°C 40°C 35°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-94 16-131 A water-to-water heat exchanger is proposed to preheat the incoming cold water by the drained hot water in a plant to save energy. The heat transfer rating of the heat exchanger and the amount of money this heat exchanger will save are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. Properties The specific heat of the hot water is given to be 4.18 kJ/kg.°C. Analysis The maximum rate of heat transfer is kW 6 . 25 C) 14 C C)(60 kJ/kg. kg/s)(4.18 60 / 8 ( ) ( , , max = ° − ° ° = − = in c in h ph h T T c m Q & & Noting that the heat exchanger will recover 72% of it, the actual heat transfer rate becomes kW 18.43 = kJ/s) 6 . 25 )( 72 . 0 ( max = = Q Q & & ε which is the heat transfer rating. The operating hours per year are The annual operating hours = (8 h/day)(5 days/week)(52 week/year) = 2080 h/year The energy saved during the entire year will be Energy saved = (heat transfer rate)(operating time) = (18.43 kJ/s)(2080 h/year)(3600 s/h) = 1.38x108 kJ/year Then amount of fuel and money saved will be ar therms/ye 1677 kJ 500 , 105 therm 1 78 . 0 kJ/year 10 38 . 1 efficiency Furnace saved Energy saved Fuel 8 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × = = Money saved = (fuel saved)(the price of fuel) = (1677 therms/year)($1.00/therm) = $1677/year Hot water 60°C 8 kg/s Cold Water 14°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-95 16-132 A shell-and-tube heat exchanger is used to heat water with geothermal steam condensing. The rate of heat transfer, the rate of condensation of steam, and the overall heat transfer coefficient are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The heat of vaporization of geothermal water at 120°C is given to be hfg = 2203 kJ/kg and specific heat of water is given to be cp = 4180 J/kg.°C. Analysis (a) The outlet temperature of the water is C 74 = C 46 C 120 46 out h, out c, ° ° − ° = − = T T Then the rate of heat transfer becomes kW 847.7 = C) 22 C C)(74 kJ/kg. kg/s)(4.18 9 . 3 ( )] ( [ water in out ° − ° ° = − = T T c m Q p & & (b) The rate of condensation of steam is determined from kg/s 0.385 = ⎯→ ⎯ = = m m h m Q fg & & & & ) kJ/kg 2203 ( kW 7 . 847 ) ( steam geothermal (c) The heat transfer area is 2 m 3.378 = m) m)(3.2 024 . 0 ( 14π π = = L D n A i i The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 98 = C 22 C 120 C 46 = C 74 C 120 in c, out h, 2 out c, in h, 1 ° ° − ° = − = Δ ° ° − ° = − = Δ T T T T T T C 8 . 68 ) 98 / 46 ln( 98 46 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 1 0 22 74 120 120 53 . 0 22 120 22 74 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P Then the overall heat transfer coefficient is determined to be C . W/m 3650 2 ° = ° = Δ = ⎯→ ⎯ Δ = C) 8 . 68 )( 1 )( m 378 . 3 ( W 700 , 847 2 CF lm, CF lm, T F A Q U T F A U Q i i i i & & Steam 120°C 22°C Water 3.9 kg/s 14 tubes 120°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-96 16-133 Water is heated by geothermal water in a double-pipe counter-flow heat exchanger. The mass flow rate of the geothermal water and the outlet temperatures of both fluids are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the geothermal water and the cold water are given to be 4.25 and 4.18 kJ/kg.°C, respectively. Analysis The heat capacity rates of the hot and cold fluids are C kW/ 5.016 = C) kJ/kg. kg/s)(4.18 (1.2 4.25 = C) kJ/kg. (4.25 ° ° = = ° = = pc c c h h ph h h c m C m m c m C & & & & C kW/ 016 . 5 min ° = = c C C and h h m m C C c & & 1802 . 1 25 . 4 016 . 5 max min = = = The NTU of this heat exchanger is 392 . 2 C kW/ 016 . 5 ) m C)(25 . kW/m 480 . 0 ( 2 2 min = ° ° = = C UA NTU s Using the effectiveness relation, we find the capacity ratio [ ] [ ] [ ] [ ] 494 . 0 ) 1 ( 392 . 2 exp 1 ) 1 ( 392 . 2 exp 1 823 . 0 ) 1 ( NTU exp 1 ) 1 ( NTU exp 1 = ⎯→ ⎯ − − − − − − = ⎯→ ⎯ − − − − − − = c c c c c c c ε Then the mass flow rate of geothermal water is determined from kg/s 2.39 = ⎯→ ⎯ = ⎯→ ⎯ = h h h m m m c & & & 1802 . 1 494 . 0 1802 . 1 The maximum heat transfer rate is kW 9 . 290 C) 17 -C C)(75 kW/ (5.016 ) ( in c, in h, min max = ° ° ° = − = T T C Q & Then the actual rate of heat transfer rate becomes kW 4 . 239 kW) 0.9 (0.823)(29 max = = = Q Q & & ε The outlet temperatures of the geothermal and cold waters are determined to be C 64.7° = ⎯→ ⎯ − ° ⎯→ ⎯ − = out c, out c, in c, out c, ) 17 C)( kW/ (5.016 = kW 4 . 239 ) ( T T T T C Q c & C 51.4° = ⎯→ ⎯ − ° − = out h, out h, out h, in h, ) C)(75 kJ/kg. kg/s)(4.25 (2.39 = kW 4 . 239 ) ( T T T T c m Q ph h & & Geothermal water 75°C Cold Water 17°C 1.2 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-97 16-134 Air is to be heated by hot oil in a cross-flow heat exchanger with both fluids unmixed. The effectiveness of the heat exchanger, the mass flow rate of the cold fluid, and the rate of heat transfer are to be determined. .Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of the air and the oil are given to be 1.006 and 2.15 kJ/kg.°C, respectively. Analysis (a) The heat capacity rates of the hot and cold fluids are c c pc c c c c ph h h m m c m C m m c m C & & & & & & 1.006 = C) kJ/kg. (1.006 1.075 = C) kJ/kg. (2.15 5 . 0 ° = = ° = = Therefore, c c m C C & 006 . 1 min = = and 936 . 0 075 . 1 006 . 1 max min = = = c c m m C C c & & The effectiveness of the heat exchanger is determined from 0.645 = − − = − − = = 18 80 18 58 ) ( ) ( in c, in h, in c, out c, max T T C T T C Q Q c c & & ε (b) The NTU of this heat exchanger is expressed as c c s m m C UA NTU & & 7455 . 0 006 . 1 C) kW/ 750 . 0 ( min = ° = = The NTU of this heat exchanger can also be determined from [ ] [ ] 724 . 3 936 . 0 1 ) 645 . 0 1 ln( 936 . 0 ln 1 ) 1 ln( ln = + − × − = + − − = c c NTU ε Then the mass flow rate of the air is determined to be kg/s 0.20 = ⎯→ ⎯ ° = ⎯→ ⎯ = c c s m m C UA & & 006 . 1 C) kW/ 750 . 0 ( 724 . 3 NTU min (c) The rate of heat transfer is determined from kW 8.05 = ° ° = − = C 18) -C)(58 kJ/kg. 6 kg/s)(1.00 (0.20 ) ( in c, out c, T T c m Q pc c & & Air 18°C Oil 80°C 58°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-98 16-135 A water-to-water counter-flow heat exchanger is considered. The outlet temperature of the cold water, the effectiveness of the heat exchanger, the mass flow rate of the cold water, and the heat transfer rate are to be determined. .Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. Properties The specific heats of both the cold and the hot water are given to be 4.18 kJ/kg.°C. Analysis (a) The heat capacity rates of the hot and cold fluids are c c pc c c c c ph h h m m c m C m m c m C & & & & & & 4.18 = C) kJ/kg. (4.18 6.27 = C) kJ/kg. (4.18 5 . 1 ° = = ° = = Therefore, c c m C C & 18 . 4 min = = and 667 . 0 27 . 6 18 . 4 max min = = = c c m m C C C & & The rate of heat transfer can be expressed as ) 20 )( 18 . 4 ( ) ( out c, in c, out c, − = − = T m T T C Q c c & & [ ] ) 80 )( 27 . 6 ( ) 15 ( 95 ) 27 . 6 ( ) ( out c, out c, out h, in h, T m T m T T C Q c c h − = + − = − = & & & Setting the above two equations equal to each other we obtain the outlet temperature of the cold water C 56° = ⎯→ ⎯ − = − = out c, out c, out c, ) 80 ( 27 . 6 ) 20 ( 18 . 4 T T m T m Q c c & & & (b) The effectiveness of the heat exchanger is determined from 0.48 = − − = − − = = ) 20 95 ( 18 . 4 ) 20 56 ( 18 . 4 ) ( ) ( in c, in h, in c, out c, max c c c c m m T T C T T C Q Q & & & & ε (c) The NTU of this heat exchanger is determined from 805 . 0 1 667 . 0 48 . 0 1 48 . 0 ln 1 667 . 0 1 1 1 ln 1 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − = c c NTU ε ε Then, from the definition of NTU, we obtain the mass flow rate of the cold fluid: kg/s 0.416 = ⎯→ ⎯ ° = ⎯→ ⎯ = c c s m m C UA NTU & & 18 . 4 C kW/ 400 . 1 805 . 0 min (d) The rate of heat transfer is determined from kW 62.6 = ° − ° = − = C ) 20 56 )( C kJ/kg. 18 . 4 )( kg/s 416 . 0 ( ) ( in c, out c, T T c m Q pc c & & Hot water 95°C Cold Water 20°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-99 16-136 Oil is cooled by water in a 2-shell passes and 4-tube passes heat exchanger. The mass flow rate of water and the surface area are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There is no fouling. Properties The specific heat of oil is given to be 2 kJ/kg.°C. The specific heat of water is taken to be 4.18 kJ/kg.°C. Analysis The logarithmic mean temperature difference for counter-flow arrangement and the correction factor F are C 30 = C 25 C 55 C 79 = C 46 C 125 , , 2 , , 1 ° ° − ° = − = Δ ° ° − ° = − = Δ in c out h out c in h T T T T T T C 61 . 50 ) 30 / 79 ln( 30 79 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T CF lm 97 . 0 3 . 0 125 55 46 25 7 . 0 125 25 125 55 1 2 2 1 1 1 1 2 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = − − = − − = F t t T T R t T t t P (Fig. 16-18) The rate of heat transfer is kW 1400 C ) 55 125 ( C) kJ/kg kg/s)(2.0 10 ( ) ( , , = ° − ° ⋅ = − = out h in h h h T T c m Q & & The mass flow rate of water is kg/s 15.9 = ° − ° ° = − = C) 25 C C)(46 kJ/kg. (4.18 kW 1400 ) ( in out p w T T c Q m & & The surface area of the heat exchanger is determined to be 2 m 31.7 = ° ⋅ = Δ = s s lm A A T UAF Q ) C 61 . 50 )( 97 . 0 ( ) C kW/m 9 . 0 ( kW 1400 2 & Water 25°C Oil 125°C 55°C 2 shell passes 4 tube passes 46°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-100 16-137 A polymer solution is heated by ethylene glycol in a parallel-flow heat exchanger. The rate of heat transfer, the outlet temperature of polymer solution, and the mass flow rate of ethylene glycol are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There is no fouling. Properties The specific heats of polymer and ethylene glycol are given to be 2.0 and 2.5 kJ/kg.°C, respectively. Analysis (a) The logarithmic mean temperature difference is C 15 C 40 = C 20 C 60 , , 2 , , 1 ° = − = Δ ° ° − ° = − = Δ out c out h in c in h T T T T T T C 49 . 25 ) 15 / 40 ln( 15 40 ) / ln( 2 1 2 1 , ° = − = Δ Δ Δ − Δ = Δ T T T T T PF lm Th e rate of heat transfer in this heat exchanger is W 4894 = ° ° = Δ = C) 49 . 25 ( ) m C)(0.8 . W/m 240 ( 2 2 lm s T UA Q & (b) The outlet temperatures of both fluids are C 43.2° = ° + ° = + Δ = ° = ° ⋅ ° = + = → − = C 2 . 28 C 15 C 2 . 28 C) J/kg kg/s)(2000 3 . 0 ( W 4894 + C 20 ) ( , , , , , , out c out out h c c in c out c in c out c c c T T T c m Q T T T T c m Q & & & & (c) The mass flow rate of ethylene glycol is determined from kg/s 0.117 = ° − ° ° = − = C) 2 . 43 C C)(60 kJ/kg. (2500 W 4894 ) ( in out p ethylene T T c Q m & & ethylene 60°C Polymer 20°C 0.3 kg/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-101 16-138 The inlet and exit temperatures and the volume flow rates of hot and cold fluids in a heat exchanger are given. The rate of heat transfer to the cold water, the overall heat transfer coefficient, the fraction of heat loss, the heat transfer efficiency, the effectiveness, and the NTU of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 Changes in the kinetic and potential energies of fluid streams are negligible. 3 Fluid properties are constant. Properties The densities of hot water and cold water at the average temperatures of (38.9+27.0)/2 = 33.0°C and (14.3+19.8)/2 = 17.1°C are 994.8 and 998.6 kg/m3, respectively. The specific heat at the average temperature is 4178 J/kg.°C for hot water and 4184 J/kg.°C for cold water (Table A-15). Analysis (a) The mass flow rates are kg/s 04145 . 0 /s) m 0 )(0.0025/6 kg/m 8 . 994 ( 3 3 = = = h h h m V & & ρ kg/s 07490 . 0 /s) m 0 )(0.0045/6 kg/m 6 . 998 ( 3 3 = = = c c c m V & & ρ The rates of heat transfer from the hot water and to the cold water are W 2061 = C) 0 . 27 C C)(38.9 kJ/kg. kg/s)(4178 04145 . 0 ( )] ( [ h ° − ° ° = − = out in p h T T c m Q & & W 1724 = C) 3 . 14 C C)(19.8 kJ/kg. kg/s)(4184 07490 . 0 ( )] ( [ c ° − ° ° = − = in out p c T T c m Q & & (b) The logarithmic mean temperature difference and the overall heat transfer coefficient are C 1 . 19 C 8 . 19 C 9 . 38 , , 1 ° = ° − ° = − = Δ out c in h T T T C 7 . 12 C 3 . 14 C 0 . 27 , , 2 ° = ° − ° = − = Δ in c out h T T T C 68 . 15 7 . 12 1 . 19 ln 7 . 12 1 . 19 ln 2 1 2 1 ° = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ Δ Δ − Δ = Δ T T T T Tlm C W/m 3017 2 ⋅ = ° + = Δ = ) C 68 . 15 )( m 4 0 . 0 ( W 2 / ) 2061 1724 ( 2 , lm m hc T A Q U & Note that we used the average of two heat transfer rates in calculations. (c) The fraction of heat loss and the heat transfer efficiency are 83.6% 16.4% = = = = = = − = − = 836 . 0 2061 1724 164 . 0 2061 1724 2061 h c h c h loss Q Q Q Q Q f & & & & & η (d) The heat capacity rates of the hot and cold fluids are C W/ 4 . 313 C) kJ/kg. kg/s)(4184 (0.07490 C W/ 2 . 173 C) kJ/kg. kg/s)(4178 (0.04145 ° = ° = = ° = ° = = pc c c ph h h c m C c m C & & Therefore C W/ 2 . 173 min ° = = h C C which is the smaller of the two heat capacity rates. Then the maximum heat transfer rate becomes W 4261 = C) 14.3 -C C)(38.9 W/ (173.2 ) ( , , min max ° ° ° = − = in c in h T T C Q & The effectiveness of the heat exchanger is 44.4% = = + = = 444 . 0 kW 4261 kW 2 / ) 2061 1724 ( max Q Q & & ε One again we used the average heat transfer rate. We could have used the smaller or greater heat transfer rates in calculations. The NTU of the heat exchanger is determined from 0.697 = ° ⋅ = = C W/ 2 . 173 ) m 04 . 0 )( C W/m 3017 ( 2 2 min C UA NTU Hot water 38.9°C Cold water 14.3°C 19.8°C 27.0°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 16-102 16-139 . . . 16-143 Design and Essay Problems 16-143 A counter flow double-pipe heat exchanger is used for cooling a liquid stream by a coolant. The rate of heat transfer and the outlet temperatures of both fluids are to be determined. Also, a replacement proposal is to be analyzed. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There is no fouling. Properties The specific heats of hot and cold fluids are given to be 3.15 and 4.2 kJ/kg.°C, respectively. Analysis (a) The overall heat transfer coefficient is .K W/m 1185 10 2 8 1 600 2 1 600 2 8 . 0 8 . 0 8 . 0 8 . 0 = + = + = h c m m U & & The rate of heat transfer may be expressed as ) 10 )( 4200 )( 8 ( ) ( , , , − = − = out c in c out c c c T T T c m Q & & (1) ) 90 )( 3150 )( 10 ( ) ( , , , out h out h in h h h T T T c m Q − = − = & & (2) It may also be expressed using the logarithmic mean temperature difference as ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − − − = Δ Δ Δ − Δ = Δ = 10 90 ln ) 10 ( ) 90 ( ) 9 )( 1185 ( ) / ln( 2 1 2 1 h c h c lm T T T T T T T T UA T UA Q & (3) We have three equations with three unknowns, solving an equation solver such as EES, we obtain C 69.6 C, 29.1 W, 10 6.42 5 ° = ° = × = out h out c T T Q , , & (b) The overall heat transfer coefficient for each unit is .K W/m 5 . 680 5 2 4 1 600 2 1 600 2 8 . 0 8 . 0 8 . 0 8 . 0 = + = + = h c m m U & & Then ) 10 )( 4200 )( 4 2 ( ) ( , , , − × = − = out c in c out c c c T T T c m Q & & (1) ) 90 )( 3150 )( 5 2 ( ) ( , , , out h out h in h h h T T T c m Q − × = − = & & (2) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − − − × = Δ Δ Δ − Δ = Δ = 10 90 ln ) 10 ( ) 90 ( ) 5 2 )( 5 . 680 ( ) / ln( 2 1 2 1 h c h c lm T T T T T T T T UA T UA Q & (3) Once again, we have three equations with three unknowns, solving an equation solver such as EES, we obtain C 75.7 C, 23.4 W, 10 4.5 5 ° = ° = × = out h out c T T Q , , & Discussion Despite a higher heat transfer area, the new heat transfer is about 30% lower. This is due to much lower U, because of the halved flow rates. So, the vendor’s recommendation is not acceptable. The vendor’s unit will do the job provided that they are connected in series. Then the two units will have the same U as in the existing unit. KJ Cold 10°C 8 kg/s Hot 90°C 10 kg/s
9219	https://news.ycombinator.com/item?id=39862278	Also, while fooling around with my c... \| Hacker News Hacker Newsnew \| past \| comments \| ask \| show \| jobs \| submitlogin fuzztesteron March 29, 2024 \| parent \| context \| favorite \| on: Thirty Seven Also, while fooling around with my calculator in school, years ago, I had noticed this pattern: 37 x 3 = 111 37 x 6 = 222 37 x 9 = 333 and so on. I tried it again just now, all the way up to 37 x 51. It works perfectly upto 37 x 27 = 999 After that the pattern changes somewhat, but still seems to have some sort of regularity to it. jamesharton March 29, 2024 \| next[–] At that point you’re really just exploring the integer multiples of 111, where it’s perhaps not surprising to find patterns in base 10, since 111 is 10^0+10^1+10^2, and two digit integers you multiply it by are going to be of the form A10^0+B10^1, so the resulting numbers are all going to be A10^0+(A+B)10^1+(A+B)10^2+B10^3. When A+B < 10, that’s going to form a pleasing pattern as you increment through As and Bs. The pattern should actually be fine up to 3754 = (373)18 = 1998; it breaks at 3757 because that’s (373)19 and 1+9 isn’t less than 10. Similarly interesting patterns will crop up when multiplying 1111 or 11111 by integers - and similarly in other bases as well (identically, in fact - multiples of 0x111 include 0x1221, 0x1332, etc) So this sort of reduces to ‘I guess It’s funny that 111 is 373’.. but on some level, that’s just how numbers work. Every number has a prime factorization. 1111’s prime factors are 11 and 101. 0x111’s prime factors are 0x3, 0x7 and 0xD; 111’s just happen to be 3 and 37. That said… it turns out 37 crops up as a prime factor of 111, 111111, 111111111, 111111111111… so there’s something a little more going on here… Although even that is really just because each of those numbers is 111 a number of the form 1001…001001001 - 1000^0+1000^1+…+1000^n folkravon March 29, 2024 \| parent \| next[–] The fact you have to lay it out like you did kind of points towards the pattern being the result of prime factorization not being as evident as you seem to think it is lol jamesharton March 29, 2024 \| root \| parent \| next[–] It’s neat that every third multiple of 37 forms a pattern. It’s just not significant. Every 271st multiple of 41 also forms a nice pattern: 11111 22222 33333 44444 … But that somehow doesn’t seem as interesting. jamesharton March 29, 2024 \| root \| parent \| prev \| next[–] Apologies if I came off as dismissive here - was not my intent at all - there’s obviously tons of fascinating little patterns to dig into here. Like, factorizing numbers of the form 111…111 is a whole thing… If there’s a composite number of digits in the number, then you can always factor it into a smaller 11…11 number, and a number consisting of 1s separated by repeating 0s - a 100…00100…001 type number. And you can do the same to that number if it has a composite number of 1s in it (turns out the 1111 case is just a ‘number with a bunch of ones separated by sequences of no zeroes’). To be concrete: 111111111111 (12 1s) = 111 1001001001 (3 1s, 4 1s) = 111 1001 1000001 (3 1s, 2 1s, 2 1s) Or = 1111 100010001 (4 1s, 3 1s) = 11 101 100010001 (2 1s, 2 1s, 3 1s) And it’s not a coincidence that the numbers of ones in those factors match the prime factors of 12. So numbers in this form tend to have factors like 11, 101, and 111… and of course that means they tend to have 37 as a factor as a result. Like, in the above example, it means 100010001 must have 37 as a factor too. Lot of weird patterns to be found in here, for sure. fuzztesteron March 29, 2024 \| root \| parent \| prev \| next[–] Wow, that's one slightly complex sentence. I did a double-take at first, then had to scan the sentence a second time, to figure out what it meant, because of the consecutive, multiple clauses in it. Such a number of puns not intended originally, but noticed and italicized ;) Okay, will confess, I added one or two puns after the fact, to enhance the result, like the "one" and "complex" at the top, and the words in this sentence. fuzztesteron March 29, 2024 \| parent \| prev \| next[–] Interesting, thanks. >That said… it turns out 37 crops up as a prime factor of 111, 111111, 111111111, 111111111111… so there’s something a little more going on here… There sure is. Paging Ramanujan ... :) fuzztesteron March 29, 2024 \| prev \| next[–] Also, not 37, but close: That was a novel I enjoyed reading around the same time, in college. sans_souseon March 29, 2024 \| parent \| next[–] he took two too many steps fuzztesteron March 29, 2024 \| root \| parent \| next[–] Ha ha, clever :) rep_lodsbon March 29, 2024 \| prev \| next[–] As the other commenter pointed out, it's just multiples of 111. You can find all sorts of patterns playing around with a calculator, I particularly like this one: 1² = 1 11² = 121 111² = 12321 1111² = 1234321 ... 111111111² = 12345678987654321 It works in other number bases too, and once you figure out the math it no longer seems that surprising. fuzztesteron March 30, 2024 \| parent \| next[–] > 1111² = 1234321 ... Yes, I remember seeing that, though only up to 4 levels or so. Right, I don't think the number base has much to do with it. These are just properties of numbers. The individual cases may vary per base, that's all. atomlibon March 29, 2024 \| prev[–] 37 × 14 = 518 Guidelines \| FAQ \| Lists \| API \| Security \| Legal \| Apply to YC \| Contact Search:
9220	https://personalpages.manchester.ac.uk/staff/david.d.apsley/lectures/hydraulics2/t4.pdf	Hydraulics 2 T4-1 David Apsley TOPIC T4: PUMPS AND TURBINES AUTUMN 2025 Objectives (1) Understand the role of pumps and turbines as energy-conversion devices and use, appropriately, the terms head, power and efficiency. (2) Be aware of the main types of pumps and turbines and the distinction between impulse and reaction turbines and between radial, axial and mixed-flow devices. (3) Match pump characteristics and system characteristics to determine the duty point. (4) Calculate characteristics for pumps in series and parallel and use the hydraulic scaling laws to calculate pump characteristics at different speeds. (5) Select the type of pump or turbine on the basis of specific speed. (6) Understand the mechanics of a centrifugal pump and an impulse turbine. (7) Recognise the problem of cavitation and how it can be avoided. 1. Energy conversion 1.1 Energy transfer in pumps and turbines 1.2 Power 1.3 Efficiency 2. Types of pumps and turbines 2.1 Impulse and reaction turbines 2.2 Positive-displacement and dynamic pumps 2.3 Radial, axial and mixed-flow devices 2.4 Common types of turbine 3. Pump and system characteristics 3.1 Pump characteristics 3.2 System characteristics 3.3 Finding the duty point 3.4 Pumps in parallel and in series 4. Hydraulic scaling 4.1 Dimensional analysis 4.2 Change of speed 4.3 Specific speed 5. Mechanics of rotodynamic devices 5.1 Centrifugal pump 5.2 Pelton wheel 6. Cavitation Hydraulics 2 T4-2 David Apsley 1. Energy Conversion 1.1 Energy Transfer in Pumps and Turbines Pumps and turbines are energy conversion devices: pumps turn electrical or mechanical energy into fluid energy; turbines turn fluid energy into electrical or mechanical energy. The energy per unit weight is the total head, 𝐻: 𝐻= 𝑝 𝜌𝑔+ 𝑧+ 𝑉2 2𝑔 The first two terms on the RHS comprise the piezometric head. The last term is the dynamic head. 1.2 Power Power = rate of conversion of energy. If a mass 𝑚 is raised through a height 𝐻 it gains energy 𝑚𝑔𝐻. If it does so in time 𝑡 then the rate of conversion is 𝑚𝑔𝐻/𝑡. For a fluid in motion the mass flow rate (𝑚/𝑡) is 𝜌𝑄. The rate of conversion to or from fluid energy when the total head is changed by H is, therefore, 𝜌𝑄× 𝑔𝐻, or power = 𝜌𝑔𝑄𝐻 1.3 Efficiency Efficiency, 𝜂, is given by 𝜂= (power)out (power)in where power refers to the rate of doing “useful” work (i.e. what the device was intended to do.) For turbines: 𝜂= (power)𝑜𝑢𝑡 𝜌𝑔𝑄𝐻 For pumps: 𝜂= 𝜌𝑔𝑄𝐻 (power)𝑖𝑛 PUMP pipeline TURBINE total head total head pipeline Hydraulics 2 T4-3 David Apsley Example. A pump lifts water from a large tank at a rate of 30 L s–1. If the input power is 10 kW and the pump is operating at an efficiency of 40%, find: (a) the head developed across the pump; (b) the maximum height to which it can raise water if the delivery pipe is vertical, with diameter 100 mm and friction factor 𝜆= 0.015. (a) Given: 𝑄= 0.03 m3 s−1 𝐼= 10000 W 𝜂= 0.4 Then, 𝜂= output power input power = 𝜌𝑔𝑄𝐻 𝐼  0.4 = 1000 × 9.81 × 0.03 × 𝐻 10000  𝐻= 13.59 m Answer: 13.6 m (b) Given: 𝐷= 0.1 m 𝜆= 0.015 The pump must lift water by height ℎ𝑠 and overcome pipe friction over a length 𝐿: 𝐻= ℎ𝑠+ 𝜆𝐿 𝐷 𝑉2 2𝑔 In this instance, 𝐿= ℎ𝑠, as the pipe is vertical: 𝐻= ℎ𝑠(1 + 𝜆 𝐷 𝑉2 2𝑔) The velocity 𝑉 is given by 𝑉= flow rate area = 𝑄 π𝐷2/4 = 4 × 0.03 π × 0.12 = 3.820 m s−1 Hence, 13.59 = ℎ𝑠(1 + 0.015 0.1 × 3.8202 2 × 9.81)  ℎ𝑠= 12.23 m Answer: 12.2 m Hydraulics 2 T4-4 David Apsley 2. Types of Pumps and Turbines 2.1 Impulse and reaction turbines In a pump or turbine a change in fluid head 𝐻= 𝑝 𝜌𝑔+ 𝑧+ 𝑉2 2𝑔 may be brought about by a change in pressure or velocity or both. • An impulse turbine (e.g. Pelton wheel; water wheel) is one where the change in head is brought about primarily by a change in velocity. This usually involves unconfined free jets of water (at atmospheric pressure) impinging on moving vanes. • A reaction turbine (e.g. Francis turbine; Kaplan turbine; windmill) is one where the change in head is brought about primarily by a change in pressure. 2.2 Positive-Displacement and Dynamic pumps Positive-displacement pumps operate by a change in volume; energy conversion is intermittent. Examples in the human body include the heart (diaphragm pump) and the intestines (peristaltic pump). In a reciprocating pump (e.g. a bicycle pump) fluid is sucked in on one part of the cycle and expelled (at higher pressure) in another. In dynamic pumps there is no change in volume and energy conversion is continuous. Most pumps are rotodynamic devices where fluid energy is exchanged with the mechanical energy of a rotating element (called a runner in turbines and an impeller in pumps), with a further conversion to or from electrical energy. This course focuses entirely on rotary devices. Note that, for gases, pumps are usually referred to as fans (for low pressures), blowers or compressors (for high pressures). 2.3 Radial, Axial and Mixed-Flow Devices The terms radial and axial refer to the change in direction of flow through a rotodynamic device (pump or turbine): Radial Axial Mixed Hydraulics 2 T4-5 David Apsley In a centrifugal pump flow enters along the axis and is expelled radially. (The reverse is true for a turbine.) An axial-flow pump is like a propeller; the direction of the flow is unchanged after passing through the device. A mixed-flow device is a hybrid device, used for intermediate heads. In many cases – notably in pumped-storage power stations – a device can be run as either a pump or a turbine. Inward-flow reaction turbine  centrifugal pump (high head / low discharge) (e.g. Francis turbine) Propeller turbine  axial-flow pump (low head / high discharge) (e.g. Kaplan turbine; windmill) volute "eye" (intake) impeller vane flow rotation blades guide vanes hub inlet outlet Hydraulics 2 T4-6 David Apsley 2.4 Common Types of Turbine Pelton wheels are impulse turbines used in hydroelectric plant where there is a very high head of water. Typically, 1 to 6 high-velocity jets of water impinge on buckets mounted around the circumference of a runner. A modern variant, which can handle a greater water flow and is popular in microhydropower is called a Turgo turbine. Francis turbines are used in many large hydropower projects (e.g. the Hoover Dam), with an efficiency in excess of 90%. Such moderate- to high-head turbines are also used in pumped-storage power stations (e.g. Dinorwig and Ffestiniog in Wales; Foyers in Scotland), which pump water uphill during periods of low energy demand and then run the system in reverse to generate power during the day. This smooths the power demands on fossil-fuelled and nuclear power stations which are not easily brought in and out of operation. Francis turbines are like centrifugal pumps in reverse. Kaplan turbines are axial-flow (propeller) turbines. In the Kaplan design the blade angles are adjustable to ensure efficient operation for a range of discharges. Wells turbines were specifically developed for wave-energy applications. They have the property that they rotate in the same direction irrespective of the flow direction. Bulb generators are large-diameter variants of the Kaplan propeller turbine, which are suitable for the low-head, high-discharge applications in tidal barrages (e.g. La Rance in France). Flow passes around the bulb, which contains the electrical generator. The Archimedes screw has been used since ancient times to raise water. It is widely used in water treatment plants because it can accommodate submerged debris. Recently, several devices have been installed beside weirs in the north of England to run in reverse as a turbine and generate power. This one is a mile away from my home! jet bucket spear valve Hydraulics 2 T4-7 David Apsley 3. Pump and System Characteristics 3.1 Pump characteristics Pump characteristics are the head, 𝐻, input power, 𝐼, and efficiency, 𝜂, as functions of discharge, 𝑄. The most important is the 𝐻 vs 𝑄 relationship. Typical shapes of these characteristics are sketched below for centrifugal and axial-flow pumps. Many pumps are variable-speed devices. Given pump characteristics at one rotation rate, 𝑁, those at different rotation rates may be determined using hydraulic scaling laws (Section 4). Ideally, one would like to operate the pump: • as close as possible to the design point (point of maximum efficiency); • in a region where the 𝐻 vs 𝑄 relationship is steep; (otherwise there are significant fluctuations in discharge for small changes in head). 3.2 System characteristics In general the pump has to supply enough energy to: • lift water through a certain height – the static lift, ℎ𝑠; • overcome losses dependent on the discharge, 𝑄. Thus the system head or system characteristic is 𝐻= ℎ𝑠+ ℎlosses Typically, losses (whether frictional or due to pipe fittings) are proportional to 𝑄2, so that the system characteristic is often quadratic: 𝐻= ℎ𝑠+ 𝛼𝑄2 The static lift is often decomposed into the rise from sump to the level of the pump (the suction head, ℎ𝑠1) and that between the pump and the delivery point (ℎ𝑠2). The first of these is limited by the maximum suction height (approximately 10 m, corresponding to 1 atmosphere) and will be discussed later in the context of cavitation. Q I H  Q I H  centrifugal pump axial-flow pump Sump delivery reservoir suction main delivery main static lift hs suction head Pump Hydraulics 2 T4-8 David Apsley 3.3 Finding the Duty Point The pump operates at a duty point where the head supplied by the pump precisely matches the head requirements of the system at the same discharge; i.e. where the pump and system characteristics intersect. Example. A water pump was tested at a rotation rate of 1500 rpm. The following data was obtained. (𝑄 is quantity of flow; 𝐻 is head of water; 𝜂 is efficiency). 𝑄 (L s–1) 0 10 20 30 40 50 𝐻 (m) 10.0 10.5 10.0 8.5 6.0 2.5 𝜂 0.0 0.40 0.64 0.72 0.64 0.40 It is proposed to use this pump to draw water from an open sump to an elevation 5.5 m above. The delivery pipe is 20.0 m long and 100 mm diameter and has a friction factor of 0.005. If operating at 1500 rpm, find: (a) the maximum discharge that the pump can provide; (b) the pump efficiency at this discharge; (c) the input power required. (a) Given: ℎ𝑠= 5.5 m 𝐿= 20 m 𝐷= 0.1 m 𝜆= 0.005 The pump characteristic (head available for a given flow) is given in the table; the system characteristic (head required for a given flow) must be found. Since the pump has to raise water through height ℎ𝑠 and overcome pipe friction over length 𝐿: 𝐻sys = ℎ𝑠+ 𝜆𝐿 𝐷 𝑉2 2𝑔 , where 𝑉= 𝑄 π𝐷2/4 Q H system characteristic pump characteristic duty point hs Hydraulics 2 T4-9 David Apsley  𝐻sys = ℎ𝑠+ 8𝜆𝐿 π2𝑔𝐷5 𝑄2  𝐻sys = 5.5 + 826.3𝑄2 (𝐻 in m, 𝑄 in m3 s−1) or, more conveniently here, if 𝑄 is in L s−1, 𝐻sys = 5.5 + 826.3(𝑄/1000)2  𝐻sys = 5.5 + 0.0008263𝑄2 (𝐻 in m, 𝑄 in L s−1) To find where the head available matches the head required, i.e. where the pump head matches the system head, it is convenient to add the system head to the table and then plot a graph of 𝐻 vs 𝑄 for pump and system characteristics. 𝑄 (L s–1) 0 10 20 30 40 50 𝐻pump (m) 10.0 10.5 10.0 8.5 6.0 2.5 𝜂 0.0 0.40 0.64 0.72 0.64 0.40 𝐻sys (m) 5.5 5.58 5.83 6.24 6.82 7.57 The graphs intersect at 𝑄= 37.7 L s−1 Answer: 37.7 L s–1 (b) The efficiency can be read off a graph at this value of 𝑄. Hydraulics 2 T4-10 David Apsley At 𝑄= 37.7 L s−1 the graph gives 𝜂= 0.673. Answer: 0.673 (c) Input and output power are related via the efficiency: 𝜂= output power input power = 𝜌𝑔𝑄𝐻 𝐼 To find 𝐻 either read it directly from the intersection point on the 𝐻 vs 𝑄 graph or substitute the already-found value of 𝑄 into the expression for the system characteristic. Both give 𝐻= 6.67 m Then, (remembering that 𝑄 must be in consistent units, 37.7 L s−1 = 0.0377 m3 s−1), 𝐼= 𝜌𝑔𝑄𝐻 𝜂 = 1000 × 9.81 × 0.0377 × 6.67 0.673 = 3670 W Answer: 3.67 kW In practice, it is desirable to run the pump at a speed where the duty point is close to that of maximum efficiency. To do this we need to determine how the pump characteristic varies with rotation rate, 𝑁 – see later. Hydraulics 2 T4-11 David Apsley 3.4 Pumps in Parallel and in Series Pumps in Parallel For the combined pump complex: – same head: 𝐻 – total discharge: 𝑄= 𝑄1 + 𝑄2 Advantages of pumps in parallel are: • high capacity: permits a large total discharge; • flexibility: pumps can be brought in and out of service if the required discharge varies; • redundancy: pumping can continue if one is not operating due to failure or planned maintenance. Pumps in Series For the combined pump complex: – same discharge: 𝑄 – total head: 𝐻= 𝐻1 + 𝐻2 Pumps in series may be used to generate high overall head, or to provide regular “boosts” along long pipelines without large pressures at any particular point. Q H single pump double the flow two pumps in parallel single pump double the head two pumps in series H Q Q1 Q2 Q H1 Q H2 Hydraulics 2 T4-12 David Apsley Example. A rotodynamic pump, having the characteristics tabulated below, delivers water from a river at elevation 102 m to a reservoir with a water level of 135 m, through a pipe of length 1 km and diameter 350 mm. The friction factor of the pipe may be taken as 𝜆= 0.035 and minor losses from valves and fittings can be described by a loss coefficient 𝐾= 9. 𝑄 (m3 s–1) 0 0.05 0.10 0.15 0.20 𝐻 (m) 60 58 52 41 25 𝜂 (%) 0 44 65 64 48 (a) Calculate the discharge and head in the pipeline (at the duty point). If the discharge is to be increased by the installation of a second identical pump: (b) determine the unregulated discharge and head produced by connecting the pump: (i) in parallel; (ii) in series; (c) determine the power demand at the duty point in the case of parallel operation; (d) in the case of parallel operation, if the total flow is throttled by a valve to 0.12 m3 s−1, calculate the head lost across the valve. (a) Given: ℎ𝑠= 135 −102 = 33 m 𝐿= 1000 m 𝐷= 0.35 m 𝜆= 0.035 𝐾= 9 The discharge 𝑄 and head 𝐻 are determined from where the pump head intersects the system head. For the system head, 𝐻sys = ℎ𝑠+ (𝜆𝐿 𝐷+ 𝐾) 𝑉2 2𝑔 , where 𝑉= 𝑄 π𝐷2/4  𝐻sys = ℎ𝑠+ (𝜆𝐿 𝐷+ 𝐾) 8 π2𝑔𝐷4 𝑄2  𝐻sys = 33 + 600.2𝑄2 (𝐻 in m, 𝑄 in m3 s−1) We need to find where this intersects the tabulated pump characteristic. We do so graphically, first, for convenience, adding system curve values to the given table. 𝑄 (m3 s–1) 0 0.05 0.10 0.15 0.20 𝐻pump (m) 60 58 52 41 25 𝐻sys (m) 33 34.5 39.0 46.5 57.0 Hydraulics 2 T4-13 David Apsley From the intersection point, 𝑄= 0.137 m3 s−1 From either the intersection point or by simply substituting this value of 𝑄 into the system characteristic, 𝐻= 44.3 m Answer: 𝑄= 0.137 m3 s−1; 𝐻= 44.3 m (b) For pumps in parallel, the combined assembly drives the same flow through each of two pipes and hence, for the same head, would pass twice the flow. For pumps in series, one particular flow gets a boost in head twice and hence, for the same flow, would create twice the overall head. These new pump-assembly characteristics are summarised in the tables below. Single pump 𝑄 (m3 s–1) 0 0.05 0.10 0.15 0.20 𝐻 (m) 60 58 52 41 25 Parallel pumps (𝑄→2𝑄) 𝑄 (m3 s–1) 0 0.10 0.20 0.30 0.40 𝐻 (m) 60 58 52 41 25 Q1 Q2 Q Hydraulics 2 T4-14 David Apsley Series pumps (𝐻→2𝐻) 𝑄 (m3 s–1) 0 0.05 0.10 0.15 0.20 𝐻 (m) 120 116 104 82 50 The system characteristic, defining the head required from the pump assembly, is unchanged (assuming that the more complex pipework doesn’t introduce significant extra minor losses). So, series and parallel pump characteristics can be plotted on a graph and the intersections with the system curve give new duty points. (i) Parallel pumps: 𝑄= 0.184 m3 s−1 𝐻= 53.3 m (Note that 𝑄 here is the total over two pumps, equal to the actual flow in the main pipeline.) (ii) Series pumps: 𝑄= 0.193 m3 s−1 𝐻= 55.4 m (c) Input power can be obtained by dividing the output power 𝜌𝑔𝑄𝐻 by the efficiency 𝜂. For a single pump the tabulated data gives the following efficiency graph. H1 Q H2 Hydraulics 2 T4-15 David Apsley The efficiency of each pump depends on the flow rate through that pump, not the total for the combination. Here, both pumps carry the same flow 𝑄1 = 0.184/2 = 0.092 m3 s−1, which, from the graph above, gives efficiency – individual pumps or combination – of 𝜂= 63.2% (= 0.632) Thus we can get the input power for the combination of pumps as either that for both pumps (total flow 𝑄= 0.184 m3s−1, head 𝐻= 53.3 m): output power efficiency = 𝜌𝑔𝑄𝐻 𝜂 = 1000 × 9.81 × 0.184 × 53.3 0.632 = 152200 W or doubling the input power for a single pump (𝑄1 = 0.092 m3s−1, head 𝐻= 53.3 m): 2 × 𝜌𝑔𝑄𝐻 𝜂 = 2 × 1000 × 9.81 × 0.092 × 53.3 0.632 = 152200 W Both ways work here because both pumps happen to carry the same flow. In the general case it is safer to compute the input powers of individual pumps and add them up, because different pumps may be operating at different flow rates and, hence, different efficiencies. Answer: 152 kW (d) At 𝑄= 0.12 m3 s−1 the total pump head for parallel operation is 57.1 m (see the graph), whilst the system head is only 41.6 m (from either the graph or the system curve). The difference between these, 57.1 −41.6 = 15.5 m constitutes the head lost at the valve. Answer: 15.5 m Hydraulics 2 T4-16 David Apsley 4 Hydraulic Scaling 4.1 Dimensional Analysis Provided that the mechanical efficiency is the same, the performance of a particular geometrically-similar family of pumps or turbines (“homologous series”) may be expected to depend on: discharge 𝑄 [L3T−1] pressure change 𝜌𝑔𝐻 [ML−1T−2] power 𝑃 [ML2T−3] (input for pumps; output for turbines) rotor diameter 𝐷 [L] rotation rate 𝑁 [T−1] fluid density 𝜌 [ML−3] fluid viscosity 𝜇 [ML−1T−1] (Rotor diameter may be replaced by any characteristic length, since geometric similarity implies that length ratios remain constant. Rotation rate is typically expressed in either rad s–1 or rpm.) Since there are 7 variables and 3 independent dimensions, Buckingham’s Pi Theorem yields a relationship between 4 independent groups, which may be rearranged as (exercise): Π1 = 𝑄 𝑁𝐷3 , Π2 = 𝑔𝐻 𝑁2𝐷2 , Π3 = 𝑃 𝜌𝑁3𝐷5 , Π4 = 𝜌𝑁𝐷2 𝜇 = Re For fully-turbulent flow the dependence on molecular viscosity, 𝜇, and hence the Reynolds number, Π4, vanishes. Then, for geometrically-similar pumps with different sizes, 𝐷, and rotation rates, 𝑁: ( 𝑄 𝑁𝐷3) 1 = ( 𝑄 𝑁𝐷3) 2 , ( 𝑔𝐻 𝑁2𝐷2) 1 = ( 𝑔𝐻 𝑁2𝐷2) 2 , ( 𝑃 𝜌𝑁3𝐷5) 1 = ( 𝑃 𝜌𝑁3𝐷5) 2 For pumps (input power 𝑃, output power 𝜌𝑔𝑄𝐻), any one of Π1, Π2, Π3 may be replaced by Π1Π2 Π3 = 𝜌𝑔𝑄𝐻 𝑃 = 𝜂 (efficiency) The reciprocal of this would be used for turbines. Hydraulics 2 T4-17 David Apsley Example. A ¼-scale model centrifugal pump is tested under a head of 7.5 m at a speed of 500 rpm. It was found that 7.5 kW was needed to drive the model. Assuming similar mechanical efficiencies, calculate: (a) the speed and power required by the prototype when pumping against a head of 44 m; (b) the ratio of the discharge in the model to that in the prototype. (a) Given: 𝐿model (𝑚)/𝐿prototype (𝑝) = 1/4 𝐻𝑚= 7.5 m 𝑄𝑚= 500 rpm 𝑃 𝑚= 7.5 kW From the head scaling: ( 𝑔𝐻 𝑁2𝐷2) 𝑚 = ( 𝑔𝐻 𝑁2𝐷2) 𝑝 As 𝑔 is the same at both scales, (𝑁𝑝 𝑁𝑚 ) 2 = (𝐻𝑝 𝐻𝑚 ) (𝐷𝑚 𝐷𝑝 ) 2 = 44 7.5 × (1 4) 2 = 0.3667  𝑁𝑝 𝑁𝑚 = 0.6056  𝑁𝑝= 0.6056𝑁𝑚 = 0.6056 × (500 rpm) = 302.8 rpm From the power scaling: ( 𝑃 𝜌𝑁3𝐷5) 𝑝 = ( 𝑃 𝜌𝑁3𝐷5) 𝑚 Assuming the same working fluid, 𝜌 is the same at both scales. Hence, 𝑃 𝑝 𝑃 𝑚 = (𝑁𝑝 𝑁𝑚 ) 3 (𝐷𝑝 𝐷𝑚 ) 5 = 0.60563 × 45 = 227.4  𝑃 𝑝= 227.4𝑃 𝑚 = 227.4 × (7.5 kW) = 1706 kW Answer: 𝑁𝑝= 303 rpm; 𝑃 𝑝= 1710 kW (b) From the discharge scaling, ( 𝑄 𝑁𝐷3) 𝑚 = ( 𝑄 𝑁𝐷3) 𝑝 Hydraulics 2 T4-18 David Apsley  𝑄𝑚 𝑄𝑝 = (𝑁𝑚 𝑁𝑝 ) (𝐷𝑚 𝐷𝑝 ) 3 = 1 0.6056 × (1 4) 3 = 0.02580 Answer: 0.0258 Hydraulics 2 T4-19 David Apsley 4.2 Change of Speed For the same pump (i.e. same 𝐷) operating at different speeds 𝑁1 and 𝑁2 the constancy of the dimensionless groups 𝑄 𝑁𝐷3 , 𝑔𝐻 𝑁2𝐷2 , 𝑃 𝜌𝑁3𝐷5 , 𝜂 gives 𝑄∝𝑁, 𝐻∝𝑁2, 𝑃∝𝑁3 , 𝜂 constant (As an aid to memory, this might be expected, since 𝑄∝velocity, whilst 𝐻∝energy ∝ velocity2). These are called the hydraulic scaling laws or affinity laws. Speed-scaling laws for a single pump: 𝑄2 𝑄1 = 𝑁2 𝑁1 , 𝐻2 𝐻1 = (𝑁2 𝑁1 ) 2 , 𝑃 2 𝑃 1 = (𝑁2 𝑁1 ) 3 , 𝜂1 = 𝜂2 Given pump characteristics at one speed one can use the hydraulic scaling laws to deduce characteristics at a different speed. 4.2.1 Finding the Duty Point at a New Pump Speed Scale each (𝑄, 𝐻) pair on the original characteristic at speed 𝑁1 to get the new characteristic at speed 𝑁2; i.e. 𝑄2 = (𝑁2 𝑁1 ) 𝑄1 , 𝐻2 = (𝑁2 𝑁1 ) 2 𝐻1 Where this scaled characteristic intercepts the system curve gives the duty point at speed 𝑁2. Q system characteristic new duty point hs N1 N2 H Hydraulics 2 T4-20 David Apsley 4.2.2 Finding the Pump Speed For a Given Duty Point (Harder) To find the pump speed for a given discharge or head plot a hydraulic-scaling curve back from the required duty point (𝑄2, 𝐻2) on the system curve, at unknown speed 𝑁2: 𝐻 𝐻2 = ( 𝑄 𝑄2 ) 2 Very important: the hydraulic scaling curve is not the same as the system curve, although, since they are both quadratic, they can be quite close. Where the hydraulic scaling curve cuts the original characteristic gives a scaled duty point (𝑄1, 𝐻1) and thence the ratio of pump speeds from either the ratio of discharges or the ratio of heads: 𝑁2 𝑁1 = 𝑄2 𝑄1 or (𝑁2 𝑁1 ) 2 = 𝐻2 𝐻1 Example. Water from a well is pumped by a centrifugal pump which delivers water to a reservoir in which the water level is 15.0 m above that in the sump. When the pump speed is 1200 rpm its pipework has the following characteristics: Pipework characteristics: Discharge (L s–1): 20 30 40 50 60 Head loss in pipework (m): 1.38 3.11 5.52 8.63 12.40 Pump characteristics: Discharge (L s–1): 0 10 20 30 40 Head (m): 22.0 21.5 20.4 19.0 17.4 A variable-speed motor drives the pump. (a) Plot the graphs of the system and pump characteristics and determine the discharge at a speed of 1200 rpm. (b) Find the pump speed in rpm if the discharge is increased to 40 L s–1. Q system characteristic new duty point hs N1 N2 scaling curve H (Q , H ) 2 2 (Q , H ) 1 1 Hydraulics 2 T4-21 David Apsley (a) The system head is (static lift) + (pipe head loss). Hence, including a static lift of 15 m, the system head is given in the following table. Discharge (L s–1): 20 30 40 50 60 System head (m): 16.38 18.11 20.52 23.63 27.40 Pump and system heads can then be plotted on the same graph: These cross at 𝑄= 32.4 L s−1. Answer: 32.4 L s–1 (b) The original speed is 𝑁1 = 1200 rpm. At some new (unknown) speed 𝑁2 the discharge is 𝑄2 = 40 L s−1 The system curve is unchanged, so we can read off the graph (or, in this case, directly from the table) the head at that speed: 𝐻2 = 20.52 m So, given point (𝑄2, 𝐻2) what point (𝑄1, 𝐻1) on the original characteristic did it scale from? To find this we derive the scaling curve through (𝑄2, 𝐻2) and follow it back until it crosses the 𝑁1 pump characteristic. (See the graph below). The equation of the scaling curve is found as follows. As the speed 𝑁 changes, Hydraulics 2 T4-22 David Apsley 𝑄 𝑄2 = 𝑁 𝑁2 , 𝐻 𝐻2 = ( 𝑁 𝑁2 ) 2 Eliminating 𝑁/𝑁2, 𝐻 𝐻2 = ( 𝑄 𝑄2 ) 2  𝐻= 𝐻2 𝑄2 2 𝑄2  𝐻= 0.0128𝑄2 Taking a few values of 𝑄 this curve can be plotted until it crosses the 𝑁1 characteristic. It is shown in red below. At the crossing point on the 𝑁1 characteristic, 𝑄1 = 37.3 L s−1 Then, 𝑁2 𝑁1 = 𝑄2 𝑄1 = 40 37.3 = 1.072  𝑁2 = 1.072𝑁1 = 1.072 × (1200 rpm) = 1286 rpm Answer: 1290 rpm Hydraulics 2 T4-23 David Apsley Example. (Exam 2020) A variable-speed pump draws water from a reservoir to an elevated tank. The difference in water levels between the reservoir and the tank is 10 m. The pipe between them has length 𝐿= 150 m, diameter 𝐷= 150 mm and friction factor 𝜆= 0.02. Minor losses can be lumped into an overall minor loss coefficient, 𝐾, which is unknown. The characteristics of the pump at the operational speed are given in the table below. Pump characteristics at 2400 rpm Discharge (L s–1) 16 26 36 47 57 Head (m) 39.0 36.4 31.3 22.9 12.1 Efficiency (%) 50.8 65.9 70.0 60.5 38.0 (a) Determine the head loss due to friction as a function of discharge, giving numerical values of the function coefficients and stating the units used for head and discharge. The discharge at the duty point is 46 L s–1. (b) Find the pump head and power consumption at the duty point. (c) Determine the overall minor loss coefficient, 𝐾. After a rearrangement of facilities, the elevated tank is raised by 15 m and the pipe lengthened by 70 m. Through careful engineering, minor losses have been significantly reduced and can be assumed to be negligible (𝐾≈0). (d) If the same discharge is to be maintained, find the new rotation speed of the pump. ℎ𝑠= 10 m 𝐿= 150 m 𝐷= 0.15 m 𝜆= 0.02 𝐾=? (a) The head loss due to friction is ℎ𝑓 = 𝜆𝐿 𝐷(𝑉2 2𝑔) where 𝑉= 𝑄 π𝐷2/4 = 8𝜆𝐿 π2𝑔𝐷5 𝑄2 With the data above, this gives ℎ𝑓= 3264𝑄2 (ℎ𝑓 in m, 𝑄 in m3 s−1) or ℎ𝑓= 0.003264𝑄2 (ℎ𝑓 in m, 𝑄 in L s−1) (b) The discharge 𝑄= 46 L s−1 (= 0.046 m3 s−1). Draw graphs of 𝐻 vs 𝑄 and 𝜂 vs 𝑄 to determine pump head and efficiency at the duty point. Hydraulics 2 T4-24 David Apsley At the duty point, 𝐻= 23.8 m 𝜂= 62.0% (= 0.62) The power consumption is then Hydraulics 2 T4-25 David Apsley input power = output power efficiency = 𝜌𝑔𝑄𝐻 𝜂 = 1000 × 9.81 × 0.0460 × 23.8 0.62 = 17320 W Answer: 𝐻= 23.8 m; 𝑃= 17.3 kW (c) At the duty point, pump head equals the system requirement (static lift + frictional losses + minor losses): 𝐻= ℎ𝑠+ ℎ𝑓+ 𝐾(𝑉2 2𝑔) where 𝑉= 𝑄 π𝐷2/4  23.8 = 10 + 6.907 + 0.3454𝐾  𝐾= 19.96 Answer: 20.0 (d) In the new arrangement, ℎ𝑠= 25 m 𝐿= 220 m 𝐷= 0.15 m 𝜆= 0.02 𝐾= 0 The system head is 𝐻sys = ℎ𝑠+ 𝜆𝐿 𝐷(𝑉2 2𝑔) where 𝑉= 𝑄 π𝐷2/4 = ℎ𝑠+ 8𝜆𝐿 π2𝑔𝐷5 𝑄2 With the data above and 𝑄= 0.046 m3 s−1, this gives 𝐻sys = 35.1 m This must match the pump head. Hence, at an unknown speed 𝑁2 the pump parameters are 𝑄2 = 46.0 L s−1 𝐻2 = 35.1 m Plot a speed-scaling curve through this point until it reaches the characteristic for the original speed 𝑁1 = 2400 rpm. This scaling curve is Hydraulics 2 T4-26 David Apsley 𝐻 𝐻2 = ( 𝑁 𝑁2 ) 2 = ( 𝑄 𝑄2 ) 2 i.e., for 𝐻 in m and 𝑄 in L s−1, 𝐻 35.1 = ( 𝑁 𝑁2 ) 2 = ( 𝑄 46.0) 2 or 𝐻= 0.0166𝑄2 This is illustrated below. The scaling curve meets the 𝑁1 characteristic at (𝑄1, 𝐻1) = (41.0,27.9) and, hence, using the ratio of discharges (say), 𝑁2 𝑁1 = 𝑄2 𝑄1  𝑁2 2400 = 46.0 41.0  𝑁2 = 2693 rpm Answer: 2693 rpm Hydraulics 2 T4-27 David Apsley 4.3 Specific speed The specific speed (or type number) is a guide to the type of pump or turbine required for a particular role. 4.3.1 Specific Speed for Pumps The specific speed, 𝑁𝑠, is the rotational speed needed to discharge 1 unit of flow against 1 unit of head. (For what “unit” means in this instance, see below.) For a given pump, the hydraulic scaling laws give Π1 ≡ 𝑄 𝑁𝐷3 = constant , Π2 = 𝑔𝐻 𝑁2𝐷2 = constant Eliminating 𝐷, and choosing an exponent that will make the combination proportional to 𝑁: (Π1 2 Π2 3) 1/4 = 𝑄1/2𝑁 (𝑔𝐻)3/4 or, since 𝑔 is constant, then at any given (e.g. maximum) efficiency: 𝑄1/2𝑁 𝐻3/4 = (dimensional) constant This constant is the specific speed, 𝑁𝑠, occurring when 𝑄 and 𝐻 in specified units (see below) are numerically equal to 1.0: Specific speed (pump): 𝑁 𝑆= 𝑄1/2𝑁 𝐻3/4 Notes. • The specific speed is a single value calculated at the “normal” operating point (usually 𝑄 and 𝐻 at the maximum efficiency point for the anticipated rotation rate 𝑁). • With the commonest definition (in the UK and Europe), 𝑁 is in rpm, 𝑄 in m3 s–1, 𝐻 in m, but this is far from universal, so be careful. • In principle, the units of 𝑁𝑠 are the same as those of 𝑁, which doesn’t look correct from the definition but only because that has been shortened from (1 m3 s−1)1/2 × 𝑁𝑠 (1 m)3/4 = 𝑄1/2𝑁 𝐻3/4 • Because of the omission of 𝑔 the definition of 𝑁𝑠 depends on the units of 𝑄 and 𝐻. A less-common (but, IMHO, more mathematically correct) quantity is the dimensionless specific speed 𝐾𝑛 given by Hydraulics 2 T4-28 David Apsley 𝐾𝑛= 𝑄1/2𝑁 (𝑔𝐻)3/4 If the time units are consistent then 𝐾𝑛 has the same angular units as 𝑁 (rev or rad). • High specific speed ↔ large discharge / small head (axial-flow device). Low specific speed ↔ small discharge / large head (centrifugal device). Approximate ranges of 𝑁𝑠 are (from Hamill, 2011): Type 𝑁𝑠 (rpm) Radial (centrifugal) 10 – 70 large head Mixed flow 70 – 170 Axial > 110 small head Example. A pump is needed to operate at 3000 rpm (i.e. 50 Hz) with a head of 6 m and a discharge of 0.2 m3 s–1. By calculating the specific speed, determine what sort of pump is required. Given (in the required units): 𝑁= 3000 rpm 𝐻= 6 m 𝑄= 0.2 m3 s−1 Specific speed: 𝑁𝑠= 𝑄1/2𝑁 𝐻3/4 = 0.21/2 × 3000 63/4 = 350.0 (rpm) This is substantially larger than 110 rpm. Hence, choose an axial-flow pump. (In terms of typical applications, 200 L s–1 is quite a high flow and 6 m a fairly low head.) Answer: 350 rpm; axial-flow pump Hydraulics 2 T4-29 David Apsley 4.3.2 Specific Speed for Turbines For turbines the output power, 𝑃, is more important than the discharge, 𝑄. The relevant dimensionless groups are Π2 ≡ 𝑔𝐻 𝑁2𝐷2 , Π3 ≡ 𝑃 𝜌𝑁3𝐷5 Eliminating 𝐷, (Π3 2 Π2 5) 1/4 = (𝑃/𝜌)1/2𝑁 (𝑔𝐻)5/4 or, since 𝜌 and 𝑔 are usually taken as constant (there does seem to be a presumption that turbines are always operating in fresh water) then at any given efficiency: 𝑃1/2𝑁 𝐻5/4 = (dimensional) constant The specific speed of a turbine, 𝑁𝑠, is the rotational speed needed to develop 1 unit of power for a head of 1 unit. (For what “unit” means in this instance, see below.) Specific speed (turbine): 𝑁 𝑆= 𝑃1/2𝑁 𝐻5/4 Notes. • With the commonest definition (in the UK and Europe), 𝑁 is in rpm, 𝑃 in kW (note), 𝐻 in m, but, again, this is not a universal convention. As with pumps, the units of 𝑁𝑠 are the same as those of 𝑁. • As with pumps, a less commonly used, but mathematically more acceptable, quantity is the dimensionless specific speed 𝐾𝑛, which retains the 𝜌 and 𝑔 dependence: 𝐾𝑛= (𝑃/𝜌)1/2𝑁 (𝑔𝐻)5/4 𝐾𝑛 has the angular units of 𝑁 (revs or radians) – see Massey (2011). • High specific speed ↔ small head (axial-flow device) Low specific speed ↔ large head (centrifugal or impulse device). Approximate ranges are (from Hamill, 2011): Type 𝑁𝑠 (rpm) Pelton wheel (impulse) 12 – 60 very large head Francis turbine (radial-flow) 60 – 500 large head Kaplan turbine (axial-flow) 280 – 800 small head Hydraulics 2 T4-30 David Apsley 5. Mechanics of Rotodynamic Devices 5.1 Centrifugal Pump Fluid enters at the eye of the impeller and flows outward. As it does so it picks up the tangential velocity of the impeller vanes (or blades) which increases linearly with radius (𝑢= 𝑟𝜔). At exit the fluid is expelled nearly tangentially at high velocity. Kinetic energy is subsequently converted to pressure energy in the expanding volute. The analysis makes use of rotational dynamics: power = torque × angular velocity torque = rate of change of angular momentum where angular momentum is given by “tangential component of momentum  radius”. The absolute velocity of the fluid is the vector sum of: impeller velocity (tangential) + velocity relative to the impeller (parallel to the vanes) Write: u for the impeller velocity (𝑢= 𝑟𝜔) w for the fluid velocity relative to the impeller v = u + w for the absolute velocity The radial component of absolute velocity is determined primarily by the flow rate: 𝑣𝑟= 𝑄 𝐴 where 𝐴 is the effective outlet area. The tangential part (also called the whirl velocity) is a combination of impeller speed (𝑢= 𝑟𝜔) and tangential component relative to the vanes: 𝑣𝑡= 𝑢−𝑤cos 𝛽 Only 𝑣𝑡 contributes to the angular momentum. With subscripts 1 and 2 denoting inlet and outlet respectively, torque, 𝑇= 𝜌𝑄(𝑣𝑡2𝑟2 −𝑣𝑡1𝑟 1) power = 𝑇𝜔= 𝜌𝑄(𝑣𝑡2𝑟2𝜔−𝑣𝑡1𝑟 1𝜔) But head 𝐻= power/(𝜌𝑔𝑄) , whilst 𝑟𝜔= 𝑢. Hence, we have:  w1 w2 u =r  2 2 u =r  1 1 2 vane resultant, v u = r w  v vt v r Hydraulics 2 T4-31 David Apsley Euler’s turbomachinery equation: 𝐻= 1 𝑔(𝑣𝑡2𝑢2 −𝑣𝑡1𝑢1) The pump is usually designed so that the initial angular momentum is small; i.e. 𝑣𝑡1 ≈0. Then 𝐻= 1 𝑔𝑣𝑡2𝑢2 Effect of Blade Angle Because in the frame of the impeller the fluid leaves the blades in a direction parallel to their surface, forward-facing blades would be expected to increase the whirl velocity 𝑣𝑡 whilst backward-facing blades would diminish it. Radial and tangential components of velocity: 𝑣𝑟= 𝑤sin 𝛽 , 𝑣𝑡= 𝑢−𝑤cos 𝛽 Eliminating 𝑤: 𝑣𝑡= 𝑢−𝑣𝑟cot 𝛽 Hence, if inlet whirl can be ignored, 𝐻= 𝑢2 𝑔(𝑢2 −𝑣𝑟2 cot 𝛽) Since 𝑣𝑟 is also 𝑄/𝐴 (where 𝐴 is exit area of the impeller): 𝐻= 𝑟2𝜔 𝑔(𝑟2𝜔−𝑄 𝐴cot 𝛽) This is of the form 𝐻= 𝑎−𝑏𝑄, where 𝐻 initially decreases with 𝑄 for backward-facing blades (𝛽< 90°; cot 𝛽> 0) 𝐻 initially increases with 𝑄 for forward-facing blades (𝛽> 90°; cot 𝛽< 0) This gives rise to the pump characteristics shown. Backward-facing blades are usually preferred because, although forward-facing blades might be expected to increase whirl velocity and hence output head, the shape of the characteristic is such that small changes in head cause large changes in discharge and the pump tends to “hunt” for its operating point (pump surge). Non-Ideal Behaviour The above is a very ideal analysis. There are many sources of losses. These include: • leakage back from the high-pressure volute to the low-pressure impeller eye; • frictional losses; • “shock” or flow-separation losses at entry; • non-uniform flow at inlet and outlet of the impeller; • cavitation (when the inlet pressure is small). Q H  90 β  = 90 β  90 β (backward-facing blades) (forward-facing blades) Hydraulics 2 T4-32 David Apsley Example. A centrifugal pump is required to provide a head of 40 m. The impeller has outlet diameter 0.5 m and inlet diameter 0.25 m and rotates at 1500 rpm. The flow approaches the impeller radially at 10 m s–1 and the radial velocity falls off as the reciprocal of the radius. Calculate the required vane angle at the outlet of the impeller. The head is given by Euler’s formula 𝐻= 𝑢2 𝑔(𝑢2 −𝑣𝑟2 cot 𝛽) where 𝑢2 = 𝑟2𝜔 is the outlet blade velocity and 𝑣𝑟2 is the outlet radial velocity. Here, 𝐻= 40 m 𝜔= 1500 rev min = 1500 × 2π rad 60 s = 157.1 rad s−1 𝑢2 = 𝑟2𝜔 = 0.5 2 × 157.1 = 39.28 m s−1 The radial velocity is stated to fall off as the reciprocal of (i.e. in inverse proportion to) the radius. As the outlet radius is twice the inlet radius the outlet (radial) velocity is half the inlet velocity; i.e. 𝑣𝑟2 = 5 m s−1 Hence, in Euler’s formula, 40 = 39.28 9.81 (39.28 −5 cot 𝛽)  cot 𝛽= 5.858  𝛽= 9.687° Answer: 9.69° Hydraulics 2 T4-33 David Apsley 5.2 Pelton Wheel A Pelton wheel is the most common type of impulse turbine. One or more jets of water impinge on buckets arranged around a turbine runner. The deflection of water changes its momentum and imparts a force to rotate the runner. The power (per jet) 𝑃 is given by: power = force (on bucket) × velocity (of bucket) Force 𝐹 on the bucket is equal and opposite to that on the jet; by the momentum principle: force (on fluid) = mass flux × change in velocity Because the absolute velocity of water leaving the bucket is the vector resultant of the runner velocity (𝑢) and the velocity relative to the bucket, the change in velocity is most easily established in the frame of reference of the moving bucket. Assuming that the relative velocity leaving the buckets is 𝑘 times the relative velocity of approach, 𝑣−𝑢 (where 𝑘 is slightly less than 1 due to friction): change in 𝑥−velocity = −𝑘(𝑣−𝑢) cos(180° −𝜃) −(𝑣−𝑢) = −(𝑣−𝑢)(1 −𝑘cos 𝜃) where 𝜃 is the total angle turned (here, greater than 90°). The maximum force would be obtained if the flow was turned through 180°, but the necessity of deflecting it clear of the next bucket means that 𝜃 is typically about 165°. From the momentum principle: −𝐹= −𝜌𝑄(𝑣−𝑢)(1 −𝑘cos 𝜃) The power transferred in each jet is then 𝑃= 𝐹𝑢= 𝜌𝑄(𝑣−𝑢)𝑢(1 −𝑘cos 𝜃) The velocity part of the power may be written (𝑣−𝑢)𝑢= 1 4 𝑣2 −(1 2 𝑣−𝑢)2 Hence, for a given jet (𝑄 and 𝑣), the power is a maximum when the runner speed 𝑢 is such that 𝑢= 1 2 𝑣, or the runner speed is half the jet speed. (At this point the absolute velocity leaving the runner at 180° would be 0 if 𝑘= 1, corresponding to the case where all kinetic energy of the fluid is transferred to the runner.) In practice, the runner speed 𝑢 is often fixed by the need to synchronise the generator to the electricity grid, so it is usually the jet which is controlled (by a spear valve). Because of other losses the speed ratio is usually slightly less than ½, a typical value being 0.46. The jet velocity is given by Bernoulli’s equation, with a correction for non-ideal flow: 𝑣= 𝑐𝑣√2𝑔𝐻 where 𝐻 is the head upstream of the nozzle (= original head minus any losses in the pipeline) bucket jet u v k(v-u) v-u  Hydraulics 2 T4-34 David Apsley and 𝑐𝑣 is an orifice coefficient with typical values in the range 0.97–0.99. Example. In a Pelton wheel, 6 jets of water, each with a diameter of 75 mm and carrying a discharge of 0.15 m3 s–1 impinge on buckets arranged around a 1.5 m diameter Pelton wheel rotating at 180 rpm. The water is turned through 165° by each bucket and leaves with 90% of the original relative velocity. Neglecting mechanical and electrical losses within the turbine, calculate the power output. Consider individual jets first, multiplying by 6 later to get total power. In each jet, calculate transferred power as force (on the bucket) × (velocity of bucket) with the force on the bucket inferred from the change of momentum (in the frame of the bucket). Bucket absolute velocity: 𝑢= 𝑅𝜔 = (1.5/2) × (180 × 2π 60) = 14.14 m s−1 Jet absolute velocity: 𝑣= 𝑄 π𝑑2/4 = 4 × 0.15 π × 0.0752 = 33.95 m s−1 Hence, in the frame of the moving bucket (i.e. subtract 𝑢 from all 𝑥-velocity components) we have, accounting for the speed reduced to 90% and angle turned on reflection leaving water flowing at 180 −165 = 15° to the negative 𝑥 axis: 𝑣𝑥1 = 𝑣−𝑢 = 19.81 m s−1 𝑣𝑥2 = −0.9 × (𝑣−𝑢) × cos 15° = −17.22 m s−1 Hence, using “force = rate of change of momentum” (on the fluid): −𝐹= 𝜌𝑄(𝑣𝑥2 −𝑣𝑥1) = 1000 × 0.15 × (−17.22 −19.81)  𝐹= 5555 N Then, power (per jet) = force × velocity = 𝐹𝑢 = 5555 × 14.14 = 78550 W The total power (six jet/bucket combinations) is 6 × 78550 = 471300 W Answer: 471 kW Hydraulics 2 T4-35 David Apsley 6. Cavitation Cavitation is the formation, growth and rapid collapse of vapour bubbles in flowing liquids. Bubbles form at low (sub-atmospheric) pressures when the absolute pressure drops to the vapour pressure and the liquid spontaneously boils. (Bubbles may also arise from dissolved gases coming out of solution.) When the bubbles are subsequently swept into higher-pressure regions they collapse very rapidly, with large radial velocities and enormous transient pressures, potentially leading to surface damage. Cavitation may cause performance loss, vibration, noise, surface pitting and, occasionally, major structural damage. Besides the inlet to pumps, the phenomenon is prevalent in marine-current turbines, ship and submarine propellers and on reservoir spillways. The best way of preventing cavitation in a pump is to ensure that the inlet (suction) pressure is not too low. The net positive suction head (NPSH) is defined as the difference between the (stagnation) pressure head and that corresponding to the vapour (or cavitation) pressure: NPSH = 𝑝0 −𝑝cav 𝜌𝑔 = (𝑝+ 1 2 𝜌𝑉2) −𝑝cav 𝜌𝑔 It is, in length units, the margin by which the stagnation pressure 𝑝0 (the pressure when the fluid is brought to rest) exceeds that at which cavitation may occur. The net positive suction head must be kept well above zero to allow for further pressure loss in the impeller. A key measure is the available net positive suction head, NPSHa, which is the NPSH at pump inlet. This should be compared with the required net positive suction head, NPSHr, which is a manufacturer-specified quantity with a margin of safety to prevent cavitation, including an allowance for further pressure loss within the pump and any dissolved gases. The inlet pressure may be determined from Bernoulli’s equation, measuring 𝑧 relative to the level in the sump: 𝐻pump inlet = 𝐻sump −head loss  ( 𝑝+ 𝜌𝑔𝑧+ 1 2 𝜌𝑉2 𝜌𝑔 ) inlet = 𝑝atm 𝜌𝑔−ℎ𝑓 Hence, (𝑝0 𝜌𝑔) inlet = 𝑝atm 𝜌𝑔−𝑧inlet −ℎ𝑓 whence NPSHa ≡(𝑝0 −𝑝cav 𝜌𝑔 ) inlet = 𝑝atm −𝑝cav 𝜌𝑔 −𝑧inlet −ℎ𝑓 sump pump inlet zinlet p atm Hydraulics 2 T4-36 David Apsley To avoid cavitation one should aim to keep NPSHa as large as possible by: • keeping 𝑧inlet small or, better still, negative (i.e. below the level of water in the sump); • keeping ℎ𝑓 small (short, large-diameter pipes). The first also assists in pump priming. Example. (Exam 2022) A variable-speed pump is used to supply water from a reservoir to an elevated tank. The difference in water levels between the reservoir and the tank is 7 m. The pipe between them has length 𝐿= 300 m, diameter 𝐷= 200 mm and friction factor 𝜆= 0.03. Minor losses can be lumped into an overall minor loss coefficient 𝐾= 25. The characteristics of the pump at the operational speed are given in the table below. Pump characteristics at 2900 rpm: Discharge, 𝑄 (L s–1 ) 10 25 40 55 70 85 Head, 𝐻 (m) 37.6 35.3 30.9 24.4 15.5 4.2 Efficiency, 𝜂 (%) 22 54 78 86 71 25 (a) Find the system characteristic (head as a function of discharge), giving numerical values and the units that you have chosen to use for head and discharge. The rotational speed of the pump is adjusted to supply a discharge of 45 L s–1. (b) Determine the rotational speed and power consumption of the pump. (c) Explain what is meant by cavitation and why it can cause damage in hydraulic systems. The pump inlet is located 5 m above the reservoir level and the suction pipe has length 10 m. The diameter and friction factor of the suction pipe are the same as those provided above. Minor losses on suction side can be accommodated by a loss coefficient 𝐾𝑢= 4.5. (d) Assuming that atmospheric pressure is 101.2 kPa and the vapour pressure of water at the operating temperature 1.7 kPa, determine the Net Positive Suction Head available at pump inlet (NPSHa) for the given discharge. (e) Is the pump expected to cavitate if the Net Positive Suction Head required (NPSHr) for the given discharge is 2.7 m? Given: ℎ𝑠= 7 m 𝐿= 300 m 𝐷= 0.2 m 𝜆= 0.03 𝐾= 25 (a) Hydraulics 2 T4-37 David Apsley head = static lift + frictional (and other) losses 𝐻= ℎ𝑠+ (𝜆𝐿 𝐷+ 𝐾) 𝑉2 2𝑔 where 𝑉= 𝑄 π𝐷2/4 whence 𝐻= ℎ𝑠+ (𝜆𝐿 𝐷+ 𝐾) 8 π2𝑔𝐷4 𝑄2 With the data above, this gives 𝐻= 7 + 3615𝑄2 (𝐻 in m, 𝑄 in m3 s−1) or 𝐻= 7 + 0.003615𝑄2 (𝐻 in m, 𝑄 in L s−1) (b) The pump characteristics are given in the question at a speed 𝑁1 = 2900 rpm. The pump speed is to be changed to 𝑁2, which is unknown. The duty point here is 𝑄2 = 45 L s−1 This must still lie on the system curve, so that the head is, using the formula from part (a), 𝐻2 = 14.32 m We can now plot a speed-scaling curve (𝑄∝𝑁, 𝐻∝𝑁2) back to meet the original pump characteristic: 𝐻 𝐻2 = ( 𝑁 𝑁2 ) 2 = ( 𝑄 𝑄2 ) 2  𝐻= 𝐻2 𝑄2 2 𝑄2  𝐻= 0.007072𝑄2 (𝐻 in m, 𝑄 in L s−1) (Obviously, you may work in different units for 𝑄 if you prefer.) Plot this on an 𝐻 vs 𝑄 graph, along with original pump characteristic and system curve. Hydraulics 2 T4-38 David Apsley This scales to the following point on the 𝑁1 characteristic (note that 𝑁2 is the lower speed in this instance): 𝑄1 = 57.3 L s−1 (= 0.0573 m3 s−1) 𝐻1 = 23.2 m The ratio of 𝑄 values (or 𝐻 values, if you prefer) then gives the ratio of speeds: 𝑁2 𝑁1 = 𝑄2 𝑄1 = 45.0 57.3 = 0.7853 Then 𝑁2 = 0.7853 × (2900 rpm) = 2277 rpm The efficiency (which is unchanged by speed scaling) can be read off the original pump characteristic at 𝑄1 = 57.3 L s−1. Hydraulics 2 T4-39 David Apsley This gives 𝜂2 = 𝜂1 = 85.4% (= 0.854) The power consumption(at the new, 𝑁2, speed) is then input power = output power efficiency = 𝜌𝑔𝑄2𝐻2 𝜂2 = 1000 × 9.81 × 0.0450 × 14.32 0.854 = 7402 W Answer: 𝑁= 2280 rpm; 𝑃= 7.40 kW (c) Cavitation is the formation of vapour bubbles at very low (sub-atmospheric) pressures. The problem occurs when the bubbles are subsequently swept into a higher-pressure zone, where they collapse rapidly, causing high local shock pressures that can lead to surface pitting. (d) The net positive suction head available is the amount by which the inlet stagnation pressure exceeds the cavitation pressure, re-expressed in head units; i.e. NPSHa = (𝑝+ 1 2 𝜌𝑉2) inlet −𝑝cav 𝜌𝑔 By Bernoulli, Hydraulics 2 T4-40 David Apsley (𝑝+ 1 2 𝜌𝑉2 + 𝜌𝑔𝑧) inlet = (𝑝+ 1 2 𝜌𝑉2 + 𝜌𝑔𝑧) reservoir −upstream losses and hence (𝑝+ 1 2 𝜌𝑉2) inlet = 𝑝atm + 𝜌𝑔(𝑧reservoir −𝑧inlet) −(𝜆𝐿𝑢 𝐷+ 𝐾𝑢) (1 2 𝜌𝑉2) Here, 𝑃 atm = 101200 Pa (absolute), 𝑧reservoir −𝑧inlet = −5 m, 𝐿𝑢= 10 m, 𝐾𝑢= 4.5 and the pipe bulk velocity is 𝑉= 𝑄 π𝐷2/4 = 4 × 0.045 π × 0.22 = 1.432 m s−1 Hence, as an absolute pressure, (𝑝+ 1 2 𝜌𝑉2) inlet = 46000 Pa Then, in terms of head rather than pressure, NPSHa = 46000 −1700 1000 × 9.81 = 4.516 m Answer: 4.52 m (e) NPSHa > NPSHr, so the pump is safe from cavitation
9221	https://www.droracle.ai/articles/242202/109-extra-pyramidal-side-effects-mechanism-a-blockage-at-d2-in-mesocortical-pathway-b-blockage-at-d2-in-nigrostriatal-pathway-c-cant-remember-other-options-110	What is the mechanism behind extra pyramidal side effects (EPS)? Select Language▼ What is the mechanism behind extra pyramidal side effects (EPS)? Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: August 4, 2025 • View editorial policy Mechanism of Extrapyramidal Side Effects (EPS) The primary mechanism behind extrapyramidal side effects (EPS) is the blockade of dopamine D2 receptors in the nigrostriatal pathway.1 Pathophysiology of EPS Extrapyramidal side effects occur due to dopamine blockade or depletion in the basal ganglia, specifically in the nigrostriatal pathway. This pathway is critical for normal motor function, and interference with dopaminergic transmission here leads to various movement disorders 2. The nigrostriatal pathway connects the substantia nigra to the striatum and is primarily involved in motor control. When antipsychotic medications block D2 receptors in this pathway, they disrupt the normal balance of dopaminergic activity, resulting in EPS 1. Key Mechanisms: Primary Mechanism: D2 receptor blockade in the nigrostriatal pathway 1, 3 Secondary Factors: Rate of drug association with D2 receptors (faster association correlates with higher EPS risk) 4 Relative selectivity for different dopamine receptor subtypes 5 Balance between dopaminergic and cholinergic activity in the basal ganglia 1 Types of Extrapyramidal Side Effects Acute Dystonia: Sudden spastic contractions of muscle groups Typically occurs within first few days of treatment Often affects neck muscles, eyes (oculogyric crisis), or torso Risk factors: young age, male gender, high-potency antipsychotics 1 Drug-induced Parkinsonism: Symptoms include bradykinesia, tremors, and rigidity Mimics idiopathic Parkinson's disease Usually develops within first 3 months of treatment 6 Akathisia: Severe restlessness and inability to sit still Often misinterpreted as anxiety or psychotic agitation Common reason for medication noncompliance 1 Tardive Dyskinesia: Late-onset involuntary movements Can be permanent even after medication discontinuation Requires regular monitoring with standardized scales 3 Medication Factors Affecting EPS Risk High vs. Low Risk Antipsychotics: High-potency conventional antipsychotics (e.g., haloperidol) have greater EPS risk due to stronger D2 receptor blockade in the nigrostriatal pathway 1 Low-potency conventional antipsychotics (e.g., chlorpromazine, thioridazine) have more anticholinergic effects but fewer EPS 1 Atypical antipsychotics generally have lower EPS risk due to: Combined serotonin-dopamine receptor antagonism Faster dissociation from D2 receptors Selective action on limbic rather than striatal dopamine pathways 5 Recent Research Insights: Recent research has revealed that the association rate (how quickly drugs bind to D2 receptors) may be more important than dissociation rate in determining EPS risk 4. This challenges the traditional "fast-off" theory that focused primarily on dissociation rates. Management of EPS Acute Dystonia: Anticholinergic medications (benztropine 1-2 mg) or antihistamines Consider prophylaxis in high-risk patients 3, 7 Drug-induced Parkinsonism: Reduce antipsychotic dose if clinically feasible Add anticholinergic agent (benztropine 1-2 mg/day) Consider amantadine as alternative Evaluate switching to atypical antipsychotic 3, 7 Akathisia: Reduce antipsychotic dose if possible Beta-blockers (propranolol) are most effective Benzodiazepines may provide relief Anticholinergics are less consistently effective 3, 6 Clinical Implications Understanding the mechanism of EPS is crucial for: Medication selection: Choose antipsychotics with lower EPS risk when possible Dose optimization: Use lowest effective dose to minimize D2 blockade Early recognition: Identify and address EPS before they become severe or persistent Appropriate management: Select targeted interventions based on specific EPS type Common Pitfalls and Caveats Misdiagnosis: Akathisia is frequently misinterpreted as anxiety or worsening psychosis, leading to inappropriate dose increases that worsen the condition 1 Overlooking EPS in atypical antipsychotics: Even atypical agents can cause EPS, especially at higher doses or in vulnerable patients 8 Prolonged anticholinergic use: While effective for acute management, long-term anticholinergic use for EPS prophylaxis is controversial due to cognitive side effects 6 Ignoring non-antipsychotic causes: Other medications including some antidepressants, antiemetics, and anticonvulsants can also cause EPS through similar mechanisms 2 Understanding that EPS primarily results from D2 receptor blockade in the nigrostriatal pathway helps guide both prevention and management strategies for these potentially serious medication side effects. References 1 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 2 Research Extrapyramidal symptoms are serious side-effects of antipsychotic and other drugs. The Nurse practitioner, 1992 3 Guideline Management of Parkinson's Disease Patients Requiring Pain Management Praxis Medical Insights: Practical Summaries of Clinical Guidelines, 2025 4 Research Extrapyramidal side effects of antipsychotics are linked to their association kinetics at dopamine D2 receptors. Nature communications, 2017 5 Research Novel antipsychotics and extrapyramidal side effects. Theory and reality. Pharmacopsychiatry, 2000 6 Research Management of acute extrapyramidal effects induced by antipsychotic drugs. American journal of health-system pharmacy : AJHP : official journal of the American Society of Health-System Pharmacists, 1997 7 Drug Official FDA Drug Label For benztropine (PO) FDA, 2025 8 Research Clozapine-related extrapyramidal side effects: a case report. Rivista di psichiatria, 2017 Related Questions Do extrapyramidal symptoms (EPS) resolve after discontinuing the medication that caused them?How are extrapyramidal side effects managed?How to treat extrapyramidal symptoms (EPS)?Can Stematil (prochlorperazine) be used for extrapyramidal side effects?What is the typical onset of extrapyramidal symptoms (EPS) after initiating antipsychotic medication?How to manage hypoglycemia (low blood sugar) in a patient with Type 2 Diabetes Mellitus (T2DM) who is not taking any diabetes medications?What is the recommended anticoagulation regimen for an 80-year-old patient with Deep Vein Thrombosis (DVT) in a nursing home?What is multiple myeloma?What are the guidelines for using haloperidol (antipsychotic medication) in the acute setting?How often should I check International Normalized Ratio (INR) for patients on warfarin?What is the significance of pleocytosis in cerebrospinal fluid (CSF)? 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9222	https://en.wikipedia.org/wiki/Internal_fertilization	Jump to content Internal fertilization العربية Bikol Central Български Català Dansk Español فارسی 한국어 Bahasa Indonesia Latina Bahasa Melayu 日本語 Norsk bokmål Polski Português Türkçe Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Union of an egg and sperm to form a zygote within the female body \| \| \| Part of a series on \| \| Sex \| \| \| \| Biological terms \| \| Sexual dimorphism Sexual differentiation + Feminization + Virilization Sex-determination system + XY + XO + ZW + ZO + Temperature-dependent + Haplodiploidy Heterogametic sex Homogametic sex Sex chromosome + X chromosome + Y chromosome + Sex chromosome anomalies Testis-determining factor Hermaphrodite + Sequential hermaphroditism + Simultaneous hermaphroditism Intersex (biology) Mating type \| \| Sexual reproduction \| \| Evolution of sexual reproduction + Anisogamy + Isogamy Germ cell Meiosis Gametogenesis + Spermatogenesis + Oogenesis Gamete + spermatozoon + ovum Fertilization + External fertilization + Internal fertilization Sexual selection Plant reproduction Fungal reproduction Sexual reproduction in animals + Sexual intercourse - Penile–vaginal intercourse + Copulation + Human reproduction + Lordosis behavior + Pelvic thrust \| \| Sexuality \| \| Plant sexuality Animal sexuality Human sexuality + Mechanics + Differentiation + Activity \| \| Sexual system \| \| Dioecy Gonochorism Hermaphrodite Intersex \| \| v t e \| Internal fertilization is the union of an egg and sperm cell during sexual reproduction inside the female body. Internal fertilization, unlike its counterpart, external fertilization, brings more control to the female with reproduction. For internal fertilization to happen there needs to be a method for the male to introduce the sperm into the female's reproductive tract. Most taxa that reproduce by internal fertilization are gonochoric.: 124–125 Male mammals, reptiles, and certain other vertebrates transfer sperm into the female's vagina or cloaca through an intromittent organ during copulation. In most birds, the cloacal kiss is used, the two animals pressing their cloacas together while transferring sperm. Salamanders, spiders, some insects and some molluscs undertake internal fertilization by transferring a spermatophore, a bundle of sperm, from the male to the female. Following fertilization, the embryos are laid as eggs in oviparous organisms, or continue to develop inside the reproductive tract of the mother to be born later as live young in viviparous organisms. Evolution of internal fertilization [edit] Main article: Evolution of sexual reproduction Internal fertilization evolved many times in animals.: 2 According to David B. Dusenbery all the features with internal fertilization were most likely a result from oogamy. It has been argued that internal fertilization evolve because of sexual selection through sperm competition. In amphibians, internal fertilization evolved from external fertilization. Methods of internal fertilization [edit] Fertilization which takes place inside the female body is called internal fertilization in animals is done through the following different ways: Copulation, which involves the insertion of the penis or other intromittent organ into the vagina (in most mammals) or to the cloaca in monotremes, most reptiles, some birds, the tailed frog, some fish, the disappeared dinosaurs, as well as in other non-vertebrate animals. Cloacal kiss, which consists in that the two animals touch their cloacae together in order to transfer the sperm of the male to the female. It is used in most birds and in the tuatara, that do not have an intromittent organ. Via spermatophore, a sperm-containing cap placed by the male in the female's cloaca. Usually, the sperm is stored in spermathecae on the roof of the cloaca until it is needed at the time of oviposition. It is used by some salamander and newt species, by the Arachnida, some insects and some mollusks. In sponges, sperm cells are released into the water to fertilize ova that are retained by the female. Some species of sponge participate in external fertilization where the ova is released. Processes unique to internal fertilization [edit] Internal fertilization involves different processes than external fertilization. Some of these unique processes include the following: sperm capacitation, building reservoirs of sperm/ sperm pockets within the oviduct, sperm migration within the oviduct, acrosomal reactions, and sperm competition. Sperm capacitation has been studied primarily in mammals such as humans, rodents, and bovines. This is an especially important process for internal fertilization because it ensures that sperm are activated at the right time and place to promote the greatest success with fertilization The prevention of polyspermy is also an important process of internal fertilization. Internal fertilization utilizes a slow block, or cortical reaction, that uses different mechanisms and components than the mechanisms and components used during external fertilization. While external fertilization also utilizes a fast block system, little is known about a fast block response associated with species that fertilize internally. Expulsion [edit] Main article: Modes of reproduction At some point, the growing egg or offspring must be expelled. There are several possible modes of reproduction. These are traditionally classified as follows: Oviparity, as in most invertebrates and reptiles, monotremes, dinosaurs and all birds which lay eggs that continue to develop after being laid, and hatch later. Viviparity, as in almost all mammals (such as whales, kangaroos and humans) which bear their young live. The developing young spend proportionately more time within the female's reproductive tract. The young are later released to survive on their own, with varying amounts of help from the parent(s) of the species. Ovoviviparity, as in the garter snake, most vipers, and the Madagascar hissing cockroach, which have eggs (with shells) that hatch as they are laid, making it resemble live birth. Advantages to internal fertilization [edit] Internal fertilization allows for: Female mate choice, which gives the female the ability to choose her partner before and after mating. The female cannot do this with external fertilization because she may have limited control of who is fertilizing her eggs, and when they are being fertilized. Making a decision for the conditions of reproduction, like location and time. In external fertilization a female can only choose the time in which she releases her eggs, but not when they are fertilized. This is similar, in ways, to cryptic female choice. Egg protection on dry land. While oviparous animals either have a jelly like ovum or a hard shell enclosing their egg, internally fertilizing animals grow their eggs and offspring inside themselves. This offers protection from predators and from dehydration on land. This allows for a higher chance of survival when there is a regulated temperature and protected area within the mother. Disadvantages to internal fertilization [edit] Gestation can and will add additional risks for the mother. The additional risks from gestation come from extra energy demands. Along with internal fertilization comes sexual reproduction, in most cases. Sexual reproduction comes with some risks as well. The risks with sexual reproduction are with intercourse, it is infrequent and only works well during peak fertility. While animals which externally fertilize are able to release egg and sperm, usually into the water, not needing a specific partner to reproduce. Fewer offspring are produced through internal fertilization in comparison to external fertilization. This is both because the mother cannot hold and grow as many offspring as eggs, and the mother cannot provide and obtain enough resources for a larger amount of offspring. Fish [edit] Some species of fish, such as guppies, have the ability to internally fertilize. This process involves the male guppy inserting a tubular fin into the female's reproductive opening, and then it will deposit sperm into the female guppy's reproductive tract. Internal fertilization in cartilaginous fishes contains the same evolutionary origin as reptiles, birds, and mammals that internally fertilize. Additionally, there is no noticeable difference in tonality for species of fish that fertilize internally. Amphibians [edit] Most amphibians utilize external fertilization, however, there are some exceptions, such as the salamander. Salamanders mainly utilize internal fertilization. Salamanders do not copulate because the male salamander does not have an external penis. Rather, the male salamander produces an encased capsule of sperm and nutrients called a spermatophore. The male deposits a spermatophore on the ground, and the female will pick it up with her cloaca (a combined urinary and genital opening) and fertilize her eggs with it. Because the female is not expelling the eggs to be fertilized, this is a form of internal fertilization. Over time, an increasing number of amphibians have been discovered transitioning to a more internalist mode of fertilization. [citation needed] This transition is likely an effect of the transition from water to land during vertebrate evolution. There is an advantage for the amphibians who utilize internal fertilization because it allows for a greater selection of a time and place for reproduction. Birds [edit] Most birds do not have penises but achieve internal fertilization via cloacal contact (or "cloaca kiss"). In these birds, males and females contact their cloacas together, typically briefly, and transfer sperm to the female. However, water fowls such as ducks and geese have penises and are able to use them for internal fertilization. While birds have internal fertilization, most species no longer have phallus structures. This makes them the only vertebrate taxon to fall into both categories of lacking the phallus but participating in internal fertilization. Mammals [edit] Mammals are ideal model organisms for studying internal fertilization because all species within the mammalian class reproduce via internal fertilization processes. Mammals copulate as their method of reproduction. Internal fertilization for all mammals involves recognition events of the sperm and oocyte, acrosome reactions and associations, piercings of the oocyte zona pellucida by sperm, and reactions such as the cortical and zona reactions. Sperm capacitation is a process more common in mammalian species than any other internally fertilizing species, due to the complex female reproductive system, requiring the sperm to travel farther and have a more significant signal recognition with the egg. 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(2018-07-27). "Molecular Basis of Human Sperm Capacitation". Frontiers in Cell and Developmental Biology. 6. doi:10.3389/fcell.2018.00072. hdl:11336/161327. ISSN 2296-634X. PMC 6078053. ^ Evans, Janice P. (March 2020). "Preventing polyspermy in mammalian eggs—Contributions of the membrane block and other mechanisms". Molecular Reproduction and Development. 87 (3): 341–349. doi:10.1002/mrd.23331. ISSN 1040-452X. PMID 32219915. ^ Wozniak, Katherine L.; Carlson, Anne E. (March 2020). "Ion channels and signaling pathways used in the fast polyspermy block". Molecular Reproduction and Development. 87 (3): 350–357. doi:10.1002/mrd.23168. ISSN 1040-452X. PMC 6851399. PMID 31087507. ^ Lodé T (2001). Les stratégies de reproduction des animaux [Reproduction Strategies in the Animal Kingdom] (in French). Paris: Dunod Sciences. ^ Blackburn DG (January 2000). "Classification of the reproductive patterns of amniotes". Herpetological Monographs. 14: 371–7. doi:10.2307/1467051. 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Journal of Sex Research. 41 (1): 101–12. doi:10.1080/00224490409552218. PMC 1255935. PMID 15216429. ^ Parker, G. A. (1970). "Sperm Competition and Its Evolutionary Consequences in the Insects". Biological Reviews. 45 (4): 525–567. doi:10.1111/j.1469-185X.1970.tb01176.x. ISSN 1469-185X. S2CID 85156929. ^ Engel, Kathrin M; Dzyuba, Viktoriya; Ninhaus-Silveira, Alexandre; Veríssimo-Silveira, Rosicleire; Dannenberger, Dirk; Schiller, Jürgen; Steinbach, Christoph; Dzyuba, Borys (February 2020). "Sperm Lipid Composition in Early Diverged Fish Species: Internal vs. External Mode of Fertilization". Biomolecules. 10 (2): 172. doi:10.3390/biom10020172. PMC 7072473. PMID 31979037. ^ Jump up to: a b Cotner, Sehoya; Wassenberg, Deena (2020). The Evolution and Biology of Sex. ^ Brennan, Patricia L. R.; Birkhead, Tim R.; Zyskowski, Kristof; Van Der Waag, Jessica; Prum, Richard O. (10 September 2008). "Independent evolutionary reductions of the phallus in basal birds". Journal of Avian Biology. 39 (5): 487–492. doi:10.1111/j.0908-8857.2008.04610.x. \| v t e Modes of reproduction \| \| \| --- \| Modes \| Oviparity Viviparity (Histotrophic, Hemotrophic) Ovoviviparity \| \| \| Fertilisation \| External fertilization Internal fertilization \| \| Parental care \| birds fish (mouthbrooding) humans mammals \| \| Related topics \| Adelphophagy Gastric-brooding frog Live-bearing aquarium fish Male pregnancy Matrotrophy Oophagy Pregnancy in fish Trophic egg \| Retrieved from " Category: Reproduction in animals Hidden categories: CS1 French-language sources (fr) Articles with short description Short description is different from Wikidata
9223	https://www.mathcelebrity.com/inclusnumwp.php	Inclusive Number Word Problems Calculator Crop Image How does the Inclusive Number Word Problems Calculator work? Free Inclusive Number Word Problems Calculator - Given an integer A and an integer B, this calculates the following inclusive word problem questions:1) The Average of all numbers inclusive from A to B2) The Count of all numbers inclusive from A to B3) The Sum of all numbers inclusive from A to BThis calculator has 2 inputs. What 5 formulas are used for the Inclusive Number Word Problems Calculator? Count = B - A + 1Average = (A + B)/2Sum of inclusive integers is the Average of inclusive numbers x Count of Inclusive Numbers Take the Quiz Take the Quiz Related Calculators Consecutive Integer Word Problems2 number Word ProblemsSum of the First (n) Numbers What 5 concepts are covered in the Inclusive Number Word Problems Calculator? average : A number expressing the central value of a setSum of Values/n count : determine the total number of a collection of items inclusive numbers : includes the first and last number and all numbers in between. sum : the total amount resulting from the addition of two or more numbers, amounts, or items word problem : Math problems involving a lengthy description and not just math symbols Example calculations for the Inclusive Number Word Problems Calculator sum of inclusive numbers 10 to 50 What is the average of all numbers from 30 to 80 Sum of inclusive numbers from 60 to 90 How many integers are there from 20 to 90 Inclusive Number Word Problems Calculator Video
9224	https://www.themathdoctors.org/adding-or-subtracting-decimals-how-and-why/	Adding or Subtracting Decimals: How and Why – The Math Doctors Typesetting math: 100% Skip to content Main Menu Home Ask a Question Blog FAQ About The Math Doctors Contact Search Search for: Adding or Subtracting Decimals: How and Why December 6, 2024 December 5, 2024 / Arithmetic / Decimals / By Dave Peterson Recently a teacher (Hi, Edite!) asked for help teaching how to divide decimals; in particular, she wanted to be able to provide a deeper understanding of the process, giving a good reason for what we do. Here I want to start a long-delayed series on operations with decimals, doing exactly this for all four basic operations. We’ll start with addition and subtraction, then cover multiplication next time, and get to division last. Adding decimals We’ll start with addition, using this question from 1998: Adding Decimals I can't figure this out! 894.56 + 4563.5 The way it’s typed suggests a likely issue! What to do: Line them up Two of us answered. The first was Doctor Barrus: Hi, Jennifer! Adding decimals isn't that different from adding whole numbers. There are just a few extra steps. I'm going to work a problem that's similar to yours, and then I hope you'll see how to do yours. Okay! Here's what we do. Say I have the problem: 992.536 + 473.9 ? The first thing to do is to line up the decimal points in a straight vertical line. So I would rewrite my problem as 992.536 + 473.9 with all the decimal points in a line. As we’ll see later, this puts all the ones (units) in a column, all the tenths in another, and so on. Next, add as many zeroes as you need to "fill in the blanks" in the decimal parts. For example, I would make my problem 992.536 + 473.900 See how I filled in the empty spaces from step 1? This works because 473.9 = 473.900, just the way 1 = 1.0. Adding zeroes to the end of the decimal part of a number doesn't change the number. With experience, you don’t need to actually write the zeroes; you can just no add anything in those columns. But this does make things clearer. Now add normally. The decimal point stays in that straight line, but the numbers just add following the normal rules, like carrying and stuff. So in my problem, I'd have 992.536 + 473.900 1466.436 I added this by just following the same steps I would to add 992536 + 473900 --------- , and I put the decimal point in the answer underneath all the other decimal points. That’s the process; next, we’ll see in more detail what it means. Why: Modeling it with money Later, Doctor Mateo answered, doing the original problem with more explanation: Hello Jennifer, To add (or subtract decimals) the first thing you should do is line up the decimal points. 894.56 + 4563.5 It is important for you to line up the decimals vertically the way I did above so that you can keep track of the place-value position of the digits. We want to add ones to ones, tenths to tenths, and so on. You could think about it like this. Suppose you have a WHOLE LOT of coins: pennies, nickels, dimes, and quarters. There are lots of ways to count the money, but what is probably the fastest way to do it since you have a whole lot of change to count? Probably separating the coins into stacks of pennies, stacks of nickels, stacks of dimes, and stacks of quarters, and then counting up the value of each stack. If you have enough of a certain coin you might even collect it in a coin holder. Putting all the coins of each kind into a stack saves us from having to add different values together (or from accidentally counting a dime as a penny). So when you add (or subtract) decimals, we like to put our digits into "stacks" of similar values too. If you remember, we like to put the digits into columns or "bundles" of powers of tens just the way we would put dimes with other dimes. So in this example we place the numbers into their place-value holder: thousands // hundreds // tens // ones // . // tenths // hundredths // 8 // 9 // 4 // . // 5 // 6 + 4 // 5 // 6 // 3 // . // 5 // We discussed place value concepts in Place Value: Whole Numbers and in Place Value: Decimals. After you line up the decimal points, you might want to put zeros in the empty spaces as space-holders, like this: 894.56 + 4563.50 One of the reasons we place zeros in the empty spaces after the decimal point is that we can have a record of the number of "coins" in that position. If you have no hundredths then you record a "0" to show that you had no hundredths to count. Now you can do regular addition. Just make sure you bring the decimal point down in the same position it is in the problem. Have fun adding. Finishing up: 11 1 894.56 + 4563.50 5458.06 By putting the decimal point in the same column, we say that the sum of the tenths is a number of tenths! Mixing whole numbers with decimals What if some numbers don’t have a decimal point? Here’s a question from 2002: Hidden Decimal Points I get confused when subtracting 4 from 24.98. I keep coming up with 24.94. Could you explain why whole numbers do not show their decimal point? Michael presumably did this, just lining up the ends of the numbers: 24.98 - 4 24.94 I answered: Hi, Michael. The first thing to do is to keep in mind, as you are aware, that the decimal point of a whole number is hidden to the right of the number. Why? Because in a whole number like 24, the rightmost digit is always the ones place; when we add tenths and hundredths, and so on, to the right of it, we need to mark where the ones place is. So we use a decimal point to separate the whole part from the fractional part, and the units place is just to the left. Whole numbers don't show it, simply because it's not necessary, and we don't like to waste ink. (It can also look confusing, as in my next sentence.) The decimal point at the end does look odd in a sentence: But we CAN show it; here we want to subtract 4. from 24.98. When you subtract whole numbers, you line up corresponding places by lining up the right side, so that you subtract ones from ones, tens from tens, and so on. With decimals, you do the same thing, but you do it by lining up the decimal points. That means we have to subtract 24.98 - 4. Now we are subtracting 4 ones from 4 ones, not 4 ones from 8 hundredths, which wouldn't make sense. Fraction or whole, what we are really doing is lining up the place values, so that the “ones” are in one column, whether that is indicated by a decimal point or not. But writing it definitely helps. To make it clearer, we can now fill in the empty places with zeros if it makes you feel better: 24.98 - 4.00 Now just do the subtraction as if the decimal points weren't there: 24.98 - 4.00 20.98 There's the answer. The only trick is lining up the decimal points, as if they were holes you put a pin through to hold everything in place. Where the pin pokes through the bottom, you put a decimal point in the answer. That’s an interesting analogy. Sometimes we line up the decimal points, sometimes we don’t Finally, here is a question from 2003, comparing addition and subtraction to multiplication and division: Lining Up Decimal Points Why do addition and subtraction have to be lined up by the decimals but division and multiplication don't have to be? This will get us into topics we’ll cover more fully next time; if you don’t want spoilers, stop when we move beyond addition. Adding and subtracting Doctor Ian answered, starting with the positive story about addition: Hi Rashonda, That's a really good question! It's easy to just take it for granted, without ever stopping to think about it. Suppose I have 3 cherries, 2 apples, 3 pears and you have 4 apples, 1 pear, 3 watermelons and we want to know what we have together. Could we do this? 3 cherries 2 apples 3 pears 4 apples 1 pear 3 watermelons ---------- -------- ------------- 7 3 6 That doesn't make any sense at all, does it? We'd have to make sure that we're adding the same kinds of things: 3 cherries 2 apples 3 pears 4 apples 1 pear 3 watermelons ---------- -------- ------- ------------- 3 cherries 6 apples 4 pears 3 watermelons This is the same as the money analogy above; but fruits don’t have “place value”, so we have to put them in the same order, making up a set of piles. With money, or with numbers, there is a natural order. This is also what's going on when we add decimals. When we write a number like '123.4', that's really just a very compact way of writing the sum 1 100 + 2 10 + 3 1 + 4 1/10 And when we write a number like 5.432, we're just writing the sum 5 1 + 4 1/10 + 3 1/100 + 2 1/1000 Now, following the example of the fruit, can we do this? 1 100 2 10 3 1 4 1/10 5 1 4 1/10 3 1/100 2 1/1000 ------- -------- --------- -------- 6 6 6 6 Does that make any more sense than adding cherries to apples? No! So we need to line up terms where the digits have the same meaning: 1 100 2 10 3 1 4 1/10 5 1 4 1/10 3 1/100 2 1/1000 ------- ------ ----- -------- --------- ---------- 1 100 2 10 8 1 8 1/10 3 1/100 2 1/100 But how do we do this? By lining up the decimal points! So lining up the decimal points is just an easy way of making sure that we're not adding apples to pears, so to speak. Does that make sense? So we don’t add, as in the first example, 1 2 3.4 + 5.4 3 2 6 6 6 6 (where would the decimal point even go?) but rather 1 2 3.4 + 5.4 3 2 1 2 8.8 3 2 Now we’re adding apples to apples. Hidden in that explanation is a deeper mathematical explanation. When we add, say, the tenths column in that example, we are adding (4×1 10)+(4×1 10)=(8×1 10) This amounts to “combining like terms” as if the place values were a variable, or factoring out the common factor 1 10. That’s why the sum of tenths is a number of tenths! We’ll see below how this doesn’t happen with the other operations, so that lining up place values doesn’t help. What follows is a preview of the next two posts, so stop reading if you don’t want a spoiler: Multiplying Now we move on to the negative story: what makes the other operations different. Okay, so much for addition and subtraction. What about multiplication and division? Let's consider the product 3.21 23.4 Now, what we'd really like is to get rid of those decimal points entirely, or at least temporarily. We can use a trick to do that. Note that 3.21 = 32.1 1/10 = 321 1/100 and 23.4 = 234 1/10 So 3.21 23.4 = (321 1/100) 23.4 'Save' two decimal places = (321 1/100) (234 1/10) 'Save' one decimal place = (321 234) (1/100 1/10) = (321 234) 1/1000 = 75114 1/1000 = 75.114 'Restore' three decimal places Notice that when you multiply two digits, you don’t get a “like digit” with the same place value; instead, you combine the two place values (by multiplying). The product of tenths and hundredths has a new place value, thousandths. Moreover, in multiplication, we don’t combine all the digits in one column; we multiply every digit in one number with every digit in the other. So we can pull out all the place information about each factor, and combine them in the final answer: So we can just forget about the decimal points while we're doing the multiplication, so long as we know where to put the decimal point when we're done. In practice, we do this: 3.21 <-- 'save' two decimal places 23.4 <-- 'save' one decimal place 321 234 75114 1/1000 <-- 'restore' three decimal places to get 75.114 So in a sense, the answer to the question "Why don't we have to line up the decimal points when we multiply?" is that we remove them until we're done. You can't line up what isn't there. Or, if you prefer, you don’t need to line up what doesn’t matter yet. We’ll see more about this next time. Dividing What about division? _____ 3.21 ) 23.4 We do a slightly different trick here, which is essentially the same thing we do when we find equivalent fractions: 23.4 23.4 100 ---- = ---- --- 3.21 3.21 100 2340 = ---- 321 So now we have __ 321 ) 2340.0 Once again, a little extra work up front lets us get rid of the decimal point in the divisor. So once again, there's nothing to line up. Here he multiplied each number by 100, which eliminates the decimal point in the divisor, and then just divide by the resulting whole number. (But in a sense we do line up some decimal points – in the dividend and the quotient.) We’ll see all this in more detail in two weeks. Post navigation ← Previous Post Next Post → 3 thoughts on “Adding or Subtracting Decimals: How and Why” Pingback: Multiplying Decimals: How and Why – The Math Doctors Pingback: Dividing Decimals: How and Why – The Math Doctors Uncle Jim March 30, 2025 at 9:10 pm As the answer you quoted from Doctor Mateo noted, the purpose for “lining up the decimal points” in addition or subtraction problems is to ensure that the digits in each column have similar place values. To emphasize this point, let’s consider what happens in situations involving numbers written in scientific notation, such as adding 5.7412×10 8+9.263×10 6. If we just “line up the decimal points” like this +5.7412 9.263××10 8 10 6 as a sort of ritual without really thinking about the purpose, then we will have a “5” representing 5×10 8 in the same column with a “9” representing 9×10 6, a “4” representing 4×10 7 in the same column with a “2” representing 2×10 5, etc. Instead, we must first rewrite one or both operands to put the problem in a form that involves multiples of the same power of 10, for example 5.4712×10 8+0.09263×10 8. Then when we line up the decimal points +5.7412 0.09263××10 8 10 8 we’ll have digits with similar place values in the same columns (for example, “9” representing 9×10 6 under 4 representing 4×10 6) where they belong.Reply Leave a Comment Cancel Reply Your email address will not be published.Required fields are marked Type here.. Name Email Website Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. Have a question? Ask it here! 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9225	https://www.examples.com/physics/units-of-acceleration.html	Units of Acceleration - Definition, List of Units, Conversion Chart Loading [MathJax]/extensions/MathMenu.js Subjects English Maths Chemistry Physics Biology Docs Design Business Education Resources English Maths Chemistry Physics Biology Docs Design Business Education AP Practice Tests AP English AP English Language AP English Literature AP Sciences AP Chemistry AP Biology AP Environmental Science AP Physics 1 AP® Physics 2: Algebra-Based AP® Physics C: Electricity and Magnetism AP® Physics C: Mechanics AP® Psychology AP Math and Computer Science AP Calculus AB AP Calculus BC AP® Computer Science A AP® Computer Science Principles AP® Precalculus AP® Statistics AP History and Social Sciences AP® European History AP® Human Geography AP® Macroeconomics AP® Microeconomics AP® United States Government and Politics AP® United States History AP® World History: Modern Accounting CPA CMA CIA Digital Sat Finance CFA CMT More Grad School MCAT Legal MBE Other ACT Digital Sat HomeUnits of Acceleration Physics Units and Measurements – 20+ Examples, Types, Uses Force – 20+ Examples, Formula, Units, Types, Characteristics Inertia – 19+ Examples, Formula, Units, Types Light Energy – 30+ Examples, Formula, Units, Types, Uses Ampere – Definition, Units, Conversion, Types, FAQ’S Units of Electric Flux – Examples, Definition, Units, Conversion Units of Electric Field – Examples, Definition, Units, Conversion Units of Displacement – Examples, Definition, Units, Conversion Units of Wavelength- Examples, Definition, Units, Conversion Units of Vibration – Examples, Definition, Units, Conversion Chart Units of Distance – Definition, Units, Formula, Conversion Units of Moment of Inertia – Definition, Formula, Units, Examples, Conversion Units of Torque – Examples, Definition, Units, Conversion Chart Units of Sound – Definition, Characteristics, Units, Conversion Units of Inductance – Definition, Formula, Units, Conversion Units of Radioactivity – Definition, Formula, Units, Conversion Units of Viscosity – Definition, List of Units, Conversion Chart Units of Current – Examples, Definition, Units, Conversion Chart Units of Magnetic Flux – Definition, List of Units, Conversion Chart Units of Temperature – Examples, Definition, Units, Conversion Home Physics Units of Acceleration – Definition, List of Units, Conversion Chart Created by:Team Physics - Examples.com,Last Updated:July 15, 2024 Notes 1 save Save Notes Units of Acceleration – Definition, List of Units, Conversion Chart What is the Unit of Acceleration? The unit of acceleration is meters per second squared (m/s²) in the International System of Units (SI). This unit represents the change in velocity per unit of time, specifically the change in velocity of one meter per second every second. SI Unit of Acceleration The SI unit of acceleration is meters per second squared(m/s²). It measures the rate at which the velocity of an object changes over time, specifically indicating how much the velocity increases or decreases by each second. It is a fundamental measure in physics and engineering. CGS Unit of Acceleration In the CGS (Centimeter-Gram-Second) system of units, the unit of acceleration is gal, named after Galileo Galilei. One gal is defined as one centimeter per second squared (cm/s²). This unit measures how quickly an object’s velocity changes by one centimeter per second, every second. Key Points: Gal is primarily used in the field of geophysics, especially in measuring the acceleration due to gravity, which is also expressed as 980 gal on Earth’s surface. More commonly, acceleration in the CGS system can be expressed as centimeters per second squared (cm/s²), which directly corresponds to the standard unit in the MKS system, meters per second squared (m/s²), but on a smaller scale. (1 m/s² = 100 cm/s²) Units of Acceleration in MKS System The MKS (Meter-Kilogram-Second) system is a metric system where the unit of acceleration is meters per second squared (m/s²). This unit measures the rate at which an object’s velocity changes in terms of meters per second, every second. Detailed Explanation: Meters per second squared (m/s²): This is the standard unit of acceleration in the International System of Units (SI), widely used in physics, engineering, and various other scientific disciplines. Additional Derived Units of Acceleration in MKS: Kilometers per second squared (km/s²): Though less commonly used, this unit measures larger accelerations, such as in astronomical contexts. (1 km/s² = 1,000 m/s²) Usage: Meters per second squared is universally accepted for describing acceleration in scientific studies, technological applications, and even in everyday contexts like automotive performance. List of Acceleration Units \| Unit of Acceleration \| Symbol \| Conversion to Meters per Second Squared (m/s²) \| Common Usage \| --- --- \| \| Meters per second squared \| m/s² \| 1 m/s² = 1 m/s² \| Standard scientific unit worldwide \| \| Feet per second squared \| ft/s² \| 1 ft/s² ≈ 0.3048 m/s² \| Used primarily in the United States \| \| Gal \| Gal \| 1 Gal = 0.01 m/s² \| Used in geophysics, especially for measuring gravitational acceleration \| \| Kilometers per second squared \| km/s² \| 1 km/s² = 1,000 m/s² \| Less common, used for large-scale accelerations such as in astrophysics \| \| G-force \| G \| 1 G ≈ 9.80665 m/s² (Earth’s gravity) \| Used to describe accelerations relative to Earth’s gravity \| Conversion of Acceleration Units \| Acceleration Conversion \| Conversion Factor \| --- \| \| Meters per second squared to Feet per second squared \| 1 m/s² = 3.28084 ft/s² \| \| Feet per second squared to Meters per second squared \| 1 ft/s² = 0.3048 m/s² \| \| Meters per second squared to G-force \| 1 m/s² = 0.10197 G \| \| G-force to Meters per second squared \| 1 G = 9.80665 m/s² \| \| Gal to Meters per second squared \| 1 Gal = 0.01 m/s² \| \| Meters per second squared to Gal \| 1 m/s² = 100 Gal \| Meters per second squared to Feet per second squared 1 Meters per second squared = 3.28084Feet per second squared ne meter per second squared (m/s²) equals approximately 3.28084 feet per second squared (ft/s²). This conversion facilitates acceleration calculations in various fields. Feet per second squared to Meters per second squared 1 Feet per second squared = 0.3048Meters per second squared One foot per second squared (ft/s²) is approximately equal to 0.3048 meters per second squared (m/s²). This conversion simplifies acceleration measurements in engineering and physics. Meters per second squared to G-force 1 Meters per second squared = 0.10197G-force One meter per second squared (m/s²) corresponds to approximately 0.10197 G-force (G). This conversion is essential for understanding acceleration in terms of gravitational forces. G-force to Meters per second squared 1 G-force = 9.80665Meters per second squared One G-force (G) is equivalent to approximately 9.80665 meters per second squared (m/s²). This conversion is crucial in fields such as aerospace engineering and automotive safety. Gal to Meters per second squared 1 Gal = 0.01Meters per second squared One Gal (Gal) equals 0.01 meters per second squared (m/s²). This conversion simplifies acceleration measurements, especially in geophysics and seismology. Meters per second squared to Gal 1 Meters per second squared = 100Gal One meter per second squared (m/s²) is equivalent to 100 Gals (Gal). This conversion facilitates acceleration calculations in seismic analysis and gravitational studies. FAQ’s How is acceleration calculated? Acceleration (𝑎) can be calculated using the formula: 𝑎=Δ𝑣/Δ𝑡, where Δ𝑣 is the change in velocity and Δ𝑡 is the change in time. What is the difference between acceleration and velocity? Velocity is the rate of change of displacement with respect to time, while acceleration is the rate of change of velocity with respect to time. In simpler terms, velocity tells you how fast an object is moving and in what direction, while acceleration tells you how quickly the velocity is changing. What are the types of acceleration? Linear acceleration: Change in velocity along a straight line. Tangential acceleration: Change in the magnitude of velocity in circular motion. Centripetal acceleration: Acceleration directed towards the center of a circular path. Can acceleration be negative? Yes, acceleration can be negative. Negative acceleration, also known as deceleration or retardation, indicates that an object is slowing down. It simply means that the velocity is decreasing over time. Share : Set content Set content Set content Set content Set content Save Download Units of Acceleration – Definition, List of Units, Conversion Chart What is the Unit of Acceleration? The unit of acceleration is meters per second squared (m/s²) in the International System of Units (SI). This unit represents the change in velocity per unit of time, specifically the change in velocity of one meter per second every second. SI Unit of Acceleration The SI unit of acceleration is meters per second squared(m/s²). It measures the rate at which the velocity of an object changes over time, specifically indicating how much the velocity increases or decreases by each second. It is a fundamental measure in physics and engineering. CGS Unit of Acceleration In the CGS (Centimeter-Gram-Second) system of units, the unit of acceleration is gal, named after Galileo Galilei. One gal is defined as one centimeter per second squared (cm/s²). This unit measures how quickly an object’s velocity changes by one centimeter per second, every second. Key Points: Gal is primarily used in the field of geophysics, especially in measuring the acceleration due to gravity, which is also expressed as 980 gal on Earth’s surface. More commonly, acceleration in the CGS system can be expressed as centimeters per second squared (cm/s²), which directly corresponds to the standard unit in the MKS system, meters per second squared (m/s²), but on a smaller scale. (1 m/s² = 100 cm/s²) Units of Acceleration in MKS System The MKS (Meter-Kilogram-Second) system is a metric system where the unit of acceleration is meters per second squared (m/s²). This unit measures the rate at which an object’s velocity changes in terms of meters per second, every second. Detailed Explanation: Meters per second squared (m/s²): This is the standard unit of acceleration in the International System of Units (SI), widely used in physics, engineering, and various other scientific disciplines. Additional Derived Units of Acceleration in MKS: Kilometers per second squared (km/s²): Though less commonly used, this unit measures larger accelerations, such as in astronomical contexts. (1 km/s² = 1,000 m/s²) Usage: Meters per second squared is universally accepted for describing acceleration in scientific studies, technological applications, and even in everyday contexts like automotive performance. List of Acceleration Units \| Unit of Acceleration \| Symbol \| Conversion to Meters per Second Squared (m/s²) \| Common Usage \| --- --- \| \| Meters per second squared \| m/s² \| 1 m/s² = 1 m/s² \| Standard scientific unit worldwide \| \| Feet per second squared \| ft/s² \| 1 ft/s² ≈ 0.3048 m/s² \| Used primarily in the United States \| \| Gal \| Gal \| 1 Gal = 0.01 m/s² \| Used in geophysics, especially for measuring gravitational acceleration \| \| Kilometers per second squared \| km/s² \| 1 km/s² = 1,000 m/s² \| Less common, used for large-scale accelerations such as in astrophysics \| \| G-force \| G \| 1 G ≈ 9.80665 m/s² (Earth’s gravity) \| Used to describe accelerations relative to Earth’s gravity \| Conversion of Acceleration Units \| Acceleration Conversion \| Conversion Factor \| --- \| \| Meters per second squared to Feet per second squared \| 1 m/s² = 3.28084 ft/s² \| \| Feet per second squared to Meters per second squared \| 1 ft/s² = 0.3048 m/s² \| \| Meters per second squared to G-force \| 1 m/s² = 0.10197 G \| \| G-force to Meters per second squared \| 1 G = 9.80665 m/s² \| \| Gal to Meters per second squared \| 1 Gal = 0.01 m/s² \| \| Meters per second squared to Gal \| 1 m/s² = 100 Gal \| Meters per second squared to Feet per second squared 1 Meters per second squared = 3.28084Feet per second squared ne meter per second squared (m/s²) equals approximately 3.28084 feet per second squared (ft/s²). This conversion facilitates acceleration calculations in various fields. Feet per second squared to Meters per second squared 1 Feet per second squared = 0.3048Meters per second squared One foot per second squared (ft/s²) is approximately equal to 0.3048 meters per second squared (m/s²). This conversion simplifies acceleration measurements in engineering and physics. Meters per second squared to G-force 1 Meters per second squared = 0.10197G-force One meter per second squared (m/s²) corresponds to approximately 0.10197 G-force (G). This conversion is essential for understanding acceleration in terms of gravitational forces. G-force to Meters per second squared 1 G-force = 9.80665Meters per second squared One G-force (G) is equivalent to approximately 9.80665 meters per second squared (m/s²). This conversion is crucial in fields such as aerospace engineering and automotive safety. Gal to Meters per second squared 1 Gal = 0.01Meters per second squared One Gal (Gal) equals 0.01 meters per second squared (m/s²). This conversion simplifies acceleration measurements, especially in geophysics and seismology. Meters per second squared to Gal 1 Meters per second squared = 100Gal One meter per second squared (m/s²) is equivalent to 100 Gals (Gal). This conversion facilitates acceleration calculations in seismic analysis and gravitational studies. FAQ’s How is acceleration calculated? Acceleration (𝑎) can be calculated using the formula: 𝑎=Δ𝑣/Δ𝑡, where Δ𝑣 is the change in velocity and Δ𝑡 is the change in time. What is the difference between acceleration and velocity? Velocity is the rate of change of displacement with respect to time, while acceleration is the rate of change of velocity with respect to time. In simpler terms, velocity tells you how fast an object is moving and in what direction, while acceleration tells you how quickly the velocity is changing. What are the types of acceleration? Linear acceleration: Change in velocity along a straight line. Tangential acceleration: Change in the magnitude of velocity in circular motion. Centripetal acceleration: Acceleration directed towards the center of a circular path. Can acceleration be negative? Yes, acceleration can be negative. Negative acceleration, also known as deceleration or retardation, indicates that an object is slowing down. It simply means that the velocity is decreasing over time. AI Generator Text prompt Add Tone Select a Tone Friendly Formal Casual Instructive Professional Empathetic Humorous Serious Optimistic Neutral Generate 10 Examples of Public speaking 20 Examples of Gas lighting Practice Test What is the SI unit of acceleration? Choose the correct answer m/s² km/h² m/min² cm/s² of 10 Convert 9.8 m/s² to cm/s². Choose the correct answer 980 cm/s² 9.8 cm/s² 98 cm/s² 9800 cm/s² of 10 What is the CGS unit of acceleration? Choose the correct answer m/s² cm/s² mm/s² km/h² of 10 How many m/s² are in 100 cm/s²? Choose the correct answer 1 m/s² 0.1 m/s² 10 m/s² 0.01 m/s² of 10 If a car accelerates at 5 m/s², what is its acceleration in km/h²? Choose the correct answer 18000 km/h² 15000 km/h² 16200 km/h² 14400 km/h² of 10 An object accelerates at 3 m/s². What is this acceleration in mm/s²? Choose the correct answer 300 mm/s² 3000 mm/s² 30 mm/s² 30000 mm/s² of 10 What is the relationship between m/s² and g (acceleration due to gravity)? Choose the correct answer 1 g = 9.8 m/s² 1 g = 10 m/s² 1 g = 0.98 m/s² 1 g = 1 m/s² of 10 Convert 5 g to m/s². Choose the correct answer 45 m/s² 46 m/s² 48 m/s² 49 m/s² of 10 An object has an acceleration of 20 m/s². What is its acceleration in km/h²? Choose the correct answer 259200 km/h² 129600 km/h² 216000 km/h² 72000 km/h² of 10 If acceleration is given in ft/s², what is the SI unit equivalent? Choose the correct answer 0.3048 m/s² 0.9144 m/s² 3.2808 m/s² 1.6093 m/s² of 10 Submit Test Free Download school Ready to Test Your Knowledge? close Before you leave, take our quick quiz to enhance your learning! assessment Assess Your Mastery emoji_events Boost Your Confidence speed Instant Results memory Enhance Retention event_available Prepare for Exams repeat Reinforce Learning 👉 Start the Quiz Now Free Interactive Resources ©2025 Examples.com. All rights reserved. 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9226	https://www.sciencedirect.com/science/article/abs/pii/S0305054820301805	Typesetting math: 100% Skip to main contentSkip to article Sign in Access through your organization Purchase PDF We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. You may also be able to exercise your privacy choices as described in our Privacy Policy Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Contextual Advertising Cookies These cookies are used for properly showing banner advertisements on our site and associated functions such as limiting the number of times ads are shown to each user.
9227	https://www.edboost.org/math-skills/factoring-out-constant	Skip to main content Factoring out a constant Polynomials can often be simplified by factoring. Factoring means breaking a polynomial down into its "factors": variables or terms that are multiplied together to create the polynomial. The most basic form of factoring is to "factor out a variable," which means to divide each term in the polynomial by a common factor. You can think of this as a sort of "reverse distributive property" process. Rather than multiplying each term by something, you're dividing each term by something. The main rule you want to remember is that, in order to factor out a variable, you need to be able to divide every term in the expression by the same factor. So if you have x2+5x, you can factor out x because both terms, x2 and 5x can be divided by x. But, if you have x2+6x+3 you can't factor by anything. x2 and 6x can be divided by x and 6x and 3 can be divided by 3, but nothing goes into all of the terms. Once you find a term that you can factor by, you divide each term by the same factor and put the factor outside the expressin (which will be in parentheses now). Look at your final product and you'll see that what you're doing is essentially the reverse of distributive property! Example: Simplify: 2x2+4x+12 What can you divide each term by? 2! 2(x2+2x+6) You end up with a distributive property expression (for more, see lesson Distributive Property). To test if you factored correctly, distribute and see if you get your original expression: 2(x2+2x+6)=(2×x2)+(2×2x)+(2×6)=2x2+4x+12 Yes! The process worked! You will often use factoring out a constant in conjunction with other types of factoring, but it's always a good first thing to think about when you want to simplify a polynomial expression. Practice Problems: Factoring out a Constant In the following quadratic equations, factor out a constant, if possible: 8x2+4x−10=0 2x2−6x+9=0 4x2−4x=−20 8x2=4x−8 3x2+6=−36x 2x2−6x+6=0 −6x+18=12x2 5x2−2x+10=0 6x2+6x+6=0 3x2−12x=−12 #### Answer Key: Factoring out a Constant Practice AK Skill: Quadratic Equations Common Core Grade Level/Subject Algebra I EdBoost Test: Algebra I Share This Page This site uses cookies. By continuing to browse the site you are agreeing to our use of cookies.
9228	https://www.sciencing.com/how-to-find-the-original-price-13712233/	How To Find The Original Price Science- [x] Astronomy Biology Chemistry Geology Nature Physics Math- [x] Algebra Geometry Technology- [x] Electronics Features About Editorial Policies Privacy Policy Terms of Use © 2025 Static Media. All Rights Reserved How To Find The Original Price ScienceMathTechnologyFeatures Technology How To Find The Original Price ByMary LougeeUpdated: Jan. 7, 2025 6:34 pm EST Rob Dobi/Getty Images Do you always strive to get the best deals on items you purchase? Knowing the original price of an item on sale can help you determine if that discount is worth considering. Some retailers mark prices upward and then take a discount off, so it looks like a great sale price, and learning how to do sales and sales tax calculations lets you find out just how good a deal you are getting. How to Find the OriginalPrice of a Discount To calculate the original price of a discounted or sale item, you need to know the sale price and the discount percentage. The calculations include a simple formula that divides the sale price by the result of 1 minus the discount in percentage form. Use this formula to calculate the original or list price of an item. (text{OP} = frac{ text{ Price}}{1 – text{ Discount}}) OP represents the original price, Price is the sale price and Discount is the percentage of discount. First, calculate 1 – Discount, and then divide the sale price by this number. For example, if you have a sale price of $40 and a discount of 30 percent: (begin{aligned} text{OP} &= frac{40}{1 – 0.30} , &= frac{40}{0.7} , &= 57.14 text{ rounded to two decimals} end{aligned}) So, the original price of the sale item was $57.14. How to Find theOriginal Price After Tax? You may want to calculate the original price of an item after (or with) sales tax so you know what the price will be when the sale is over, because you may not be able to make the purchase during the sale. In this instance, you have the sales price with tax and you want to find the original price with sales tax. You find the original price as above and add the sales tax to it. Building on the previous example: If the original price of the sale item is $57.14 and the tax rate is 8 percent, you would use this formula: (begin{aligned} text{OP with sales tax} &= text{OP} × (text{ tax rate in decimal form} + 1)) (&= text{OP} × (1.08) &= 57.14 × 1.08 &= $61.71 end{aligned}) How Do YouReverse a Percentage? This is another method of finding the original price of an item in which you know the sale price and the percentage of the original price that is discounted. This calculation helps you to find the original price after a percentage decrease. Subtract the discount from 100 to get the percentage of the original price. 2. Multiply the final price by 100. 3. Divide by the percentage in Step One. For example, if the sale price of an item is $200 and it was discounted by 30 percent, then: (100 – 30 = 70) (200 × 100 = 20,000) (20,000 ÷ 70 = 285.71) $285.71 was the original or list price of the item. OtherConsiderations Convert percentages to decimal form by moving the decimal point two places to the left. Always round your answer to two digits following the period when you are finding values with money. References Exceljet: Get Original Price From Percentage Discount Varsity Tutors: Algebra 1: How to Find the Sale Price Math Salamanders: Reverse Percentages Calculator Recommended
9229	https://www.britannica.com/science/atomic-mass	SUBSCRIBE Ask the Chatbot Games & Quizzes History & Society Science & Tech Biographies Animals & Nature Geography & Travel Arts & Culture ProCon Money Videos Related Topics: : atom : atomic mass unit On the Web: : Chemistry LibreTexts - Atomic Mass (Aug. 04, 2025) See all related content atomic mass, the quantity of matter contained in an atom of an element. The observed atomic mass is slightly less than the sum of the mass of the protons, neutrons, and electrons that make up the atom. The difference, called the mass defect, is accounted for during the combination of these particles by conversion into binding energy, according to an equation in which the energy (E) released equals the product of the mass (m) consumed and the square of the velocity of light in vacuum (c); thus, E = mc2. See also atomic weight. Atomic mass is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units. In this scale, 1 atomic mass unit (amu) corresponds to 1.660539040 × 10−24 gram. The atomic mass unit is also called the dalton (Da), after English chemist John Dalton. The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by Melissa Petruzzello. atom Table of Contents Introduction & Top Questions Atomic model Basic properties Atomic number Atomic mass and isotopes The electron Charge, mass, and spin Orbits and energy levels Electron shells Atomic bonds Conductors and insulators Magnetic properties The nucleus Nuclear forces Nuclear shell model Radioactive decay Nuclear energy Development of atomic theory The atomic philosophy of the early Greeks The emergence of experimental science The beginnings of modern atomic theory Experimental foundation of atomic chemistry Atomic weights and the periodic table Kinetic theory of gases Studies of the properties of atoms Size of atoms Electric properties of atoms Light and spectral lines Discovery of electrons Identification of positive ions Discovery of radioactivity Models of atomic structure Rutherford’s nuclear model Moseley’s X-ray studies Bohr’s shell model The laws of quantum mechanics Schrödinger’s wave equation Antiparticles and the electron’s spin Advances in nuclear and subatomic physics Structure of the nucleus Quantum field theory and the standard model References & Edit History Related Topics Images & Videos For Students atom summary Quizzes Facts You Should Know: The Periodic Table Quiz Related Questions How is the atomic number of an atom defined? When are isotopes stable? atom matter Written by James Trefil Clarence J. Robinson Professor of Physics, George Mason University, Fairfax, Virginia. Author of Science Matters: Achieving Scientific Literacy; Other Worlds: The Solar System and Beyond; and Encyclopedia... James Trefil , Sharon Bertsch McGrayne Science writer. Author of Nobel Prize Women in Science, Prometheans in the Lab, The Theory That Would Not Die, and others. Sharon Bertsch McGrayne •All Fact-checked by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Article History Key People: : Ernest Rutherford : Niels Bohr : Lev Davidovich Landau : Steven Chu : William D. Phillips Related Topics: : subatomic particle : allotrope : line spectrum : particle radiation : atomic nucleus See all related content Top Questions What is an atom? An atom is the basic building block of chemistry. It is the smallest unit into which matter can be divided without the release of electrically charged particles. It also is the smallest unit of matter that has the characteristic properties of a chemical element. Are all atoms the same size? All atoms are roughly the same size, whether they have 3 or 90 electrons. Approximately 50 million atoms of solid matter lined up in a row would measure 1 cm (0.4 inches). A convenient unit of length for measuring atomic sizes is the angstrom, defined as 10−10 meters. What does the mass of an atom consist of? The mass of an atom consists of the mass of the nucleus plus that of the electrons. That means the atomic mass unit is not exactly the same as the mass of the proton or neutron. How is the atomic number of an atom defined? The single most important characteristic of an atom is its atomic number (usually denoted by the letter Z), which is defined as the number of units of positive charge (protons) in the nucleus. For example, if an atom has a Z of 6, it is carbon, while a Z of 92 corresponds to uranium. News • Scientists capture the secret quantum dance of atoms for the first time • Aug. 11, 2025, 12:20 AM ET (ScienceDaily) atom, the basic building block of all matter and chemistry. Atoms can combine with other atoms to form molecules but cannot be divided into smaller parts by ordinary chemical processes. Most of the atom is empty space. The rest consists of three basic types of subatomic particles: protons, neutrons, and electrons. The protons and neutrons form the atom’s central nucleus. (The ordinary hydrogen atom is an exception; it contains one proton but no neutrons.) As their names suggest, protons have a positive electrical charge, while neutrons are electrically neutral—they carry no charge; overall, then, the nucleus has a positive charge. Circling the nucleus is a cloud of electrons, which are negatively charged. Like opposite ends of a magnet that attract one another, the negative electrons are attracted to a positive force, which binds them to the nucleus. The nucleus is small and dense compared with the electrons, which are the lightest charged particles in nature. The electrons circle the nucleus in orbital paths called shells, each of which holds only a certain number of electrons. An ordinary, neutral atom has an equal number of protons (in the nucleus) and electrons (surrounding the nucleus). Thus the positive and negative charges are balanced. Some atoms, however, lose or gain electrons in chemical reactions or in collisions with other particles. Ordinary atoms that either gain or lose electrons are called ions. If a neutral atom loses an electron, it becomes a positive ion. If it gains an electron, it becomes a negative ion. These basic subatomic particles—protons, neutrons, and electrons—are themselves made up of smaller substances, such as quarks and leptons. More than 90 types of atoms exist in nature, and each kind of atom forms a different chemical element. Chemical elements are made up of only one type of atom—gold contains only gold atoms, and neon contains only neon atoms--and they are ranked in order of their atomic number (the total number of protons in its nucleus) in a chart called the periodic table. Accordingly, because an atom of iron has 26 protons in its nucleus, its atomic number is 26 and its ranking on the periodic table of chemical elements is 26. Because an ordinary atom has the same number of electrons as protons, an element’s atomic number also tells how many electrons its atoms have, and it is the number and arrangement of the electrons in their orbiting shells that determines how one atom interacts with another. The key shell is the outermost one, called the valence shell. If this outermost shell is complete, or filled with the maximum number of electrons for that shell, the atom is stable, with little or no tendency to interact with other atoms. But atoms with incomplete outer shells seek to fill or to empty such shells by gaining or losing electrons or by sharing electrons with other atoms. This is the basis of an atom’s chemical activity. Atoms that have the same number of electrons in the outer shell have similar chemical properties. This article opens with a broad overview of the fundamental properties of the atom and its constituent particles and forces. Following this overview is a historical survey of the most influential concepts about the atom that have been formulated through the centuries. Britannica Quiz Facts You Should Know: The Periodic Table Quiz Atomic model Most matter consists of an agglomeration of molecules, which can be separated relatively easily. Molecules, in turn, are composed of atoms joined by chemical bonds that are more difficult to break. Each individual atom consists of smaller particles—namely, electrons and nuclei. These particles are electrically charged, and the electric forces on the charge are responsible for holding the atom together. Attempts to separate these smaller constituent particles require ever-increasing amounts of energy and result in the creation of new subatomic particles, many of which are charged. As noted in the introduction to this article, an atom consists largely of empty space. The nucleus is the positively charged centre of an atom and contains most of its mass. It is composed of protons, which have a positive charge, and neutrons, which have no charge. Protons, neutrons, and the electrons surrounding them are long-lived particles present in all ordinary, naturally occurring atoms. Other subatomic particles may be found in association with these three types of particles. They can be created only with the addition of enormous amounts of energy, however, and are very short-lived. Access for the whole family! Bundle Britannica Premium and Kids for the ultimate resource destination. Subscribe All atoms are roughly the same size, whether they have 3 or 90 electrons. Approximately 50 million atoms of solid matter lined up in a row would measure 1 cm (0.4 inch). A convenient unit of length for measuring atomic sizes is the angstrom (Å), defined as 10−10 metre. The radius of an atom measures 1–2 Å. Compared with the overall size of the atom, the nucleus is even more minute. It is in the same proportion to the atom as a marble is to a football field. In volume the nucleus takes up only 10−14 metres of the space in the atom—i.e., 1 part in 100,000. A convenient unit of length for measuring nuclear sizes is the femtometre (fm), which equals 10−15 metre. The diameter of a nucleus depends on the number of particles it contains and ranges from about 4 fm for a light nucleus such as carbon to 15 fm for a heavy nucleus such as lead. In spite of the small size of the nucleus, virtually all the mass of the atom is concentrated there. The protons are massive, positively charged particles, whereas the neutrons have no charge and are slightly more massive than the protons. The fact that nuclei can have anywhere from 1 to nearly 300 protons and neutrons accounts for their wide variation in mass. The lightest nucleus, that of hydrogen, is 1,836 times more massive than an electron, while heavy nuclei are nearly 500,000 times more massive. Basic properties Atomic number The single most important characteristic of an atom is its atomic number (usually denoted by the letter Z), which is defined as the number of units of positive charge (protons) in the nucleus. For example, if an atom has a Z of 6, it is carbon, while a Z of 92 corresponds to uranium. A neutral atom has an equal number of protons and electrons so that the positive and negative charges exactly balance. Since it is the electrons that determine how one atom interacts with another, in the end it is the number of protons in the nucleus that determines the chemical properties of an atom. Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style The Editors of Encyclopaedia Britannica. "atomic mass". Encyclopedia Britannica, 4 Aug. 2025, Accessed 30 August 2025. Share Share to social media Facebook X External Websites San Jose State University - The Semiempirical Formula for Atomic Masses Montgomery College - Atomic Mass (PDF) Chemistry LibreTexts - Atomic Mass and Atomic Number Westfield State University - Department of Chemical and Physical Sciences - Atomic Mass Open Library Publishing Platform - Enhanced Introductory College Chemistry - Atomic Mass Chemistry LibreTexts - Atomic Mass Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. print Print Please select which sections you would like to print: verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Trefil, James, McGrayne, Sharon Bertsch, Bertsch, George F.. "atom". Encyclopedia Britannica, 21 Apr. 2025, Accessed 30 August 2025. Share Share to social media Facebook X URL External Websites Institute of Physics - The atom Energy Education - Atom Open Oregon Educational Resources - Elements and Atoms: The Building Blocks of Matter Mustansiriyah University - Fundamental particles of an atom (PDF) Space.com - Atoms: What are they and how do they build the elements? Annenberg Learner - The Behavior of Atoms: Phases of Matter and the Properties of Gases UEN Digital Press with Pressbooks - The Structure of the Atom United States Nuclear Regulatory Commission - What is an atom ? Projects at Harvard - Atomic Structure & Chemical Bonding Chemistry LibreTexts - The Atom Live Science - What Is an Atom? 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9230	https://www.chegg.com/homework-help/questions-and-answers/buckingham-pi-theorem-class-problem-3-model-propeller-tested-determine-thrust-characterist-q33063999	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: Buckingham Pi Theorem Class Problem #3 A model propeller is being tested to determine thrust characteristics. The variables that are deemed important to the analysis are thrust, T, density, ρ, rotational speed ω, speed of advance, V, propeller diameter, d, shaft depth ,h and viscosity, H. Find the non-dimensionless groups that define this experiment. Hint: 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
9231	https://www.econgraphs.org/textbooks/econ50fall24/week2/lecture4/elasticity_of_substitution	Note: This work is under development and has not yet been professionally edited. If you catch a typo or error, or just have a suggestion, please submit a note here. Thanks! Chapter 4 / Monday, September 30 \| Production Functions 4.7 Elasticity of Substitution Thinking about the different kinds of production functions, one important feature concerns how substitutable capital and labor are. For example, with a linear production function, capital and labor are what we might call perfect substitutes: if $f(L,K) = L + K$, for example, you can always replace one hour of labor with one unit of capital. On the other hand, with a Cobb-Douglas production function, as we just saw, the MRTS changes as you move along an isoquant. One measure of the substitutability of capital and labor is called the elasticity of substitution. A third production function, called CES (or “constant elasticity of substitution”), takes the form (f(L,K) = (aL^\rho + bK^\rho)^{1 \over \rho},)where the Greek letter $\rho$ is a parameter related to the elasticity of substitution. If you drag the blue dot along the isoquant in the graph below, you can see that the MRTS changes; and by using the slider, you can see how the elasticity of substitution affects the behavior of the MRTS: See interactive graph online here. In this case, it diminishes, or gets smaller in absolute value, as you move to the right. This means that as you use more and more labor, the amount of capital it would take to replace one unit of labor gets smaller. The elasticity of substitution is defined as the percentage change in the MRTS due to a $1\%$ change in the ratio of capital to labor, $K/L$, as one moves along an isoquant. Use the slider to change the elasticity of substitution. As you can see, if you drag the elasticity all the way to the left, the isoquants are extremely “bendy,” even to the point of being almost L-shaped; in the middle, the isoquants have the “bowed” shape of the Cobb-Douglas production function; and as it becomes perfectly elastic, the isoquants become linear, indicating that capital and labor are perfectly substitutable. This substitutability of labor and capital has profound political and moral implications. A centerpiece of Andrew Yang’s 2020 Presidential campaign centered around solving the problem of structural job loss caused by automation — an example of companies shifting their production processes away from labor and toward capital. Semi-autonomous truck convoys may be able to deliver goods more safely, more quickly, and with lower emissions than human-driven trucks; but such automation threatens thousands of long-haul truck driving jobs, which are some of the best-paying jobs for workers without a college degree. So the elasticity of substitution isn’t just a dry mathematical formula: at its core, it’s a measurement of how much people’s jobs are at risk of being lost to automation. That doesn’t make the concept “bad,” any more than an earthquake is “bad;” but it does mean that if you’re interested in solving problems of income inequality, understanding the substitutability of labor and capital has got to figure prominently in your analysis. Previous: Cobb-Douglas Production Functions Next: Scaling Production in the Short Run Copyright (c) Christopher Makler / econgraphs.org
9232	https://arxiv.org/pdf/2407.07029	Constructing k-ary Orientable Sequences with Asymptotically Optimal Length Daniel Gabri´ c University of Guelph, Canada Joe Sawada University of Guelph, Canada Abstract An orientable sequence of order n over an alphabet {0, 1, . . . , k −1} is a cyclic sequence such that each length-n substring appears at most once in either direction . When k = 2 , efficient algorithms are known to construct binary orientable sequences, with asymptotically optimal length, by applying the classic cycle-joining technique. The key to the construction is the definition of a parent rule to construct a cycle-joining tree of asymmetric bracelets. Unfortunately, the parent rule does not generalize to larger alphabets. Furthermore, unlike the binary case, a cycle-joining tree does not immediately lead to a simple successor-rule when k ≥ 3 unless the tree has certain properties. In this paper, we derive a parent rule to derive a cycle-joining tree of k-ary asymmetric bracelets. This leads to a successor rule that constructs asymptotically optimal k-ary orientable sequences in O(n) time per symbol using O(n) space. In the special case when n = 2 , we provide a simple construction of k-ary orientable sequences of maximal length. 2012 ACM Subject Classification Mathematics of computing → Discrete mathematics Keywords and phrases orientable sequence, de Bruijn sequence, concatenation tree, cycle-joining, universal cycle Digital Object Identifier 10.4230/LIPIcs... 1 Introduction Given a set S of k-ary strings of length n, a universal cycle is a cyclic sequence of length \|S\| that contains each string in S as a substring exactly once. When S consists of all k-ary strings of length n, universal cycles are known as de Bruijn sequences . Universal cycles have been studied for many fundamental objects including permutations, subsets, and graphs [ 3, 6]. Universal cycles do not exist directly for permutations; however, efficient constructions exist using a shorthand representation [ 21 , 34 ]. Universal cycles for n-subsets of a k-set must satisfy the following necessary condition: k divides (kn ), or equivalently n divides (k−1 n−1 ). For, if 3-subsets of a 6-set, S will contain exactly one of {123 , 132 , 213 , 231 , 312 , 321 }. Universal cycles for subsets, are only known to exist for small values of n [ 23 , 24 , 33 , 39 ]. This gives rise to studying subset packings , that is, cyclic sequences that contains each n-subset at most once as a substring [8, 10, 39]. In this paper we are interested in a set S that does not contain both a string and its reversal. Similar to the problem of subset packings, determining a maximal set S that admits a universal cycle is extremely challenging. A universal cycle for such a k-ary set of length-n strings is known as an orientable sequence of order n (an OS k(n)). By definition, an orientable sequence does not contain a length-n substring that is a palindrome. Orientable sequences were introduced for binary strings by Dai, Martin, Robshaw, and Wild [ 9] with an application related to robotic position sensing. In particular, consider an autonomous robot with limited sen-sors. To determine its location on a cyclic track labeled with coloured squares, the robot scans a window of n squares directly beneath it (see the graphic on the right). For the position and orientation to be uniquely determined, the track is designed with the property that each length n window can appear at most once in either direction . © ; licensed under Creative Commons License CC-BY 4.0 Leibniz International Proceedings in Informatics Schloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany arXiv:2407.07029v1 [cs.DM] 9 Jul 2024 XX:2 Orientable sequences Example 1 Consider the sequence S = 012013023123 over the alphabet {0, 1, 2, 3}. In the forward direction, including the wraparound, S contains 012 , 120 , 201 , 013 , 130 , 302 , 023 , 231 , 312 , 123 , 230 , 301 as substrings; in the reverse direction S contains 321 , 213 , 132 , 320 , 203 , 031 , 310 , 102 , 021 , 210 , 103 , 032 .Since each substring is unique, S is an OS 4(3) with length 12. For biological applications, {0, 1, 2, 3} represents the four nucleotide bases {A, C, G, T} of a DNA strand – see the discussion at the end of the section. Recently, Mitchell and Wild developed a recursive algorithm to construct long orientable sequences for a binary alphabet . Subsequently, efficient constructions of OS 2(n)s with asymptotically optimal length have been developed based on cycle-joining [ 18 ], and a previously known existence proof [ 9]. However, there is no known construction of long orientable sequences for k > 2. In this paper we demonstrate that the binary cycle-joining approach does not naturally generalize to larger alphabets. In particular, there are special cases for k = 3 that arise for n ≥ 12 (see Example 6 in Section 5), and there are additional challenges in deriving successor rules from cycle-joining trees for k ≥ 3 (see Section 2.2). However, by deriving a new parent rule that satisfies the Chain Property (see Section 2.2), we are able to obtain the first efficient construction of long OS k(n)s. Main result : For k ≥ 3, we develop a successor rule to construct an OS k(n) of asymptotically optimal length in O(n) time per symbol using O(n) space. For n = 2 , we construct OS k(2) s of maximal length in O(1) time per symbol. Let Mk(n) denote the maximum length of an OS k(n). When k = 2 , the maximum length of an orientable sequence is known only for n ≤ 7 [9, 18 ]. For n = 2 and k ≥ 3, we demonstrate that Mk(2) = k⌊(k − 1) /2⌋ by a simple construction (see Section 3). For n ≥ 3, exhaustive search demonstrates that M3(3) = 9 , M4(3) = 20 , and M3(4) = 30 . Search also reveals an OS 5(3) of length 50 which attains the upper bound stated in [ 2 ], and thus M5(3) = 50 . Orientable sequences that admit these maximal lengths are given below: n = 3 , k = 3 : 001120122 (9), n = 3 , k = 4 : 00112012230130231233 (20), n = 3 , k = 5 : 00112003102210320331140142042132143043144223342344 (50), n = 4 , k = 3 : 000102001201112022101121022212 (30). Since the number of palindromes of length n is k⌊(n+1) /2⌋, a trivial upper bound on Mk(n) is (kn − k⌊(n+1) /2⌋)/2. A deeper analysis on upper bounds is given in . Recall the problem of determining a robot’s position and orientation on a track. Suppose now that we allow the track to be non-cyclic. The corresponding sequence that allows one to determine orientation and position is called an acyclic orientable sequence . One can construct an acyclic OS k(n) from a cyclic OS k(n) by taking the cyclic OS k(n) and appending its prefix of length n−1 to the end. See the paper by Burns and Mitchell [ 5] for more on binary acyclic orientable sequences, which they call aperiodic 2-orientable window sequences . Gabric and Sawada provide some long acyclic orientable sequences in [ 17 ]. Rampersad and Shallit [ 32 ] showed that for every alphabet size k ≥ 2 there is an infinite sequence such that for every sufficiently long substring, the reversal of the substring does not appear in the sequence. Fleischer and Shallit [ 15 ] later reproved the results of the previous paper using theorem-proving software. See [ 7 , 29 ] for more work on sequences avoiding reversals of substrings. Families of strings related to orientable sequences also appear in DNA computing. Two single strands of DNA can bind to each other if they are “reverse complements” of each other, where A is the complement of T and C of G. The binding of DNA strands allows for the creation of secondary structures, which are useful in certain DNA computing XX:3 techniques [ 31 ]. For example, a stem-loop , also known as a hairpin , is a DNA secondary structure that has applications in DNA computing [ 11 , 28 , 41 ]. Roughly speaking, a string of symbols u contains a hairpin if it has substring v and θ(v),where θ is an antimorphic involution. Hairpin-free strings, that is, strings that do not contain a hairpin of sufficient length, have been studied with the motivation of creating a large collection of DNA molecules that do not bind to themselves in undesirable ways [ 27 ]. In the case that θ is the mirror involution (i.e., reversal), hairpin-free strings are essentially orientable sequences without the restriction that every substring of a specified length occurs at most once. Thus, every orientable sequence is a hairpin-free sequence. See [26, 40] for more on applications of long hairpin-free sequences. Outline. In Section 2, we provide background definitions and notation, including a review of the cycle-joining technique, and k-ary successor rules. In Section 3, we present a simple construction for OS k(2) s and demonstrate they are of maximal length. In Section 4, we define k-ary symmetric/asymmetric necklaces and bracelets, and provide some useful properties of these objects. In Section 5, we provide a parent rule for constructing a cycle-joining tree composed of asymmetric bracelets. This leads to an O(n) time per symbol successor-rule construction of OS k(n)s that we demonstrate has asymptotically optimal length in Section 6. An implementation of our construction is available for download at . 2 Preliminaries Let Σ = {0, 1, 2, . . . , k −1} be an alphabet of size k ≥ 2. Let Σn denote the set of all length-n strings over Σ. Let α = a1a2 · · · an ∈ Σn and β = b1b2 · · · bm ∈ Σm for some m, n ≥ 0, Throughout this paper, we use lexicographic order when comparing two strings. More specifically, α < β if α is a prefix of β or if ai < bi for the smallest i such that ai̸ = bi.Let αR denote the reversal anan−1 · · · a1 of α; α is a palindrome if α = αR. For j ≥ 1, let αj denote j copies of α concatenated together. If α = γj for some non-empty string γ and some j > 1, then α is said to be periodic ; otherwise, α is said to be aperiodic (or primitive ). Let ap( α) denote the shortest string γ such that α = γt for some positive integer t;we say γ is the aperiodic prefix of α.A necklace class is an equivalence class of strings under rotation. Let [α] denote the set of strings in α’s necklace class. We say α is a necklace if it is the lexicographically smallest string in [α]. Let ˜α denote the necklace in [α]. For example, if α = 0201 , then [α] = {0201 , 2010 , 0102 , 1020 } and ˜α = 0102 . Let Nk(n) denote the set of k-ary necklaces of length n. A bracelet class is an equivalence class of strings under rotation and reversal. We say α is a bracelet if it is the lexicographically smallest string in [α] ∪ [αR]. A bracelet is always a necklace, but a necklace need not be a bracelet. Given S ⊆ Σn, a universal cycle U for S is a cyclic sequence of length \|S\| that contains each string in S as a substring (exactly once). An orientable sequence is a universal cycle where if α ∈ S, then αR /∈ S. If S = Σ n then U is known as a de Bruijn sequence . Given a universal cycle U for a set S ⊆ Σn, a successor rule for U is a function f : S → Σ such that f (α) is the symbol following α in U . 2.1 Cycle-joining trees In this section we review how two universal cycles can be joined to obtain a larger universal cycle. Cycle joining is perhaps the most fundamental technique applied to construct universal cycles; it has graph-theoretic underpinnings related to Hierholzer’s algorithm for constructing Euler cycles . For some applications, see [12, 13, 14, 16, 19, 22, 25, 36, 38]. Let x, y be distinct symbols in Σ. If α = xa 2 · · · an and ˆα = ya 2 · · · an, then α and ˆα are said to be conjugates of each other, and (α, ˆα) is called a conjugate pair . We say γ belongs to a conjugate pair (α, ˆα) if either γ = α or γ = ˆ α. The following well-known result (see for instance Lemma 3 in [ 37 ]) based on conjugate pairs is the crux of the cycle-joining approach. ▶ Theorem 1. Let S1 and S2 be disjoint subsets of Σn such that α = xa 2 · · · an ∈ S1 and ˆα = ya 2 · · · an ∈ S2; (α, ˆα) is a conjugate pair. If U1 is a universal cycle for S1 with suffix α and U2 is a universal cycle for S2 with suffix ˆα then U = U1U2 is a universal cycle for S1 ∪ S2.XX:4 Orientable sequences Let Ui denote a universal cycle for Si ⊆ Σn. Two universal cycles U1 and U2 are said to be disjoint if S1 ∩ S2 = ∅.Theorem 1 states that two disjoint universal cycles can be joined to form a single universal cycle if they each contain one string of a conjugate pair as a substring. Note that necklaces correspond to disjoint cycles that partition the set Σn.A cycle-joining tree T is an unordered tree where the nodes correspond to a disjoint set of universal cycles U1, U 2, . . . , U t;an edge between Ui and Uj is defined by a conjugate pair (α, ˆα) such that α ∈ Si and ˆα ∈ Sj . For our purposes, we consider cycle-joining trees to be rooted. Example 2 Let n = 3 and k = 4 . Consider necklace classes S1 = , S2 = , and S3 = 210 110 310 with corresponding universal cycles U1 = 210 , U2 = 110 and U3 = 310 . The three cycles can be joined via conjugate pairs ( 210 , 110 ) and ( 210 , 310 ) to form the cycle-joining tree on the right. Joining U1 and U2 we obtain the larger cycle 210110 ; joining U3 to this cycle we obtain the universal cycle 110210310 for (S1 ∪ S2) ∪ S3. If we join U1 and U3 first, we obtain a different universal cycle 310210110 .Many universal cycle constructions have a corresponding cycle-joining tree that can be defined by a rather simple parent rule . For example, when S = Σ n and α = a1a2 · · · an ∈ Nk(n), the following are four of the simplest parent rules that define how to construct cycle-joining trees with nodes corresponding to Nk(n) : firstSymbol( α) = the necklace of [( a1−1) a2 · · · an] with root (k−1) n, lastSymbol( α) = the necklace of [a1a2 · · · an−1(an+1)] with root 0n, firstNonMin( α) = 0 i−1(ai−1) ai+1 · · · an with root 0n, and lastNonMax( α) = a1 · · · aj−1(aj +1)( k−1) n−j with root (k−1) n,where i denotes the index of the first non-zero in α and j denotes the index of the last non-( k−1) in α. Addition on the symbols is modulo k. From the definition of a necklace, it is straightforward to see that if α is a (non-root) necklace, then both 0i−1(ai−1) ai+1 · · · an and a1 · · · aj−1(aj +1)( k−1) n−j are also necklaces. Note that if a necklace α is periodic, the corresponding universal cycle for α is its aperiodic prefix. However, for simplicity of understanding, we use the full necklace to represent a node in our cycle-joining trees. For instance, when n = 3 , we use 000 instead of 0. As an example, Figure 1 illustrates the cycle-joining trees induced by the four parent rules above with nodes N3(3) . Each node α and its parent β are joined by a conjugate pair, where the highlighted bit in α is the first bit in one of the conjugates. When k > 3,we apply the last three of the four parent rules to construct an OS k(n) with length that is asymptotically optimal. When k = 3 , we must introduce one additional function. 000 001 002 012 022 122 222 011 111 021 112 lastSymbol( α)000 001 002 012 022 122 222 011 111 021 112 firstNonMin( α)000 001 002 012 022 122 222 011 111 021 112 lastNonMax( α)000 001 002 012 022 122 222 011 111 112 021 firstSymbol( α) Figure 1 Cycle-joining trees for N3(3) induced by four different parent rules. XX:5 2.2 Successor rules In this section we outline how to derive a successor-rule from a cycle-joining tree T for an underlying set S. In the binary case, each cycle-joining tree corresponds to a unique universal cycle; however, when k > 2, this is not necessarily the case. Uniqueness Property : The cycles in a cycle-joining tree T are joined such that no two conjugate pairs that have a string in common, i.e., there are no two conjugate pairs of the form ( xa 2 · · · an, ya 2 · · · an) and ( xa 2 · · · an, za 2 · · · an). The Uniqueness Property is always satisfied when k = 2 , but is not necessarily the case for k > 2. For example, none of the trees in Figure 1 satisfy the Uniqueness Property. If T has the Uniqueness Property, then a successor rule for the corresponding unique universal cycle is given by f (α), where α = a1a2 · · · an: f (α) = { y if α belongs to some conjugate pair (α, ya 2 · · · an); a1 otherwise. When the Uniqueness Property is not satisfied, a universal cycle derived from a cycle-joining tree depends on the order that the cycles are joined together, as illustrated in Example 2. See [ 35 ] for a deeper analysis on different universal cycles that can be obtained from the same cycle-joining tree. Our challenge is to create a relatively simple successor rule that defines a universal cycle derived from a cycle-joining tree T that does not satisfy the Uniqueness Property. Ultimately, we require our cycle-joining tree to be defined with the following property: Chain Property : If a node in a cycle-joining tree T has two children joined via conjugate pairs (xa 2 · · · an, ya 2 · · · an) and (x′b2 · · · bn, y′b2 · · · bn), then a2 · · · an̸ = b2 · · · bn.Let α1, α 2, . . . , α m denote a maximal-length path of nodes in T such that for each 1 ≤ i < m , the node αi is the parent of αi+1 and they are joined via a conjugate pair of the form (xiβ, xi+1 β), where β is the same in each conjugate pair. We call such a path a chain of length m. Any node with j children will belong to at least j chains. Given a chain let NEXT (xiβ) = xi+1 , where xm+1 = x1. Example 3 The nodes α1 = 112 , α2 = 011 , and α3 = 111 in the first tree of Figure 1 form a chain with length m = 3 joined by conjugate pairs (211,011) and (011,111). N EXT (211) = 0 , N EXT (011) = 1 , and N EXT (111) = 2 .If T is a cycle-joining tree with the Chain Property for an underlying set S, then the following function g is a successor rule for a universal cycle of S (based on theory in ): g(α) = { NEXT (α) if α belongs to some conjugate pair; a1 otherwise. When k = 2 , f = g. As stated, this successor rule requires exponential space to store the conjugate pairs. In our application (see Section 5.2), the corresponding successor rule will run in O(n) time per symbol and use O(n) space. 3 A maximal-length construction for OS k(2) s In this section we consider the case when n = 2 . There does not exist an OS 1(2) or an OS 2(2) , so we assume k ≥ 3.Since there are no substrings of the form xx in any OS 2(2) , each symbol in any OS k(2) can appear at most ⌊(k − 1) /2⌋ times. Thus, Mk(2) ≤ k⌊(k − 1) /2⌋. In fact, we show that this bound is tight via a simple construction that depends on the parity of k.XX:6 Orientable sequences For each pair of distinct symbols x and y, an OS k(2) can contain either xy or yx , but not both. If k is odd, our OS k(2) will contain exactly one such substring for each pair of symbols. If k is even, we must remove k/ 2 such pairs to meet the upper bound. In particular, we remove the pairs {0, 1}, {2, 3}, . . . , {k−2, k −1}. We outline our choices from each pair of symbols, and illustrate OS k(2) s in Figure 2 for k = 7 and k = 8 ; it is straightforward to generalize the construction depending on the parity of k, as follows. k odd . Let σ3 = 012 . For k ≥ 5, let σk = s1s2 · · · s2k−3 where s1s3s5 · · · s2k−5 = 012 · · · (k−3) and s2s4s6 · · · s2k−4s2k−3 = (( k−2)( k−1)) (k−1) /2. k even . Let σ4 = 0213 . For k ≥ 6, let τk = t1t2 · · · t2k−4 where t1t3t5 · · · t2k−5 = 012 · · · (k−3) and t2t4t6 · · · t2k−4 = (( k−2)( k−1)) k/ 2−1.Let Uk = σ3σ5 · · · σk for k odd, and let Uk = τ4τ6 · · · τk for k even. 10 02 21 30 04 50 06 13 41 15 61 32 24 52 26 43 35 63 54 46 65 70 17 72 37 74 57 76 U7 = 012031423405162536456 U8 = 021304152435061726374657 01 20 12 03 40 05 60 31 14 51 16 23 42 25 62 34 53 36 45 64 56 Figure 2 Illustrating the construction of U7and U8. Example 4 The following illustrates Uk for 3 ≤ k ≤ 8: U3 = 012 U5 = 012 0314234 U7 = 012 0314234 05162536456 U4 = 0213 U6 = 0213 04152435 U8 = 0213 04152435 061726374657 ▶ Theorem 2. For k > 2, Uk is an OS k(2) with length k⌊(k − 1) /2⌋ that can be generated in O(1) -time per symbol. Proof. We provide a high-level proof outline. Clearly U3 is an OS 3(2) with length 3, and U4 is an OS 4(2) with length 4. Suppose k ≥ 5 is odd. The sequences σ3, σ 5, · · · , σ k are disjoint OS k(2) s that are cycle-joined to obtain Uk. Simple math shows Uk has length k⌊(k − 1) /2⌋. A similar analysis can be applied for even k. It is straightforward to generate Uk in constant time per symbol. ◀ An immediate consequence of this theorem is Mk(2) = k⌊(k − 1) /2⌋.Observe that when n is odd, Uk is a universal cycle for the 2-subsets of a k-set. Previously, such universal cycles were claimed to exist in [ 6], with justification in [ 24 ]. No construction was previously provided. When n is even, Uk is a maximal 2-subset packing of a k-set. 4 Symmetric and asymmetric necklaces and bracelets In this section, we present properties for symmetric/asymmetric necklaces and bracelets that are necessary for our main results in Section 5. A necklace α is symmetric if it belongs to the same necklace class as αR, i.e., both α and αR belong XX:7 to [α]. By this definition, a symmetric necklace is a bracelet. If a necklace or bracelet is not symmetric, it is said to be asymmetric . Let Ak(n) denote the set of all k-ary asymmetric bracelets of length n. Table 1 lists all 70 necklaces in N4(4) partitioned into asymmetric necklace pairs and symmetric necklaces. The asymmetric necklace pairs belong to the same bracelet class, and the first string in each pair is an asymmetric bracelet. Thus, \|A4(4) \| = 15 . In general, \|Ak(n)\| is equal to the number of k-ary necklaces of length n minus the number of k-ary bracelets of length n. For more background on asymmetric bracelets, see [1, 18]. Asymmetric necklace pairs Symmetric necklaces 0012 , 0021 0133 , 0331 0000 0102 0222 1113 1323 0013 , 0031 0213 , 0312 0001 0103 0232 1122 1333 0023 , 0032 0223 , 0322 0002 0111 0303 1133 2222 0112 , 0211 0233 , 0332 0003 0121 0313 1212 2223 0113 , 0311 1123 , 1132 0011 0131 0323 1213 2233 0122 , 0221 1223 , 1322 0022 0202 0333 1222 2323 0123 , 0321 1233 , 1332 0033 0203 1111 1232 2333 0132 , 0231 0101 0212 1112 1313 3333 Table 1 The 70 necklaces in N4(4) partitioned into 15 asymmetric necklace pairs and 40 symmetric necklaces. The first (highlighted) necklace in each pair is an asymmetric bracelet in A4(4) . Let α1, α 2, . . . , α m denote the asymmetric bracelets in Ak(n). Let Sk(n) denote the set [α1] ∪ [α2] ∪ · · · ∪ [αm]. Important property: If α ∈ Sk(n), then αR /∈ Sk(n).Let Lk(n) = \|Sk(n)\|. Our main result defines a cycle-joining tree with nodes Ak(n), producing an OS k(n) of length Lk(n). In Section 6, we provide a formula for Lk(n) and demonstrate it to be an asymptotically optimal lower bound for Mk(n) (the maximal length of an OS k(n)). The reminder of this section is devoted to properties of symmetric/asymmetric necklaces and bracelets required to prove our main results in Section 5. ▶ Lemma 3 () . A necklace α is symmetric if and only if there exists palindromes β1 and β2 such that α = β1β2. ▶ Lemma 4. Let α = a1a2 · · · an̸ = 0 n be a bracelet, where i is the index of the first non-zero symbol. If ai > 1, then firstNonMin( α) is a bracelet; moreover, if α ∈ Ak(n), then firstNonMin( α) ∈ Ak(n). Proof. Suppose ai > 1. It is straightforward to apply the definition of a bracelet to verify that β = firstNonMin( α) = 0i−1(ai−1) ai+1 · · · an is a bracelet, given α is a bracelet. Let α ∈ Ak(n). Suppose β is symmetric. From Lemma 3, β = β1β2 for palindromes β1 and β2. Clearly \|β1\| ≥ i − 1. Since α is a bracelet ai ≤ an, and thus \|β1\|̸ = i − 1. If \|β1\| = i,then i = 1 which implies that α is symmetric; otherwise, since β1 has prefix 0i−1, it must have length at least 2i − 1. If \|β1\| = 2 i − 1, then α is symmetric; if \|β1\| > 2i − 1 then α is not a bracelet. Each case contradicts that α ∈ Ak(n). Thus, β is an asymmetric bracelet. ◀ The following lemma follows by applying the definition of a bracelet. ▶ Lemma 5. If α̸ = ( k−1) n is a bracelet, then lastNonMax( α) is a bracelet. Note that lastSymbol( α) = lastNonMax( α) when an < k −1. ▶ Lemma 6. Let α = a1a2 · · · an ∈ Ak(n), where j is the index of the last non-( k−1) and ℓ is the index of the second last non-( k−1). If (i) an < k −1 or (ii) aj < k −2 or (iii) a1 · · · aℓ̸ = ( a1 · · · aℓ)R, then lastNonMax( α) ∈ Ak(n);otherwise, lastNonMax( α) is a symmetric bracelet. XX:8 Orientable sequences Proof. From Lemma 5, β = lastNonMax( α) is a bracelet. If β is symmetric, then from Lemma 3 it can be written as β1β2 for some palindromes β1 and β2. Since a1 < an because α ∈ Ak(n), we clearly have β1 and β2 are non-empty. (i) Let an < k −1. Suppose β is symmetric. Let γ be β2 with its last symbol decremented; α = β1γ. Then (γβ 1)R ≤ α,contradicting that α ∈ Ak(n).(ii) Let aj < k −2. Suppose β is symmetric. If \|β1\| > j , then it must start and end with (k−1) and include aj . This contradicts that β is a bracelet. If \|β1\| = j, then since β1 is a palindrome, the length j prefix of α is greater than its reversal, contradicting that α is a bracelet. Thus β2 has suffix (aj +1) aj+1 · · · an. If \|β2\| = 2( n − j) + 1 , then the suffix of α with the same length is also a palindrome, which implies α is symmetric, contradiction. Otherwise, \|β2\| > 2( n − j) + 1 . Let γ denote the suffix of α of length \|β2\|. It must be that γ > γ R, which implies that (γβ 1)R < α ,contradicting that α is a bracelet. Thus, β ∈ Ak(n).(iii) Let a1 · · · aℓ̸ = ( a1 · · · aℓ)R. By the definition of a bracelet, a1 · · · aℓ < (a1 · · · aℓ)R. If aj < k −2, then β is in Ak(n) by the previous case. Consider aj = k−2. Suppose β = a1 · · · aℓ(k−1) n−ℓ is symmetric. We must have \|β1\| < ℓ .Since aℓ < (k−1) , \|β2\| ≥ 2( n − ℓ) + 1 . But this implies that α is not a bracelet, contradiction. Thus, β ∈ Ak(n).Finally, if an = k−1, aj = k−2, and a1 · · · aℓ = ( a1 · · · aℓ)R, then lastNonMax( α) = a1 · · · aℓ(k−1) n−ℓ, which is symmetric by Lemma 3. ◀▶ Lemma 7. If α ∈ Ak(n) such that lastNonMax( α) /∈ Ak(n), lastSymbol( α) /∈ Ak(n) and firstNonMin( α) ∈ Ak(n), then lastNonMax(firstNonMin( α)) ∈ Ak(n) or lastSymbol(firstNonMin( α)) ∈ Ak(n). Proof. Consider α = a1a2 · · · an ∈ Ak(n) where i is the index of the first non-zero, j is the index of the last non-(k−1) ,and ℓ is the index of the second-last non-(k−1) . Since lastNonMax( α) /∈ Ak(n), from Lemma 6, aj = k−2, an = k−1,and a1 · · · aℓ = ( a1 · · · aℓ)R. Since α ∈ Ak(n), α = a1 · · · aℓ(k−1) j−ℓ−1(k−2)( k−1) n−j where j − ℓ − 1 < n − j.This implies that lastSymbol( α) = 0 a1 · · · an−1 has prefix 0i; it is a necklace since α has no 0i substring. Since lastSymbol( α) /∈ Ak(n), ai · · · an−1 ≥ (ai · · · an−1)R, which means that ai ≥ an−1 ≥ k−2. Let β = firstNonMin( α).If i ≥ n − 1, then α = 0 n−2(k−2)( k−1) and lastNonMax( β) = α. Thus, assume i < n − 1 and consider two cases. Suppose i > ℓ . Then i = ℓ + 1 and α = 0 ℓ(k−1) j−i(k−2)( k−1) n−j . Since j − i < n − j and ai · · · an−1 ≥ (ai · · · an−1)R, j − i = n − j − 1 > 1. Thus, lastSymbol( β) = 0 ℓ(k−2)( k−1) n−i which is in Ak(n).Suppose i ≤ ℓ. If the first ℓ symbols of β are not a palindrome, then by Lemma 6, lastNonMax( β) ∈ Ak(n).Otherwise, ℓ is odd, i = ( ℓ + 1) /2, and α has prefix 0i−1ai0i−1. If ai = k−2, since ai · · · an−1 ≥ (ai · · · an−1)R, we have an−1 = k−2, α = 0 i−1(k−2)0 i−1(k−2)( k−1) , and lastSymbol( β) = 0 i(k−3)0 i−1(k−2) which is in Ak(n).Otherwise, ai = k−1 and an−1 = k−1 which means i > 1, j < n − 1 and an−2 ≥ k−2. But this contradicts that ai · · · an−1 ≥ (ai · · · an−1)R. ◀ 5 A cycle-joining construction for orientable sequences In this section we provide a parent rule to create a cycle-joining tree with nodes Ak(n). We then apply the tree to derive an O(n)-time successor rule for a corresponding universal cycle, which is an OS k(n) of length Lk(n). 5.1 A simple parent rule The parent rule for the binary case defined in [ 18 ] uses 0n−41011 as the root, and the parent of each non-root node α ∈ A2(n) is the first string in the list ⟨firstNonMin( α), lastSymbol( α), lastNonMax( α)⟩ that is also in A2(n). However, there are several issues when generalizing to a larger alphabet. In particular, the rule is not well-defined for k = 3 , and the corresponding cycle-joining tree does not have the Chain Property. We will demonstrate each of these short-comings before deriving a new parent rule for alphabets of arbitrary size. Assume n, k ≥ 3. Let rn,k = 0 n−2(k−2)( k−1) denote the root of our upcoming cycle-joining tree. The following example is for this specific root; however, similar examples exist for any arbitrary root. XX:9 Example 5 Consider any parent rule with root rn,k where the the parent of α ∈ Ak(n) is the first string in a list starting with ⟨firstNonMin( α), lastSymbol( α), . . . ⟩ that is also in Ak(n). Let α = 012202 and β = 010122 .Then the parent of α is lastSymbol( α) = 000122 and the parent of β is firstNonMin( β) = 000122 ; α is joined via conjugate pair ( 201220 , 001220 ) and β is joined via conjugate pair ( 101220 , 001220 ). The two conjugate pairs share a string, and thus the corresponding cycle-joining tree does not have the Chain Property. The next example, and following lemma, demonstrate that the four functions firstSymbol , firstNonMin , lastSymbol ,and lastNonMax alone are not sufficient to define a parent rule with root rn,k when k = 3 . Example 6 Consider α = 001022010012 which is in A3(12) . No matter how we change the first, last, first non-zero, or last non-( k−1) symbol in α, the resulting string is not in A3(12) . In particular: firstSymbol( α) = 001220102201 is not a bracelet, lastSymbol( α) = 000102201001 is not a bracelet, firstNonMin( α) = 0000022010012 is not a bracelet, and lastNonMax( α) = 001022010022 is symmetric. Such strings are uncommon. There are only 82 such strings in A3(20) and they all have suffix 0012. ▶ Lemma 8. Let α = 00102 (n−10) 010012 for n ≥ 12 and k = 3 . Then α is in Ak(n) and each of firstSymbol( α), lastSymbol( α), firstNonMin( α), and lastNonMax( α) is not in Ak(n). Proof. It is straightforward to observe that α is a bracelet by definition and is asymmetric by Lemma 3. Applying the definitions, firstSymbol( α) = 001220102 (n−10) 01 , lastSymbol( α) = 000102 (n−10) 01001 , and firstNonMin( α) = 000002 (n−10) 010012 are all not bracelets, and lastNonMax( α) = 00102 (n−10) 010022 is symmetric. ◀ Lemma 8, demonstrates that for k = 3 , no parent rule exists for A3(n) that applies only a combination of the four rules from Section 2.1. Thus, we define an additional function in order to define our parent rule for A3(n). Let ℓ denotes the second last symbol in α that is not (k−1) , and define secondLastNonMax( α) = a1 · · · aℓ−1(aℓ+1) aℓ+1 · · · an.This function is well-defined, since each α ∈ Ak(n) must contain at least two symbols that are not k−1. Recall Example 6, where α = 001022010012 . Observe that secondLastNonMax( α) = 001022010112 , which is in A3(12) . Parent rule for cycle-joining Ak(n) with root rn,k . If α = a1a2 · · · an ∈ Ak(n) \ { rn,k }, then define par( α) to be the first string that is an asymmetric bracelet in the list ⟨ lastNonMax( α), lastSymbol( α), firstNonMin( α), secondLastNonMax( α) ⟩. In the upcoming Lemma 9, we demonstrate that par( α) is well-defined and interestingly, that secondLastNonMax( α) is only necessary for k = 3 . The upcoming Theorem 10 demonstrates that the above parent rule induces a cycle-joining tree, which we denote by Tk(n). Moreover, we demonstrate Tk(n) has the Chain Property in Theorem 12. In proving these results, we do not consider the case when n = 3 and k = 3 , since A3(3) = {012 }. Figure 3 illustrates T3(6) . ▶ Lemma 9. Let α = a1a2 · · · an ∈ Ak(n) \ { rn,k } for some n ≥ 3, k ≥ 4 or n ≥ 4, k = 3 such that lastNonMax( α) and lastSymbol( α) are not in Ak(n). If k ≥ 4, then firstNonMin( α) ∈ Ak(n). Furthermore, if k = 3 and firstNonMin( α) /∈ A3(n), then α = 0 γ012 where γ is a palindrome, and secondLastNonMax( α) ∈ A3(n).XX:10 Orientable sequences 000012 0001 1200012 20010 120012 020112 02011012 0020 1200101 10122 0201012 2001 122 0120 1200 2122 0011 120101 120001 0200122 2001 022 0111 120011 0200112 101 1222 0012 1201222 2021 122 011 122 0122 120112 1202122 201122 1012 122 11212 20121 1201112 1002 022 00102 1012 022 Figure 3 The cycle-joining tree T3(6) . Each node differs from its parent (cyclically) at the highlighted symbol. The symbols highlighted in blue indicate that par( α) = lastNonMax( α); the symbols highlighted in red indicate that par( α) = lastSymbol( α);the symbols highlighted in bold black indicate that par( α) = firstNonMin( α). There are no nodes in this tree such that par( α) = secondLastNonMax( α); the first instance of such a case arises when n = 12 . Proof. Consider α where i is the index of the first non-zero, i′ is the index of the second non-zero, j is the index of the last non-(k−1) , and ℓ is the index of the second-last non-(k−1) . Since lastNonMax( α) /∈ Ak(n), from Lemma 6, aj = k−2, an = k−1, and a1 · · · aℓ = ( a1 · · · aℓ)R. Thus, since α ∈ Ak(n), α = a1 · · · aℓ(k−1) j−ℓ−1(k−2)( k−1) n−j where j − ℓ − 1 < n − j. Thus, if j = n − 1, then ℓ = n − 2. This implies that lastSymbol( α) = 0 a1 · · · an−1 with prefix 0i, which is a necklace since α has no 0i substring. Since lastSymbol( α) /∈ Ak(n), ai · · · an−1 ≥ (ai · · · an−1)R,which means that ai ≥ an−1 ≥ k−2. If k ≥ 4, then ai > 1, and thus by Lemma 4 firstNonMin( α) ∈ Ak(n). If k = 3 ,suppose firstNonMin( α) /∈ Ak(n). Then by Lemma 4, ai = 1 , and ai′ · · · an ≥ (ai′ · · · an)R. Since 1 = ai ≥ an−1, j = n − 1 and an−1 = ( k−2) = 1 . If i = 1 then α does not contain any 0s, and thus i′ = 2 . Thus a2 = an = 2 . The only bracelet that both starts and ends with 12 is of the form (12) s which is asymmetric, contradicting that α ∈ Ak(n).Thus i > 1 and a1 = aℓ = 0 . Recall j = n − 1 which implied ℓ = n − 2. Thus, α = 0 γ012 where γ is a palindrome. Let β = secondLastNonMax( α); it has suffix 112. Observe β is a bracelet by applying the definition. Suppose β is symmetric. By Lemma 3, β = β1β2 where β1 and β2 are both palindromes. Clearly, \|β2\| > 3. Note that α does not have prefix 02, since α is a necklace containing 012 as a substring. If β2 = 2112 , then α has prefix 02 , contradiction. If β2 = 21112 , then α starts and ends with 012. Since α is a necklace, this implies that α = (012) t for some integer t, which contradicts the form of β2. Thus \|β\| > 5. Let δ denote the suffix of α of length \|β2\|. Then α is not a bracelet since δ > δ R and thus αR < α . Contradiction. Thus, β is asymmetric and in A3(n). ◀ Given a node α in Tk(n), we say that γ is an ancestor of α if γ = α or γ = par t(α) for some t ≥ 1. ▶ Theorem 10. For n, k ≥ 3, the parent rule par( α) for Ak(n) induces a cycle-joining tree Tk(n) with nodes Ak(n) rooted at rn,k . Proof. Let α = a1a2 · · · an ∈ A(n){ rn,k }, where i is the index of the first non-zero, j is the index of the last non-(k−1) ,and ℓ is the index of the second last non-(k−1) symbol in α. We demonstrate that rn,k is an ancestor of α. If i = n − 1 then par (k−1−an)+( k−2−an−1)(α) = rn,k = 0 n−2(k−2)( k−1) , where each application of par uses lastNonMax . For i ≤ n − 2, we demonstrate that α has an ancestor β such that β < α , and thus must have an ancestor where i = n − 1.If i = j, then α = 0 i−1ai(k−1) n−i and par k−1−ai (α) = 0 i(k−2)( k−2) n−i−1 < α . Otherwise i < j . If k = 3 , then from Lemma 9, par( α) = secondLastNonMax( α) occurs only when α has suffix 012. Thus, repeated applications of only lastNonMax and secondLastNonMax to α will never change any of the first i symbols. Since both operations only increment symbols to at most (k−1) , one of lastSymbol or firstNonMin must eventually be applied by repeated XX:11 application of par starting with α. If firstNonMin is applied then it will decrement ai leading to an ancestor that is less than α. Similarly, lastSymbol will change the last symbol from (k−1) to 0 leading to a node with prefix 0i which is also less than α. ◀ By repeatedly joining cycles from Tk(n) via conjugate pairs, we can construct an OS k(n) (universal cycle) of length Lk(n) using exponential time per symbol (to search for the conjugates) and exponential space. Table 2 illustrates the lengths of the OS k(n)s constructed for small n, k . In Section 6, we present an exact formula for Lk(n). ▶ Theorem 11. There exists a universal cycle for Sk(n), which is an OS k(n), of length Lk(n). nk 3 4 5 6 7 83 3 12 30 60 105 168 4 12 60 180 420 840 1512 5 60 360 1260 3360 7560 15120 6 225 1608 6750 21150 54831 124320 7 819 7308 36890 135450 403389 1034264 8 2676 30300 187980 821940 2844408 8315496 9 8778 126516 962580 5003970 20101326 66961608 10 27180 511680 4836300 30097620 140902440 536135040 11 84579 2074644 24328150 181141950 988016337 4293525544 12 257205 8327808 121790490 1087414170 6917824935 34352668560 13 782964 33447960 609843780 6528527460 48439152216 274864275504 14 2361177 133931952 3050119450 39175228260 339088485771 2198957209792 15 7125423 536379792 15255860130 235079896440 2373737520945 17592060218208 16 21419076 2146175580 76284577980 1410507942900 16616280850008 140736884449896 17 64402800 8587706400 381453125040 8463244062000 116314913988000 1125898765992000 18 193357350 34353845664 1907295914700 50779660926240 814205346309138 9007193819084160 19 580569795 137428992036 9536650390670 304679295576630 5699444909171768 72057583837380680 20 1742213832 549729612720 47683422899280 1828077103852860 39896121850134479 576460703985782112 Table 2 Lower bounds Lk (n) on the maximal length of an OS k (n) for n ≤ 20 and k ≤ 8. ▶ Theorem 12. Tk(n) has the chain property. Proof. By contradiction. Suppose Tk(n) has a node γ with two children α and β joined via conjugate pairs (xσ, yσ) and (xσ, y′σ), respectively, for some string σ. Note that α, β, γ ∈ Ak(n). Since α̸ = β, the functions they apply to obtain their parent γ cannot both increment a symbol. Without loss of generality, assume par( β) = firstNonMin( β) and par( α) applies one of lastNonMax , lastSymbol , or secondLastNonMax . Let α = a1a2 · · · an and consider the three possible cases for par( α). par( α) = lastNonMax( α). Let j denote the index of the last non-(k−1) in α. Since γ = a1 · · · aj−1(aj +1) aj+1 · · · an, β = a1 · · · aj−1(aj +2) aj+1 · · · an where aj +2 ≤ (k−1) . However, from Lemma 4, either par( β) = lastNonMax( β),or β is not in Ak(n). Contradiction. par( α) = lastSymbol( α). As noted in the proof of Lemma 9, γ = 0 a1 · · · an−1. Therefore β = 1 a1 · · · an−1.Since β ∈ Ak(n), 1a1 · · · an−2 < (1 a1 · · · an−2)R; otherwise, β is not a bracelet or it is symmetric (see Lemma 3). Since par( β)̸ = lastNonMax( β), from Lemma 6, an−1 = k−1. However, this means that lastSymbol( β) = 01 a1 · · · an−2 ∈ Ak(n), which contradicts that par( β) = firstNonMin( β). par( α) = secondLastNonMax( α). From Lemma 9, α has suffix 012 and thus γ = a1 · · · an−3112 . Therefore β = a1 · · · an−3212 with its first non-0 at index n − 2, which means β is symmetric. Contradiction. ◀XX:12 Orientable sequences 5.2 An O(n)-time successor rule In this section, we apply the generic successor rule g(α) defined in Section 2.2 to the cycle-joining tree Tk(n) that admits the Chain Property. In particular, we determine whether or not α belongs to a conjugate pair, and if so, how to efficiently compute the function N EXT (α).Given α = a1a2 · · · an ∈ Sk(n), let i be the largest index such that ai > 0, let j be the smallest index greater than 1 such that aj < k −1, and let ℓ be the second smallest index greater than 1 such that aℓ < k −1. Recall that ˜α denotes the necklace of [α]. If α belongs to some conjugate pair (possibly more than one) used to create Tk(n), then we consider the possibilities for α depending on whether or not it belongs to a parent or child node joined by a given conjugate pair. If α belongs to the child, let β = ˜ α. If par( β) = lastNonMax( β), then β = aj aj+1 · · · ana1(k−1) j−2. In other words, a1 corresponds to the last non-( k−1) symbol in β. If α belongs to the parent, let γ denote the child node. If par( γ) = lastNonMax( γ) = ˜ α, then it must be that γ = aj aj+1 · · · an(a1−1)( k−1) j−2. A similar analysis holds for the other three cases of the parent rule par( α) giving rise to the definitions of the following eight strings: β1=ajaj+1 · · · ana1(k−1) j−2γ1=ajaj+1 · · · an(a1−1)( k−1) j−2(lastNonMax ) β2=a2a3· · · ana1γ2=a2a3· · · an(a1−1) (lastSymbol ) β3= 0 n−ia1a2· · · aiγ3= 0 n−i(a1+1)a2· · · ai(firstNonMin ) β4=aℓaℓ+1 · · · ana1a2· · · aℓ−1γ4=aℓaℓ+1 · · · an(a1−1) a2· · · aℓ−1(secondLastNonMax ). Assume addition on the symbols is modulo k; i.e., (k−1) + 1 = 0 and 0 − 1 = ( k−1) . The above strings can be tested to determine whether or not α belongs to a conjugate pair. For instance if γ1 ∈ Ak(n) and par( γ1) = lastNonMax( γ1),then α belongs to the conjugate pair (α, (a1−1) a2 · · · an). If α is found to belong to some conjugate pair, then the second issue is to efficiently compute the function NEXT (α). If α belongs to some parent node of a conjugate pair, then NEXT (α) is simply the incremented or decremented value of a1 defined by the parent rule for the corresponding γi. If α does not belong to a parent in any conjugate pair, then it belongs to the last node in its corresponding chain which contains some βi;to compute NEXT (α) we must determine the first node in the chain. Naïvely, we can repeatedly check the ancestors α until we reach the top of the chain. In the worst case this will take O(kn ) time. With a deeper analysis of the four cases, we can remove the factor k. Let σ = aj aj+1 · · · an(k−2)( k−1) j−2.Suppose β1 ∈ Ak(n) and par( β1) = lastNonMax( β1). Then Lemma 6 implies that σ ∈ Ak(n) and NEXT (α) is either (k−2) or (k−1) , depending on whether or not par( σ) = lastNonMax( σ).Suppose β2 ∈ Ak(n) and par( β2) = lastSymbol( β2) = δ. Since lastNonMax( β2) /∈ Ak(n), from Lemma 6, a1 = k−1. Furthermore, δ = 0 a2 · · · an since from Lemma 6, an is either (k−2) or (k−1) . It is easy to see that δ (if not the root) is not joined to par( δ) via a conjugate pair containing 0a2 · · · an. Thus, N EXT (α) = 0 .Suppose β3 ∈ Ak(n) and par( β3) = firstNonMin( β3) = δ. Then Lemma 7 implies that par( δ) = lastNonMax( δ) or par( δ) = lastSymbol( δ). Thus, δ is not joined to par( δ) via a conjugate pair containing the string (a1−1) a2 · · · an because it implies par( δ) = ˜ α. Thus N EXT (α) = a1 − 1.Suppose β4 ∈ Ak(n) and par( β4) = secondLastNonMax( β4) = δ. Then k = 3 and by Lemma 9, β4 starts with 0 and has suffix 012 . This implies δ starts with 0 and has suffix 112. By Lemma 6, lastNonMax( δ) ∈ A3(n) and hence par( δ) = lastNonMax( δ). Thus, N EXT (α) = a1 + 1 .The above analysis gives rise to the following successor rule h(α) = g(α) based on Tk(n). Successor-rule based on Tk(n) to construct an OS k(n) of length Lk(n) Apply the conditions top down: XX:13 h(α) =  a1 − 1 if γ1 ∈ Ak (n) and par( γ1) = lastNonMax( γ1); a1 − 1 if γ2 ∈ Ak (n) and par( γ2) = lastSymbol( γ2); a1 + 1 if γ3 ∈ Ak (n) and par( γ3) = firstNonMin( γ3); a1 − 1 if γ4 ∈ Ak (n) and par( γ4) = secondLastNonMax( γ4); k−1 if β1 ∈ Ak (n) and par( β1) = lastNonMax( β1) and par( σ) = lastNonMax( σ); k−2 if β1 ∈ Ak (n) and par( β1) = lastNonMax( β1); 0 if β2 ∈ Ak (n) and par( β2) = lastSymbol( β2); a1 − 1 if β3 ∈ Ak (n) and par( β3) = firstNonMin( β3); a1 + 1 if β4 ∈ Ak (n) and par( β4) = secondLastNonMax( β4); a1 otherwise. ▶ Theorem 13. For n, k ≥ 3, h(α) is a successor rule for an OS k(n) of length Lk(n) that runs in O(n) time and uses O(n) space. Proof. Let α = a1a2 · · · an ∈ Sk(n). Our previous analysis demonstrates that h(α) = g(α) for the cycle joining tree Tk(n). Determining whether or not a string is a symmetric/asymmetric necklace or bracelet can be computed in O(n) time and O(n) space [ 4, 18 ]. Thus, all of the membership tester and functions required in the definition of h(α) can be computed in O(n) time using O(n) space. ◀ In the next section, we demonstrate that length of the OS k(n) generated by our successor rule h(α) is asymptotically optimal. 6 Bounds on the maximal length of an OS k(n) Recall that Lk(n) = \|Sk(n)\| and Mk(n) is the maximal length of an OS k(n). Our construction in Section 5 demonstrates that Lk(n) ≤ Mk(n) or k ≥ 3. Dai et al. [ 9] provide the following lower bound L2(n) for M2(n), where μ is the Möbius function: L2(n) = 12 2n − ∑ d\|n μ(n/d ) nd H2(d)  , where H2(d) = 12 ∑ i\|d i ( 2⌊ i+1 2 ⌋ + 2 ⌊ i 2⌋+1 ) . Applying the same techniques, this formula can be generalized to Lk(n), Lk(n) = 12 kn − ∑ d\|n μ(n/d ) nd Hk(d)  , where Hk(d) = 12 ∑ i\|d i ( k⌊ i+1 2 ⌋ + k⌊ i 2⌋+1 ) . Exact values of Lk(n) for some small n, k are given in Table 2. ▶ Theorem 14. For n ≥ 3 and k ≥ 3, lim n→∞ Mk(n) − Lk(n) Lk(n) = 0 . Proof. Since μ(n/d ) ≤ 1 and Hk(d) = 12 ∑ i\|d i ( k⌊ i+1 2 ⌋ + k⌊ i 2⌋+1 ) ≤ d ∑ i=1 ik ⌊ i 2⌋+1 ≤ d2k⌊ d 2⌋+1 , we have ∑ d\|n μ(n/d ) nd Hk(d) ≤ n ∑ d=1 nd d2k⌊ d 2⌋+1 ≤ n3k⌊ n 2⌋+1 .XX:14 Orientable sequences Thus, Lk(n) = 12 kn − ∑ d\|n μ(n/d ) nd Hk(d)  ≥ 12 ( kn − n3k⌊ n 2⌋+1 ) . Recalling from Section 1 that Mk(n) ≤ 12 (kn − k⌊(n+1) /2⌋), we have Mk(n) − Lk(n) Lk(n) ≤ 12 (kn − k⌊(n+1) /2⌋) − 12 (kn − n3k⌊ n 2⌋+1 ) 12 (kn − n3k⌊ n 2⌋+1 ) = n3k⌊ n 2⌋+1 − k⌊(n+1) /2⌋ kn − n3k⌊ n 2⌋+1 . The result follows. ◀▶ Corollary 15. The OS k(n) generated by the successor rule h(α) has asymptotically optimal length. References 1 ADAMSON , D., G USEV , V. V., P OTAPOV , I., AND DELIGKAS , A. Ranking bracelets in polynomial time. In 32nd Annual Symposium on Combinatorial Pattern Matching (CPM 2021) (Dagstuhl, Germany, 2021), P. Gawrychowski and T. 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In Combinatorial Algorithms -24th International Workshop, IWOCA 2013, Rouen, France, July 10-12, 2013, LNCS 8288 (2013), pp. 388–401. 38 SAWADA , J., AND WONG , D. Efficient universal cycle constructions for weak orders. Discrete Math. 343 , 10 (2020), 112022. 39 STEVENS , B., B USKELL , P., E CIMOVIC , P., I VANESCU , C., M ALIK , A. M., S AVU , A., V ASSILEV , T. S., V ERRALL , H., YANG , B., AND ZHAO , Z. Solution of an outstanding conjecture: the non-existence of universal cycles with k = n − 2. Discrete Mathematics 258 , 1 (2002), 193–204. 40 TAKAHASHI , N., K AMEDA , A., Y AMAMOTO , M., AND OHUCHI , A. Aqueous computing with DNA hairpin-based RAM. In DNA Computing (2005), C. Ferretti, G. Mauri, and C. Zandron, Eds., Springer Berlin Heidelberg, pp. 355–364. 41 TAKINOUE , M., AND SUYAMA , A. Hairpin-DNA memory using molecular addressing. Small 2 , 11 (2006), 1244–1247.
9233	https://www.quora.com/When-2-2017-is-divided-by-13-what-is-the-remainder	When 2^2017 is divided by 13, what is the remainder? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Modular Exponentiation Integer Properties Remainder Mathematics Homework Ques... Divisibility Rules Laws of Exponents Division Algorithms Arithmetic 5 When 2^2017 is divided by 13, what is the remainder? All related (47) Sort Recommended Abhi Technology Analyst at Goldman Sachs (company) (2019–present) ·8y Originally Answered: What is the remainder when 2^2017 is divided by 13? · The answer is 2 2^2017 % 13 is equivalent to (2(2^4)^504) % 13 (2(2^4)^504) % 13 is equivalent to (2(3^504)) % 13 [ since 16 mod 13 = 3] (2(3^504)) % 13 is equivalent to (2((3^3)^168)) % 13 (2((3^3)^168)) % 13 is equivalent to 21 = 2 [ since 27 mod 13 = 1] So, the final result is 2. Upvote · 9 5 Sponsored by Google Cloud Skill up, stand out, and lead with Generative AI. Your journey from learner to Gen AI leader starts now, with skills that shape the future. Learn More 99 51 Ravi Handa Teaching Number Theory for 10 years on www.handakafunda.com · Author has 758 answers and 11.1M answer views ·8y Process: Rem [2^2017/13] = Rem [264^336/13] = Rem [21/13] = 2 Explanation: First of all, we need to think of a power of 2 that is close to a multiple of 13 2^6 = 64 = 65 - 1 = 135 - 1 Now, we should try to convert 2^2017 as a power of 64 2^2017 = 2 2^2016 = 2 64^336 Now, 64 leaves a remainder of -1 from 13 so 64^336 will leave a remainder of (-1)^336, which is 1 This means, the remainder of the 2^2016 from 13 would be 1 and the remainder of 2^2017 from 13 would be 2. Upvote · 99 11 9 1 Vijay Mankar HoD (Electronics) at Government Polytechnic Nagpur · Author has 7K answers and 11.2M answer views ·4y As [math]13[/math] is prime and [math]gcd(2,13)=1[/math], we can apply Fermat's Little theorem [math]2^{12}\equiv 1\pmod{13}[/math] [math]2^{2017}=2^{12×168+1}\equiv 1^{168}×2^1\equiv\boxed{2}\pmod{13}[/math] Hence, the remainder is [math]\boxed{2}.[/math] Upvote · 9 2 Related questions More answers below What is the remainder when 2^2017 is divided by 31? What is the remainder when [math]2^{2016}[/math] is divided by [math]13[/math]? How do I get the remainder of math ^4[/math] divided by 2000? Divided by 16, the quotients was 35 and the remainder was 13. What is the remainder? What is the remainder when we divide sizeof13! By 13? Kishan Panaganti Badrinath Former Engineer at Qualcomm (2017–2018) · Author has 566 answers and 2.5M answer views ·8y From Fermat's little theorem we get [math]2^{12} \equiv 1 \pmod{13}.[/math] So, [math]2^{2017} = 2^{12168 + 1} = 22^{16812} \equiv 2 \pmod{13}.[/math] Upvote · 9 4 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 623 Related questions What is the remainder when 2^2017 is divided by 31? What is the remainder when [math]2^{2016}[/math] is divided by [math]13[/math]? How do I get the remainder of math ^4[/math] divided by 2000? Divided by 16, the quotients was 35 and the remainder was 13. What is the remainder? What is the remainder when we divide sizeof13! By 13? What is the remainder when 2 to the power 5555 is divided by 13? How do you you find the remainder when 2^62 is divided by 10? Related questions What is the remainder when 2^2017 is divided by 31? What is the remainder when [math]2^{2016}[/math] is divided by [math]13[/math]? How do I get the remainder of math ^4[/math] divided by 2000? Divided by 16, the quotients was 35 and the remainder was 13. What is the remainder? What is the remainder when we divide sizeof13! By 13? What is the remainder when 2 to the power 5555 is divided by 13? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
9234	https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf?srsltid=AfmBOoqD05Ni7AJvr_C-Anvdxvp2ToMVuop6DFyGiC7tV2R8-BSO62pN	Olympiad Inequalities Thomas J. Mildorf December 22, 2005 It is the purpose of this document to familiarize the reader with a wide range of theorems and techniques that can be used to solve inequalities of the variety typically appearing on mathematical olympiads or other elementary proof contests. The Standard Dozen is an exhibition of twelve famous inequalities which can be cited and applied without proof in a solution. It is expected that most problems will fall entirely within the span of these inequalities. The Examples section provides numerous complete solutions as well as remarks on inequality-solving intuition, all intended to increase the reader’s aptitude for the material covered here. It is organized in rough order of difficulty. Finally, the Problems section contains exercises without solutions, ranging from straightforward to quite difficult, for the purpose of practicing techniques contained in this document. I have compiled much of this from posts by my peers in a number of mathematical communities, particularly the Mathlinks-Art of Problem Solving forums, 1 as well as from various MOP lectures, 2 Kiran Kedlaya’s inequalities packet, 3 and John Scholes’ site. 4 I have tried to take note of original sources where possible. This work in progress is distributed for personal educational use only. In particular, any publication of all or part of this manuscript without prior consent of the author, as well as any original sources noted herein, is strictly prohibited. Please send comments - suggestions, corrections, missing information, 5 or other interesting problems - to the author at tmildorf@mit.edu .Without further delay... 1 and respectively, though they have merged into a single, very large and robust group. The forums there are also host to a considerable wealth of additional material outside of inequalities. 2 Math Olympiad Program. Although some people would try to convince me it is the Math Olympiad Summer Program and therefore is due the acronym MOSP, those who know acknowledge that the traditional “MOP” is the preferred appellation. 3 The particularly diligent student of inequalities would be interested in this document, which is available online at Further ma-terial is also available in the books Andreescu-Cartoaje-Dospinescu-Lascu, Old and New Inequalities , GIL Publishing House, and Hardy-Littlewood-P´ olya, Inequalities , Cambridge University Press. (The former is elementary and geared towards contests, the latter is more technical.) 4 where a seemingly inexhaustible supply of Olympiads is available. 5 Such as the source of the last problem in this document. 11 The Standard Dozen Throughout this lecture, we refer to convex and concave functions. Write I and I′ for the intervals [ a, b ] and ( a, b ) respectively. A function f is said to be convex on I if and only if λf (x) + (1 − λ)f (y) ≥ f (λx + (1 − λ)y) for all x, y ∈ I and 0 ≤ λ ≤ 1. Conversely, if the inequality always holds in the opposite direction, the function is said to be concave on the interval. A function f that is continuous on I and twice differentiable on I′ is convex on I if and only if f ′′ (x) ≥ 0 for all x ∈ I (Concave if the inequality is flipped.) Let x1 ≥ x2 ≥ · · · ≥ xn; y1 ≥ y2 ≥ · · · ≥ yn be two sequences of real numbers. If x1 + · · · + xk ≥ y1 + · · · + yk for k = 1 , 2, . . . , n with equality where k = n, then the sequence {xi} is said to majorize the sequence {yi}. An equivalent criterion is that for all real numbers t, \|t − x1\| + \|t − x2\| + · · · + \|t − xn\| ≥ \| t − y1\| + \|t − y2\| + · · · + \|t − yn\| We use these definitions to introduce some famous inequalities. Theorem 1 (Jensen) Let f : I → R be a convex function. Then for any x1, . . . , x n ∈ I and any nonnegative reals ω1, . . . , ω n, ω1f (x1) + · · · + ωnf (xn) ≥ (ω1 + · · · + ωn) f (ω1x1 + · · · + ωnxn ω1 + · · · + ωn ) If f is concave, then the inequality is flipped. Theorem 2 (Weighted Power Mean) If x1, . . . , x n are nonnegative reals and ω1, . . . , ω n are nonnegative reals with a postive sum, then f (r) := (ω1xr 1 · · · + ωnxrn ω1 + · · · + ωn )1 r is a non-decreasing function of r, with the convention that r = 0 is the weighted geometric mean. f is strictly increasing unless all the xi are equal except possibly for r ∈ (−∞ , 0] ,where if some xi is zero f is identically 0. In particular, f (1) ≥ f (0) ≥ f (−1) gives the AM-GM-HM inequality. Theorem 3 (H¨ older) Let a1, . . . , a n; b1, . . . , b n; · · · ; z1, . . . , z n be sequences of nonnegative real numbers, and let λa, λ b, . . . , λ z positive reals which sum to 1. Then (a1 + · · · + an)λa (b1 + · · · + bn)λb · · · (z1 + · · · + zn)λz ≥ aλa 1 bλb 1 · · · zλz 1 · · · + aλz n bλb n · · · zλz n This theorem is customarily identified as Cauchy when there are just two sequences. Theorem 4 (Rearrangement) Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. Then, for any permutation π of {1, 2, . . . , n }, we have a1b1 + a2b2 + · · · + anbn ≥ a1bπ(1) + a2bπ(2) + · · · + anbπ(n) ≥ a1bn + a2bn−1 + · · · + anb1 with equality on the left and right holding if and only if the sequence π(1) , . . . , π (n) is de-creasing and increasing respectively. 2Theorem 5 (Chebyshev) Let a1 ≤ a2 ≤ · · · ≤ an; b1 ≤ b2 ≤ · · · ≤ bn be two nondecreas-ing sequences of real numbers. Then a1b1 + a2b2 + · · · + anbn n ≥ a1 + a2 + · · · + an n ·b1 + b2 + · · · + bn n ≥ a1bn + a2bn−1 + · · · + anb1 n Theorem 6 (Schur) Let a, b, c be nonnegative reals and r > 0. Then ar(a − b)( a − c) + br(b − c)( b − a) + cr(c − a)( c − b) ≥ 0 with equality if and only if a = b = c or some two of a, b, c are equal and the other is 0. Remark - This can be improved considerably. (See the problems section.) However, they are not as well known (as of now) as this form of Schur, and so should be proven whenever used on a contest. Theorem 7 (Newton) Let x1, . . . , x n be nonnegative real numbers. Define the symmetric polynomials s0, s 1, . . . , s n by (x + x1)( x + x2) · · · (x + xn) = snxn + · · · + s1x + s0, and define the symmetric averages by di = si/(ni ). Then d2 i ≥ di+1 di−1 Theorem 8 (Maclaurin) Let di be defined as above. Then d1 ≥ √d2 ≥ 3 √d3 ≥ · · · ≥ n √dn Theorem 9 (Majorization) Let f : I → R be a convex on I and suppose that the sequence x1, . . . , x n majorizes the sequence y1, . . . , y n, where xi, y i ∈ I. Then f (x1) + · · · + f (xn) ≥ f (y1) + · · · + f (yn) Theorem 10 (Popoviciu) Let f : I → R be convex on I, and let x, y, z ∈ I. Then for any positive reals p, q, r , pf (x) + qf (y) + rf (z) + (p + q + r)f (px + qy + rz p + q + r ) ≥ (p + q)f (px + qy p + q ) ( q + r)f (qy + rz q + r ) ( r + p)f (rz + px r + p ) Theorem 11 (Bernoulli) For all r ≥ 1 and x ≥ − 1, (1 + x)r ≥ 1 + xr 3Theorem 12 (Muirhead) Suppose the sequence a1, . . . , a n majorizes the sequence b1, . . . , b n.Then for any positive reals x1, . . . , x n, ∑ sym xa1 1 xa2 2 · · · xan n ≥ ∑ sym xb1 1 xb2 2 · · · xbn n where the sums are taken over all permutations of n variables. Remark - Although Muirhead’s theorem is a named theorem, it is generally not favor-ably regarded as part of a formal olympiad solution. Essentially, the majorization criterion guarantees that Muirhead’s inequality can be deduced from a suitable application of AM-GM. Hence, whenever possible, you should use Muirhead’s inequality only to deduce the correct relationship and then explicitly write all of the necessary applications of AM-GM. For a particular case this is a simple matter. We now present an array of problems and solutions based primarily on these inequalities and ideas. 2 Examples When solving any kind of problem, we should always look for a comparatively easy solu-tion first, and only later try medium or hard approaches. Although what constitutes this notoriously indeterminate “difficulty” varies widely from person to person, I usually con-sider “Dumbassing,” AM-GM (Power Mean), Cauchy, Chebyshev (Rearrangement), Jensen, H¨ older, in that order before moving to more clever techniques. (The first technique is de-scribed in remarks after example 1.) Weak inequalities will fall to AM-GM, which blatantly pins a sum to its smallest term. Weighted Jensen and H¨ older are “smarter” in that the effect of widely unequal terms does not cost a large degree of sharpness 6 - observe what happens when a weight of 0 appears. Especially sharp inequalities may be assailable only through clever algebra. Anyway, I have arranged the following with that in mind. 1. Show that for positive reals a, b, c (a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥ 9a2b2c2 Solution 1. Simply use AM-GM on the terms within each factor, obtaining (a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥ ( 3 3 √a3b3c3 ) ( 3 3 √a3b3c3 ) = 9 a2b2c2 6The sharpness of an inequality generally refers to the extent to which the two sides mimic each other, particularly near equality cases. 4Solution 2. Rearrange the terms of each factor and apply Cauchy, (a2b + b2c + c2a) ( bc 2 + ca 2 + ab 2) ≥ (√a3b3c3 + √a3b3c3 + √a3b3c3 )2 = 9 a2b2c2 Solution 3. Expand the left hand side, then apply AM-GM, obtaining (a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) = a3b3 + a2b2c2 + a4bc ab 4c + b3c3 + a2b2c2 a2b2c2 + abc 4 + a3c3 ≥ 9 9 √a18 b18 c18 = 9 a2b2c2 We knew this solution existed by Muirhead, since (4 , 1, 1) , (3 , 3, 0), and (2 , 2, 2) all majorize (2 , 2, 2). The strategy of multiplying out all polynomial expressions and ap-plying AM-GM in conjunction with Schur is generally knowing as dumbassing because it requires only the calculational fortitude to compute polynomial products and no real ingenuity. As we shall see, dumbassing is a valuable technique. We also remark that the AM-GM combining all of the terms together was a particularly weak inequality, but the desired was a multiple of a2b2c2’s, the smallest 6th degree symmetric polynomial of three variables; such a singular AM-GM may not always suffice. 2. Let a, b, c be positive reals such that abc = 1. Prove that a + b + c ≤ a2 + b2 + c2 Solution. First, we homogenize the inequality. that is, apply the constraint so as to make all terms of the same degree. Once an inequality is homogenous in degree d, we may scale all of the variables by an arbitrary factor of k, causing both sides of the inequality to scale by the factor kd. This is valid in that it does not change the correctness of an inequality for any positive k, and if d is even, for any nonzero k. Hence, we need consider a nonhomogenous constraint no futher. In this case, we multiply the left hand side by 3 √abc , obtaining a43 b13 c13 + a13 b43 c13 + a13 b13 c43 ≤ a2 + b2 + c2 As abc = 1 is not homogenous, the above inequality must be true for all nonnegative a, b, c . As (2 , 0, 0) majorizes (4 /3, 1/3, 1/3), we know it is true, and the necessary AM-GM is 2a2 3 + b2 6 + c2 6 = a2 + a2 + a2 + a2 + b2 + c2 6 ≥ 6 √a8b2c2 = a43 b13 c13 Let P (x) be a polynomial with positive coefficients. Prove that if P ( 1 x ) ≥ 1 P (x)5holds for x = 1, then it holds for all x > 0. Solution. Let P (x) = anxn + an−1xn−1 + · · · + a1x + a0. The first thing we notice is that the given is P (1) ≥ 1. Hence, the natural strategy is to combine P (x) and P ( 1 x ) into P (1) in some fashion. The best way to accomplish this is Cauchy, which gives P (x)P ( 1 x ) = (anxn + · · · + a1x + a0) ( an 1 xn + · · · + a1 1 x + a0 ) ≥ (an + · · · + a1 + a0)2 = P (1) 2 ≥ 1as desired. This illustrates a useful means of eliminating denominators - by introducing similar factors weighted by reciprocals and applying Cauchy / H¨ older. 4. (USAMO 78/1) a, b, c, d, e are real numbers such that a + b + c + d + e = 8 a2 + b2 + c2 + d2 + e2 = 16 What is the largest possible value of e? Solution. Observe that the givens can be effectively combined by considering squares: (a − r)2 + ( b − r)2 + ( c − r)2 + ( d − r)2 + ( e − r)2 = (a2 + b2 + c2 + d2 + e2) − 2r(a + b + c + d + e) + 5 r2 = 16 − 16 r + 5 r2 Since these squares are nonnegative, e ≤ √5r2 − 16 r + 16 + r = f (r) for all r. Since equality e = f (r) can be achieved when a = b = c = d = r, we need only compute the smallest value f (r). Since f grows large at either infinity, the minimum occurs when f ′(r) = 1 + 10 r−16 2√5r2−16 r+16 = 0. The resultant quadratic is easily solved for r = 65 and r = 2, with the latter being an extraneous root introduced by squaring. The largest possible e and greatest lower bound of f (r) is then f (6 /5) = 16 /5, which occurs when a = b = c = d = 6 /5 and e = 16 /5. Alternatively, proceed as before except write a = b = c = d = 8−e 4 since the maximum e must occur when the other four variables are equal. The second condition becomes a quadratic, and the largest solution is seen to be e = 16 5 .The notion of equating a, b, c, d is closely related to the idea of smoothing and Jensen’s inequality. If we are working with S1 = f (x1) + · · · + f (xn) under the constraint of a fixed sum x1 + · · · + xn, we can decrease S1 by moving several xi in the same interval I together (that is, replacing xi1 < x i2 with x′ i1 = xi1 + ≤ < x i2 − ≤ = x′ i2 for any sufficiently small ≤) for any I where f is convex. S1 can also be decreased by spreading xi in the same interval where f is concave. When seeking the maximum of S1, we proceed in the opposite fashion, pushing xi on the concave intervals of f together and moving xi on the convex intervals apart. 65. Show that for all positive reals a, b, c, d ,1 a + 1 b + 4 c + 16 d ≥ 64 a + b + c + d Solution. Upon noticing that the numerators are all squares with √1 + √1 + √4 + √16 = √64, Cauchy should seem a natural choice. Indeed, multiplying through by a + b + c + d and applying Cauchy, we have (a + b + c + d) (12 a + 12 b + 22 c + 42 d ) ≥ (1 + 1 + 2 + 4) 2 = 64 as desired. 6. (USAMO 80/5) Show that for all non-negative reals a, b, c ≤ 1, ab + c + 1 + bc + a + 1 + ca + b + 1 + (1 − a)(1 − b)(1 − c) ≤ 1 Solution. Let f (a, b, c ) denote the left hand side of the inequality. Since ∂2 ∂a 2 f = 2b (c+a+1) 3 2c (a+b+1) 3 ≥ 0, we have that f is convex in each of the three variables; hence, the maximum must occur where a, b, c ∈ { 0, 1}. Since f is 1 at each of these 8 points, the inequality follows. Second derivative testing for convexity/concavity is one of the few places where the use of Calculus is not seriously loathed by olympiad graders. It is one of the standard techniques in inequalities and deserves to be in any mental checklist of inequality solving. In this instance, it led to an easy solution. 7. (USAMO 77/5) If a, b, c, d, e are positive reals bounded by p and q with 0 < p ≤ q,prove that (a + b + c + d + e) ( 1 a + 1 b + 1 c + 1 d + 1 e ) ≤ 25 + 6 (√ pq − √ qp )2 and determine when equality holds. Solution. As a function f of five variables, the left hand side is convex in each of a, b, c, d, e ; hence, its maximum must occur when a, b, c, d, e ∈ { p, q }. When all five variables are p or all five are q, f is 25. If one is p and the other four are q, or vice versa, f becomes 17 + 4( pq + qp ), and when three are of one value and two of the other, f = 13 + 6( pq + qp ). pq + qp ≥ 2, with equality if and only if p = q. Clearly, equality holds where p = q. Otherwise, the largest value assumed by f is 13 + 6( pq + qp ), which is obtained only when two of a, b, c, d, e are p and the other three are q, or vice versa. In such instances, f is identically the right hand side. This is a particular case of the Schweitzer inequality, which, in its weighted form, is sometimes known as the Kantorovich inequality. 78. a, b, c, are non-negative reals such that a + b + c = 1. Prove that a3 + b3 + c3 + 6 abc ≥ 14 Solution. Multiplying by 4 and homogenizing, we seek 4a3 + 4 b3 + 4 c3 + 24 abc ≥ (a + b + c)3 = a3 + b3 + c3 + 3 (a2(b + c) + b2(c + a) + c2(a + b)) + 6 abc ⇐⇒ a3 + b3 + c3 + 6 abc ≥ a2(b + c) + b2(c + a) + c2(a + b)Recalling that Schur’s inequality gives a3 +b3 +c3 +3 abc ≥ a2(b+c)+ b2(c+a)+ c2(a+b), the inequality follows. In particular, equality necessitates that the extra 3 abc on the left is 0. Combined with the equality condition of Schur, we have equality where two of a, b, c are 12 and the third is 0. This is a typical dumbass solution. Solution 2. Without loss of generality, take a ≥ b ≥ c. As a+b+c = 1, we have c ≤ 13 or 1 −3c ≥ 0. Write the left hand side as ( a+b)3 −3ab (a+b−2c) = ( a+b)3 −3ab (1 −3c). This is minimized for a fixed sum a + b where ab is made as large as possible. As by AM-GM ( a + b)2 ≥ 4ab , this minimum occurs if and only if a = b. Hence, we need only consider the one variable inequality 2 (1−c 2 )3 + c3 + 6 (1−c 2 )2 c = 14 · (9 c3 − 9c2 + 3 c + 1). Since c ≤ 13 , 3 c ≥ 9c2. Dropping this term and 9 c3, the inequality follows. Particularly, 9c3 = 0 if and only if c = 0, and the equality cases are when two variables are 12 and the third is 0. 9. (IMO 74/5) If a, b, c, d are positive reals, then determine the possible values of aa + b + d + bb + c + a + cb + c + d + da + c + d Solution. We can obtain any real value in (1 , 2). The lower bound is approached by a → ∞ , b = d = √a, and c = 1. The upper bound is approached by a = c → ∞ , b = d = 1. As the expression is a continuous function of the variables, we can obtain all of the values in between these bounds. Finally, these bounds are strict because aa + b + d + bb + c + a + cb + c + d + da + c + d >aa + b + c + d + ba + b + c + d + ca + b + c + d + da + b + c + d = 1 and aa + b + d + bb + c + a + cb + c + d + da + c + d <aa + b + ba + b + cc + d + dc + d = 2 Whenever extrema occur for unusual parameterizations, we should expect the need for non-classical inequalities such as those of this problem where terms were completely dropped. 810. (IMO 95/2) a, b, c are positive reals with abc = 1. Prove that 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) ≥ 32 Solution 1. Let x = 1 a , y = 1 b , and z = 1 c . We perform this substitution to move terms out of the denominator. Since abc = xyz = 1, we have 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) = x2 y + z + y2 x + z + z2 x + y Now, multiplying through by ( x + y)( y + z)( z + x), we seek x4 + y4 + z4 + x3y + x3z + y3z + xy 3 + xz 3 + yz 3 + x2yz + xy 2z + xyz 2 ≥ 3 √xyz · ( 3xyz + 32 · (x2y + x2z + y2x + xy 2 + xz 2 + yz 2)) which follows immediately by AM-GM, since x2yz +xy 2z+xyz 2 ≥ 3 3 √x4y4z4, x3y+xy 3+x3z 3 ≥ 3 √x7y4z and 7x4+4 y4+z4 12 ≥ 3 √x7y4z - as guaranteed by Muirhead’s inequality. Solution 2. Substitute x, y, z as before. Now, consider the convex function f (x) = x−1 for x > 0. ( f (x) = xc is convex for c < 0 and c ≥ 1, and concave for 0 < c ≤ 1, verify this with the second derivative test.) Now, by Jensen, x2 y + z + y2 z + x + z2 x + y = xf (y + zx ) yf (z + xy ) zf (x + yz ) ≥ (x + y + z)f ((y + z) + ( z + x) + ( x + y) x + y + z ) = x + y + z 2But x + y + z ≥ 3 3 √xyz = 3, as desired. Solution 3. Perform the same substitution. Now, multiplying by ( x + y + z) and applying Cauchy, we have 12 (( y + z) + ( z + x) + ( x + y)) ( x2 y + z + y2 z + x + z2 x + y ) ≥ 12(x + y + z)2 Upon recalling that x+y +z ≥ 3 we are done. Incidentally, the progress of this solution with Cauchy is very similar to the weighted Jensen solution shown above. This is no coincidence, it happens for many convex f (x) = xc. Solution 4. Apply the same substitution, and put x ≥ y ≥ z. Simultaneously, xy+z ≥ yz+x ≥ zx+y . Hence, by Chebyshev, x · ( xy + z ) y · ( yz + x ) z · ( zx + y ) ≥ x + y + z 3 ( xy + z + yx + z + zx + y ) Again, x + y + z ≥ 3. But now we have Nesbitt’s inequality, xy+z + yx+z + zx+y ≥ 32 . This follows immediately from AM-HM upon adding 1 to each term. 911. Let a, b, c be positive reals such that abc = 1. Show that 2(a + 1) 2 + b2 + 1 + 2(b + 1) 2 + c2 + 1 + 2(c + 1) 2 + a2 + 1 ≤ 1 Solution. The previous problem showed the substitution offers a way to rewrite an inequality in a more convenient form. Substitution can also be used to implicity use a given. First, expand the denominators and apply AM-GM, obtaining 2(a + 1) 2 + b2 + 1 = 2 a2 + b2 + 2 a + 2 ≤ 1 ab + a + 1 Now, write a = xy , b = yz , c = zx . We have 1 ab +a+1 = 1 xz+xy+1 = yz xy +yz +zx . It is now evident that the sum of the new fractions is 1. 12. (USAMO 98/3) Let a0, . . . , a n real numbers in the interval (0 , π 2 ) such that tan ( a0 − π 4 ) tan ( a1 − π 4 ) · · · + tan ( an − π 4 ) ≥ n − 1Prove that tan( a0) tan( a1) · · · tan( an) ≥ nn+1 Solution 1. Let yi = tan (x − π 4 ). We have tan( xi) = tan ((xi − π 4 ) + π 4 ) = yi+1 1−yi .Hence, given s = y0 + · · · + yn ≥ n − 1 we seek to prove ∏ni=0 1+ yi 1−yi ≥ nn+1 . Observe that for any a > b and fixed sum a + b, the expression ( 1 + a 1 − a ) · ( 1 + b 1 − b ) = 1 + 2( a + b)(1 − a)(1 − b)can be decreased by moving a and b any amount closer together. Hence, for any sequence y0, . . . , y n, we can replace any yi > sn+1 and yj < sn+1 with y′ i = sn+1 and y′ j = yi + yj − sn+1 , decreasing the product. Now we have n ∏ i=0 1 + yi 1 − yi ≥ ( 1 + sn+1 1 − sn+1 )n+1 ≥ ( 2nn+1 2 n+1 )n+1 = nn+1 Where the last inequality follows from the fact that 1+ x 1−x is an increasing function of x. Solution 2. Perform the same substitution. The given can be written as 1 + yi ≥ ∑ j6=i (1 − yj ), which by AM-GM gives 1+ yn n ≥ ∏ j6=i (1 − yj ) 1 n . Now we have n ∏ i=0 1 + yi n ≥ n ∏ i=0 ∏ j6=i (1 − yj ) 1 n = n ∏ i=0 (1 − yi)as desired. 10 13. Let a, b, c be positive reals. Prove that 1 a(1 + b) + 1 b(1 + c) + 1 c(1 + a) ≥ 31 + abc with equality if and only if a = b = c = 1. Solution. Multiply through by 1+ abc and add three to each side, on the left obtaining 1 + a + ab + abc a(1 + b) + 1 + b + bc + abc b(1 + c) + 1 + c + ac + abc c(1 + a)= (1 + a) + ab (1 + c) a(1 + b) + (1 + b) + bc (1 + a) b(1 + c) + (1 + c) + ac (1 + b) c(1 + a)which is at least 6 by AM-GM, as desired. In particular, this AM-GM asserts the equivalence of (1+ a) a(1+ b) and a(1+ b)1+ a , or that they are both one. Likewise, all of the other terms must be 1. Now, (1 + a)2 = a2(1 + b)2 = a2b2(1 + c)2 = a2b2c2(1 + a)2, so the product abc = 1. Hence, 1+ aa(1+ b) = bc (1+ a)1+ b = bc (1+ a) b(1+ c) so that 1 + b = b + bc = b + 1 a . It is now easy to see that equality holds if and only if a = b = c = 1. 14. (Romanian TST) Let a, b, x, y, z be positive reals. Show that xay + bz + yaz + bx + zax + by ≥ 3 a + b Solution. Note that ( a + b)( xy + yz + xz ) = ( x(ay + bz ) + y(az + bx ) + z(ax + by )). We introduce this factor in the inequality, obtaining (x(ay + bz ) + y(az + bx ) + z(ax + by )) ( xay + bz + yaz + bx + zax + by ) ≥ (x + y + z)2 ≥ 3( xy + yz + xz )Where the last inequality is simple AM-GM. The desired follows by simple algebra. Again we have used the idea of introducing a convenient factor to clear denominators with Cauchy. 15. The numbers x1, x 2, . . . , x n obey −1 ≤ x1, x 2, . . . , x n ≤ 1 and x 31 + x 32 + · · · + x 3 n = 0. Prove that x1 + x2 + · · · + xn ≤ n 3 Solution 1. Substitute yi = x3 i so that y1 + · · · + yn = 0. In maximizing 3 √y1 + · · · + 3 √yn, we note that f (y) = y 13 is concave over [0 , 1] and convex over [ −1, 0], with \|f ′(y1)\| ≥ \| f ′(y2)\| ⇐⇒ 0 < \|y1\| ≤ \| y2\|. Hence, we may put y1 = · · · = yk = −1; −1 ≤ yk+1 < 0, and yk+2 = · · · = yn = k−yk+1 n−k−1 . We first show that yk+1 leads to a maximal sum of 3 √yi if it is -1 or can be made positive. If \|yk+1 \| < \|yk+2 \|, we set 11 y′ k+1 = y′ k+2 = yk+1 +yk+2 2 , increasing the sum while making yk+1 positive. Otherwise, set y′ k+1 = −1 and y′ k+2 = 1 − yk+1 − yk+2 , again increasing the sum of the 3 √yi. Now we may apply Jensen to equate all positive variables, so that we need only show k 3 √−1 + ( n − k) 3 √ kn − k ≤ n 3But we have (n + 3 k)3 − 27( n − k)2k = n3 − 18 n2k + 81 nk 2 = n(n − 9k)2 ≥ 0as desired. Particularly, as k is an integer, equality can hold only if 9 \|n and then if and only if one ninth of the variables yi are -1 and the rest are 1/8. Solution 2. Let xi = sin( αi), and write 0 = x31 + · · · + x3 n = sin 3(α1) + · · · + sin 3(αn) = 14 ((3 sin( α1) − sin(3 α1)) + · · · + (3 sin( αn) − sin(3 αn))). It follows that x1 + · · · + xn =sin( α1) + · · · + sin( αn) = sin(3 α1)+ ··· +sin(3 αn)3 ≤ n 3 . The only values of sin( α) which lead to sin(3 α) = 1 are 12 and -1. The condition for equality follows. 16. (Turkey) Let n ≥ 2 be an integer, and x1, x 2, . . . , x n positive reals such that x21 + x22 + · · · + x2 n = 1. Determine the smallest possible value of x51 x2 + x3 + · · · + xn x52 x3 + · · · + xn + x1 · · · + x5 n x1 + · · · + xn−1 Solution. Observe that ∑ni=1 xi ∑ j6=i xj ≤ n − 1, so that ( n∑ i=1 xi (∑ j6=i xj )) ( n∑ i=1 x5 i ∑ j6=i xi ) ≥ (x31 + · · · + x3 n )2 = n2 (x31 + · · · + x3 n n )2 ≥ n2 (x21 + · · · + x2 n n )3 = 1 n Leads to n∑ i=1 x5 i ∑ j6=i xi ≥ 1 n(n − 1) with equality if and only if x1 = · · · = xn = 1√n .17. (Poland 95) Let n be a positive integer. Compute the minimum value of the sum x1 + x22 2 + x33 3 + · · · + xnn n 12 where x1, x 2, . . . , x n are positive reals such that 1 x1 1 x2 · · · + 1 xn = n Solution. The given is that the harmonic mean of x1, . . . , x n is 1, which implies that the product x1x2 · · · xn is at least 1. Now, we apply weighted AM-GM x1 + x22 2 + x33 3 + · · · + xnn n ≥ ( 1 + 12 + 13 + · · · + 1 n ) 1+ 12 +··· + 1 n √x1x2 · · · xn = 1 + 12 + 13 + · · · + 1 n Prove that for all positive reals a, b, c, d , a4b + b4c + c4d + d4a ≥ abcd (a + b + c + d) Solution. By AM-GM, 23 a4b + 7 b4c + 11 c4d + 10 ad 4 51 ≥ 51 √a102 b51 c51 d51 = a2bcd from which the desired follows easily. Indeed, the most difficult part of this problem is determining suitable weights for the AM-GM. One way is to suppose arbitrary weights x1, x 2, x 3, x 4 for a4b, b 4c, c 4d, ad 4 respectively, and solve the system x1 + x2 + x3 + x4 = 14x1 + x2 = 24x2 + x3 = 14x3 + x4 = 119. (USAMO 01/3) Let a, b, c be nonnegative reals such that a2 + b2 + c2 + abc = 4 Prove that 0 ≤ ab + bc + ca − abc ≤ 2 Solution [by Tony Zhang.] For the left hand side, note that we cannot have a, b, c > Suppose WLOG that c ≤ 1. Then ab +bc +ca −abc ≥ ab +bc +ca −ab = c(a+b) ≥ 0. For the right, 4 = a2 + b2 + c2 + abc ≥ 4( abc )43 =⇒ abc ≤ 1. Since by the pigeon hole principle, among three numbers either two exceed 1 or two are at most 1. Hence, we assume WLOG that ( a − 1)( b − 1) ≥ 0, which gives ab + 1 ≥ a + b ⇐⇒ abc + c ≥ ac + bc ⇐⇒ c ≥ ac + bc − abc . Now, we have ab + bc + ca − abc ≤ ab + c. Either we are done or ab +c > 2. But in the latter case, 4 = ( a2 +b2)+ c(c+2 ab ) > 2ab +2 c = 2( ab +c) > 4, a contradiction. 13 20. (Vietnam 98) Let x1, . . . , x n be positive reals such that 1 x1 + 1998 + 1 x2 + 1998 + · · · + 1 xn + 1998 = 11998 Prove that n √x1x2 · · · xn n − 1 ≥ 1998 Solution. Let yi = 1 xi+1998 so that y1 + · · · + yn = 11998 and xi = 1 yi − 1998. Now n ∏ i=1 xi = n ∏ i=1 ( 1 yi − 1998 ) = e Pni=1 ln “1 yi−1998 ” Hence, to minimize the product of the xi, we equivalently minimize the sum of ln ( 1 yi − 1998 ) .In particular, ddy ( ln ( 1 y − 1998 )) = 1 ( 1 y − 1998 )2 · −1 y2 = −1 y − 1998 y2 d2 dy 2 ( ln ( 1 y − 1998 )) = 1 − 3996 y (y − 1998 y2)2 So ln ( 1 y − 1998 ) is convex on [0 , 1/3996]. If we had 0 < y i ≤ 1/3996 for all i we could apply Jensen. Since yi + yj ≤ 1/1998 for all i, j , we consider ( 1 a − 1998 ) ( 1 b − 1998 ) ≥ ( 2 a + b − 1998 )2 ⇐⇒ 1 ab − 1998 ( 1 a + 1 b ) ≥ 4(a + b)2 − 4 · 1998 a + b ⇐⇒ (a + b)2 − 1998( a + b)3 ≥ 4ab − 4ab (a + b) · 1998 ⇐⇒ (a − b)2 ≥ 1998( a + b)( a − b)2 which incidentally holds for any a + b ≤ 11998 . Hence, any two yi and yj may be set to their average while decreasing the sum in question; hence, we may assume yi ∈ (0 , 13996 ]. Now Jensen’s inequality shows that the minimum occurs when yi = 11998 n for all i, or when xi = 1998( n − 1) for all i. It is easy to see that this yields equality. 21. (Romania 99) Show that for all positive reals x1, . . . , x n with x1x2 · · · xn = 1, we have 1 n − 1 + x1 · · · + 1 n − 1 + xn ≤ 114 Solution. First, we prove a lemma: the maximum of the sum occurs when n − 1 of the xi are equal. Consider f (y) = 1 k+ey for an arbitrary nonnegative constant k. We have f ′(y) = −ey (k+ey)2 and f ′′ (y) = ey (ey −k)(k+ey )3 . Evidently f ′′ (y) ≥ 0 ⇐⇒ ey ≥ k. Hence, f (y) has a single inflexion point where y = ln( k), where f (y) is convex over the interval ((ln( k), ∞). Now, we employ the substitution yi = ln( xi) so that y1 + · · · + yn = 0 and n ∑ i=1 1 n − 1 + xi = n ∑ i=1 f (yi)We take k = n − 1 and write k0 = ln( n − 1). Suppose that y1 ≥ · · · ≥ ym ≥ k0 ≥ ym+1 ≥ · · · xn for some positive m. Then by, Majorization, f (y1) + · · · + f (ym) ≤ (m − 1) f (k0) + f (y1 + · · · + ym − (m − 1) k0)But then, also by Majorization, (m − 1) f (k0) + f (ym+1 ) + · · · + f (yn) ≤ (n − 1) f ((m − 1) k0 + ym+1 + · · · + yn n − 1 ) Otherwise, all of the yi are less than k0. In that case we may directly apply Majorization to equate n − 1 of the yi whilst increasing the sum in question. Hence, the lemma is valid. 7 N Applying the lemma, it would suffice to show kk + x + 1 k + 1 xk ≤ 1Clearing the denominators, ( k2 + kxk ) ( k + x) ≤ k2 + k ( x + 1 xk ) x1−k −xk + x + k ≤ x1−k But now this is evident. We have Bernoulli’s inequality, since x1−k = (1 + ( x − 1)) 1−k ≥ 1 + ( x − 1)(1 − k) = x + k − xk . Equality holds only where x = 1 or n = 2. 22. (Darij Grinberg) Show that for all positive reals a, b, c , √b + ca + √c + ab + √a + bc ≥ 4( a + b + c) √(a + b)( b + c)( c + a) 7This n−1 equal value principle is particularly useful. If a differentiable function has a single inflexion point and is evaluated at narbitrary reals with a fixed sum, any minimum or maximum must occur where some n−1 variables are equal. 15 Solution 1. By Cauchy, we have √(a + b)( a + c) ≥ a + √bc . Now, ∑ cyc √b + ca ≥ 4( a + b + c) √(a + b)( b + c)( c + a) ⇐⇒ ∑ cyc b + ca √(a + b)( a + c) ≥ 4( a + b + c)Substituting our result from Cauchy, it would suffice to show ∑ cyc (b + c) √bc a ≥ 2( a + b + c)WLOG a ≥ b ≥ c, implying b + c ≤ c + a ≤ a + b and √bc a ≤ √ca b ≤ √ab c . Hence, by Chebyshev and AM-GM, ∑ cyc (b + c) √bc a ≥ (2( a + b + c)) (√bc a + √ca b + √ab c ) 3 ≥ 2( a + b + c)as desired. Solution 2. Let x = √b + c, y = √c + a, z = √a + b. Then x, y, z are the sides of acute triangle XY Z (in the typical manner), since x2 + y2 = a + b + 2 c > a + b = z2.The inequality is equivalent to ∑ cyc xy2 + z2 − x2 ≥ x2 + y2 + z2 xyz Recalling that y2 + z2 − x2 = 2 yz cos( X), we reduce this to the equivalent ∑ cyc x2 cos( X) ≥ 2( x2 + y2 + z2)WLOG, we have x ≥ y ≥ z, implying 1cos( X) ≥ 1cos( Y ) ≥ 1cos( Z) , so that applying Chebyshev to the left reduces the desired to proving that the sum of the reciprocals of the cosines is at least 6. By AM-HM, 1cos( X) + 1cos( Y ) + 1cos( Z) ≥ 9cos( X) + cos( Y ) + cos( Z)But recall from triangle geometry that cos( X) + cos( Y ) + cos( Z) = 1 + rR and R ≥ 2r.The desired is now evident. 16 23. Show that for all positive numbers x1, . . . , x n, x31 x21 + x1x2 + x22 x32 x22 + x2x3 + x23 · · · + x3 n x2 n xnx1 + x21 ≥ x1 + · · · + xn 3 Solution. Observe that 0 = ( x1 −x2)+( x2 −x3)+ · · · +( xn −x1) = ∑ni=1 x3 i−x3 i+1 x2 i+xixi+1 +x2 i+1 .Hence, (where xn+1 = x1) n ∑ i=1 x3 i x2 i xixi+1 x2 i+1 = 12 n ∑ i=1 x3 i x3 i+1 x2 i xixi+1 + x2 i+1 But now a3 + b3 ≥ 13 a3 + 23 a2b + 23 ab 2 + 13 b3 = 13 (a + b)( a2 + ab + b2). Hence, 12 n ∑ i=1 x3 i x3 i+1 x2 i xixi+1 + x2 i+1 ≥ 12 n ∑ i=1 xi + xi+3 3 = 13 n ∑ i=1 xi as desired. 24. Let a, b, c be positive reals such that a + b ≥ c; b + c ≥ a; and c + a ≥ b, we have 2a2(b + c) + 2 b2(c + a) + 2 c2(a + b) ≥ a3 + b3 + c3 + 9 abc Solution. After checking that equality holds for ( a, b, c ) = ( t, t, t ) and (2 t, t, t ), it is apparent that more than straight AM-GM will be required. To handle the condition, put a = y + z, b = z + x, c = x + y with x, y, z ≥ 0. Now, the left hand side becomes 4x3 + 4 y3 + 4 z3 + 10 x2(y + z) + 10 y2(z + x) + 10 z2(x + y) + 24 xyz while the right hand side becomes 2 x3 + 2 y3 + 2 z3 + 12 x2(y + z) + 12 y2(z + x) + 12 z2(x + y) + 18 xyz .The desired is seen to be equivalent to x3 + y3 + z3 + 3 xyz ≥ x2(y + z) + y2(z + x) + z2(x + y), which is Schur’s inequality. Equality holds where x = y = z, which gives ( a, b, c ) = ( t, t, t ), or when two of x, y, z are equal and the third is 0, which gives (a, b, c ) ∈ { (2 t, t, t ), (t, 2t, t ), (t, t, 2t)}.25. Let a, b, c be the lengths of the sides of a triangle. Prove that a √2b2 + 2 c2 − a2 + b √2c2 + 2 a2 − b2 + c √2a2 + 2 b2 − c2 ≥ √3 Solution 1. Again write a = y + z, b = z + x, and c = x + y, noting that x, y, z are positive. (Triangles are generally taken to be non-degenerate when used in inequalities.) We have ∑ cyc a √2b2 + 2 c2 − a2 = ∑ cyc y + z √4x2 + 4 xy + 4 xz + y2 + z2 − 2yz 17 Consider the convex function f (x) = 1√x . (As we shall see, Jensen almost always provides a tractable means of eliminating radicals from inequalities.) Put x+y +z = 1. We have ∑ cyc (y + z)f (4x2 + 4 xy + 4 xz + y2 + z2 − 2yz ) ≥ (( y + z) + ( z + x) + ( x + y)) f (∑ cyc (y + z) (4 x2 + 4 xy + 4 xz + y2 + z2 − 2yz )(y + z) + ( z + x) + ( x + y) ) = 2√2 √∑ cyc 4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2 Noting that ∑ cyc 4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2 = ∑ cyc 2x3 + 7 x2(y + z) + 8 xyz ,8( x + y + z)3 ≥ 3 ∑ cyc 2x3 + 7 x2(y + z) + 8 xyz ⇐⇒ ∑ sym 4x3 + 24 x2y + 8 xyz ≥ ∑ sym 3x3 + 21 x2y + 12 xyz ⇐⇒ 2x3 + 2 y3 + 2 z3 + 3 (x2(y + z) + y2(z + x) + z2(x + y)) ≥ 24 xyz which follows by AM-GM. As a follow up on an earlier mentioned connection, oberserve the similarity of the above application of Jensen and the following inequality (which follows by H¨ older’s inequality) (∑ i αiβi ) ( ∑ i αi 1 √βi )2 ≥ (∑ i αi )3 Solution 2 [by Darij Grinberg.] Let ABC be a triangle of side lengths a, b, c in the usual order. Denote by ma, m b, m c the lengths of the medians from A, B, C respectively. Recall from triangle goemetry that ma = 12 √2b2 + 2 c2 − a2, so that we need only show ama + bmb + cmc ≥ 2√3. But a triangle with side lengths ma, m b, m c, in turn, has medians of length 3a 4 , 3b 4 , and 3c 4 . The desired inequality is therefore equivalent to 43 ma a 43 mb b 43 mc c ≥ 2√3 where we refer to the new triangle ABC . Recalling that 23 ma = AG , where G is the centroid, the desired is seen to be equivalent to the geometric inequality AG a + BG b + CG c ≥ √3. But we are done as we recall from triangle geometry that AM a + BM b + CM c ≥ √3 holds for any point inside triangle ABC .8 8For a complete proof of this last inequality, see post #14. 18 26. (IMO 99/2) For n ≥ 2 a fixed positive integer, find the smallest constant C such that for all nonnegative reals x1, . . . , x n, ∑ 1≤i<j ≤n xixj (x2 i x2 j ) ≤ C ( n∑ i=1 xi )4 Solution. The answer is C = 18 , which is obtained when any two xi are non-zero and equal and the rest are 0. Observe that by AM-GM, (x1 + · · · + xn)4 = ( n∑ i=1 x2 i 2 ∑ 1≤i<j ≤n xixj )2 ≥ 4 ( n∑ i=1 x2 i ) ( 2 ∑ 1≤i<j ≤n xixj ) = 8 ∑ 1≤i<j ≤n xixjn∑ k=1 x2 k But x21 + · · · + x2 n ≥ x2 i x2 j with equality iff xk = 0 for all k 6 = i, j . It follows that (x1 + · · · + xn)4 ≥ 8 ∑ 1≤i<j ≤n xixj (x2 i x2 j ) as desired. 27. Show that for nonnegative reals a, b, c ,2a6 + 2 b6 + 2 c6 + 16 a3b3 + 16 b3c3 + 16 c3a3 ≥ 9a4(b2 + c2) + 9 b4(c2 + a2) + 9 c4(a2 + b2) Solution 1. Consider ∑ cyc (a − b)6 = ∑ cyc a6 − 6a5b + 15 a4b2 − 20 a3b3 + 15 a2b4 − 6ab 5 + b6 ≥ 0and ∑ cyc ab (a − b)4 = ∑ cyc a5b − 4a4b2 + 6 a3b3 − 4a2b4 + ab 5 ≥ 0Adding six times the latter to the former yields the desired result. Solution 2. We shall prove a6 − 9a4b2 + 16 a3b3 − 9a2b4 + b6 ≥ 0. We have a6 − 2a3b3 + b6 = (a3 − b3)2 = ((a − b)( a2 + ab + b2))2 ≥ (a − b)2(3 ab )2 = 9 a4b2 − 18 a3b3 + 9 a2b4 19 As desired. The result now follows from adding this lemma cyclicly. The main difficulty with this problem is the absence of a5b terms on the right and also the presence of a4b2 terms on the right - contrary to where Schur’s inequality would generate them. Evidently AM-GM is too weak to be applied directly, since a6 + 2 a3b3 ≥ 3a4b2 cannot be added symmetrically to deduce the problem. By introducing the factor ( a − b)2,however, we weight the AM-GM by a factor which we “know” will be zero at equality, thereby increasing its sharpness. 28. Let 0 ≤ a, b, c ≤ 12 be real numbers with a + b + c = 1. Show that a3 + b3 + c3 + 4 abc ≤ 932 Solution. Let f (a, b, c ) = a3 + b3 + c3 + 4 abc and g(a, b, c ) = a + b + c = 1. Because f and g are polynomials, they have continuous first partial derivatives. Moreover, the gradient of g is never zero. Hence, by the theorem of Lagrange Multipliers ,any extrema occur on the boundary or where ∇f = λ∇g for suitable scalars λ. As ∇f =< 3a2 + 4 bc, 3b2 + 4 ca, 3c2 + 4 ab > and ∇g =< 1, 1, 1 >, we have λ = 3a2 + 4 bc = 3b2 + 4 ca = 3c2 + 4 ab g(a, b, c ) = a + b + c = 1 We have 3 a2 + 4 bc = 3 b2 + 4 ca or ( a − b)(3 a + 3 b − 4c) = ( a − b)(3 − 7c) = 0 for any permutation of a, b, c . Hence, without loss of generality, a = b. Now, 3 a2 + 4 ac =3c2 + 4 a2 and a2 − 4ac + 3 c2 = ( a − c)( a − 3c) = 0. The interior local extrema therefore occur when a = b = c or when two of {a, b, c } are three times as large as the third. Checking, we have f (13 , 13 , 13 ) = 7 /27 < 13 /49 = f (17 , 37 , 37 ). Recalling that f (a, b, c ) is symmetric in a, b, c , the only boundary check we need is f (12 , t, 12 −t) ≤ 932 for 0 ≤ t ≤ 12 .We solve h(t) = f (12, t, 12 − t ) = 18 + t3 + (12 − t )3 2 t (12 − t ) = 14 + t 4 − t2 2 h(t) is 14 at either endpoint. Its derivative h′(t) = 14 − t is zero only at t = 14 . Checking, h(14 ) = f (12 , 14 , 14 ) = 932 . Since h(t) has a continuous derivative, we are done. (As a further check, we could observe that h′′ (t) = −1 < 0, which guarantees that h(14 ) is a local minimum.) 20 Usage Note. The use of Lagrange Multipliers in any solution will almost certainly draw hostile review, in the sense that the tiniest of errors will be grounds for null marks. If you consider multipliers on Olympiads, be diligent and provide explicit, kosher remarks about the continuous first partial derivatives of both f (x1, . . . , x n) and the constraint g(x1, . . . , x n) = k, as well as ∇g 6 = 0, before proceeding to solve the system ∇f = λ∇g. The main reason this approach is so severely detested is that, given sufficient computational fortitude (if you are able to sort through the relevant algebra and Calculus), it can and will produce a complete solution. The example provided here is included for completeness of instruction; typical multipliers solutions will not be as clean or painless. 9 (Vascile Cartoaje) Let p ≥ 2 be a real number. Show that for all nonnegative reals a, b, c , 3 √ a3 + pabc 1 + p + 3 √ b3 + pabc 1 + p + 3 √ c3 + pabc 1 + p ≤ a + b + c Solution. By H¨ older, (∑ cyc 3 √ a3 + pabc 1 + p )3 ≤ (∑ cyc 11 + p ) ( ∑ cyc a ) ( ∑ cyc a2 + pbc ) But a2 + b2 + c2 ≥ ab + bc + ca (proven by AM-GM, factoring, or a number of other methods) implies that ∑ cyc a2 + pbc ≤ (p + 1) ∑ cyc a2 + 2 bc 3 = p + 1 3 (a + b + c)2 From which we conclude (∑ cyc 3 √ a3 + pabc 1 + p )3 ≤ (a + b + c)3 as desired. 30. Let a, b, c be real numbers such that abc = −1. Show that a4 + b4 + c4 + 3( a + b + c) ≥ a2 b + a2 c + b2 c + b2 a + c2 a + c2 b Solution. First we homogenize, obtaining a4 + b4 + c4 + a3(b + c) + b3(c + a) + c3(a + b) − 3abc (a + b + c) ≥ 0. As this is homogenous in the fourth degree, we can scale a, b, c 9Just how painful can the calculations get? Most multipliers solutions will tend to look more like than this solution. 21 by any real k and hence may now ignore abc = −1. Equality holds at a = b = c = 1, but also at a = b = 1 , c = −2, a = 1 , b = 0 , c = −1, and a number of unusual locations with the commonality that a + b + c = 0. Indeed, c = −a − b is a parametric solution, and we discover the factorization ( a + b + c)2(a2 + b2 + c2 − ab − bc − ca ) ≥ 0. (We are motivated to work with factorizations because there are essentially no other inequalities with a + b + c = 0 as an equality condition.) 31. (MOP 2003) Show that for all nonnegative reals a, b, c , a4(b2 + c2) + b4(c2 + a2) + c4(a2 + b2) +2abc (a2b + a2c + b2c + b2a + c2a + c2b − a3 − b3 − c3 − 3abc ) ≥ 2a3b3 + 2 b3c3 + 2 c3a3 Solution. As was suggested by the previous problem, checking for equality cases is important when deciding how to solve a problem. We see that setting a = b produces equality. As the expression is symmetric, this certainly implies that b = c and c = a are equality cases. Hence, if P (a, b, c ) is the difference LHS - RHS, then ( a − b)( b − c)( c − a)\|P (a, b, c ). Obviously, if the problem is going to be true, ( a−b) must be a double root of P , and accordingly we discover the factorization P (a, b, c ) = ( a − b)2(b − c)2(c − a)2.The result illustrated above was no accident. If ( x−y) divides a symmetric polynomial P (x, y, z ), then ( x − y)2 divides the same polynomial. If we write P (x, y, z ) = ( x − y)Q(x, y, z ), then ( x − y)Q(x, y, z ) = P (x, y, z ) = P (y, x, z ) = ( y − x)Q(y, x, z ), which gives Q(x, y, z ) = −Q(y, x, z ). Hence Q(x, x, z ) = 0, and ( x − y) also divides Q(x, y, z ). 32. (Cezar Lupu) Let a, b, c be positive reals such that a + b + c + abc = 4. Prove that a √b + c + b √c + a + c √a + b ≥√22 · (a + b + c) Solution. By Cauchy (∑ cyc a√b + c ) ( ∑ cyc a √b + c ) ≥ (a + b + c)2 But, also by Cauchy, √(a + b + c) ( a(b + c) + b(c + a) + c(a + b)) ≥ ∑ cyc a√b + c Hence, ∑ cyc a √b + c ≥√22 · (a + b + c) · √ a + b + cab + bc + ca 22 And we need only show a + b + c ≥ ab + bc + ca . Schur’s inequality for r = 1 can be expressed as 9abc a+b+c ≥ 4( ab + bc + ca ) − (a + b + c)2. Now, we suppose that ab + bc + ca > a + b + c, and have 9abc a + b + c ≥ 4( ab + bc + ca ) − (a + b + c)2 (a + b + c) (4 − (a + b + c)) = abc (a + b + c)Hence, a + b + c < 3. But then abc < 1, which implies 4 = a + b + c + abc < 4. Contradiction, as desired. 33. (Iran 1996) Show that for all positive real numbers a, b, c ,(ab + bc + ca ) ( 1(a + b)2 + 1(b + c)2 + 1(c + a)2 ) ≥ 94 Solution. Fearless courage is the foundation of all success. 10 When everything else fails, return to the sure-fire strategy of clearing all denominators. In this case, we obtain 4( a + b)2(b + c)2(c + a)2(ab + bc + ca ) ( 1(a + b)2 + 1(b + c)2 + 1(c + a)2 ) = ∑ sym 4a5b + 8 a4b2 + 10 a4bc + 6 a3b3 + 52 a3b2c + 16 a2b2c2 on the left, and on the right, 9( a + b)2(b + c)2(c + a)2 = ∑ sym 9a4b2 + 9 a4bc + 9 a3b3 + 54 a3b2c + 15 a2b2c2 Canceling like terms, we seek ∑ sym 4a5b − a4b2 + a4bc − 3a3b3 − 2a3b2c + a2b2c2 Sure enough, this is true, since 3a5b+ab 5 4 ≥ a4b2 and a4b2+a2b4 2 ≥ a3b3 by AM-GM, and abc (a3 + b3 + c3 − a2(b + c) + b2(c + a) + c2(a + b) + 3 abc ) ≥ 0 by Schur. 34. (Japan 1997) Show that for all positive reals a, b, c ,(a + b − c)2 (a + b)2 + c2 + (b + c − a)2 (b + c)2 + a2 + (c + a − b)2 (c + a)2 + b2 ≥ 35 10 Found on a fortune cookie by Po-Ru Loh while grading an inequality on 2005 Mock IMO Day 2 that was solved by brutal force. 23 Solution. Put a + b + c = 3 so that equality will hold at a = b = c = 1 and suppose that there exists some k for which (b + c − a)2 (b + c)2 + a2 = (3 − 2a)2 (3 − a)2 + a2 ≥ 15 + ka − k for all positive a, b, c ; such an inequality would allow us to add cyclicly to deduce the desired inequality. As the inequality is parametrically contrived to yield equality where a = 1, we need to find k such that a = 1 is a double root. At a = 1, the derivative on the left is (2(3 −2a)·− 2)((3 −a)2+a2)−((3 −2a)2)(2(3 −a)·− 1+2 a)((3 −a)2+a2)2 = −18 25 . The derivative on the right is k, so we set k = −18 25 . But for this k we find (3 − 2a)2 − (15 + ka − k ) ((3 − a)2 + a2) = 18 25 − 54 a2 25 + 36 a3 25 = 18 25 (a − 1) 2(2 a + 1) ≥ 0as desired. Alternatively, we could have used AM-GM to show a3 + a3 + 1 ≥ 3a2. As hinted at by a previous problem, inequalities are closely linked to polynomials with roots of even multiplicity. The isolated manipulation idea used in this solution offers a completely different approach to the inequalities which work with every term. 35. (MOP 02) Let a, b, c be positive reals. Prove that ( 2ab + c )23 + ( 2bc + a )23 + ( 2ca + b )23 ≥ 3 Solution. Suppose that there exists some r such that ( 2ab + c )23 ≥ 3ar ar + br + cr We could sum the inequality cyclicly to deduce what we want. Since equality holds at a = b = c = 1, we use derivatives to find a suitable r. At the said equality case, on the left, the partial derivative with respect to a is 23 , while the same derivative on the right is 23 r. Equating the two we have r = 1. (This is necessary since otherwise the inequality will not hold for either a = 1 + ≤ or a = 1 − ≤.) 11 Now, 3aa + b + c ≤ 3a 3 3 √ a · (b+c 2 )2 11 Actually, even this is a special case of the general sense that the convexity of one side must exceed the convexity of the other. More precisely, we have the following result: Let fand gfunctions over the domain Dwith continuous partial derivatives. If f(ν)≥g(ν) for all ν∈D, then at every equality case ν0, ∇(f−g)( ν0) = 0and every component of ∇2(f−g) ( ν0) is nonnegative. 24 = a23 (b+c 2 )23 = ( 2ab + c )23 by AM-GM, as desired. 36. (Mildorf) Let n ≥ 2 be an integer. Prove that for all reals a1, a 2, . . . , a n > 0 and reals p, k ≥ 1, ( a1 + a2 + · · · + an ap 1 ap 2 · · · + apn )k ≥ ak 1 ak 2 · · · + akn apk 1 apk 2 · · · + apk n where inequality holds iff p = 1 or k = 1 or a1 = a2 = · · · = an, flips if instead 0 < p < 1, and flips (possibly again) if instead 0 < k < 1. Solution. Taking the kth root of both sides, we see that the inequality is equivalent to n∑ i=1 k √ aki ak 1 ak 2 · · · + akn ≥ n ∑ i=1 k √ apk i apk 1 apk 2 · · · apk n WLOG, suppose that a1 ≥ a2 ≥ · · · ≥ an. We prove a lemma. Let Si = api ap 1+··· +apn and Ti = aqi aq 1+··· +aqn for i = 1 , 2, . . . , n where 0 < q < p . Then the sequence S1, S 2, . . . , S n majorizes the sequence T1, T 2, . . . , T n.To prove the claim, we note that S1 ≥ · · · ≥ Sn and T1 ≥ · · · ≥ Tn and have, for m ≤ n, m ∑ i=1 Si ≥ m ∑ i=1 Ti ⇐⇒ (ap 1 · · · + apm) ( aq 1 · · · + aqn) ≥ (aq 1 · · · + aqm) ( ap 1 · · · + apn) ⇐⇒ (ap 1 · · · + apm) (aqm+1 + · · · + aqn ) ≥ (aq 1 · · · + aqm) (apm+1 + · · · + apn ) ⇐⇒ ∑ (i,j )\| { 1≤i≤m<j ≤n} api aqj − aqi apj ≥ 0Which is obvious. In particular, m = n is the equality case, and the claim is established. But now the desired is a direct consequence of the Majorization inequality applied to the sequences in question and the function f (x) = k √x.37. (Vascile Cartoaje) Show that for all real numbers a, b, c,(a2 + b2 + c2)2 ≥ 3 (a3b + b3c + c3a) 25 Solution. We will be content to give the identity (a2 + b2 + c2)2 − 3( a3b + b3c + c3a) = 12 ∑ cyc (a2 − 2ab + bc − c2 + ca )2 Any Olympiad partipant should be comfortable constructing various inequalities through well-chosen squares. Here, we could certainly have figured we were summing the square of a quadratic that is 0 when a = b = c such that no term a2bc is left uncancelled. A good exercise is to show that equality actually holds iff a = b = c or, for some cyclic permutation, a : b : c ≡ sin 2 (4π 7 ) : sin 2 (2π 7 ) : sin 2 (π 7 ).38. (Anh-Cuong) Show that for all nonnegative reals a, b, c , a3 + b3 + c3 + 3 abc ≥ ab √2a2 + 2 b2 + bc √2b2 + 2 c2 + ca √2c2 + 2 a2 Solution. Upon observing that this inequality is stronger than Schur’s inequality for r = 1, we are inspired to prove a sharp lemma to eliminate the radical. Knowing that √2x2 + 2 y2 ≥ x + y ≥ 2xy x+y , we seek a combination of the latter two that exceeds the former. We find 3x2 + 2 xy + 3 y2 2( x + y) ≥ √2x2 + 2 y2 This follows from algebra, since (3 x2 + 2 xy + 3 y2)2 = 9 x4 + 12 x3y + 22 x2y2 + 12 xy 3 +9y4 ≥ 8x4 + 16 x3y + 16 x2y2 + 16 xy 3 + 8 y4 = 4( x + y)2(2 x2 + 2 y2), so that (3 x2 + 2 xy +3y2)2 − 4( x + y)2(2 x2 + 2 y2) = x4 − 4x3y + 6 x2y2 − 4xy 3 + y4 = ( x − y)4 ≥ 0. Now, ∑ cyc ab √2a2 + 2 b2 ≤ ∑ cyc (3 a2 + 2 ab + 3 b2)ab 2( a + b)So it would suffice to show ∑ cyc a(a − b)( a − c) = ∑ cyc (a3 + abc − ab (a + b)) ≥ ∑ cyc (3 a2 + 2 ab + 3 b2)ab 2( a + b) − ab (a + b)= ∑ cyc 3a3b + 2 a2b2 + 3 ab 3 − 2a3b − 4a2c2 − 2ab 3 2( a + b)= ∑ cyc ab (a − b)2 2( a + b)But ∑ cyc (b + c − a)( b − c)2 = 2 ∑ cyc a(a − b)( a − c)26 so that the desired is ∑ cyc ( b + c − a − bc b + c ) (b − c)2 ≥ 0which is evident, since without loss of generality we may assume a ≥ b ≥ c and find ( a + b − c − ab a + b ) (a − b)2 ≥ 0 ( c + a − b − ac a + c ) ((a − c)2 − (b − c)2) ≥ 0 ( b + c − a − bc b + c ) (b − c)2 + ( c + a − b − ac a + c ) (b − c)2 ≥ 0The key to this solution was the sharp upper bound on the root-mean-square. At first glance our lemma seems rather arbitrary and contrived. Actually, it is a special case of a very sharp bound on the two variable power mean that I have conjectured and proved. Mildorf’s Lemma 1 Let k ≥ − 1 be an integer. Then for all positive reals a and b, (1 + k)( a − b)2 + 8 ab 4( a + b) ≥ k √ak + bk 2 with equality if and only if a = b or k = ±1, where the power mean k = 0 is interpreted to be the geometric mean √ab . Moreover, if k < −1, then the inequality holds in the reverse direction, with equality if and only if a = b. Usage Note. As of early November 2005, I have proven an extension of this lemma to additional values of k.12 Thus, you may rest assured that the result stated above is true. I was unable to get this result published, so I have instead posted the proof here as “ASharpBound.pdf.” However, the proof is rather difficult (or at least so I think, being as though it took me nearly half a year) and the lemma is far from mainstream. Thus, should you require it on an Olympiad, you should prove it for whatever particular value of k you are invoking. This is not terribly difficult if k is a small integer. One simply takes the kth power of both sides and factors the difference of the two sides as (a − b)4 · P (a, b ), etc. For x ≥ y ≥ 1, prove that x √x + y + y √y + 1 + 1 √x + 1 ≥ y √x + y + x √x + 1 + 1 √y + 1 12 In particular, the inequality holds for all kin ( −∞ ,−1) ,{− 1,0,1},(1 ,3/2] ,[2 ,∞) with the signs ≤,≥,≤ ,≥respectively, with equality iff a=bor k=±1. 27 Solution. By observation, equality holds when y = 1 and when x = y. Combining this with the restriction, it makes sense to write x = y + a and y = 1 + b where a, b ≥ 0. Now we can write x − y √x + y + y − 1 √y + 1 + 1 − x √1 + x ≥ 0 ⇐⇒ a √2 + a + 2 b + b √2 + b ≥ a + b √2 + a + b But this is evident by Jensen’s inequality applied to the convex function f (x) = 1√x ,since af (2 + a + 2 b) + bf (2 + b) ≥ (a + b)f (a(2 + a + 2 b) + b(2 + b) a + b ) = (a + b)f ((a + b)2 + 2( a + b) a + b ) = a + b √2 + a + b as desired. 40. (MOP) For n ≥ 2 a fixed positive integer, let x1, . . . , x n be positive reals such that x1 + x2 + · · · + xn = 1 x1 1 x2 · · · + 1 xn Prove that 1 n − 1 + x1 1 n − 1 + x2 · · · + 1 n − 1 + xn ≤ 1 Solution. We will prove the contrapositive. (We are motivated to do this for two good reasons: 1) it is usually difficult the show that the sum of some reciprocals is bounded above, and 2) the given relation in its current form is an abomination.) Take yi = 1 n−1+ xi , and for the sake of contradiction assume y1 + · · · + yn > 1. Since the yi are too large, the xi are too small and we shall prove 1 x1 · · · + 1 xn x 1 + · · · + xn.Since xiyi = 1 − (n − 1) yi, we have (n − 1) yi > (n − 1) ( yi + 1 − n ∑ j=1 yj ) = (n − 1) yi − 1 + n ∑ j=1 (1 − (n − 1) yj )= −xiyi + n ∑ j=1 xj yj (∗)=⇒ n − 1 xi −1 + n ∑ j=1 xj yj xiyi (∗∗ )28 Summing () over i,(n − 1) ( 1 x1 · · · + 1 xn ) n ∑ i=1 xiyi (( n∑ j=1 1 xj yj ) − 1 xiyi ) But by Cauchy and (), we have ( n∑ j=1 1 xj yj ) − 1 xiyi ≥ (n − 1) 2 (∑nj=1 xj yj ) − xiyi (n − 1) 2 (n − 1) yi = n − 1 yi Hence, (n − 1) ( 1 x1 · · · + 1 xn ) n ∑ i=1 xiyi (n − 1 yi ) = ( n − 1)( x1 + · · · + xn)as desired. 41. (Vascile Cartoaje) Show that for positive reals a, b, c ,14a2 − ab + 4 b2 + 14b2 − bc + 4 c2 + 14c2 − ca + 4 a2 ≥ 97( a2 + b2 + c2) Solution. Upon expansion, we see that it is equivalent to ∑ sym 56 a6 − 28 a5b + 128 a4b2 + 44 a3b3 + 95 2 a4bc + 31 a3b2c − 45 2 a2b2c2 ≥ 0We conjure up the following inequalities: ∑ sym a6 − 2a5b + a4bc ≥ 0 (1) ∑ sym a5b − 4a4b2 + 3 a3b3 ≥ 0 (2) ∑ sym a4b2 − a4bc − a3b3 + 2 a3b2c − a2b2c2 ≥ 0 (3) ∑ sym a4bc − 2a3b2c + a2b2c2 ≥ 0 (4) (1) and (4) follow from Schur’s inequality for r = 4 and r = 1 (multiplied by abc )respectively. (2) is the result of expanding ∑ cyc ab (a − b)4 ≥ 0, and (3) is the expanded form of the famous ( a − b)2(b − c)2(c − a)2 ≥ 0. The desired now follows by subtracting 56 times (1), 84 times (2), 208 times (3), 399 2 times (4), and then simple AM-GM to clear the remaining a2b2c2.29 This is about as difficult as a dumbass solution can get. A good general strategy is to work with the sharpest inequalities you can find until you reduce a problem to something obvious, starting with the most powerful (most bunched, in this case ∑ sym a6) term and work your way down to the weak terms while keeping the most powerful term’s coefficient positive. My solution to this problem starts with (1), Schur with r = 4 (Schur is stronger for larger r), which is almost certainly sharper than the inequality in question. Next, inequality (2) is a sharp cyclic sum to use the a5b terms. In particular, it relates terms involving only two of the three variables. Most of the time, the only inequality that can “pull up” symmetric sums involving three variables to stronger ones involving just two is Schur, although it does so at the expense of a very strong term with only one variable. Hence, we made a logical choice. Inequality (3) is extremely sharp, and allowed us to obtain more a4bc and a3b3 terms simultaneously. In particular, it was necessary to cancel the a3b3 terms. I’ll note that this inequality is peculiar to sixth degree symmetry in three variables - it does not belong to a family of similar, nice inequalities. Finally, inequality (4), which is a handy corollary to (3), is another Schur. Every inequality we have used so far is quite sharp, and so it is no surprise that the leftovers are the comparatively weak AM-GM. 42. (Reid Barton, IMO Shortlist 03/A6.) Let n ≥ 2 be a positive integer and x1, x 2, . . . , x n,y1, y 2, . . . , y n a sequence of 2 n positive reals. Suppose z2, z 3, . . . , z 2n is such that z2 i+j ≥ xiyj for all i, j ∈ { 1, . . . , n }. Let M = max {z2, z 3, . . . , z 2n}. Prove that (M + z2 + z3 + · · · + z2n 2n )2 ≥ (x1 + · · · + xn n ) ( y1 + · · · + yn n ) Reid’s official solution. Let max( x1, . . . , x n) = max( y1, . . . , y n) = 1. (We can do this by factoring X from every xi, Y from every yj , and √XY from every zi+j without changing the sign of the inequality.) We will prove M + z2 + · · · + z2n ≥ x1 + x2 + · · · + xn + y1 + y2 + · · · + yn, after which the desired follows by AM-GM. We will show that the number of terms on the left which are greater than r is at least as large as the number of terms on the right which are greater than r, for all r ≥ 0. For r ≥ 1, the claim is obvious, since all terms on the right are at most 1. Now take r < 1. Let A and B denote the set of i for which xi > r and the set of j for which yj > r respectively, and write a = \|A\|, b = \|B\|. Evidently, from our scaling, a, b ≥ 1. Now, xi > r and yj > r implies zi+j ≥ √xiyj ≥ r. Hence, if C is the set of k for which zk > r , we have \|C\| ≥ \| A + B\|, where the set addition is defined by the set of possible values if we take an element of A and add it to an element of B. How-ever, \|A + B\| ≥ \| A\| + \|B\| − 1, since if A and B consist of the values p1 < · · · < p a and q1 < · · · < q b respectively we have all of the values p1 +q1 < . . . < p a +q1 < · · · < p a +qb in A + B. Hence, \|C\| ≥ a + b − 1. Since \|C\| ≥ 1, there is some zk > r , and hence, M > r . Therefore, the left side of the inequality in question has at least a + b terms which exceed r, as desired. • 30 The preponderance of difficulty here stemmed from dealing with the superabundance of givens, especially the mysterious M . Scaling allowed us to introduce some degree of control and, with marked audacity, a profoundly clever idea. As it turned out, the in-equality was no sharper than simple AM-GM! It is my opinion that it is highly unlikely that a problem as staggeringly pernicious as this one will appear on an Olympiad - at least in the foreseeable future. Nevertheless, I have included it here for the purpose of illustrating just how unusual and creative a solution can be. 3 Problems (MOP 04) Show that for all positive reals a, b, c , ( a + 2 ba + 2 c )3 + ( b + 2 cb + 2 a )3 + (c + 2 ac + 2 b )3 ≥ 32. (MOP) Show that if k is a positive integer and a1, a 2, . . . , a n are positive reals which sum to 1, then n∏ i=1 1 − aki aki ≥ (nk − 1)n Let a1, a 2, . . . , a n be nonnegative reals with a sum of 1. Prove that a1a2 + a2a3 + · · · + an−1an ≤ 144. (Ukraine 01) Let a, b, c, x, y, z be nonnegative reals such that x + y + z = 1. Show that ax + by + cz + 2 √(ab + bc + ca )( xy + yz + zx ) ≤ a + b + c Let n > 1 be a positive integer and a1, a 2, . . . , a n positive reals such that a1a2 . . . a n = 1. Show that 11 + a1 · · · + 11 + an ≤ a1 + · · · + an + n (Aaron Pixton) Let a, b, c be positive reals with product 1. Show that 5 + ab + bc + ca ≥ (1 + a)(1 + b)(1 + c)7. (Valentin Vornicu 13 ) Let a, b, c, x, y, z be arbitrary reals such that a ≥ b ≥ c and either x ≥ y ≥ z or x ≤ y ≤ z. Let f : R → R+0 be either monotonic or convex, and let k be a positive integer. Prove that f (x)( a − b)k(a − c)k + f (y)( b − c)k(b − a)k + f (z)( c − a)k(c − b)k ≥ 0 13 This improvement is more widely known than the other one in this packet, and is published in his book, Olimpiada de Matematica... de la provocare la experienta , GIL Publishing House, Zalau, Romania. (In English, “The Math Olympiad... from challenge to experience.”) 31 8. (IMO 01/2) Let a, b, c be positive reals. Prove that a √a2 + 8 bc + b √b2 + 8 ca + c √c2 + 8 ab ≥ 19. (USAMO 04/5) Let a, b, c be positive reals. Prove that (a5 − a2 + 3 ) ( b5 − b2 + 3 ) ( c5 − c2 + 3 ) ≥ (a + b + c)3 (Titu Andreescu) Show that for all nonzero reals a, b, c , a2 b2 + b2 c2 + c2 a2 ≥ ac + cb + ba (IMO 96 Shortlist) Let a, b, c be positive reals with abc = 1. Show that ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤ 112. Let a, b, c be positive reals such that a + b + c = 1. Prove that √ab + c + √bc + a + √ca + b ≥ 1 + √ab + √bc + √ca (APMO 2005/2) Let a, b, c be positive reals with abc = 8. Prove that a2 √(a3 + 1) ( b3 + 1) + b2 √(b3 + 1) ( c3 + 1) + c2 √(c3 + 1) ( a3 + 1) ≥ 4314. Show that for all positive reals a, b, c , a3 b2 − bc + c2 + b3 c2 − ca + a2 + c3 a2 − ab + b2 ≥ a + b + c (USAMO 97/5) Prove that for all positive reals a, b, c ,1 a3 + b3 + abc + 1 b3 + c3 + abc + 1 c3 + a3 + abc ≤ 1 abc (Mathlinks Lore) Show that for all positive reals a, b, c, d with abcd = 1, and k ≥ 2, 1(1 + a)k + 1(1 + b)k + 1(1 + c)k + 1(1 + d)k ≥ 22−k (IMO 05/3) Prove that for all positive a, b, c with product at least 1, a5 − a2 a5 + b2 + c2 + b5 − b2 b5 + c2 + a2 + c5 − c2 c5 + a2 + b2 ≥ 032 18. (Mildorf) Let a, b, c, k be positive reals. Determine a simple, necessary and sufficient condition for the following inequality to hold: (a + b + c)k (akbk + bkck + ckak) ≤ (ab + bc + ca )k(ak + bk + ck)19. Let a, b, c be reals with a + b + c = 1 and a, b, c ≥ − 34 . Prove that aa2 + 1 + bb2 + 1 + cc2 + 1 ≤ 910 20. (Mildorf) Show that for all positive reals a, b, c , 3 √4a3 + 4 b3 + 3 √4b3 + 4 c3 + 3 √4c3 + 4 a3 ≤ 4a2 a + b + 4b2 b + c + 4c2 c + a Let a, b, c, x, y, z be real numbers such that (a + b + c)( x + y + z) = 3 , (a2 + b2 + c2)( x2 + y2 + z2) = 4 Prove that ax + by + cz ≥ 022. (Po-Ru Loh) Let a, b, c be reals with a, b, c > 1 such that 1 a2 − 1 + 1 b2 − 1 + 1 c2 − 1 = 1 Prove that 1 a + 1 + 1 b + 1 + 1 c + 1 ≤ 123. (Weighao Wu) Prove that (sin x)sin x < (cos x)cos x for all real numbers 0 < x < π 4 .24. (Mock IMO 05/2) Let a, b, c be positive reals. Show that 1 < a √a2 + b2 + b √b2 + c2 + c √c2 + a2 ≤ 3√2225. (Gabriel Dospinescu) Let n ≥ 2 be a positive integer. Show that for all positive reals a1, a 2, . . . , a n with a1a2 . . . a n = 1, √a21 + 1 2 + · · · + √a2 n 1 2 ≤ a1 + · · · + an 33 26. Let n ≥ 2 be a positive integer, and let k ≥ n−1 n be a real number. Show that for all positive reals a1, a 2, . . . , a n, ( (n − 1) a1 a2 + · · · + an )k + ( (n − 1) a2 a3 + · · · + an + a1 )k · · · + ( (n − 1) an a1 + · · · + an−1 )k ≥ n (Mildorf) Let a, b, c be arbitrary reals such that a ≥ b ≥ c, and let x, y, z be nonnegative reals with x + z ≥ y. Prove that x2(a − b)( a − c) + y2(b − c)( b − a) + z2(c − a)( c − b) ≥ 0and determine where equality holds. 28. (USAMO 00/6) Let n ≥ 2 be an integer and S = {1, 2, . . . , n }. Show that for all nonnegative reals a1, a 2, . . . , a n, b 1, b 2, . . . , b n, ∑ i,j ∈S min {aiaj , b ibj } ≤ ∑ i,j ∈S min {aibj , a j bi} (Kiran Kedlaya) Show that for all nonnegative a1, a 2, . . . , a n, a1 + √a1a2 + · · · + n √a1 · · · an n ≤ n √ a1 · a1 + a2 2 · · · a1 + · · · + an n (Vascile Cartoaje) Prove that for all positive reals a, b, c such that a + b + c = 3, aab + 1 + bbc + 1 + cca + 1 ≥ 3231. (Gabriel Dospinescu) Prove that ∀a, b, c, x, y, z ∈ R+\| xy + yz + zx = 3, a(y + z) b + c + b(z + x) c + a + c(x + y) a + b ≥ 332. (Mildorf) Let a, b, c be non-negative reals. Show that for all real k, ∑ cyc max( ak, b k)( a − b)2 2 ≥ ∑ cyc ak(a − b)( a − c) ≥ ∑ cyc min( ak, b k)( a − b)2 2(where a, b, c 6 = 0 if k ≤ 0) and determine where equality holds for k > 0, k = 0, and k < 0 respectively. 33. (Vascile Cartoaje) Let a, b, c, k be positive reals. Prove that ab + ( k − 3) bc + ca (b − c)2 + kbc + bc + ( k − 3) ca + ab (c − a)2 + kca + ca + ( k − 3) ab + bc (a − b)2 + kab ≥ 3( k − 1) k (Taiwan? 02) Show that for all positive a, b, c, d ≤ k, we have abcd (2 k − a)(2 k − b)(2 k − c)(2 k − d) ≤ a4 + b4 + c4 + d4 (2 k − a)4 + (2 k − b)4 + (2 k − c)4 + (2 k − d)4 34
9235	https://grams-to-pounds.appspot.com/454-grams-to-pounds.html	454 Grams To Pounds Converter \| 454 g To lbs Converter Grams To Pounds 454 g to lbs 454 Grams to Pounds 454 Gram to Pound converter g = lbs How to convert 454 grams to pounds? 454 g 0.0022046226 lbs=1.0008986703 lbs 1 g A common question isHow many gram in 454 pound?And the answer is 205930.93598 g in 454 lbs. Likewise the question how many pound in 454 gram has the answer of 1.0008986703 lbs in 454 g. How much are 454 grams in pounds? 454 grams equal 1.0008986703 pounds (454g = 1.0008986703lbs). Converting 454 g to lb is easy. Simply use our calculator above, or apply the formula to change the length 454 g to lbs. Convert 454 g to common mass \| Unit \| Mass \| --- \| \| Microgram \| 454000000.0 µg \| \| Milligram \| 454000.0 mg \| \| Gram \| 454.0 g \| \| Ounce \| 16.0143787251 oz \| \| Pound \| 1.0008986703 lbs \| \| Kilogram \| 0.454 kg \| \| Stone \| 0.0714927622 st \| \| US ton \| 0.0005004493 ton \| \| Tonne \| 0.000454 t \| \| Imperial ton \| 0.0004468298 Long tons \| What is 454 grams in lbs? To convert 454 g to lbs multiply the mass in grams by 0.0022046226. The 454 g in lbs formula is [lb] = 454 0.0022046226. Thus, for 454 grams in pound we get 1.0008986703 lbs. 454 Gram Conversion Table Further grams to pounds calculations 444 Grams to lbs 445 Grams to Pound 446 g to Pound 447 Grams to lbs 448 Grams to lb 449 g to Pound 450 Grams to lb 451 Grams to lbs 452 Grams to lb 453 Grams to lbs 454 Grams to lbs 455 Grams to Pounds 456 g to lb 457 g to Pounds 458 Grams to Pounds 459 Grams to Pound 460 g to lb 461 Grams to lb 462 g to lb 463 Grams to Pound 464 Grams to lb Alternative spelling 454 g to Pounds, 454 g in Pounds, 454 Gram to Pounds, 454 Gram in Pounds, 454 Grams to lb, 454 Grams in lb, 454 Grams to lbs, 454 Grams in lbs, 454 Grams to Pound, 454 Grams in Pound, 454 Gram to Pound, 454 Gram in Pound, 454 g to lb, 454 g in lb, 454 Grams to Pounds, 454 Grams in Pounds, 454 Gram to lb, 454 Gram in lb Further Languages ‎454 Grams To Pounds ‎454 Грам в Фунт ‎454 Gram Na Libra ‎454 Gram Til Pund ‎454 Gramm In Pfund ‎454 Γραμμάριο σε λίμπρα ‎454 Gramo En Libra ‎454 Gramm Et Nael ‎454 Gramma Pauna ‎454 Gramme En Livre ‎454 Gram U Funta ‎454 Gramm Font ‎454 Grammo In Libbra ‎454 Gramas Iki Svaras ‎454 Gramma Fil Lira ‎454 Gram Naar Pond ‎454 Gram Na Funt ‎454 Grama Em Libra ‎454 Gram în livră ‎454 Gram Na Libra ‎454 Gram Till Pund ‎454 Gram In Pond ‏454 رطل إلى غرام ‎454 Qram Narınlamaq ‎454 গ্রাম মধ্যে পাউন্ড ‎454 Gram A Lliura ‎454 ग्राम से पाउण्ड ‎454 Gram Ke Pon ‎454 グラムからポンド ‎454 그램 파운드 ‎454 Gram Til Libra ‎454 Грамм в Фунт ‎454 Gram V Funt ‎454 Grami Në Funta ‎454 กรัมปอนด์ ‎454 ગ્રામ પાઉન્ડ ‎454 Gram Pound ‎454 грам в Фунт ‎454 Gam Sang Pound ‎454 克为磅 ‎454 克至磅 ‎454 Gramme To Pound Sitemap 0.1 - 100 Sitemap 101 - 1000 Sitemap 1010 - 10000 Impressum © Meta Technologies GmbH
9236	https://mathcircle.wustl.edu/files/2023/02/20191103-mo-arml-fancy-factoring.pdf	FANTASTIC FACTORING [Thanks to my friend Harold Reiter of North Carolina for much of this material.] Following are some factoring patterns, formulas, and a theorem that you might already recognize. Difference of squares Pascal’s Triangle x2 – y2 = (x – y)(x + y) Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2) Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2) Vieta’s Formulas connect the coefficients of a polynomial to sums and products of its roots. This quartic example suggests Viete’s formulas [but be careful to correctly assign plus and minus signs]. (x – p)(x – q)(x – r)(x – s) = x4 – (p+q+r+s) x3 + (pq+pr+ps+qr+qs+rs) x2– (pqr+pqs+prs+qrs) x + pqrs THEOREM: Given function f(x) and constant x = a, the following four statements are equivalent. 1. f(a) = 0 2. x – a is a factor of f(x). 3. x = a is a zero of f(x) 4. (a, 0) is an x-intercept of graph of f(x) Factor the following expressions. 1. 16x4 – 81y4 2. 125x3 + 64y3 3. 8x3 – 27 4. x4 + 4x3y + 6x2y2 + 4xy3 + y4 5. x5 – 10x4 + 40x3 – 80x2 + 80x – 32 6. There are two distinct ways to “attack” the factorization of x6 – y6 . Try both ways and then convince yourself that they are equivalent. A. x6 – y6 = B. x6 – y6 = 7. By studying the patterns for Sum and Difference of Cubes, above, can you determine how to factor each of these? A. x7 + y7 = B. x10 – y10 = C. Explain why x8 + y8 cannot be factored in a similar way. 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 8. One technique for factoring expressions with 4 or more terms is factor by grouping. Try these. A. 4ab – 8b2 + 3a3 – 6a2b B. xy + x + y + 1 C. Can you expand (x + 1)(y + 1)(z + 1) in “one step”? To solve most polynomial equations, you set an expression equal to zero and factor it. However, if you are told that the solutions are integers, other methods are possible. 9. If x is a positive integer, solve: x(x + 1)(x + 2)(x + 3) + 1 = 3792 Simon’s Favorite Factoring Trick (SFFT) is a great tool for solving certain math contest problems. I will present this example both algebraically and geometrically! I will make each of the three terms represent the AREA of one of the four rectangles in this diagram. Then, I will be “complete the rectangle”. EXAMPLE: Given that x and y are positive integers, solve: x2 + 5x2y2 + 20y2 = 269 x2 + 5x2y2 + 20y2 + _ = 269 + __ 10. p and q are non-zero integers. How many ordered pairs (p, q) satisfy 2pq + 2p + 3q = 18 ? Note: SFFT also works when some terms are negative. 11. Twice the area of a non-square rectangle equals triple its perimeter. If the dimensions are positive integers, what is the area of the rectangle? 12. Compute all integer value of n between 90 and 100 inclusive that cannot be written in the form n = a + b + ab, where a and b are positive integers. 13. A, M, and C are positive integers such that A > M > C and A + M + C = 12. If AMC + AM + AC + CM = 71, what is the maximum possible value of A? 14. If x5 + 5x4 + 10x3 + 10x2 – 5x = 9 and x ≠ – 1, compute the numerical value of (x + 1)4 . 15. Find the number of ordered pairs of integers (m, n) for which mn > 0 and m3 + n3 + 99mn = 333 ANSWERS TO FANTASTIC FACTORING If need assistance in the solutions of any of these problems, please email me at rickarmstrongpi@gmail.com or ask friends or a math teacher. Rick Armstrong 1. (4x2 + 9y2) (2x – 3y)(2x + 3y) 2. (5x + 4y)(25x2 – 20xy + 16y2) 3. (2x – 3)(4x2 + 6x + 9) 4. (x + y)4 5. (x – y)5 6A. Diff. of Squares: (x3 – y3)(x3 + y3) = (x – y)(x + y)(x2 + xy + y2) (x2 – xy + y2) 6B. Diff. of Cubes: (x2 – y2)(x4 + x2 y2 + y4) 6. Proof: It is very difficult to produce the factorization of (x4 + x2 y2 + y4) from 6B into the two quadratic factors of 6A: (x2 + xy + y2) (x2 – xy + y2). But you can check it by expanding: (x2 + xy + y2) (x2 – xy + y2). 7A. x7 + y7 = (x + y)(x6 – x5 y + x4 y2 – x3 y3 + x2 y4 – x5 y + y6) 7B. x10 – y10 = (x – y)( x9 + x8 y + x7 y2 + x6 y3 + …… + x y8 + y9) 7C. Using the given theorem, since y = – x does not make x10 + y10 equal to zero, (x + y) is not a factor of x10 + y10 so we cannot use the given patterns to factor x10 + y10 . 8 A. 4ab – 8b2 + 3a3 – 6a2b = 4b(a – 2b) + 3a2 (a – 2b) = (a – 2b)(4b + 3a2) 8B. (x + 1)(y + 1) 8C. xyz + xy + xz + yz + x + y + z + 1 9. 198 ARML, Individual #1 x(x + 1)(x + 2)(x + 3) + 1 = 3792 OR x(x + 1)(x + 2)(x + 3) = 3792 – 12 = 378 380. x(x + 1)(x + 2)(x + 3) requires that the 4 factors be consecutive integers. With that clue, factor: 380 378 = 19 20 18 21 and x = 18 EXAMPLE: Given that x and y are positive integers, solve: x2 + 5x2y2 + 20y2 = 269 x2 + 5x2y2 + 20y2 + 4 = 269 + 4 (x2 + 4)(5y2 + 1) = 273 = 3 7 13 By inspection, the only solution with positive integers requires x2 + 4 = 13 while 5y2 + 1 = 21 with x = 3 and y = 2 10. SIX: (2, 2); ( – , 20); ( – 12, – 2); ( –5, –4); ( –3, –-8); ( –2, –21 ) 11. 48 12. 96 and 100 [1990 ARML, Team #7] 13. 13 [adapted from 2000 AMC, #12] 14. 10 [1994 ARML, Team #1 ] 15. 35 [1999 AHSME, #30] HINT: Set s = m + n so that s3 = (m + n)3 = m3 + n3 + 3mn(m + n)And subtract the given equation from this equation. After factoring, replace s with m + n. Good Luck! x2 4 1 5y2 5y2 20y2 5x2 y2 4 TEAM ROUND – 20 MINUTES 1. The number (96 + 1) is the product of three primes. Compute the largest of these 3 primes. 2. Of the integers between 1 and 2310, how many are divisible by exactly three of the five primes 2, 3, 5, 7, and 11. 3. If x and y are positive integers such that x2 = y2 + 61, find x(x + 2) + y(y + 3). 4. The graph of xy + 3x + 2y = 0 can be produced by translating the graph of y = k/x to the left and down for some constant value k. Find k. 5. Let f(x) = x2 + bx + 9 and g(x) = x2 + dx + e. If f(x) has zeroes r and s, and g(x) has zeroes –r and –s, compute the two roots of f(x) + g(x) = 0. 6. How many ordered pairs of integers (x, y) with 1 < x < 100 and 1 < y < 100 make the quantity xy – x – y a multiple of 5? 7. If three of the roots of x4 + ax2 + bx + c = 0 are 1, 2, and 3, find the value of a + c . 8. x and y are real numbers that satisfy equations x – y = 1 and x5 – y5 = 2016. Calculate 𝑥5+ 𝑦5 𝑥+𝑦−(𝑥4 + 𝑦4 ) 9. How many ordered pairs of positive integers (a, b) are that such that 1 𝑎− 1 𝑏= 1 143 ? 10. Suppose that a, b, c, d are real numbers such that: ab + 3a + 3b = 216; bc + 3b + 3c = 96; and cd + 3c + 3d = 40. Determine the maximum possible value of ad + 3a + 3d . ANSWERS TO TEAM ROUND If need assistance in the solutions of any of these problems, please email me at rickarmstrongpi@gmail.com or ask friends or a math teacher. Rick Armstrong 1. 6481 [1992 NYSML, Team #5] 6. 1600 [mathleague.org 11202, Large Team #7] 2. 186 [mathleague.org 11207, Large Team #4] 7. – 61 [AHSME 1966, #30] 3. 2013 [mathleague.org 11607, Team #2] 8. – 403 [mathleague.org 11607, Target #6] 4. 6 [mathleague.org 11301, Sprint #10] 9. 4 [mathleague.org 11308, Sprint #28] 5. ± 3i [1991 NYSML, Individual #2] 10. 96 [mathleague.org 11307, Sprint #11]
9237	https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_General_Chemistry/20%3A_Organic_Chemistry/20.3%3A_Aldehydes_Ketones_Carboxylic_Acids_and_Esters	Skip to main content 20.3: Aldehydes, Ketones, Carboxylic Acids, and Esters Last updated : Jun 5, 2019 Save as PDF 20.2: Alcohols and Ethers 20.4: Amines and Amides Page ID : 105689 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Skills to Develop Describe the structure and properties of aldehydes, ketones, carboxylic acids and esters Another class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The trigonal planar carbon in the carbonyl group can attach to two other substituents leading to several subfamilies (aldehydes, ketones, carboxylic acids and esters) described in this section. Aldehydes and Ketones Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively. In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms. As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–. In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond. Like the C=O bond in carbon dioxide, the C=O bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar C=O bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure 20.3.1). Figure 20.3.1: The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar. The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms: Example 20.3.1: Oxidation and Reduction in Organic Chemistry Methane represents the completely reduced form of an organic molecule that contains one carbon atom. Sequentially replacing each of the carbon-hydrogen bonds with a carbon-oxygen bond would lead to an alcohol, then an aldehyde, then a carboxylic acid (discussed later), and, finally, carbon dioxide: CH4⟶CH3OH⟶CH2O⟶HCO2H⟶CO2(20.3.1) What are the oxidation numbers for the carbon atoms in the molecules shown here? Solution In this example, we can calculate the oxidation number (review the chapter on oxidation-reduction reactions if necessary) for the carbon atom in each case (note how this would become difficult for larger molecules with additional carbon atoms and hydrogen atoms, which is why organic chemists use the definition dealing with replacing C–H bonds with C–O bonds described). For CH4, the carbon atom carries a –4 oxidation number (the hydrogen atoms are assigned oxidation numbers of +1 and the carbon atom balances that by having an oxidation number of –4) For the alcohol (in this case, methanol), the carbon atom has an oxidation number of –2 (the oxygen atom is assigned –2, the four hydrogen atoms each are assigned +1, and the carbon atom balances the sum by having an oxidation number of –2; note that compared to the carbon atom in CH4, this carbon atom has lost two electrons so it was oxidized) For the aldehyde, the carbon atom’s oxidation number is 0 (–2 for the oxygen atom and +1 for each hydrogen atom already balances to 0, so the oxidation number for the carbon atom is 0) For the carboxylic acid, the carbon atom’s oxidation number is +2 (two oxygen atoms each at –2 and two hydrogen atoms at +1) For carbon dioxide, the carbon atom’s oxidation number is +4 (here, the carbon atom needs to balance the –4 sum from the two oxygen atoms). Exercise 20.3.1 Indicate whether the marked carbon atoms in the three molecules here are oxidized or reduced relative to the marked carbon atom in ethanol: There is no need to calculate oxidation states in this case; instead, just compare the types of atoms bonded to the marked carbon atoms: Answer a : reduced (bond to oxygen atom replaced by bond to hydrogen atom); Answer b : oxidized (one bond to hydrogen atom replaced by one bond to oxygen atom); Answer c : oxidized (2 bonds to hydrogen atoms have been replaced by bonds to an oxygen atom) Aldehydes are commonly prepared by the oxidation of alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol: Alcohols that have their –OH groups in the middle of the chain are necessary to synthesize a ketone, which requires the carbonyl group to be bonded to two other carbon atoms: An alcohol with its –OH group bonded to a carbon atom that is bonded to no or one other carbon atom will form an aldehyde. An alcohol with its –OH group attached to two other carbon atoms will form a ketone. If three carbons are attached to the carbon bonded to the –OH, the molecule will not have a C–H bond to be replaced, so it will not be susceptible to oxidation. Formaldehyde, an aldehyde with the formula HCHO, is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance. Dimethyl ketone, CH3COCH3, commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals. Carboxylic Acids and Esters The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (Figure 20.3.2). Figure 20.3.2: Esters are responsible for the odors associated with various plants and their fruits. Both carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples): The functional groups for an acid and for an ester are shown in red in these formulas. The hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt: Carboxylic acids are weak acids, meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution. We prepare carboxylic acids by the oxidation of aldehydes or alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol: Esters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, CH3CO2CH2CH3, is formed when acetic acid reacts with ethanol: The simplest carboxylic acid is formic acid, HCO2H, known since 1670. Its name comes from the Latin word formicus, which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests. Acetic acid, CH3CO2H, constitutes 3–6% vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon. The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure 20.3.3). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C3H5(OH)3, with large carboxylic acids, such as palmitic acid, CH3(CH2)14CO2H, stearic acid, CH3(CH2)16CO2H, and oleic acid, CH3(CH2)7CH=CH(CH2)7CO2H. Oleic acid is an unsaturated acid; it contains a C=C double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds. Figure 20.3.3: Over 350 different volatile molecules (many members of the ester family) have been identified in strawberries. (credit: Rebecca Siegel) Summary Functional groups related to the carbonyl group include the –CHO group of an aldehyde, the –CO– group of a ketone, the –CO2H group of a carboxylic acid, and the –CO2R group of an ester. The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom. All of these compounds contain oxidized carbon atoms relative to the carbon atom of an alcohol group. Glossary aldehyde : organic compound containing a carbonyl group bonded to two hydrogen atoms or a hydrogen atom and a carbon substituent carbonyl group : carbon atom double bonded to an oxygen atom carboxylic acid : organic compound containing a carbonyl group with an attached hydroxyl group ester : organic compound containing a carbonyl group with an attached oxygen atom that is bonded to a carbon substituent ketone : organic compound containing a carbonyl group with two carbon substituents attached to it Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at 20.2: Alcohols and Ethers 20.4: Amines and Amides
9238	https://stackoverflow.com/questions/19818762/most-efficient-way-to-calculate-ncr-modulo-142857	java - Most efficient way to calculate nCr modulo 142857 - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Most efficient way to calculate nCr modulo 142857 Ask Question Asked 11 years, 10 months ago Modified11 years, 10 months ago Viewed 1k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I want to calculate nCr modulo 142857. Following is my code in Java: java private static int nCr2(int n, int r) { if (n == r \|\| r == 0) { return 1; } double l = 1; if (n - r < r) { r = n - r; } for (int i = 0; i < r; i++) { l = (n - i); l /= (i + 1); } return (int) (l % 142857); } This gives nCr in O(r) time. I want an algorithm to get the result in less time than this. Is there such an algorithm? java algorithm 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 6, 2013 at 17:35 Adam Stelmaszczyk 19.9k 4 4 gold badges 72 72 silver badges 111 111 bronze badges asked Nov 6, 2013 at 17:30 JainendraJainendra 25.2k 30 30 gold badges 131 131 silver badges 172 172 bronze badges 13 commons.apache.org/proper/commons-math/apidocs/org/apache/…jrey –jrey 2013-11-06 17:36:48 +00:00 Commented Nov 6, 2013 at 17:36 What is your typical range for n and r? Also do you tend to compute many similar n,r pairs in groups, I.e. could you benefit from having a small cache of recently computed values?Jason C –Jason C 2013-11-06 17:42:31 +00:00 Commented Nov 6, 2013 at 17:42 2 You might want to check out this. I'm concerned that if you're using a double to do the calculation, you will lose precision and get the wrong answer if n is large enough, when an exact answer should be available using modular arithmetic. Using BigInteger would be better, but the link may give a more efficient way.ajb –ajb 2013-11-06 17:46:48 +00:00 Commented Nov 6, 2013 at 17:46 @JasonC 1 <= n <= 1000000000Jainendra –Jainendra 2013-11-06 17:48:39 +00:00 Commented Nov 6, 2013 at 17:48 1 I can’t find any input value for which the computation will take significant time before infinity is reached. So optimizing is pointless. Especially as Math.ulp(l) will be greater than 142857 even earlier so the modulo does not provide useful results then.Holger –Holger 2013-11-06 18:01:13 +00:00 Commented Nov 6, 2013 at 18:01 \|Show 8 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. You can precompute results for given n and r pairs and hard-code them in the table int t[][]. Later, during run-time, when you need nCr(n, r), you just make a look-up to this table: t[n][r]. This is O(1) during run-time. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Nov 6, 2013 at 17:41 Adam StelmaszczykAdam Stelmaszczyk 19.9k 4 4 gold badges 72 72 silver badges 111 111 bronze badges 1 Comment Add a comment Bernhard Barker Bernhard BarkerOver a year ago My guess is that the range of values is too large for this to be viable. 2013-11-06T17:46:36.723Z+00:00 1 Reply Copy link This answer is useful 1 Save this answer. Show activity on this post. As your number is no prime, this answer doesn't apply. But you could easily decompose 142857 into primes, compute the corresponding moduli, and use the Chinese Remainder Theorem to get your result. This may or may not make sense for numbers you're working with. In any case you must avoid double, unless you can be sure that all your intermediate results can be represented exactly with only 53 bits (otherwise you lose precision and get a non-sense out). Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited May 23, 2017 at 11:51 CommunityBot 1 1 1 silver badge answered Nov 7, 2013 at 9:05 maaartinusmaaartinus 46.7k 40 40 gold badges 175 175 silver badges 342 342 bronze badges 1 Comment Add a comment PlsWork PlsWorkOver a year ago Excellent answer! Could you elaborate on how to use the Chinese remainder theorem to solve this problem, do you have any sources on how to use it to solve this problem? 2017-06-29T17:09:01.54Z+00:00 0 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. You already have most of the answer in the function that you mention. If n is fixed and r is variable, you can use nCr = nC(r-1) (n - r + 1) / r. So you can use a table for nCr and build it incrementally (unlike what the other answer mentions where precomputation is not incremental). So your new function can be made recursive with a table being passed. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Nov 6, 2013 at 20:20 user1952500user1952500 6,781 5 5 gold badges 29 29 silver badges 40 40 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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9239	https://www.mathtutordvd.com/public/Argon-Ar.cfm	Argon (Ar) Atomic Number: 18 Atomic Weight: 39.948 Melting Point: 83.80 K (-189.35°C or -308.83°F) Boiling Point: 87.30 K (-185.85°C or -302.53°F) Density: 0.0017837 grams per cubic centimeter Phase at Room Temperature: Gas Element Classification: Non-metal Period Number: 3 Group Number: 18 Group Name: Noble Gas Number of Stable Isotopes: 3 Ionization Energy: 15.760 eV Oxidation States: 0 Electron Shell Configuration: 1s2 2s2 2p6 3s2 3p6 Member Benefits: View All Courses Online 24/7 Access Over 1000 Lessons 500+ Hours of Video New Lessons Weekly Worksheets Message Boards Homework Help Join Now Learn More Excellent "Buying your algebra, calculus, and physics tutor dvds has been the best educational investment." "I went from a 'C' to an 'A' student last semester!" "Math Tutor DVDs are fantastic! Jason presents the material in a clear and well-organized form. I was completely terrified of physics, but just after the first lecture I felt at ease." "Your methods are so clear that my seven year old son was grasping the trigonometry lessons. I'm picking up some new stuff too." Gary G. "Watching your math help videos is wonderful because, as you work the problems, you show and explain every step." M. Dalrymple Lancaster, CA "All of the instruction and examples on the math tutor DVD's are very clearly explained, and Jason's style of teaching definitely makes the viewer very comfortable with the material being presented." "I found the lectures very clear, straight to the point, and the pace was just right for me, who has not seen any calculus or trig in the past 10 years and need to come up to speed fast." "Just wanted to let you know that, thanks to the foundation I got from your math help DVD's (especially the precalculus DVD) I was able to pass my precalculus course this semester with an A!" "You have a serious teaching gift. Proof is, I'm watching your DVDs when I would normally be out and about. Never thought I could learn math. I'm blasting straight through to calculus and then physics. I really enjoy this, and am thinking about a career change. Great job!" Member Benefits: View All Courses Online 24/7 Access Over 1000 Lessons 500+ Hours of Video New Lessons Weekly Worksheets Message Boards Homework Help Join Now Learn More Excellent "Buying your algebra, calculus, and physics tutor dvds has been the best educational investment." "I went from a 'C' to an 'A' student last semester!" "Math Tutor DVDs are fantastic! Jason presents the material in a clear and well-organized form. I was completely terrified of physics, but just after the first lecture I felt at ease." "Your methods are so clear that my seven year old son was grasping the trigonometry lessons. I'm picking up some new stuff too." Gary G. "Watching your math help videos is wonderful because, as you work the problems, you show and explain every step." M. Dalrymple Lancaster, CA "All of the instruction and examples on the math tutor DVD's are very clearly explained, and Jason's style of teaching definitely makes the viewer very comfortable with the material being presented." "I found the lectures very clear, straight to the point, and the pace was just right for me, who has not seen any calculus or trig in the past 10 years and need to come up to speed fast." "Just wanted to let you know that, thanks to the foundation I got from your math help DVD's (especially the precalculus DVD) I was able to pass my precalculus course this semester with an A!" "You have a serious teaching gift. Proof is, I'm watching your DVDs when I would normally be out and about. Never thought I could learn math. I'm blasting straight through to calculus and then physics. I really enjoy this, and am thinking about a career change. Great job!"
9240	https://www.tiger-algebra.com/en/solution/scientific-notation-conversion/0.000001/	Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Scientific notation/Standard form Other Ways to Solve Step-by-step explanation 1. Make the number a new number between 1 and 10 Move the decimal point to make 0.000001 a new number between 1 and 10. Because our original number is less than one, we move the decimal point to the right. Drop any zeroes in front of the number. Keep track of how many times we move the decimal point. 0.000001 -> 1 Our new number is 1. We moved the decimal point 6 times. 2. Define the power of 10 Because our original number was less than one, the exponent defining the power of 10 is negative. Remember, we moved the decimal point 6 times, so the exponent is negative 6 10−6 3. Final result 1⋅10−6 How did we do? Why learn this Scientific notation, or standard form, makes things easier when working with very small or very big numbers, both of which come up frequently in the fields of science and engineering. It is used in science, for example, to convey the mass of the heavenly bodies: Jupiter’s mass is 1.898⋅1027kg, which is easier to comprehend than writing the number 1,898 followed by 24 zeroes. Scientific notation also makes solving problems that use such high or low numbers more straightforward. Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
9241	https://gwern.net/doc/statistics/probability/2009-froelich.pdf	Teacher’s Corner Does Your iPod Really Play Favorites? Amy G. FROELICH, William M. DUCKWORTH, and Jessica CULHANE Since the introduction of the ﬁrst iPod portable music player (MP3 player) by Apple, Inc., users have questioned the ran-domness of the shufﬂe feature. Most evidence cited by users claiming to show nonrandom behavior in the shufﬂe feature is anecdotal in nature and not based on any systematic analysis of its randomness. This article reports on our attempt to investi-gate the shufﬂe feature on the iPod and to test its randomness through the use of probability and statistical modeling. We be-gin by reviewing the research on people’s inability to perceive and understand both random and nonrandom behavior. Prob-ability models are then developed, under the assumption of a random shufﬂe, for several of the most common types of events cited as evidence of a nonrandom shufﬂe. Under this null hy-pothesis of a random shufﬂe, several goodness-of-ﬁt tests of one of the probability models are conducted using data col-lected from real iPods. No evidence to support user claims of a nonrandom shufﬂe was found. Finally, we conclude with some reﬂections on and ideas for incorporating these examples into undergraduate probability and statistics courses. KEY WORDS: Goodness-of-ﬁt test; Probability models; Ran-domness. 1. INTRODUCTION Since the introduction of the ﬁrst iPod portable music player by Apple, Inc. in October 2001, the small device has become a huge social phenomenon. At the original iPod product launch, Steve Jobs, CEO of Apple, Inc. stated “... iPod, a thousand songs in your pocket. This is a major, major breakthrough.” (Levy 2006, p. 9). One of the amazing aspects about storing 1000 songs in your pocket is the ability to become your own disc jockey. This aspect is further enhanced by a feature built into the iPod software called “shufﬂe.” The shufﬂe feature takes Amy G. Froelich is Associate Professor, Department of Statistics, Iowa State University, 3109 Snedecor Hall, Ames, IA 50011-1210 (E-mail: amyf@iastate.edu). William M. Duckworth is Associate Professor of Decision Sciences, Creighton University, College of Business, 2500 California Plaza, Omaha, NE 68178 (E-mail: wmd@creighton.edu). Jessica Culhane is Actuarial Analyst, Milliman, 1301 Fifth Avenue, Suite 3800, Seattle, WA 98101 (E-mail: jessica.culhane@milliman.com). Portions of this manuscript were completed as part of the third author’s senior honors project under the direction of the ﬁrst author at Iowa State University. The authors wish to thank Dr. W. Robert Stephenson for his assistance on the probabilities in Section 3.3 and the Editor, Associate Editor, and two anonymous reviewers for their helpful comments on earlier versions of this manuscript. a list of songs, called a playlist, and rearranges them in a ran-dom order. Since its introduction, users have questioned the randomness of the shufﬂe on the iPod. Most notable is an article by Steven Levy from Newsweek magazine titled “Does Your iPod Play Fa-vorites?” (Levy 2005). In his article and subsequent book (Levy 2006), he reported anecdotal evidence of potential nonrandom behavior in the shufﬂe feature when using his iPod. A study of many iPod and technology related websites shows the same re-sults; people believe the random shufﬂe feature on the iPod is not really random. Adding to the controversy is the refusal of Apple, Inc. to release the code used to produce these random permutations. On some websites, this controversy has become a full-blown conspiracy. After reading the Newsweek article and other sources, we were very skeptical of the reported evidence of nonrandom be-havior of the shufﬂe feature. As statisticians, we are very famil-iar with people’s inability to understand randomness. There are many examples in the literature of how people’s intuitive ideas of probabilities do not match reality. Songs also evoke emo-tions, which play a role in our inability to recognize random behavior. With this background, we decided to look speciﬁcally at some of the reported evidence of nonrandom behavior and de-velop probability models for these events under the assumption of a random shufﬂe. We then used data collected from our own iPods to conduct goodness-of-ﬁt tests under the null hypothe-sis of a random shufﬂe for one of these probability models. In Section 2 of this article, we provide a summary of the current research on the psychology of understanding random and non-random events. In Section 3, we report on the development of the probability models and then study some of the probabili-ties for events reported in the Newsweek article and book by Steven Levy. The results of several goodness-of-ﬁt tests for one of these probability models are included in Section 4 of this ar-ticle, along with the collected data. In Section 5 below, we show how these examples can be used and incorporated into under-graduate probability and statistics courses, and in Section 6, we offer ﬁnal conclusions. 2. THE PSYCHOLOGY OF UNDERSTANDING RANDOMNESS The concept of randomness is not easily understood. Most people feel that any randomly produced series should contain very few, if any, extended runs of the same event and should represent the long run expected frequencies of events. In actu-ality, randomly generated series may not display either of these properties in the short run. When iPod users notice that their shufﬂe feature seems to violate these properties by choosing © 2009 American Statistical Association DOI: 10.1198/tast.2009.07073 The American Statistician, August 2009, Vol. 63, No. 3 263 three songs in a row from the same album or ﬁve songs out of ten from the same artist, they conclude the shufﬂe is not ran-dom. An aversion to runs is shown in studies where participants are asked to generate random binary series, like coin tosses. The series they create tend to have too many alterations and too few runs of two or three successes than what would be ex-pected from a random process (Bakan 1960; Orr, Federspiel, and Maxwell 1972; Diener and Thompson 1985). For an inﬁnite number of coin tosses, the limiting proportion of heads equals the limiting proportion of tails. Many people also expect this to be true over the short run, that is, over relatively few tosses of a coin. The belief is that occurrences in the short run different from this expectation should quickly correct themselves. This is an example of the gambler’s fallacy, the belief that past occur-rences of a random event will inﬂuence its future occurrences. So, if a gambler has tossed ten heads in a row, he may believe that tails are “due,” or conversely, he may feel that heads are “hot.” When a few series do not represent the expected frequen-cies of events, people tend to label the event a “coincidence.” The more personally meaningful or seemingly improbable a co-incidence is, the more surprising it will seem (Falk 1989). With music, “personally meaningful” could mean anything from the ﬁrst song at your wedding to the least favorite song in your li-brary. So when a personally meaningful song is played in the ﬁrst ten songs of a shufﬂe, it is more surprising than if the song had no personal meaning. If an event is more surprising, we have a better chance of remembering it, and, therefore, these surprising events will seem to occur too often to be random. One example of the relative surprise of coincidences is com-monly referred to as the birthday paradox. When asked, “What is the probability that two or more people out of a group of 25 will share the same birthday?” most people ﬁnd it hard to be-lieve the answer is more than 50%. This is so surprising because we tend to focus on our own birthday, which has personal mean-ing. Therefore, the question becomes “What is the probability that someone in a group of 25 people will share my birthday?” In this case, the chance is much smaller–less than 7% (Bennett 1998). Because randomness and probability are counterintuitive to many people, it is a very difﬁcult concept to teach. People can-not help but remember surprising coincidences, such as a shuf-ﬂe with six songs in a row that begin with the letter “D.” As Paul Kocher, CEO of Cryptography Research, notes in Levy’s book, “Our brains aren’t wired to understand randomness– there’s even a huge industry that takes advantage of people’s in-ability to deal with random distributions. It’s called gambling.” (Levy 2006, p. 191). 3. PROBABILITY MODELS FROM A RANDOM SHUFFLE In this section, we develop several probability models for dif-ferent outcomes from a shufﬂe under the assumption the shuf-ﬂe feature is truly random. Most of the probability models can be found in standard textbooks like Ross (2006) and Wackerly, Mendenhall, and Scheaffer (2007). The shufﬂe feature of any digital music player works by tak-ing a collection of songs, called a playlist, and producing a ran-dom permutation. Each song will appear in the shufﬂed playlist only once and each shufﬂe of a playlist is independent from all others. This type of shufﬂe contrasts with a physical shufﬂing of objects (like a rifﬂe shufﬂe of a deck of cards). See Baker and Diaconis (1992) for results on the randomness of the rifﬂe shufﬂe. Let N denote the number of songs in a playlist. The number of songs in the shufﬂed playlist the user listens to before se-lecting a different playlist or reshufﬂing the same playlist will be denoted as n. While n ≤N, n is almost always strictly less than N for longer playlists. Each song in a N song playlist can be classiﬁed according to one of several groupings, including Artist, Album, Date When Song was Added to iTunes, whether or not Song was purchased from the iTunes Store, etc. For a par-ticular grouping (Artist, Album, etc.), let Gi denote the number of songs in the playlist belonging to the ith group and g denote the number of groups. Assuming a random shufﬂe, the probability that any of the N! shufﬂes will occur is 1/N!. However, if the user only listens to the ﬁrst n songs in the shufﬂed playlist, the probability that any of the N!/(N −n)! shufﬂes will occur is equal to (N −n)!/N!. 3.1 Length of Time in One Shufﬂe Before a Particular Group Occurs Both in the Newsweek article and in his book, Steven Levy describes a phenomenon he calls the Length of Time Before Steely Dan (LTBSD) Factor. At the time, Levy’s iPod con-tained approximately 3,000 songs with approximately 50 of these songs belonging to the artist Steely Dan. To him, his ﬁrst iPod seemed to prefer this musical group over other artists. He noted this by observing the ﬁrst Steely Dan song appeared to occur very early in a shufﬂe, much earlier than he would have expected. Assuming a random shufﬂe, the probability the ﬁrst song of a shufﬂe would belong to the ith artist is equal to Gi/N. Let Xj = 1 indicate the jth song in the shufﬂe is from the ith artist and 0 otherwise and let T be deﬁned as the minimum value of j so that Xj = 1. Then the distribution of T is negative hyper-geometric (with r = 1) (Miller and Fridell 2007) so that P(T = t) = N−t Gi−1 N Gi , t = 1,2,...,N −Gi + 1. (1) Assuming 50 songs from Steely Dan in a playlist of N = 3,000 songs, in approximately 11% of all shufﬂes, the ﬁrst Steely Dan song will occur at or before the seventh song in the shufﬂed playlist and in approximately 50% of all shufﬂes, the ﬁrst Steely Dan song will occur at or before the 41st song in the shufﬂed playlist. Contrary to Levy’s ﬁrst impressions, the ﬁrst song from Steely Dan will occur fairly early in a large percentage of shufﬂes. 3.2 Number of Songs From One Group in One Shufﬂe Many comments from Levy and from other users of the iPod claim nonrandomness in the shufﬂe feature based on the num-ber of songs from a particular grouping, such as Artist or Al-bum, that occur during the ﬁrst n songs in a shufﬂed playlist. Assuming a random shufﬂe, for a particular grouping, the num-264 Teacher’s Corner Table 1. Probabilities of hearing 0, 1, 2, or 3 songs from Steely Dan in the ﬁrst 20, 40, or 60 songs in a shufﬂed playlist of N = 3,000 songs. n P(Yi = 0) P(Yi = 1) P(Yi = 2) P(Yi = 3) 20 0.7138 0.2435 0.0387 0.0038 40 0.5083 0.3492 0.1146 0.0239 60 0.3611 0.3747 0.1873 0.0601 ber of songs Yi from the ith group that appears in the ﬁrst n songs of a N song playlist has a hypergeometric distribution with probability distribution function P(Yi = y) = Gi y N−Gi n−y N n y = max(0,Gi + n −N),...,min(Gi,n). (2) In the Newsweek article, Levy points again to the overrep-resentation of songs from Steely Dan “whose songs always seemed to pop up two or three times in the ﬁrst hour of play.” Assuming Gi = 50 songs from Steely Dan out of a playlist of N = 3,000, the probabilities of getting zero, one, two, or three songs from Steely Dan in the ﬁrst n = 20,40, and 60 songs (corresponding to roughly 1, 2, and 3 hours of play) in the shuf-ﬂed playlist are summarized in Table 1. For each value of n, the probabilities of hearing two or three songs from Steely Dan are all fairly low. In repeated shufﬂes, this outcome should not happen with any great regularity. Unfortunately, Levy did not perform a systematic study of this outcome, making it possible these “coincidences” did not occur as often as stated. This pos-sibility is further reinforced by Levy’s own description of the group’s music as “terse, jazzy, and sometimes lyrically incom-prehensible” (Levy 2006, p. 177). Although Levy’s conclusions may have been inﬂuenced by a combination of “coincidences” and personally meaningful music, we cannot immediately re-fute his claims based on the probabilities. 3.3 Numbers of Songs From All Groups in One Shufﬂe In 2004, Apple, Inc. introduced a new low-priced iPod called Shufﬂe, built entirely around the shufﬂe feature. The ﬁrst Shuf-ﬂe held approximately 120 songs. Since most users have many more than 120 songs in their digital music libraries, Apple uses a feature called Autoﬁll to randomly select enough songs from a user’s digital library to ﬁll the capacity of the Shufﬂe. In his Newsweek article and book, Levy tested the Autoﬁll feature and reported “The ﬁrst few times..., I found some disturbing clus-ters in the songs chosen. More than once the ‘random’ playlist included three tracks from the same album! Since there are more than 3,000 tunes in my library, this seemed to defy the odds.” One way to look at this statement is to choose an album, look at the 120 songs selected to ﬁll the Shufﬂe, and count how many songs are selected from the chosen album. The num-ber of songs selected from this album follows the hyperge-ometric distribution of Equation (2). From a Gi = 12 song album with a total library of N = 3,000 songs, the probabil-ity of obtaining three or more songs from this particular al-bum out of the n = 120 songs selected to ﬁll the Shufﬂe is indeed small, approximately 1%. Looking at Levy’s statement in this manner is equivalent to looking at the probability that someone in a room will share your birthday. Focusing on one possible outcome makes the probability of this event fairly small. However, this is not the event Levy is describing. The event he describes is the event where there are three or more songs from any album in the n = 120 songs selected by the Autoﬁll feature. This is the same as grouping the library of N = 3,000 songs by Album and looking at the maximum number of songs from any one album in the n = 120 songs selected to ﬁll the Shufﬂe. In terms of the birthday example, his statement is the equivalent of looking at the probability that someone in a room will share any birthday. Deﬁne random variables Y1,Y2,...,Yg to be the numbers of songs selected to ﬁll the n = 120 song Shufﬂe from each of the g Albums in the playlist. The joint distribution of Y1,Y2,...,Yg follows a multivariate hypergeometric distribution (Johnson and Kotz 1969) with probabilities P(Y1 = y1,Y2 = y2,...,Yg = yg) = G1 y1 G2 y2 ··· Gg yg N n . (3) The random variable described in Levy’s statement is the max-imum observation from a multivariate hypergeometric distribu-tion (max(Y1,Y2,...,Yg)) and the event described is the prob-ability this maximum value will be three or more. Published research on the multivariate hypergeometric dis-tribution is focused on the values of (Y1,Y2,...,Yg) produc-ing the largest probability value, the mode of the distribution. A survey of the literature failed to produce any results for the probability distribution of the maximum observation from this distribution. The reason could be tied to the difﬁculty of directly calculating probabilities for the maximum observation from this distribution. Looking at a simpliﬁed case, assume the number of songs in the playlist is N = 3,000 and each album in the li-brary contains Gi = 12 songs for a total of g = 250 albums. In this case, the maximum number of songs from any one album can range from 1 to 12. To calculate the desired probability that the maximum number of songs from any one album is three or more, we calculate the complement event; the probability the maximum number of songs selected from any one album is one or two. For the maximum number of songs to be one, each of the n = 120 songs selected must each come from only one of the different g = 250 albums. This probability is P(max(Y1,Y2,...,Y250) = 1) = 250 120 12 1 120 3000 120 = 9.856 × 10−15. (4) For the maximum number of songs to be two, two songs could come from one album, or two songs could be selected from each of two albums, or two songs could be selected from each of three albums, etc. ending with two songs could come from each of 60 albums. The remaining songs must each come The American Statistician, August 2009, Vol. 63, No. 3 265 from a different album not already selected. This probabil-ity is P(max(Y1,Y2,...,Y250) = 2) = 60 x=1 250 x 250−x 120−2x 12 2 x12 1 120−2x 3000 120 = 0.05554727. (5) This makes the desired probability P(max(Y1,Y2,...,Y250) ≥3) = 1 −P(max(Y1,Y2,...,Y250) ≤2) = 0.9445. (6) Using this simpliﬁed example, the situation described in Levy’s article would happen in 94.45% of all Autoﬁlls of the iPod Shufﬂe. Far from defying the odds, the event of observing a cluster of three of more songs from any album when using the Autoﬁll feature is very much expected to occur. Just as in the birthday example, going from one particular outcome to any possible outcome leads to a much larger probability than ex-pected. Determining the probabilities of other maximum values be-comes very difﬁcult for even this simpliﬁed case, requiring more knowledge in combinatorics than most students in un-dergraduate probability and statistics courses generally possess. Instead of calculating these probabilities directly, a simulation program was written to estimate the probability distribution of the maximum observation from a multivariate hypergeometric distribution. The estimated probabilities based on 100,000 trials are listed in Table 2 below. In reality, the digital music libraries of iPod users will not contain equal numbers of songs per albums or artists. This is especially true given the option of purchasing through iTunes a few favorite songs from a given album instead of owning the entire album. To determine how much the probability distrib-ution of the maximum observation from the multivariate hy-pergeometric distribution would vary in more realistic settings, another simulation program was written to estimate this proba-bility distribution function for a general user’s library. For ex-ample, using the ﬁrst author’s library of g = 81 albums and N = 1017 songs, the estimated probability of getting three or more songs from any one album is approximately 1, the es-timated probability of getting 8 or more songs from any one album is 0.06817, and the estimated probability of getting 9 or more songs from any one album is 0.01825. If you group songs by Artist instead of Album, the effect becomes more dra-matic. Using the second author’s library of g = 109 artists and Table 2. Estimated probabilities for the maximum observation of simpliﬁed multivariate hypergeometric distribution. x P(max Yi ≥x) x P(max Yi ≥x) 1 1.0000 5 0.0142 2 1.0000 6 0.0006 3 0.9453 7 0.0003 4 0.2116 8 0.0000 N = 856 songs, the estimated probability of getting 3 or more songs from any one artist is approximately 1, the estimated probability of getting 17 or more songs from any one artist is 0.05996 and the estimated probability of getting 18 or more songs from any one artist is 0.02977. 3.4 Number of Shufﬂes Required in Order to Hear One Particular Song In his article in Newsweek, Levy contrasted the perceived fa-voritism shown by his iPod to Steely Dan songs versus the lack of favoritism for a particular song he purchased online. “Months after I bought Wild Thing from the iTunes store, I’m still wait-ing for my iPod to cue it up.” Clearly, if you listen to an entire shufﬂed playlist, you will hear each song on the playlist just once. However, this sce-nario almost never happens since users will listen only to the ﬁrst n songs in an N song playlist before reshufﬂing the same playlist or choosing another. Assuming a random shufﬂe, the probability a particular song occurs in the ﬁrst n songs of a N song playlist is n/N. Assuming each shufﬂe is independent, the number of shufﬂes S that occur until a particular song ap-pears in the ﬁrst n songs of the shufﬂe has a geometric distrib-ution with probability n/N. Table 3 gives the 10th, 25th, 50th, 75th, and 90th percentiles for the random variable S assuming a N = 3,000 song playlist with varying values of n (30, 60, 90, 120). Depending on his listening habits, it would be entirely possible under a random shufﬂe to need many shufﬂes in order to hear Wild Thing. 4. GOODNESS–OF–FIT TESTS FOR NUMBER OF SONGS FROM ONE GROUP IN ONE SHUFFLE Using several probability models, we are able to explain much of the anecdotal evidence of a nonrandom shufﬂe men-tioned in Levy’s Newsweek article and book. However, the seeming overrepresentation of a particular Artist early in the shufﬂed playlist cannot be refuted based on the probabilities alone. This perceived overrepresentation of certain groups early in a shufﬂed playlist has been reported not only by Levy, but by iPod users on several different websites. Usually, the groupings are by Artist or Album, but some users have claimed the shufﬂe favors songs more recently added to their iPods or songs pur-chased through Apple’s iTunes store over songs from their CD collection. To determine if there is any merit to these claims, we devel-oped several goodness-of-ﬁt tests for the number of songs from Table 3. Percentiles from the distribution of the number of shufﬂes required to hear one song when listening to the ﬁrst n = 30,60,90, or 120 songs from a N = 3,000 song playlist. Percentile n 10th 25th 50th 75th 90th 30 10 28 68 137 229 60 5 14 34 68 113 90 3 9 22 45 75 120 2 7 16 33 56 266 Teacher’s Corner one group appearing in the ﬁrst n songs of a N song playlist. We then conducted each test using real data collected from iPods purchased by the authors. In order to develop and conduct these test, we assumed (a) the software on all computers and iPods tested have no defects and are no different than any other iPods available for purchase, and (b) to avoid listening to each song in its entirety and to complete the tests in a reasonable amount of time, the playlist must be shufﬂed as soon as the user selects it. The act of clicking past a song in the shufﬂed playlist with-out listening to the entire song does not change the order of the songs in the current shufﬂed playlist. Under the null hypothesis of a random shufﬂe, the number of songs Yi from a particular group appearing in the ﬁrst n songs of a N song playlist has the hypergeometric distribution in Equa-tion (2). Under this hypothesis, in s shufﬂes of the playlist, the expected number of times y songs will appear in the ﬁrst n songs of the playlist is Ei = s ∗P(Yi = y). Pearson’s chi-square goodness-of-ﬁt test statistic is: X2 = c i=1 (Oi −Ei)2 Ei . (7) Several outcomes y are combined in Equation (7) above so that the expected number Ei for each outcome i is at least 5. The p-value of the test is P(χ2 c−1 > X2) and we will reject the null hypothesis when the p-value is less than α = 5%. We conducted ﬁve separate goodness-of-ﬁt tests. The ﬁrst three tests were all conducted using the same N = 240 song playlist on the ﬁrst author’s iPod. For each of the three tests, the number of songs by a particular artist out of the ﬁrst n = 60 songs in the playlist was recorded. The artists were selected based on the number of songs on the playlist. There were Gi = 3 songs in the playlist by Faith Hill, Gi = 14 songs by Queen and Gi = 31 songs by Jimmy Buffett, the most of any artist in the playlist. For each of the three tests, s = 100 shufﬂes were performed. The data collected in each of these three tests are given in Table 4. Combined outcomes are marked with one asterisk or with two asterisks. Included in the table are the test statistic Equation (7) and p-value for each test. In each case, we will fail to reject the null hypothesis and conclude the shufﬂe is random. To test the perception of favoritism of recently added songs or songs purchased through Apple, the third author’s iPod was used. For the ﬁrst test, a playlist of N = 40 songs was created. Twenty of these songs were chosen randomly from a large num-ber that were added on a single day shortly after the iPod was purchased, and the remaining 20 were chosen randomly from a large number that were added on a single day over a year later. For the second test, a different playlist of N = 40 songs was cre-ated, with 20 songs randomly chosen from all songs purchased from Apple’s iTunes Store and 20 songs randomly chosen from all songs from the third author’s CD collection. For each test, the number of songs appearing in the ﬁrst n = 10 songs in the two groups (recently added songs for the ﬁrst test and purchased songs for the second test) was recorded for s = 200 shufﬂes. The data collected in each of these two tests are given in Ta-ble 5. Combined outcomes are marked with one asterisk or with two asterisks. Included in the table are the test statistic found in Equation (7) and p-value for each test. In each case, we again fail to reject the null hypothesis and conclude the shufﬂe is ran-dom. Thus, we failed to ﬁnd any evidence to support the claim of users like Steven Levy of favoritism of certain groups in the shufﬂe. 5. CLASSROOM USES Several of these examples have been used in the undergrad-uate probability and statistics courses at Iowa State University. These examples were well received by students, and several of them stayed after class or visited ofﬁce hours to discuss ideas related to testing the shufﬂe feature or their personal impres-sions of the randomness of the shufﬂe. After teaching these courses for many years, the ﬁrst author can state without reser-vation that no other examples or material has elicited this kind of response from students in these courses. (See Stefanski 2007 for thought-provoking examples to teach variable selection in linear regression courses.) Table 4. Number of songs appearing in the ﬁrst n = 60 songs of a N = 240 song playlist from three different artists in each of s = 100 shufﬂes. Faith Hill (Gi = 3 songs) Queen (Gi = 14 songs) Jimmy Buffett (Gi = 31 songs) # of songs # of shufﬂes # of songs # of shufﬂes # of songs # of shufﬂes 0 51 0 1 0, 1, or 2 0 1 32 1 7 3 1 2 12 2 20 4 6 3 5 3 20 5 9 4 23 6 14 5 20 7 20 6 7 8 16 7 2 9 17 8 or more 0 10 11 11 3 12 2 13 1 14 or more 0 X2 = 4.6539 X2 = 3.31200 X2 = 3.57408 p-value = 0.0976 p-value = 0.6520 p-value = 0.8934 The American Statistician, August 2009, Vol. 63, No. 3 267 Table 5. Number of songs appearing in the ﬁrst n = 10 songs of two N = 40 song playlists (songs recently added and songs purchased through iTunes) in each of s = 100 shufﬂes. Recently added Purchased # of songs # of trials # of trials 0 0 0 1 1 2 2 4 7 3 19 18 4 40 47 5 61 52 6 43 42 7 24 25 8 6 6 9 2 1 10 0 0 X2 2.1154 2.9092 p-value 0.9088 0.8202 If the examples in this article are to be used in the class-room, the general idea of a random shufﬂe feature should be discussed ﬁrst, so that all students have a general understand-ing of the topic. The statistical tests could be used directly from this article, or students could be asked to design and perform their own tests on a particular aspect of the random shufﬂe fea-ture. If iPods are not available, the same type of study could be conducted using the shufﬂing code of a statistical package. Finally, the analyses in this article could be used to start a dis-cussion of several aspects of hypothesis tests. Setting the α level of a hypothesis test is based on prior belief in the truth of the null hypothesis, and is open to interpretation. From our view-point, there is no reason to think the software engineers at Ap-ple, Inc. made an error in the shufﬂe code. Further, people are notoriously poor at detecting random and nonrandom patterns and determining probabilities of events intuitively. These rea-sons, coupled with the fact that music can be personally mean-ingful and cause a emotional response in the listener, would lead us to discuss setting the α level of any test of the random shufﬂe feature very low, at most 1%. One could also use these results to discuss the power of a hypothesis test. The statisti-cal tests presented in this article are, of course, not a deﬁnitive proof of the randomness of the shufﬂe. Using a statistical pack-age, students could estimate the power of a statistical test to detect a nonrandom shufﬂe by simulating using probabilities different than what is indicated by a random shufﬂe. 6. CONCLUSIONS Random behavior is a difﬁcult concept. The lack of under-standing of random behavior often leads to misconceptions about what constitutes nonrandom behavior. Music also evokes strong emotions. Songs on a person’s iPod are personally mean-ingful to them; they own the songs. The controversy about the shufﬂe feature combines a difﬁcult and often misunderstood concept with personally meaningful events. Consequently, the existence of this controversy is not at all surprising. Much of the evidence of nonrandom behavior reported by Steven Levy and others does not hold up when the probability models of the events are determined. Our results show the prob-ability models for a random shufﬂe in many cases do not match the intuition of users. In addition, our statistical tests show the long-term occurrences of these events are within expectations under the assumption of a random shufﬂe. [Received April 2007. Revised April 2009.] REFERENCES Bakan, P. (1960), “Response-Tendencies in Attempts to Generate Random Bi-nary Series,” The American Journal of Psychology, 73 (1), 127–131. Baker, D., and Diaconis, P. (1992), “Trailing the Dovetail Shufﬂe to Its Lair,” The Annals of Applied Probability, 2 (2), 294–313. Bennett, D. (1998), Randomness, London: Harvard University Press. Diener, D., and Thompson, W. B. (1985), “Recognizing Randomness,” The American Journal of Psychology, 98 (3), 433–447. Falk, R. (1989), “Judgment of Coincidences: Mine versus Yours,” The Ameri-can Journal of Psychology, 102 (4), 477–493. Johnson, N. L., and Kotz, S. (1969), Discrete Distributions, New York: Wiley. Levy, S. (2005), “Does Your iPod Play Favorites?” Newsweek, CXLV (5). (2006), The Perfect Thing, New York: Simon & Schuster. Miller, G. K., and Fridell, S. L. (2007), “A Forgotten Discrete Distribution? Reviving the Negative Hypergeometric Model,” The American Statistician, 61 (4), 347–350. Orr, W. F., Federspiel, C. F., and Maxwell, O. D. (1972), “Patterns in Random Chance,” The American Journal of Psychology, 85 (1), 99–115. Ross, S. (2006), A First Course in Probability, New Jersey: Pearson Prentice Hall. Stefanski, L. A. (2007), “Residual (Sur)Realism,” The American Statistician, 61 (2), 163–177. Wackerly, D., Mendenhall, W., and Scheaffer, R. (2007), Mathematical Statis-tics With Applications, California: Thomson Brooks/Cole. 268 Teacher’s Corner
9242	https://medtube.net/angiology/medical-videos/8616-platelet-aggregation	Platelet Aggregation • Video • MEDtube.net English Deutsch Français Español Polski Menu Menu Main page Specialties Videos Images Events Publications MEDtube eLearning eBooks MEDtube LIVE Specialties Aesthetic medicine Allergology Anatomy Anesthesiology Angiology Audiology Balneology Bariatric surgery Cardiac Surgery Cardiology Clinical Genetics Clinical Immunology Clinical Pharmacology Clinical nutrition Colorectal surgery Conservative Dentistry... 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9243	https://www.studocu.com/in/document/delhi-technological-university/digital-circuits-and-systems/morris-mano-digital-design-2/45214808	Morris Mano Digital Design-2 - 128 Chapter 4 Combinational Logic If we want to pursue the - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. 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Report Document Students also viewed Microwave Engineering Unit-3 Microwave Engineering Unit-5 Microwave Engineering Unit-IV Microwave Engineering Unit-2 AE Lecture Notes-3 - Complete notes on multistage and power amplifiers Setting up a fibrer optic analog Link Related documents Discrete Structures Important practise questions Algorithm analysis and design UNIT-4 UHV Handout 3-Harmony in the Family UHV Handout 5-Harmony in the Nature and Existence Unit-1 Introduction of Engineering Analysis and Design Microwave Engineering Unit-1 Preview text 128 Chapter 4 Combinational Logic If we want to pursue the investigation and determine the information transformation task achieved by this circuit, we can draw the circuit from the derived Boolean expres- sions and try to recognize a familiar operation. The Boolean functions for F 1 and F 2 implement a circuit discussed in Section 4. Merely finding a Boolean representation of a circuit doesn’t provide insight into its behavior, but in this example we will observe that the Boolean equations and truth table for F 1 and F 2 match those describing the functionality of what we call a full adder. The derivation of the truth table for a circuit is a straightforward process once the output Boolean functions are known. To obtain the truth table directly from the logic diagram without going through the derivations of the Boolean functions, we proceed as follows: 1. Determine the number of input variables in the circuit. For n inputs, form the 2n possible input combinations and list the binary numbers from 0 to (2n - 1) in a table. 2. Label the outputs of selected gates with arbitrary symbols. 3. Obtain the truth table for the outputs of those gates which are a function of the input variables only. 4. Proceed to obtain the truth table for the outputs of those gates which are a func- tion of previously defined values until the columns for all outputs are determined. A B A B C A B C A C B C F 2 F 1 T 3 T 2 T 1 F 2 FIGURE 4. Logic diagram for analysis example Section 4 Design Procedure 129 This process is illustrated with the circuit of Fig. 4. In Table 4, we form the eight possible combinations for the three input variables. The truth table for F 2 is determined directly from the values of A, B, and C, with F 2 equal to 1 for any com- bination that has two or three inputs equal to 1. The truth table for F 2 is the comple- ment of F 2. The truth tables for T 1 and T 2 are the OR and AND functions of the input variables, respectively. The values for T 3 are derived from T 1 and F 2 :T 3 is equal to 1 when both T 1 and F 2 are equal to 1, and T 3 is equal to 0 otherwise. Finally, F 1 is equal to 1 for those combinations in which either T 2 or T 3 or both are equal to 1. Inspection of the truth table combinations for A, B, C, F 1 , and F 2 shows that it is identical to the truth table of the full adder given in Section 4 for x, y, z, S, and C, respectively. Another way of analyzing a combinational circuit is by means of logic simulation. This is not practical, however, because the number of input patterns that might be needed to generate meaningful outputs could be very large. But simulation has a very practical application in verifying that the functionality of a circuit actually matches its specification. In Section 4, we demonstrate the logic simulation and verification of the circuit of Fig. 4, using Verilog HDL. 4. 4 D E S I G N P R O C E D U R E The design of combinational circuits starts from the specification of the design objective and culminates in a logic circuit diagram or a set of Boolean functions from which the logic diagram can be obtained. The procedure involves the following steps: 1. From the specifications of the circuit, determine the required number of inputs and outputs and assign a symbol to each. 2. Derive the truth table that defines the required relationship between inputs and outputs. Table 4. Truth Table for the Logic Diagram of Fig. 4. A B C F 2 F 2 T 1 T 2 T 3 F 1 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 0 1 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 1 1 0 1 1 0 1 0 0 0 1 1 0 1 0 1 0 0 0 1 1 1 1 0 1 1 0 1 Section 4 Design Procedure 131 be four input variables and four output variables. We designate the four input binary variables by the symbols A, B, C, and D, and the four output variables by w, x, y, and z. The truth table relating the input and output variables is shown in Table 4. The bit combinations for the inputs and their corresponding outputs are obtained directly from Section 1. Note that four binary variables may have 16 bit combinations, but only 10 are listed in the truth table. The six bit combinations not listed for the input variables are don’t-care combinations. These values have no meaning in BCD and we assume that they will never occur in actual operation of the circuit. Therefore, we are at liberty to assign to the output variables either a 1 or a 0, whichever gives a simpler circuit. The maps in Fig. 4 are plotted to obtain simplified Boolean functions for the outputs. Each one of the four maps represents one of the four outputs of the circuit as a function of the four input variables. The 1’s marked inside the squares are obtained from the minterms that make the output equal to 1. The 1’s are obtained from the truth table by going over the output columns one at a time. For example, the column under output z has five 1’s; therefore, the map for z has five 1’s, each being in a square corresponding to the minterm that makes z equal to 1. The six don’t-care minterms 10 through 15 are marked with an X. One possible way to sim- plify the functions into sum-of-products form is listed under the map of each variable. (See Chapter 3.) A two-level logic diagram for each output may be obtained directly from the Boolean expressions derived from the maps. There are various other possibilities for a logic diagram that implements this circuit. The expressions obtained in Fig. 4 may be manipulated algebraically for the purpose of using common gates for two or more outputs. This manip- ulation, shown next, illustrates the flexibility obtained with multiple-output systems when Table 4. Truth Table for Code Conversion Example Input BCD Output Excess-3 Code A B C D w x y z 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 0 0 1 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1 0 0 1 0 1 1 1 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 1 1 1 0 0 132 Chapter 4 Combinational Logic implemented with three or more levels of gates: z = D y = CD + CD = CD + 1 C + D 2 x = BC + BD + BCD = B 1 C + D 2 + BCD = B 1 C + D 2 + B 1 C + D 2 w = A + BC + BD = A + B 1 C + D 2 The logic diagram that implements these expressions is shown in Fig. 4. Note that the OR gate whose output is C + D has been used to implement partially each of three outputs. Not counting input inverters, the implementation in sum-of-products form requires seven AND gates and three OR gates. The implementation of Fig. 4 requires four AND gates, four OR gates, and one inverter. If only the normal inputs are available, the first 00 1 01 11 10 ABCD 00 01 11 10 1 1 1 X X X X X 1 X D # m 0 m 1 m 3 m 2 # m 4 m 5 m 7 m 6 # m 12 m 13 m 15 m 14 # m 8 m 9 m 11 m 10 C B A 00 01 11 10 AB 00 01 11 10 CD 1 X X X X X 1 C # m 0 m 1 m 3 m 2 # m 4 m 5 m 7 m 6 # m 12 m 13 m 15 m 14 # m 8 m 9 m 11 m 10 1 1 1 X D B A 00 01 11 10 ABCD 00 01 11 1 1 1 1 X X X X X 1 1 X # m 0 m 1 m 3 m 2 # m 4 m 5 m 7 m 6 # m 12 m 13 m 15 m 14 # m 8 m 9 m 11 m 10 C B A D # m 0 m 1 m 3 m 2 # m 4 m 5 m 7 m 6 # m 12 m 13 m 15 m 14 # m 8 m 9 m 11 m 10 00 01 11 10 00 01 11 10 B AB CD A 1 1 1 1 X X X X X 1 X D C z D y CD CD x BC BD BCD w A BC BD FIGURE 4. Maps for BCD-to-excess-3 code converter 134 Chapter 4 Combinational Logic A binary adder–subtractor is a combinational circuit that performs the arithmetic operations of addition and subtraction with binary numbers. We will develop this circuit by means of a hierarchical design. The half adder design is carried out first, from which we develop the full adder. Connecting n full adders in cascade produces a binary adder for two n-bit numbers. The subtraction circuit is included in a complementing circuit. Half Adder From the verbal explanation of a half adder, we find that this circuit needs two binary inputs and two binary outputs. The input variables designate the augend and addend bits; the output variables produce the sum and carry. We assign symbols x and y to the two inputs and S (for sum) and C (for carry) to the outputs. The truth table for the half adder is listed in Table 4. The C output is 1 only when both inputs are 1. The S output represents the least significant bit of the sum. The simplified Boolean functions for the two outputs can be obtained directly from the truth table. The simplified sum-of-products expressions are S = xy + xy C = xy The logic diagram of the half adder implemented in sum of products is shown in Fig. 4(a). It can be also implemented with an exclusive-OR and an AND gate as shown in Fig. 4(b). This form is used to show that two half adders can be used to construct a full adder. x y x y x y S C x y S C (a) S xy xy C xy (b) S x y C xy FIGURE 4. Implementation of half adder Table 4. Half Adder x y C S 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 0 Section 4 Binary Adder–Subtractor 135 Full Adder Addition of n-bit binary numbers requires the use of a full adder, and the process of addi- tion proceeds on a bit-by-bit basis, right to left, beginning with the least significant bit. After the least significant bit, addition at each position adds not only the respective bits of the words, but must also consider a possible carry bit from addition at the previous position. A full adder is a combinational circuit that forms the arithmetic sum of three bits. It consists of three inputs and two outputs. Two of the input variables, denoted by x and y, represent the two significant bits to be added. The third input, z, represents the carry from the previous lower significant position. Two outputs are necessary because the arithmetic sum of three binary digits ranges in value from 0 to 3, and binary representation of 2 or 3 needs two bits. The two outputs are designated by the symbols S for sum and C for carry. The binary variable S gives the value of the least significant bit of the sum. The binary variable C gives the output carry formed by adding the input carry and the bits of the words. The truth table of the full adder is listed in Table 4. The eight rows under the input variables designate all possible combinations of the three variables. The output variables are determined from the arithmetic sum of the input bits. When all input bits are 0, the output is 0. The S output is equal to 1 when only one input is equal to 1 or when all three inputs are equal to 1. The C output has a carry of 1 if two or three inputs are equal to 1. The input and output bits of the combinational circuit have different interpretations at various stages of the problem. On the one hand, physically, the binary signals of the inputs are considered binary digits to be added arithmetically to form a two-digit sum at the output. On the other hand, the same binary values are considered as variables of Boolean functions when expressed in the truth table or when the circuit is implemented with logic gates. The maps for the outputs of the full adder are shown in Fig. 4. The simplified expressions are S = xyz + xyz + xyz + xyz C = xy + xz + yz The logic diagram for the full adder implemented in sum-of-products form is shown in Fig. 4. It can also be implemented with two half adders and one OR gate, as shown Table 4. Full Adder x y z C S 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 Section 4 Binary Adder–Subtractor 137 Addition of n-bit numbers requires a chain of n full adders or a chain of one-half adder and n 9 1 full adders. In the former case, the input carry to the least significant position is fixed at 0. Figure 4 shows the interconnection of four full-adder (FA) circuits to provide a four-bit binary ripple carry adder. The augend bits of A and the addend bits of B are designated by subscript numbers from right to left, with subscript 0 denoting the least significant bit. The carries are connected in a chain through the full adders. The input carry to the adder is C 0 , and it ripples through the full adders to the output carry C 4. The S outputs generate the required sum bits. An n-bit adder requires n full adders, with each output carry connected to the input carry of the next higher order full adder. To demonstrate with a specific example, consider the two binary numbers A = 1011 and B = 0011. Their sum S = 1110 is formed with the four-bit adder as follows: Subscript i: 3 2 1 0 Input carry 0 1 1 0 Ci Augend 1 0 1 1 Ai Addend 0 0 1 1 Bi Sum 1 1 1 0 Si Output carry 0 0 1 1 Ci + 1 The bits are added with full adders, starting from the least significant position (subscript 0), to form the sum bit and carry bit. The input carry C 0 in the least significant position must be 0. The value of Ci + 1 in a given significant position is the output carry of the full adder. This value is transferred into the input carry of the full adder that adds the bits one higher significant position to the left. The sum bits are thus generated starting from the rightmost position and are available as soon as the corresponding previous carry bit is generated. All the carries must be generated for the correct sum bits to appear at the outputs. The four-bit adder is a typical example of a standard component. It can be used in many applications involving arithmetic operations. Observe that the design of this circuit x y z S C x y xy (x y) z xy (x y) z (x y) z FIGURE 4. Implementation of full adder with two half adders and an OR gate 138 Chapter 4 Combinational Logic by the classical method would require a truth table with 2 9 = 512 entries, since there are nine inputs to the circuit. By using an iterative method of cascading a standard func- tion, it is possible to obtain a simple and straightforward implementation. Carry Propagation The addition of two binary numbers in parallel implies that all the bits of the augend and addend are available for computation at the same time. As in any combinational circuit, the signal must propagate through the gates before the correct output sum is available in the output terminals. The total propagation time is equal to the propagation delay of a typical gate, times the number of gate levels in the circuit. The longest propa- gation delay time in an adder is the time it takes the carry to propagate through the full adders. Since each bit of the sum output depends on the value of the input carry, the value of Si at any given stage in the adder will be in its steady-state final value only after the input carry to that stage has been propagated. In this regard, consider output S 3 in Fig. 4. Inputs A 3 and B 3 are available as soon as input signals are applied to the adder. However, input carry C 3 does not settle to its final value until C 2 is available from the previous stage. Similarly, C 2 has to wait for C 1 and so on down to C 0. Thus, only after the carry propagates and ripples through all stages will the last output S 3 and carry C 4 settle to their final correct value. The number of gate levels for the carry propagation can be found from the circuit of the full adder. The circuit is redrawn with different labels in Fig. 4 for convenience. The input and output variables use the subscript i to denote a typical stage of the adder. The signals at Pi and Gi settle to their steady-state values after they propagate through their respective gates. These two signals are common to all half adders and depend on only the input augend and addend bits. The signal from the input carry Ci to the output carry Ci + 1 propagates through an AND gate and an OR gate, which constitute two gate levels. If there are four full adders in the adder, the output carry C 4 would have 2 4 = 8 gate levels from C 0 to C 4. For an n-bit adder, there are 2n gate levels for the carry to propagate from input to output. B 3 C 4 S 3 A 3 FA B 2 C 3 S 2 A 2 FA B 1 C 2 S 1 A 1 FA B 0 C 1 S 0 A 0 FA C 0 FIGURE 4. Four-bit adder 140 Chapter 4 Combinational Logic C 2 = G 1 + P 1 C 1 = G 1 + P 1 1 G 0 + P 0 C 02 = G 1 + P 1 G 0 + P 1 P 0 C 0 C 3 = G 2 + P 2 C 2 = G 2 + P 2 G 1 + P 2 P 1 G 0 = P 2 P 1 P 0 C 0 Since the Boolean function for each output carry is expressed in sum-of-products form, each function can be implemented with one level of AND gates followed by an OR gate (or by a two-level NAND). The three Boolean functions for C 1 , C 2 , and C 3 are imple- mented in the carry lookahead generator shown in Fig. 4. Note that this circuit can add in less time because C 3 does not have to wait for C 2 and C 1 to propagate; in fact, C 3 is propagated at the same time as C 1 and C 2. This gain in speed of operation is achieved at the expense of additional complexity (hardware). The construction of a four-bit adder with a carry lookahead scheme is shown in Fig. 4. Each sum output requires two exclusive-OR gates. The output of the first exclusive-OR gate generates the Pi variable, and the AND gate generates the Gi variable. The carries are propagated through the carry lookahead generator (similar to that in Fig. 4) and applied as inputs to the second exclusive-OR gate. All output carries are generated after C 3 C 2 C 1 P 2 G 2 P 1 G 1 P 0 G 0 C 0 FIGURE 4. Logic diagram of carry lookahead generator Section 4 Binary Adder–Subtractor 141 a delay through two levels of gates. Thus, outputs S 1 through S 3 have equal propagation delay times. The two-level circuit for the output carry C 4 is not shown. This circuit can easily be derived by the equation-substitution method. Binary Subtractor The subtraction of unsigned binary numbers can be done most conveniently by means of complements, as discussed in Section 1. Remember that the subtraction A - B can be done by taking the 2’s complement of B and adding it to A. The 2’s complement can be obtained by taking the 1’s complement and adding 1 to the least significant pair of bits. The 1’s complement can be implemented with inverters, and a 1 can be added to the sum through the input carry. B 3 P 3 G 3 P 2 G 2 P 1 G 1 P 0 G 0 C 0 A 3 P 3 C 3 C 4 S 3 C 4 P 2 C 2 S 2 P 1 C 1 S 1 P 0 S 0 B 2 A 2 B 1 A 1 B 0 A 0 C 0 Carry Lookahead Generator FIGURE 4. Four-bit adder with carry lookahead Section 4 Binary Adder–Subtractor 143 Overflow When two numbers with n digits each are added and the sum is a number occupying n + 1 digits, we say that an overflow occurred. This is true for binary or decimal num- bers, signed or unsigned. When the addition is performed with paper and pencil, an overflow is not a problem, since there is no limit by the width of the page to write down the sum. Overflow is a problem in digital computers because the number of bits that hold the number is finite and a result that contains n + 1 bits cannot be accommodated by an n-bit word. For this reason, many computers detect the occurrence of an overflow, and when it occurs, a corresponding flip-flop is set that can then be checked by the user. The detection of an overflow after the addition of two binary numbers depends on whether the numbers are considered to be signed or unsigned. When two unsigned numbers are added, an overflow is detected from the end carry out of the most signifi- cant position. In the case of signed numbers, two details are important: the leftmost bit always represents the sign, and negative numbers are in 2’s-complement form. When two signed numbers are added, the sign bit is treated as part of the number and the end carry does not indicate an overflow. An overflow cannot occur after an addition if one number is positive and the other is negative, since adding a positive number to a negative number produces a result whose magnitude is smaller than the larger of the two original numbers. An overflow may occur if the two numbers added are both positive or both negative. To see how this can happen, consider the following example: Two signed binary numbers, +70 and +80, are stored in two eight-bit registers. The range of numbers that each register can accom- modate is from binary +127 to binary - 128. Since the sum of the two numbers is +150, it exceeds the capacity of an eight-bit register. This is also true for - 70 and - 80. The two additions in binary are shown next, together with the last two carries: carries: 0 1 carries: 1 0 + 70 0 1000110 - 70 1 0111010 + 80 0 1010000 - 80 1 0110000 + 150 1 0010110 - 150 0 1101010 Note that the eight-bit result that should have been positive has a negative sign bit (i., the eighth bit) and the eight-bit result that should have been negative has a positive sign bit. If, however, the carry out of the sign bit position is taken as the sign bit of the result, then the nine-bit answer so obtained will be correct. But since the answer cannot be accommodated within eight bits, we say that an overflow has occurred. An overflow condition can be detected by observing the carry into the sign bit position and the carry out of the sign bit position. If these two carries are not equal, an overflow has occurred. This is indicated in the examples in which the two carries are explicitly shown. If the two carries are applied to an exclusive-OR gate, an overflow is detected when the output of the gate is equal to 1. For this method to work correctly, the 2’s comple- ment of a negative number must be computed by taking the 1’s complement and adding 1. This takes care of the condition when the maximum negative number is complemented. 144 Chapter 4 Combinational Logic The binary adder–subtractor circuit with outputs C and V is shown in Fig. 4. If the two binary numbers are considered to be unsigned, then the C bit detects a carry after addition or a borrow after subtraction. If the numbers are considered to be signed, then the V bit detects an overflow. If V = 0 after an addition or subtraction, then no overflow occurred and the n-bit result is correct. If V = 1, then the result of the operation contains n + 1 bits, but only the rightmost n bits of the number fit in the space available, so an overflow has occurred. The 1 n + 12 th bit is the actual sign and has been shifted out of position. 4. 6 D E C I M A L A D D E R Computers or calculators that perform arithmetic operations directly in the decimal number system represent decimal numbers in binary coded form. An adder for such a computer must employ arithmetic circuits that accept coded decimal numbers and present results in the same code. For binary addition, it is sufficient to consider a pair of significant bits together with a previous carry. A decimal adder requires a minimum of nine inputs and five outputs, since four bits are required to code each decimal digit and the circuit must have an input and output carry. There is a wide variety of possible decimal adder circuits, depending upon the code used to repre- sent the decimal digits. Here we examine a decimal adder for the BCD code. (See Section 1.) BCD Adder Consider the arithmetic addition of two decimal digits in BCD, together with an input carry from a previous stage. Since each input digit does not exceed 9, the output sum cannot be greater than 9 + 9 + 1 = 19, the 1 in the sum being an input carry. Sup- pose we apply two BCD digits to a four-bit binary adder. The adder will form the sum in binary and produce a result that ranges from 0 through 19. These binary numbers are listed in Table 4 and are labeled by symbols K, Z 8 , Z 4 , Z 2 , and Z 1. K is the carry, and the subscripts under the letter Z represent the weights 8, 4, 2, and 1 that can be assigned to the four bits in the BCD code. The columns under the binary sum list the binary value that appears in the outputs of the four-bit binary adder. The output sum of two decimal digits must be represented in BCD and should appear in the form listed in the columns under “BCD Sum.” The problem is to find a rule by which the binary sum is converted to the correct BCD digit representation of the number in the BCD sum. In examining the contents of the table, it becomes apparent that when the binary sum is equal to or less than 1001, the corresponding BCD number is identical, and therefore no conversion is needed. When the binary sum is greater than 1001, we obtain an invalid BCD representation. The addition of binary 6 (0110) to the binary sum converts it to the correct BCD representation and also produces an output carry as required. 146 Chapter 4 Combinational Logic adder can be ignored, since it supplies information already available at the output carry terminal. A decimal parallel adder that adds n decimal digits needs n BCD adder stages. The output carry from one stage must be connected to the input carry of the next higher order stage. 4. 7 B I N A R Y M U L T I P L I E R Multiplication of binary numbers is performed in the same way as multiplication of decimal numbers. The multiplicand is multiplied by each bit of the multiplier, starting from the least significant bit. Each such multiplication forms a partial product. Succes- sive partial products are shifted one position to the left. The final product is obtained from the sum of the partial products. To see how a binary multiplier can be implemented with a combinational circuit, consider the multiplication of two 2-bit numbers as shown in Fig. 4. The multiplicand Output carry Carry out Carry in Addend Augend 4-bit binary adder 4-bit binary adder K 0 Z 8 Z 4 Z 2 Z 1 S 8 S 4 S 2 S 1 FIGURE 4. Block diagram of a BCD adder Section 4 Binary Multiplier 147 bits are B 1 and B 0 , the multiplier bits are A 1 and A 0 , and the product is C 3 C 2 C 1 C 0. The first partial product is formed by multiplying B 1 B 0 by A 0. The multiplication of two bits such as A 0 and B 0 produces a 1 if both bits are 1; otherwise, it produces a 0. This is iden- tical to an AND operation. Therefore, the partial product can be implemented with AND gates as shown in the diagram. The second partial product is formed by multiply- ing B 1 B 0 by A 1 and shifting one position to the left. The two partial products are added with two half-adder (HA) circuits. Usually, there are more bits in the partial products and it is necessary to use full adders to produce the sum of the partial products. Note that the least significant bit of the product does not have to go through an adder, since it is formed by the output of the first AND gate. A combinational circuit binary multiplier with more bits can be constructed in a similar fashion. A bit of the multiplier is ANDed with each bit of the multiplicand in as many levels as there are bits in the multiplier. The binary output in each level of AND gates is added with the partial product of the previous level to form a new partial prod- uct. The last level produces the product. For J multiplier bits and K multiplicand bits, we need 1 J K 2 AND gates and 1 J - 12 K-bit adders to produce a product of (J + K) bits. As a second example, consider a multiplier circuit that multiplies a binary number represented by four bits by a number represented by three bits. Let the multiplicand be represented by B 3 B 2 B 1 B 0 and the multiplier by A 2 A 1 A 0. Since K = 4 and J = 3, we need 12 AND gates and two 4-bit adders to produce a product of seven bits. The logic diagram of the multiplier is shown in Fig. 4. HA HA C 3 B 1 B 1 C 3 C 2 C 1 C 0 B 0 A 1 A 1 B 1 A 1 B 0 A 0 B 1 A 0 B 0 A 0 A 1 A 0 B 0 B 1 B 0 C 2 C 1 C 0 FIGURE 4. Two-bit by two-bit binary multiplier Morris Mano Digital Design-2 Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 1 0 Save Morris Mano Digital Design-2 Course: Digital Circuits and Systems (EE204) 3 documents University: Delhi Technological University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 1 0 Save 128 Chapter 4 Combinational Logic If we want to pursue the investigation and determine the information transformation task achieved by this circuit, we can dra w the circuit fr om the derived Boolean expres- sions and try to recognize a familiar oper ation.T he Boolean functions for F 1 and F 2 implement a circuit discussed in Section 4.5. Merely finding a Boolean r epresentation of a circuit doesn’t pr ovide insight into its behavior, but in this example we will observe that the Boolean equations and truth table for F 1 and F 2 match those describing the functionality of what we call a full adder. T he deriv ation of the truth table for a circuit is a str aightforward pr ocess once the output Boolean functions are known.T o obtain the truth table dir ectly from the logic diagram without going thr ough the derivations of the Boolean functions, we proceed as follows: 1.Determine the number of input variables in the cir cuit. F or n inputs, form the 2 n possible input combinations and list the binary numbers from 0 to (2 n-1) in a table. 2.Label the outputs of selected gates with arbitr ary symbols. 3.Obtain the truth table for the outputs of those gates which are a function of the input variables only. 4.Proceed to obtain the truth table for the outputs of those gates which are a func- tion of previously defined v alues until the columns for all outputs are determined. A B A B C A B C A C B C F 2 F 1 T 3 T 2 T 1 F⬘2 FIGURE 4.2 Logic diagram for analysis example Section 4.4 Design Procedure 129 T his process is illustrated with the circuit of F ig. 4.2 .In T able 4.1 ,we form the eight possible combinations for the three input v ariables.T he truth table for F 2 i s determined directly fr om the values of A , B, and C ,with F 2 equal to 1 for any com- bination that has two or three inputs equal to 1.T he truth table for F⬘ 2 is the comple- ment of F 2. The truth tables for T 1 and T 2 are the OR and AND functions of the input variables, respectively.T he v alues for T 3 are derived from T 1 and F⬘ 2:T 3 is equal to 1 when both T 1 and F⬘ 2 are equal to 1, and T 3 is equal to 0 otherwise. F inally,F 1 i s equal to 1 for those combinations in which either T 2 o r T 3 or both are equal to 1. Inspection of the truth table combinations for A, B, C,F 1, and F 2 shows that it is identical to the truth table of the full adder given in Section 4.5 for x, y, z, S, and C , respectively. Another way of analyzing a combinational cir cuit is by means of logic simulation. T his is not pr actical, however, because the number of input patterns that might be needed to generate meaningful outputs could be very lar ge. But simulation has a very practical application in verifying that the functionality of a cir cuit actually matches its specification. In Section 4.12, we demonstr ate the logic simulation and verification of the circuit of F ig. 4.2 ,using V erilog HDL. 4.4 DESIGN PROCEDURE T he design of combinational cir cuits starts from the specification of the design objective and culminates in a logic circuit diagr am or a set of Boolean functions from which the logic diagram can be obtained.The pr ocedur e involves the following steps: 1.Fr om the specifications of the circuit, determine the requir ed number of inputs and outputs and assign a symbol to each. 2.Derive the truth table that defines the requir ed r elationship between inputs and outputs. T able 4.1 T ruth T able for the Logic Diagram of Fig. 4.2 A B C F 2 F⬘ 2 T 1 T 2 T 3 F 1 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 0 1 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 1 1 0 1 1 0 1 0 0 0 1 1 0 1 0 1 0 0 0 1 1 1 1 0 1 1 0 1 130 Chapter 4 Combinational Logic 3.Obtain the simplified Boolean functions for each output as a function of the input variables. 4.Draw the logic diagr am and verify the corr ectness of the design (manually or by simulation). A truth table for a combinational circuit consists of input columns and output columns.T he input columns ar e obtained from the 2 n binary numbers for the n input variables.T he binary values for the outputs are determined fr om the stated specifica- tions.T he output functions specified in the truth table give the exact definition of the combinational circuit. It is important that the verbal specifications be interpreted correctly in the truth table, as they ar e often incomplete, and any wrong interpr etation may r esult in an incorrect truth table. T he output binary functions listed in the truth table are simplified by any available method, such as algebraic manipulation, the map method, or a computer-based sim- plification progr am. F requently, there is a variety of simplified expr essions from which to choose. In a particular application, certain criteria will serve as a guide in the process of choosing an implementation.A practical design must consider such constraints as the number of gates, number of inputs to a gate, propagation time of the signal through the gates, number of interconnections, limitations of the driving capability of each gate (i.e., the number of gates to which the output of the circuit may be connected), and various other criteria that must be taken into consider ation when designing integrated cir cuits. Since the importance of each constr aint is dictated by the particular application, it is difficult to make a gener al statement about what constitutes an acceptable implementation. In most cases, the simplification begins by satisfying an elementary objective, such as producing the simplified Boolean func- tions in a standard form.Then the simplification pr oceeds with further steps to meet other performance criteria. Code Conversion Example T he av ailability of a lar ge variety of codes for the same discr ete elements of information results in the use of dif ferent codes by dif ferent digital systems. It is sometimes necessary to use the output of one system as the input to another.A conversion circuit must be inserted between the two systems if each uses differ ent codes for the same information. T hus, a code converter is a circuit that makes the two systems compatible even though each uses a differ ent binary code. T o convert from binary code A to binary code B, the input lines must supply the bit combination of elements as specified by code A and the output lines must gener- ate the corresponding bit combination of code B.A combinational circuit performs this transformation by means of logic gates.T he design procedure will be illustr ated by an example that converts binary coded decimal (BCD) to the excess-3 code for the decimal digits. T he bit combinations assigned to the BCD and excess-3 codes ar e listed in T able 1.5 (Section 1.7). Since each code uses four bits to represent a decimal digit, there must Too long to read on your phone? Save to read later on your computer Save to a Studylist Section 4.4 Design Procedure 131 be four input variables and four output v ariables.W e designate the four input binary variables by the symbols A, B, C, and D, and the four output variables by w, x, y ,and z.T he truth table r elating the input and output variables is shown in T able 4.2 .The bit combinations for the inputs and their corresponding outputs ar e obtained directly from Section 1.7. Note that four binary variables may have 1 6 bit combinations, but only 10 are listed in the truth table.T he six bit combinations not listed for the input variables ar e don’t-care combinations.T hese values have no meaning in BCD and we assume that they will never occur in actual operation of the cir cuit.T herefore, we ar e at liberty to assign to the output variables either a 1 or a 0, whichever gives a simpler circuit. T he maps in F ig.4.3 ar e plotted to obtain simplified Boolean functions for the outputs. Each one of the four maps r epresents one of the four outputs of the cir cuit as a function of the four input variables.T he 1’s marked inside the squar es are obtained from the minterms that mak e the output equal to 1.The 1’s ar e obtained from the truth table by going over the output columns one at a time. F or example, the column under output z has five 1’s; therefore, the map for z has five 1’s, each being in a square corr esponding to the minterm that makes z equal to 1.T he six don’t-car e minterms 10 thr ough 1 5 are mark ed with an X. One possible way to sim- plify the functions into sum-of-products form is listed under the map of each v ariable. (See Chapter 3 .) A two-level logic diagram for each output may be obtained directly fr om the Boolean expressions derived fr om the maps.T her e are v arious other possibilities for a logic diagr am that implements this circuit.The expr essions obtained in Fig.4.3 may be manipulated algebraically for the purpose of using common gates for two or mor e outputs.T his manip- ulation, shown next, illustr ates the flexibility obtained with multiple-output systems when T able 4.2 T ruth T able for Code Conversion Example Input BCD Output Excess-3 Code A B C D w x y z 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 0 0 1 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1 0 0 1 0 1 1 1 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 1 1 1 0 0 132 Chapter 4 Combinational Logic implemented with three or mor e levels of gates: z=D⬘ y=CD+C⬘D⬘=CD+1 C+D 2⬘ x=B⬘C+B⬘D+BC⬘D⬘=B⬘1 C+D 2+BC⬘D⬘ =B⬘1 C+D 2+B 1 C+D 2⬘ w=A+BC+BD=A+B 1 C+D 2 The logic diagr am that implements these expr essions is shown in Fig.4.4 . Note that the OR gate whose output is C+D has been used to implement partially each of three outputs. Not counting input inverters, the implementation in sum-of-products form requires seven AND gates and three OR gates.T he implementation of F ig.4.4 requires four AND gates, four OR gates, and one inverter. If only the normal inputs are av ailable, the first 1 00 01 11 10 00 01 11 10 AB CD 1 1 1 XX X X X 1 X D m 0 m 1 m 3 m 2 m 4 m 5 m 7 m 6 m 12 m 13 m 15 m 14 m 8 m 9 m 11 m 10 C B A 00 01 11 10 00 01 11 10 AB CD 1 XX X XX 1 C m 0 m 1 m 3 m 2 m 4 m 5 m 7 m 6 m 12 m 13 m 15 m 14 m 8 m 9 m 11 m 10 1 1 1 X D B A 00 01 11 10 00 01 11 10 AB CD 11 1 1X X X XX 1 1 X m 0 m 1 m 3 m 2 m 4 m 5 m 7 m 6 m 12 m 13 m 15 m 14 m 8 m 9 m 11 m 10 C B A D m 0 m 1 m 3 m 2 m 4 m 5 m 7 m 6 m 12 m 13 m 15 m 14 m 8 m 9 m 11 m 10 00 01 11 10 00 01 11 10 B AB CD A 11 11X X X XX 1 X D C z⫽D⬘y⫽CD⫹C⬘D⬘ x⫽B⬘C⫹B⬘D⫹BC⬘D⬘w⫽A⫹BC⫹BD FIGURE 4.3 Maps for BCD-to-excess-3 code converter Section 4.5 Binary Adder–Subtractor 133 implementation will requir e inverters for v ariables B, C, and D, and the second implementation will requir e inverters for v ariables B and D.T hus, the three-level logic circuit r equires fewer gates, all of which in turn require no mor e than two inputs. 4.5 BINARY ADDER–SUBTRACTOR Digital computers perform a variety of information-processing tasks.Among the func- tions encountered ar e the various arithmetic oper ations.T he most basic arithmetic operation is the addition of two binary digits.T his simple addition consists of four pos- sible elementary operations:0+0=0, 0+1=1, 1+0=1, and 1+1=10. T he first three oper ations produce a sum of one digit, but when both augend and addend bits are equal to 1, the binary sum consists of two digits.T he higher significant bit of this result is called a carry.When the augend and addend numbers contain more significant digits, the carry obtained fr om the addition of two bits is added to the next higher order pair of significant bits.A combinational circuit that performs the addition of two bits is called a half adder. One that performs the addition of three bits (two significant bits and a previous carry) is a full adder.T he names of the circuits stem fr om the fact that two half adders can be employed to implement a full adder. x z y w D C⫹D (C⫹D)⬘ CD D⬘ C B A FIGURE 4.4 Logic diagram for BCD-to-excess-3 code converter Document continues below Discover more from: Digital Circuits and SystemsEE204Delhi Technological University 3 documents Go to course 145 Morris Mano Digital Design-3 Digital Circuits and Systems Lecture notes None 130 Morris Mano Digital Design-4 Digital Circuits and Systems Lecture notes None Discover more from: Digital Circuits and SystemsEE204Delhi Technological University3 documents Go to course 145 Morris Mano Digital Design-3 Digital Circuits and Systems None 130 Morris Mano Digital Design-4 Digital Circuits and Systems None 134 Chapter 4 Combinational Logic A binary adder–subtractor is a combinational circuit that performs the arithmetic operations of addition and subtr action with binary numbers.W e will develop this circuit by means of a hier archical design.T he half adder design is carried out first, from which we develop the full adder. Connecting n full adders in cascade produces a binary adder for two n-bit numbers.T he subtr action circuit is included in a complementing circuit. Half Adder F rom the verbal explanation of a half adder, we find that this circuit needs two binary inputs and two binary outputs.T he input v ariables designate the augend and addend bits; the output variables produce the sum and carry.W e assign symbols x and y to the two inputs and S (for sum) and C (for carry) to the outputs.T he truth table for the half adder is listed in T able 4.3 .T he C output is 1 only when both inputs are 1.T he S output repr esents the least significant bit of the sum. The simplified Boolean functions for the two outputs can be obtained dir ectly from the truth table.The simplified sum-of-pr oducts expr essions are S=x⬘y+xy⬘ C=xy The logic diagr am of the half adder implemented in sum of products is shown in F ig.4.5(a) . It can be also implemented with an exclusive-OR and an AND gate as shown in Fig.4.5(b) .T his form is used to show that two half adders can be used to construct a full adder. x y ⬘ x⬘ y x y S C x y S C (a) S ⫽ xy⬘ ⫹ x⬘y C ⫽ xy (b) S ⫽ x 丣 y C ⫽ xy FIGURE 4.5 Implementation of half adder T able 4.3 Half Adder x y C S 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 0 1 out of 145 Share Download Download More from:Digital Circuits and Systems(EE204) More from: Digital Circuits and SystemsEE204Delhi Technological University 3 documents Go to course 145 Morris Mano Digital Design-3 Digital Circuits and Systems Lecture notes None 130 Morris Mano Digital Design-4 Digital Circuits and Systems Lecture notes None More from: Digital Circuits and SystemsEE204Delhi Technological University3 documents Go to course 145 Morris Mano Digital Design-3 Digital Circuits and Systems None 130 Morris Mano Digital Design-4 Digital Circuits and Systems None More from:Rishabh Chandra AE46 More from: Rishabh Chandra AE46 impact 999+ impact 999+ Delhi Technological University Discover more 63 Anand Kumar Sn S - Complete notes on signals and systems Signal and Systems Lecture notes 100% (4) 60 SK Bhattacharya-1 - Complete notes on basic concepts Engineering Analysis and Design ECE Lecture notes 100% (3) 30 AE Lecture Notes-1 - Complete notes on diode circuits Analog Electronics Lecture notes 100% (3) 9 Regenerative Rankine Cycle Thermal Engineering Lecture notes 100% (3) More from: Rishabh Chandra AE46impact 999+ impact 999+ Delhi Technological University Discover more 63 Anand Kumar Sn S - Complete notes on signals and systems Signal and Systems 100% (4) 60 SK Bhattacharya-1 - Complete notes on basic concepts Engineering Analysis and Design ECE 100% (3) 30 AE Lecture Notes-1 - Complete notes on diode circuits Analog Electronics 100% (3) 9 Regenerative Rankine Cycle Thermal Engineering 100% (3) 17 Fibonacci and Golden Search Method Numerical and Engineering Optimization Methods 100% (3) 69 Fitzgerald Electric Machinery 6th ed-1 Asynchronous and Synchronous Machines 100% (3) Students also viewed Microwave Engineering Unit-3 Microwave Engineering Unit-5 Microwave Engineering Unit-IV Microwave Engineering Unit-2 AE Lecture Notes-3 - Complete notes on multistage and power amplifiers Setting up a fibrer optic analog Link Related documents Discrete Structures Important practise questions Algorithm analysis and design UNIT-4 UHV Handout 3-Harmony in the Family UHV Handout 5-Harmony in the Nature and Existence Unit-1 Introduction of Engineering Analysis and Design Microwave Engineering Unit-1 Get homework AI help with the Studocu App Open the App English India Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. 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9244	https://blog.tcea.org/coin-flip-tools-gone-digital-online-and-free/	Five Free Online Coin-Flipping Tools – TCEA TechNotes Blog Top Posts Why It May Be Time to Dump Bloom’s... An Online Wheel Spinner for Every Occasion CER Resources for the Science Classroom Human Bingo: A Fun and Effective Icebreaker Personality Test for Professional Development Easily Share Websites with Chrome’s QR Code Generator Create AI-Powered YouTube Quizzes in Seconds 14 Editable SEL Check-In Templates Five Interactive Tools for Learning the Periodic Table Email Writing for Students AI for Educators Conference All Events All Events AI for Educators Conference 2024-25 Lunch and Learn Webinars TCEA Convention SysAdmin Conference Courses Search HomeContent AreasMathFive Free Online Coin-Flipping Tools MathOnline Tools and Apps Five Free Online Coin-Flipping Tools by Miguel GuhlinNovember 17, 2021 written by Miguel GuhlinNovember 17, 2021 22.4K Have to make a choice between Mad Greens or ModPizza for Austin eats? Want to encourage friendly conflict resolution in your teaching and learning environment? Need your math students to hone their probability skills? Well, grab a virtual or digital coin flipper and keep charges of cheating, trick coins, and more at bay. Let’s explore several digital alternatives to the old “heads or tails” coin toss that you might find beneficial. Note: Here are a variety of text and shape wheel spinners if you need even more complex choice makers. 1. Google’s Flip a Coin This is a simple flip located on top of Google search results. The Steps: Visit or type “flip a coin” into any Google Search. Wait for the flip. Get the result. Click Flip Again to start the process over. Tip: Tap on the down arrow below the coin graphic to see more tools, like a Metronome, Roll a Die, Spinner, Calculator, Meditate, and Color Picker. 2. Just Flip a Coin This is a screen-sized flip that auto-starts upon arrival at the website. The Steps: Visit The coin will flip upon arrival. You will also see how many times the coin has been tossed. You can click Flip Again to instigate a new flip. Tap Facebook or Twitter buttons to share the coin button. Tip: Scroll down and change the coin’s color from grey to white, red, purple, blue, or green. 3. FlipSimu This is a simple heads or tails simulator with statistics tracking and random flipping. You can also long-press and then release to simulate flipping energy. The Steps: Visit Adjust the settings to customize text, images, quantity, colors, and sound. Choose which flip type you want. There are three: 1) just flip, 2) test your intuition, and 3) test your luck. Click the Flip It button or press and hold to simulate flipping energy. Tip: Customize the coins you flip. You can change the text and colors for heads and/or tails. You can see below how I customized FlipSimu to reflect my wife and I. What a great way to win an argument, eh? 4. Random.org’s Coin Flipper This is a virtual coin flipper that derives the randomness of the flip from atmospheric noise. Visit Select a coin type. Random.orgoffers novelty coins, antique coins, Austrian, Australian, and Brazilian coins, among others. Decide how many times you want to flip the coins (1-200 times). Click the Flip Coin(s) button. Tip: Want to see your own coins tossed? Send high quality coin pictures to Random.org. I’m considering sending them a Panamanian penny, known as a centesimo. 5. ESL Kids Games’ Classroom Coin Flip This is another flipper that can go full screen and features the U.S. 1998 quarter. Visit Enlarge to full screen. Click the Flip the Coin button. The interface is simple enough for anyone to use. An added bonus? It keeps track of how many times the coin came up heads or tails. This record can be handy in a disagreement or in a math lesson. Bonus! RandomWordGenerator’s Coin Flip has an extra option that provides background information on randomness. You can also read tidbits of information about coin flipping you may not know about. And, Zac in the Comments, mentions this easy to use Flip a Coin Online tool: I hope these digital tools assist you in finding your way through any situation. If you happen to have another resource, email me at mguhlin@tcea.org or drop it in the comments. Coincoin flipFlippersmathprobabilityrandom 5 comments 0FacebookTwitterPinterestLinkedinEmail Miguel Guhlin Transforming teaching, learning and leadership through the strategic application of technology has been Miguel Guhlin’s motto. Learn more about his work online at blog.tcea.org, mguhlin.org, and mglead.org/mglead2.org. Catch him on Mastodon @mguhlin@mastodon.education Areas of interest flow from his experiences as a district technology administrator, regional education specialist, and classroom educator in bilingual/ESL situations. Learn more about his credentials online at mguhlin.net. previous post ##### Resources for Building a Digital Citizenship Program next post ##### Eight Chrome Extensions and Sites You May Not Know (November 2021) You may also like ClassLink’s Assessment Insights: Make Data-Driven Decisions Without the... September 23, 2025 The Educator’s Guide to Creating Perfect AI-Generated Images... August 25, 2025 Harnessing Visual Learning with Free Tools July 8, 2025 Kiwix: Offline Learning for K12 June 26, 2025 Five Techie Things Teachers Should Try Over the... May 28, 2025 Tackle Word Problems in STRIDE April 10, 2025 App Vetting: It’s Time To Take Control of... April 9, 2025 Embedding Interactive Lessons with Desmos and PhET in... April 8, 2025 Padlet’s AI Tools Supercharge Digital Collaboration April 4, 2025 Got GRITS? A Coaching Strategy for Lesson Study... March 20, 2025 5 comments zak hollowayJune 16, 2023 - 7:52 am I am using and found it awesome. Reply Miguel GuhlinJune 16, 2023 - 8:01 am Zak, great addition! Thanks so much for making us aware of it. Miguel Guhlin Reply mostamb November 20, 2024 - 4:15 pm Miguel, this is a fantastic list of coin-flipping tools! I’d like to recommend another excellent option: . It’s a sleek and intuitive coin flip simulator that’s perfect for quick decisions or probability exercises. It even offers a smooth, ad-free experience, making it a great addition to this lineup. Check it out and let me know what you think! Reply Miguel GuhlinNovember 20, 2024 - 4:53 pm That is an amazing suggestion! Thank you. Reply Alax TigerDecember 2, 2024 - 1:03 am I highly recommend adding the coin-flipping tool from to your post. It’s a simple yet effective tool that provides a realistic coin-flipping experience. Perfect for decision-making, games, or just casual fun, it enhances engagement and is easy to integrate. Visit website to explore this fantastic tool! 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9245	https://www.comp.nus.edu.sg/~warut/ijcai24-tutorial.pdf	Tournaments in Computational Social Choice Warut Suksompong National University of Singapore 33rd International Joint Conference on Artiﬁcial Intelligence August 3, 2024 Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 1 / 57 Outline Social choice theory How to choose a socially desirable outcome from a set of alternatives. Origins in mathematics, economics, and political science. Tournaments model scenarios in which decisions are made based on pairwise comparisons. Sports, elections, webpage ranking, biological interactions, . . . This tutorial Part 1: Tournament solutions (methods for choosing winners from a given tournament) Part 2: Single-elimination tournaments (setting up the bracket to help a certain player, bribery issues) Based partially on my survey published at IJCAI 2021. For work before 2016, see Chapters 3 and 19 in the Handbook of Computational Social Choice [Brandt et al. ’16] W. Suksompong. “Tournaments in computational social choice: Recent developments”, IJCAI 2021 F. Brandt, V. Conitzer, U. Endriss, J. Lang, A. D. Procaccia. “Handbook of Computational Social Choice”, 2016 Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 2 / 57 Part 1: Tournament Solutions Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 3 / 57 Tournaments a b c d e 1: a ≻1 b ≻1 c ≻1 e ≻1 d 2: d ≻2 c ≻2 a ≻2 b ≻2 e 3: e ≻3 d ≻3 b ≻3 c ≻3 a Tournament T = (A, ≻), where A is the set of alternatives and ≻is the dominance relation. In this example, A = {a, b, c, d, e} and a ≻b, b ≻c, d ≻b, e ≻d, etc. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 4 / 57 Tournament Solutions A tournament solution is a method for choosing the “winners” of any tournament. Copeland set (CO): Alternatives with the highest outdegree. Top cycle (TC): Alternatives that can reach every other alternative via a directed path (which by deﬁnition has length ≤n −1). Uncovered set (UC): Alternatives that can reach every other alternative via a directed path of length ≤2. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 5 / 57 Example a b c d e f All omitted edges point from right to left. Outdegrees: a: 0, b: 2, c: 3, d: 3, e: 3, f : 4 CO = {f } TC = {b, c, d, e, f } UC = {c, d, e, f } No Condorcet winner (alternative that dominates all other alternatives), but a Condorcet loser (a) Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 6 / 57 Top Cycle Equivalent deﬁnition of the top cycle: (Unique) smallest nonempty set B of alternatives such that all alternatives in B dominate all alternatives outside B. T x y Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 7 / 57 Top Cycle Equivalent deﬁnition of TC: (Unique) smallest nonempty set B of alternatives such that all alternatives in B dominate all alternatives outside B. T B p q r Proof of equivalence: p ̸∈B cannot reach q ∈B, so p does not belong to TC. q ∈B can reach p ̸∈B directly. If q ∈B could not reach r ∈B, all alternatives that could reach r would form a smaller subset in the deﬁnition of TC, contradiction. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 8 / 57 Uncovered Set Covering relation: An alternative x covers another alternative y if x dominates y. For any z, if y dominates z, then x also dominates z. Strong indicator that x is better than y. Equivalent deﬁnition of UC: The set of all uncovered alternatives. Proof: x can reach y in ≤2 steps ⇐ ⇒y does not cover x. T x y z CO ⊆UC ⊆TC always holds. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 9 / 57 Axioms Condorcet-consistency: If there is a Condorcet winner x, then x is uniquely chosen. CO , TC , UC Monotonicity: If x is chosen, then it should remain chosen when it is strengthened against another alternative. CO , TC , UC Stability: A set is chosen from two diﬀerent sets of alternatives if and only if it is chosen from the union of these sets. CO , TC , UC Composition-consistency: It chooses the “best” alternatives from the “best” components. CO , TC , UC Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 10 / 57 Computation The input has size O(n2), where n = number of alternatives. Copeland set: Compute all outdegrees in time O(n2). Top cycle: Find the strongly connected components of the tournament, and output the unique one that dominates the rest. Can be done in time O(n2) by Tarjan’s or Kosaraju’s algorithm. Uncovered set: Use the “can reach everything else in ≤2 steps” deﬁnition. Multiply the adjacency matrix with itself to check reachability in 2 steps. Can be done in O(n2.37) using matrix multiplication. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 11 / 57 Tournament Solutions Tournament solution containment diagram [Brandt/Brill/Harrenstein ’16] Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 12 / 57 Tournament Solutions Banks set (BA): Alternatives that appear as the maximal element of some maximal transitive subtournament. Transitive tournament: The alternatives can be ordered as a1, . . . , ak so that ai dominates aj for all i < j. Slater set (SL): Alternatives that are maximal elements in some transitive tournament that can be obtained by inverting as few edges as possible. Bipartisan set (BP): Alternatives that are chosen with nonzero probability in the (unique) Nash equilibrium of the zero-sum game formed by the tournament matrix. Markov set (MC): Alternatives that stay most often in the “winner-stays” competition corresponding to the tournament. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 13 / 57 Separation Index Can two given tournament solutions return disjoint sets of alternatives? If so, what is the smallest tournament size for which this can happen? Table of separation indices [Brandt/Dau/Seedig ’15] Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 14 / 57 Separation Index Whether BA and BP always overlap was resolved recently . . . Brandt/Grundbacher showed that BA and BP are disjoint for this tournament of size 36. The separation index of BA and BP is between 11 and 36. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 15 / 57 Query Complexity Unlike computational complexity, the query complexity is never higher than Θ(n2). However, it is Θ(n2) for several tournament solutions [Maiti/Dey ’24] For CO, the algorithm may need to query all edges. Idea: Consider when all alternatives have the same outdegree. Proof for TC: Consider two sets A, B with 2k + 1 alternatives each. A B In each set, every alternative can reach every other alternative. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 16 / 57 Query Complexity Proof for TC: Consider two sets A, B with 2k + 1 alternatives each. A B If a query is within A or B, answer as in the ﬁgure. Else, answer that a ∈A dominates b ∈B. Claim: All edges between A and B must be queried. If all a ∈A dominate all b ∈B, then TC ⊆A. If at least one b ∈B dominates at least one a ∈A, then TC ̸⊆A. A similar idea works for UC. If TC is small, all of these tournament solutions can be computed with fewer queries. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 17 / 57 Query Complexity Bonus: What about deciding whether there is a Condorcet winner? The query complexity is exactly 2n −⌊log2 n⌋−2. [Balasubramanian/Raman/Srinivasaragavan ’97, Procaccia ’08] Algorithm: Stage 1: Let alternatives compete in a balanced single-elimination tournament. Suppose the winner is x. Stage 2: Let x compete against the alternatives that it has not competed against in Stage 1. Output Yes if x beats all of them. 1 2 3 4 5 6 7 8 2 3 6 8 3 8 3 Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 18 / 57 Random Tournaments Several tournament solutions, including TC and UC, tend to select all alternatives in large random tournaments. Consider the uniform random model, where each edge is oriented in either direction with probability 1/2, independently of other edges. Proof for UC (which implies one for TC): The probability that x cannot reach y in two steps ≤(3/4)n−2 x y z By union bound over all pairs of alternatives, the probability that there exists such a pair x, y is at most n2 · (3/4)n−2 →0 as n →∞ When no such pair exists, UC is the set of all alternatives. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 19 / 57 Condorcet Random Model Condorcet random model: There exists an underlying linear order of players. In general, a stronger player wins against a weaker player, but the weaker player upsets the stronger player with uniform probability p ≤1/2. The uniform random model corresponds to the case p = 1/2. Luczak/Ruci´ nski/Gruszka showed that the top cycle selects all alternatives with high probability when p = ω(1/n), and this is tight. The Condorcet random model is still rather unrealistic for two important reasons. In tournaments in the real world, the orientations of diﬀerent edges are typically determined by diﬀerent probabilities. Not all probabilities of the orientation of the edges necessarily respect the ordering: “bogey teams”. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 20 / 57 Generalized Random Model Generalized random model: The orientation of each edge is determined by probabilities within the range [p, 1 −p] for some parameter p, independently of other edges. These probabilities are allowed to vary across edges. Question: What is the least p such that the tournament solution selects all alternatives with high probability? For TC we need p ∈ω(1/n), while for UC we only need p ∈Ω( p log n/n) [Saile/S. ’20] k-kings: Alternatives that can reach every other alternative via a directed path of length ≤k. 2-kings (uncovered set) ⊆3-kings ⊆· · · ⊆(n −1)-kings (top cycle) Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 21 / 57 Generalized Random Model We determine how the probability threshold changes as we move from 2-kings to (n −1)-kings [Manurangsi/S. ’22] k-kings Threshold p k = 2 Ω( p log n/n) 3 ≤k ≤4 Ω(log n/n) k = 5 Ω(log log n/n) 6 ≤k ≤n −2 ω(1/n) k = n −1 ω(1/n) The case 6 ≤k ≤n −2 strengthens the previous result for k = n −1. All bounds are asymptotically tight, except for k = 5, where the gap is between Ω(log log n/n) and ω(1/n). ω(1/n) is tight for the Condorcet random model. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 22 / 57 Generalized Random Model We determine how the probability threshold changes as we move from 2-kings to (n −1)-kings [Manurangsi/S. ’22] k-kings Threshold p k = 2 Ω( p log n/n) 3 ≤k ≤4 Ω(log n/n) k = 5 Ω(log log n/n) 6 ≤k ≤n −2 ω(1/n) k = n −1 ω(1/n) The uncovered set is clearly more selective than k-kings for k ≥3. 3-kings and 4-kings are slightly more selective than higher-order kings. There is virtually no diﬀerence from k = 5 all the way to k = n −1. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 23 / 57 Margin of Victory How can we diﬀerentiate between the winning alternatives? Brill/Schmidt-Kraepelin/S. proposed using the margin of victory (MoV). Similar concepts have been applied in voting, sports modeling, political districting, etc. MoV(x) = minimum number of edges that need to be reversed so that x drops out of the winner set. Can also deﬁne a weighted version with weighted edges. The weights may represent the amount of bribe needed to change the match outcomes. The MoV of CO, TC, UC can be computed in polynomial time. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 24 / 57 Margin of Victory Theorem [Brill/Schmidt-Kraepelin/S. ’20] The MoV for CO, TC, UC can be as high as ⌊n/2⌋, but no higher. Upper bound: MoV(x) ≤⌊n/2⌋ Take y ̸= x with the highest outdegree. y has outdegree at least n−1 2 Can make y a Condorcet winner using (n −1) − n−1 2 = ⌊n/2⌋ reversals. Lower bound: Possible that MoV(x) ≥⌊n/2⌋ Since CO ⊆UC ⊆TC, suﬃces to prove for CO. Suppose x is a Condorcet winner (outdegree n −1), while the maximum outdegree of y ̸= x is n−1 2 Each reversal decreases outdeg(x) −outdeg(y) by ≤1 (except the reversal between x and y, which decreases the diﬀerence by 2) Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 25 / 57 Margin of Victory Brill et al. conducted an axiomatic analysis of the MoV. Cover-consistency: x covers y ⇒MoV(x) ≥MoV(y) CO , TC , UC MoV is typically aligned with the covering relation. Degree-consistency: outdeg(x) > outdeg(y) ⇒MoV(x) ≥MoV(y) CO , TC , UC Strong deg.-cons.: outdeg(x) ≥outdeg(y) ⇒MoV(x) ≥MoV(y) CO , TC , UC MoV often provides information beyond simply the outdegrees. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 26 / 57 Margin of Victory Does MoV really distinguish among winners? Brill et al. ran experiments to answer this question. On average, the number of alternatives with maximum MoV is a small fraction of the winners. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 27 / 57 Randomized Tournament Solutions Randomized tournament solution: Returns a probability distribution over the alternatives Condorcet-consistency: A Condorcet winner should receive probability 1 k-strongly-non-manipulable-α: No group of size k can increase their combined probability by more than α Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 28 / 57 Randomized Tournament Solutions Observation: No Condorcet-consistent randomized tournament solution can be 2-SNM-α for any α < 1/3. a b c For any randomized tournament solution, some pair of players receive a combined probability of at most 2/3. This pair of players can reverse their match outcome and increase their probability to 1. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 29 / 57 Randomized Tournament Solutions Theorem [Schneider/Schvartzman/Weinberg ’17] A uniformly random SE bracket is 2-SNM-1/3. Coupling argument: For a bracket where a pair of players could gain by manipulating, tie it with two other brackets with no manipulation potential for this pair Many other rules are 2-SNM-1/2 or worse! Randomized King-of-the-Hill: If there is a Condorcet winner, declare it as the winner. Else, select a player uniformly at random, and remove it along with all players that it beats. Repeat the previous step. Theorem [Schvartzman/Weinberg/Zlatin/Zuo ’20] Randomized King-of-the-Hill is 2-SNM-1/3 and cover-consistent. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 30 / 57 Randomized Tournament Solutions Schvartzman/Weinberg/Zlatin/Zuo (2020): Assume Condorcet-consistency There exists a rule that is k-SNM-2/3 for all k No rule can be k-SNM-1/2 for large enough k Ding/Weinberg (2021): Outcomes of matches are randomized Randomized Death Match: Pick two uniformly random players, eliminate the loser, and repeat This rule and Random SE Bracket perform optimally for 2-SNM Dinev/Weinberg (2022): RDM is 3-SNM-31/60, and this is tight for this rule Dale/Fielding/Ramakrishnan/Sathyanarayanan/Weinberg (2022): Multiple prizes according to full ranking Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 31 / 57 Part 2: Single-Elimination Tournaments Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 32 / 57 Single-Elimination Tournaments An alternative is said to be a single-elimination winner if it wins a balanced single-elimination tournament under some bracket. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 33 / 57 Single-Elimination Tournaments a b c d e f g h b c f h c h c Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 34 / 57 Tournament Fixing Problem The winner of a given SE tournament can depend signiﬁcantly on the initial bracket! The Tournament Fixing Problem (TFP): Given A set of players Information for each pair of players (x, y) about whether x or y would win in a head-to-head matchup (“tournament graph”) A player of interest v Does there exist a bracket such that v wins the tournament? Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 35 / 57 Complexity Theorem [Aziz/Gaspers/Mackenzie/Mattei/Stursberg/Walsh ’18] TFP is NP-complete. Kim/Vassilevska Williams : The problem remains NP-complete even when the player of interest v is a king that beats n/4 other players in the tournament graph. A king is a player who can reach any other player via at most 2 edges in the tournament graph. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 36 / 57 Algorithms Theorem [Kim/Vassilevska Williams ’15] TFP can be solved in time O(2npoly(n)). In fact, the algorithm can also count the number of brackets under which v wins the tournament. Idea: Consider all possible ways of partitioning the set of players S into two subsets T and S \ T of equal size such that v ∈T. Iterate over all players w ∈S \ T beaten by v. Compute the number of winning brackets of v in T and w in S \ T. Use a fast subset convolution subroutine of Bj¨ orklund/Husfeldt/ Kaski/Koivisto . Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 37 / 57 Algorithms Players can be partitioned into a constant number of types: Polynomial-time solvable [Aziz et al. ’18] Let k be the size of a smallest feedback arc set (a set of edges whose removal leaves the tournament acyclic) Aziz et al. : nO(k) via dynamic programming Ramanujan/Szeider : Fixed-parameter tractable (FPT) algorithm running in time 2O(k2 log k)nO(1) Translate TFP into an algebraic system of equations and feeding it into an integer linear programming (ILP) solver Gupta et al. : 2O(k log k)nO(1) via combinatorial algorithm Zehavi : Same running time for feedback vertex set Open direction: Other parameters, e.g., directed treewidth? Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 38 / 57 Tournament Value Maximization What if the organizers want to maximize the proﬁt/popularity of the tournament? Tournament Value Maximization problem: Given A set of players Information for each pair of players (x, y) about whether x or y would win in a head-to-head matchup (“tournament graph”), and their value if they meet in a certain round Find a bracket that maximizes the sum of values across all matches. The values are round-oblivious if the value of every pair is independent of the round in which they meet. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 39 / 57 Tournament Value Maximization This problem was studied by Chaudhary/Molter/Zehavi The problem is NP-hard (and APX-hard) when All values are in {0, 1} There are 3 distinct values and the values are round-oblivious 1/ log n approximation based on maximum-weight matching If the total value of a tournament can be determined by the number of wins of each player, there exists an nO(log n) algorithm. If players can be classiﬁed as “popular” or “unpopular”, and the value of each match equals the popularity of the winning player, there is a linear-time greedy algorithm. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 40 / 57 Structural Results Let v be a king. Suppose v beats A and loses to B. j + / ? S S w QQ Q s + ? QQ Q s ? QQ Q s PPPPP P q? + ) v A B Theorem (King who beats half the players) [Vassilevska Williams ’10] If \|A\| ≥n/2, then v is a SE winner. Theorem (Superking) [Vassilevska Williams ’10] If every player in B loses to at least log2 n players from A, then v is a SE winner.p Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 41 / 57 Structural Results Let v be a king. Suppose v beats A and loses to B. j + / ? S S w QQ Q s + ? QQ Q s ? QQ Q s PPPPP P q? + ) v A B H I J Theorem [Kim/S./Vassilevska Williams ’17] Suppose that B is a disjoint union of three sets H, I, J such that 1 \|H\| < \|A\| 2 Each player in I loses to at least log2 n players in A. 3 outdeg(j) ≤\|A\| for all j ∈J. Then v is a SE winner.p Superking: H = J = ∅, King who beats n/2: I = J = ∅ Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 42 / 57 Structural Results Let v be a king. Suppose v beats A and loses to B. j + / ? S S w QQ Q s + ? QQ Q s ? QQ Q s PPPPP P q? + ) v A B H I J Theorem [Kim/S./Vassilevska Williams ’17] Suppose that B is a disjoint union of three sets H, I, J such that 1 \|H\| < \|A\| 2 Each player in I loses to at least log2 n players in A. 3 outdeg(j) ≤\|A\| for all j ∈J. Then v is a SE winner.p Idea: Match players to maintain all invariants, and apply induction Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 42 / 57 Structural Results 3-king: A player who can reach any other player via at most 3 edges in the tournament graph. abcde j + / ? S S w QQ Q s + ? QQ Q s + ? QQ Q s ? QQ Q s PPPPP P q? + ) ? QQ Q s PPPPP P q? + ) v A B C A 3-king might not be a SE winner even if it beats n −3 players! [Kim/Vassilevska Williams ’15] Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 43 / 57 Structural Results Suﬃcient conditions for a 3-king to win a SE tournament. Kim/Vassilevska Williams : 1 \|A\| ≥n/3 2 Each b ∈B beats no more players than v does 3 There is a perfect matching from B onto C (in particular, \|B\| ≥\|C\|) Kim/S./Vassilevska Williams : 1 \|A\| ≥n/2 2 Every a ∈A beats every b ∈B 3 \|B\| ≥\|C\| Any two of these three conditions are insuﬃcient. Open direction: To what extent can we weaken these conditions? Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 44 / 57 Bribery Bribery-TFP (BTFP): The organizers are allowed to bribe up to b players to lose a match they would otherwise win. If b = 0, BTFP reduces to TFP, which is NP-hard. If b = log2 n, the tournament can be trivially rigged. Theorem [Kim/Vassilevska Williams ’15] For any constant ε > 0, BTFP is NP-hard when b ≤(1 −ε) log2 n. Gupta/Saurabh/Sridharan/Zehavi : Algorithm running in time 2O(k2 log k)nO(1), where k = size of a smallest feedback arc set Obfuscation operations which can take in one bribery solution and output another solution in polynomial time Russell/Walsh , Mattei/Goldsmith/Klapper/Mundhenk : Bracket given in advance, but bribery is allowed Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 45 / 57 Probabilistic Approaches Players have varying strengths, so not all tournament graphs are equally likely to occur. Condorcet random model: There exists an underlying linear order of players. In general, a stronger player beats a weaker player, but the weaker player upsets the stronger player with probability p ≤1/2. x1 x2 x3 x4 0.8 0.8 0.8 0.8 0.8 0.8 Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 46 / 57 Probabilistic Approaches Observation: If p ∈o(log n/n), the weakest player is expected to win o(log n) matches, which is insuﬃcient to be a SE winner. If p ∈Ω( p log n/n), with high probability, every player can win under some bracket [Vassilevska Williams ’10] In fact, p ∈Θ(log n/n) is the sharp threshold! [Kim/S./Vassilevska Williams ’17] With bribery, for any p, it suﬃces to bribe the top O(log n) players in the linear ordering to make any player win [Konicki/Vassilevska Williams ’19] Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 47 / 57 Probabilistic Approaches Generalized random model: For each pair i, j, player i beats player j with probability pi,j, independently of other pairs No linear ordering of players! x1 x2 x3 x4 0.6 0.7 0.8 0.2 0.5 0.35 Theorem [Manurangsi/S. ’22] If pi,j ∈Ω(log n/n) for all i, j, then with high probability, every player can win a SE tournament under some bracket. p Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 48 / 57 Probabilistic Approaches Probabilistic TFP (PTFP): Player i beats player j with probability qi,j. The bracket must be chosen before this uncertainty is resolved. Theorem [Chatterjee/Ibsen-Jensen/Tkadlec ’16] There exists a deterministic tournament graph such that: For one winning bracket of a player, the winning probability can drop by Θ(εn) through ε-perturbations. For another winning bracket of this player, the drop is only Θ(ε log n). The robustness can vary signiﬁcantly across brackets! Open question: Suppose that the probability matrix is monotonic, i.e., qi,j ≥qi,j−1 for all i ≤j −2. What is the complexity of PTFP? If bribery is allowed, it is NP-hard [Konicki/Vassilevska Williams ’19] Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 49 / 57 Seeded Setting Results on TFP so far assume that any bracket can be chosen. Many real-world tournaments assign seeds to players to prevent top players from meeting too early. For example, in ATP tennis tournaments with 32 players: 8 players are assigned seeds 1, 2, . . . , 8. The top 2 seeds cannot meet until the ﬁnal. The top 4 seeds cannot meet until the semiﬁnals. The top 8 seeds cannot meet until the quarterﬁnals. Let n = number of players, s = number of seeds Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 50 / 57 Seeded Setting Theorem [Manurangsi/S. ’23] For any n ≥4 and any s, a king who beats n −2 players may not be able to win a SE tournament.q Proof: Let x be our desired winner, and y and z be the top two seeds. A x y z Only y can beat z, so z makes the ﬁnal in every bracket. Even if x makes the ﬁnal, x will lose to z. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 51 / 57 Seeded Setting Superking: A king x such that for any y who beats x, there exist at least log2 n players who lose to x but beat y. Superkings can win a knockout tournament [Vassilevska Williams ’10] This remains true if s = 2, but not if s ≥4 [Manurangsi/S. ’23] Ultraking: A king x such that for any y who beats x, there exist at least n/2 players who lose to x but beat y. Theorem [Manurangsi/S. ’23] For any s, an ultraking can win a knockout tournament. This result would no longer hold if we replace n/2 by n/2 −1 in the ultraking deﬁnition. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 52 / 57 SE winners & Tournament Solutions The set of SE winners can also be viewed as a tournament solution. Theorem [Kim/S./Vassilevska Williams ’17] Every player in the Copeland set is a SE winner. Proof: Since CO ⊆UC, any player in CO is a king who beats at least n/2 other players. Theorem [Kim/S./Vassilevska Williams ’17] For any 0 < r < 1, there exists a tournament such that 1 the proportion of kings who are SE winners is less than r, and 2 the proportion of SE winners who are kings is also less than r. Kings and SE winners are largely disjoint! Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 53 / 57 SE winners & Tournament Solutions Theorem [Kim/S./Vassilevska Williams ’17] For any 0 < r < 1, there exists a tournament such that 1 the proportion of kings who are SE winners is less than r, and 2 the proportion of SE winners who are kings is also less than r. A B x y The set of kings is A ∪{x, y}. If \|B\| ≫\|A\| and the players in B are of roughly equal strength, then all players in B are SE winners while none of the players in A is. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 54 / 57 SE winners & Tournament Solutions A SE winner may have a low Copeland score (= outdegree). Theorem [Hulett ’19] There exists a tournament graph such that the SE winner according to a uniformly random bracket has Copeland score n · 2−Θ(√log n). n · 2−Θ(√log n) is lower than, say, Θ(n/ log n). Choosing a uniformly random alternative already yields Copeland score (n −1)/2. As an indicator of strength, the ability to win a SE tournament does not necessarily align with the Copeland score. Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 55 / 57 Future Directions Study other tournament formats Double-elimination [Stanton/Vassilevska Williams ’13, Aziz et al. ’18] Round-robin Stepladder/challenge-the-champ [Mattei/Goldsmith/Klapper/ Mundhenk ’15, Yang/Dimitrov ’21, Chaudhary/Molter/Zehavi ’24] Swiss-system [F¨ uhrlich/Cseh/Lenzner ’24] Multi-stage tournaments Promotion/relegation features Perform empirical studies on real-world tournaments, e.g., using data from sports competitions [Russell/van Beek ’11, Mattei/Walsh ’16] Examine the eﬀects of the tournament structure on fairness [Ryvkin/Ortmann ’08, S. ’16, Arlegi/Dimitrov ’20] Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 56 / 57 Let’s make tournaments great again! Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 57 / 57 References Surveys F. Brandt, M. Brill, P. Harrenstein. “Tournament solutions”, Handbook of Computational Social Choice (2016), chapter 3. V. Vassilevska Williams. “Knockout tournaments”, Handbook of Computational Social Choice (2016), chapter 19. W. Suksompong. “Tournaments in computational social choice: Recent developments”, IJCAI (2021) Research papers R. Arlegi, D. Dimitrov. “Fair elimination-type competitions”, European Journal of Operational Research (2020) H. Aziz, S. Gaspers, S. Mackenzie, N. Mattei, P. Stursberg, T. Walsh. “Fixing balanced knockout and double elimination tournaments”, Artiﬁcial Intelligence (2018) R. Balasubramanian, V. Raman, G. Srinivasaragavan. “Finding scores in tournaments”, Journal of Algorithms (1997) A. Bj¨ orklund, T. Husfeldt, P. Kaski, M. Koivisto. “Fourier meets m¨ obius: Fast subset convolution”, STOC (2007) F. Brandt, A. Dau, H. G. Seedig. “Bounds on the disparity and separation of tournament solutions”, Discrete Applied Mathematics (2015) F. Brandt, F. Grundbacher. “The Banks set and the bipartisan set may be disjoint”, arXiv (2023) M. Brill, U. Schmidt-Kraepelin, W. Suksompong. “Reﬁning tournament solutions via margin of victory”, AAAI (2020) M. Brill, U. Schmidt-Kraepelin, W. Suksompong. “Margin of victory in tournaments: Structural and experimental results”, AAAI (2021) K. Chatterjee, R. Ibsen-Jensen, J. Tkadlec. “Robust draws in balanced knockout tournaments”, IJCAI (2016) J. Chaudhary, H. Molter, M. Zehavi. “How to make knockout tournaments more popular?”, AAAI (2024) J. Chaudhary, H. Molter, M. Zehavi. “Parameterized analysis of bribery in challenge the champ tournaments”, arXiv (2024) Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 1 / 3 References Research papers (cont.) E. Dale, J. Fielding, H. Ramakrishnan, S. Sathyanarayanan, S. M. Weinberg. “Approximately strategyproof tournament rules with multiple prizes”, EC (2022) A. Dinev, S. M. Weinberg. “Tight bounds on 3-team manipulations in randomized death match”, WINE (2022) K. Ding, S. M. Weinberg. “Approximately strategyproof tournament rules in the probabilistic setting”, ITCS (2021) P. F¨ uhrlich, A. Cseh, P. Lenzner. “Improving ranking quality and fairness in Swiss-system chess tournaments”, Journal of Quantitative Analysis in Sports (2024) S. Gupta, S. Roy, S. Saurabh, M. Zehavi. “When rigging a tournament, let greediness blind you”, IJCAI (2018) S. Gupta, S. Roy, S. Saurabh, M. Zehavi. “Winning a tournament by any means necessary”, IJCAI (2018) S. Gupta, S. Saurabh, R. Sridharan, M. Zehavi. “On succinct encodings for the tournament ﬁxing problem”, IJCAI (2019) R. Hulett. “Single-elimination brackets fail to approximate Copeland winner”, APPROX (2019) M. P. Kim, W. Suksompong, V. Vassilevska Williams. “Who can win a single-elimination tournament?”, SIAM Journal on Discrete Mathematics (2017) M. P. Kim, V. Vassilevska Williams. “Fixing tournaments for kings, chokers, and more”, IJCAI (2015) C. Konicki, V. Vassilevska Williams. “Bribery in balanced knockout tournaments”, AAMAS (2019) T. Luczak, A. Ruci´ nski, J. Gruszka. “On the evolution of a random tournament”, Discrete Mathematics (1996) A. Maiti, P. Dey. “Query complexity of tournament solutions”, Theoretical Computer Science (2024) P. Manurangsi, W. Suksompong. “Generalized kings and single-elimination winners in random tournaments”, Autonomous Agents and Multi-Agent Systems (2022) P. Manurangsi, W. Suksompong. “Fixing knockout tournaments with seeds”, Discrete Applied Mathematics (2023) Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 2 / 3 References Research papers (cont.) N. Mattei, J. Goldsmith, A. Klapper, M. Mundhenk. “On the complexity of bribery and manipulation in tournaments with uncertain information”, Journal of Applied Logic (2015) N. Mattei, T. Walsh. “Empirical evaluation of real world tournaments”, arXiv (2016) A. D. Procaccia. “A note on the query complexity of the Condorcet winner problem”, Information Processing Letters (2008) M. S. Ramanujan, S. Szeider. “Rigging nearly acyclic tournaments is ﬁxed-parameter tractable”, AAAI (2017) T. Russell, P. van Beek. “An empirical study of seeding manipulations and their prevention”, IJCAI (2011) T. Russell, T. Walsh. “Manipulating tournaments in cup and round robin competitions”, ADT (2009) D. Ryvkin, A. Ortmann. “The predictive power of three prominent tournament formats”, Management Science (2008) C. Saile, W. Suksompong. “Robust bounds on choosing from large tournaments”, Social Choice and Welfare (2020) J. Schneider, A. Schvartzman, S. M. Weinberg. “Condorcet-consistent and approximately strategyproof tournament rules”, ITCS (2017) A. Schvartzman, S. M. Weinberg, E. Zlatin, A. Zuo. “Approximately strategyproof tournament rules: on large manipulating sets and cover-consistence”, ITCS (2020) I. Stanton, V. Vassilevska Williams. “The structure, eﬃcacy, and manipulation of double-elimination tournaments”, Journal of Quantitative Analysis in Sports (2013) W. Suksompong. “Scheduling asynchronous round-robin tournaments”, Operations Research Letters (2016) V. Vassilevska Williams. “Fixing a tournament”, AAAI (2010) Y. Yang, D. Dimitrov. “Weak transitivity and agenda control for extended stepladder tournaments”, Economic Theory Bulletin (2021) M. Zehavi. “Tournament ﬁxing parameterized by feedback vertex set number is FPT”, AAAI (2023) Warut Suksompong (NUS) Tournaments in COMSOC August 3, 2024 3 / 3
9246	https://antonhaugen.medium.com/r-squared-or-the-coefficient-of-determination-64c0d3c2241a	R-Squared, or the Coefficient of Determination \| by Anton Haugen \| Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in R-Squared, or the Coefficient of Determination Anton Haugen Follow 3 min read · Jan 19, 2021 25 Listen Share From: To prepare for some of my job interviews, I’ve been reviewing essential concepts from my Data Science Immersive course at the Flatiron School. Lately, I’ve been reviewing some ideas essential to early on in the program when we focused on Statistical Theory, Probability, and Linear Regression. Linear Regression models are not just used for prediction, but can help you understand the relationship between different attributes of your data. One of my instructors at the Flatiron School would build a OLS model as part of his exploratory data analysis to understand how much of his dependent variable could be explained from the data at hand. One concept essential to understanding this relation is R², or the Coefficient of Determination. R² is a way of understanding a model’s goodness fit, or how much of the variance in the data of the dependent variable is explained through the independent variable. To save you a google search, the dependent variable is also known as the target or what you’re trying to predict based on the other parts of your data. For example, if you were trying to predict annual salary based on age, annual salary would be your dependent variable while age would be your independent variable. Get Anton Haugen’s stories in your inbox Join Medium for free to get updates from this writer. Subscribe Subscribe R-Squared compares the performance of your linear regression model based on the performance of a baseline model. The baseline model is the mean of the observed values irrespective of the values of X. So if the mean value of annual salary was 50,000, this baseline model, or mean model, would only give 50,000 whether a person was 16 or 1600. Press enter or click to view image in full size Calculating R-Squared involves finding the quotient from the sum of squared errors from your linear regression model (SSres) and the sum of squared error from this simple baseline model (SStot) and subtracting this value from 1. The closer this value is to 1, i.e. the more similar these two sums of squares are to each other, the lower your R-squared value will be and thus the less amount of variance in the relationship of these two variables your linear regression model is able to describe. A way to express an R-Squared of 0.85 would be “85% of the variance in my data is accounted for in this linear regression.” Though one might initially want a high R-Squared value, one should be wary of R-Squared values higher than 0.9, and frightened of R-squared values of 1. Often this means there was some kind of data leakage, or your dependent variable somehow snuck into your independent variables. During a project predicting the length of time a dog would stay in the Austin Animal Shelter, one of our models had a very good R-Squared value, only to find that this model had the dog’s age upon leaving the shelter as one of its features. Fortunately, we caught this immediately and continued to engineer other features. Depending your project’s purpose, there are situations in which a high R-squared might not even be relevant. For example, in a study on the relationship of religiosity and health, one would expect a low R-Squared because there are clearly other, but an R-Squared of 0.10 to 0.15 can let you know that there is some kind of relationship. This blog post goes into more detail: You might be wondering, as I did, how similar RMSE is to R-Squared. R-Squared is a relative measure of fit according to a baseline model while RMSE is the square root of the variance of the residuals from the observed data and predicted values in your test set, given in the same units as the dependent variable. While R-Squared is a measure of the amount of variance your linear regression model is able to explain, RMSE accounts for the errors in the prediction from the test set and is considered the most important metric if your aim is a predictive model. That’s all for this week! I’ll be returning to more PySpark information next week. 25 25 Follow Written by Anton Haugen ----------------------- 34 followers ·18 following Data Scientist and Writer Follow No responses yet Write a response What are your thoughts? Cancel Respond More from Anton Haugen Anton Haugen Feature Selection with PySpark ------------------------------ ### This past week I was doing my first machine learning project for Layla AI’s PySpark for Data Scientist course on Udemy. 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9247	https://dictionary.cambridge.org/us/example/english/allure	Examples of allure Word of the Day Victoria sponge Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio a soft cake made with eggs, sugar, flour, and a type of fat such as butter. It is made in two layers with jam or cream, or both, between them Blog Calm and collected (The language of staying calm in a crisis) New Words vibe coding © Cambridge University Press & Assessment 2025 © Cambridge University Press & Assessment 2025 Learn more with +Plus Learn more with +Plus {{message}} {{message}} There was a problem sending your report. {{message}} {{message}} There was a problem sending your report.
9248	https://math.stackexchange.com/questions/3649534/expression-for-the-bound-on-the-error-of-a-poisson-approximation	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. current community your communities more stack exchange communities Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Expression for the bound on the error of a Poisson approximation? I'm hoping that somebody can help clarify a question I had regarding the Poisson approximation and its applications. My textbook presents the following theorem, which I'm having trouble making sense of. My assumption is that the first capital $N$ is meant to be a lowercase $n$, and that all subsequent $N$'s represent a discrete random variable with a binomial distribution: Theorem 2.5. Consider independent events $A_i$, $i=1,2,...,n$, with probabilities $p_i=P(A_i)$. Let $N$ be the number of events that occur, let $\lambda=p_1+···+p_n$, and let $Z$ have a Poisson distribution with parameter $\lambda$. Then, for any set of integers $B$,$$\left\vert P(N\in B)-P(Z\in B)\right\vert\leq\sum_{i=1}^n p_i^2\tag{2.14}$$ We can simplify the right-hand side by noting $$\sum_{i=1}^n p_i^2\leq \max_i p_i \sum_{i=1}^n p_i=\lambda\max_ip_i$$ This says that if all the $p_i$ are small then the distribution of $N$ is close to a Poisson with parameter $\lambda$. Taking $B={k}$, we see that the individual probabilities $P(N=k)$ are close to $P(Z=k)$, but this result says more. The probabilities of events suchas $P(3\leq N\leq 8)$ are close to $P(3\leq N\leq 8)$ and we have an explicit bound on the error. The text then refers back to an example comparing the exact probability of obtaining exactly one double $6$ in twelve rolls of a pair of dice to the corresponding Poisson approximation: Suppose we roll two dice $12$ times and we let $D$ be the number of times a double $6$ appears. Here, $n=12$ and $p=1/36$, so $np=1/3$. We now compare $P(D=k)$ with the Poisson approximation for $k=1$. $$k=1 \text{ exact answer:}\,\,\,\,\,\,P(D=1)=\left(1-\frac{1}{36}\right)^{12} =0.7132$$ $$\text{Poisson approximation:}\,\,\,\,\,\,P(D=1)=e^{-1/3}=0.7165$$ For a concrete situation, consider [the example above], where $n=12$ and all the $p_i=1/36$. In this case the error bound is $$\sum_{i=1}^{12} p_i^2 =12\left(\frac{1}{36}\right) ^2=\frac{1}{108}=0.00926$$ while the error for the approximation for $k=1$ is $0.0057$. From this context, my thought is that the error bound they allude to is specifically that for the Poisson approximation to the binomial distribution, and not the Poisson approximation to some other type of distribution. Can somebody with a more complete understanding of the Poisson distribution (and its relationship to the binomial) confirm or refute this assertion? I'd also be curious to know where the proof of this theorem comes from, as my text doesn't offer any obvious reference. Citation: Elementary Probability for Applications, Rick Durrett. 0 You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Explore related questions See similar questions with these tags. Related Hot Network Questions Subscribe to RSS To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Mathematics Company Stack Exchange Network Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.9.29.34589
9249	https://www.youtube.com/watch?v=eTiOQBkbsE4	UNIFORM Probability Distribution for Discrete Random Variables (9-5) Research By Design 153000 subscribers 13 likes Description 555 views Posted: 15 Dec 2022 Uniform Probability Distribution: (i.e., a rectangular distribution) is a probability distribution involving one random variable with a constant probability. Each potential outcome is equally likely, such as flipping coin and getting heads is always 50/50. On Chaos Night, Dante experiments with randomly assigning patrons to one of the nine circles of comedy. Louis Cipher arrives one evening hoping to have three friends for dinner. What is the probability that Louie will be placed in the ninth circle? 2 comments Transcript: [Music] what is the probability of rolling a six on a fair six-sided die the answer is one in six what is the probability of rolling a three or a two or a one or a five every side of the die has an equal probability of coming up this is what is called a uniform probability distribution in a uniform distribution the variable has a constant or equal probability for every outcome it's also called a rectangular distribution and here is the formula that we would use for calculating this uniform probability distribution it's pretty straightforward each frequency is 1 over n in which n is the number of values that the random variable may assume the values the random variable are equally likely if we roll a single die a hundred thousand times we would get a probability distribution that would look pretty close to uniform however we know that theoretical probability distributions differ from experimental probability distributions the experimental role of the die would not be exactly like the theoretical distribution but it would be quite close here's what the theoretical distribution would look like every option has exactly the same probability rolling one or three or five or two has a probability of 0.166 which has been calculated as 1 over n where n equals six for a six sided die to compute the average we would find the score in the middle by adding up all potential outcomes one through six and dividing by 6. 21 divided by 6 is 3.5 making the average of all of the numbers rolled 3.5 this same logic would hold true if we're using more exotic dice like this four-sided die the probability of any number coming up is one in four the expected value the average of all of the rolls would be two for a ten sided die every option every potential outcome has a one in ten chance of occurring and the expected value would be five on a 12 sided die the same logic would would apply every outcome has a 1 in 12 chance of occurring and the expected value would be a six for our business of the week Dante decides to shake things up down at the Divine Comedy Club and create chaos night in which you are randomly assigned where you will sit in the theater on chaos night Dante experiments with randomly assigning patrons to one of the Nine Circles of Comedy Louis Cipher arrives one evening hoping to have three friends for dinner what is the probability that Louis will be placed in the Ninth Circle calculating that probability involves knowing how many total circles there are there are nine so the probability of being in any one of those circles is 1 over 9 or 0.11 and as I've done previously I have created an Excel spreadsheet for how to calculate expected values and variance for this uniform probability distribution we're going to take a look at that in our next video foreign [Music]
9250	https://www.youtube.com/watch?v=CZ5ne_mX5_I	Multiplying & dividing powers (integer exponents) \| Mathematics I \| High School Math \| Khan Academy Khan Academy 9090000 subscribers 1091 likes Description 409590 views Posted: 29 Mar 2017 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: These are worked examples for using these properties with integer exponents. Watch the next lesson: Missed the previous lesson? High School Math on Khan Academy: Did you realize that the word "algebra" comes from Arabic (just like "algorithm" and "al jazeera" and "Aladdin")? And what is so great about algebra anyway? This tutorial doesn't explore algebra so much as it introduces the history and ideas that underpin it. About Khan Academy: Khan Academy is a nonprofit with a mission to provide a free, world-class education for anyone, anywhere. We believe learners of all ages should have unlimited access to free educational content they can master at their own pace. We use intelligent software, deep data analytics and intuitive user interfaces to help students and teachers around the world. Our resources cover preschool through early college education, including math, biology, chemistry, physics, economics, finance, history, grammar and more. We offer free personalized SAT test prep in partnership with the test developer, the College Board. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide every year. For more information, visit www.khanacademy.org, join us on Facebook or follow us on Twitter at @khanacademy. And remember, you can learn anything. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academyâ s High School Math channel: Subscribe to Khan Academy: 60 comments Transcript: let's get some practice with our exponent properties especially when we have integer exponents so let's think about what 4 to the -3 4 to the 5th power is going to be equal to and I encourage you to pause the video and think about it on your own well there's a couple of ways to do this one you say look I'm multiplying two things that have the same base so this is going to be that base four and then I add the exponents 4 to the -3 + 5 power which is equal to 4 to the 2 power and that's just a straightforward exponent property but you can also think about why does that actually make sense 4 the -3 power that is 1 over 4 the 3 power or you could view that is 1 over 4 4 4 and then 4 5th that's 54s being multiplied together so it's times 4 4 4 4 4 and so notice when you multiply this out you're going to have five fours in the numerator and three fours in the denominator and so three of these in the denominator going to cancel out with three of these in the numerator and so you're going to be left with 5 - 3 or -3 + 5 fours so this 4 4 is the same thing as 4 squared now let's do one with variables so let's say that you have a to the -4 power time a to the let's say a sar what is that going to be well once again you have the same base in this case it's a and so and since I'm multiplying them you can just add the exponents so it's going to be a to the -4 + 2 power which is equal to a to the -2 power and once again it should make sense this right over here here that is 1 over a a a a and then this is time a a so that cancels with that that cancels with that and you're still left with 1 over a a which is the same thing as a to the -2 power now let's do it with some quotients so what if I were to ask you what is what is 12 to the -7 / 12 to -5 power well when you're dividing you subtract exponents if you have the same base so this is going to be equal to 12 to the -7 minus -5 power you're subtracting the bottom exponent and so this is going to be equal to 12 to the well that a Nega subtracting a negative is the same thing as adding the positive a 12 to the -2 power and once again we just have to think about why does this actually make sense well you can actually rewrite this 12-7 / 12 -5 that's the same thing as 12 -7 time 12 to the 5th power if we take the reciprocal of if we take the reciprocal of this right over here you would make the exponent positive and then you get exactly what we were doing in those previous examples with products and so let's just do one more with variables for good measure let's say I have x to the 20th power ided x to the 5th power well once again we have the same base and we're taking a quotient so this is going to be x to the -20 minus 5 because we have this one right over here in the denominator so this is going to be equal to x to the - 25th power and once again you could view our original expression as X to - 20th and having an x to the 5th in the denominator dividing by X the 5th is the same thing as multiplying by X to the5 and so here you just add the exponents and once again you would get x to the - 25th power
9251	https://www.medcomhk.com/hkdvb/pdf/2003v11n03-03.pdf	H.K. Dermatol. Venereol. Bull. (2003) 11, 130-138 Review Article Cutaneous tuberculosis: clinical features, diagnosis and management SCK Ho Social Hygiene Service, Department of Health, Hong Kong SCK Ho, FHKCP , FHKAM(Medicine) Correspondence to: Dr. SCK Ho Yaumatei Dermatology Clinic, 12/F , Yaumatei Specialist Clinic, 143 Battery Street, Kowloon, Hong Kong The clinicopathological manifestations of cutaneous tuberculosis are protean. Compared to pulmonary tuberculosis, cutaneous tuberculosis is uncommon and more difficult to diagnose. The use of polymerase chain reaction (PCR) has enhanced detection of cutaneous tuberculosis. The clinicopathological features, diagnosis, and treatment of cutaneous tuberculosis are reviewed. Exclusion of concurrent tuberculosis at other sites is important. Keywords: Cell-mediated immunity, cutaneous tuberculosis, tuberculids Introduction Tuberculosis is caused by Mycobacterium tuberculosis and Mycobacterium bovis. An attenuated strain of Mycobacterium bovis, Bacille Calmette Guerin (BCG) may occasionally be responsible. The most common manifestation is lung infection, while cutaneous tuberculosis is relatively uncommon and accounts for 1% of extra-pulmonary tuberculosis. The skin may be infected via haematogenous spread, direct inoculation or auto-inoculation. Pathogenesis The presentation of cutaneous tuberculosis depends on the pathogenicity of the mycobacteria, route of infection, and level of cell-mediated immunity (CMI) of the host. Following infection, the mycobacteria are phagocytosed by macrophages, which circulate to lymph nodes and then by haematogenous spread to other parts of the body.1 Macrophages act as antigen presenting Cutaneous tuberculosis – review 131 cells and interact with T lymphocytes. Memory T lymphocytes are generated during initial sensitisation within three to ten weeks and circulate in the blood and internal organs. As CMI may deteriorate as a result of illness, immunosuppression, ageing, HIV infection and malnutrition, there is a long-term risk of reactivation. This risk is greatest two to three years after primary infection.2 Pathology In early lesions, there is non-specific inflammation consisting of polymorphs and macrophages. Tubercles appear after three to six weeks as immunity develops.3 The tubercle consists of a collection of epithelioid cells at the centre with a variable number of Langhans giant cells surrounded by a rim of lymphocytes. As the inflammation progresses, the centre of the granuloma undergoes caseation necrosis, which is a distinctive feature of tuberculous infection. In patients with adequate CMI, the granuloma is able to contain the infection with progressive fibrosis and finally calcification. In 10% cases, tuberculosis infection will lead to active disease.4 In lupus vulgaris (LV), there is effective CMI in the host. Granulomas are formed with little caseation and acid-fast bacilli (AFB) are infrequently found. There are tuberculoid granulomas in the upper dermis. In scrofuloderma (SFD), there is moderate cell-mediated immunity in the host. The granulomas are less well formed and are located at the periphery of the lesion with more caseation necrosis at the centre. In addition, some AFBs can be found. In tuberculosis verrucosa cutis (TVC), there are often epidermal changes such as warty hyperkeratosis or pseudoepitheliomatous hyperplasia with epithelioid and giant cells in the mid-dermis. Tubercles are less common and AFBs are occasionally seen. At the other end of the immunological spectrum, there is miliary tuberculosis with poor or absent granuloma formation and marked tissue necrosis. Numerous AFBs are present. The histological picture is similar in primary tuberculous chancre, but as CMI develops, tuberculoid granulomas are formed. In patients with poor CMI such as in acute disseminated miliary tuberculosis, mycobacteria are present. There is massive necrosis and non-specific inflammation. Differentiation from other types of mycobacteria can only be achieved by cultures. Sarcoidosis may be distinguished from mycobacterial infection by the paucity of lymphocytes around the granuloma; so called naked granulomas. Tertiary syphilis is characterised by plasma cell infiltration and vascular changes. In tuberculoid leprosy, the granulomas tend to be located around cutaneous nerves. Clinical subtypes of cutaneous tuberculosis A classification of cutaneous tuberculosis is shown in Table 1.3,5 The clinical presentation of cutaneous tuberculosis depends on whether the patient has been previously sensitised. In primary infection, there is no previous sensitisation and tuberculous chancre, acute miliary disseminated tuberculosis are the main presentations. Re-infection may result in LV, TVC, and re-activation can lead to tuberculosis cutis orificialis or SFD. Secondary infection comprises more than 95% of cutaneous tuberculosis infections. I. Inoculation tuberculosis 1. Primary inoculation tuberculosis (tuberculous chancre) Incidence Primary inoculation tuberculosis (PIT) mainly affects SCK Ho 132 children in endemic areas and accounts for 1-2% of cutaneous tuberculosis.2 Clinical features PIT is due to direct inoculation of M. tuberculosis into the skin in a non-sensitised patient.3,6 This may result from minor abrasions, tattooing, ear-piercing, minor surgical procedures or injections. A red/brown nodule develops two to four weeks after inoculation, which erodes to form an ulcer. There is an indurated base and the edges may be undermined. There is lymphatic spread of the bacilli to produce painless lymphadenopathy in the next three to four weeks. The combination of the chancre and regional lymphadenopathy is equivalent to the Ghon focus in pulmonary tuberculosis. The lesions often affect the face but may involve the mucous membranes as well. This may present as ulceration or oedema if the eyelids are affected. It usually resolves after three to 12 weeks as immunity develops, leaving an atrophic scar. PIT may occasionally evolve into LV or TVC. Alternatively, the regional lymph nodes may erode, leading to SFD. The tuberculin test is initially negative and becomes positive as CMI develops. 2. Lupus vulgaris Incidence Lupus vulgaris has declined since a post war peak of 775 per million in the 1950s.6 It is found in cool, moist environments and is still the most common form of cutaneous tuberculosis in Europe. There is a higher incidence in women and all age groups are affected. Clinical features Lupus vulgaris may develop as a result of inoculation or it may follow primary inoculation tuberculosis or BCG vaccination. Some cases of LV are due to spread of tuberculosis from elsewhere in the body (often lung or cervical lymph nodes) via the lymphatic system or direct spread. It may also follow SFD or tuberculous infection of the mucous membranes. Rarely, it may follow haematogenous dissemination. LV has been associated with tuberculous lymphadenitis in 40% cases, SFD in 30% of cases, and tuberculosis of the lungs or bones in 10 to 20% cases.3,6,7 The lesions are usually solitary and progress steadily, affecting the head and neck in most cases. Multiple lesions may occur when the immune response has been lowered especially after measles. The earlobe, nose or cheeks are most often affected. In tropical countries, the buttocks and lower limbs are more frequently affected.2 Table 1. The classification of cutaneous tuberculosis Route Disease Immunity I. Inoculation tuberculosis • Lupus vulgaris • Present (exogenous source) • Tuberculosis verrucosa cutis • Present • Primary inoculation tuberculosis • Absent II. Secondary tuberculosis • Scrofuloderma • Equivocal (endogenous) • Orificial tuberculosis • Absent III. Haematogenous tuberculosis • Acute miliary tuberculosis (AMT) • Absent • Tuberculous gumma • Absent • Some cases of lupus vulgaris • Present IV. Tuberculids • Papulonecrotic tuberculid Micropapular • Lichen scrofulosorum Papular • Erythema induratum Nodular Cutaneous tuberculosis – review 133 A brownish-red, soft plaque is formed as the initial papule enlarges or by coalescence of several smaller papules. Tubercles can be seen on diascopy as "red currant jelly" nodules on the surface of the plaque. Spontaneous involution in one area may be followed by progression in another area. Involvement of the mucous membranes is more likely to result in deformity such as destruction of the nasal bridge or laryngeal stenosis. There are five clinical variants: i) Plaque forms are flat plaques in which the surface may be smooth or scaly. There is minimal central scarring or infiltration. ii) Hypertrophic forms can be associated with lymphoedema and limb deformity. The surface is nodular and soft, with a tendency to ulcerate. iii) Ulcerative forms can lead to severe mutilation if the nasal or auricular cartilage is affected. iv) Tumour-like forms present as a collection of soft nodules or as plaques with deep infiltration and respond poorly to treatment. Large tumours may affect the ear lobe and lymphoedema may be present. v) LV may also be present in a papular or nodular form often as multiple lesions. Although there may be periods of inactivity, LV runs a chronic course without treatment. There is a 10% risk of developing squamous cell carcinoma from LV and this may delayed from 10 to 15 years. 3. Tubeculosis verrucosa cutis Incidence Tuberculosis verrucosa cutis is less common in western countries. It is more common in the tropical regions and was the most common form of cutaneous tuberculosis reported in Hong Kong in 1968 (46% of cases).8 The age of onset was before ten years in over half of the cases.9 Clinical features TVC is due to direct inoculation into the skin in a previously sensitised patient. This has been associated with certain professions in the past, namely, pathologists (prosector's wart), laboratory workers and medical students. TVC is often found on the hands and areas prone to trauma as single lesions, although multiple lesions are occasionally seen. In tropical areas, the buttocks and lower extremities are commonly affected sites. This affects children in particular as they play on the pavement and become infected via contaminated sputum.9 The lesion begins as a papule or papulopustule and slowly enlarges to form a plaque. Occasionally, the plaque is psoriasiform or keloidal and deformity of the limbs may result from papillomatous or sclerotic forms. It is frequently misdiagnosed as a wart. Spontaneous healing may occur at the centre and the entire lesion may resolve after several months or years. II. Secondary tuberculosis 1. Scrofuloderma Incidence Scrofuloderma was common before anti-tuberculous therapy was available. It is now more common in immigrants from developing countries.6 In U.K, most cases in the indigenous population affect patients older than 50 years old, while in the Asian community most cases are between 10 to 50 years of age.7 Clinical features SFD is due to reactivation of dormant tuberculosis. There is contiguous involvement of overlying skin from an underlying tuberculous focus such as tuberculous lymphadenitis or tuberculous bone disease. Tuberculin test is usually positive. A cold abscess is formed and the overlying skin is eroded. SCK Ho 134 SFD from tuberculous lymphadenitis often affects the parotid, submandibular, supraclavicular and both sides of the neck. The lesions begin as subcutaneous nodules, which become doughy in consistency. With progressive liquefaction, a cold abscess is formed and the skin erodes to form a discharging sinus. There may be healing with scarring and recurrence of disease over several years. 2. Oroficial tuberculosis (Tuberculosis Ulcerosa Cutis et Mucosae) Incidence Oroficial tuberculosis a rare condition affecting approximately 0.2% of tuberculosis.2 It is more common in males and with advancing age. Clinical features This is due to auto-inoculation at mucosal orifices adjacent to organs (especially lung, gastro-intestinal tract, genito-urinary tract) infected with tuberculosis. Commonly affected sites include the mouth, oropharynx, soft or hard palate, larynx, perianal area, and genitalia. The tuberculin test is weak or negative. A small red or yellow nodule develops which then breaks down to form an ulcer. There may be a pseudo-membrane with surrounding inflammation and oedema. Severe pain may interfere with eating, micturition, or defaecation. The ulcers may enlarge and persist if not treated properly. III. Haematogenous tuberculosis 1. Acute disseminated miliary tuberculosis Clinical features Acute disseminated miliary tuberculosis is a rare condition, affecting mainly infants with a decreased immune response resulting in haematogenous spread of bacilli to the rest of the body.3 This may follow viral infections such as measles. The patient is often ill with widespread tuberculosis (the focus of infection is often the lung or the meninges). Skin lesions consist of discrete, pin-head sized, red-blue papules that are topped by small vesicles. These rupture and form a crust, leaving a depressed scar after resolution. The tuberculin test is negative. 2. Tuberculous abscess (tuberculous gumma) Incidence Tuberculous gummata are more common in immigrants from endemic areas. Clinical features Tu b e r c u l o u s g u m m a t a a r e d u e t o haematogenous spread of tuberculosis from a primary source as a result of either breakdown of an old healed tubercle, or due to reduced immunity.3 The lesions present as solitary or multiple subcutaneous abscesses that break down to form a discharging sinus.2 In immunocompetent patients, the lesions are solitary and may resolve spontaneously, but in immunocompromised cases, multiple lesions may occur and are associated with a poorer prognosis. IV. The tuberculids The tuberculids are believed to be hyper-sensitivity reactions to blood spread of tuberculosis.3,6 However, this definition has been used freely in the past and many conditions like acne agminata, rosacea-like tuberculid, lichenoid tuberculid had been included under this group. This has led to confusion and controversy regarding the relation between tuberculosis and the tuberculids. i) The evidence against a tuberculous patho-genesis is as follows:-• Mycobacteria frequently cannot be found in these lesions. Cutaneous tuberculosis – review 135 • The tuberculoid features, which are seen histologically in tuberculids, are also seen with other conditions such as deep fungal infection and therefore does not provide definitive evidence of a tuberculous origin. ii) The evidence suggesting a tuberculous origin include: • Tuberculids have become uncommon where there has been a decline in tuberculosis but not in areas where it has persisted.10 In areas where there has been a resurgence of tuberculosis, the tuberculids have also increased in incidence. • Tuberculids respond to anti-tuberculous therapy.11 The tuberculids have been divided into those which have good evidence in association with tuberculosis and facultative tuberculids. The former include lichen scrofulosorum and papulonecrotic tuberculid. Erythema induratum has been classified as a facultative tuberculid. Tuberculosis may be one of several aetiological factors in erythema induratum, but up to-date, there have been no other agents implicated in its pathogenesis. 1. Papulonecrotic tuberculid Incidence Papulonecrotic tuberculid (PNT) was common in the past and remains relatively common in areas with a high incidence of tuberculosis. In one series, 91 cases were reported in South Africa over a 17-year period.12 In areas with low incidence of tuberculosis, it has become rare. Clinical features The most commonly affected sites are the buttocks, extensor aspects of the knees and elbows and lower trunk. There are recurrent crops of red papules in a symmetrical distribution, persisting over months or years. Central necrosis and a crust may develop on the top of the papules. Spontaneous resolution of the lesion or removal of the crust will leave a depressed scar. PNT has also been associated with erythema induratum.12,13 In addition, LV developed in four cases from pre-existing PNT lesions. Histopathology There is wedge-shaped necrosis of the upper dermis and epidermis resulting from a leuco-cytoclastic vasculitis. This area may be surrounded by histiocytes and blood vessel involvement may lead to endothelial damage and thrombosis. Differential diagnosis PNT may resemble pityriasis lichenoides et varioliformis acuta (PLEVA), although the latter is more widespread. Other conditions including leucocytoclastic vasculitis, nodular prurigo and secondary syphilis also need to be distinguished from PNT. 2. Lichen scrofulosorum Incidence Lichen scrofulosorum (LS) was first described by von Hebra and has been an uncommon condition, even in the past. It usually affects children and young adults and is often associated with tuberculosis of the bone, lymph nodes or pleura. It has been reported after BCG vaccination.14 Clinical features There is a lichenoid eruption often in children with tuberculosis. The lesions are perifollicular and are yellow-brown or pink papules with a scaly or hyperkeratotic top. These are seen mainly on the trunk with a lichenoid distribution and persist for months before resolving spontaneously. Discoid plaques may be formed as the papules coalesce. The lesions resolve with anti-tuberculous therapy within weeks to months.10 Histopathology Superficial granulomata are present within or near hair follicles and sweat ducts. There is no caseation and mycobacteria are not found. Differential diagnosis This includes other lichenoid conditions such as SCK Ho 136 lichen nitidus, lichen planus, lichenoid secondary syphilis and micropapular sarcoidosis. 3. Erythema induratum Incidence Erythema induratum (EI) was the most common form of tuberculid in Hong Kong.15 It is found mainly in females and affects all ages, although there are two peaks at adolescence and menopause. It is also more prevalent in the spring and autumn. Clinical features Affected patients present with erythrocyanotic changes in the lower limbs. The blood vessels react abnormally to the cold and perniosis may be present.16 There has been a debate about the tuberculous origin of EI as active tuberculosis is not always present. However, M. tuberculosus DNA has been detected by PCR,17 which suggests that at least some cases are associated with tuberculosis. The backs of the legs are commonly affected in a symmetrical pattern. Affected patients tend to have plump heavy legs. There are indolent, ill-defined nodules, which may improve in the summer and are exacerbated by cold. These may regress or ulcerate over several months. The resultant ulcers are irregular and may have bluish borders. Spontaneous resolution with scarring occurs after several months. Histopathology A nodular vasculitis with areas of fat necrosis, lobular panniculitis and foreign body reaction may be seen. Tuberculoid granulomas can also be found, although there is no caseation. BCG vaccination and tuberculosis of the skin BCG vaccination has been associated with cutaneous tuberculosis infections including: i) LV , SFD may develop at the injection site and the clinical features, treatment and course are similar to ordinary LV .18 ii) Severe regional lymphadenitis is the most common complication and affects younger patients. iii) Generalised tuberculid-like reactions have been reported on rare occasions.19 Diagnosis I. Absolute criteria Culture The only absolute criteria in confirming a diagnosis of cutaneous tuberculosis is a positive culture of M. tuberculosus from the biopsy material on Lowenstein Jensen's media. However, culture of M. tuberculosus requires up to four to six weeks, leading to considerable delay in diagnosis. However, the incidence of positive cultures for cutaneous tuberculosis is low,20 and diagnosis frequently relies on relative criteria. II. Relative criteria In the absence of positive cultures, relative criteria are used for diagnosis as follows:3,6,20 i) Evidence or history of active tuberculosis at other sites. ii) Clinical history and physical appearance. iii) The presence of acid-fast bacilli. iv) Tuberculous granulomas seen on histology. v) Positive Mantoux test. vi) Response to anti-tuberculosis therapy. III. Polymerase chain reaction in the diagnosis of cutaneous tuberculosis The polymerase chain reaction (PCR) can aid in the diagnosis of cutaneous tuberculosis.21 Primers targeting the IS 6110 repetitive insertion sequence of M. tuberculosus DNA have been used.22 PCR has proved useful in the diagnosis of various forms of cutaneous tuberculosis including inoculation tuberculosis,21 LV ,23 and SFD.24 However, PCR is not always positive in paucibacillary cases (LV , TVC).25 Cutaneous tuberculosis – review 137 M. tuberculosus DNA has been detected in some cases of EI,26-29 providing evidence to support the role of tuberculosis in these lesions, although it is possible that the mycobacterial DNA detected may have been from previous tuberculous infection. The association between tuberculosis and EI remains unclear. Although PCR does not distinguish between current and past infections, it does distinguish between M. tuberculosus DNA and atypical mycobacterial DNA.27 Possible reasons for these discrepancies may be found at the various stages in the PCR metho-dology. Many studies were done on archival DNA, resulting in suboptimal DNA extraction and false negatives. In addition, suboptimal DNA extraction may be due to the resistant lipid-rich wall of mycobacteria. Conversely, false positives may be due to contamination. Therefore the diagnosis should not be based on PCR alone. Treatment Principles of chemotherapy The aim is to eradicate all viable mycobacteria in the patient, which can be divided into three groups:30 i) Freely dividing extracellular bacilli. ii) Dormant bacilli within cells and caseous material. iii) Slowly dividing bacilli within the macrophages and in inflammatory lesions. The treatment of cutaneous tuberculosis is similar to that of pulmonary tuberculosis (Table 2). This consists of two phases:2,3 • Phase I targets rapidly dividing bacilli and consists of an initial phase of intensive therapy with three or four drugs for two months.21 • Phase II is directed at the remaining dormant bacilli (maintenance therapy) and consists of isoniazid and rifampicin for four further months. The advantage of the initial two-month intensive treatment is that a significant proportion of bacilli will be eradicated even if the patient defaults after this period. The sensitivities may vary with the locality and need to be adjusted accordingly. The regimens recommended by the WHO are given in Table 3.21 In patients with HIV , treatment with isoniazid and rifampicin is continued for seven months after the initial two months of quadruple therapy. Surgical measures may complement medical therapy. For example, localised lesions of LV or TVC may be excised while continuing on medical therapy and surgical correction of deformity may be required. It is important in the initial assessment of cutaneous tuberculosis, to look for any concurrent tuberculosis at other sites. Table 2. Dosage regime of the first-line anti-tuberculous drugs Drug Child Adult Isoniazid 5 mg/kg/day 300 mg/day Rifampicin 10 mg/kg/day 450 mg/day (<50 kg) 600 mg/day (>50 kg) Ethambutol 15 mg/kg/day 15 mg/kg/day Pyrazinamide 30 mg/kg/day 1.5 g/day (<50 kg) 2 g/day (>50 kg) Table 3. WHO recommended drug regimens for treatment of tuberculosis Phase I (Intensive): Phase II (Maintenance): 2 months 4 months Standard regimen: daily treatment INH, RIF, PYR INH, RIF Regimen suggested when drug resistance suspected: INH, RIF, PYR, STP INH, RIF Intermittent regimens: three times/week INH, RIF, ETH, PYR INH, RIF, PYR INH, RIF, STP, PYR INH, RIF, PYR INH=isoniazid; RIF=rifampicin; PYR=pyazinamide; ETH=ethambutol; STP=streptomycin SCK Ho 138 Conclusion Cutaneous tuberculosis can be difficult to diagnose, as cultures are often negative. Polymerase chain reaction may aid in the diagnosis although it should be interpreted in the light of the clinical and histological features. If in doubt, an empirical course of anti-tuberculous treatment may be required to confirm the diagnosis. References 1. Sehgal VN, Bhattacharya SN, Jain S, Logani K. Cutaneous tuberculosis: the evolving scenario. Int J Dermatol 1994;33:97-104. 2. MacGregor RR. Cutaneous tuberculosis. Clin Dermatol 1995;13:245-55. 3. Gawkrodger DJ. Mycobacterial Infections. In: Champion RH, Burton JL, Burns DA and Breathnach SM (Editors). Rook/Wilkinson/Ebling. Textbook of Dermatology, 6th ed. Oxford, London: Blackwell Scientific; 1998. 1181-1206. 4. Barnes PF, Bloch AB, Davidson PT, Snider DE Jr. Tuberculosis in patients with human immunodeficiency virus infection. N Engl J Med 1991;324:1644-50. 5. Beyt BE Jr, Ortbals DW, Santa Cruz DJ, Kobayashi GS, Eisen AZ, Medoff G. Cutaneous mycobacteriosis: analysis of 34 cases with a new classification of the disease. Medicine (Baltimore) 1981;60:95-109. 6. Fitzpatrick TB, Eisen AZ, Wolff K, Freedberg IM, Austen KF . Tuberculosis and other mycobacterial infections. In: Dermatology in general medicine. 5th Ed. New York: McGraw-Hill; 1999:2370-95. 7. Sehgal VN, Wagh SA. Cutaneous tuberculosis. Current concepts. Int J Dermatol 1990;29:237-52. 8. Wong KO, Lee KP , Chiu SF . Tuberculosis of the skin in Hong Kong. (A review of 160 cases). Br J Dermatol 1968;80:424-9. 9. Mitchell PG. Tuberculosis verrucosa cutis among Chinese in Hong Kong. Br J Dermatol 1954;66:444-8. 10. Smith NP, Ryan TJ, Sanderson KV, Sarkany I. Lichen scrofulosorum. Lichen scrofulosorum. A report of four cases. Br J Dermatol 1976;94:319-25. 11. Forstrom L, Hannuksela M. Antituberculous treatment of erythema induratum Bazin. Acta Derm Venereol 1970;50:143-7. 12. Morrison JG, Fourie ED. The papulonecrotic tuberculide. From Arthus reaction to lupus vulgaris. Br J Dermatol 1974;91:263-70. 13. Jordaan HF , Van Niekerk DJ, Louw M. Papulonecrotic tuberculid. A clinical, histopathological, and immunohistochemical study of 15 patients. Am J Dermatopathol 1994;16:474-85. 14. Curtis HM, Leck I, Bamford FN. Incidence of childhood tuberculosis after neonatal BCG vaccination. Lancet 1984;1:145-8. 15. Chong LY, Lo KK. Cutaneous tuberculosis in Hong Kong: a 10-year retrospective study. Int J Dermatol 1995;34: 26-9. 16. Sehgal VN. Cutaneous tuberculosis. Dermatol Clin 1994;12:645-53. 17. Schneider JW, Jordaan HF , Geiger DH, Victor T, Van Helden PD, Rossouw DJ. Erythema induratum of Bazin. A clinicopathological study of 20 cases and detection of Mycobacterium tuberculosis DNA in skin lesions by polymerase chain reaction. Am J Dermatopathol 1995; 17:350-6. 18. Izumi AK, Matsunaga J. BCG vaccine-induced lupus vulgaris. Arch Dermatol 1982;118:171-2. 19. Dostovsky A, Sagher F. Dermatological complication of BCG vaccination. Br J Dermatol 1963;75:181-92. 20. Sehgal VN, Srivastava G, Khurana VK, Sharma VK, Bhalla P, Beohar PC. An appraisal of epidemiologic, clinical, bacteriologic, histopathologic, and immunologic parameters in cutaneous tuberculosis. Int J Dermatol 1987;26:521-6. 21. Penneys NS, Leonardi CL, Cook S, Blauvelt A, Rosenberg S, Eells LD, et al. Identification of Mycobacterium tuberculosis DNA in five different types of cutaneous lesions by the polymerase chain reaction. Arch Dermatol 1993;129:1594-8. 22. Seckin D, Akpolat T, Ceyhan M, Tuncer S, Turanli AY. Polymerase chain reaction in cutaneous tuberculosis. Int J Dermatol 1997;36:51-4. 23. Faizal M, Jimenez G, Burgos C, Del Portillo P , Romero RE, Patarroyo ME. Diagnosis of cutaneous tuberculosis by polymerase chain reaction using a species-specific gene. Int J Dermatol 1996;35:185-8. 24. Taniguchi S, Chanoki M, Hamada T. Scrofuloderma: the DNA analysis of mycobacteria by the polymerase chain reaction. Arch Dermatol 1993;129:1618-9. 25. Tan SH, Tan BH, Goh CL, Tan KC, Tan MF , Ng WC, et al. Detection of Mycobacterium tuberculosis DNA using polymerase chain reaction in cutaneous tuberculosis and tuberculids. Int J Dermatol 1999;38:122-7. 26. Schneider JW, Jordaan HF , Geiger DH, Victor T, Van Helden PD, Rossouw DJ. Erythema induratum of Bazin. A clinicopathological study of 20 cases and detection of Mycobacterium tuberculosis DNA in skin lesions by polymerase chain reaction. Am J Dermatopathol 1995; 17:350-6. 27. Degitz K, Steidl M, Thomas P , Plewig G, Volkenandt M. Aetiology of tuberculids. Lancet 1993;341:239-40. 28. Schneider JW, Geiger DH, Rossouw DJ, Jordaan HF, Victor T, van Helden PD. Mycobacterium tuberculosis DNA in erythema induratum of Bazin. Lancet 1993;342: 747-8. 29. Degitz K, Messer G, Schirren H, Classen V, Meurer M. Successful treatment of erythema induratum of bazin following rapid detection of mycobacterial DNA by polymerase chain reaction. Arch Dermatol 1993;129: 1619-20. 30. Grange JM. Therapy of Mycobacterial Disease. In: Mycobacteria and Human Disease, 2nd ed. London: Arnold 1996;204-24.
9252	https://www.degruyterbrill.com/document/doi/10.1515/bgsl-2020-0016/html?lang=en&srsltid=AfmBOorD0y2Zzz6w-OYRNtPDxiclj84LleIfcVTKURcvD4WzfnjRyG0K	pbb 2020; 142(2): 267–305 Besprechungen Agnes Jäger : Vergleichskonstruktionen im Deutschen . Diachroner Wandel und synchrone Variation, Berlin u. Boston: de Gruyter 2018, XIII, 568 S. (Linguisti-sche Arbeiten 569) Dr. Julia Bacskai‑Atkari : Universität Konstanz, Fachbereich Linguistik, Universitätsstraße 10, D-78457 Konstanz, Deutschland, E-Mail: julia.bacskai -atkari@uni -konstanz.de Das Buch untersucht Vergleichskonstruktionen im Deutschen, insbesondere deren historische Entwicklung und regionale Verteilung. In Kap. 1 (S. 137) wird ein Überblick über den Stand der Forschung, der im Buch untersuchten Konstruk-tionen, die angewendeten Methoden sowie den Aufbau des Buches gegeben. Ver-gleichskonstruktionen sind in der Fachliteratur insgesamt gut vertreten, jedoch gibt es noch viele offene Fragestellungen, auf die bisher noch nicht eingegangen wurde. Das von der Autorin betrachtete Spektrum ist auffällig breit: Neben Arbei-ten in der generativen Syntax und sprachgeschichtlichen Arbeiten werden auch synchrone und typologische Untersuchungen berücksichtigt und miteinander in Verbindung gebracht. Das theoretische Spektrum wird auch durch das empirische Spektrum ergänzt: Die Untersuchung berücksichtigt nämlich nicht nur Kompara-tivvergleiche, wie es in der Literatur üblich ist, sondern auch Äquativvergleiche und darunter sogar Nicht-Gradäquative. In Kap. 2 (S. 3896) diskutiert die Autorin die Vergleichsarten im Althochdeut -schen. Es wird gezeigt, dass sich alle Typen aus dem Germanischen respektive aus dem Indogermanischen ableiten lassen. Bei den Vergleichskonstruktionen wird gezeigt, dass mit der Vergleichspartikel ( thanne ) und dem Vergleichskasus (Dativ) zwei typologische Muster im gleichen System zu finden sind. Die Autorin argumentiert, dass Vergleichskonstruktionen mit Vergleichskasus auf keinen Fall als lateinische Lehnsyntax, sondern als genuin altgermanisch zu betrachten sind, womit die Analyse einen wichtigen Beitrag für die Ähnlichkeiten zwischen Sprachtypologie und Sprachwandel leistet. Bei den Äquativen wird gezeigt, dass sämtliche komplexe Vergleichspartikel (z. B. soso , samaso , also ) von der Kombination einer ursprünglichen Matrixpartikel und einer Vergleichspartikel abstammen. Kap. 3 (S. 97 148) untersucht die Vergleichsarten im Mittelhochdeutschen. Es lässt sich eine Verschiebung bei der Distribution der Formen feststellen, wobei die meisten zumindest schon im Althochdeutschen nachzuweisen sind. Bei den Äquativen gibt es signifikante Unterschiede zwischen den beiden Haupttypen: In
9253	https://bcmath.ca/competition-math-formulas.pdf	Formulas for Pre-Olympiad Competition Math eashang1 August 16, 2017 This is a compilation of various formulas, theorems, lemmas, and facts that are useful for competition math. I’ve ordered them by topic (geometry, number theory, algebra, and count-ing/probability). It is designed to be a reference - not a study guide. The starred () formulas are ones you must know for competition math, as they are very useful and come up in nearly every competition. The others listed are good to know, fun to learn, and are used occasionally, but aren’t necessary for scoring well. Thanks to all the AoPSers (especially mathwiz0803) who contributed through their time and suggestions! For more in depth explanations for each of these, visit AoPS Wiki or search for explanations on Youtube. I hope this helps! Page 1 eashang1 1 Geometry 1.1 Area of a Triangle A = bh 2 = rs = 1 2ab sin θ = abc 4R Where A is the area, b is the base, and h is the height. In the second equation, r is the inradius and s is the semiperimeter (which is half the perimeter). In the third equation, θ is the angle between two sides a, b of the triangle. In the ﬁnal equation, a, b, c are the sides of the triangle with circumradius R. 1.2 Area of a Square (and kite/rhombus) A = bh = s2 or alternatively d1·d2 2 Where A is the area, b is the base, h is the height, s is the side length, and d1,2 is the length of a diagonal. The prior equation only applies to squares. The latter formula applied to any quadrilateral with perpendicular diagonals (such as kites and rhombi). 1.3 Area of a Rectangle A = bh Where A is the area, b is the base, and h is the height. 1.4 Area of a Trapezoid A = (b1+b2)(h) 2 Where A is the area, b1 and b2 are bases, and h is the height. 1.5 Area of a Regular Hexagon A = 3 √ 3s2 2 Where A is the area and s is the side length. Deriving this by breaking the hexagon into six equilateral triangles and then 12 right triangles is a useful exercise. 1.6 Area of a Regular Polygon A = ap 2 or ns2 4 tan( 180 n ) Where A is the area, a is the apothem, p is the perimeter, n is the number of sides, and s is the side length. 1.7 Volume/Surface Area of a Cone V = πr2h 3 , SA = πr2 + πrl Where V is the volume, SA is the surface area, r is the radius of the circular base, h is the height, and l is the slant height. Page 2 eashang1 1.8 Volume/Surface Area of a Sphere V = 4πr3 3 , SA = 4πr2 Where V is the volume, SA is the surface area, and r is the radius of the sphere (which is radius of the central cross section/the base of the semisphere). 1.9 Volume/Surface Area of a Cube V = s3, SA = 6s2 Where V is the volume, SA is the surface area, and s is the length of a side. 1.10 Volume/Surface Area of a Pyramid V = 1 3bh, SA = 2sl + b Where V is the volume, SA is the surface area, b is the area of the base, h is the height, l is the slant height, and s is the length of a side of the base. Note that a pyramid can have a base of any polygon, but if none is speciﬁed, assume a square base. A pyramid with a triangular base is known as a tetrahedron. 1.11 Volume/Surface Area of a Cylinder V = πr2h, SA = 2πr2 + 2πrh Where V is the volume, SA is the surface area, r is the radius of the circular base, and h is the height. 1.12 Volume/Surface Area of a Prism V = lwh, SA = 2(lw + lh + wh) Where V is the volume, SA is the surface area, l is the length, w is the width, and h is the height. 1.13 Pythagorean Theorem and Right Triangles a2 + b2 = c2 Where c is the hypotenuse of a right triangle with legs a and b. Note that there are some “special” right triangles. These include right triangles with angle measures 45◦−90◦−45◦and 30◦− 60◦−90◦. The prior type of right triangle has the property that if either leg (they are identical) has length x, the hypotenuse has length x √ 2. Similarly, the latter type of right triangle has the property that if the side opposite the 30◦angle has length x, the side opposite the 60◦angle has length x √ 3, and the side opposite the 90◦angle has length 2x. You should also memorize some common Pythagorean triples (if a triangle has these side lengths, or these side lengths multiplied by some factor, it is a right triangle): 3 −4 −5, 5 −12 −13, 7 −24 −25, and 8 −15 −17. Page 3 eashang1 1.14 Distance Formula d = p (x2 −x1)2 + (y2 −y1)2 Where (x1, y1) and are points a coordinate plane and d is the distance between them. This is essentially the Pythagorean Theorem restated for points on a plane. 1.15 Heron’s Formula A = p (s)(s −a)(s −b)(s −c) Where A is the area and s is the semiperimeter of the triangle with sides a, b, c. 1.16 Cyclic Quadrilaterals A quadrilateral is cyclic if and only if the quadrilateral can be inscribed in a circle. Here are some of the fundamental properties of cyclic quadrilaterals. 1. Opposite angles add to 180◦. 2. A convex quadrilateral is cyclic if and only if the four perpendicular bisectors to the sides are concurrent. This common point is the circumcenter. 3. In cyclic quadrilateral ABCD, ̸ ABD = ̸ ACD, ̸ BCA = ̸ BDA, ̸ BAC = ̸ BDC, ̸ CAD = ̸ CBD 1.17 Ptolemy’s Theorem ab + cd = ef Where ABCD is a cyclic quadrilateral with side lengths a, b, c, d and diagonals e, f (with a opposite b and c opposite d). 1.18 Brahmagupta’s Formula K = p (s −a)(s −b)(s −c)(s −d) Where K is the area and s is the semiperimeter of the quadrilateral with sides a, b, c, d. For this formula to work, the quadrilateral must be cyclic. Page 4 eashang1 1.19 Power of a Point There are three cases for this theorem: This theorem states that in the leftmost diagram (two intersecting internal chords), AE · EC = DE · EB, in the middle diagram (a tangent and a secant that meet at a point) AB 2 = BC · CD, and in the last diagram (secants that intersect outside the circle) CB · CA = CD · CE. 1.20 Ceva’s Theorem Ceva’s theorem states that in with points D, E, F on sides BC, AC, AB respectively, are concurrent if, and only if, BD DC · CE EA · AF F B = 1 1.21 Menelaus’ Theorem This conﬁguration shows up from time to time, though not often. The theorem states that given a conﬁguration as shown below, BP · CQ · PC = QA · RB · AR 1.22 Arcs and Angles in a Circle An arc is a portion of the circle’s circumference measured in degrees. The measure of an angle formed by the center of a circle and two point on the circumference is equal to the measure of the intercepted (subtended) arc. An angle formed by three points on the circumference of the circle is equal to 1 2 the measure of the subtended arc. 1.23 Angle Bisector Theorem The Angle Bisector Theorem states that given △ABC and angle bisector AD, where D is on side BC, then c m = b n. Likewise, the converse of this theorem holds as well Page 5 eashang1 1.24 Trigonometric Identities Note that tan θ = sin θ cos θ and cot θ = cos θ sin θ . Also sec θ = 1 cos θ and csc θ = 1 sin θ. Therefore, the identities for tan, cot, sec, csc are easily derived from the identities for sin and cos. Double Angle: sin 2θ = 2 sin θ cos θ cos 2θ = cos2 θ −sin2 θ Negative Angles: sin cos −θ = sin cos θ cos sin −θ = cos −sin θ = cos sin θ Pythagorean Identities: sin2 θ + cos2 θ = 1 cot2 θ + 1 = csc2 θ tan2 θ + 1 = sec2 θ Addition/Subtraction Identities: sin a ± B = sin a cos B ± sin B cos a cos a ± B = cos a cos B ± sin a sin B Half Angle Identities: sin θ 2 = ± q 1−cosθ 2 cos θ 2 = ± q 1+cosθ 2 1.25 Triangle Inequality The Triangle Inequality says that in nondegenerate △ABC: AB + BC > AC, BC + AC > AB, AC + AB > BC 1.26 Pick’s Theorem A = I + 1 2B −1 Where A is the area, I is the number of lattice points in the interior, and B is the number of lattice points on the boundary of a ﬁgure in the coordinate plane. 1.27 Stewart’s Theorem Take △ABC with sides of length a, b, c opposite vertices A, B, C respectively. If cevian AD is drawn so that BD = m, DC = n, AD = d we have that man + dad = bmb + cnc, which can be remembered using the mnemonic device, “A man and his dad put a bomb in the sink.” This theorem isn’t used very often in competition math, but it can trivialize otherwise diﬃcult problems when it does come up. 1.28 (Extended) Law of Sines a sin A = b sin B = c sin C = 2R Where a, b, c are sides of a triangle, each opposite its respective angle A, B, C. R is the circumradius. 1.29 Law of Cosines For a triangle with sides a, b, c and opposite angles A, B, C respectively, the Law of Cosines states c2 = a2 + b2 −2ab cos C Page 6 eashang1 1.30 Shoelace Theorem Shoelace Theorem Suppose the polygon has vertices (a1, b1), (a2, b2), ..., (an, bn) listed in clockwise order. Then the area of P is 1 2\|(a1b2 + a2b3 + ... + anb1) −(b1a2 + b2a3 + ... + bna1)\| The Shoelace Theorem gets its name because if one lists the coordinates in a column and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes. (a1, b1) (a2, b2) ... (an, bn) (a1, b1) 2 Number Theory 2.1 Sum of an Arithmetic Series Sn = n 2 (a1 + an) Where n is the number of terms, Sn is the sum, and a1 is the ﬁrst term. 2.2 Sum of the ﬁrst n terms of a Geometric Series S = a1(1−rn) 1−r Where S is the sum, a1 is the ﬁrst term, and r is the common ratio. 2.3 Sum of an Inﬁnite Geometric Series S = a1 1−r Where r is the common ratio, S is the sum, and a1 is the ﬁrst term. Note that \|r\| < 1 for this formula to work (otherwise the sum doesn’t converge). 2.4 Sum of an Arithmetic Series The sum of the ﬁrst odd numbers is simply n2. The sum of the ﬁrst n numbers n(n+1) 2 . The sum of the ﬁrst n even numbers is n(n + 1). 2.5 Sum of the First n Integers/Odd Integers The sum of the ﬁrst n odd numbers is simply n2. The sum of the ﬁrst n positive integers is given by n(n+1) 2 . 2.6 Number/Sum of Divisors Let px 1py 2...pn m be the prime factorization of sum number a. The number of divisors a has is given by (x + 1)(y + 1)...(n + 1). Similarly, the sum of the divisors of a is given by (p0 1 + p1 1...px 1)(p0 2 + p1 2...py 2)...(p0 m + p1 m...pn m). 2.7 Chinese Remainder Theorem The Chinese Remainder Theorem (or CRT) allows you to solve a system of linear congruences. Let m1, m2, ..., mr be a collection of pairwise relatively prime integers. Then the system of simultane-ous congruences x ≡a1( mod m1), x ≡a2( mod m2), ..., x ≡ar( mod mr) has a unique solution modulo M = m1m2, ∆∆∆mr, for any given integers a1, a2, ..., ar. This is easier to understand in an example and is very well explained in this video: Page 7 eashang1 2.8 Chicken McNugget Theorem The Chicken McNugget Theorem states that for any two relatively prime positive integers, m, n, the greatest integer that cannot be written in the form am + bn (the greatest number that can not be expressed as a sum of the two numbers) where a, b are positive integers is mn −m −n. It follows that there are (m−1)(n−1) 2 positive integers which cannot be expressed as the sum of some number of s and s. 2.9 Euler’s Totient/Phi Function φ(n) = n(1 −1 p1 )(1 −1 p2 )...(1 −1 pn ) Where n is any positive integer and pn are prime divisors of n. This gives the number of relatively prime positive integers less than or equal to some number n. 2.10 Wilson’s Theorem Wilson’s Theorem is rather uncommon, but it is very powerful when you use it since it’s a bijection. Also, note that Wilson’s Theorem provides a primality test. However, there is no quick way to compute p!. It states that for any prime p, (p −1)! ≡−1 mod p 2.11 Trivial Inequality Yes, this is a real thing. It states that x2 ≥0 for all real x (now you know why it is called “trivial”). You don’t really need this for anything, but many other well known theorems and inequalities are based on this. 2.12 Fibonacci Numbers Deﬁne a sequence Fn such that F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2. The ﬁrst few terms look like 0, 1, 1, 2, 3, 5, 8, . . . The quotient of two consecutive terms approaches the golden ratio: 1+ √ 5 2 2.13 Pigeonhole Principle If we distribute n balls into k boxes such that n > k then at least one box must have multiple balls. 3 Algebra 3.1 Logarithm Rules Logarithmic to Exponential: loga b = x ⇒ax = b Addition: loga b + loga c = loga bc Subtraction: loga b −loga c = loga b c Exponent Reducing: loga bn = n loga b Change of Base: loga b = logc b logc a Reciprocals: loga b = 1 logb a 3.2 Vieta’s Formulas Vieta’s formulas relate the coeﬃcients of a polynomial to its roots. This set of formulas is one of the most useful in competition math. These state that the sum of the roots for a quadratic polynomial ax2 + bx + c is −b a and that the product of the roots is c a. This extends to higher degree polynomials as well. For example, if the cubic polynomial ax3 +bx2 +cx+d has roots r, s, t, r + s + t = −b a , rst = −d a , and rs + rt + st = c a. Notice the signs alternate (with b being negative and alternating thereafter). This takes some practice to get used to and I recommend you visit the AoPS page on Vieta’s Formulas for further details. Page 8 eashang1 3.3 Common Factorizations and SFFT a2 −b2 = (a + b)(a −b) a3 −b3 = (a −b)(a2 + ab + b2) a3 + b3 = (a + b)(a2 −ab + b2) ab + a + b + 1 = (a + 1)(b + 1) ab −a −b + 1 = (a −1)(b −1) SFFT, or Simon’s Favorite Factoring, refers to the strategy of adding some number to both sides of a polynomial so that it will factor and then breaking the side that is an integer into its prime factors to ﬁnd posible solutions. 3.4 Quadratic Formula and Discriminant x = b± √ b2−4ac 2a d = b2 −4ac Where a, b, c are the coeﬃcients of the x2, x1, x0 terms respectively. If the discriminant is , the quadratic will have one real solution. If the discriminant is negative, the quadratic will have no real solutions. If the discriminant is positive, the quadratic will have two real solutions. 3.5 RMS-AM-GM-HM q x2 1+...+xn n n ≥x1+...xn n ≥ n √x1...xn ≥ n 1 x1 +...+ 1 xn This essentially states that the root-mean square is greater than or equal to the arithmetic mean is greater than or equal to the geometric mean is greater than or equal to the harmonic mean. The most important part of this inequality is AM-GM, which states that the arithmetic mean is greater than or equal to the geometric mean. Equality holds when all the s are the same. It is often used for maximization and minimization problems. Here is an interesting “proof” for this from AoPS Proofs Without Words gallery: 3.6 Location of Roots (DeMoivre’s Theorem Let cos(θ) + i sin θ = cis(θ). DeMoivre’s Theorem allows complex numbers in polar form - that is r · cis(θ) - to be raised to raised to a power. It states that for a rational x and integral n, cis(x)n = cis(nx). This can be used to ﬁnd the nth root of a number/polynomial. See here for more information: 4 Counting and Probability 4.1 Permutations nPk = n! (n−k)! Where n is the total number of objects from which you are choosing k objects (order does matter). Page 9 eashang1 4.2 Combinations n k = n! k!(n−k)! Where n is the total number of objects from which you are choosing k objects (order doesn’t matter). 4.3 Pascals Identity n k = n−1 k−1 + n−1 k Where n is the total number of objects from which you are choosing k objects. Note that this can be easily observed from Pascal’s Triangle. This identity can be extended to sums of three or more binomial coeﬃcients 4.4 Binomial Theorem and Sum of Row in Pascal’s Triangle The binomial theorem states that (a + b)n = n P k=0 n k an−kbk. This is easier to understand by doing practice expansions. The coeﬃcients of the powers of a and b are given in Pascals Triangle. The sum of the nth row in the triangle is 2n−1 (as shown below). 4.5 Fundamental Theorem of Counting If one event has n possible outcomes and another event has m possible outcomes, the total number of events if both events occur is m × n. We can recursively prove this for multiple events. 4.6 Burnside’s Lemma Burnside’s Lemma is a way to count objects if you need to account for rotations and reﬂections. It has appeared on Olympiads and even AMCs (it can be used for a variety of problems, though it may not always be the fastest method). The video below provides a great explanation and example for this lemma and I recommend you watch it. 4.7 Stars and Bars The number of ways to distribute n indistinguishable items into k distinguishable boxes, where each box must receive at least one item, is n−1 k−1 Alternatively, the number of ways to distribute indistinguishable balls in distinguishable boxes, where some box(es) may remain empty, is n+k−1 k−1 . 4.8 Expected Value If the expected value is E(X), E(X) = P i P(Xi)V (Xi),where P(Xi) is the probability of event Xi occurring, and V (Xi) is the value of outcome Xi. If X1, X2, ...Xk are several events (independent or not), then E(X1 + X2 + ... + Xk) = E(X1) + E(X2) + ... + E(Xk). Basically, you can sum expected values (weighted averages). For clarity, here is an example: You roll two dice. What is the expected sum? We can compute the expected value of a dice roll. Since all outcomes are equally likely, the weighted average for one roll of a dice is simply the average, or 3.5. Since we can sum expected values, the expected value of the sum is 3.5 + 3.5 = 7 Page 10 eashang1 Resources/Practice Recommendations -AoPS Volumes 1+2 for AMC and AIME preparation -EGMO by Evan Chen for AIME to Olympiad geometry -For AMC/AIME/USA(J)MO practice, visit the AoPS past competitions pages -AoPS community/forums, videos, and AoPS Wiki for reference (make an AoPS account!!!!!!) -This may seem redundant, but the best way to improve is to practice problems from the competitions you want to improve for -Handouts by Evan Chen (for very high level AIME to Olympiad): -Shameless self advertising:)
9254	https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:prob-comb/x9e81a4f98389efdf:combinations/e/combinations_1	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
9255	https://math.libretexts.org/Courses/Borough_of_Manhattan_Community_College/MAT_206.5/03%3A_Functions/3.08%3A_Inverse_Functions	Skip to main content 3.8: Inverse Functions Last updated : Nov 11, 2020 Save as PDF Section 3.7E: Exercises Section 3.8E: Exercises Page ID : 33059 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Skills to Develop Verify inverse functions. Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one. Find or evaluate the inverse of a function. Use the graph of a one-to-one function to graph its inverse function on the same axes. A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating. If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. Figure provides a visual representation of this question. In this section, we will consider the reverse nature of functions. Figure : Can a function “machine” operate in reverse? Verifying That Two Functions Are Inverse Functions Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula and substitutes 75 for to calculate Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. Figure : A forecast of Monday’s through Thursday’s weather. At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for after substituting a value for . For example, to convert 26 degrees Celsius, she could write After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature. The formula for which Betty is searching corresponds to the idea of an inverse function, a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function. Given a function , we represent its inverse as , read as “ inverse of .” The raised is part of the notation. It is not an exponent; it does not imply a power of . In other words, does not mean because is the reciprocal (or multiplicative inverse) of and not the function inverse. The “exponent-like” notation comes from an analogy between function composition and multiplication: just as (1 is the identity element for multiplication) for any nonzero number , so equals the identity function; that is, if , then This holds for all in the domain of . Informally, this means that an inverse function “undoes” the original function. However, just as zero does not have a reciprocal, some functions do not have inverses, as we will see later in this section. For example, and are inverse functions. and A few coordinate pairs from the graph of the function are , , and . A few coordinate pairs from the graph of the function are , , and . If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function. Definition: Inverse Function For any one-to-one function , a function is an inverse function of if . This can also be written as for all in the domain of . It also follows that for all in the domain of . Note that if a function has an inverse , then the inverse of is the function . The notation is read “ inverse.” Like any other function, we can use any variable name as the input for , so we will often write , which we read as “ inverse of .” Keep in mind that and not all functions have inverses. Example : Identifying Inverse Function Values for a Given Input-Output Pair If for a particular one-to-one function and , what are the corresponding input and output values for the inverse function? Solution The inverse function reverses the input and output quantities, so if . Alternatively, if we name the inverse function , then and . Analysis Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table . Table \| \| \| --- \| \| \| \| \| \| \| Given that , what are the corresponding input and output values of the original function ? Answer Given two functions and , test whether the functions are inverses of each other Determine whether . Determine whether . If both statements are true, then and . If either statement is false, then both are false, and and . Example : Testing Inverse Relationships Algebraically If and , is ? Solution so This is enough to answer yes to the question, but we can also verify the other formula. Analysis Notice the inverse operations are in reverse order of the operations from the original function. If and , is ? Answer : Yes Example : Determining Inverse Relationships for Power Functions If (the cube function) and , is ? Solution No, the functions are not inverses. Analysis The correct inverse to the cube is, of course, the cube root ; that is, the one-third is an exponent, not a multiplier. If and , is ? Answer : Yes Finding Domain and Range of Inverse Functions The outputs of the function are the inputs to , so the range of is the domain of . Likewise, because the outputs of are the inputs to , the domain of is the range of . We can visualize the situation as in Figure . Figure : Domain and range of a function and its inverse. However, not every function has an inverse. Consider the toolkit quadratic function . If we want to construct an inverse to this function, the "wannabe" function is . However, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it cannot have an inverse function. In order for a function to have an inverse, it must be a one-to-one function. If a function is not one-to-one, it is possible to restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function with its range limited to , which is now a one-to-one function (it passes the horizontal line test) and which has an inverse; namely, the square-root function . If on , then the inverse function is . The domain of = range of . The domain of = range of . Is it possible for a function to have more than one inverse? No. If two supposedly different functions, say and , both meet the definition of being inverses of another function , then you can prove that . We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse. Note: Domain and Range of Inverse Functions The range of a function is the domain of the inverse function . The domain of is the range of . Given a function, find the domain and range of its inverse If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function. The domain of a one-to-one function is and the range of function is . Find the domain and range of the inverse function. Answer : The domain of function is and the range of function is . Example : Finding the Inverses of Toolkit Functions Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are listed in Table . We must restrict the domain in such a fashion that the function assumes all -values exactly once. Table \| Constant \| Identity \| Quadratic \| Cubic \| Reciprocal \| \| \| \| \| \| \| \| Reciprocal squared \| Cube Root \| Square Root \| Absolute Value \| \| \| \| \| \| \| \| Solution The identity, cubic, reciprocal, cube root and square root functions are all one-to-one. The constant function is not one-to-one, and the only domain on which it could be one-to-one is a set containing a single point, so the constant function has an inverse only if it has a restricted domain of a single input value. The absolute value function can be restricted to the domain , where it is equal to the identity function. The reciprocal-squared function can be restricted to the domain . Analysis We can see that and (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure . They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse. Figure : (a) Absolute value (b) Reciprocal squared Finding and Evaluating Inverse Functions Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. Inverting Tabular Functions Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range. Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Example : Interpreting the Inverse of a Tabular Function A function is given in Table , showing distance in miles that a car has traveled in minutes. Find and interpret Table \| (minutes) \| 30 \| 50 \| 70 \| 90 \| \| (miles) \| 20 \| 40 \| 60 \| 70 \| The inverse function takes an output of and returns an input of . So in the expression , 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function , 90 minutes, so . The interpretation of this is that, to drive 70 miles, it took 90 minutes. Alternatively, recall that the definition of the inverse was that if , then . By this definition, if we are given , then we are looking for a value so that . In this case, we are looking for a so that , which is when . Using Table , find and interpret (a) ,and (b) . Table \| (minutes) \| 30 \| 50 \| 60 \| 70 \| 90 \| \| (miles) \| 20 \| 40 \| 50 \| 60 \| 70 \| Answer : . In 60 minutes, 50 miles are traveled. . To travel 60 miles, it will take 70 minutes. Evaluating the Inverse of a Function, Given a Graph of the Original Function We saw in Section 3.2 Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. Given the graph of a function, evaluate its inverse at specific points Find the desired input on the -axis of the given graph. Read the inverse function’s output from the -axis of the given graph. Example : Evaluating a Function and Its Inverse from a Graph at Specific Points A function is given in Figure . Find and . . Figure : Graph of Solution To evaluate , we find 3 on the -axis and find the corresponding output value on the -axis. The point tells us that . To evaluate , recall that by definition, means the value of for which . By looking for the output value 3 on the vertical axis, we find the point on the graph, which means , so by definition, See Figure . Figure : Graph of . Using the graph in Figure , (a) find , and (b) estimate . Answer : (a) 3 (b) about 5.6 Finding Inverses of Functions Represented by Formulas Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula — for example, as a function of — we can often find the inverse function by solving to obtain as a function of . Given a function represented by a formula, find the inverse. Make sure is a one-to-one function. Solve for . Interchange and . Example : Inverting the Fahrenheit-to-Celsius Function Given the function that gives Celsius temperature as a function of Fahrenheit temperature, find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. Solution By solving in general, we have uncovered the inverse function. If then In this case, we introduced a function to represent the conversion because the input and output variables are descriptive, and writing could get confusing. Solve for in terms of , given . Answer Example : Solving a Formula to Find an Inverse Function Find the inverse of the function . Solution So or . Extra Credit: The domain of is , while the domain of is . What is the range of , and what is the range of ? If you think you know the answer, explain your reasoning, and hand it in to your instructor. Example : Solving a Formula to Find an Inverse with Radicals Find the inverse of the function . Solution So . The domain of is . Notice that the range of is , so this means that the domain of the inverse function is also . Analysis The formula we found for looks like it would be valid for all real . However, itself must have an inverse (namely, ) so we have to restrict the domain of to in order to make a one-to-one function. This domain of is exactly the range of . What is the inverse of the function ? State the domains of both the function and the inverse function. Answer : ; domain of : ; domain of : Finding Inverse Functions and Their Graphs Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function restricted to a domain , on which this function is one-to-one, and consider its graph, in Figure . Figure : Quadratic function with domain restricted to . Restricting the domain to makes the function one-to-one (it obviously passes the horizontal line test), so it has an inverse on this restricted domain. We already know that the inverse of the toolkit quadratic function is the square root function; that is, . What happens if we graph both and on the same set of axes, using the -axis for the input to both and ? We notice a distinct relationship: The graph of is the graph of reflected across the diagonal line , which we will call the identity line, shown in Figure . . Figure : Square and square-root functions on the non-negative domain This graphical relationship holds true for all one-to-one functions and their inverses, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes. Example : Finding the Inverse of a Function Using Reflection about the Identity Line Given the graph of in Figure , sketch a graph of . Figure : Graph of . This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of and range of so the inverse will have a domain of and range of . If we reflect this graph over the line , the point reflects to and the point reflects to . Sketching the inverse on the same axes as the original graph gives Figure . Figure : The function and its inverse, showing reflection about the identity line Draw graphs of the functions and from Example . Answer : Figure : Graph of and . Is there any function that is equal to its own inverse? Yes. If , then , and there are several functions that have this property. The identity function does, and so does the reciprocal function, because Any function , where is a constant, is also equal to its own inverse. Note: Can you show that for ? If you can, show your work carefully, and hand it in to your instructor for Extra Credit. Key Concepts If is the inverse of , then . For a function to have an inverse, it must be one-to-one (pass the horizontal line test). A function that is not one-to-one over its entire domain may be one-to-one on part of its domain. For a one-to-one tabular function, exchange the input and output rows to obtain the inverse. To find the inverse of a formula, solve the equation for as a function of . Then exchange the labels and . The graph of an inverse function is the reflection of the graph of the original function across the line . Contributors Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at Section 3.7E: Exercises Section 3.8E: Exercises
9256	https://brainly.com/question/23896265	[FREE] In 15–20, write the number positioned at each point. - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +62,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +21,9k Ace exams faster, with practice that adapts to you Practice Worksheets +6,1k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified In 15–20, write the number positioned at each point. 2 See answers Explain with Learning Companion NEW Asked by AF286984 • 06/02/2021 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 18780807 people 18M 4.3 1 Upload your school material for a more relevant answer A number line is a horizontal line that has equally spread number increments. The points at A, B, C, D, E and F are -3.25, -4.5, 1.25, -5.75, 0.5 and -2.5 What is Number system? A number system is defined as a system of writing to express numbers. It is the mathematical notation for representing numbers of a given set by using digits or other symbols in a consistent manner. A number line is a horizontal line that has equally spread number increments. The points at A, B, C, D, E and F are 15.) -3.25 16.) -4.5 17.) 1.25 18.) -5.75 19.) 0.5 20.) -2.5 Therefore the points at A, B, C, D, E and F are -3.25, -4.5, 1.25, -5.75, 0.5 and -2.5. To learn more on Number System click: brainly.com/question/19532363 SPJ1 Answered by pukhrajvt •14.4K answers•18.8M people helped Thanks 1 4.3 (4 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 18780807 people 18M 3.5 1 Astronomy 2e - Andrew Fraknoi, David Morrison, Sidney Wolff Astronomy - Andrew Fraknoi, David Morrison, Sidney C. Wolff Statistics - Barbara Illowsky, Susan Dean Upload your school material for a more relevant answer The numbers positioned at points 15 to 20 on the number line are -3.25, -4.5, 1.25, -5.75, 0.5, and -2.5 respectively. Understanding the distances of these points from zero helps in visualizing their locations. This reinforces the concept of positive and negative numbering on a number line. Explanation To determine the number positioned at each point on a number line, we first need to understand how a number line functions. A number line is essentially a horizontal line where numbers are spaced evenly at regular intervals. Let’s label the points from 15 to 20: 15.) -3.25 16.) -4.5 17.) 1.25 18.) -5.75 19.) 0.5 20.) -2.5 Here’s how we arrive at these points: The point at 15 corresponds to the value -3.25, which suggests it lies 3.25 units to the left of zero on the number line. For point 16, we see a value of -4.5, indicating it's 4.5 units to the left of zero. At point 17, the value is 1.25, meaning it is situated 1.25 units to the right of zero. Point 18 has a value of -5.75, marking its place 5.75 units to the left. Point 19 is at 0.5, positioned 0.5 units to the right of zero. The final point 20 corresponds with -2.5, which is also to the left of zero by 2.5 units. This method of labeling allows us to visualize and understand the positions of various numbers relative to zero, enhancing our comprehension of real number placement on a number line. Examples & Evidence For example, if we add another point at 21, we might say it is positioned at 3.0, indicating that it is 3 units to the right of zero. Similarly, if we wanted to add point 22, we could place it at -1.5, demonstrating how negative values affect the positioning. The explanation of how each number is represented on the number line adheres to the rules of number placement based on their distance from zero, which is a fundamental principle in mathematics. Thanks 1 3.5 (4 votes) Advertisement Community Answer This answer helped 998150 people 998K 3.2 4 See below. Explanation 15.) -3.25 16.) -4.5 17.) 1.25 18.) -5.75 19.) 0.5 20.) -2.5 Answered by SDCInfinity •1.4K answers•998.1K people helped Thanks 4 3.2 (5 votes) 2 Advertisement ### Free Mathematics solutions and answers Community Answer 5.0 In 15–20, write the number positioned at each point in fractions Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics If PR=4 x−2 and RS=3 x−5, which expression represents PS? A. x−7 B. x−3 C. 7 x−7 D. $7 x+3 What can you conclude about a polynomial if its graph has 4 turning points? A. The polynomial has a degree of 4 B. The polynomial has a degree of 5 C. The polynomial cannot be factored D. The polynomial has only 3 zeros Suppose that f(x)=l n(x 3+3)8. Find f′(3). A random group of high school and college students were asked if they study with or without listening to music. The frequency table shows the results of the poll. Music No Music Total High school 68 65 133 College 85 51 136 Total 153 116 269 Which statements are true? Choose two correct answers. A. The conditional relative frequency that someone is in high school, given that the person prefers to study without music, is about 0.56 . B. The table shows conditional relative frequencies by column. An equation that defines y as a function of x is given. a. Solve for y in terms of x and replace y with the function notation f(x). b. Find f(3). The equation is: y+2 x 2=4−4 x Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
9257	https://www.ijpediatrics.com/index.php/ijcp/article/view/2196	Intermittent clobazam prophylaxis in simple febrile convulsions: a randomised controlled trial \| International Journal of Contemporary Pediatrics Skip to main contentSkip to main navigation menuSkip to site footer Open Menu International Journal of Contemporary Pediatrics Home About About the Journal Submissions Editorial Team Privacy Statement Contact Login Register Articles In Press Current Archives Author Guidelines Search Home/ Archives/ Vol. 6 No. 2 (2019): March-April 2019/ Original Research Articles Intermittent clobazam prophylaxis in simple febrile convulsions: a randomised controlled trial Authors Vinod Kumar Department of Pediatrics, Dr. S. N. Medical College, Jodhpur, Rajasthan, India Ashish Gupta Department of Pediatrics, Dr. S. N. Medical College, Jodhpur, Rajasthan, India DOI: Keywords: Febrile convulsions, Febrile seizure recurrence, Fever, Intermittent oral clobazam therapy Abstract Background: Febrile seizure (FS) is the most common type of childhood seizure disorder with a prevalence of 2-5% in children less than 5 years. Although the prognosis of febrile seizure is usually good, however, the possibility of recurrence keeps many parents and families in a state of anxiety and concerned, for years after the first seizure. Thus, intermittent prophylactic treatment might be advised in children with high risk of recurrence. Methods: The study was a prospective randomized, double blind, placebo-controlled trial conducted at Department of Pediatrics, Umaid Hospital, Dr S N Medical College, Jodhpur on neurologically normal children aged from 6 months to 5 years with a history of simple febrile seizures and normal electroencephalogram without any evidence of acute central nervous system infection. Subjects were randomly prescribed oral clobazam according to weight of child and placebo when they developed a febrile disease during the first 48 h of the onset of fever. Temperature reduction measures with paracetamol and tepid sponging were also advised. Patients were followed up for the frequency and time of febrile seizure recurrence, febrile episodes and side effects of drugs for 12 months. Results: Ten (3.8%) of 257 episodes in clobazam group and 38 (14.07%) episodes in placebo group had seizure recurrence (p value <0.001). The two groups were not significantly different in terms of side effects. (p >0.05). Conclusions: Intermittent oral clobazam therapy is a very effective measure in preventing recurrence of febrile seizures. Metrics PDF views 9,301 monthly \| yearly Twitter 1 References Engel J. Report of the ILAE classification core group. Epilepsia. 2006; 47(9):1558-68. Mikati MA, Hani AJ. Seizures in childhood. In: Kliegman RM, Stanton BF, St. Geme JW, Schor NF. Nelson textbook of Pediatrics. 20th Edition, Philadelphia: Elsevier; 2016; 2823-2857. Karande S. Febrile seizures: a review for family physicians. Indian J Med Sci. 2007;61:161-72. Shinnar S. Febrile seizures. In: Swaiman KF, Ashwal S, Ferriero DM, eds. Pediatric neurology principles & practice. 4th ed. Philadelphia: Mosby; 2006; 1079-82. Wallace SJ. Febrile seizures. In: Wallace SJ, Farrel K, eds. Epilepsy in children. London:A RNOLD;2004;123-1306. Pavlidou E, Tzitiridou M, Kontopoulos E, Panteliadis CP. Which factors determine febrile seizure recurrence? A prospective study. Brain Dev. 2008; 30(1):7-13. Knudsen FU. Febrile seizures:treatment and prognosis. Epilepsia. 2000;41(1):2-9. Verity CM, Greenwood R, Golding J. Long term intellectual and behavioural outcomes of children with febrile convulsion. N Engl J Med. 1998; 338:1723-8. Annegers JF, Hauser WA, Shirts SB, Kurland LT. Factors prognostic of unprovoked seizure after febrile convulsions. N Engl J Med. 1987;316:493-8. Amouian S, Jalili Akbarian M, Arabi M. Comparing Clobazam with Diazepam in Preventing Febrile Seizure in Children: A Randomized Clinical Trial. J Mazand Univ Med Sci. 2014; 24(111): 23-2. Verrotti A, Latini G, di Corcia G, Giannuzzi R, Salladini C, Trotta D, et al. Intermittent oral diazepam prophylaxis in febrile convulsions: its effectiveness for febrile seizure recurrence. Eur J Paediatr Neurol. 2004;8(3):131-4. Gupta S. Febrile seizures. An overview and use of clobazam as intermitant therapy. Pediatr Today. 2002;7:244-9. Rose W, Kirubakaran C, Scott JX. Intermittent clobazam therapy in febrile seizures. Indian J Pediatr. 2005;72(1):31-3. Bajaj AS, Bajaj BK, Puri V, Tayal G. Intermittent clobazam in febrile seizures: an Indian experience. J Pediatr Neurol. 2005;3(1):19-23. Akman CI. Febrile seizures: the role of intermittent prophylaxis. J Pediatr Neurol. 2005; 3:1-3. Gulati S, Saini D, Pandey RM, Kalra V. Randomized controlled trial to compare efficacy of oral clobazam with oral diazepam for prophylaxis of febrile seizures. Neuropediatr. 2006;37:13. Manreza MLG, Gherpelli JLD, Machado HLR, Pedreire CCC, Diament A, Heise CO. Treatment of Febrile seizures with intermittent clobazam. Arq. Neuro-psiquiatr. 2007; 55:757-61. Downloads PDF Published 2019-02-23 How to Cite Kumar, V., & Gupta, A. (2019). Intermittent clobazam prophylaxis in simple febrile convulsions: a randomised controlled trial. International Journal of Contemporary Pediatrics, 6(2), 732–735. More Citation Formats ACM ACS APA ABNT Chicago Harvard IEEE MLA Turabian Vancouver Download Citation Endnote/Zotero/Mendeley (RIS) BibTeX Issue Vol. 6 No. 2 (2019): March-April 2019 Section Original Research Articles Make a Submission Make a Submission Information For Readers For Authors For Librarians Keywords Current Issue International Journal Of Contemporary Pediatrics. Copyright © 2025. 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9258	https://labuladong.online/en/games/one-stroke-puzzle/	One Stroke Puzzle \| labuladong.online Labuladong Algo Notes RoadmapProblemsGamesVisualizationsTutorial 中 Login One Stroke Puzzle Connect all edges with one stroke, each edge can only be traversed once.Click any node to start, click adjacent nodes to connect edges, click current node (orange) to undo. Hints:Eulerian path exists📍 - must start from odd-degree nodes to complete Node Degree Details(6 nodes，7 edges，2 odd-degree nodes) 0 4°1 2°2 1°3 2°4 2°5 3° New Game Restart Game Progress 0/7 (0%) 🎉 Congratulations! You successfully connected all edges with one stroke! Completed in -1 steps, traversed 7 edges Play Again Continue Viewing Question Solution The One Stroke Puzzle is a classic puzzle game based on graph theory and Eulerian paths. You need to implement an algorithm to analyze the adjacency list of a graph and calculate the operation sequence to complete the one-stroke drawing. Game Rules Connect all points with one continuous line Each edge can only be passed once The goal is to traverse all edges in the graph Mathematical Principle The one-stroke problem is essentially finding an Eulerian path in a graph: If all vertices in the graph have even degree, then an Eulerian circuit exists (can start from any point) If exactly two vertices in the graph have odd degree, then an Eulerian path exists (must start from an odd-degree point) If more than two vertices have odd degree, then no Eulerian path exists Your Task Implement a solveOneStroke function that takes a graph's adjacency list as input and outputs an operation sequence for user reference. Example Output Step 1: Click node 0 to start the game Step 2: Click node 1 (connect from node 0 to node 1) Step 3: Click node 3 (connect from node 1 to node 3) ... Input Format The adjacency list is an object where keys are node numbers and values are lists of adjacent nodes. Example: { "0": ["1", "2"], "1": ["0", "3"], "2": ["0", "3"], "3": ["1", "2"] } This represents a square graph where node 0 connects to nodes 1 and 2, node 1 connects to nodes 0 and 3, and so on. Requirements Determine if the graph has an Eulerian path/circuit If it exists, use Hierholzer's algorithm to calculate the path Output detailed operation steps Handle edge cases (disconnected graph, no Eulerian path, etc.) Good luck! Editor Submit Reset 1 2 3 4 5 6 7 8 9 10 / @param{Object}adjacencyList-Graph adjacency list,format like{0:[1,2],1:[0,3],...} / function solveOneStroke(adjacencyList){ console.log(adjacencyList); //TODO:Output specific operation steps to help y ou solve the one-stroke puzzle }
9259	https://pmc.ncbi.nlm.nih.gov/articles/PMC9262923/	Mode-pairing quantum key distribution - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Nat Commun . 2022 Jul 7;13:3903. doi: 10.1038/s41467-022-31534-7 Search in PMC Search in PubMed View in NLM Catalog Add to search Mode-pairing quantum key distribution Pei Zeng Pei Zeng 1 Center for Quantum Information, Institute for Interdisciplinary Information Sciences, Tsinghua University, Beijing, 100084 China Find articles by Pei Zeng 1, Hongyi Zhou Hongyi Zhou 1 Center for Quantum Information, Institute for Interdisciplinary Information Sciences, Tsinghua University, Beijing, 100084 China Find articles by Hongyi Zhou 1, Weijie Wu Weijie Wu 1 Center for Quantum Information, Institute for Interdisciplinary Information Sciences, Tsinghua University, Beijing, 100084 China Find articles by Weijie Wu 1, Xiongfeng Ma Xiongfeng Ma 1 Center for Quantum Information, Institute for Interdisciplinary Information Sciences, Tsinghua University, Beijing, 100084 China Find articles by Xiongfeng Ma 1,✉ Author information Article notes Copyright and License information 1 Center for Quantum Information, Institute for Interdisciplinary Information Sciences, Tsinghua University, Beijing, 100084 China ✉ Corresponding author. Received 2021 Nov 18; Accepted 2022 Jun 9; Collection date 2022. © The Author(s) 2022 Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons license, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this license, visit PMC Copyright notice PMCID: PMC9262923 PMID: 35798740 Abstract Quantum key distribution — the establishment of information-theoretically secure keys based on quantum physics — is mainly limited by its practical performance, which is characterised by the dependence of the key rate on the channel transmittance R(η). Recently, schemes based on single-photon interference have been proposed to improve the key rate to 𝑅=𝑂⁡(√𝜂) by overcoming the point-to-point secret key capacity bound with interferometers. Unfortunately, all of these schemes require challenging global phase locking to realise a stable long-arm single-photon interferometer with a precision of approximately 100 nm over fibres that are hundreds of kilometres long. Aiming to address this problem, we propose a mode-pairing measurement-device-independent quantum key distribution scheme in which the encoded key bits and bases are determined during data post-processing. Using conventional second-order interference, this scheme can achieve a key rate of 𝑅=𝑂⁡(√𝜂) without global phase locking when the local phase fluctuation is mild. We expect this high-performance scheme to be ready-to-implement with off-the-shelf optical devices. Subject terms: Quantum information, Quantum optics Measurement-device-independent QKD schemes suffer from a trade-off between ease of implementation (avoiding the need for global phase locking) and high rates (quadratic improvement in rate). Here, the authors propose a protocol which offers both simple implementation and strong performances. Introduction Quantum key distribution (QKD)1,2 is currently the most successful application of quantum information science and serves as the first stepping stone towards a future quantum communication network3. A core advantage of QKD compared to other quantum communication tasks is that it is ready to implement with current commercially available off-the-shelf optical devices. However, two major characteristics of QKD—its practical security and key-rate performance—limit its real-life implementation. The key generation speed suffers heavily from transmission loss in the optical channel. Fundamentally, the asymptotic key rate for point-to-point QKD schemes is upper bounded by the repeaterless rate-transmittance bounds4,5, which are approximately linear functions of the transmittance, R ≤ O(η). For example, when η is small, the PLOB repeaterless rate-transmittance bound5 is about 1.44 η. Quantum repeaters6–8 have been proposed as a radical solution to this problem. Unfortunately, none of the quantum repeater proposals is easy to implement in the near term. In real-life use, the deviation of the realistic behaviour of physical devices from their ideal ones gives rise to critical issues in practical security. There are many quantum attacks that can take advantage of the loopholes introduced by device imperfections9. A typical QKD system can be divided into three parts: source, channel, and measurement. The security of the channel has been well addressed in the security proofs for QKD10–12. The source is relatively simple and can be well characterised13. In contrast, the measurement device is complicated and difficult to calibrate. Moreover, an adversary could manipulate the measurement device by sending unexpected signals14,15. To solve this implementation security problem, measurement-device-independent quantum key distribution (MDI-QKD) schemes have been proposed to close the detection loopholes once and for all16. Various experimental systems have been successfully demonstrated17–20, with extension to a communication network21. A generic MDI-QKD setup is shown in Fig.1a. Each of the two communicating parties, Alice and Bob, holds a quantum light source, encodes random bits into quantum pulses, and sends these pulses to a measurement site through lossy channels. Measurement devices are possessed by an untrusted party, Charlie, who is supposed to correlate Alice’s and Bob’s signals via interference detection. Based on the detection results announced by Charlie, Alice and Bob sift the local random bits encoded in the pulses to generate secure key bits. Note that the security of MDI-QKD schemes does not rely upon the physical implementation of the detection devices. Alice and Bob need to trust only their own locally encoded quantum sources. Since neither Alice nor Bob receives quantum signals from the channel during key distribution, any hacker’s attempt to manipulate the users’ devices becomes extremely difficult compared to regular QKD schemes14,15. Fig. 1. Comparison of two-mode, one-mode and mode-pairing MDI-QKD schemes. Open in a new tab a Schematic diagram of a generic MDI-QKD scheme. The solid and dashed pulses yield successful and unsuccessful detection, respectively, at the measurement site. For b, c and d, each wave packet in the diagram represents two independent pulses emitted simultaneously by Alice and Bob. b In two-mode MDI-QKD schemes, the pairing of the blue pulses (as phase references) and orange pulses (as signals) is predetermined, necessitating coincidence detection. c In one-mode MDI-QKD schemes (e.g., twin-field quantum key distribution and its variants), there is no phase reference pulse, necessitating global phase locking. d In the mode-pairing MDI-QKD scheme, in accordance with the detection results, Alice and Bob pair the clicked pulses and assign them to be either reference or signal pulses, such that neither coincidence detection nor global phase locking is required. Strictly speaking, MDI-QKD is not a point-to-point scheme, as there is an interference site between Alice and Bob. Consequently, it is not necessarily limited by the repeaterless rate-transmittance bound. Nevertheless, the original MDI-QKD scheme16, in which Alice and Bob both encode a ‘dual-rail’ qubit into a single-photon subspace on two polarization modes, unfortunately, cannot overcome this bound. Later, alternative schemes were proposed22,23 in which the qubit is encoded into two optical time bins. We refer to schemes of this type as two-mode MDI-QKD, in the sense that the single-side key information is encoded in the relative phase of the coherent states in the two orthogonal optical modes, i.e., second-quantized electromagnetic fields. To correlate Alice’s and Bob’s encoded information in a two-mode scheme, a successful two-photon interference measurement is required. If either Alice or Bob’s emitted photon is lost in transmission, there will be no conclusive detection result. For example, in the time-bin encoding scheme23 shown in Fig.1b, Alice and Bob each emit a qubit encoded in two time-bin modes, with Alice emitting A 1 and A 2 and Bob emitting B 1 and B 2. Only when both the interference between modes A 1 and B 1 and that between A 2 and B 2 yield successful detection can Alice restore Bob’s raw key information. Thus, successful interference requires a coincidence detection. Due to this coincidence-detection requirement, rounds with only a single detection are discarded, resulting in a relatively low key generation rate—one that is a linear function of the transmittance, O(η). From the perspective of practical implementation, however, coincidence detection also has certain merits. This approach can ensure stable optical interference, while Alice and Bob need only to stabilise the relative phases between the two modes. Coincidence detection is the essential factor that prevents MDI-QKD from overcoming the linear key-rate bound. To eliminate this requirement, a new type of MDI-QKD scheme called twin-field quantum key distribution (TF-QKD) based on encoding information into a single-optical mode have been proposed24, illustrated in Fig.1c. Later on, variants of TF-QKD have been proposed, among which the key information in encoded in either the phase25,26 (known as phase-matching QKD) or the intensity27 (known as sending-or-not-sending TF-QKD) of coherent states. In this work, we refer to these twin-field-type schemes as one-mode MDI-QKD schemes for a conceptual comparison to the traditional two-mode MDI-QKD schemes, since the single-side information in these schemes is encoded into a single-optical mode in each round. We remark that the single-optical-mode encoding MDI-QKD scheme was first proposed in ref. 28 as “MDI-B92” scheme. Similar to the Duan-Lukin-Cirac-Zoller-type repeater design29, such one-mode schemes use single-photon interference instead of coincidence detection, hence yielding a quadratic improvement in key rate compared to two-mode schemes24–26. As a result, they can overcome the point-to-point linear key-rate bound4,5. Unfortunately, one-mode schemes are more challenging to implement due to the unstable optical interference resulting from the lack of global phase references. For example, in the phase-matching QKD (PM-QKD) scheme25, the key information is encoded into the global phase of Alice’s and Bob’s coherent states. The phases of the coherent states generated by two remote and independent lasers need to be matched at the measurement site. A small phase drift or fluctuation caused by the lasers and/or channels is hazardous for key generation. At first glance, it seems that we cannot simultaneously enjoy the advantages of one-mode schemes (i.e., quadratic improvement in successful detection) and two-mode schemes (i.e., stable optical interference), due to an intrinsic trade-off between the information-encoding efficiency and robustness. On the one hand, the relative information among different optical modes is more difficult to retrieve when the channel loss is large. On the other hand, the global phase of a coherent state is not as stable as the relative phase between two coherent states travelling through the same quantum channel. In a typical 200-km fibre with a telecommunication frequency of 1550 nm, the phase of a coherent state is susceptible to small fluctuations in the optical transmission time (~10−15 s), optical length (~200 nm) and light frequency (~100 kHz). Recently, experimentalists have made great efforts to demonstrate high-performance in one-mode schemes, utilising high-end technologies to perform a precise control operation to stabilise the global phase by locking the frequency and phase of the coherent states30–37. However, this increases the experimental difficulty and undermines the applicability of one-mode schemes in real life. In this work, we propose a mode-pairing MDI-QKD scheme that aims to offer both—simple implementation and high performance. Hereafter, we refer to this scheme as the mode-pairing scheme for simplicity. By observing that the majority of detection events are single-clicks and are discard in the two-mode MDI-QKD schemes, we try to recycle the discarded single-click in the mode-pairing scheme. To do that, the coherent states in the transmitted modes are initially prepared independently with randomly encoded information. Based on the fact that the two detection events used to read out the encoded information do not need to occur at two predetermined locations, the key is extracted from two paired detection events rather than coincidence detection, as shown in Fig.1d. This offers a quadratic improvement akin to that of one-mode schemes when the local phases can be stabilized using currently available phase stabilization techniques. Moreover, key information about the mode-pairing scheme is encoded in the relative phases or intensities, whose stability relies only upon the conditions of the local phase references and optical paths. Therefore, the technical complexity is similar to that of two-mode schemes, which have been widely implemented both in the laboratory17–19,38 and in the field21,39. Notably, to adapt to different hardware conditions, the mode-pairing scheme can be freely tuned between the one-mode and two-mode schemes by adjusting a pulse-interval parameter (as discussed later in Results’ subsection “Pairing strategy”) during data postprocessing to optimise the system performance. Results Mode-pairing scheme In the mode-pairing scheme, Alice and Bob first prepare coherent states with independently and randomly chosen intensities and phases in each emitted optical mode. These coherent states are sent to the untrusted measurement site, Charlie. Based on Charlie’s announced measurement results, Alice and Bob pair the optical modes with successful detection and determine the key bits and bases for each mode pair locally. They then sift the bases and generate secure key bits via postprocessing. The scheme is introduced in Box1 and illustrated in Fig.2a. For simplicity of the introduction of the main protocol design, we omit the details of the decoy-state method40 and discrete phase randomisation here. A complete description of the mode-pairing scheme is given in the Methods’ subsection “Mode-pairing scheme with decoy states”. Fig. 2. Schematic diagram of the mode-pairing MDI-QKD scheme and the simple pairing strategy with maximal-pairing interval l. Open in a new tab The solid and dashed pulses are those with and without successful detection, respectively. Orange and blue pulses are, respectively, the front and rear pulses that succeed in pairing within l pulses, while grey pulses are the ones fail in pairing. a In the mode-pairing MDI-QKD scheme, Alice and Bob, first prepare coherent pulses with random intensities chosen from {0, μ} and random phases and send them to Charlie. After interference measurement, Charlie announces the detection results, based on which Alice and Bob pair the pulses and determine their encoding bases. For X-pairs, they announce the alignment angles θ a and θ b and keep data for which θ a = θ b. They use Z-pairs to generate keys and other data for parameter estimation. b We set l = 2 in the simple pairing strategy for example. The labels F k and R k represent the front and rear pulses, respectively, in the k-th successful pair. In the mode-pairing scheme, we mainly consider the keys generated from the Z-pair data, since they have a much lower quantum bit error rate than the X-pair data. The encoding of the mode-pairing scheme in Box1 originates from the time-bin encoding MDI-QKD scheme23. If Alice’s two paired optical modes {A i, A j} are assigned to the Z-basis, then the state of the two optical modes is either or , where and are two independent random phases. We can write the encoded states in a unified form: 1 where κ a is the encoded key information and is the inverse of κ. In the other case, in which the two optical modes {A i, A j} are assigned to the X-basis, we can rewrite their two independent random phases and as 2 In this way, the phase ϕ a becomes a global random phase on the pulse pair, while is the relative phase for quantum information ‘encoding’. Due to the independence of and , the phases ϕ a and are also independent of each other and uniformly range from [0, 2 π). By definition, we have . Then, the X-pair state can be written as, 3 where μ a ∈ {0, μ}. When θ = 0 or π/2, Alice emits X-basis or Y-basis states, respectively, as used in the time-bin encoding MDI-QKD scheme23. We remark that in either the Z-pair state in Eq. (1) or the X-pair state in Eq. (3), there is a global random phase ϕ a, which will not be revealed publicly. With this (global coherent state) phase randomisation, the emitted Z- and X-pair states can be regarded as a mixture of photon number states40. Then, Alice and Bob can estimate the detections caused by the pairs where they both emit single photons and use them to generate secure keys, in a manner similar to traditional two-mode schemes. Therefore, the security of the mode-pairing scheme is similar to that of two-mode schemes. Nevertheless, the mode-pairing scheme in Box1 has the following unique features. The emitted states in different optical modes {A i} are independent and identically distributed (i.i.d.). Therefore, the information encoded in different optical modes is completely decoupled. Based on the postselection of clicked signals, different optical modes are paired afterwards. The relative information between the two modes is then converted into raw key data. In the mode-pairing scheme, the key information is determined not in the state preparation step, but by the detection location, sharing some similarities with the differential-phase-shifting QKD scheme41,42. It is the untrusted measurement site that determines the location of successful detection and thereby affects the pairing setting. The ‘dual-rail’ qubits encoded on the single photons are ‘postselected’ on the basis of this detection. By virtual of the independence of the optical modes, the information encoded in the ‘postselected’ qubits cannot be revealed from other optical pulses. For another comparison, the sending-or-not-sending (SNS) TF-QKD scheme27 also uses a Z-basis time-bin encoding, whereby either Alice or Bob emits an optical mode to generate key bits. The state preparation of the mode-pairing scheme shares similarities with the SNS-TFQKD scheme. However, the information of the mode-pairing scheme is encoded into the relative information between the two optical modes. As a result, the basis-sifting and key mapping of the mode-pairing scheme follow different logic originated from the time-bin encoding MDI-QKD scheme23. Note that in the SNS scheme, bits 0 and 1 are highly biased in the Z basis, whereas in the mode-pairing scheme, they are evenly distributed. A critical issue in the security analysis of the mode-pairing scheme is to maintain the flexibility to determine in which two optical modes to perform the overall photon number measurement until Charlie announces the detection results. Note that, in the original two-mode QKD schemes, the encoders can always be assumed to perform an overall photon number measurement and post-select the single-photon components as good ‘dual-rail’ qubits before they emit their signals to Charlie. In the mode-pairing scheme, however, this is not viable because the optical pulse pair, for which the single-photon component is defined, is postselected based on Charlie’s detection announcement. To solve this problem, we introduce source replacement for the random phases in the coherent states to purify them as ancillary qudits and define an indirect overall photon number measurement on them. The source-replacement procedure can be found in the Methods’ subsection “Source replacement of the encoding state”. Conditioned on the indirect overall photon number measurement result to be single-photon states, the X-basis error rate fairly estimates the Z-basis phase error rate for the signals for which Alice and Bob both emit single photons. In Supplementary Note2, we provide a detailed security proof based on entanglement distillation. The main idea is to introduce a ‘fixed-pairing’ scheme, in which the pairing setting, i.e., which locations are paired together, is predetermined and hence independent of Charlie’s announcement. We first prove that, with any given pairing setting, the fixed-pairing scheme is secure, as it can be reduced to a two-mode MDI-QKD scheme. Afterwards, we examine the private state generated by the mode-pairing scheme and prove that it is the same as that of a fixed-pairing scheme under all possible measurements that Charlie could perform and announcement methods. In this way, we prove the equivalence of the mode-pairing scheme to a group of fixed-pairing schemes with different pairing settings. Box 1 Mode-pairing scheme. State preparation: In the i-th round (i = 1, 2, . . . , N), Alice prepares a coherent state in optical mode A i with an intensity randomly chosen from {0, μ} and a phase uniformly chosen from [0, 2 π). Similarly, Bob randomly chooses and and prepares in mode B i. Measurement: Alice and Bob send modes A i and B i to Charlie, who performs single-photon interference measurements. Charlie announces the click patterns for both detectors L and R. Alice and Bob repeat the above two steps for N rounds. Then, they postprocess the data as follows. Mode pairing: For all rounds with successful detection, in which one and only one of the two detectors clicks, Alice and Bob apply a strategy of grouping two clicked rounds as a pair. The encoded phases and intensities in these two rounds form a data pair. A simple pairing strategy is introduced in Box2. Basis sifting: Based on the intensities of the two grouped rounds indexed by i and j, Alice labels the ‘basis’ of the data pair as Z if the intensities are (0, μ) or (μ, 0), as X if the intensities are (μ, μ), or as ‘0’ if the intensities are (0, 0). Bob sets the basis using the same method. Alice and Bob announce the basis of each data pair; if they both announce the basis X or Z, they maintain the data pairs, whereas otherwise, the data pairs are discarded. Key mapping: For each Z-basis pair (Z-pair for simplicity) at locations i and j, Alice sets her key as κ a = 0 if and κ a = 1 if . For each X-basis pair (X-pair for simplicity) at locations i and j, the key is extracted from the relative phase , where the raw key bit is and the alignment angle is . In a similar way, Bob assigns his raw key bit κ b and determines θ b. The difference in the key mapping for Z-pairs is that, Bob sets the raw key bit κ b as 0 if and κ b = 1 if . As an extra step on the X-pairs, if Charlie’s detection announcement is (L, L) or (R, R), Bob keeps the bit κ b; otherwise, if Charlie’s announcement is (L, R) or (R, L), Bob flips κ b. For the X-pairs, Alice and Bob announce the alignment angles θ a and θ b. If θ a = θ b, then the data pairs are kept; otherwise, the data pairs are discarded. Parameter estimation: Alice and Bob estimate the fraction of clicked signals q(1, 1) and the corresponding phase error rate of Z-pairs where Alice and Bob both emit a single photon at locations i and j, using the data of the Z-pairs and X-pairs. They also estimate the quantum bit error rate E(μ, μ),Z of the Z-pairs. Key distillation: Alice and Bob use the Z-pairs to generate a key. They perform error correction and privacy amplification on the basis of q(1, 1), E(μ, μ),Z and . Pairing strategy The pairing strategy mentioned in Step 3 lies at the core of the mode-pairing scheme in Box1, which correlates two independent signals and determines their bases and key bits. Note that the relative phase between two paired quantum signals determines the key information on the X basis. When the time interval between these two pulses becomes too large, the key information suffers from phase fluctuation, which is characterised by the laser coherence time. Therefore, Alice and Bob should establish a maximal pairing interval l, such that the number of pulses between the two paired signals should not exceed l. In practice, l can be estimated by multiplying the laser coherence time by the system repetition rate. Here, we consider a simple pairing strategy in which Alice pairs adjacent detection pulses together if the time interval between them is not too large (≤l). The details are shown in the simple pairing strategy in Box2 and illustrated in Fig.2b. Charlie’s announcement in the i-th round is denoted by a Boolean variable C i that indicates whether the detection is successful. That is, C i = 1 implies that either the detector L or R clicks. Otherwise, there is no click or double clicks. To check the efficiency of this pairing strategy, let us calculate the pairing rate r p (i.e. the average number of pairs generated per pulse). We assume that Alice and Bob choose intensities 0 and μ with equal probability, maximising the number of successful pairs in the Z basis. With a typical QKD channel model, the pairing rate r p is calculated as shown in the Methods’ subsection “Mode-pairing-efficiency calculation”, 4 where p is the probability that the emitted pulses result in a click event, given approximately by η s μ. Here, η s and η denote the channel transmittance from Alice to Charlie and the total transmittance from Alice to Bob, respectively. When the channel is symmetric for Alice and Bob, we have . An explicit simulation formula for p in a pure-loss channel is given in Supplementary Note4. Note that both the pairing ratio r p and the detection probability p can be directly obtained by experimentation. The raw key rate mainly depends on the pairing rate r p. Now, let us check the scaling of r p with the channel transmittance in the symmetric-channel case. If the local phase reference is sufficiently stable, then the maximal interval can be set to l → +∞. In this case, 5 where the optimal intensity is μ = O(1), as evaluated in Supplementary Note5. On the other hand, if the local phase reference is not at all stable, one must set l = 1; then, 6 In this case, the experimental requirements for the mode-pairing scheme are close to those of the existing time-bin MDI-QKD scheme23. Now, if we consider a finite value of l, the dependence of r p(p, l) on η will be decided by how the denominator of the first term in Eq. (4), p[1 − (1−p)l], depends on p ≈ η s μ. When p l ≫ 1, r p(p, l) scales with p linearly, hence ; when p l ≪ 1, it scales with p 2, resulting in r p = O(η). Around p l = 1, there will be a performance transition from to r p = O(η). In practice, l can be adjusted in accordance with the laser quality and quantum-channel fluctuations. Note that l can also be adjusted during data postprocessing, offering flexibility for various environmental changes in real time. Generally, the whole pairing strategy can be adjusted through different realisations. Box 2 Simple pairing strategy. Practical issues and simulation The key rate of the mode-pairing scheme, as rigorously analysed in the Supplementary Note2, has a decoy-state MDI-QKD form: 7 where r p is the pairing rate contributed by each block, r s is the proportion of Z-pairs among all the generated location pairs (~1/8), q(1, 1) is the fraction of Z-pairs caused by single-photon-pair states ρ(1, 1) in which both Alice and Bob send single-photon states in the two paired modes, is the phase error rate of the detection caused by ρ(1, 1), f is the error-correction efficiency, and E(μ, μ),Z is the bit error rate of the sifted raw data. The fraction q(1, 1) and the phase error can be estimated using the decoy-state method40,43,44. A detailed estimation procedure for q(1, 1) and with the vacuum + weak decoy-state method is introduced in Supplementary Note3. During the key mapping step in Box1, the X-pair sifting condition θ a = θ b is impossible to fulfil exactly. This results in insufficient data for X-basis error rate estimation. To solve this problem, one can apply discrete phase randomisation45 such that θ a and θ b are chosen from a discrete set. We expect the discretisation effect to be negligible when the number of discrete phases is reasonably large, such as D = 16, similar to the situation in previous works on one-mode MDI-QKD46. Based on the above analysis, we simulate the asymptotic performance of the mode-pairing scheme under a typical symmetric quantum-channel model, using practical experimental parameter settings. We assign the maximal pairing interval l of the mode-pairing scheme as a value between 1 and 1 × 10 6, aiming to illustrate the dependence of the key rate on l. We also compare the key rate of the mode-pairing scheme with those of a typical two-mode scheme, time-bin encoding MDI-QKD23, and two one-mode schemes — PM-QKD46 and SNS-TFQKD47. The simulation results are shown in Fig.3. We set the misalignment error rate of the mode-pairing scheme to be the same as the one-mode schemes for a fair comparison. In Supplementary Note5, we show that the key-rate performance of the mode-pairing scheme is robust against misalignment errors. Even with a misalignment error rate of 15%, the mode-pairing scheme is able to surpass the repeaterless rate-transmittance bound with l = 2000. Here, we compare the asymptotic key-rate performance of all the schemes under the scenario of one-way local-operation and classical communication. The simulation formulas for these schemes are listed in Supplementary Note4. Recently, researches48,49 show that the key-rate performance of SNS-TFQKD can be further improved by introducing the two-way classical communication50,51. We will leave the advanced key distillation for future studies. Fig. 3. Asymptotic key-rate performance of the mode-pairing scheme. Open in a new tab The horizontal axis representing the total communication distance with a fibre loss of 0.2 dB/km and the vertical axis representing the key generation rate. a Main Panel: Performance comparison of the mode-pairing scheme (denoted by MP-QKD in the plot) with the decoy-state Bennett-Brassard 1984 (BB84)1,13,40, MDI-QKD16, PM-QKD25,46, SNS-TFQKD27,47 schemes and the repeaterless rate-transmittance bound (PLOB bound)5. Inset: The simulation parameters used in the key-rate plot, which are mainly from ref. 32. b Main Panel: The rate-distance dependence of the mode-pairing scheme with different maximal-pairing intervals l. Inset: The key rate with respect to the pairing interval l for a communication distance of 400 km. As shown in Fig.3a, the mode-pairing scheme with only neighbour pairing, l = 1, show a performance comparable to that of the original two-mode scheme. These two schemes have the same scaling property, i.e., R = O(η). The deviation is caused by an extra sifting factor in the mode-pairing scheme as a result of independent encoding. When the maximal pairing interval l is increased to 1 × 10 3, the key rate is significantly enhanced by 3 orders of magnitude compared to the l = 1 case, making it able to surpass the linear key-rate bound. If we further increase l above 1 × 10 5, then the mode-pairing scheme has a similar key rate to PM-QKD and SNS-TFQKD and a scaling property given by . In Fig.3b, we further compare the key-rate performance of the mode-pairing scheme under different settings for l. When l falls within the range of 1 to 1 × 10 6, the key rate of the mode-pairing scheme lies between the two extreme cases of O(η) and . The key-rate behaviour is dominated by the pairing rate given in Eq. (4). In typical optical experiments, the typical line width of a common commercial laser is 3 kHz (see for example, ref. 32). Hence, the coherence time of the laser is around 333 μs. In practice, the frequency fluctuation of the lasers will affect the stabilization of the phase. To test the feasibility of the mode-pairing scheme, we perform an interference experiment using a commercial optical communication system with a repetition rate of 625 MHz. The experiment detail is shown in Supplementary Note6. Based on the experimental data, we find that the phase coherence can be maintained well in a time interval of 5 μs, corresponding to l = 3000 ~ 4000. If we apply the state-of-the-art optical communication system with the repetition rate of 4 GHz37, we can realize a pairing interval over l = 20000. As an extra remark, our current discussion on the implementation of the mode-pairing scheme is based on the multiplexing of optical time-bin modes. Nonetheless, the proposed mode-pairing design is generic for the multiplexing of other optical degrees of freedom. For example, we can introduce frequency multiplexing. The optical modes with different frequencies are first prepared and interfered independently, i.e., only the pulses with the same frequency will be interfered. After the announcement of detection results, Alice and Bob then pair the locations with different frequencies during the post-processing. This can be used to increase the effective maximal pairing interval to an even larger value without the global phase locking. From Fig.3b we can see that the key rate of the mode-pairing scheme with l = 1 × 10 4 remains when η is smaller than 30 dB, corresponding to a communication distance of 300 km. The asymptotic key rate of the mode-pairing scheme is 3 to 5 orders of magnitude higher than that of the two-mode scheme. We remark that the decoherence effect caused by the optical-fibre channel is negligible compared to the laser coherence time. When the fibre length is around 500 km, the velocity of phase drift in the fibre is <10 rad/ms32, which can be calibrated using strong laser pulses without the need for real-time feedback control. As a result, the value of l depends only upon the local phase reference and not the communication distance. One advantage of the mode-pairing scheme is that it can be adapted to specific hardware conditions. In practice, optical systems may be unstable, causing the local phase reference to fluctuate rapidly. In this case, we can reduce the maximal pairing interval l and search for the optimal pairing strategy during the postprocessing procedure. As shown in the inset plot of Fig.3b, the key rate of the mode-pairing scheme first increases linearly with increasing l before saturating when l is larger than . In this case, Alice and Bob find successful detection within l locations with a high probability. Even when the optical system is unstable, the key rate can be nearly l times higher than that of the original time-bin MDI-QKD scheme when the value of l does not exceed . We remark that, with the original experimental apparatus used in time-bin MDI-QKD, one can directly enhance the key rate by a factor of ~100 using the mode-pairing scheme. On the other hand, we note that for a given communication distance, l does not need to be very large to reach the maximal key-rate performance. For example, when the distance reaches 200 km, a maximal pairing interval of l = 1000 is sufficient to achieve the optimal key-rate performance. We leave a detailed evaluation for future research. Discussion Based on a re-examination of the conventional two-mode MDI-QKD schemes and the recently proposed one-mode MDI-QKD schemes, we have developed a mode-pairing MDI-QKD scheme that retains the advantages of both, namely, achieving a high key rate with easy implementation. Since MDI-QKD schemes have the highest practical security level among the currently feasible QKD schemes, we expect the mode-pairing scheme paves the way for an optimal design for QKD, simultaneously enjoying high practicality, implementation security, and performance. There remain several interesting directions for future work. Natural follow-up questions lie in the statistical analysis of the mode-pairing scheme in the finite-data-size regime and efficient parameter estimation. Due to the photon-number-based property of the mode-pairing scheme, previous studies of the statistical analysis of two-mode MDI-QKD schemes52–54 can be readily extended to analyse the mode-pairing scheme. To improve the efficiency of data usage, Alice and Bob may perform parameter estimation before basis sifting in order to use all signals that were originally discarded. On the other hand, one could design a mode-pairing scheme using the X-basis for key generation and the Z-basis for parameter estimation. In this work, we employ a simple mode-pairing strategy based on pairing adjacent detection pulses. A more sophisticated pairing method might make bit and basis sifting more efficient. To improve the pairing strategy, Alice and Bob could reveal parts of the encoded intensity and phase information. For example, in the simple pairing strategy introduced in Box2, Alice and Bob reveal the bases of the generated data pairs immediately after locations i and j are paired. If their basis choices differ, Alice and Bob ‘unpair’ locations i and j, and seek the next good pairing location for location i until the basis choices match. To further enhance the performance, we could extend the mode-pairing design to other optical degrees of freedom, such as angular momentum and spectrum mode. Meanwhile, we could multiplex the usage of different degrees of freedom to enhance the repetition rate and extend the pairing interval l. Such multiplexing techniques would have additional benefits for the mode-pairing scheme. Suppose that we multiplex m quantum channels for a QKD task. In a normal setting, the key generation speed would be improved by a factor of m. For the mode-pairing scheme, in addition to this m-fold improvement, multiplexing would also introduce a larger pairing interval m l, since Alice and Bob would be able to pair quantum signals from different channels. A larger pairing interval m l would result in more paired signals and, hence, more key bits. Especially in the high-channel-loss regime where the distance between two clicked signals is large, the number of successful pairs becomes proportional to the maximum pairing interval m l. Thus, the key generation rate is proportional to m 2 in the high-channel-loss regime. Meanwhile, entanglement-based MDI-QKD schemes are essentially based on entanglement-swapping, which is the core design feature of quantum repeaters. The mode-pairing technique may help design a robust quantum repeater against a lossy channel. Note that our work shares similarities with the memory-assisted MDI-QKD protocol55 with quantum memories in the middle and with the all-photonic intercity MDI-QKD protocol56 with adaptive Bell-state measurement on the postselected photons. It is interesting to discuss the possibility of combining the mode-pairing design with an adaptive Bell-state measurement to tolerate more losses. Moreover, the mode-pairing scheme has a unique feature in that the key bits are determined not in the encoding or measurement steps but upon postprocessing, which is an approach that can be further explored in other quantum communication tasks, including continuous-variable schemes. Methods Source replacement of the encoding state The main idea of the security proof for the mode-pairing scheme is to introduce an entanglement-based scheme and reduce the security of the scheme to that of a traditional two-mode MDI-QKD scheme. To realise this, we perform a systematic source-replacement procedure57,58. Without loss of generality, in this subsection, we always assume the paired locations (i, j) to be (1, 2) to simplify the notations. For convenience in the security proof, we slightly modify the scheme described in Box1. First, we assume that the random phase of each mode is discretely chosen from a set of D phases, evenly distributed in [0, 2 π). We expect the corresponding correction term in the security analysis due to the discretisation effect to be negligible45,46. Second, in the security proof, we modify the phase encoding and postprocessing procedures, as shown in Table1. In the original scheme, Alice modulates A 1 and A 2 based on two random phases and , respectively. During the X-basis processing, she calculates the relative phase difference and splits it into an alignment angle θ a in the range of [0, π) and a raw key bit κ a. We modify these procedures as follows: in addition to the two random phases and , Alice also generates two bits and and applies extra phase modulations of and to A 1 and A 2, respectively. During the X-basis processing, she calculates the relative phase difference and directly announces it for alignment-angle sifting. In theSupplementary Information, we prove the equivalence of these two encoding methods. Table 1. Comparison of the phase encoding and postprocessing procedures of the mode-pairing scheme presented in the main text and the modified scheme considered in the security proof. \| \| Modulated phase \| X-basis postprocessing \| Sifting condition \| --- --- \| \| Original scheme \| \| \| θ a = θ b \| \| Modified scheme \| \| \| θ a − θ b = 0 or π \| Open in a new tab In the modified scheme, Alice introduces an extra π-phase modulation for the storage of a bit . This helps to decouple the phase randomisation and phase encoding analysis. With the modification above, Alice further generates a random bit and a random dit (d = D) j 1 in the first round. Based on the values of , and , she prepares the state 8 with . As shown in Fig.4, we substitute the encoding of random encoded information into the introduction of extra ancillary qubit and qudit systems labelled as , and . The purified encoding state is 9 Fig. 4. Source-replacement procedure for the mode-pairing scheme. Open in a new tab We substitute the encoding of all random encoded information into the introduction of purified ancillary systems. In Fig.4, we provide a specific state preparation procedure. The initial state is 10 Here Alice applies a controlled-phase gate with from the qudit to optical mode A 1. The controlled-phase gate is defined as 11 where a† and a are the creation and annihilation operators, respectively, of mode A 1. Alice also applies a controlled-phase gate from to A 1. In the entanglement-based mode-pairing scheme, Alice and Bob generate the composite encoding state defined in Eq. (9) in each round. They emit the optical modes to Charlie for interference. Based on Charlie’s announcement, they pair the locations and perform global operations on the corresponding ancillaries to generate raw key bits and useful parameters. In Fig.5, we list the global operations performed on Alice’s paired locations. Among them, the relative encoded intensity is used to determine the basis choice. The encoded intensity and the relative encoded phase are the raw key bits in the Z-basis and X-basis postprocessing, respectively. Fig. 5. The quantum operations and usage of Alice’s encoding states on two paired locations (1, 2). Open in a new tab There are 8 systems based on Alice's two paired locations. Among them, the two qudits and are measured to obtain the overall photon number k a and the relative phase θ a of two optical modes A 1 and A 2. The two qubits and are measured to obtain the relative phase, which is the raw key bit in the X-basis. Another two qubits and are measured to obtain the encoded intensity in A 1 and the relative encoded intensity, which are used for the key mapping on the Z-basis and the basis assignment, respectively. A key point in our security proof is that we replace the random phases and register them into purified systems and . This enables us to define a global measurement M(k, θ) on and to simultaneously obtain the overall photon number and the relative phase information encoded in optical modes A 1 and A 2. The construction of M(k, θ) is described in Supplementary Note1. With the introduction of the purified systems and and the existence of the global measurement M(k, θ), Alice (same for Bob) is able to determine at which two locations to perform the global photon number measurement after Charlie’s announcement. With this measurement, Alice and Bob can further reduce the encoding state to a two-mode scheme. The detailed security proof is provided in Supplementary Note2. Mode-pairing scheme with decoy states Here, we present the mode-pairing scheme with an extra decoy intensity ν to estimate the parameters q 11 and . Of course, more decoy intensities can be applied in a similar manner. State preparation: In the i-th round (i = 1, 2, . . . , N), Alice prepares a coherent state in optical mode A i with an intensity randomly chosen from {0, ν, μ} (0 <ν<μ< 1) and a phase uniformly chosen from the set . She records and for later use. Likewise, Bob chooses and randomly and prepares in mode B i. Measurement: (Same as Step 2 in Box1.) Alice and Bob send modes A i and B i to Charlie, who performs the single-photon interference measurement. Charlie announces the clicks of the detectors L and/or R. Alice and Bob repeat the above two steps N times; then, they perform the following data postprocessing procedures: Mode pairing: (Same as Step 3 in Box1.) For all rounds with successful detection (L or R clicks), Alice and Bob establish a strategy for grouping two clicked rounds as a pair. A specific pairing strategy is introduced in Box2. Basis sifting: Based on the intensities of two grouped rounds, Alice labels the ‘basis’ of the data pair as: Z if one of the intensities is 0 and the other is nonzero; X if both of the intensities are the same and nonzero; or ‘0’ if the intensities are (0, 0), which will be reserved for decoy estimation of both the Z and X bases; or ‘discard’ when both intensities are nonzero and not equal. See also Table2 for the basis assignment. Alice and Bob announce the basis (X, Z, ‘0’, or ‘discard’) and the sum of the intensities for each location pair i, j. If the announced bases are the same and no ‘discard’ state occurs, they record the pair basis and maintain the data pairs; if one of the announced bases is ‘0’ and the other one is X(Z), they record the pair basis as X(Z) and keep the data pairs; if both of the announced bases are ‘0’, they record the pair basis as ‘0’ and maintain the data pairs; and otherwise, they discard the data. See also Table3 for the basis-sifting strategy. Key mapping: (Same as Step 5 in Box1) For each Z-pair at locations i and j, Alice sets her key to κ a = 0 if the intensity of the i-th pulse is and to κ a = 1 if . For each X-pair at locations i and j, the key is extracted from the relative phase , where the raw key bit is and the alignment angle is . Similarly, Bob also assigns his raw key bit κ b and determines θ b. For the X-pairs, Alice and Bob announce the alignment angles θ a and θ b. If θ a = θ b, they keep the data pairs; otherwise, they discard them. Parameter estimation: Alice and Bob estimate the quantum bit error rate of the raw key data in Z-pairs with overall intensities of . They use Z-pairs with different intensity settings to estimate the clicked single-photon fraction q 11 using the decoy-state method, and the X-pairs are used to estimate the single-photon phase error rate . Specially, q 11 and are estimated via the decoy-state method introduced in Supplementary Note3. Key distillation: (Same as Step 7 in Box1.) Alice and Bob use the Z-pairs to generate a key. They perform error correction and privacy amplification in accordance with q 11, and . Table 2. Alice’s (or Bob’s) basis assignment on the paired locations i and j. \| \| μ i \| \| \| \| --- --- \| \| \| 0 \| ν \| μ \| \| μ j \| \| \| \| \| \| 0 \| \| ‘0’ \| Z \| Z \| \| ν \| \| Z \| X \| ‘discard’ \| \| μ \| \| Z \| ‘discard’ \| X \| Open in a new tab Based on the intensities μ i and μ j on the i-th and j-th location, Alice (or Bob) assign the basis to be either X, Z, ‘0’, or ‘discard’. Table 3. Alice and Bob’s basis sifting procedure on the paired locations i and j. \| \| Alice \| \| \| \| --- --- \| \| \| ‘0’ \| X \| Z \| \| Bob \| \| \| \| \| \| ‘0’ \| \| ‘0’ \| X \| Z \| \| X \| \| X \| X \| ‘discard’ \| \| Z \| \| Z \| ‘discard’ \| Z (key generation) \| Open in a new tab Based on the assigned basis, Alice and Bob decide whether to keep the data for Z-basis key generation, Z(X)-basis parameter estimation, or discard the data. Mode-pairing-efficiency calculation We calculate the expected pairing number r p(p, l) that corresponds to the simple mode-pairing strategy in Box2, which is related to the average click probability p during each round, and the maximal pairing interval l. For calculation convenience, we assume that in addition to the front and rear locations (F k, R k) of the k-th pair, Alice and Bob also record the starting location S k, which indicates the location at which the first successful detection signal occurs during the pairing procedure for the k-th pair. If the second successful detection signal R k is found within the next l locations, then F k = S k; otherwise, F k will be larger than S k. Let G k ≔ S k+1 − S k denote a random variable that reflects the location gap between the k-th and (k + 1)-th starting pulses. Then the expected pairing number per pulse is given by 12 Hence, we need to calculate only the expectation value of G k. First, we split it into two parts, 13 where H k: = R k − S k and . Hence, 14 It is easy to show that obeys a geometric distribution, 15 Then, the expectation value is . The calculation of the pulse interval H k is more complex. Suppose that we already know the expectation value ; now we calculate the expectation value conditioned on the distance between the starting point and the following click. We have 16 Therefore, 17 We have 18 therefore, 19 Note added to proof After we submitted our work for reviewing, we became aware of a relevant work by Xie et al.59, who consider a similar MDI-QKD protocol that match the clicked data to generate key information. Under the assumption that the single-photon distributions in all the Charlie’s successful detection events are independent and identically distributed, the authors simulate the performance of the protocol and show its ability to break the repeaterless rate-transmittance bound. Supplementary information Supplementary Information (1.6MB, pdf) Acknowledgements We especially thank Norbert Lütkenhaus for the helpful discussions on the security analysis, thorough proofreading, and beneficial suggestions on the manuscript presentation. We thank Yizhi Huang, Guoding Liu, Zhenhuan Liu, Tian Ye, Junjie Chen, Minbo Gao, and Xingjian Zhang for the helpful discussion on the pairing rate calculation and general comments on the presentation. We especially thank Hao-Tao Zhu and Teng-Yun Chen for providing us with some preliminary results showing the phase stabilization after removing phase-locking in the mode-pairing scheme. This work was supported by the National Natural Science Foundation of China Grants No. 11875173 and No. 12174216 and the National Key Research and Development Program of China Grants No. 2019QY0702 and No. 2017YFA0303903. Author contributions X.M. conceived the research. P.Z., H.Z. and X.M. designed the protocol. X.M., P.Z., W.W. and H.Z. finished the security analysis. P.Z. and W.W. performed the protocol analysis and numerical simulation. All authors contributed extensively to the preparation of this manuscript. Peer review Peer review information Nature Communications thanks the anonymous reviewer(s) for their contribution to the peer review of this work. Data availability The methods to generate the data in the plots are provided inSupplementary Information. The data that support the plots within this paper and other findings of this study are available from the corresponding authors upon reasonable request. Code availability The detailed simulation methods for the plots are provided inSupplementary Information. The specific code that support the plots within this paper and other findings of this study are available from the corresponding authors upon reasonable request. Competing interests The authors declare no competing interests. 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9260	https://mmids-textbook.github.io/chap02_ls/03_orthog/roch-mmids-ls-orthog.html	2.3. Geometry of least squares: the orthogonal projection — MMiDS Textbook Skip to main content Back to top - [x] - [x] Ctrl+K Search Ctrl+K MATHEMATICAL METHODS in DATA SCIENCE 1. Introduction: a first data science problem- [x] 1.1. Motivating example: identifying penguin species 1.2. Background: quick refresher of matrix algebra, differential calculus, and elementary probability 1.3. Clustering: an objective, an algorithm and a guarantee 1.4. Some observations about high-dimensional data 1.5. Exercises 1.6. Online supplementary material 2. Least squares: geometric, algebraic, and numerical aspects- [x] 2.1. Motivating example: predicting sales 2.2. Background: review of vector spaces and matrix inverses 2.3. Geometry of least squares: the orthogonal projection 2.4. QR decomposition and Householder transformations 2.5. Application: regression analysis 2.6. Exercises 2.7. Online supplementary material 3. Optimization theory and algorithms- [x] 3.1. Motivating example: analyzing customer satisfaction 3.2. Background: review of differentiable functions of several variables 3.3. Optimality conditions 3.4. Convexity 3.5. Gradient descent and its convergence analysis 3.6. Application: logistic regression 3.7. Exercises 3.8. Online supplementary material 4. Singular value decomposition- [x] 4.1. Motivating example: visualizing viral evolution 4.2. Background: review of matrix rank and spectral decomposition 4.3. Approximating subspaces and the SVD 4.4. Power iteration 4.5. Application: principal components analysis 4.6. Further applications of the SVD: low-rank approximations and ridge regression 4.7. Exercises 4.8. Online supplementary material 5. Spectral graph theory- [x] 5.1. Motivating example: uncovering social groups 5.2. Background: basic concepts in graph theory 5.3. Variational characterization of eigenvalues 5.4. Spectral properties of the Laplacian matrix 5.5. Application: graph partitioning via spectral clustering 5.6. Erdős-Rényi random graph and stochastic blockmodel 5.7. Exercises 5.8. Online supplementary material 6. Probabilistic models: from simple to complex- [x] 6.1. Motivating example: tracking location 6.2. Background: introduction to parametric families and maximum likelihood estimation 6.3. Modeling more complex dependencies 1: using conditional independence 6.4. Modeling more complex dependencies 2: marginalizing out an unobserved variable 6.5. Application: linear-Gaussian models and Kalman filtering 6.6. Exercises 6.7. Online supplementary material 7. Random walks on graphs and Markov chains- [x] 7.1. Motivating example: discovering mathematical topics 7.2. Background: elements of finite Markov chains 7.3. Limit behavior 1: stationary distributions 7.4. Limit behavior 2: convergence to equilibrium 7.5. Application: random walks on graphs and PageRank 7.6. Further applications: Gibbs sampling and generating images 7.7. Exercises 7.8. Online supplementary material 8. Neural networks, backpropagation and stochastic gradient descent- [x] 8.1. Motivating example: classifying natural images 8.2. Background: Jacobian, chain rule, and a brief introduction to automatic differentiation 8.3. Building blocks of AI 1: backpropagation 8.4. Building blocks of AI 2: stochastic gradient descent 8.5. Building blocks of AI 3: neural networks 8.6. Exercises 8.7. Online supplementary material Repository Open issue .ipynb .pdf Geometry of least squares: the orthogonal projection Contents 2.3.1. A key concept: orthogonality 2.3.2. Orthogonal projection 2.3.3. Orthogonal complement 2.3.4. Overdetermined systems 2.3. Geometry of least squares: the orthogonal projection# We consider the following problem: we are given A∈R n×m an n×m matrix and b∈R n a vector. We are looking to solve the system A x≈b. In the special case where A is invertible, a unique exact solution exists. In general, however, a solution may not exist or may not be unique. We focus here on the over-determined case where the former situation generically occurs. We begin by rewieving the concept of orthogonality. 2.3.1. A key concept: orthogonality# Orthogonality plays a key role in linear algebra for data science thanks to its computational properties and its connection to the least-squares problem. DEFINITION(Orthogonality) Vectors u and v in R n (as column vectors) are orthogonal if their inner product is zero ⟨u,v⟩=u T v=∑i=1 n u i v i=0. ♮ Orthogonality has important implications. The following classical result will be useful below. Throughout, we use ‖u‖ for the Euclidean norm of u. THEOREM(Pythagoras) Let u,v∈R n be orthogonal. Then ‖u+v‖2=‖u‖2+‖v‖2. ♯ Proof: Using ‖w‖2=⟨w,w⟩, we get ‖u+v‖2=⟨u+v,u+v⟩=⟨u,u⟩+2⟨u,v⟩+⟨v,v⟩=‖u‖2+‖v‖2. ◻ An application of Pythagoras’ Theorem is a proof of the Cauchy-Schwarz Inequality. Proof: (Cauchy-Schwarz) Let q=v‖v‖ be the unit vector in the direction of v. We want to show \|⟨u,q⟩\|≤‖u‖. Decompose u as follows: u=⟨u,q⟩q+{u−⟨u,q⟩q}. The two terms on the right-hand side are orthogonal: ⟨⟨u,q⟩q,u−⟨u,q⟩q⟩=⟨u,q⟩2−⟨u,q⟩2⟨q,q⟩=0. So Pythagoras gives ‖u‖2=‖⟨u,q⟩q‖2+‖u−⟨u,q⟩q‖2≥‖⟨u,q⟩q‖2=⟨u,q⟩2. Taking a square root gives the claim. ◻ Orthonormal basis expansion To begin to see the power of orthogonality, consider the following. A list of vectors {u 1,…,u m} is an orthonormal list if the u i’s are pairwise orthogonal and each has norm 1, that is, for all i and all j≠i, we have ‖u i‖=1 and ⟨u i,u j⟩=0. Alternatively, ⟨u i,u j⟩={1 if i=j 0 if i≠j. LEMMA(Properties of Orthonormal Lists) Let {u 1,…,u m} be an orthonormal list. Then: for any α j∈R, j=1,…,m, we have ‖∑j=1 m α j u j‖2=∑j=1 m α j 2, the vectors {u 1,…,u m} are linearly independent. ♭ Proof: For 1., using that ‖x‖2=⟨x,x⟩, we have ‖∑j=1 m α j u j‖2=⟨∑i=1 m α i u i,∑j=1 m α j u j⟩=∑i=1 m α i⟨u i,∑j=1 m α j u j⟩=∑i=1 m∑j=1 m α i α j⟨u i,u j⟩=∑i=1 m α i 2 where we used orthonormality in the last equation, that is, ⟨u i,u j⟩ is 1 if i=j and 0 otherwise. For 2., suppose ∑i=1 m β i u i=0, then we must have by 1. that ∑i=1 m β i 2=0. That implies β i=0 for all i. Hence the u i’s are linearly independent. ◻ Given a basis {u 1,…,u m} of U, we know that: for any w∈U, w=∑i=1 m α i u i for some α i’s. It is not immediately obvious in general how to find the α i’s – one must solve a system of linear equations. In the orthonormal case, however, there is a formula. We say that the basis {u 1,…,u m} is orthonormal if it forms an orthonormal list. THEOREM(Orthonormal Expansion) Let q 1,…,q m be an orthonormal basis of U and let w∈U. Then w=∑j=1 m⟨w,q j⟩q j. ♯ Proof: Because w∈U, w=∑i=1 m α i q i for some α i. Take the inner product with q j and use orthonormality: ⟨w,q j⟩=⟨∑i=1 m α i q i,q j⟩=∑i=1 m α i⟨q i,q j⟩=α j. Hence, we have determined all α j’s in the basis expansion of w. ◻ EXAMPLE: Consider again the linear subspace W=span(w 1,w 2,w 3), where w 1=(1,0,1), w 2=(0,1,1), and w 3=(1,−1,0). We have shown that in fact span(w 1,w 2,w 3)=span(w 1,w 2), as w 1,w 2 form a basis of W. On the other hand, ⟨w 1,w 2⟩=(1)(0)+(0)(1)+(1)(1)=0+0+1=1≠0 so this basis is not orthonormal. Indeed, an orthonormal list is necessarily an independent list, but the opposite may not hold. To produce an orthonormal basis of W, we can first proceed by normalizing w 1 q 1=w 1‖w 1‖=w 1 1 2+0 2+1 2=1 2 w 1. Then ‖q 1‖=1 since, in general, by absolute homogeneity of the norm ‖w 1‖w 1‖‖=1‖w 1‖‖w 1‖=1. We then seek a second basis vector. It must satisfy two conditions in this case: it must be of unit norm and be orthogonal to q 1; and w 2 must be a linear combination of q 1 and q 2. The latter condition guarantees that span(q 1,q 2)=span(w 1,w 2). (Formally, that would imply only that span(w 1,w 2)⊆span(q 1,q 2). In this case, it is easy to see that the containment must go in the opposite direction as well. Why?) The first condition translates into 1=‖q 2‖2=q 21 2+q 22 2+q 23 2, where q 2=(q 21,q 22,q 23), and 0=⟨q 1,q 2⟩=1 2[1⋅q 21+0⋅q 22+1⋅q 23]=1 2[q 21+q 23]. That is, simplifying the second display and plugging into the first, q 23=−q 21 and q 22=1−2 q 21 2. The second condition translates into: there is β 1,β 2∈R such that w 2=(0 1 1)=β 1 q 1+β 2 q 2=β 1 1 2(1 0 1)+β 2(q 21 1−2 q 21 2−q 21). The first entry gives β 1/2+β 2 q 21=0 while the third entry gives β 1/2−β 2 q 21=1. Adding up the equations gives β 1=1/2. Plugging back into the first one gives β 2=−1/(2 q 21). Returning to the equation for w 2, we get from the second entry 1=−1 2 q 21 1−2 q 21 2. Rearranging and taking a square, we want the negative solution to 4 q 21 2=1−2 q 21 2, that is, q 21=−1/6. Finally, we get q 23=−q 21=1/6 and q 22=1−2 q 21 2=1−1/3=2/3=2/6. To summarize, we have q 1=1 2(1 0 1),q 2=1 6(−1 2 1). We confirm that ⟨q 1,q 2⟩=1 2 6[(1)(−1)+(0)(2)+(1)(1)]=0 and ‖q 2‖2=(−1 6)2+(2 6)2+(1 6)2=1 6+4 6+1 6=1. We can use the Orthonormal Expansion Theorem to write w 2 as a linear combination of q 1 and q 2. The inner products are ⟨w 2,q 1⟩=0(1 2)+1(0 2)+1(1 2)=1 2, ⟨w 2,q 2⟩=0(−1 6)+1(2 6)+1(1 6)=3 6. So w 2=1 2 q 1+3 6 q 2. Check it! Try w 3. ⊲ Gram-Schmidt We have shown that working with orthonormal bases is desirable. What if we do not have one? We could try to construct one by hand as we did in the previous example. But there are better ways. We review the Gram-Schmidt algorithm in an upcoming section, which will imply that every linear subspace has an orthonormal basis. That is, we will prove the following theorem. THEOREM(Gram-Schmidt) Let a 1,…,a m be linearly independent. Then there exists an orthonormal basis q 1,…,q m of span(a 1,…,a m). ♯ But, first, we will need to define the orthogonal projection, which will play a key role in our applications. This is done next. 2.3.2. Orthogonal projection# To solve the overdetermined case, i.e., when n>m, we consider the following more general problem first. We have a linear subspace U⊆R n and a vector v∉U. We want to find the vector p in U that is closest to v in Euclidean norm, that is, we want to solve min p∈U‖p−v‖. EXAMPLE: Consider the two-dimensional case with a one-dimensional subspace, say U=span(u 1) with ‖u 1‖=1. The geometrical intuition is in the following figure. The solution p=v∗ has the property that the difference v−v∗ makes a right angle with u 1, that is, it is orthogonal to it. Letting v∗=α∗u 1, the geometrical condition above translates into 0=⟨u 1,v−v∗⟩=⟨u 1,v−α∗u 1⟩=⟨u 1,v⟩−α∗⟨u 1,u 1⟩=⟨u 1,v⟩−α∗ so v∗=⟨u 1,v⟩u 1. By Pythagoras’ Theorem, we then have for any α∈R ‖v−α u 1‖2=‖v−v∗+v∗−α u 1‖2=‖v−v∗+(α∗−α)u 1‖2=‖v−v∗‖2+‖(α∗−α)u 1‖2≥‖v−v∗‖2, where we used that v−v∗ is orthogonal to u 1 (and therefore (α∗−α)u 1) on the third line. That confirms the optimality of v∗. The argument in this example carries through in higher dimension, as we show next. ⊲ DEFINITION(Orthogonal Projection on an Orthonormal List) Let q 1,…,q m be an orthonormal list. The orthogonal projection of v∈R n on {q i}i=1 m is defined as proj{q i}i=1 m v=∑j=1 m⟨v,q j⟩q j. ♮ THEOREM(Orthogonal Projection) Let U⊆V be a linear subspace and let v∈R n. Then: a) There exists a unique solution p∗ to min p∈U‖p−v‖. We denote it by p∗=proj U v and refer to it as the orthogonal projection of v onto U. b) The solution p∗∈U is characterized geometrically by (∗)⟨v−p∗,u⟩=0,∀u∈U. c) For any orthonormal basis q 1,…,q m of U, proj U v=proj{q i}i=1 m v. ♯ Proof: Let p∗ be any vector in U satisfying (∗). We show first that it necessarily satisfies (∗∗)‖p∗−v‖≤‖p−v‖,∀p∈U. Note that for any p∈U the vector u=p−p∗ is also in U. Hence by (∗) and Pythagoras, ‖p−v‖2=‖p−p∗+p∗−v‖2=‖p−p∗‖2+‖p∗−v‖2≥‖p∗−v‖2. Furthermore, equality holds only if ‖p−p∗‖2=0 which holds only if p=p∗ by the point-separating property of the Euclidean norm. Hence, if such a vector p∗ exists, it is unique. Next, we show that any minimizer must satisfy (∗). Let p∗ be a minimizer and suppose, for contradiction, that (∗) does not hold. Then there exists u∈U with ⟨v−p∗,u⟩=c≠0. Consider p t=p∗+t u for small t. Then: ‖p t−v‖2=‖(p∗−v)+t u‖2=‖p∗−v‖2+2 t⟨v−p∗,u⟩+t 2‖u‖2=‖p∗−v‖2+2 t c+t 2‖u‖2. For small t with appropriate sign, this is smaller than ‖p∗−v‖2, contradicting minimality. It remains to show that there is at least one vector in U satisfying (∗). By the Gram-Schmidt Theorem, the linear subspace U has an orthonormal basis q 1,…,q m. By definition, proj{q i}i=1 m v∈span({q i}i=1 m)=U. We show that proj{q i}i=1 m v satisfies (∗). We can write any u∈U as ∑i=1 m α i q i with α i=⟨u,q i⟩. So, using this representation, we get ⟨v−∑j=1 m⟨v,q j⟩q j,∑i=1 m α i q i⟩=∑i=1 m⟨v,q i⟩α i−∑j=1 m∑i=1 m α i⟨v,q j⟩⟨q j,q i⟩=∑i=1 m⟨v,q i⟩α i−∑j=1 m α j⟨v,q j⟩=0, where we used the orthonormality of the q j’s on the second line. ◻ EXAMPLE:(continued) Consider again the linear subspace W=span(w 1,w 2,w 3), where w 1=(1,0,1), w 2=(0,1,1), and w 3=(1,−1,0). We have shown that q 1=1 2(1 0 1),q 2=1 6(−1 2 1), is an orthonormal basis. Let w 4=(0,0,2). It is immediate that w 4∉span(w 1,w 2) since vectors in that span are of the form (x,y,x+y) for some x,y∈R. We can however compute the orthogonal projection w 4 onto W. The inner products are ⟨w 4,q 1⟩=0(1 2)+0(0 2)+2(1 2)=2 2, ⟨w 4,q 2⟩=0(−1 6)+0(2 6)+2(1 6)=2 6. So proj W w 4=2 2 q 1+2 6 q 2=(2/3 2/3 4/3). As a sanity check, note that w 4∈W since its third entry is equal to the sum of its first two entries. ⊲ The map proj U is linear, that is, proj U(α x+y)=α proj U x+proj U y for all α∈R and x,y∈R n. Indeed, proj U(α x+y)=∑j=1 m⟨α x+y,q j⟩q j=∑j=1 m{α⟨x,q j⟩+⟨y,q j⟩}q j=α proj U x+proj U y. Any linear map from R n to R n can be encoded as an n×n matrix P. Let Q=(\|\|q 1…q m\|\|) and note that computing Q T v=(⟨v,q 1⟩⋯⟨v,q m⟩) lists the coefficients in the expansion of proj U v over the basis q 1,…,q m. Hence we see that P=Q Q T. Indeed, for any vector v, P v=Q Q T v=Q[Q T v]. So the output is a linear combination of the columns of Q (i.e., the q i’s) where the coefficients are the entries of the vector in square brackets Q T v. EXAMPLE:(continued) Consider again the linear subspace W=span(w 1,w 2,w 3), where w 1=(1,0,1), w 2=(0,1,1), and w 3=(1,−1,0), with orthonormal basis q 1=1 2(1 0 1),q 2=1 6(−1 2 1). Then orthogonal projection onto W can be written in matrix form as follows. The matrix Q is Q=(1/2−1/6 0 2/6 1/2 1/6). Then Q Q T=(1/2−1/6 0 2/6 1/2 1/6)(1/2 0 1/2−1/6 2/6 1/6)=(2/3−1/3 1/3−1/3 2/3 1/3 1/3 1/3 2/3). So the projection of w 4=(0,0,2) is (2/3−1/3 1/3−1/3 2/3 1/3 1/3 1/3 2/3)(0 0 2)=(2/3 2/3 4/3), as previously computed. ⊲ The matrix P=Q Q T is not to be confused with Q T Q=(⟨q 1,q 1⟩⋯⟨q 1,q m⟩⟨q 2,q 1⟩⋯⟨q 2,q m⟩⋮⋱⋮⟨q m,q 1⟩⋯⟨q m,q m⟩)=I m×m where I m×m denotes the m×m identity matrix. This follows from the fact that the q i’s are orthonormal. EXAMPLE: Let q 1,…,q n be an orthonormal basis of R n and form the matrix Q=(\|\|q 1…q n\|\|). We show that Q−1=Q T. We just pointed out that Q T Q=(⟨q 1,q 1⟩⋯⟨q 1,q n⟩⟨q 2,q 1⟩⋯⟨q 2,q n⟩⋮⋱⋮⟨q n,q 1⟩⋯⟨q n,q n⟩)=I n×n where I n×n denotes the n×n identity matrix. In the other direction, we claim that Q Q T=I n×n as well. Indeed the matrix Q Q T is the orthogonal projection on the span of the q i’s, that is, R n. By the Orthogonal Projection Theorem, the orthogonal projection Q Q T v finds the closest vector to v in the span of the q i’s. But that span contains all vectors, including v, so we must have Q Q T v=v. Since this holds for all v∈R n, the matrix Q Q T is the identity map and we have proved the claim. ⊲ Matrices that satisfy Q T Q=Q Q T=I n×n are called orthogonal matrices. DEFINITION(Orthogonal Matrix) A square matrix Q∈R m×m is orthogonal if Q T Q=Q Q T=I m×m. ♮ KNOWLEDGE CHECK: Let Z be a linear subspace of R n and let v∈R n. Show that ‖proj Z v‖2≤‖v‖2. [Hint: Use the geometric characterization.] ✓ 2.3.3. Orthogonal complement# Before returning to overdetermined systems, we take a little detour to derive a consequence of the orthogonal projection that will be useful later. The Orthogonal Projection Theorem implies that any v∈R n can be decomposed into its orthogonal projection onto U and a vector orthogonal to it. DEFINITION(Orthogonal Complement) Let U⊆R n be a linear subspace. The orthogonal complement of U, denoted U⊥, is defined as U⊥={w∈R n:⟨w,u⟩=0,∀u∈U}. ♮ EXAMPLE: Continuing a previous example, we compute the orthogonal complement of the linear subspace W=span(w 1,w 2,w 3), where w 1=(1,0,1), w 2=(0,1,1), and w 3=(1,−1,0). One way to proceed is to find all vectors that are orthogonal to the orthonormal basis q 1=1 2(1 0 1),q 2=1 6(−1 2 1). We require 0=⟨u,q 1⟩=u 1(1 2)+u 2(0 2)+u 3(1 2)=u 1+u 3 2, 0=⟨u,q 2⟩=u 1(−1 6)+u 2(2 6)+u 3(1 6)=−u 1+2 u 2+u 3 6. The first equation implies u 3=−u 1, which after replacing into the second equation and rearranging gives u 2=u 1. So all vectors of the form (u 1,u 1,−u 1) for some u 1∈R are orthogonal to all of W. This is a one-dimensional linear subspace. We can choose an orthonormal basis by finding a solution to 1=(u 1)2+(u 1)2+(−u 1)2=3 u 1 2, Take u 1=1/3, that is, let q 3=1 3(1 1−1). Then we have W⊥=span(q 3). ⊲ LEMMA(Orthogonal Decomposition) Let U⊆R n be a linear subspace and let v∈R n. Then v can be decomposed as proj U v+(v−proj U v) where proj U v∈U and (v−proj U v)∈U⊥. Moreover, this decomposition is unique in the following sense: if v=u+u⊥ with u∈U and u⊥∈U⊥, then u=proj U v and u⊥=v−proj U v. ♭ Proof: The first part is an immediate consequence of the Orthogonal Projection Theorem. For the second part, assume v=u+u⊥ with u∈U and u⊥∈U⊥. Subtracting v=proj U v+(v−proj U v), we see that (∗)0=w 1+w 2 with w 1=u−proj U v∈U,w 2=u⊥−(v−proj U v)∈U⊥. If w 1=w 2=0, we are done. Otherwise, they must both be nonzero by (∗). Further, by the Properties of Orthonormal Lists, w 1 and w 2 must be linearly independent. But this is contradicted by the fact that w 2=−w 1 by (∗). ◻ Figure: Orthogonal decomposition (Source) ⋈ Formally, the Orthogonal Decomposition Lemma states that R n is a direct sum of any linear subspace U and of its orthogonal complement U⊥: that is, any vector v∈R n can be written uniquely as v=u+u⊥ with u∈U and u⊥∈U⊥. This is denoted R n=U⊕U⊥. Let a 1,…,a ℓ be an orthonormal basis of U and b 1,…,b k be an orthonormal basis of U⊥. By definition of the orthogonal complement, the list L={a 1,…,a ℓ,b 1,…,b k} is orthonormal, so it forms a basis of its span. Because any vector in R n can be written as a sum of a vector from U and a vector from U⊥, all of R n is in the span of L. It follows from the Dimension Theorem that n=ℓ+k, that is, dim(U)+dim(U⊥)=n. 2.3.4. Overdetermined systems# In this section, we discuss the least-squares problem. Let again A∈R n×m be an n×m matrix with linearly independent columns and let b∈R n be a vector. We are looking to solve the system A x≈b. If n=m, we can use the matrix inverse to solve the system (provided A is nonsingular of course). But we are interested in the overdetermined case, i.e. when n>m: there are more equations than variables. We cannot use the matrix inverse then. Indeed, because the columns do not span all of R n, there is a vector b∈R n that is not in the column space of A. A natural way to make sense of the overdetermined problem is to cast it as the linear least-squares problem min x∈R m‖A x−b‖2. In words, we look for the best-fitting solution under the squared Euclidean norm. Equivalently, writing A=(\|\|a 1…a m\|\|)=(a 11⋯a 1 m a 21⋯a 2 m⋮⋱⋮a n 1⋯a n m)and b=(b 1⋮b n) we seek a linear combination of the columns of A that minimizes the objective ‖∑j=1 m x j a j−b‖2=∑i=1 n(∑j=1 m a i j x j−b i)2. We have already solved a closely related problem when we introduced the orthogonal projection. We make the connection explicit next. THEOREM(Normal Equations) Let A∈R n×m be an n×m matrix with n≥m and let b∈R n be a vector. A solution x∗ to the linear least-squares problem min x∈R m‖A x−b‖2 satisfies the normal equations A T A x∗=A T b. If further the columns of A are linearly independent, then there exists a unique solution x∗. ♯ Proof idea: Apply our characterization of the orthogonal projection onto the column space of A. Proof: Let U=col(A)=span(a 1,…,a m). By the Orthogonal Projection Theorem, the orthogonal projection p∗=proj U b of b onto U is the unique, closest vector to b in U, that is, p∗=arg⁡min p∈U‖p−b‖=arg⁡min p∈U‖p−b‖2, where we used the Composing with a Non-Decreasing Function Lemma to justify taking a square in the rightmost expression. Because p∗ is in U=col(A), it must be of the form p∗=A x∗. This establishes that x∗ is a solution to the linear least-squares problem in the statement. By the Orthogonal Projection Theorem, it must satisfy ⟨b−A x∗,u⟩=0 for all u∈U. Because the columns a i are in U, that implies that 0=⟨b−A x∗,a i⟩=a i T(b−A x∗),∀i∈[m]. Stacking up these equations gives in matrix form A T(b−A x∗)=0, as claimed (after rearranging). Important observation: While we have shown that p∗ is unique (by the Orthogonal Projection Theorem), it is not clear at all that x∗ (i.e., the linear combination of columns of A corresponding to p∗) is unique. We have seen in a previous example that, when A has full column rank, the matrix A T A is invertible. That implies the uniqueness claim. ◻ NUMERICAL CORNER: To solve a linear system in NumPy, use numpy.linalg.solve. As an example, we consider the overdetermined system with A=(1 0 0 1 1 1)and b=(0 0 2). We use numpy.ndarray.T for the transpose and @ for matrix-matrix or matrix-vector product. w1 = np.array([1., 0., 1.]) w2 = np.array([0., 1., 1.]) A = np.stack((w1, w2),axis=-1) b = np.array([0., 0., 2.]) x = LA.solve(A.T @ A, A.T @ b) print(x) [0.66666667 0.66666667] We can also use numpy.linalg.lstsq directly on the overdetermined system to compute the least-square solution. x = LA.lstsq(A, b) print(x) [0.66666667 0.66666667] ⊴ Self-assessment quiz (with help from Claude, Gemini, and ChatGPT) 1 Let q 1,…,q m be an orthonormal list of vectors in R n. Which of the following is the orthogonal projection of a vector v∈R n onto span(q 1,…,q m)? a) ∑i=1 m q i b) ∑i=1 m⟨v,q i⟩ c) ∑i=1 m⟨v,q i⟩q i d) ∑i=1 m⟨q i,q i⟩v 2 According to the Normal Equations Theorem, what condition must a solution x∗ to the linear least squares problem satisfy? a) A T A x∗=b b) A T A x∗=A T b c) A x∗=A T b d) A x∗=b 3 Which property characterizes the orthogonal projection proj U v of a vector v onto a subspace U? a) v−proj U v is a scalar multiple of v. b) v−proj U v is orthogonal to v. c) v−proj U v is orthogonal to U. d) proj U v is always the zero vector. 4 What is the interpretation of the linear least squares problem A x≈b in terms of the column space of A? a) Finding the exact solution x such that A x=b. b) Finding the vector x that makes the linear combination A x of the columns of A as close as possible to b in Euclidean norm. c) Finding the orthogonal projection of b onto the column space of A. d) Finding the orthogonal complement of the column space of A. 5 Which matrix equation must hold true for a matrix Q to be orthogonal? a) Q Q T=0. b) Q Q T=I. c) Q T Q=0. d) Q T=Q. Answer for 1: c. Justification: This is the definition of the orthogonal projection onto an orthonormal list given in the text. Answer for 2: b. Justification: The Normal Equations Theorem states that a solution x∗ to the linear least squares problem satisfies A T A x∗=A T b. Answer for 3: c. Justification: The text states that the orthogonal projection proj U v has the property that “the difference v−proj U v is orthogonal to U.” Answer for 4: b. Justification: The text defines the linear least squares problem as seeking a linear combination of the columns of A that minimizes the distance to b in Euclidean norm. Answer for 5: b. Justification: The text states that an orthogonal matrix Q must satisfy Q Q T=I. previous 2.2. Background: review of vector spaces and matrix inversesnext 2.4. QR decomposition and Householder transformations Contents 2.3.1. A key concept: orthogonality 2.3.2. Orthogonal projection 2.3.3. Orthogonal complement 2.3.4. Overdetermined systems By Sebastien Roch © Copyright 2025.
9261	https://www.mytutor.co.uk/answers/58700/GCSE/English-Literature/How-does-Shakespeare-present-the-theme-of-Guilt-in-Macbeth-Use-Act-5-Scene-1-as-a-reference-in-your-answer/	+44 (0) 203 773 6024 How does Shakespeare present the theme of Guilt in 'Macbeth'. Use Act 5 Scene 1 as a reference in your answer Throughout Act 5 scene 5, Shakespeare presents Lady Macbeth's giult through her sleepwalking, particulary using the symbol of blood. The character is seen to repeatedly wash her hands, while saying "Out damned spot!" The use of imperative here lets the audience into LM's desparation to be free of her guilt. This hallucination of the metaphorical 'spot' symbolises the blood of Duncan after to the terrible deed she forced Macbeth to commit. She admittedly notes that "All the perfumes of Arabia will not sweeten this little hand". Shakespear's use of hyperbole here to emphesise LM's angiush and heartache after what she's done. She feels as though she will never be clean of the murder of Duncan. This motif of blood also manifests itself earlier in Act 3 scene 4, after Macbeth has Banquo killed. He starts to hallucinate, much like Lady Macbeth, and commands the Ghost to "Never shake thy gory locks at me". Again, an imperative is used for a similar effect and the symbolism of blood expressed in the adjective "gory". Both the forms of guilt are a result of secondary acts of murder. Like LM, Macbeth didn't physically kill Banquo, but ordering someone else to do it is enough to riddle these characters with guilt. Shakespeare uses the symbol of Hallicination to present the theme of Guilt in Act 5, scene 1. Lady Macbeth is hallucinating a "spot" of blood on her hands, Duncan's blood, and prehaps even the blood of the other characters killed as a result of Macbeth's killing spree, like Banquo. She also seems to hallucinate the night that she manipulated Macbeth into killing Duncan, and she jumps in her conversation from "come come give me your hand" to "put on your nightgown". Her replaying that night in her head signifies that she is constantly reliving it, in a nightmarish cycle. Hallucination is also apparent in Act 3 scene 4, where Macbeth sees the Ghost of Banquo at the banquet. Uses the supernatural is a reuccuring motif in Shakespeare's plays, one of them also being Hamlet with the Ghost of Old Hamlet, and again in the beginning of Macbeth with the witches. In the Jacobean era, the supernatural were thought to be an unnatural evil which people who were God fearing belived. The fact that LM and Macbeth both hallucinate, show the extent of their crimes against God. Treason being the biggest one, as following the Great Chain of Beings, the monarch was right after God in terms of power, and murder of course. 54295 Views Need help with English Literature? One-to-one online tuition can be a great way to brush up on your English Literature 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 English Literature GCSE answers How do I approach an unseen poem in an exam? How do I structure this paragraph when they're asking about use of language in this extract of Jane Eyre? How do you write a strong essay under timed conditions? Discuss the importance of the Beast in Goulding's Lord of the Flies 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
9262	https://www.statology.org/cdf-vs-pdf/	CDF vs. PDF: What’s the Difference? by Zach Bobbitt Posted on This tutorial provides a simple explanation of the difference between a PDF (probability density function) and a CDF (cumulative distribution function) in statistics. Random Variables Before we can define a PDF or a CDF, we first need to understand random variables. A random variable, usually denoted as X, is a variable whose values are numerical outcomes of some random process. There are two types of random variables: discrete and continuous. Discrete Random Variables A discrete random variable is one which can take on only a countable number of distinct values like 0, 1, 2, 3, 4, 5…100, 1 million, etc. Some examples of discrete random variables include: The number of times a coin lands on tails after being flipped 20 times. The number of times a dice lands on the number 4after being rolled 100 times. Continuous Random Variables A continuous random variable is one which can take on an infinite number of possible values. Some examples of continuous random variables include: Height of a person Weight of an animal Time required to run a mile For example, the height of a person could be 60.2 inches, 65.2344 inches, 70.431222 inches, etc. There are an infinite amount of possible values for height. Rule of Thumb: If you can count the number of outcomes, then you are working with a discrete random variable (e.g. counting the number of times a coin lands on heads). But if you can measurethe outcome, you are working with a continuous random variable (e.g. measuring, height, weight, time, etc.) Probability Density Functions A probability density function(pdf) tells us the probability that a random variable takes on a certain value. For example, suppose we roll a dice one time. If we let xdenote the number that the dice lands on, then the probability density function for the outcome can be described as follows: P(x < 1) : 0 P(x = 1) : 1/6 P(x = 2) : 1/6 P(x = 3) : 1/6 P(x = 4) : 1/6 P(x = 5) : 1/6 P(x = 6) : 1/6 P(x > 6) : 0 Note that this is an example of a discrete random variable, since xcan only take on integer values. For a continuous random variable, we cannot use a PDF directly, since the probability that xtakes on any exact value is zero. For example, suppose we want to know the probability that a burger from a particular restaurant weighs a quarter-pound (0.25 lbs). Since weightis a continuous variable, it can take on an infinite number of values. For example, a given burger might actually weight 0.250001 pounds, or 0.24 pounds, or 0.2488 pounds. The probability that a given burger weights exactly .25 pounds is essentially zero. Cumulative Distribution Functions A cumulative distribution function (cdf) tells us the probability that a random variable takes on a value less than or equal to x. For example, suppose we roll a dice one time. If we let xdenote the number that the dice lands on, then the cumulative distribution function for the outcome can be described as follows: P(x ≤ 0) : 0 P(x ≤ 1) : 1/6 P(x ≤ 2) : 2/6 P(x ≤ 3) : 3/6 P(x ≤ 4) : 4/6 P(x ≤ 5) : 5/6 P(x ≤ 6) : 6/6 P(x > 6) : 0 Notice that the probability that xis less than or equal to 6is 6/6, which is equal to 1. This is because the dice will land on either 1, 2, 3, 4, 5, or 6 with 100% probability. This example uses a discrete random variable, but a continuous density function can also be used for a continuous random variable. Cumulative distribution functions have the following properties: The probability that a random variable takes on a value less than the smallest possible value is zero. For example, the probability that a dice lands on a value less than 1 is zero. The probability that a random variable takes on a value less than or equal to the largest possible value is one. For example, the probability that a dice lands on a value of 1, 2, 3, 4, 5, or 6 is one. It must land on one of those numbers. The cdf is always non-decreasing. That is, the probability that a dice lands on a number less than or equal to 1 is 1/6, the probability that it lands on a number less than or equal to 2 is 2/6, the probability that it lands on a number less than or equal to 3 is 3/6, etc. The cumulative probabilities are always non-decreasing. Related:You can use an ogive graph to visualize a cumulative distribution function. The Relationship Between a CDF and a PDF In technical terms, a probability density function (pdf) is the derivative of a cumulative distribution function (cdf). Furthermore, the area under the curve of a pdf between negative infinity and xis equal to the value of xon the cdf. For an in-depth explanation of the relationship between a pdf and a cdf, along with the proof for why the pdf is the derivative of the cdf, refer to a statistical textbook. Zach Bobbitt Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations. 4 Replies to “CDF vs. PDF: What’s the Difference?” Thanks sir.Useful Reply 2. Practically cummulative distribution for descrete variableP(x <1) : cannot be 1/6 is will be 0P(x <2) : cannot be 2/6 is will be 1and so on, because in descrete varible less than 1 is 0 only and less than 2 is 1 only and so on. Reply it was less than or equal to… Reply 3. with simple example to let me understand,how cdf form equation of pdf? Reply Leave a Reply Cancel reply
9263	https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/vegetarian-diet	Vegetarian Diet - an overview \| ScienceDirect Topics Skip to Main content Journals & Books Vegetarian Diet In subject area:Agricultural and Biological Sciences A vegetarian diet is defined as a dietary regimen that excludes all types of meat, with variations that may include or exclude animal-derived products like eggs and milk. This diet can be adopted for various reasons, including ethical concerns, environmental considerations, food allergies, religious beliefs, or health motivations. AI generated definition based on:Nutrition and Bariatric Surgery, 2021 How useful is this definition? Press Enter to select rating, 1 out of 3 stars Press Enter to select rating, 2 out of 3 stars Press Enter to select rating, 3 out of 3 stars About this page Add to MendeleySet alert Also in subject area: Medicine and Dentistry Discover other topics 1. On this page On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Vegetarian Diets 2016, Encyclopedia of Food and HealthN. Burkholder, ... J. Sabaté Conclusion A vegetarian diet could present a marked advantage over meat-based diets, and ample research has demonstrated increased longevity and several chronic diseases in vegetarians. A vegetarian diet is lower in total fat, saturated fat, cholesterol, and animal protein but higher in complex carbohydrates, fiber, antioxidants, and unsaturated fats and thus provides a potential dietary option for preventing diabetes, CVD, obesity, and select cancers. Despite lower intakes of certain nutrients for strict vegetarians, careful dietary planning can ensure that current recommended intakes for essential nutrients are met. Additionally, a plant-based diet is abundant protective food components that benefit health. The reported health benefits of a vegetarian diet as well as the economic and environmental ramifications of meat consumption should be considered when assessing the health effects of a vegetarian diet. View chapterExplore book Read full chapter URL: Reference work 2016, Encyclopedia of Food and HealthN. Burkholder, ... J. Sabaté Review article The influence of dietary patterns on gut microbiome and its consequences for nonalcoholic fatty liver disease 2020, Trends in Food Science & TechnologyQing-Song Zhang, ... Wei Chen 2.3 Vegetarian diet The Vegetarian diet, characterized by consumption of a very high proportion of plant-based foods, such as vegetables, fruits, nuts, seeds, whole grains, beans and cereals. Depending on the degree of animal product avoidances, vegetarian diets can be classified into several categories including lacto-ovo-vegetarian, lacto-vegetarian and vegan. Vegetarian diets have many important health benefits, such as reduced blood pressure, lower blood cholesterol, and a lower risk of type 2 diabetes (Key, Appleby, & Rosell, 2006). However, there seems to be no direct and reliable evidence of the benefits of Vegetarian diets in the prevention of NAFLD. Vegetarian diets are usually low in saturated fat and cholesterol. The diet provides more dietary fiber, polyphenol, folate and carotenoids (Duda-Chodak, Tarko, Satora, & Sroka, 2015). Long-term increased intake of these nutrients may explain why vegetarian diets are associated with such health benefits. Vegetarian food is usually rich in dietary fiber, mainly derived from whole grains, legumes, vegetables and fruits. The major source of dietary fiber in a vegetarian diet is grain-based foods which account for 17.8% of total intake, followed by “fruits, vegetables, and bean foods” which contribute to 14.9%, 13.7% and 6.3% of the total intake, respectively (Reicks, Jonnalagadda, Albertson, & Joshi, 2014). Dietary fiber has also been shown to increase antioxidant enzyme activity, decrease erythrocyte lipid peroxidation and alter hepatic detoxifying enzymes activity. Increasing dietary fiber intake can reduce serum zonulin levels, decrease liver enzymes and improve hepatic steatosis in NAFLD patients (Krawczyk et al., 2018). Soluble fiber has been reported to decrease cholesterol, which may inhibit the progression of simple steatosis into NASH (Ioannou, Haigh, Thorning, & Savard, 2013). Polyphenols are secondary metabolites of plants that are ubiquitous in plant-based foods, including various fruits, vegetables, nuts, beans, teas and grains. Regular consumption of polyphenols can reduce the risk of many metabolic disorders such as obesity, insulin resistance, and NAFLD. Numerous studies have shown that polyphenols can reduce fatty acids and high glucose induced accumulation of triacylglycerol in the liver by inhibiting lipid synthesis and promoting fatty acids catabolism (Rodriguez-Ramiro, Vauzour, & Minihane, 2016). Polyphenols also improve NAFLD through their anti-inflammatory and indirect antioxidant properties. Furthermore, polyphenols have been shown to reduce the expression of SREBP-1c by activating the adenosine monophosphate-activated protein kinase (AMPK) signaling pathway, which in turn inhibits the fatty acids synthesis and triacylglycerol metabolism (Liu et al., 2011). The upregulation of peroxisome proliferator-activated alpha (PPARα) by polyphenols promotes the β-oxidation of fatty acids and reduces hepatic fat storage (Berlanga, Guiu-Jurado, Porras, & Auguet, 2014). In addition, polyphenols also can activate nuclear factor-erythroid 2-related factor 2 (Nrf2) to reduce oxidative stress (Rodriguez-Ramiro, Ramos, Bravo, Goya, & Martin, 2012) (Fig. 3). Sign in to download hi-res image Fig. 3. Effects of dietary fiber and Polyphenol on NAFLD in a vegetarian diet. AMPK, adenosine monophosphate-activated protein kinase; Nrf2, Nuclear factor-erythroid 2-related factor 2. A vegetarian diet is also abundant in folate, which is widely found in a variety of foods, especially in dark green leafy vegetables and citrus fruits. Folate is a co-enzymatic substrate of one-carbon metabolism, which was found to play an important role in hepatic steatosis susceptibility. Animal studies have indicated that folate deficiency affects hepatic lipid storage and metabolism, which further deteriorates NAFLD (Champier, Claustrat, Nazaret, Montange, & Claustrat, 2012). Dietary supplementation with folic acid has been reported to reduce liver fat content in a mouse model (Christensen et al., 2010). Clinical trials indicated a negative correlation between serum folate levels and body mass index (BMI) in patients with obesity, NAFLD and type 2 diabetes (Hirsch et al., 2005). Furthermore, in a cohort of Chinese patients with NAFLD, low serum concentration of folate was identified as an independent risk factor for NAFLD (Xia et al., 2018). Christensen et al. demonstrated that hepatic lipid transport by very low-density lipoprotein (VLDL) was impaired in folate-deficient mice, as folate is a precursor of VLDL synthesis in the hepatocytes, folate deficiency may affect the lipid output by VLDL and induce the accumulation of fat in the liver (Christensen et al., 2010). In addition, carotenoids, another bioactive component high in Vegetarian diets, have been reported to be potential against NAFLD. The exact mechanisms of the protective effects of carotenoids on NAFLD have not been thoroughly understood, but the antioxidant and anti-inflammatory effects of these bioactive components may be partly responsible. A previous study showed that plasma levels of carotenoids including lycopene, β-carotene, lutein, zeaxanthin and β-cryptoxanthin were significantly decreased in NASH patients compared to healthy subjects (Erhardt et al., 2011). Bahcecioglu et al. investigated the preventative effect of lycopene on high-fat diet-induced NASH in rats and found lower levels of hepatic ALT, plasma triglycerides, serum TNF-α, steatosis and inflammation in rats supplemented with lycopene compared to high-fat diet fed rats (Bahcecioglu et al., 2010). Show more View article Read full article URL: Journal2020, Trends in Food Science & TechnologyQing-Song Zhang, ... Wei Chen Chapter Vegetarian Diets Across the Lifecycle 2015, Advances in Food and Nutrition ResearchMeika Foster, Samir Samman 2 Definitions of Vegetarian Diets In classic terms (Table 1), an individual is considered a vegetarian if he/she abstains from eating all flesh foods (meat, poultry, fish, shellfish); those who follow a total vegetarian or “vegan” diet consume only plant-derived foods, excluding all foods of animal origin including eggs and dairy products. Eating patterns that are similar to a vegetarian diet include the macrobiotic diet, which is low in meat and dairy products, and the pescetarian diet, in which fish/shellfish is the only animal flesh consumed. Motivations for following a vegetarian diet in Western cultures commonly include a combination of animal rights and welfare, environmental, religious, and health considerations. In Western societies, women are more likely to be vegetarian than men (Beardsworth & Bryman, 1999; McLennan & Podger, 1995; White & Frank, 1994), which is consistent with findings that some nonvegetarian women avoid eating meat and poultry (Fayet, Flood, Petocz, & Samman, 2013), eat less meat than their male counterparts (Beardsworth et al., 2002), and are more likely than men to be decreasing their meat consumption (Beardsworth et al., 2002; Fargerli & Wandel, 1999; Ruby, 2012). Table 1. Classifications of vegetarian eating patterns \| Type of vegetarian \| Definition \| --- \| \| Classic \| \| Ovo-lacto-/lacto-ovo-, ovo-, lacto-vegetarian \| Diet is devoid of all flesh foods, but includes egg (ovo) and/or dairy (lacto) products \| \| Vegan \| Diet excludes all animal products \| \| New \| \| Meat reductionist \| Diet includes only limited amounts of animal flesh \| \| Semi-vegetarian \| Fish/shellfish and poultry are the only animal flesh consumed \| \| Pesco-vegetarian \| Fish/shellfish is the only animal flesh consumed \| \| Pollo-vegetarian \| Poultry is the only animal flesh consumed \| View chapterExplore book Read full chapter URL: Book series2015, Advances in Food and Nutrition ResearchMeika Foster, Samir Samman Chapter Impact of cardiometabolic disease on cognitive function 2021, Nutraceuticals in Brain Health and BeyondBradley J. McEwen Vegetarian and vegan diet Vegetarian diets encompass several diet types. These diets include semi-vegetarian (flexitarian), pesco-vegetarian, lacto-vegetarian, ovo-vegetarian, lacto-ovo-vegetarian, and vegan diet. Individuals who follow vegan diets exclude all meats and animal products . Vegetarian diets, including vegan diets, have numerous benefits for cardiovascular health , reduced blood pressure [76–78], improved glycemic control (reduce glycated hemoglobin) , reduced glucose levels , reduced plasma lipids [76,77], and lower total cholesterol and LDL-cholesterol . View chapterExplore book Read full chapter URL: Book 2021, Nutraceuticals in Brain Health and BeyondBradley J. McEwen Review article Effects of vegetarian diet-associated nutrients on gut microbiota and intestinal physiology 2022, Food Science and Human WellnessWei Xiao, ... Qixiao Zhai 1 Introduction Vegetarian diet is one that strictly or partially excludes products of animal origin . Vegetarianism can be further subdivided into 7 categories according to whether seafood, eggs or dairy are also excluded: veganism, lacto-ovo vegetarianism, ovo-vegetarianism, lacto-vegetarianism, semi-vegetarianism, raw veganism, and pescetarianism [2,3]. Table 1 presents the characteristics of 8 dietary patterns and their total scores for the Healthy Eating Index 2010 (HEI) and Mediterranean Diet Score (MDS). Table 1. Characteristics of the eight dietary patterns and their total scores for HEI and MDS. \| Dietary pattern \| Characteristics \| HEI 2010 \| MDS \| --- --- \| \| Vegan \| Excludes all products of animal origin. \| 65.4 \| 5.8 \| \| Lacto-ovo vegetarian \| Excludes red meat, poultry, and fish, but consumes dairy and eggs. \| 58.7 \| 4.6 \| \| Ovo-vegetarian \| Excludes red meat, dairy, poultry, and fish, but consumes eggs. - \| \| Raw vegan \| Based vegetables, fruits, nuts, seeds, legumes, and grains. Unprocessed foods account for 75%–100% of diet. - \| \| Pescatarian \| Excludes red meat and poultry, consumes fish, dairy and eggs. \| 58.7 \| 5.5 \| \| Lacto-vegetarian \| Excludes eggs, red meat, poultry, and fish but consumes dairy products. - \| \| Semi-vegetarian \| Consumes red meat, poultry, and fish no more than once a week. \| 59.4 \| 5.2 \| \| Omnivore \| Consumes a variety of foods, eats fish or meat almost every day. \| 54.2 \| 4.1 \| Note: The horizontal line indicates no reports of relevant scores. The possible scores for the HEI range from 0 to 100, with higher scores being indicative that the foods consumed are in line with the dietary recommendations issued by the US Department of Agriculture. The possible scores for the MDS range from 1 to 10, with higher scores being closer to the Mediterranean diet pattern . With the increasing popularity of vegetarianism, vegetarian diets have become increasingly regarded as healthy and potentially medically beneficial dietary option . Several epidemiological studies support the benefits of vegetarian diets. For example, compared to omnivores, vegetarians have lower body mass index, lower serum cholesterol levels, reduced incidence of diabetes, and lower blood pressure [1,4,5]. Furthermore, the increased intake of plant-derived nutrients, such as dietary fiber, antioxidants, and so on, has been related to a low incidence of cardiometabolic risk and other chronic diseases[1,5,6]. The American Dietetic Association has reported that a well-planned vegetarian diet can not only lead to significant health benefits, but also meet the nutritional needs of all age groups . Table 2 shows the average daily nutritional intake across both omnivorous and vegetarian diets. Total carbohydrates intake was of the same magnitude in the two diets, but the vegetarian diets contained more non-carbohydrates nutrients than the omnivorous diet . In addition, compared with omnivorous diet, vegetarian diets consume lower total fat and higher unsaturated fat. Furthermore, vitamin C intake by omnivores is lower than that of vegetarians . Similar results were found for vitamin A, folate, calcium, and magnesium, with a high intake being seen in vegetarian diets [2,8]. Additionally, the intake of phytochemicalphytosterols in vegetarian diets was higher than omnivorous diet [9,10]. Interestingly, Burkholder-Cooley et al. observed high flavonoids intake in the vegetarian diets, while total polyphenols intake was not as good as that of omnivorous diet. It may be expected that vegetarians consume more fruits and vegetables that were good sources of flavonoids, while non-vegetarians consume more coffee which is another source of phenolic acids belonging to another main type of polyphenols . Table 2. Comparison of average daily nutritional intake of vegetarians and omnivores. \| Macro- and micronutrients \| Vegetarian \| Empty Cell \| Omnivore \| Empty Cell \| Subjects \| Reference \| --- --- \| Total carbohydrates (g/day) \| 343 \| \| 322 \| \| 1475 people were recruited to the study: 369 males, and 1 106 females, of which 155 were omnivores and 573 were vegetarians. \| \| \| Dietary fibers (g/day) \| 34 \| \| 27 \| \| \| \| \| Protein (g/day) \| 93 \| \| 112 \| \| \| \| \| Total fat (g/day) \| 96 \| \| 122 \| \| \| \| \| Poly-unsaturated fat (g/day) \| 24 \| \| 22 \| \| \| \| \| Sodium (mg/day) \| 2228 \| \| 3296 \| \| \| \| \| Calcium (mg/day) \| 1465 \| \| 1199 \| \| \| \| \| Iron (mg/day) \| 20 \| \| 17 \| \| \| \| \| Phytosterols (mg/day) \| 347.6 \| \| 263.2 \| \| 542 people were recruited to the study: 333 were omnivores and 209 were vegetarians. \| \| \| Isoflavone (mg/day) \| 10.1 \| \| 0.23 \| \| 37643 women were recruited to the study: 25912 were non-vegetarians and 11731 were vegetarians. \| \| \| \| Man \| Menopausal women \| Man \| Menopausal women \| 3501 people were recruited to the study: 1253 males and 2248 females, of which 2230 were omnivores and 1271 were vegetarians. \| \| \| Vitamin C (mg/day) \| 176 \| 165 \| 165 \| 164 \| \| \| \| Vitamin A (mg/day) \| 2.7 \| 2.45 \| 2.05 \| 2.18 \| \| \| \| Folate (μg/day) \| 493 \| 488 \| 413 \| 433 \| \| \| \| Magnesium (mg/day) \| 325 \| 290 \| 286 \| 258 \| \| \| \| Vitamin B 12 (μg/day) \| 1.1 \| 1.2 \| 4.1 \| 2.9 \| \| \| \| Niacin (mg/day) \| 19 \| 20 \| 20 \| 22 \| \| \| \| Riboflavin (mg/day) \| 1.1 \| 1.1 \| 1.1 \| 1.2 \| \| \| \| Thiamine (mg/day) \| 1.5 \| 1.8 \| 1.2 \| 1.3 \| \| \| \| Zinc (mg/day) \| 8.8 \| 11.0 \| 8.3 \| 10.4 \| \| \| \| \| Coffee consumers \| Non-coffee consumers \| Coffee consumers \| Non-coffee consumers \| 57519 people were recruited to the study: 21666 were omnivores and 35853 were vegetarians. \| \| \| Polyphenol (mg/day) \| 929 \| 463 \| 1119 \| 412 \| \| \| A vegetarian diet significantly impacts the relative abundance of gut microbiota at the genus, species, and even strain levels [12–15]. Moreover, a vegetarian diet may be a potential nutritional therapy for the regulation of intestinal health and protection against inflammatory responses. In recent years, mostly reviews have focused on the vegetarian diets only investigated the impacts of adopting plant-rich diets on the prevention or treatment of chronic diseases or discussed the nutritional value of vegetarian diets [3,16,17]. Few researchers pay attention to the impact of vegetarian diets on the intestinal tract itself. Therefore, this review summarizes the impact of vegetarian diet-derived nutrients on gut microbial abundance. The role of the nutrients and microbial metabolites of the nutrients in the protection against intestinal disease and maintenance of intestinal physiology is also reviewed. Show more View article Read full article URL: Journal2022, Food Science and Human WellnessWei Xiao, ... Qixiao Zhai Chapter Impact of Nutrition on the Gut Microbiota 2018, Gut MicrobiotaEdward Ishiguro, ... Kristina Campbell Vegetarian Diet Pattern Vegetarianism is a dietary pattern that is based on the consumption of plants rather than meats. It includes different types of diets that vary on whether they include animal-derived foods such as milk and eggs (do Rosario et al., 2016). Consumption of a vegetarian diet is associated with a number of health benefits such as a significantly lower risk for ischemic heart disease mortality, cancer (Huang et al., 2012), and type 2 diabetes, compared with nonvegetarians (Satija et al., 2016). The health benefits appear to derive from the increased consumption of polyphenols and fibers, combined with restriction of meat and/or animal products; it has recently been hypothesized that this combination creates a specific bacterial niche and leads to the production of distinct metabolites that have diverse abilities to metabolize certain nutrients (do Rosario et al., 2016), leading to robust health. Two intervention trials have examined the impact of vegetarian diets on the human microbiota (Kim et al., 2013; David et al., 2013). In the first, obese volunteers that followed a strict vegetarian diet for 1 month showed significant changes in the composition of the microbiota compared with baseline (Kim et al., 2013): a significant reduction in the Firmicutes to Bacteroidetes ratio; a reduction in pathobionts; and an enhanced growth of bacteria from the Lachnospiraceae, Ruminococcaceae, and Erysipelotrichaceae families. The subjects also showed reduced body weight, improved markers of metabolic health, and reduced gut inflammation. The second study was a crossover trial where each subject was provided with an animal-based diet (with no fiber) and a plant-based diet (high in fiber), with a washout period between the two diets. The animal-based diet had the greatest impact on the subjects’ microbiota, with increased diversity and abundance of bile-tolerant microorganisms (Alistipes, Bilophila, and Bacteroides) and decreased levels of bacteria that metabolize dietary fiber (Roseburia, E. rectale, and Ruminococcus bromii). On the plant-based diet, subjects showed increased populations of Prevotella (David et al., 2013). These trials support the notion of increased dietary fiber having positive effects on the gut microbiota and on health. Several observational studies that examined the gut microbiota of subjects consuming a vegetarian diet versus an omnivorous diet support these diet intervention trials (Liszt et al., 2009; Matijasic et al., 2014; Ruengsomwong et al., 2014; Zimmer et al., 2012; Kabeerdoss et al., 2012; Ferrocino et al., 2015; Reddy et al., 1998). Table 6.1 summarizes findings on the impact of a vegetarian diet on the microbiota as described in observational studies. The different results obtained from these studies may be attributable to methodological differences, but most of the changes observed have been linked with some form of health benefit. Further large-scale interventional studies over longer periods are warranted to confirm these results. Table 6.1. Findings of observational studies on the impact of a vegetarian diet on the microbiota \| Common findings in individuals consuming a vegetarian diet \| \| Higher bacterial diversity (Liszt et al., 2009) \| \| Increased levels of Prevotella (Matijasic et al., 2014; Ruengsomwong et al., 2014; De Filippo et al., 2010) \| \| Increased levels of Bacteroides (Liszt et al., 2009; Matijasic et al., 2014; De Filippo et al., 2010) \| \| Decreased pathobionts, including members of the family Enterobacteriaceae (Kim et al., 2013; De Filippo et al., 2010) \| \| Increased counts of Faecalibacterium prausnitzii (Liszt et al., 2009; Matijasic et al., 2014) \| \| Reduced stool pH (Wu et al., 2011; Zimmer et al., 2012) \| Modified from do Rosario, V.A., Fernandes, R., de Trindade, E.B.S.M., 2016. Vegetarian diets and gut microbiota: Important shifts in markers of metabolism and cardiovascular disease. Nutr. Rev. 74 (7), 444–454. Show more View chapterExplore book Read full chapter URL: Book 2018, Gut MicrobiotaEdward Ishiguro, ... Kristina Campbell Chapter Vegetarianism and Veganism 2014, Reference Module in Biomedical SciencesE.H. Haddad, P. Faed Nutrient Adequacy of Vegetarian Diets In a position paper, experts from the Academy of Nutrition and Dietetics concluded that well-planned and executed vegetarian diets have adequate nutrients for all stages of life (Craig and Mangels, 2009). Generally, vegetarian diets are rich in carbohydrates, n-6 fatty acids, dietary fiber, carotenoids, folate, vitamin C, potassium, and magnesium. At the same time, these diets may be low in energy, protein, long-chain n-3 fatty acids, vitamin B-12, vitamin D, calcium, iron, and zinc (Key et al.,2006). View chapterExplore book Read full chapter URL: Reference work2014, Reference Module in Biomedical SciencesE.H. Haddad, P. Faed Chapter Volume 2 2023, Encyclopedia of Human Nutrition (Fourth Edition)J. Dwyer, J. Harvey Vegetarian patterns and practices The term “vegetarian diet” does not fully describe the variety in dietary and nutrient intakes and associated health status of followers of such eating patterns in Western countries. Today, myriad vegetarian eating patterns exist that are not easily described by focusing on a single dimension, such as animal food intake (see Table 1). The impact of these patterns on nutritional status and health requires more complete characterization of diet and other aspects of lifestyle than a simple description of which animal foods are left over to eat after others have been omitted from the diet. It is also important to characterize what other foods are eaten in greater detail if one is to obtain an accurate profile of the diet's nutrient adequacy. Traditionally, vegetarian diets were poor sources of energy, protein of high biological value, vitamins A, D, B 2, B 6, and B 12, zinc, bioavailable iron, omega-3 fatty acids, choline, calcium, and sometimes iodine. Today a wide variety of plant foods are available year around in most Western countries, along with fortified foods and nutrient containing dietary supplements as well as meat, poultry, milk and egg analogs that can fill these nutrient gaps. However, for a variety of reasons not all vegetarians are willing to use these products, and so risks of inadequacy still remain for some individuals. View chapterExplore book Read full chapter URL: Reference work 2023, Encyclopedia of Human Nutrition (Fourth Edition)J. Dwyer, J. Harvey Chapter Vegetarian diets and disease outcomes 2016, Fruits, Vegetables, and HerbsMing-Chin Yeh, Marian Glick-Bauer Conclusion Vegetarian diets may provide health benefits in the treatment of diseases including cardiovascular disease and hypertension, metabolic syndrome and diabetes, obesity, cancer, renal disease, and RA, among others. Vegetarian diets are associated with lower levels of obesity, improved weight loss, a reduced risk of cardiovascular disease, and lower total mortality. Given the increasing evidence for the health benefits of plant-based diets, vegetarian diets may hold promise as adjuvant therapy in disease management. Research is increasingly looking into the role of vegetarian diets in influencing the gut microbiota, which in turn modulates the inflammatory response and impacts disease states. More research is needed to understand the mechanisms that link diet and health and whether short-term diet therapy can confer long-term health benefits. View chapterExplore book Read full chapter URL: Book 2016, Fruits, Vegetables, and HerbsMing-Chin Yeh, Marian Glick-Bauer Chapter CORONARY HEART DISEASE \| Prevention 2005, Encyclopedia of Human Nutrition (Second Edition)K. Srinath Reddy Vegetarian Diets A reduced risk of CVD has been reported in populations of vegetarians living in affluent countries and in case–control comparisons in developing countries. Reduced consumption of animal fat and increased consumption of fruit, vegetables, nuts, and cereals may underlie such a protective effect. However, ‘vegetarian diets’ per se need not be healthful. If not well planned, they can contain a large amount of refined carbohydrates and t-FAs, while being deficient in the levels of vegetable and fruit consumption. The composition of the vegetarian diet should, therefore, be defined in terms of its cardioprotective constituents. View chapterExplore book Read full chapter URL: Reference work 2005, Encyclopedia of Human Nutrition (Second Edition)K. Srinath Reddy Related terms: Bioavailability Antioxidant Vitamin B12 Blood Pressure Animal Product Dairy Product Vitamin D Vegan Diet Plant-Based Diet Omnivore View all Topics Recommended publications The American Journal of Clinical NutritionJournal AppetiteJournal Food Research InternationalJournal Clinical NutritionJournal Browse books and journals Featured Authors Pellegrini, NicolettaUniversità degli Studi di Udine, Udine, Italy Citations39,997 h-index62 Publications75 Battino, Maurizio AntonioUniversità Politecnica delle Marche, Ancona, Italy Citations35,155 h-index86 Publications179 Gue´ant, Jean LouisUniversité de Lorraine, Nancy, France Citations17,427 h-index66 Publications187 Barnard, Neal D.The George Washington University School of Medicine and Health Sciences, Washington, D.C., United States Citations8,396 h-index52 Publications61 About ScienceDirect Remote access Advertise Contact and support Terms and conditions Privacy policy Cookies are used by this site. 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9264	https://press.rebus.community/intro-to-phil-logic/chapter/chapter-5-necessary-and-sufficient-conditions/	Necessary and Sufficient Conditions – Introduction to Philosophy: Logic Skip to content Menu Primary Navigation Home Read Buy Sign in Search in book: Search Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Book Contents Navigation Contents What is an Open Textbook? Christina Hendricks How to Access and Use the Books Christina Hendricks Introduction to the Series Christina Hendricks Praise for the Book Acknowledgements Benjamin Martin and Christina Hendricks Introduction to the Book Benjamin Martin 1. What is Logic? Matthew Knachel 2. Evaluating Arguments Nathan Smith 3. Formal Logic in Philosophy Bahram Assadian 4. Informal Fallacies Cassiano Terra Rodrigues 5. Necessary and Sufficient Conditions Michael Shaffer References and Further Reading Exercise Solutions Glossary About the Contributors Feedback and Suggestions Adoption Form Licensing and Attribution Information Review Statement Accessibility Assessment Version History Introduction to Philosophy: Logic Buy 5 Necessary and Sufficient Conditions Michael Shaffer The concepts of necessary and sufficient conditions play central and vital roles in analytic philosophy. For example, being an unmarried male is a necessary condition for being a bachelor and being a bachelor is a sufficient condition for being an unmarried male. That these concepts are vital to philosophy is beyond question, and it is primarily because the orthodox account of the methodology of analytic philosophy involves the contention that philosophy aims to yield accurate specifications of sets of necessary and sufficient conditions, such as the claim that all bachelors are unmarried men. It is, then, obviously and deeply important to philosophy that we have an adequate logical grasp of these concepts. In terms of both propositional and first-order logic the concepts of necessary and sufficient conditions are intimately related to the concept of the conditional (i.e. a statement of the form “if p, then q”) as the following canonical account makes clear. Where S(p, q) means “p is a sufficient condition for q” and N(q, p) means “q is a necessary condition for p”, p→q means “if p, then q,” and p≡q means “p and q are logically equivalent,” the following two definitions are supposed to represent these two important ideas: (D1) S(p,q)≡(p→q) (D2) N(q,p)≡(p→q) In effect, D1 and D2 are then intended to be the standard logical interpretations of our ordinary language concepts of necessary and sufficient conditions framed in terms of classical propositional logic. They are based on the idea that necessary and sufficient conditions can be exhaustively defined in terms of the conditional understood as material implication and represented by the “→” of classical propositional logic with the following familiar truth conditions: Truth table for material conditionalABA→BTTTTFFFTTFFT Of course, material implication plays an important role in reasoning in general, particularly with respect to the following valid inferential forms in classical propositional logic, as we saw in Chapter 3. Affirming the Antecedent (Modus Ponens) A→B A /∴B Denying the Consequent (Modus Tollens) A→B ¬B /∴¬A These inference forms have important connections to the concepts of necessary and sufficient conditions, and to how we reason using them. In the case of affirming the antecedent, the first premise can be understood to be the claim that A is sufficient for B, and the second premise the claim that the condition A obtains. So, from these claims it validly follows that B obtains. In the case of denying the consequent, the first premise can be read as the claim that B is a necessary condition for A and the second premise as the claim that B does not obtain. From these premises it validly follows that A does not obtain. However, the following inferential forms involving material implication are invalid in classical propositional logic: Affirming the Consequent A→B B /∴A Denying the Antecedent A→B ¬A /∴¬B These invalid inference forms also are importantly related to the concepts of necessary and sufficient conditions. In the case of affirming the consequent, the first premise can be read as the claim that A is a necessary condition for B and the second premise as the claim that B is true. But, from these premises it does not validly follow that A is also true. The fact that B is necessary for A does not ensure it is also sufficient for A. In the case of denying the antecedent, the first premise can be read as the claim that A is a sufficient condition for B and the second premise as the claim that A is not true. From these premises it does not validly follow that B is not true, as some other condition that suffices for B might, in fact, obtain. Moreover, where NS(p, q) means “p is necessary and sufficient for q, and q is necessary and sufficient for p,” such jointly necessary and sufficient conditions take the following form: (D3) NS(p,q)≡[(p→q)&(q→p)] However, since the formula (p→q)&(q→p) is equivalent to the formula (p≡q) in classical propositional logic, sets of such necessary and sufficient conditions can be more compactly defined in terms of logical equivalence as follows: (D4) NS(p,q)≡(p≡q) This concept is just the idea that the truth values of p and q are always the same, and the notion of logical equivalence has the following truth conditions: Truth table for logical equivalenceABA≡BTTTTFFFTFFFT Sets of jointly necessary and sufficient conditions are, then, just definitions regimented as sentences of this sort. For example, it turns out that being a bachelor and being an unmarried male are jointly necessary and sufficient conditions for one another. Now why, specifically, are the concepts of necessary and sufficient conditions, so understood, of such central significance in contemporary analytic philosophy? Conceptual Analysis and Necessary and Sufficient Conditions The central account of the methods of contemporary analytic philosophy is predicated on the claim that philosophical methodology is intuition-driven conceptual analysis that aims to determine true sets of necessary and sufficient conditions. In fact, according to a significant number of philosophers, such conceptual analysis is the only method of philosophy. For the purposes at hand, this account of the methods of philosophy will be referred to as the standard philosophical method (SPM). Conceptual analyses take the form of specifications of the content of a pre-theoretical concept (the analysans) through the articulation of a set of necessary and sufficient conditions (the analysandum or analysanda), and here we find the locus of the connection between the concepts of necessary and sufficient conditions and philosophical methodology. This methodological account of philosophy can be more completely characterized as follows: (1) Conceptual analyses take the form of proposed definitions (i.e. sets of necessary and sufficient conditions) of analysanda. (2) The adequacy of any analysandum can be tested against concrete and/or imagined cases. (3) Whether or not a proposed analysandum is adequate with respect to a given case can be determined by the use of a priori intuition, with a priori intuition being a distinct, reliable and fallible non-sensory mental faculty. (4) Intuition allows us to reliably access knowledge about concepts. (5) The method of reflective equilibrium is the particular method by which intuitions can be used to confirm/disconfirm analysanda. According to the defenders of SPM, this is essentially the orthodox methodology of analytic philosophy, and it has been assumed to be adequate for the solution of philosophical problems by a significant number of both practicing and prominent philosophers throughout the recent history of philosophy. For example, this is the contention made by Colin McGinn in a recent book. McGinn is not in the least bit tentative in his blanket defense of SPM as the one and only method of philosophy. With this aim in mind, early in his 2012 book he makes the following extended declaration about philosophy: … it is not a species of empirical enquiry, and it is not methodologically comparable to the natural sciences (though it is comparable to the formal sciences). It seeks the discovery of essences. It operates “from the armchair”: that is, by unaided (usually solitary) contemplation. Its only experiments are thought-experiments, and its data are possibilities (or “intuitions” about possibilities). Thus philosophy seeks a priori knowledge of objective being—of non-linguistic and non-conceptual reality. We are investigating being as such, but we do so using only a priori methods. (McGinn 2012, 4) As should be immediately apparent, this is a clear, straightforward, and ringing endorsement of SPM as it has been understood here. To buttress this contention we need only take note of his other claims that “…the proper method for uncovering the essence of things is precisely conceptual analysis,” (McGinn 2012, 4) and that “philosophy, correctly conceived, simply is conceptual analysis” (McGinn 2012, 11). In effect, he believes then that we arrive at such analyses by considering possible cases and asking ourselves whether the concept applies or not in those cases—that is by consulting our “intuitions” about such cases (McGinn 2012, 5). What is also important for the purposes at hand is his acknowledgment that this account of philosophical methodology “was really the standard conception for most of the history of the subject, in one form or another” (McGinn 2012, 7). So, not only does McGinn endorse SPM as the sole methodology of contemporary philosophy, but he also claims that it is the enduring methodology of philosophical inquiry throughout its history. One important clarification regarding McGinn’s version of SPM concerns the nature of the object of analysis (the analysans) and, more importantly, the nature of the analysandum itself as they are typically understood (i.e. as definitions of a particular sort framed as sets of necessary and sufficient conditions). Carl Hempel usefully makes a crucial distinction in this regard, which we can use to illuminate the standard view of such definitions: The word “definition” has come to be used in several different senses….A real definition is conceived of as a statement of the “essential characteristics” of some entity, as when man is defined as a rational animal or a chair as a separate moveable seat for one person. A nominal definition, on the other hand, is a convention which merely introduces an alternative—and usually abbreviated—notation for a given linguistic expression, in the manner of a stipulation. (Hempel 1952, 2) Moreover, he tells us further that some real definitions are to be understood as meaning analyses, or as analytic definitions, of the term in question. The validation of such claims requires only that we know the meanings of the constituent expressions, and no empirical investigation is necessary to determine the correctness of the analysandum (Hempel 1952, 8). This is, of course, precisely what McGinn has in mind with respect to conceptual analysis. It is, then, worth making the obvious point that conceptual analysis is the operation of analyzing concepts via proposing definitions, but to point that out is not enough to fully grasp the view. It is true that SPM is a method that takes as inputs our concepts, but it involves the clear recognition that the definitions involved are to be understood as meaning analyses rather than as nominal or stipulative (i.e. “dictionary”) definitions. So, for example, the question of whether knowledge is justified true belief is just the question of the analysis of the concept of knowledge in terms of definitions constituted by sets of necessary and sufficient conditions understood as a meaning analysis. Conceptual analysis is then a method of doing something _with concepts that we already possess—_wherever they have ultimately come from. It is defining a pre-theoretical concept by offering a synonymous expression. It then appears to be the case that the defenders of SPM must believe that concepts have the form of sets of necessary and sufficient conditions, that such analyses are meaning analyses, and that analyses of our pre-analytic concepts are informative. Typical analysanda are thus kinds of decompositions of pre-analytic concepts. They are conceptual truths with the form of analytic definitions. So, for McGinn and other like-minded thinkers, analysanda have a very simple logical form, and we can see this via the example of the analysis of the concept of knowledge. Where K_x_ is “x is knowledge”, J_x_ is “x is justified”, T_x_ is “x is true” and B_x_ is “x is believed”, the standard analysis of knowledge looks like this: x is K_x_ ≡ x is J_x_ & x is T_x_ & x is B_x_ This analysis is supposed to tell us the true nature, or essence, of the concept of knowledge in terms of a finite set of defining essential features, with the logical form of a set of jointly necessary and sufficient conditions. So, providing such an analysis involves decomposing the analysans into a list of features, thus exposing in some important sense the content of the concept. A Problem with the Orthodox View and SPM Many recent critics have attacked SPM in terms of (2)-(5) by challenging the reliability of the faculty of intuition. This is the main line of criticism against SPM offered by many defenders of what is called experimental philosophy, and it is an interesting criticism of orthodox philosophy indeed. However, some critics have alternatively attacked SPM by challenging (1) on the basis of the theory of concepts it assumes; specifically, the idea that concepts can be adequately captured by sets of necessary and sufficient conditions. One version of this latter form of criticism is particularly relevant to this chapter. This criticism is based on the contention that SPM wrongly assumes that concepts take the form of necessary and sufficient conditions at all. Call this the potential vacuity problem. The Potential Vacuity Problem The problem of potential vacuity arises as follows, and is based on Ludwig Wittgenstein’s infamous remarks about the theory of concepts assumed in SPM. He addressed the matter of the reliability of SPM in his Philosophical Investigations and The Blue and Brown Books, and therein Wittgenstein attacks the foundation of the project of conceptual analysis by attempting to undermine (1) via examination of the claim that concepts have the form of sets of necessary and sufficient conditions. First, Wittgenstein rejected the notion that most, or even perhaps any, concepts can be defined precisely via the specification of sets of necessary and sufficient conditions, and that this is a problem central to orthodox philosophy. This important revelation was made by noting that philosophical attempts at conceptual analysis have systematically failed to produce the goods. He tells us explicitly that, We are unable to clearly circumscribe the concepts we use; not because we don’t know their real definition, but because there is no real “definition” to them. (Wittgenstein 1958, 25) Second, he sought to replace the notion of concepts understood as sets of necessary and sufficient conditions with an alternative theory of concepts. This alternative account of concepts is based on the notion of a “family resemblance relation.” To see the first point more clearly, let us look at Wittgenstein’s favorite example from his Philosophical Investigations. Wittgenstein specifically argued that the concept of a game cannot be correctly analyzed in terms of a set of necessary and sufficient conditions. This is because games do not share some set of defining features in common. Rather, the members of the set of games are only similar to one another in some respects, and it is these relations of similarity that constitute the family of games. As we have seen, SPM assumes the following principle: (CON) For any concept C, there exists a set of necessary and sufficient conditions that constitutes the content of C. Wittgenstein’s attack on SPM is mounted via an attack on CON, and this is the fundamental ground of the potential vacuity problem. Essentially, the gist of the problem is that if there are no (or even just very few) concepts that can be correctly regimented as sets of necessary and sufficient conditions, there can be no (or very few) correct conceptual analyses in the sense of SPM. The basis of Wittgenstein’s criticism then can be understood as follows: it is clear from the consideration of examples across the history of philosophy that most or all philosophical attempts to analyze concepts by providing sets of necessary and sufficient conditions have failed. This is because, for any proposed set of necessary or sufficient conditions intended to be the correct analysis of a concept, there are instances of that concept that do not meet the set of proposed defining conditions. Think back to Wittgenstein’s favorite example of the concept of a game. Poker and soccer are both plausibly taken to be games and so we might, for example, posit that something is a game, if and only if, that activity involves a winner and a loser. But, the game patty cake is another plausible case of a game and does not have a winner and a loser. So, this definition of a game in terms of a set of necessary and sufficient conditions fails. Wittgenstein claims that this example generalizes, and the presumptive best explanation for the failed philosophical attempts to articulate the contents of concepts in terms of sets of necessary and sufficient conditions is that the contents of concepts are not captured by sets of necessary and sufficient conditions (i.e. the denial of CON). In other words, Wittgenstein holds that for any (or, at least most) attempt(s) to specify the contents of concepts in terms of necessary and sufficient conditions, we will find counter-examples. As a replacement for CON, Wittgenstein introduces the notion of a family resemblance class. The central idea is that the cases that fall under a concept are related to one another not by a defining set of necessary and sufficient conditions, but rather by complex overlapping similarity conditions that relate groups of members of the total set of cases that fall under the concept. However, no one set of conditions holds for all and only the members that exhibit that concept. Thus, if Wittgenstein is correct, the reason that there are no correct conceptual analyses is due to the fact that concepts cannot be analysed in terms of necessary and sufficient conditions. SPM is, thus, potentially (if not actually) vacuous. Prospective Solutions to the Potential Vacuity Problem Does Wittgenstein’s criticism signal defeat of the SPM, then? Not necessarily. Colin McGinn (2012) proposes a solution to the problem. First, notice that Wittgenstein’s criticism is a direct denial of (1). McGinn responds by biting the bullet against Wittgenstein and arguing that, although they are very often difficult to articulate, concepts are properly characterized by sets of necessary and sufficient conditions. Pace Wittgenstein, our failure to articulate definitive examples of such analyses is no reason to suppose that there are no such things. More cleverly, he shows how Wittgenstein’s criticism can be effectively rebutted in the following way. As we have seen, Wittgenstein’s claim that concepts cannot be captured by sets of necessary and sufficient conditions is supposed to follow from his investigation of the concept of a game. But, as McGinn points out, from the fact that it is difficult to produce the goods in this (or any other) case, it does not necessarily follow that there are no such analyses (McGinn 2012, 21-28). Second, Wittgenstein uses this point in support of the claim that concepts actually have the structure of a set of family resemblance relations between paradigm and non-paradigm elements in the extension of a concept. What McGinn then shows is that Wittgenstein’s own theory of concepts in terms of family resemblances presupposes that concepts can be captured by a special type of necessary and sufficient conditions: for any concept C, the non-paradigmatic members of C bear a family resemblance relation to the paradigmatic case(s) of C. So, it would appear to be the case that according to Wittgenstein, something is necessarily a concept, if and only if, it is a set of entities related by family resemblance relations to one or more paradigm cases. As such, McGinn rightly claims that Wittgenstein does not reject SPM. Rather, in his treatment of the concept of game he is “favoring a particular form of it—one in which the analysis takes the form ‘family-resembles paradigm games’ (such as chess, tennis, etc.)” (McGinn 2012, 18-19). However, this response does nothing to defuse the problem that such specifications of conceptual contents cannot plausibly be necessary truths, as McGinn and other defenders of SPM typically believe. This is because family resemblance relations cannot plausibly be understood to be necessary truths. In other words, it is clearly not the case that resemblance relations between objects are such that they are true in all possible worlds. This is the case because resemblances are not purely objective relations between objects. They are perceiver relative, and so vary depending on what features one focuses on. For example, a pen resembles a pencil when one focuses on the function of writing. But, a pen and a pencil do not resemble one another when one focuses instead on the feature of containing ink. EXERCISES Exercise One For each pair, decide whether the first member of the pair is either a necessary condition for the second, a sufficient condition, or neither. Example: Bob’s car is blue/Bob’s car is coloured Answer: Bob’s car being blue is sufficient for it being coloured, as its being blue ensures that it is coloured. However, it isn’t a necessary condition, for Bob’s car could be coloured without being blue—it could be red, for example. Bob drew the eight of Spades from an ordinary deck of playing cards. Bob drew a black card from a deck of ordinary playing cards. Alice has a brother-in-law. Alice is not an only child. Alice’s daughter is married. Alice is a parent. Alice’s daughter is married. Alice is a grandmother. Some women pay taxes. Some taxpayers are women. All women pay taxes. All taxpayers are women. Being a mammal. Being warm blooded. Being warm blooded. Being a mammal. Exercise Two For each claim, rewrite it in terms of necessary and/or sufficient conditions. Example: You can’t play football without a ball Answer: Having a ball is necessary for playing football. You must pay if you want to enter. A cloud chamber is needed to observe subatomic particles. If something is an electron it is a charged particle. Your car is only cool if it’s a Honda. Being a triangle just is being a three-sided, two-dimensional shape. Exercise Three Test for yourself the traditional philosophical assumption that concepts are defined by necessary and sufficient conditions. Try to provide necessary and sufficient conditions for the following concepts, and then test these set of conditions with potential counterexamples: Spoon Garden Success Health (mental and physical) Potential counterexamples to your analysis of these concepts in terms of necessary and sufficient conditions can either take the form of: Cases that the concept should apply to, but which don’t fulfill your necessary and sufficient conditions. Cases that the concept should not apply to, but which do fulfill your necessary and sufficient conditions. As given previously in Chapter 3. ↵ See, for example, Copi, Cohen and Flage (2007, 196, 446, 449) and Fisher (2001, 241). ↵ The concept of the material conditional introduced here is just a formalization of what we were previously and informally calling “conditionals”. ↵ NS(p, q) is then equivalent to S(p, q) & S(q, p) & N(p, q) & N(q, p). ↵ A priori knowledge is knowledge totally independent of any experience. ↵ Recent defenses of SPM include: Bealer (1996), Jackson (1998), and McGinn (2012). For closely related views, see Braddon-Mitchell and Nola (2009). See Shaffer (forthcoming) for extensive discussion of this view. Reflective equilibrium is the method of bringing intuitively true cases into conformity with a rule or principle. ↵ See McGinn (2012, 4-11) for a summary of significant historical examples of the use of SPM, including some of those discussed here in more detail. ↵ Strictly speaking, conceptual analyses can also involve some degree of alteration in the content of the pre-theoretical concepts, as often happens when such analysis involves making a concept more precise. ↵ See Moore (1968) and Wittgenstein (1953), for example. Moore’s paradox of analysis appears to show that such analyses are uninformative, and Wittgenstein claims that concepts have the form of family resemblances, rather than sets of necessary and sufficient conditions. See also Brennan (2017) and Shaffer (2015) for additional worries about the nature of necessary and sufficient conditions. ↵ See Wittgenstein (1953), Lakoff (1987), Ramsey (1998), Rosch and Mervis (1998), and McGinn (2012, Ch. 3) for more on this matter. ↵ Wittgenstein’s criticism also has important additional application to views, like that of McGinn, where conceptual truths are understood to be necessary truths. This is because if concepts are not captured by sets of necessary and sufficient conditions, and only have the form of sets of cases related by family resemblances, then it is not easily understood how they could possibly be necessarily true definitions. This is simply because relations of resemblance between things appear to be contingent relations. ↵ Paradigm members of a family resemblance class are the obvious central cases, whereas non-paradigmatic cases are less central and obvious cases of that class. So, for example, a robin is a paradigmatic case of the class of birds, whereas a penguin is (plausibly) a non-paradigmatic case of a bird. ↵ This understanding of necessary truth as claims that are true in all possible worlds is the standard concept of a necessary truth. Such truths cannot be false in any consistent arrangement of what could possibly exist. ↵ definition An event or proposition which is required for another event to occur or proposition to be true. Conditionals express that the consequent is a necessary condition for the antecedent. ×Close definition An event or proposition which ensures that another event occurs or another proposition is true. Conditionals express that the antecedent is a sufficient condition for the consequent. ×Close definition Previous/next navigation Previous: Informal Fallacies Next: References Back to top License Necessary and Sufficient Conditions Copyright © 2020 by Michael Shaffer is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. Share This Book Powered by Pressbooks Guides and Tutorials \|Pressbooks Directory \|Contact Pressbooks on YouTube Pressbooks on LinkedIn
9265	https://www.homeschoolmath.net/online/pythagorean_theorem.php	\| \| \| \| --- \| You are here: Home → Online resources → Geometry → The Pythagorean theoremThe Pythagorean Theorem: online activities, tutorials, and worksheets This is a hand-picked list of online activities, tutorials, and worksheets concerning the The Pythagorean Theorem. Pythagorean Theorem Pythagorean Theorem Investigate the areas of the squares on the sides of right angled triangles using this interactive figure. www.interactive-maths.com/pythagoras-theorem-ggb.html Proof Without Words: Pythagorean Theorem Watch a dynamic, geometric "proof without words" of the Pythagorean Theorem. Can you explain the proof? illuminations.nctm.org/Activity.aspx?id=4211 Pythagorean Explorer Practice calculating the missing side length of right triangles. Includes three difficulty levels. www.shodor.org/interactivate/activities/PythagoreanExplorer A virtual manipulative: The Pythagorean Theorem Solve two puzzles that illustrate the proof of the Pythagorean Theorem. nlvm.usu.edu/en/nav/frames_asid_164_g_3_t_3.html Pythagorean Theorem Lesson This lesson introduces and explores the Pythagorean Theorem using the Pythagorean Explorer. Three computer activities give students the opportunity to observe triangles, learn and use the Pythagorean Theorem and practice different ways of determining areas of triangles. www.shodor.org/interactivate/lessons/pyth.html Pythagorean Theorem for elementary school? This is a lesson plan showing how you could introduce the Pythagorean Theorem in elementary school. I don't really recommend such because I feel it belongs to the middle school, but it seems some student-teachers are asked to write a lesson for such. Worksheets Math Mammoth Geometry Worksheets Collection Math Mammoth Geometry Worksheets Collection A collection of quality worksheets with variable problems for grades 3-8. Topics include angle relationships, triangles, quadrilaterals, congruency, similar figures, constructions, area, volume, and the Pythagorean Theorem. Price: $9.00 download. See the free samples! www.mathmammoth.com/worksheets/geometry.php Maisonet Math Free PDF worksheets where you can practice finding the unknown length of the hypotenuse of a right triangle or find the length of a missing leg! www.mrmaisonet.com/index.php?/Pythagorean-Theorem/View-category.html Let's Practice Geometry - Free Worksheets Lots of free worksheets for high school geometry, including topics such as the Pythagorean Theorem, perimeter, area, volume, angle relationships, triangle theorems, similarity and congruence, logic, proofs, trig, polygons, and circles. www.letspracticegeometry.com/free-geometry-worksheets Geometry worksheets at AGMath.com Hundreds of free PDF worksheets for high school geometry topics, including geometric constructions, triangle congruence, circle, area, the Pythagorean Theorem, solid geometry, and similarity. The worksheets are loosely based on the "Discovering Geometry" textbook by Michael Serra. Compiled by math teacher Jason Batterson. AGmath.com/1973.html Books Geometry: A Guided Inquiry An excellent choice for high school geometry. I've written an in-depth review of this book and of its supplement Home Study Companion - Geometry, latter by David Chandler. Harold Jacobs Geometry A popular choice for geometry among homeschoolers, and well written. Please read my review. Dr. Math Presents More Geometry An inexpensive companion to any high school geometry course with excellent explanations. Read my review. \| \| \| Online math resources menu Elementary Addition and subtraction Place value Clock Money Measuring Multiplication Division Math facts practice The four operations Factoring and number theory Geometry topics Middle and high school Fractions Decimals Percent Integers Exponents Statistics & Graphs Probability Geometry topics Algebra Calculus Trigonometry Logic and proof For all levels Favorite math puzzles Favorite challenging puzzles Math in real world Problem solving & math projects For gifted children Math history Math games and fun websites Interactive math tutorials Interactive math tools & activities Math help & online tutoring Assessment, review, and test prep Online math curricula \| \| Practice makes perfect. Practice math at IXL.com \| The Pythagorean Theorem: online activities, tutorials, and worksheets This is a hand-picked list of online activities, tutorials, and worksheets concerning the The Pythagorean Theorem. Pythagorean Theorem Pythagorean Theorem Investigate the areas of the squares on the sides of right angled triangles using this interactive figure. www.interactive-maths.com/pythagoras-theorem-ggb.html Proof Without Words: Pythagorean Theorem Watch a dynamic, geometric "proof without words" of the Pythagorean Theorem. Can you explain the proof? illuminations.nctm.org/Activity.aspx?id=4211 Pythagorean Explorer Practice calculating the missing side length of right triangles. Includes three difficulty levels. www.shodor.org/interactivate/activities/PythagoreanExplorer A virtual manipulative: The Pythagorean Theorem Solve two puzzles that illustrate the proof of the Pythagorean Theorem. nlvm.usu.edu/en/nav/frames_asid_164_g_3_t_3.html Pythagorean Theorem Lesson This lesson introduces and explores the Pythagorean Theorem using the Pythagorean Explorer. Three computer activities give students the opportunity to observe triangles, learn and use the Pythagorean Theorem and practice different ways of determining areas of triangles. www.shodor.org/interactivate/lessons/pyth.html Pythagorean Theorem for elementary school? This is a lesson plan showing how you could introduce the Pythagorean Theorem in elementary school. I don't really recommend such because I feel it belongs to the middle school, but it seems some student-teachers are asked to write a lesson for such. Worksheets Math Mammoth Geometry Worksheets Collection A collection of quality worksheets with variable problems for grades 3-8. Topics include angle relationships, triangles, quadrilaterals, congruency, similar figures, constructions, area, volume, and the Pythagorean Theorem. Price: $9.00 download. See the free samples! www.mathmammoth.com/worksheets/geometry.php Maisonet Math Free PDF worksheets where you can practice finding the unknown length of the hypotenuse of a right triangle or find the length of a missing leg! www.mrmaisonet.com/index.php?/Pythagorean-Theorem/View-category.html Let's Practice Geometry - Free Worksheets Lots of free worksheets for high school geometry, including topics such as the Pythagorean Theorem, perimeter, area, volume, angle relationships, triangle theorems, similarity and congruence, logic, proofs, trig, polygons, and circles. www.letspracticegeometry.com/free-geometry-worksheets Geometry worksheets at AGMath.com Hundreds of free PDF worksheets for high school geometry topics, including geometric constructions, triangle congruence, circle, area, the Pythagorean Theorem, solid geometry, and similarity. The worksheets are loosely based on the "Discovering Geometry" textbook by Michael Serra. Compiled by math teacher Jason Batterson. AGmath.com/1973.html Books Geometry: A Guided Inquiry An excellent choice for high school geometry. I've written an in-depth review of this book and of its supplement Home Study Companion - Geometry, latter by David Chandler. An excellent choice for high school geometry. I've written an in-depth review of this book and of its supplement Home Study Companion - Geometry, latter by David Chandler. Harold Jacobs Geometry A popular choice for geometry among homeschoolers, and well written. Please read my review. A popular choice for geometry among homeschoolers, and well written. Please read my review. Dr. Math Presents More Geometry An inexpensive companion to any high school geometry course with excellent explanations. 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9266	https://repositorio.uchile.cl/handle/2250/156925	Oxidation state of lithium species. XPS binding energies of lithium 1s electrons in salts, the metal and intercalated in molybdenum sulfide xmlui.mirage2.page-structure.toggleNavigation About Contact Help Sending publications How to publish Search Advanced Search xmlui.mirage2.page-structure.toggleNavigation View Item Home Facultad de Ciencias Artículos de revistas View Item Home Facultad de Ciencias Artículos de revistas View Item JavaScript is disabled for your browser. Some features of this site may not work without it. Browse byCommunities and CollectionsDateAuthorsTitlesSubjectsThis CollectionDateAuthorsTitlesSubjects My Account Login to my accountRegister Biblioteca Digital - Universidad de Chile Revistas Chilenas Repositorios Latinoamericanos Tesis LatinoAmericanas Tesis chilenas Related linksRegistry of Open Access RepositoriesOpenDOARGoogle scholarCOREBASE My Account Login to my accountRegister Oxidation state of lithium species. XPS binding energies of lithium 1s electrons in salts, the metal and intercalated in molybdenum sulfide Artículo Open/Download item_0000853332.pdf (1.690Kb) Publication date 1996 Metadata Show full item record Cómo citar Gonzalez, G. Cómo citar Oxidation state of lithium species. XPS binding energies of lithium 1s electrons in salts, the metal and intercalated in molybdenum sulfideFormato de cita . Copiar Cerrar Author Gonzalez, G.; Binder, H.; Abstract The binding energies of Li (1s) electron in a series of lithium salts, the metal, and lithium intercalated in molybdenum sulfide measured with respect to Au (4f7/2) agree with both the expected ionic character and oxidation state of lithium in the compounds. Thus, the binding energy increases with increasing charge on the observed atom. The Eb value (55.6 eV) as well as the sensitivity to air observed for lithium in the compound Li0.8MoS2 are similar to the metal. Indexation Artículo de publicación SCOPUS Identifier URI: ISSN: 03661644 Quote Item Boletin de la Sociedad Chilena de Quimica, Volumen 41, Issue 2, 2018, Pages 121-127 Collections Artículos de revistas Except where otherwise noted, this item's license is described as Attribution-NonCommercial-NoDerivs 3.0 Chile xmlui.footer.title 31 participating institutions More than 73,000 publications More than 110,000 topics More than 75,000 authors Published in the repository How to publish Definitions Copyright Frequent questions Documents Dating Guide Thesis authorization Document authorization How to prepare a thesis (PDF) Services Digital library Chilean academic journals portal Latin American Repository Network Latin American theses Chilean theses Dirección de Servicios de Información y Bibliotecas (SISIB) Universidad de Chile © 2020 DSpace Access my account
9267	https://blog.csdn.net/qq_18510183/article/details/102853712	关于函数凹凸性两种定义与二阶导数符号之间的联系证明_凹函数的二阶导数一定大于0吗-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作关于函数凹凸性两种定义与二阶导数符号之间的联系证明最新推荐文章于 2025-06-28 19:26:28 发布夜尘与夜于 2019-11-01 17:04:30 发布阅读量7.8w收藏 98 点赞数 41 CC 4.0 BY-SA版权分类专栏：高数版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。本文链接： GitCode 开源社区文章已被社区收录加入社区高数专栏收录该内容 8 篇文章订阅专栏本文深入探讨了函数的凹凸性概念，通过中点定义法和切线定义法详细解释了凹函数和凸函数的特性，并揭示了二阶导数与函数凹凸性的内在联系。什么是函数的凹凸性函数的凹凸性即对一个在某区间A上连续的函数，它的图像上凸或者上凹，则分别称为凸函数或者凹函数。而对于在某个区间内既有凹图像又有凸图像，则将凹图像所在区间称为函数的凹区间，凸图像所在区间则称为凸区间。例如， y = ln ⁡ x y=\ln x y=ln x 与 y = x 3 y=x^3 y=x 3 在 ( 0 , + ∞ ) (0,+\infty) (0,+∞)内都单调递增，但是前者为凸函数，后者为凹函数。凹凸性数学定义中点定义法随意取一个凹函数 f ( x ) f(x) f(x)，在其图像上取两点 x 1 , x 2 x_1,x_2 x 1,x 2，发现两点连线构成的直线总在两点之间的图像的上方。而两点横坐标中点 x 0 x_0 x 0的函数值 f ( x 0 ) f(x_0) f(x 0)显然小于 x 0 x_0 x 0在直线上的点的纵坐标。用数学语言来讲，就是对于一个在 [ a , b ] [a,b] [a,b]上有连续的函数 f ( x ) f(x) f(x)，总有 x 1 , x 2 ∈ [ a , b ] x_1,x_2 \in[a,b] x 1,x 2∈[a,b]，使得 f ( x 1 + x 2 2 ) < f ( x 1 ) + f ( x 2 ) 2 f(\frac{x_1+x_2}{2})<\frac{f(x_1)+f(x_2)}{2} f(2 x 1+x 2)<2 f(x 1)+f(x 2) 在 ( a , b ) (a,b) (a,b)恒成立，则称该函数为凹函数同理，凸函数的定义为：对于一个在 [ a , b ] [a,b] [a,b]上连续的函数，总有 x 1 , x 2 ∈ [ a , b ] x_1,x_2 \in[a,b] x 1,x 2∈[a,b]，使得 f ( x 1 + x 2 2 ) > f ( x 1 ) + f ( x 2 ) 2 f(\frac{x_1+x_2}{2})>\frac{f(x_1)+f(x_2)}{2} f(2 x 1+x 2)>2 f(x 1)+f(x 2) 在 ( a , b ) (a,b) (a,b)恒成立，则称该函数为凸函数。切线定义法描述性定义同样是观察凹凸函数的图像，发现凹函数的切线总在函数图像下方，而凸函数则相反。由此得出凹凸函数的描述性定义：对于在 [ a , b ] [a,b] [a,b]连续的函数，若函数切线全在函数图像下方，则其为凹函数，反之函数切线全在函数图像上方，则为凸函数。精确性定义函数图像全在切线上方，就是该点函数值总是大于等于该图像任意切线在该点所对应的纵坐标。如图：转换为数学语言，就是：对于在 ( a , b ) (a,b) (a,b)上连续的函数， ∀ x 1 , x 2 ∈ ( a , b ) \forall x_1,x_2\in(a,b) ∀x 1,x 2∈(a,b),有 f ( x 1 ) − [ f ′ ( x 2 ) ( x 1 − x 2 ) + f ( x 2 ) ] ≥ 0 f(x_1)-[f'(x_2)(x_1-x_2)+f(x_2)]≥0 f(x 1)−[f′(x 2)(x 1−x 2)+f(x 2)]≥0 则称该函数为凹函数。反之，在 ( a , b ) (a,b) (a,b)上连续的函数， ∀ x 1 , x 2 ∈ ( a , b ) \forall x_1,x_2\in(a,b) ∀x 1,x 2∈(a,b),若有 f ( x 1 ) − [ f ′ ( x 2 ) ( x 1 − x 2 ) + f ( x 2 ) ] ≤ 0 f(x_1)-[f'(x_2)(x_1-x_2)+f(x_2)]≤0 f(x 1)−[f′(x 2)(x 1−x 2)+f(x 2)]≤0 则称该函数为凸函数。二阶导数符号与函数凹凸性之间的关系前面对函数凹凸性做了简单的介绍,现在开始介绍二阶导数符号与函数凹凸性之间的关系. 观察下图凹函数的切线,切线的斜率似乎在不断增大. 实际上也确实如此,凹函数的切线斜率随着x的增大而增大,相对的,凸函数的切线斜率随着x的增大而减小.又二阶导数的几何意义正是图像切线的斜率,因此便对应起来. 即:函数为凹函数,则二阶导数大于0;函数为凸函数,则二阶导数小于零. 此时自然而然想到,如果将条件和结论倒换过来,该推论还会成立吗? 即二阶导数大于0,则函数为凹函数;二阶导数小于零,函数为凸函数这个推论,是否成立? 我们证明一下. 证明根据凹凸性的定义,有两种证明方法,下面一一介绍. 为了表达的方便,下面两种方法都是证明二阶导数大于0的函数是凹函数. 证明凸函数可自行类比. 中点定义法证明对在 ( a , b ) (a,b) (a,b)上可导,在 [ a , b ] [a,b] [a,b]上连续的函数 f ( x ) f(x) f(x), 设 x 1 , x 2 ∈ [ a , b ] x_1,x_2\in [a,b] x 1,x 2∈[a,b],则 x 0 = x 1 + x 2 2 , h = x 2 − x 0 = x 0 − x 1 x_0=\frac{x_1+x_2}{2},h=x_2-x_0=x_0-x_1 x 0=2 x 1+x 2,h=x 2−x 0=x 0−x 1 由拉格朗日定理,得: f ( x 2 ) − f ( x 0 ) = f ′ ( x 0 + θ 1 h ) h f ( x 0 ) − f ( x 1 ) = f ′ ( x 0 − θ 2 h ) h f(x_2)-f(x_0)=f'(x_0+\theta_1h)h \ f(x_0)-f(x_1)=f'(x_0-\theta_2h)h f(x 2)−f(x 0)=f′(x 0+θ 1h)h f(x 0)−f(x 1)=f′(x 0−θ 2h)h 又 x 2 = h + x 0 , x 1 = x 0 − h x_2=h+x_0,x_1=x_0-h x 2=h+x 0,x 1=x 0−h 则有: f ( x 0 + h ) − f ( x 0 ) = f ′ ( x 0 + θ 1 h ) h f ( x 0 ) − f ( x 0 − h ) = f ′ ( x 0 − θ 2 h ) h f(x_0+h)-f(x_0)=f'(x_0+\theta_1h)h \ f(x_0)-f(x_0-h)=f'(x_0-\theta_2h)h f(x 0+h)−f(x 0)=f′(x 0+θ 1h)h f(x 0)−f(x 0−h)=f′(x 0−θ 2h)h 二式相减,得到: f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) = [ f ′ ( x 0 + θ 1 h ) − f ′ ( x 0 − θ 2 h ) ] h f(x_0+h)+f(x_0-h)-2f(x_0)=[f'(x_0+\theta_1h)-f'(x_0-\theta_2h)]h f(x 0+h)+f(x 0−h)−2 f(x 0)=[f′(x 0+θ 1h)−f′(x 0−θ 2h)]h 观察到等式右边仍可用拉格朗日定理得到,于是有: ∃ ξ ∈ [ x 0 − θ 2 h , x 0 + θ 1 h ] , \exists \xi\in[x_0-\theta_2h,x_0+\theta_1h], ∃ξ∈[x 0−θ 2h,x 0+θ 1h],使得 f ′ ( x 0 + θ 1 h ) − f ′ ( x 0 − θ 2 h ) = f ′ ′ ( ξ ) ( θ 1 + θ 2 ) h f'(x_0+\theta_1h)-f'(x_0-\theta_2h)=f''(\xi)(\theta_1+\theta_2)h f′(x 0+θ 1h)−f′(x 0−θ 2h)=f′′(ξ)(θ 1+θ 2)h 成立. 两边各乘一个 h h h,得到: [ f ′ ( x 0 + θ 1 h ) − f ′ ( x 0 − θ 2 h ) ] h = f ′ ′ ( ξ ) ( θ 1 + θ 2 ) h 2 [f'(x_0+\theta_1h)-f'(x_0-\theta_2h)]h=f''(\xi)(\theta_1+\theta_2)h^2 [f′(x 0+θ 1h)−f′(x 0−θ 2h)]h=f′′(ξ)(θ 1+θ 2)h 2 显然在 f ′ ′ ( x ) > 0 f''(x)>0 f′′(x)>0 的情况下, f ′ ′ ( ξ ) ( θ 1 + θ 2 ) h 2 > 0 f''(\xi)(\theta_1+\theta_2)h^2>0 f′′(ξ)(θ 1+θ 2)h 2>0 因此 [ f ′ ( x 0 + θ 1 h ) − f ′ ( x 0 − θ 2 h ) ] h > 0 [f'(x_0+\theta_1h)-f'(x_0-\theta_2h)]h>0 [f′(x 0+θ 1h)−f′(x 0−θ 2h)]h>0 即 f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) f(x_0+h)+f(x_0-h)-2f(x_0) f(x 0+h)+f(x 0−h)−2 f(x 0) = [ f ′ ( x 0 + θ 1 h ) − f ′ ( x 0 − θ 2 h ) ] h > 0 =[f'(x_0+\theta_1h)-f'(x_0-\theta_2h)]h>0 =[f′(x 0+θ 1h)−f′(x 0−θ 2h)]h>0 将 f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) > 0 f(x_0+h)+f(x_0-h)-2f(x_0)>0 f(x 0+h)+f(x 0−h)−2 f(x 0)>0 变形,就能得到: f ( x 0 + h ) + f ( x 0 − h ) 2 > f ( x 0 ) \frac{f(x_0+h)+f(x_0-h)}{2}>f(x_0) 2 f(x 0+h)+f(x 0−h)>f(x 0) 即 f ( x 1 ) + f ( x 2 ) 2 > f ( x 1 + x 2 2 ) \frac{f(x_1)+f(x_2)}{2}>f(\frac{x_1+x_2}{2}) 2 f(x 1)+f(x 2)>f(2 x 1+x 2) 至此,命题得证. 切线定义法要根据切线定义证明二阶导数大于0.函数为凹函数, 就要证明: 对于在 ( a , b ) (a,b) (a,b)上连续的函数， ∀ x 1 , x 2 ∈ ( a , b ) \forall x_1,x_2\in(a,b) ∀x 1,x 2∈(a,b),有 f ( x 1 ) − [ f ′ ( x 2 ) ( x 1 − x 2 ) + f ( x 2 ) ] ≥ 0 f(x_1)-[f'(x_2)(x_1-x_2)+f(x_2)]≥0 f(x 1)−[f′(x 2)(x 1−x 2)+f(x 2)]≥0 反过来,就是对任何情况的 x 1 , x 2 x_1,x_2 x 1,x 2,都能够满足该式子.则依据两点的相对位置为分类标准进行分类讨论. 设 x 1 , x 2 ∈ [ a , b ] x_1,x_2\in [a,b] x 1,x 2∈[a,b] (1). x 1 = x 2 x_1=x_2 x 1=x 2时: f ( x 1 ) − [ f ′ ( x 2 ) ( x 1 − x 2 ) + f ′ ( x 2 ) ] = 0 f(x_1)-[f'(x_2)(x_1-x_2)+f'(x_2)]=0 f(x 1)−[f′(x 2)(x 1−x 2)+f′(x 2)]=0 满足定义式. (2). x 1 < x 2 x_1<x_2 x 1<x 2: 由拉格朗日定理: ∃ ξ ∈ [ x 1 , x 2 ] , \exist\xi\in[x_1,x_2], ∃ξ∈[x 1,x 2],满足 f ′ ( ξ ) = f ( x 1 ) − f ( x 2 ) x 1 − x 2 f'(\xi)=\frac{f(x_1)-f(x_2)}{x_1-x_2} f′(ξ)=x 1−x 2f(x 1)−f(x 2) 又 f ′ ′ ( x ) > 0 f''(x)>0 f′′(x)>0, f ′ ( x ) f'(x) f′(x)在 [ x 1 , x 2 ] [x_1,x_2] [x 1,x 2]上递减且 ξ < x 2 \xi<x_2 ξ<x 2 即 f ( x 1 ) − f ( x 2 ) x 1 − x 2 = f ′ ( ξ ) < f ′ ( x 2 ) \frac{f(x_1)-f(x_2)}{x_1-x_2}=f'(\xi)<f'(x_2) x 1−x 2f(x 1)−f(x 2)=f′(ξ)<f′(x 2) 又 x 1 − x 2 < 0 x_1-x_2<0 x 1−x 2<0 所以将其乘到右边需要变换不等式符号,即 f ( x 1 ) − f ( x 2 ) > f ′ ( x 2 ) ( x 1 − x 2 ) f(x_1)-f(x_2)>f'(x_2)(x_1-x_2) f(x 1)−f(x 2)>f′(x 2)(x 1−x 2) f ( x 1 ) − [ f ′ ( x 2 ) ( x 1 − x 2 ) + f ( x 2 ) ] > 0 f(x_1)-[f'(x_2)(x_1-x_2)+f(x_2)]>0 f(x 1)−[f′(x 2)(x 1−x 2)+f(x 2)]>0 满足定义式. (3) x 1 > x 2 x_1>x_2 x 1>x 2: 同(2),利用拉格朗日定理得到: x 2 < ξ , f ′ ( x 2 ) < f ′ ( ξ ) = f ( x 1 ) − f ( x 2 ) x 1 − x 2 x_2<\xi,f'(x_2)<f'(\xi)=\frac{f(x_1)-f(x_2)}{x_1-x_2} x 2<ξ,f′(x 2)<f′(ξ)=x 1−x 2f(x 1)−f(x 2) f ( x 1 ) − f ( x 2 ) x 1 − x 2 > f ′ ( x 2 ) \frac{f(x_1)-f(x_2)}{x_1-x_2}>f'(x_2) x 1−x 2f(x 1)−f(x 2)>f′(x 2) 这次因为 x 1 − x 2 > 0 x_1-x_2>0 x 1−x 2>0,所以化简时无需改变不等式符号.即: f ( x 1 ) − f ( x 2 ) > f ′ ( x 2 ) ( x 1 − x 2 ) f(x_1)-f(x_2)>f'(x_2)(x_1-x_2) f(x 1)−f(x 2)>f′(x 2)(x 1−x 2) f ( x 1 ) − [ f ′ ( x 2 ) ( x 1 − x 2 ) + f ( x 2 ) ] > 0 f(x_1)-[f'(x_2)(x_1-x_2)+f(x_2)]>0 f(x 1)−[f′(x 2)(x 1−x 2)+f(x 2)]>0 满足定义式综上,命题得证. 确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用夜尘与夜关注关注 41点赞踩 98 收藏觉得还不错? 一键收藏 3评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录函数的凹凸性 universe_1207的博客 08-07 3672 文章目录凹凸函数的定义引理定理6.14定理6.15Jesen不等式例题拐点定理6.16定理6.17 凹凸函数的定义 fff 定义在III上 ∀x1,x2∈I,∀λ∈(0,1)\forall x_1,x_2\in I,\forall\lambda\in(0,1)∀x1,x2∈I,∀λ∈(0,1) ①若f(λx1+(1−λ)x2)≤λf(x1)+(1−λ)f(x2)f(\lambda x_1+(1... 3 条评论您还未登录，请先登录后发表或查看评论 ...函数的凹凸性和拐点_ 凹曲线的一阶导数和二阶导数的关系 9-23 凹凸性的定义:若在函数中任取两点,它们的函数值和的一半大于它们中心位置的函数值,则该函数图像是凸的。它们的函数值和的一半小于它们中心位置的函数值,则该函数图像是凹的。凹曲线:一阶导数递增,二阶导数大于零。凸曲线:一阶导数递减,二阶导数小于零。(二阶导数正凹负凸) 注:拐点指的是二阶导数等于零或... 机器学习基础概念解析 9-27 在凸函数的情况下比较适用,但是在凹函数的情况下,会卡在局部最优值附近 4、RMSProp 是对AdaGrad的一个变体 RMSprop是Geoff Hinton提出的一种自适应学习率方法。 RMSprop和Adadelta都是为了解决Adagrad学习率急剧下降问题的,Hinton建议设定γ为 0.9,学习率η为 0.0 0 1。 7.解决训练样本类别不平衡问题? 现象:训练样本中,正... 函数的凹凸性证明 _ 函数不等式大招——凹凸性反转~ weixin_34078749的博客 12-28 5108 写在最前面：先说清楚，该法由海明大佬 @未灬秋色原创，大家若想更加深入了解，可浏览专栏“海明的放缩笔记”海明的放缩笔记zhuanlan.zhihu.com这篇文章相当于我对他思想的一个发展和传播（主要是当初我看他的文章期初也没搞懂这个方法，考虑到海明大佬的巨大影响力，以及很多和我一样起初没看懂的同学，或者是根本没听过这个方法的同学，我萌生了写这篇文章的想法）好久没写函数和不等式类的文章了~文章... 凸优化：凸函数定义及其常见判定方式最新发布 m0_69490017的博客 06-28 1056 若domfdomf是凸集，且fθx1−θy⩽θfx1−θfyfθx1−θy⩽θfx1−θfy对所有xy∈domf 0⩽θ⩽1xy∈domf 0⩽θ⩽1都成立，称fRn→RfRn→R是凸函数。若−f-f−f是凸函数，则称fff是凹函数。此外，当x≠y 0 θ1xy 0 θ1，上述不等式严格成立时，称fff为严凸函数。注记。凸凹函数特性解析 9-18 凸函数:斜率不断上涨,即斜率的导数大于 0,即原函数的二阶导数大于 0 凹函数:斜率不断下跌,即斜率的导数小于 0,即原函数的二阶导数小于 0 SVM详解与数学原理 9-6 SVM详解手推SVM 1-凹函数 Y=X^2是凹函数因为y的导数是2x,二阶导数是2>0 2.拉格朗日 (1)拉格朗日对偶问题一般是凹函数(求最大值),即使原问题是非凸的,变成对偶问题更容易优化求解 (2)拉格朗日对偶函数一定是凹函数 (3)凹函数一般都有极大值凹函数与凸函数 Concave and convex functions qlkaicx的博客 02-20 8435 凸函数之所以呈现凹状，与我们主观常识不太一样，是因为凸凹性的定义是基于函数在任意两点之间的线段与函数图像之间的位置关系。具体来说，如果一个函数在任意两点之间的线段都在函数图像之上，则该函数被称为凸函数；如果线段都在函数图像之下，则该函数被称为凹函数。这个定义与我们日常生活中的直观感受可能不太一致，因为我们通常是根据一个物体或形状的表面来判断它是凸还是凹的。例如，一个凸起的山丘或球面通常被认为是凸的，而一个凹陷的坑洞或碗则被认为是凹的。函数的凹凸性证明 _ 函数的凹凸性与不等式 weixin_39963255的博客 01-02 7954 一、函数的凹凸性考察抛物线的图像，如图1.1所示，它有一个明显的特征：对于图像上任意两点，可以连成一条弦，而与之间的曲线可以看成一条弧.显然，弧上所有的点都在相对应的弦的下面.严格地说，对任意 ,横坐标为的点所对应弧上点之纵坐标小于弦上点之纵坐标，即但若令则有且，于是以上不等式可以写成具有以上特征的函数，我们通常称为下凸函数。对于一般的来... 证明:二阶导函数大于零时为凹函数 _ 二阶导数大于 0 是凹还是凸 9-16 如果这个函数的二阶导数大于 0,那么这个函数就是凸函数;反之,则为凹函数(简便方法)(不一定二阶可导,)重要性质:琴生不等式机器学习中损失函数得到凹函数怎么办呢?取反即可基础概率公式两点分布:二分类问题泊松分布:类似于价格分布,比如卖盒饭,大部分的盒饭价格为2 0 元,豪华版盒饭价格更高,但是数量较少,也有精简版... 函数的单调性与曲线的凹凸性 _曲线是凸的为什么会单调递增 9-23 一阶导数表示函数图像的斜率,二阶导数表示函数图像的斜率的变化率,当二阶导数大于 0 时,表示函数图像的斜率在增加,表示函数图像是凹的,当二阶导数小于 0 时,表示函数图像的斜率在减小,表示函数图像是凸的。我们进行证明: 函数的拐点: 如图所示,在x 0 左侧的图像是凸的,在x 0 右侧的图像是凹的,我们把函数经过这一点,... 函数的凹凸性证明数学分析提高课~第8次：不等式 _证明，为赋新词强说愁 weixin_39907596的博客 01-01 1650 从中学开始，不等式的证明题目就因其突变的画风，不走寻常路的逻辑，总是让人感觉被耍的团团转。本以为上了大学，有了更高级的定理和概念，能把不等式打的丢盔弃甲，现实却PiaPia打脸，许多不等式即便是我们数分老师，也得大战三百回合，可能也还是头破血流的做不出来。我们参照一些经典的数分教材，简单总结了一下不等式在数学分析中，尤其是题目中的足迹，希望能对大家的解题有所帮助。参照北大张筑生老师的教材... 利用二次导数对函数凹凸性的证明热门推荐 silence1214的专栏 03-18 3万+ 很多人其实都知道可以利用函数的二次导数来判断函数的凹凸性，但是很多人忘记了怎么来证明的，在这里我来再次证明一下。求证：若f(x)在(a,b)内连续并且二次可导，若f''(x)>0 则函数凹，反之函数凸前序：先给出几个定理以及说明。关于函数凹凸性的说明：函数 f(x)在(a,b)内连续，对于任意的a 若f(x 0)(f(x1)+f(x2))/2，则认为函数(向上)凸二阶导数怎么表示，matlab 02-06 二阶导数是数学中描述函数变化率的变化程度的概念，在微积分中有广泛应用。它可以帮助我们了解曲线的凹凸性和拐点等特性。 ### 1. 数学表示对于一元函数 f(x)f(x)，其一阶导数记作 f′(x)f′(x) 或者 ( \frac{... 读《反脆弱性》：凹凸性资料搜集 weixin_45981643的博客 02-28 926 经济学中函数的凸凹性质问题在现代经济学的讨论中，我们经常遇到凸函数、凹函数以及拟凹函数、拟凸函数等概念，例如生产可能性边界曲线是凹函数，无差异曲线是凸函数等等，但是这些数学名词对于非专业人员来说比较抽象，有的文章或教材采取形象的说法，比如说曲线凸向原点或凹向原点、图形是凸的、上凸函数、下凸函数等等，这样一来，就将严谨的数学概念搞的不伦不类，有的教科书甚至错误地定义了凸性和凹性。一、关于凸函数... 【高等数学】【强化第二章】一元函数微分学：导数与微分、导数应用、极值最值判断 hiliang521的博客 10-25 1708 【高等数学】【强化第二章】一元函数微分学：导数与微分、导数应用、极值最值判断 xy轴函数图像的导数与微分：函数变化率的几何意义，理解导数与微分的概念，掌握函数变化率的计算导数和微分是微积分中的两个基本概念，它们描述了函数的变化率。导数是函数在某一点变化率的瞬时值，而微分是函数在某一点变化率的近似值。导数的几何意义可以从函数图像的切线来理解。切线是函数图像在某一点处的... 什么是驻点和拐点_驻点、极值点、拐点、鞍点的区别与联系 weixin_33437929的博客 02-05 8764 最近有些考研的小伙伴问到我这个问题，正好也给自己梳理一下思路，毕竟在机器学习里面这4个概念也是非常重要的，不过这里由于知识所限，就只整理跟考研部分比较相关的知识点了。既然是4种点，首先就需要将其进行大致的分类，大致来说如下。$$ \begin {cases}一元函数 \quad \begin {cases}一阶导数f'(x) \quad 驻点、极值点、鞍点 \[3ex] 二阶导数 f''(x... 双曲函数奇偶性第二百一十二夜：高考押题-双曲 _函数的性质 weixin_39628551的博客 12-20 292 “上蒸下煮，大暑。”去年的现在，重庆已进入烧烤模式。而今年，大雨如注，水天一色。似此心情非昨夜，争如人事不遂人。1 围观：一叶障目，抑或胸有成竹显然，本题是冲着16题去的，背景便是双曲函数。选项①考查奇偶性，选项②考查恒变换，选项③考查单调性，选项④考查极值与最值，选项⑤则考查反函数。知识繁多，内容庞杂，心里阴影面积略大。2 套路：手足无措，抑或从容不迫3 脑洞：浮光掠影，抑或醍醐灌顶1.命题背... 函数的凹凸性证明 _ 二阶导数与函数凹凸性证明 weixin_39746241的博客 12-20 3092 证明设f(x)在[a,b]上连续，在(a,b)内具有一阶和二阶导数，那么若在(a,b)内f"(x)>0，则f(x)在[a,b]上的图形是凹的。设x1和x2是[a,b]内任意两点，且x1，记(x1+x2)/2=x 0，并记x2-x 0=x 0-x1=h，则x1=x 0-h,x2=x 0+h，由拉格朗日中值公式得f(x 0+h)-f(x 0)=f'(x 0+θ1h)h,f(x-0 f()x 0-h)=f'(x... 函数的凹凸性证明判断复杂 _函数的凹凸性 weixin_42499800的博客 01-02 4034 判断无人机能量x关系函数的凹凸性（函数是关于v和drt的二元函数）函数有非常多的参数，极其复杂，看到就烦，我首先用画函数的方法通过图像法来观察，但是画出来的图像不忍直视（或许是我画的图像不对，反正看起来就非常low，一看就知道图像不正确）因此放弃了通过几天的烦恼，终于想到一个法子，先求带有参数的海森矩阵然后再把参数值带入所求到的式子中，分别列举v和drt的取值，将所得的值带入到海森矩阵中q求海森矩... 函数凹凸性 qq_25018077的博客 03-08 6028 1. 凸函数（向上凸函数和向下凸函数） 2. 凸曲线的几何特征分析（“弦在弧上”） 3. 向下凸函数（凸函数、严格凸函数、严格向下凸函数）的定义 4. 向上凸函数（严格向上凸函数）的定义 5. 凹曲线（凹弧）、凸曲线（凸弧）的定义及其与凸函数及向上凸函数的对应关系 6. 凸曲线与其切线的位置关系 7. 向下凸函数（严格向下凸函数）的充分必要条件 8. 凸函数的判定（一阶导数单调增加则向下凸；一阶导数单调减小向下凸；二阶导数大于等于零则向下凸；二阶导数小于等于零. 函数的凹凸性证明积累篇\|含积分式 _证明题的类型总结~~ weixin_42522669的博客 01-01 2249 关注“考研数学帝(ID：King_maths)”↑突破来自点滴的积累~含积分式证明题的类型总结~(突破含积分式的相关证明题~)1：积分不等式的证明在积分不等式的证明中，无论它如何千变万化，都必须找一个常用的不等式作为“初始不等式”，在这一点上与微分不等式的证明是一致的.下面列出最常用的几种类型的初始不等式：(1)积分估值不等式；(2)基本不等式：(3)柯西—施瓦茨不等式：它的证明方法是根... 函数的凹凸性证明微分中值定理、洛必达法则、泰勒公式以及 _函数的单调性和凹凸性... weixin_36104594的博客 12-26 3104 [大家好，我是专升本数学学霸，这次来谈论微分中值定理、洛必达法则、泰勒公式和数的单调性和凹凸性这些内容。那你知道微分中值定理、洛必达法则、泰勒公式这些内容呢?没关系！学霸来帮你来了。一、微分中值定理1.罗尔定理①(费马引理)设函数 f(x)在点x 0 的某领域U(x 0)内有定义，并且在 x 0 处可导，如果对任意的x∈U(x 0)，有那么f'(x 0)=0。 ②罗尔定理如果函数 f(x)满足 (1)在闭区间a,... 函数的凹凸性证明【最优化方法】拟凸 _函数定义及其证明 weixin_33841242的博客 01-02 4242 一、下水平集的定义函数的下水平集（）定义如下凸函数的下水平集是凸的下水平集是凸的，该函数不一定是凸函数（个人理解：原因在于下水平集描述的是定义域中的集合，无法反映值域的集合凹凸性）二、Epigraph的定义函数的定义如下凸函数是凸的如下证明：所以（由于和选取的任意性, 所以不妨令）并且所以是凸函数建立了凸集合与凸函数之间的联系（个人理解： ... 关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司夜尘与夜博客等级码龄11年 30 原创87 点赞 194 收藏 39 粉丝关注私信猜你想问如何利用中点定义判断函数凹凸性？切线定义法如何体现凸函数特性？二阶导数怎样反映函数凹凸变化？热门文章关于函数凹凸性两种定义与二阶导数符号之间的联系证明 78177 海涅定理 24066 极限唯一性的三种反证法 9649 常用三角函数恒等式 9571 投影直角三角形法求空间线段实长 6266 分类专栏微积分20篇目录1篇计算机二级英语四六级建筑制图高数8篇 python 风变课程笔记展开全部收起上一篇：投影直角三角形法求空间线段实长下一篇：关于极坐标体系思想在三维空间的延伸最新评论关于函数凹凸性两种定义与二阶导数符号之间的联系证明 weijackTAO:严格凹凸高数好题错题公式集合 CSDN-Ada助手:非常感谢CSDN博主分享的这篇博客《高数好题错题公式集合》，对学习高数的同学来说一定非常有帮助。我认为，下一篇博客可以继续围绕高数展开，例如可以写一篇《高数实战：应用场景解析》，分享高数在实际应用中的具体场景和解决方法，这样的技术文章对其他用户一定也很有启发和帮助。相信会有更多读者受益于你的分享。为了方便博主创作，提高生产力，CSDN上线了AI写作助手功能，就在创作编辑器右侧哦～（）诚邀您来加入测评，到此（积分法 BODH0x:看完了，写得很好，指出以下几点问题: 复合函数换元法3的dx应该改dfx,6的lnx应该改x。三角换元法说要给的常用三角函数积分公式没有给出。极限唯一性的三种反证法素颜女神:问一下A—B绝对值为什么an—A减an—B 关于函数凹凸性两种定义与二阶导数符号之间的联系证明 Victorycarlson:中点定义法只是一个特例，在（a，b）区间全都是成立的大家在看 gemini 模型更新、新命名为gemini-pro-latest 1107 SpringBoot+MyBatis+Vue3指南 640 2025量子计算算法开发三大焦点 112 前端数据源无缝切换实战 177 RTX4090：AI创作算力新巅峰 73 最新文章高数好题错题公式集合不错的题目关于不定积分和积分上限函数区别的简单讨论 2020年 1篇 2019年 29篇目录什么是函数的凹凸性凹凸性数学定义中点定义法切线定义法描述性定义精确性定义二阶导数符号与函数凹凸性之间的关系证明中点定义法证明切线定义法展开全部收起目录什么是函数的凹凸性凹凸性数学定义中点定义法切线定义法描述性定义精确性定义二阶导数符号与函数凹凸性之间的关系证明中点定义法证明切线定义法展开全部收起上一篇：投影直角三角形法求空间线段实长下一篇：关于极坐标体系思想在三维空间的延伸分类专栏微积分20篇目录1篇计算机二级英语四六级建筑制图高数8篇 python 风变课程笔记展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论 3 被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 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9268	https://www.youtube.com/watch?v=WCvC4vT7t1w&pp=ygULI2NhdHBvYmxlbXM%3D	Work rate problems mathonify 75200 subscribers 275 likes Description 21850 views Posted: 24 Mar 2020 In this video I go through two work rate problems, and point out why I think this can often be a confusing topic. 19 comments Transcript: in this video I'm going to go through a couple of work rate problems the problem on the screen at the moment is not a work rate problem but I'm going to use it to illustrate why work great problems can often be confusing so this problem is just about speed and distance in time now we're used to thinking about rates of change such as speed km/h things like that and work rate problems we don't have rates that we used to think about so anyways let's get straight into this problem and then I'll show you why I'm doing this one in particular so this problem says it takes 1.5 hours for John to travel one kilometer John is a snail that's why it's going so slow Tim can travel one kilometer in two hours Tim is also a snail and Tim's traveling even slower than John clearly and actually if they were snouts then they're pretty pretty fit snails to travel that far in that amount of time anyways the question says if they start at the same time how long will it take John and Tim to travel one kilometer all together hopefully you're thinking that you want the speed in terms of kilometers per hour for each one so if John travels 1.5 hours in one kilometer what's his speed well if we break this time up into thirds so if he's traveled 1.5 hours this could be thought of as 30-minute blocks so 30 minutes plus 30 minutes minutes plus 30 minutes and so we want the speed in terms of Commodus per hour so how far would he travel in one hour so two of those 30 minute blocks well that would be two thirds of his original distance right so this is going to be two thirds of a kilometer so for John we have a speed of 2/3 kilometers per hour okay for Tim his speed is a little bit easier he travels one common in two hours how many comments will he travel in one hour that's going to be half a kilometre half a kilometer per hour okay so we have the speeds of these super-fit snails and now we're wondering well how long would it take for them both to travel one kilometer all together so they start at the same time and they travel travel travel until they have all together both of them traveled one comma now John is going to cover more distance than Tim but they're going to travel for the same amount of time right now both going to stop once their collective distance equals one kilometer so for this we can think of this formula distance equals speed times time so John's distance is going to be his speed times the time that he's traveling for Tim's distance is going to be his speed times the time he's traveling for and if we add these distances together we'll have the total distance have covered in this case we know it needs to equal one kilometer so we can say that John's distance is his speed times the time he's traveled let's call that T adding Tim's distance which will be his speed times the time that he's traveled for now we know in this case they're going to be traveling for the same amount of time so this is also going to be multiplied by T the time that Tim has traveled for and the distances added together need to equal one one kilometer so I hope you followed my logic so far because now we've ended up with an equation that we can actually solve for T so we need to add these fractions together 2/3 plus 1/2 2/3 plus 1/2 you can multiply this one by 2 this would be 4 over 6 plus multiply this one by 3 this would be 3 over 6 this equals 4 plus 3 is 7 7 over 6 so adding these two fractions together we get 7 over 6 T equal to 1 and then solving for T then we'd multiply it by 6 divided by 7 T is going to be 6 on 7 hours which is approximately approximately 51 minutes but the question doesn't ask you to write it in terms of only particular units so 600 hours or 51 minutes would be fine therefore a final answer so I'm hoping that is a question that is slightly more intuitive to answer because you're thinking of speed and distance and we know distance is speed times time so now I want to look at a pretty much identical work rate problem this problem says it takes 1.5 hours for John to mow the lawn Tim can mowed the same lawn in two hours how long will it take John and Tim working together to mow the lawn so hopefully you can see these are pretty much identical problems but in this case we're not talking about distance and time we're talking about lawns and that's where I think the confusion can come in with work rate problems because it's not a typical rate that you're used to it's not commas per hour or meters per second so in this case however we can still use the exactly the same logic so the rate that John can cut the lawn is rate is two thirds of a lawn so two thirds not two twos two thirds of a lawn per hour from our previous question right we did we've already done that working for Tim his rate is a half a lawn per hour and if we want the amount of lawn they cut to equal one write the total lon we can do we can use this formula right before we said distance equals speed times time in this case we're talking about lawns the amount of lawn they cut is going to be their rate times the amount of time they're cutting the lawn for the same logic but it's not a formula that we typically think of the amount of lawn you cut is the lawn rate times the time you're cutting lawn for however the logic is still going to work here so the amount of lawn that John cuts is 2/3 times the time is cutting lon plus the amount of lawn that Tim cuts is his rate times the time he's coming on for this needs to equal why lawn in total and we've really worked out this equation as well we found T to equal six on seven hours or approximately 51 minutes okay so the reason I went through this problem which is not a work quote problem essentially is to show that often the confusion comes in with work rate problems because the rates that we are thinking about lawns per hour or you know it might be boxes per day it's the things that you're making that are not typical we're not used to thinking about rates in that way so again I'm hoping that you had a slightly more intuitive understanding of this question and that will help you understand how I'm solving this problem down here okay one more problem that I'm going to go through problem 2 it says it takes six hours for pump a used alone to fill a tank of water pump be used alone takes eight hours to fill the same tank we want to use three pumps a B and another pump C to fill the tank in two hours what should be the rate of pump see how long would it take pump C use the line to fill the tank okay so Pompeii it takes six hours to fill the tank so it's rates would be well how much of the tank would it feel in one hour it would fill a six of the tank so we've got six a sip of whatever units we're measuring the tank in so units cubed per hour for pump B it fills the tank in eight hours I'm going to convert this to a rate this would be one eighth of the tank I could actually write tank here so maybe that would be easier one eighth of the tank per hour I'll change this one to tank as well and for pump C we don't know how long it would take so let's say it takes C X hours its rate would be one on X tanks per hour okay now we want to come up with an equation similarly to how I solved from one what am I trying to do I'm trying to fill up the tank and if I was just using pump a taking six hours if I was just using pump me take me eight hours if I was just using pump C it would take me exhales but I want to use all three and I want it to take two hours so we have the rate for pump a that's a sixth if we multiply that by the time we're pumping for that will give me how much of the water I pump with pompe add on the amount of water we pump with pump B that would be its rate times two and then add on the amount we pump with pump see that would be its rate on x times 2 so that's 2 on its and this will give me one full tank of water simplifying this then I have 206 which I could write as 1/3 2 on 8 which I can write as 1/4 and 2 on X which I'll leave as is this equals 1 1/3 plus 1/4 I need to work that out I'll make some notes over here at third plus 1/4 multiply this one by 4 this will be 4 on twelve multiply this one by three this would be three on 12 this equals seven on twelve so these two fractions simplified to seven on twelve plus two on X equal to one and one take seven on twelve that leaves you with the remainder of five on twelve so two on x equals five on 12 and now you need to be able to rearrange this in terms of X if you're used to working with algebraic fractions you'll be confident with this if you're not it might take you a minute but here you can't cross multiply so 12 times 2 and then 5 times X we would have 5 X equal to 24 and then X equal to 24 on 5 and remember X was the number of hours it takes pump C to fill the tank remember that was the question what rate what should be the rate of pump C and how long would it take pump C to fill the tank essentially we've answered both of those so we could leave that as is so 24 on 5 hours or if you want to convert that to a fraction sorry a decimal if you want to convert that to a decimal that would be 4 point 8 hours okay and so then the rate would be one on 4 point 8 hours as a fraction of the tank per hour okay so there you go hopefully that gives you some strategies to approach work great problems the other thing I want to point out is you can think of these problems in terms of a graph so if you are trying to graph these rates pump a you would have a straight line because it's always pumping at the same rate this would be this would have a gradient of 1/6 pump B would have a slightly less of a gradient because it has a slower rate and pump C was the fastest one so this would be our graph here and if we're trying to work out how long it takes to fill the tank they're starting at the same time and they're finishing at the same time so this axis here would be time in order to solve this I have set time they're all pumping four so that would be let's just call that t1 and all of these values over here which is the amount of water they pump they all need to add up to the total amount of water in the tank so in this case we essentially said they all need to add up to one right because the rate was in tanks per hour so this was amount of tanks so that's another way to think about this you can think about that in terms of a graph if that helps you to make sense of it again I do believe myself that work rate problems can be confusing because they are not the types of rates we typically used to think of such as speed distance time they're often you know lawns per hour or tanks per hour which gets confusing all right hopefully you found this useful please leave a like if he did subscribe it from seeing more content and I'll see in the next one bye for now
9269	https://mathmonks.com/trigonometric-functions/domain-and-range-of-trigonometric-functions	Domain and Range of Trigonometric Functions - Table & Examples Shapes Rectangle Square Circle Triangle Rhombus Squircle Oval Hexagon Pentagon Trapezoid Kite Parallelogram Quadrilateral Polygon Nonagon Heptagon Decagon Octagon Ellipse Parallelepiped Tetrahedron Cylinder Prism Sphere Pyramid Frustum Polyhedron Dodecagon Dodecahedron Octahedron Torus Cube Cone Hyperbola Rectangular Prism Fibonacci Sequence Golden Ratio Parabola Worksheets Calculators Fraction Calculator Mixed Fraction Calculator Greatest Common Factor Calulator Decimal to Fraction Calculator Angle Arithmetic Whole Numbers Rational Numbers Place Value Irrational Numbers Natural Numbers Binary Operation Numerator and Denominator Decimal Order of Operations (PEMDAS) Scientific Notation Symmetry Fractions Triangular Number Complex Number Binary Number System Logarithm Binomial Theorem Quartic Function Mathematical Induction Group Theory Modular Arithmetic Euler’s Number Inequalities Sets De Morgan’s Laws Transcendental Numbers About Us Shapes Rectangle Square Circle Triangle Rhombus Squircle Oval Hexagon Pentagon Trapezoid Kite Parallelogram Quadrilateral Polygon Nonagon Heptagon Decagon Octagon Ellipse Parallelepiped Tetrahedron Cylinder Prism Sphere Pyramid Frustum Polyhedron Dodecagon Dodecahedron Octahedron Torus Cube Cone Hyperbola Rectangular Prism Fibonacci Sequence Golden Ratio Parabola Worksheets Calculators Fraction Calculator Mixed Fraction Calculator Greatest Common Factor Calulator Decimal to Fraction Calculator Angle Arithmetic Whole Numbers Rational Numbers Place Value Irrational Numbers Natural Numbers Binary Operation Numerator and Denominator Decimal Order of Operations (PEMDAS) Scientific Notation Symmetry Fractions Triangular Number Complex Number Binary Number System Logarithm Binomial Theorem Quartic Function Mathematical Induction Group Theory Modular Arithmetic Euler’s Number Inequalities Sets De Morgan’s Laws Transcendental Numbers About Us Search Table of Contents Sine (y = sin x) Cosine (y = cos x) Tangent (y = tan x) Cosecant (y = cosec x) Secant (y = sec x) Cotangent (y = cot x) Graphing How To Find Domain and Range For Inverse Trigonometric Functions Solved Examples Last modified on March 7th, 2025 chapter outline Sine (y = sin x) Cosine (y = cos x) Tangent (y = tan x) Cosecant (y = cosec x) Secant (y = sec x) Cotangent (y = cot x) Graphing How To Find Domain and Range For Inverse Trigonometric Functions Solved Examples Home » Trigonometry » Trigonometric Functions » Domain and Range of Trigonometric Functions Domain and Range of Trigonometric Functions Domain refers to all possible input values (usually represented by x) for which a function is true, while range represents all possible output values (usually represented by y) the function can produce. For trigonometric functions, the domain and range are closely related to their periodic nature and the unit circle. They define the properties of the functions. Sine (y = sin x) Since there is no restriction on the input x, the sine function holds true for all real numbers. Its output ranges between -1 and 1. 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% Domain: (-∞, ∞) Range: [-1, 1] Cosine (y = cos x) Like sine, cosine is defined for all real numbers. Its output also oscillates between -1 and 1. Domain: (-∞, ∞) Range: [-1, 1] Tangent (y = tan x) The tangent function is undefined at odd multiples of π 2 because the cosine of those angles is 0 (division by zero is undefined). Its output takes all real values. Domain: x∈R−{π 2+n π\|n∈Z} Range: (-∞, ∞) Cosecant (y = cosec x) For being the inverse of the sine function, the cosecant function is undefined at integer multiples of π. Its range is given below: Domain: x∈R−{n π\|n∈Z} Range = (-∞, -1] ∪ [1, ∞) Secant (y = sec x) For being the inverse of the cosine function, the secant function is undefined at odd multiples of π 2. Its range is given below: Domain: x∈R−{π 2+n π\|n∈Z} Range: (-∞, -1] ∪ [1, ∞) Cotangent (y = cot x) For being the inverse of the tangent function, the cotangent function is undefined at integer multiples of π. Its output takes all real values. Domain = x∈R−{n π\|n∈Z} Range = (-∞, ∞) Here is a summary table for the domain and range of all six trigonometric functions: Graphing Here are the graphs of the six trigonometric functions: In the graphs, the x-and-y-axis represents the domain and range of the functions, respectively. Thus, we can find the domain and range of these functions from their corresponding graphs. How To Find Domain and Range Let us find the domain and range of the function y = 3 cos x – 1 Step 1: Studying the Parent Function As we know, the domain of the cosine function is (-∞, ∞), and the range is [-1, 1] -∞ < x < ∞ …..(i) -1 ≤ cos x ≤ 1 …..(ii) Step 2: Transforming the Range of the Parent Function On multiplying (ii) by 3, we get -3 ≤ 3 cos x ≤ 3 Subtracting 1 from both sides, we get ⇒ -3 – 1 ≤ 3 cos x – 1 ≤ 3 – 1 ⇒ -4 ≤ 3 cos x – 1 ≤ 2 ⇒ -4 ≤ y ≤ 2 …..(iii) Step 3: Determining the Domain and Range Thus, from (i) and (iii), we get The domain of the function is (-∞, ∞) The range of the function is [-4, 2] For Inverse Trigonometric Functions Here is the table representing the domain and range of the inverse of the six basic trigonometric functions: Solved Examples Find the domain and range of the function: y = sec 2x + 4 Solution: As we know, the domain and the range of the secant function (sec x) are: Domain = x∈R−{π 2+n π\|n∈Z} …..(i) Range = (-∞, -1] ∪ [1, ∞) …..(ii) Here, the function is y = sec 2x + 4 From (i), we get 2 x=π 2+n π, n ∈ ℤ ⇒ x=π 4+n π 2 Thus, the domain is x∈R−{π 4+n π 2\|n∈Z} Now, from (ii), we get sec 2x ∈ (-∞, -1] ∪ [1, ∞) Adding 4 to both sides, we get ⇒ sec 2x + 4 ∈ (-∞, 3] ∪ [5, ∞) Thus, the range is (-∞, 3] ∪ 5, ∞) Find the domain and range of f(x) = tan-1 (2x – 1) Solution: As we know, the domain of tan-1 x is (-∞,∞), and the range of tan-1 x is (−π 2,π 2) The argument of the function is 2x – 1, which is a linear expression. Since there are no restrictions on x, 2x – 1 can take any real value. Thus, the domain of f(x) is (-∞,∞) Since the range of tan-1 x depends only on the output of the function, the transformation 2x – 1 does not affect the range of the inverse tangent function. Thus, the range of f(x) is (−π 2,π 2) Last modified on March 7th, 2025 [About us Contact us Privacy Policy Categories Algebra Arithmetic Geometry Statistics Trigonometry Grades 1st Grade2nd Grade3rd Grade4th Grade5th Grade6th Grade7th Grade8th Grade9th Grade10th Grade11th Grade12th Grade Join Our Newsletter Subscribe to our weekly newsletter to get latest worksheets and study materials in your email. Email Please leave this field empty. © 2025 Mathmonks.com. All rights reserved. Reproduction in whole or in part without permission is prohibited. ✕ Do not sell or share my personal information. You have chosen to opt-out of the sale or sharing of your information from this site and any of its affiliates. 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9270	https://openstax.org/books/precalculus-2e/pages/12-3-continuity	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Precalculus 2e 12.3 Continuity Precalculus 2e12.3 Continuity Search for key terms or text. Learning Objectives In this section, you will: Determine whether a function is continuous at a number. Determine the numbers for which a function is discontinuous. Determine whether a function is continuous. Arizona is known for its dry heat. On a particular day, the temperature might rise as high as and drop down only to a brisk Figure 1 shows the function where the output of is the temperature in Fahrenheit degrees and the input is the time of day, using a 24-hour clock on a particular summer day. Figure 1 Temperature as a function of time forms a continuous function. When we analyze this graph, we notice a specific characteristic. There are no breaks in the graph. We could trace the graph without picking up our pencil. This single observation tells us a great deal about the function. In this section, we will investigate functions with and without breaks. Determining Whether a Function Is Continuous at a Number Let’s consider a specific example of temperature in terms of date and location, such as June 27, 2013, in Phoenix, AZ. The graph in Figure 1 indicates that, at 2 a.m., the temperature was . By 2 p.m. the temperature had risen to and by 4 p.m. it was Sometime between 2 a.m. and 4 p.m., the temperature outside must have been exactly In fact, any temperature between and occurred at some point that day. This means all real numbers in the output between and are generated at some point by the function according to the intermediate value theorem, Look again at Figure 1. There are no breaks in the function’s graph for this 24-hour period. At no point did the temperature cease to exist, nor was there a point at which the temperature jumped instantaneously by several degrees. A function that has no holes or breaks in its graph is known as a continuous function. Temperature as a function of time is an example of a continuous function. If temperature represents a continuous function, what kind of function would not be continuous? Consider an example of dollars expressed as a function of hours of parking. Let’s create the function where is the output representing cost in dollars for parking number of hours. See Figure 2. Suppose a parking garage charges $4.00 per hour or fraction of an hour, with a $25 per day maximum charge. Park for two hours and five minutes and the charge is $12. Park an additional hour and the charge is $16. We can never be charged $13, $14, or $15. There are real numbers between 12 and 16 that the function never outputs. There are breaks in the function’s graph for this 24-hour period, points at which the price of parking jumps instantaneously by several dollars. Figure 2 Parking-garage charges form a discontinuous function. A function that remains level for an interval and then jumps instantaneously to a higher value is called a stepwise function. This function is an example. A function that has any hole or break in its graph is known as a discontinuous function. A stepwise function, such as parking-garage charges as a function of hours parked, is an example of a discontinuous function. So how can we decide if a function is continuous at a particular number? We can check three different conditions. Let’s use the function represented in Figure 3 as an example. Figure 3 Condition 1 According to Condition 1, the function defined at must exist. In other words, there is a y-coordinate at as in Figure 4. Figure 4 Condition 2 According to Condition 2, at the limit, written must exist. This means that at the left-hand limit must equal the right-hand limit. Notice as the graph of in Figure 3 approaches from the left and right, the same y-coordinate is approached. Therefore, Condition 2 is satisfied. However, there could still be a hole in the graph at . Condition 3 According to Condition 3, the corresponding coordinate at fills in the hole in the graph of This is written Satisfying all three conditions means that the function is continuous. All three conditions are satisfied for the function represented in Figure 5 so the function is continuous as Figure 5 All three conditions are satisfied. The function is continuous at . Figure 6 through Figure 9 provide several examples of graphs of functions that are not continuous at and the condition or conditions that fail. Figure 6 Condition 2 is satisfied. Conditions 1 and 3 both fail. Figure 7 Conditions 1 and 2 are both satisfied. Condition 3 fails. Figure 8 Condition 1 is satisfied. Conditions 2 and 3 fail. Figure 9 Conditions 1, 2, and 3 all fail. Definition of Continuity A function is continuous at provided all three of the following conditions hold true: Condition 1: exists. Condition 2: exists at . Condition 3: . If a function is not continuous at the function is discontinuous at . Identifying a Jump Discontinuity Discontinuity can occur in different ways. We saw in the previous section that a function could have a left-hand limit and a right-hand limit even if they are not equal. If the left- and right-hand limits exist but are different, the graph “jumps” at . The function is said to have a jump discontinuity. As an example, look at the graph of the function in Figure 10. Notice as approaches how the output approaches different values from the left and from the right. Figure 10 Graph of a function with a jump discontinuity. Jump Discontinuity A function has a jump discontinuity at if the left- and right-hand limits both exist but are not equal: . Identifying Removable Discontinuity Some functions have a discontinuity, but it is possible to redefine the function at that point to make it continuous. This type of function is said to have a removable discontinuity. Let’s look at the function represented by the graph in Figure 11. The function has a limit. However, there is a hole at . The hole can be filled by extending the domain to include the input and defining the corresponding output of the function at that value as the limit of the function at . Figure 11 Graph of function with a removable discontinuity at . Removable Discontinuity A function has a removable discontinuity at if the limit, exists, but either does not exist or the value of the function at does not equal the limit, Example 1 Identifying Discontinuities Identify all discontinuities for the following functions as either a jump or a removable discontinuity. ⓐ ⓑ Solution ⓐ Notice that the function is defined everywhere except at Thus, does not exist, Condition 2 is not satisfied. Since Condition 1 is satisfied, the limit as approaches 5 is 8, and Condition 2 is not satisfied.This means there is a removable discontinuity at 2. ⓑ Condition 2 is satisfied because Notice that the function is a piecewise function, and for each piece, the function is defined everywhere on its domain. Let’s examine Condition 1 by determining the left- and right-hand limits as approaches 2. Left-hand limit: The left-hand limit exists. Right-hand limit: The right-hand limit exists. But So, does not exist, and Condition 2 fails: There is no removable discontinuity. However, since both left- and right-hand limits exist but are not equal, the conditions are satisfied for a jump discontinuity at Try It #1 Identify all discontinuities for the following functions as either a jump or a removable discontinuity. ⓐ ⓑ Recognizing Continuous and Discontinuous Real-Number Functions Many of the functions we have encountered in earlier chapters are continuous everywhere. They never have a hole in them, and they never jump from one value to the next. For all of these functions, the limit of as approaches is the same as the value of when So There are some functions that are continuous everywhere and some that are only continuous where they are defined on their domain because they are not defined for all real numbers. Examples of Continuous Functions The following functions are continuous everywhere: \| \| \| --- \| \| Polynomial functions \| Ex: \| \| Exponential functions \| Ex: \| \| Sine functions \| Ex: \| \| Cosine functions \| Ex: \| Table 1 The following functions are continuous everywhere they are defined on their domain: \| \| \| --- \| \| Logarithmic functions \| Ex: , \| \| Tangent functions \| Ex: is an integer \| \| Rational functions \| Ex: \| Table 2 How To Given a function determine if the function is continuous at Check Condition 1: exists. Check Condition 2: exists at Check Condition 3: If all three conditions are satisfied, the function is continuous at If any one of the conditions is not satisfied, the function is not continuous at Example 2 Determining Whether a Piecewise Function is Continuous at a Given Number Determine whether the function is continuous at ⓐ ⓑ Solution To determine if the function is continuous at we will determine if the three conditions of continuity are satisfied at . ⓐ Condition 1: Does exist? Condition 2: Does exist? To the left of to the right of We need to evaluate the left- and right-hand limits as approaches 1. Left-hand limit: Right-hand limit: Because does not exist. There is no need to proceed further. Condition 2 fails at If any of the conditions of continuity are not satisfied at the function is not continuous at ⓑ Condition 1: Does exist? Condition 2: Does exist? To the left of to the right of We need to evaluate the left- and right-hand limits as approaches Left-hand limit: Right-hand limit: Because exists, Condition 3: Is Because all three conditions of continuity are satisfied at the function is continuous at Try It #2 Determine whether the function is continuous at Example 3 Determining Whether a Rational Function is Continuous at a Given Number Determine whether the function is continuous at Solution To determine if the function is continuous at we will determine if the three conditions of continuity are satisfied at Condition 1: There is no need to proceed further. Condition 2 fails at If any of the conditions of continuity are not satisfied at the function is not continuous at Analysis See Figure 12. Notice that for Condition 2 we have At there exists a removable discontinuity. See Figure 12. Figure 12 Try It #3 Determine whether the function is continuous at If not, state the type of discontinuity. Determining the Input Values for Which a Function Is Discontinuous Now that we can identify continuous functions, jump discontinuities, and removable discontinuities, we will look at more complex functions to find discontinuities. Here, we will analyze a piecewise function to determine if any real numbers exist where the function is not continuous. A piecewise function may have discontinuities at the boundary points of the function as well as within the functions that make it up. To determine the real numbers for which a piecewise function composed of polynomial functions is not continuous, recall that polynomial functions themselves are continuous on the set of real numbers. Any discontinuity would be at the boundary points. So we need to explore the three conditions of continuity at the boundary points of the piecewise function. How To Given a piecewise function, determine whether it is continuous at the boundary points. For each boundary point of the piecewise function, determine the left- and right-hand limits as approaches as well as the function value at Check each condition for each value to determine if all three conditions are satisfied. Determine whether each value satisfies condition 1: exists. Determine whether each value satisfies condition 2: exists. Determine whether each value satisfies condition 3: If all three conditions are satisfied, the function is continuous at If any one of the conditions fails, the function is not continuous at Example 4 Determining the Input Values for Which a Piecewise Function Is Discontinuous Determine whether the function is discontinuous for any real numbers. Solution The piecewise function is defined by three functions, which are all polynomial functions, on on and on Polynomial functions are continuous everywhere. Any discontinuities would be at the boundary points, and At let us check the three conditions of continuity. Condition 1: Condition 2: Because a different function defines the output left and right of does Left-hand limit: Right-hand limit: Because , Condition 3: Because all three conditions are satisfied at the function is continuous at At let us check the three conditions of continuity. Condition 2: Because a different function defines the output left and right of does Left-hand limit: Right-hand limit: Because , so does not exist. Because one of the three conditions does not hold at the function is discontinuous at Analysis See Figure 13. At there exists a jump discontinuity. Notice that the function is continuous at Figure 13 Graph is continuous at but shows a jump discontinuity at Try It #4 Determine where the function is discontinuous. Determining Whether a Function Is Continuous To determine whether a piecewise function is continuous or discontinuous, in addition to checking the boundary points, we must also check whether each of the functions that make up the piecewise function is continuous. How To Given a piecewise function, determine whether it is continuous. Determine whether each component function of the piecewise function is continuous. If there are discontinuities, do they occur within the domain where that component function is applied? For each boundary point of the piecewise function, determine if each of the three conditions hold. Example 5 Determining Whether a Piecewise Function Is Continuous Determine whether the function below is continuous. If it is not, state the location and type of each discontinuity. Solution The two functions composing this piecewise function are on and on The sine function and all polynomial functions are continuous everywhere. Any discontinuities would be at the boundary point, At let us check the three conditions of continuity. Condition 1: Because all three conditions are not satisfied at the function is discontinuous at Analysis See Figure 14. There exists a removable discontinuity at thus the limit exists and is finite, but does not exist. Figure 14 Function has removable discontinuity at 0. Media Access these online resources for additional instruction and practice with continuity. Continuity at a Point Continuity at a Point: Concept Check 12.3 Section Exercises Verbal 1. State in your own words what it means for a function to be continuous at State in your own words what it means for a function to be continuous on the interval Algebraic For the following exercises, determine why the function is discontinuous at a given point on the graph. State which condition fails. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. For the following exercises, determine whether or not the given function is continuous everywhere. If it is continuous everywhere it is defined, state for what range it is continuous. If it is discontinuous, state where it is discontinuous. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. . 35. Determine the values of and such that the following function is continuous on the entire real number line. Graphical For the following exercises, refer to Figure 15. Each square represents one square unit. For each value of determine which of the three conditions of continuity are satisfied at and which are not. Figure 15 37. 38. 39. For the following exercises, use a graphing utility to graph the function as in Figure 16. Set the x-axis a short distance before and after 0 to illustrate the point of discontinuity. Figure 16 Which conditions for continuity fail at the point of discontinuity? 41. Evaluate Solve for if 43. What is the domain of For the following exercises, consider the function shown in Figure 17. Figure 17 At what x-coordinates is the function discontinuous? 45. What condition of continuity is violated at these points? Consider the function shown in Figure 18. At what x-coordinates is the function discontinuous? What condition(s) of continuity were violated? Figure 18 47. Construct a function that passes through the origin with a constant slope of 1, with removable discontinuities at and The function is graphed in Figure 19. It appears to be continuous on the interval but there is an x-value on that interval at which the function is discontinuous. Determine the value of at which the function is discontinuous, and explain the pitfall of utilizing technology when considering continuity of a function by examining its graph. Figure 19 49. Find the limit and determine if the following function is continuous at The graph of is shown in Figure 20. Is the function continuous at Why or why not? 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Authors: Jay Abramson Publisher/website: OpenStax Book title: Precalculus 2e Publication date: Dec 21, 2021 Location: Houston, Texas Book URL: Section URL: © Jun 16, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
9271	https://arxiv.org/pdf/2005.11440	arXiv:2005.11440v2 [math.PR] 7 Aug 2020 THE MAKI-THOMPSON RUMOR MODEL ON INFINITE CAYLEY TREES VALDIVINO V. JUNIOR, PABLO M. RODRIGUEZ, AND ADALTO SPEROTO Abstract. In this paper we study the Maki-Thompson rumor model on infinite Cayley trees. The basic version of the model is defined by assuming that a population represented by a graph is subdivided into three classes of individuals: ignorants, spreaders and stiflers. A spreader tells the rumor to any of its (nearest) ignorant neighbors at rate one. At the same rate, a spreader becomes a stifler after a contact with other (nearest neighbor) spreaders, or stiflers. In this work we study this model on infinite Cayley trees, which is formulated as a continuous-times Markov chain, and we extend our analysis to the generalization in which each spreader ceases to propagate the rumor right after being involved in a given number of stifling experiences. We study sufficient conditions under which the rumor either becomes extinct or survives with positive probability. Introduction Currently, there exist a wide variety of mathematical models formulated to describe in a simple way the phenomenon of information transmission on a population. A wide range of these models is formed by epidemic-like processes inspired by the Daley-Kendal and the Maki-Thompson models. The Daley-Kendal model has been formulated in the mid 60’s as an al-ternative, to describe information spreading, to the well-known susceptible-infected-recovered epidemic model, see [11, 12]. Later the Maki-Thompson model has appeared in as a sim-plification of the Daley-Kendal model. Since both models behaves asymptotically equal the Maki-Thompson model, that we just refer as the MT-model, has been used as a basis for many generalizations. The MT-model assumes a homogeneously mixed population of size N + 1 subdivided into three classes of individuals: Ignorants (those not aware of the rumor), spreaders (who are spreading it), and stiflers (who know the rumor but have ceased communicating it after meeting somebody who has already heard it). The number of ignorants, spreaders and stiflers at time t is denoted by X(t), Y (t) and Z(t), respectively. Initially, X(0) = N , Y (0) = 1 and Z(0) = 0 , and X(t) + Y (t) + Z(t) = N + 1 for all t. The MT-model is the continuous-time Markov chain {(X(t), Y (t)) }t≥0 with transitions and corresponding rates given by transition rate (−1, 1) XY, (0 , −1) Y (N − X). This means that if the Markov chain is in state (i, j ) at time t, then the probabilities of jumping to states (i − 1, j + 1) or (i, j − 1) at time t + h are, respectively, i j h + o(h) 2010 Mathematics Subject Classification. 60K35, 60K37, 82B26. Key words and phrases. Maki-Thompson Model, Phase-Transition, Homogeneous Tree, Branching Process, Rumor Spreading. 1MAKI-THOMPSON RUMOR MODEL ON TREES 2 and j(N − i) h + o(h), where o(h) represents a function such that lim h→0 o(h)/h = 0 . In words, the rumor spreads by directed contact of spreaders with other individuals and the two possible transitions correspond to spreader-ignorant, or spreader-spreader and spreader-stifler interactions, respectively. In the first case, the spreader tells the rumor to the ignorant, who becomes a spreader, and in the other case we have the transformation of the spreader initiating the meeting into a stifler. The second transition is what we call a stifling experience and it represents the loss of interest in propagating the rumor derived from learning that it is already known by the other individual in the meeting. The first results for the MT-model are related to the asymptotic behavior of the proportion of ignorants at the end of the process. We refer the reader to [6, 14, 17–19, 24, 25, 27] and the references therein for an overview of existing results in this direction. Also we refer the reader to [10, Chapter 5] for an excellent account on the subject of rumor models. We point out that all these works deal with the case of homogeneously mixed populations, which is the same to say that the population is represented by a complete graph. For the case of a population represented by another type of graph we refer the reader to [1, 9] for rigorous results based on probabilistic methods and to [3, 22, 23, 28, 29] for approximation results based on mean-field arguments and computational simulations. Here we are interested in the MT-model with k-stifling. This is the modified version of the MT-model for which each spreader ceases to propagate the rumor right after being involved in k stifling experiences, for some fixed k ∈ N. The known results for this model are for the case of a homogeneously mixed population; mostly, limit theorems for the remaining proportion of ignorants as the population size goes to infinity. The case k = 1 is, of course, the MT-model. A Law of Large Numbers (LLN) for this proportion has been stated by and a Central Limit Theorem (CLT) by . The version k = 2 has been considered by who states a LLN. Later obtain results by mean of the deterministic version for a general k ∈ N and obtain a LLN where k is assumed to be a discrete random variable. Moreover, an interesting connection between the MT-model with k-stifling and a system of random walks has been showed by . In this work we propose by the first time studying the MT-model with k-stifling on a non-complete graph. We consider the model defined on an infinite Cayley tree, also known of Bethe lattice or homogeneous tree, which is an infinite connected cycle-free graph where the vertices all have the same degree; i.e., each vertex is connected to d neighbours and d is called the coordination number. The interest in this type of graph is twofold; on one hand its structure allows to obtain sharp results regarding survival or extinction of the rumor. The physical relevance of these results is that this type of graph represents some mean field limit of Euclidean lattices of large dimensions. On the other hand, trees are structures that usually appear in random graph models so our model may serves as an inspiration for the formulation of more general rumor models. We study the survival or not of the rumor on this graph according to the values of d and k. Our approach relies on a comparison of our rumor model with a suitable defined branching process. The paper is organized as follows. Section 2 is devoted to the formulation of the model and the statements of our main results. Section 3 contains the proofs of our theorems and is subdivided into two parts. In Subsection 3.1 we study the distribution of the number of spreaders one spreader generates which is a key quantity for our results. Also this subsection is of interest by itself because of a connection with the Coupon Collector’s Problem. Subsection 3.2 includes the construction of a underlying branching process and the proofs of our main results. MAKI-THOMPSON RUMOR MODEL ON TREES 3 The Model and Main Results In what follows we let Td = ( V, E) for an infinite Cayley tree of coordination number d + 1 , with d ≥ 2. The notation is coming from Graph Theory since the same graph is known as (d + 1) -dimensional homogeneous tree. Here V stands for the set of vertices and E ⊂ {{ u, v } : u, v ∈ V , u 6 = v} stands for the set of edges. We shall abuse notation by writing V = Td, and we identify one vertex as the root and denote it by 0. If {u, v } ∈ E , we say that u and v are neighbors, which is denoted by u ∼ v. The degree of a vertex v, denoted by deg (v), is the number of its neighbors. A path in Td is a finite sequence v0, v 1, . . . , v n of distinct vertices such that vi ∼ vi+1 for each i, and a ray in Td is a path with infinite vertices starting at 0. Since Td is a tree, there is a unique path connecting any pair of distinct vertices u and v. Therefore we may define the distance between them, which is denoted by d(u, v ), as the number of edges in such path. We point out that Td is a graph with an infinite number of vertices, without cycles and such that every vertex has degree d + 1 . For each v ∈ V define \|v\| := d(0, v ). For u, v ∈ V , we say that u ≤ v if u is one of the vertices of the path connecting 0 and v; u < v if u ≤ v and u 6 = v. We call v a descendant of u if u ≤ v and denote by T u = {v ∈ V : u ≤ v} the set of descendants of u. On the other hand, v is said to be a successor of u if u ≤ v and u ∼ v. For n ≥ 1, we denote by ∂Td,n the set of vertices at distance n from the root. That is, ∂Td,n = {v ∈ Td : \|v\| = n}.The MT-model with k-stifling on Td may be defined as a continuous-time Markov process (η(k) t )t≥0 with states space S = {− 1, 0, 1, 2, . . . , k }Td , i.e. at time t the state of the process is some function ηt : Td −→ {− 1, 0, 1, 2, . . . , k }. We assume that each vertex x ∈ Td represents an individual, which is said to be an ignorant if η(x) = −1, a spreader who experimented i stifling experiences if η(x) = i, for i ∈ { 0, 1, . . . , k − 1}, and a stifler if η(x) = k. Remember that ignorants are those who do not know about the rumor, spreaders are those who know about the rumor and they are transmitting it, and stiflers known about the rumor but they have stopped of propagating it. Then, if the system is in configuration η ∈ S , the state of vertex x changes according to the following transition rates transition rate −1 → 0, ∑k−1 i=0 ni(x, η ),i → i + 1 , ∑ki=0 ni(x, η ), (2.1) where ni(x, η ) = ∑ y∼x 1{η(y) = i} is the number of nearest neighbors of vertex x in state i for the configuration η, for i ∈{− 1, 0, . . . , k }. Formally, (2.1) means that if the vertex x is in state, say, −1 at time t then the probability that it will be in state 0 at time t + h, for h small, is ∑k−1 i=0 ni(x, η )h + o(h),where o(h) represents a function such that lim h→0 o(h)/h = 0 . Roughly speaking, the rates in (2.1) represent how the changes of states of individuals depend on the states of its neighbors. While the change of state of an ignorant is influenced by its spreader neighbors, the change of state for a spreader is influenced by the number of non-ignorant neighbors. We point out that stiflers do not interact with ignorants. We call the Markov process (η(k) t )t≥0 the Maki-Thompson rumor model with k-stifling on Td, and for the sake of simplicity we abbreviate, as before, as MT-model with k-stifling on Td,or just as MT-model when k = 1 . Since we are considering a graph with an infinite number MAKI-THOMPSON RUMOR MODEL ON TREES 4 of vertices our first task shall be to define the event of survival or extinction for the rumor process. Definition 2.1. Consider the MT-model with k-stifling on Td with initial configuration η0 such that η0(0) = 0 and η0(x) = −1 for all x 6 = 0. We say that there is survival of the rumor if there exist a sequence {(vi, t i)}i≥0, with (vi, t i) ∈ Td × R+, such that v0 = 0, t0 = 0 , vi+1 is a successor of vi, ti < t i+1 , and ηti (vi) = 0 , for all i ≥ 0. If there is not survival, we say that the rumor becomes extinct. We denote by θ(d, k ) the survival probability and we let θ(d) := θ(d, 1) .In other words, by the previous definition we have that there is survival of the rumor if we can guarantee the existence of a ray from the root of Td such that all the vertices in the ray were spreaders at some time. Let us start by analysing the occurrence or not of this event for the basic MT-model. Theorem 2.1. Consider the MT-model on Td. Then θ(d) > 0 if, and only if, d ≥ 3. Moreover, θ(d) = 1 − d+1 ∑ i=1 i! ( di − 1 ) ( ψ d + 1 )i , where ψ is the smallest non-negative root of the equation d ∑ i=0 i! (di ) ( s d + 1 )i ( i + 1 d + 1 ) = s. Corollary 2.1. lim d→∞ θ(d) = 1 . Theorem 2.1 gains in interest if we realize that the MT-model exhibit two different behaviors according to d = 2 or d ≥ 3. For d = 2 we obtain that the rumor propagates, almost surely, only to a finite set of individuals. In the other cases, for d ≥ 3, the rumor propagates to infinitely many individuals with positive probability. In other words, the MT-model exhibit a phase transition. We do not consider the case d = 1 , the path graph with infinite vertices, because it is trivial. In that case it is enough to note that the number of spreaders will be bounded from above by the sum of two random variables with geometric law. When the MT-model is considered on the complete graph, a quantity of interest is related to the number of stiflers at the end of the process. Note that this is the number of individuals who hear about the rumor at some time. By Theorem 2.1 we known that this number is finite almost surely provided d = 2 . In what follows, we give a better characterization of that number by identifying the distribution of the quantity of stiflers at the end of the process. Theorem 2.2. Consider the MT-model on T2, and let S∞ be the final number of stiflers at the end of the process. Then P(S∞ = i) = 1 9i { 3 G(i−1) i (0) (i − 1)! + 8 G(i−2) i (0) (i − 2)! + 6 G(i−3) i (0) (i − 3)! } , i ∈ N, (2.2) where Gi(s) := ( 2s2 + 4 s + 3 9 )i , s ∈ [−1, 1] ,MAKI-THOMPSON RUMOR MODEL ON TREES 5 and G(j) i (s) := dj (Gi(s)) ds j , for j ≥ 0, and G(j) i (s) := 0 other case. Moreover, E(S∞) = 18 . Another quantity useful to measure the impact of the rumor for d = 2 is what we call the range of spreading in the following theorem. We emphasize that according to Theorem 2.1, if d ≥ 3 then the rumor propagates to infinitely many individuals with positive probability. Theorem 2.3. Consider the MT-model on T2. Let R := max {n ≥ 1 : ηt(x) = 1 for some x ∈ ∂T2,n , and t ∈ R+}, (2.3) be the range of spreading. Then, for any n ≥ 03 9 α1(n) + 4 9 α1(n)2 + 2 9 α1(n)3 ≤ P(R ≤ n) ≤ 3 9 α2(n) + 4 9 α2(n)2 + 2 9 α2(n)3, (2.4) where α1(n) := (13 /9) {1 − (8 /9) n} 13 /9 − (8 /9) n , and α2(n) := (4 /3) {1 − (8 /9) n} 4/3 − (8 /9) n . Besides this, 6.144 ≤ E(R) ≤ 7.448 . Now, let us state our result related to the MT-model with k-stifling with k ≥ 2. By a standard coupling argument it is not difficult to see that θ(d, k ) is non-decreasing in k (indeed it is non-decreasing in d as well). Therefore, by Corollary 2.1 we have lim d→∞ θ(d, k ) = 1 for any k ≥ 2. In the next Theorem we prove that for the MT-model with k-stifling on Td with k ≥ 2 and d ≥ 2, differently of what happens in the MT-model (i.e., k = 1 ), the only behaviour is that the rumor propagates to infinitely many individulas with positive probability. Moreover, we localize the value of the survival probability as a function of k and d. Theorem 2.4. Consider the MT-model with k-stifling on Td with k ≥ 2 and d ≥ 2. Then θ(d, k ) > 0. Moreover, let S∗(i, k ) := i ∑ m1=1 i ∑ m2=m1 · · · i ∑ mk−1=mk−2 k−1 ∏ ℓ=1 mℓ, (2.5) and S(i, k ) := i+1 ∑ m1=1 i+1 ∑ m2=m1 · · · i+1 ∑ mk−1=mk−2 k−1 ∏ ℓ=1 mℓ. (2.6) Then θ(d, k ) = 1 − d+1 ∑ i=1 i ( ψ d + 1 )i i! (d + 1) k (d + 1 i ) S∗(i, k ) (2.7) where ψ is the smallest non-negative root of the equation d ∑ i=0 (di ) ( s d + 1 )i (i + 1)! (d + 1) k S(i, k ) = s. (2.8) MAKI-THOMPSON RUMOR MODEL ON TREES 6 We point out that Theorem 2.4 is usefull for the computation of the survival probability. By fixing k and d one can obtain this value by mean of some (computational for higher values) calculations. Table 1 exhibits the values of θ(d, k ) for d ∈ { 2, 3, 4, 5, 6, 7, 50 } and k ∈ { 1, 2}. d23456750 k= 1 0.000000 0.661289 0.869802 0.931135 0.957300 0.970887 0.999583 k= 2 0.937500 0.991439 0.997434 0.998936 0.999474 0.999708 0.999999 Table 1. The behavior of θ(d, k )for k∈ { 1,2}and some values of d. Since θ(d, k ) is non-decreasing in both k and d one can see that for k ≥ 3 we should have θ(d, k ) ≈ 1. Indeed, for k = 3 we have S∗(i, 3) = i ∑ m1=1 i ∑ m2=m1 m1m2 = 1 24 i(i + 1)(3 i2 + 7 i + 2) , and S(i, 3) := i+1 ∑ m1=1 i+1 ∑ m2=m1 m1m2 = 1 24 (i + 1)( i + 2)(3 i2 + 13 i + 12) , which together with (2.7) and (2.8) for d = 2 implies θ(2 , 3) = 0 .9964 .Remember that the original Maki-Thompson model is obtained by considering k = 1 .Consequently, our theorems contribute with the theory complementing the existing results, proved by [5, 8, 18, 25, 27], to the case of infinite Cayley trees. 3. Proofs The main idea behind our proofs is the identification of an underlying branching process related to the rumor process. Then we apply well-known results of these processes. We shall see that the offspring distribution of such a branching process is the same as the one of the number of spreaders one spreader generates. Therefore, it is enough to study the mean of this distribution to obtain results about the survival or not of the rumor and its generating function to localize the survival probability, respectively. We subdivide this section into two parts: in the first one, Subsection 3.1, we study the distribution and mean of the number of spreaders one spreader generates for the MT-model first and for the MT-model with k-stifling later. The second part, Subsection 3.2, is devoted to construct the underlying branching process whose survival is equivalent with the survival of the rumor process. Also in this subsection we prove our theorems. 3.1. The distribution of the number of spreaders one spreader generates. Let us start with the MT-model on Td. Let N be the number of spreaders generated by the initial spreader. First we are interested in the law of this discrete random variable. Lemma 3.1. P(N = i) = i! (d + 1 i ) i (d + 1) i+1 , i ∈ { 1, . . . , d + 1 }. (3.1) Proof. It is not difficult to see that N takes values in the set {1, . . . , d + 1 }. This is because up to become a stifler the root can contact at most d+1 individuals, event which happens when the stifling experience occurs only at the (d + 2) -th contact. In general, for any i ∈ { 1, . . . , d + 1 },MAKI-THOMPSON RUMOR MODEL ON TREES 7 {N = i} occurs if the first i contacts are with ignorants and the (i + 1) -th contact is a stifling experience. Note that this event has probability given by P(N = i) =  i−1 ∏ j=0 ( d + 1 − j d + 1 ) i d + 1 , which can be written as in (3.1). Remark 1. The previous result it is of interest by itself once one realize its connection with the Coupon Collector’s Problem - a classic and well-known problem in probability theory. The problem can be stated as follows: At each stage, a collector obtains a coupon which is equally likely to be any one of n types. Assuming that the results of successive stages are independent, among other results, what is the earliest stage at which all n coupons have been picked at least once? This question and many interesting generalizations have been addressed in the literature, see for example [7, 16, 26]. An alternative problem is studying the number of coupons that would be expected drawn up to seeing a duplicate; that is, a coupon that already is part of the collection. As far as we know, no attention has been paid to this quantity before. We point out that the law of such a variable is the one given by (3.1) with n = d + 1 .In what follows we consider the number of spreaders one given spreader (different from the root) generates. Let X be such a number. Lemma 3.2. P(X = i) = (di ) (i + 1)! (d + 1) i+1 , i ∈ { 0, . . . , d }. (3.2) Moreover, E(X) > 1 if, and only if, d ≥ 3.Proof. The law (3.2) may be obtained by observing that X = N − 1 in law. Now, since P(X = i) > 0 for any i ∈ { 0, . . . , d } we have E(X) = d ∑ i=0 iP(X = i) > P (X = 2) + d ∑ i=1 P(X = i), and since P(X = 2) > P(X = 0) provided d ≥ 3 we conclude E(X) > d ∑ i=0 P(X = i) = 1 . That is, E(X) > 1 provided d ≥ 3. For d = 2 , and after some calculations, we obtain E(X) = 8 /9 so the proof is complete. In what follows we consider the MT-model with k-stifling, for k ≥ 2 and for this process we denote by N (k) (or by X(k)) the number of spreaders the root (or another spreader) generates. Lemma 3.3. P ( X(k) = i ) = (di ) (i + 1)! (d + 1) i+k S(i, k ), i ∈ { 0, 1, 2, 3, . . . , d }. (3.3) Moreover, E(X(k)) > 1 for any d ≥ 2.MAKI-THOMPSON RUMOR MODEL ON TREES 8 Proof. It is not difficult to see that X(k) takes values on the set {0, 1, . . . , d }. Note that, for any i ∈ { 0, 1, . . . , d }, {X(k) = i} occurs if, and only if, we have exactly k − 1 stifling experiences between the first (i + k − 1) contacts and we have the k-th stifling experience at the (i + k)-th contact with another individual. Let 1 ≤ m1 < m 2 < · · · < m k−1 ≤ i + k − 1 and let Am1,m 2,...,m k−1 be the event of the k stifling experiences occur at the m1-th, m2-th, . . . , mk−1-th, and (i + k)-th contacts with other individuals, respectively. Thus defined we can write {X(k) = i} = ⋃ 1≤m1<m 2<··· <m k−1≤i+k−1 Am1,m 2,...,m k−1 , and since P (Am1,m 2,...,m k−1 ) = (di ) (i + 1)! (d + 1) i+kk−1∏ ℓ=1 (mℓ − ℓ + 1) , we get (3.3). In order to prove that E(X(k)) > 1 for any d ≥ 2 we shall consider first the case k = 2 . By (3.3) we have that P ( X(2) = i ) = (di ) (i + 1) ( i + 2)! 2( d + 1) i+1 , i ∈ { 0, 1, 2, 3, . . . , d }. (3.4) In particular, note that for any d ≥ 2 we have P ( X(2) = 0 ) = 1 (d + 1) 2 < 18 d(d − 1) (d + 1) 4 = P ( X(2) = 2 ) . Thus, by applying similar arguments as in Lemma (3.2) we have E(X(2) ) > P(X(2) = 2) + 1 − P(X(2) = 0) > 1. Since a tree is a graph without cycles, and since we are assuming that only the root is a spreader at time zero, a standard coupling argument allow us to conclude that given d ≥ 2 we have for any k ≥ 3 that E(X(k)) ≥ E(X(2) ). This complete the proof. Remark 2. Coming back to the Coupon Collector’s Problem, and analogously as for the case k = 1 , X(k) has the same distribution as the number of coupons that would be drawn up to seeing the kth coupon that already is part of the collection provided the collection is initially formed by one coupon. 3.2. Proof of Main Theorems. 3.2.1. The underlying branching process. Consider the MT-model on Td with k-stifling, and assume that η0(0) = 0 and η0(x) = −1 for any x 6 = 0. For any n ≥ 0 we let Bn := { v ∈ ∂Td,n +1 : ⋃ t> 0 {ηt(v) = 1 } , and we define the random variable Zn := \|B n\|. Thus defined, B0 is formed by those vertices at distance one from 0 which are spreaders at some time, B1 is formed by those vertices at distance two from 0 which are spreaders at some time, and so on. Moreover, Z0 is equal to N (k) in law, and it is not difficult to see that Zn+1 = Zn ∑ i=1 X(k) i , (3.5) MAKI-THOMPSON RUMOR MODEL ON TREES 9 where X(k)1 , X (k)2 , . . . are independent copies of X(k). Thus defined, (Zn)n≥0 is a branching process such that Z0 has a law given by N (k) and the offspring distribution is given by X(k) (see (3.3)). For a complete reference of the Theory of Branching Processes we refer the reader to . Our construction gains in interest if we realize the following connection between the rumor model and the branching process. Lemma 3.4. The MT-model on Td survives if, and only if, the branching process (Zn)n≥0 survives. Proof. It is direct by construction. 3.2.2. Proof of Theorems 2.1 and 2.4. We shall use the well-known fact that a branching process survives with positive probability if, and only, if, the mean of the offspring distribution is greater than 1. Moreover, the survival probability can be obtained as the smallest root in (0 , 1] of the equation ϕ(s) = s, where ϕ is the generating probability function of the offspring distribution. Note that for any generating probability function ϕ we can guarantee ϕ(1) = 1 .Therefore, the proof of Theorem 2.1 is a consequence of Lemma 3.4 and Lemma 3.2. Indeed, the MT-model survives with probability positive if, and only if, the underlying branching process does it, which happens if, and only if, E(X) > 1. Analogously, the proof of Theorem 2.4 is a consequence of Lemma 3.4 and Lemma 3.3. 3.2.3. Proof of Corollary 2.1. We shall verify that lim d→∞ d+1 ∑ i=1 i! (d + 1 i ) ( ψ d + 1 )i i d + 1 = 0 . Since the limit in the left side of the previous equality can be written as lim d→∞ ψ d + 1 d+1 ∑ i=1 i! (d + 1 i ) ( ψ d + 1 )i−1 i d + 1 , (3.6) and since \|ψ\| ≤ 1 so lim d→∞ ψ/ (d + 1) = 0 , it is enough if we show that the right side factor in (3.6) is bounded for any d. It is not difficult to see that d+1 ∑ i=1 i! (d + 1 i ) ( ψ d + 1 )i−1 i d + 1 < ∞ ∑ i=1 i ( dψ d + 1 )i−1 ≤ ∞ ∑ i=1 i ( d d + 1 )i−1 < ∞. Therefore the proof is complete. 3.2.4. Proof of Theorem 2.2. Consider the MT-model on T2, and let S∞ be the final number of stiflers at the end of the process. We already prove that the MT-model may be seen as the branching process given by (3.5). Therefore S∞ coincides with the total progeny of such a branching processes. In order to prove Theorem 2.2 we appeal to . Indeed, notice that P(S∞ = i) = 3 ∑ n=1 P(S∞ = i\|N = n)P(N = n), (3.7) where P(N = 1) = 3 9 , P(N = 2) = 4 9 , P(N = 3) = 2 9 , (3.8) MAKI-THOMPSON RUMOR MODEL ON TREES 10 and by [13, Main Theorem, p. 682] P(S∞ = i\|N = n) =  n i P  i ∑ j=1 Xj = i − n  , i ≥ n, 0, other wise. (3.9) Here X1, X 2, . . . denote independent and identically distributed random variables with com-mon law given by (3.2) (with d = 2 ). If Gi(s) and GXj (s) are the probability generating functions of ∑ij=1 Xj and Xj , respectively, then we have Gi(s) = i ∏ j=1 GXj (s) = ( 2s2 + 4 s + 3 9 )i . (3.10) By joining (3.7) to (3.10) we get (2.2). Now let us prove that E(S∞) = 18 . Since E(S∞) = 3 ∑ n=1 E(S∞\|N = n)P(N = n), (3.11) and μ := E(X) = 8 /9 (see Lemma 3.2 for d = 2 ) we have for n ∈ { 1, 2, 3} E(S∞\|N = n) = ( 1 + n ∞ ∑ i=0 μi ) = 1 + 9 n. (3.12) We conclude by (3.8), (3.11) and (3.12) that E(S∞) = 18 .3.2.5. Proof of Theorem 2.3. Before proving Theorem 2.3 we state an auxiliary result regarding branching processes. Lemma 3.5. Let (Zn)n≥0 be a branching processes with Z0 = 1 and offspring distribution given by (3.2) (with d = 2 ). Let GX and GZn be the probability generating functions of X and Zn, respectively. Then, 13 /45 + (128 s)/{45(5 − s)} ≤ GX (s) ≤ 1/3 + (2 s)/(4 − s), s ∈ [−1, 1] , (3.13) and, for n ≥ 0(13 /9) {1 − (8 /9) n} 13 /9 − (8 /9) n ≤ P(T ≤ n) ≤ (4 /3) {1 − (8 /9) n} 4/3 − (8 /9) n . (3.14) Proof. The spirit behind the proof of the lemma is to apply [15, Main Theorem, p. 450]. The first step is to check the condition h(s) := G(1) X (1) G(1) X (s)(1 − s)2 − (1 − GX (s)) 2 < 0, where G(1) X (s) := dG X (s)/ds . Indeed, it is not difficult to see that h(s) = 8 94( s + 1) 9 (1 − s)2 − { 1 − (2 s2 + 4 s + 3) 9 }2 , and after some calculations we obtain h(s) = − 32 81 (s − 1) 2 ( s2 + 23 4 s + 3 4 s ) .MAKI-THOMPSON RUMOR MODEL ON TREES 11 Thus h(s) < 0 for all s ∈ [0 , 1) . Therefore, by (i) from [15, Main Theorem, p. 450] we have that the best upper bounding fractional linear generating function for GX (s) is given by U (s) := 1 /3 + (2 s)/(4 − s). Analogously, (ii) from [15, Main Theorem, p. 450] implies that the best lower bounding fractional linear generating function for GX (s) is given by L(s) := 13 /45 + (128 s)/{45(5 − s)}. Therefore we get (3.13). It is well-known, see , that the inequality is preserved by compositions of the same functions. This in turns implies that for any n ≥ 0 Ln(s) ≤ GX,n (s) ≤ Un(s), where Ln, G X,n , and Un are the n-th composition of L, G X , and U , respectively. Moreover, since GX,n (s) = GZn (s), GZn (0) = P(T ≤ n), and by [2, Equation (3.1)] we have Ln(0) = (13 /9) {1 − (8 /9) n} 13 /9 − (8 /9) n , Un(0) = (4 /3) {1 − (8 /9) n} 4/3 − (8 /9) n ; we conclude (3.14) and the proof is complete. Consider the MT-model on T2 and let R be given by (2.3) the range of the spreading. Note that, for any n P(R ≤ n) = 3 ∑ i=1 P(R ≤ n\|N = i)P(N = i) = 3 ∑ i=1 P(T ≤ n)iP(N = i), where T is the extinction time of a branching process (Zn)n≥0 with Z0 = 1 and offspring distribution given by (3.2) (with d = 2 ). From the law of N , see (3.1) (with d = 2 ) we have P(R ≤ n) = 3 9 P(T ≤ n) + 4 9 P(T ≤ n)2 + 2 9 P(T ≤ n)3. (3.15) In addition, by Lemma 3.5 we get α1(n) ≤ P(T ≤ n) ≤ α2(n). In order to find a lower and upper bound for E(R) we use (3.15) to obtain P(R > n ) = 17 9 P(T > n ) − 10 9 P(T > n )2 + 2 9 P(T > n )3. Therefore E(R) = ∞ ∑ n=0 P(R > n ) = 17 9 ∞ ∑ n=0 P(T > n ) − 10 9 ∞ ∑ n=0 P(T > n )2 + 2 9 ∞ ∑ n=0 P(T > n )3. (3.16) Again, by Lemma 3.5 we have the following bounds for the series of the previous expression: 4.4619 ≤ ∞ ∑ n=0 (1 /3)(8 /9) n 4/3 − (8 /9) n ≤ ∞ ∑ n=0 P(T > n ) ≤ ∞ ∑ n=0 (4 /9)(8 /9) n 13 /9 − (8 /9) n ≤ 4.9792 , 2.0982 ≤ ∞ ∑ n=0 { (1 /3)(8 /9) n 4/3 − (8 /9) n }2 ≤ ∞ ∑ n=0 P(T > n )2 ≤ ∞ ∑ n=0 { (4 /9)(8 /9) n 13 /9 − (8 /9) n }2 ≤ 2.3592 , 1.5189 ≤ ∞ ∑ n=0 { (1 /3)(8 /9) n 4/3 − (8 /9) n }3 ≤ ∞ ∑ n=0 P(T > n )3 ≤ ∞ ∑ n=0 { (4 /9)(8 /9) n 13 /9 − (8 /9) n }3 ≤ 1.6804 .MAKI-THOMPSON RUMOR MODEL ON TREES 12 Finally, by a suitable application of the previous bounds in (3.16) we get 6.144 = 17 9 4.4619 − 10 9 2.3592 + 2 9 1.5189 ≤ E(R) ≤ 17 9 4.9792 − 10 9 2.0982 + 2 9 1.6804 = 7 .448 . Acknowledgements Part of this work has been developed during a visit of V.V.J. to the ICMC-University of São Paulo and a visit of P.M.R. to the IME-Federal University of Goiás. The authors thank these institutions for the hospitality and support. 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Probab. 16(2) (1979), 449-453. Kobza, J. E., Jacobson, S. H. and Vaughan, D. E., A Survey of the Collector’s Problem with Random Sample Sizes, Methodol. Comput. Appl. Probab. 9 (2007), 573-584. Lebensztayn, E., A large deviations principle for the Maki-Thompson rumour model, J. Math. Analysis and Applications 432 (2015), 142-155. Lebensztayn, E., Machado, F. P. and Rodríguez, P. M., On the behaviour of a rumour process with random stifling, Environ. Modell. Softw. 26 (2011), 517-522. Lebensztayn, E., Machado, F. and Rodríguez, P. M., Limit Theorems for a General Stochastic Rumour Model, SIAM J. Appl. Math. 71 (2011), 1476-1486. Lebensztayn, E. and Rodriguez, P. M., A connection between a system of random walks and rumor transmission, Physica A 392 (2013), 5793-5800. Maki, D. P. and Thompson, M., Mathematical Models and Applications. With Emphasis on the Social, Life, and Management Sciences (Prentice-Hall, Englewood Cliffs, New Jersey, 1973). Moreno, Y., Nekovee, M., and Pacheco, A. F., Dynamics of rumour spreading in complex networks, Phys. Rev. E 69 (2004), 066130. MAKI-THOMPSON RUMOR MODEL ON TREES 13 Nekovee, M., Moreno, Y., Bianconi, G., and Marsili, M., Theory of rumour spreading in complex social neworks, Physica A 374 (2007), 457-470. Pearce, C. E. M., The exact solution of the general stochastic rumour, Math. Comput. Modelling 31 (2000), 289-298. Sudbury, A., The proportion of the population never hearing a rumour, J. Appl. Probab. 22 (1985), 443-446. von Schelling, H., Coupon collecting for unequal probabilities, Amer. Math. Monthly 61 (1954), 306-311. Watson, R., On the size of a rumour, Stochastic Process. Appl. 27 (1988), 141-149. Zanette, D. H., Critical behavior of propagation on small-world networks, Phys. Rev. E 64 (2001), (R)050901. Zanette, D. H., Dynamics of rumour propagation on small-world networks, Phys. Rev. Lett. 65(4) (2002), 041908. Valdivino V. Junior. Universidade Federal de Goias, Campus Samambaia, CEP 74001-970, Goiânia, GO, Brazil. E-mail: vvjunior@ufg.br Pablo Rodriguez. Universidade Federal de Pernambuco, Av. Prof. Moraes Rego, 1235. Cidade Universitária, CEP 50670-901, Recife, PE, Brazil. E-mail: pablo@de.ufpe.br Adalto Speroto. Universidade de São Paulo, Caixa Postal 668, CEP 13560-970, São Carlos, SP, Brazil. E-mail: speroto@usp.br
9272	https://en.wikipedia.org/wiki/Equivalent_weight	Equivalent weight - 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 In history 2 Use in general chemistry 3 Use in volumetric analysis 4 Use in gravimetric analysis 5 Use in polymer chemistry 6 References Equivalent weight [x] 16 languages العربية বাংলা Català Dansk Ελληνικά Español Français Հայերեն हिन्दी Italiano ಕನ್ನಡ Polski Simple English ไทย Türkçe 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 Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Mass of one equivalent In chemistry, equivalent weight (more precisely, equivalent mass) is the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance. The equivalent weight of an element is the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. The corresponding unit of measurement is sometimes expressed as "gram equivalent". The equivalent weight of an element is the mass of a mole of the element divided by the element's valence. That is, in grams, the atomic weight of the element divided by the usual valence. For example, the equivalent weight of oxygen is 16.0/2 = 8.0 grams. For acid–base reactions, the equivalent weight of an acid or base is the mass which supplies or reacts with one mole of hydrogen cations(H+ ). For redox reactions, the equivalent weight of each reactant supplies or reacts with one mole of electrons(e−) in a redox reaction. Equivalent weight has the units of mass, unlike atomic weight, which is now used as a synonym for relative atomic mass and is dimensionless. Equivalent weights were originally determined by experiment, but (insofar as they are still used) are now derived from molar masses. The equivalent weight of a compound can also be calculated by dividing the molecular mass by the number of positive or negative electrical charges that result from the dissolution of the compound. In history [edit] Jeremias Benjamin Richter (1762–1807), one of the first chemists to publish tables of equivalent weights, and also the coiner of the word "stoichiometry". The first equivalent weights were published for acids and bases by Carl Friedrich Wenzel in 1777. A larger set of tables was prepared, possibly independently, by Jeremias Benjamin Richter, starting in 1792. However, neither Wenzel nor Richter had a single reference point for their tables, and so had to publish separate tables for each pair of acid and base. John Dalton's first table of atomic weights (1808) suggested a reference point, at least for the elements: taking the equivalent weight of hydrogen to be one unit of mass. However, Dalton's atomic theory was far from universally accepted in the early 19th century. One of the greatest problems was the reaction of hydrogen with oxygen to produce water. One gram of hydrogen reacts with eight grams of oxygen to produce nine grams of water, so the equivalent weight of oxygen was defined as eight grams. Since Dalton supposed (incorrectly) that a water molecule consisted of one hydrogen and one oxygen atom, this would imply an atomic weight of oxygen equal to eight. However, expressing the reaction in terms of gas volumes following Gay-Lussac's law of combining gas volumes, two volumes of hydrogen react with one volume of oxygen to produce two volumes of water, suggesting (correctly) that the atomic weight of oxygen is sixteen. The work of Charles Frédéric Gerhardt (1816–56), Henri Victor Regnault (1810–78) and Stanislao Cannizzaro (1826–1910) helped to rationalise this and many similar paradoxes, but the problem was still the subject of debate at the Karlsruhe Congress (1860). Nevertheless, many chemists found equivalent weights to be a useful tool even if they did not subscribe to atomic theory. Equivalent weights were a useful generalisation of Joseph Proust'slaw of definite proportions (1794) which enabled chemistry to become a quantitative science. French chemist Jean-Baptiste Dumas (1800–84) became one of the more influential opponents of atomic theory, after having embraced it earlier in his career, but was a staunch supporter of equivalent weights. Insofar as the atomic tables have been drawn up in part following the laws of Wenzel and Richter, in part by simple speculations, they have left plenty of doubts in the best of minds. It was to escape this problem that it was attempted to deduce the atomic weights from the density of the elements in the vapour state, from their specific heat, from their crystalline form. But one must not forget that the value of the figures deduced from these properties is not in the least absolute… To sum up, what have left from this ambitious excursion that we have allowed ourselves in the realm of the atoms? Nothing, nothing necessary at the very least. What we have left is the conviction that chemistry got itself lost there, as it always does when it abandons experiment, it tried to walk without a guide through the shadows. With experiment as a guide, you find Wenzel's equivalents, Mitscherlich's equivalents, they are nothing else but molecular groups. If I had the power, I would erase the word 'atom' from science, persuaded that it oversteps the evidence of experiment; and, in chemistry, we must never overstep the evidence of experiment. — Jean-Baptiste Dumas, lecture at the Collège de France, 1843/44 Equivalent weights were not without problems of their own. For a start, the scale based on hydrogen was not particularly practical, as most elements do not react directly with hydrogen to form simple compounds. However, one gram of hydrogen reacts with 8 grams of oxygen to give water or with 35.5 grams of chlorine to give hydrogen chloride: hence 8 grams of oxygen and 35.5 grams of chlorine can be taken to be equivalent to one gram of hydrogen for the measurement of equivalent weights. This system can be extended further through different acids and bases. Much more serious was the problem of elements which form more than one oxide or series of salts, which have (in today's terminology) different oxidation states. Copper will react with oxygen to form either brick red cuprous oxide (copper(I) oxide, with 63.5 g of copper for 8 g of oxygen) or black cupric oxide (copper(II) oxide, with 32.7 g of copper for 8 g of oxygen), and so has two equivalent weights. Supporters of atomic weights could turn to the Dulong–Petit law (1819), which relates the atomic weight of a solid element to its specific heat capacity, to arrive at a unique and unambiguous set of atomic weights. Most supporters of equivalent weights - which included the great majority of chemists prior to 1860 — simply ignored the inconvenient fact that most elements exhibited multiple equivalent weights. Instead, these chemists had settled on a list of what were universally called "equivalents" (H = 1, O = 8, C = 6, S = 16, Cl = 35.5, Na = 23, Ca = 20, and so on). However, these nineteenth-century "equivalents" were not equivalents in the original or modern sense of the term. Since they represented dimensionless numbers that for any given element were unique and unchanging, they were in fact simply an alternative set of atomic weights, in which the elements of even valence have atomic weights one-half of the modern values. This fact was not recognized until much later. The final death blow for the use of equivalent weights for the elements was Dmitri Mendeleev's presentation of his periodic table in 1869, in which he related the chemical properties of the elements to the approximate order of their atomic weights. However, equivalent weights continued to be used for many compounds for another hundred years, particularly in analytical chemistry. Equivalent weights of common reagents could be tabulated, simplifying analytical calculations in the days before the widespread availability of electronic calculators: such tables were commonplace in textbooks of analytical chemistry. Use in general chemistry [edit] The use of equivalent weights in general chemistry has largely been superseded by the use of molar masses. Equivalent weights may be calculated from molar masses if the chemistry of the substance is well known: sulfuric acid has a molar mass of 98.078(5)g mol−1, and supplies two moles of hydrogen ions per mole of sulfuric acid, so its equivalent weight is 98.078(5)g mol−1/2 eq mol−1= 49.039(3)g eq−1. potassium permanganate has a molar mass of 158.034(1)g mol−1, and reacts with five moles of electrons per mole of potassium permanganate, so its equivalent weight is 158.034(1)g mol−1/5 eq mol−1= 31.6068(3)g eq−1. Historically, the equivalent weights of the elements were often determined by studying their reactions with oxygen. For example, 50 g of zinc will react with oxygen to produce 62.24 g of zinc oxide, implying that the zinc has reacted with 12.24 g of oxygen (from the Law of conservation of mass): the equivalent weight of zinc is the mass which will react with eight grams of oxygen, hence 50 g×8 g/12.24 g= 32.7 g. Some contemporary general chemistry textbooks make no mention of equivalent weights. Others explain the topic, but point out that it is merely an alternate method of doing calculations using moles. Use in volumetric analysis [edit] Burette over a conical flask with phenolphthalein indicator used for acid–base titration When choosing primary standards in analytical chemistry, compounds with higher equivalent weights are generally more desirable because weighing errors are reduced. An example is the volumetric standardisation of a solution of sodium hydroxide which has been prepared to approximately 0.1 mol dm−3. It is necessary to calculate the mass of a solid acid which will react with about 20 cm 3 of this solution (for a titration using a 25 cm 3burette): suitable solid acids include oxalic acid dihydrate, potassium hydrogen phthalate and potassium hydrogen iodate. The equivalent weights of the three acids 63.04 g, 204.23 g and 389.92 g respectively, and the masses required for the standardisation are 126.1 mg, 408.5 mg and 779.8 mg respectively. Given that the measurement uncertainty in the mass measured on a standard analytical balance is ±0.1 mg, the relative uncertainty in the mass of oxalic acid dihydrate would be about one part in a thousand, similar to the measurement uncertainty in the volume measurement in the titration. However the measurement uncertainty in the mass of potassium hydrogen iodate would be five times lower, because its equivalent weight is five times higher: such an uncertainty in the measured mass is negligible in comparison to the uncertainty in the volume measured during the titration (see example below). As an example, assume that 22.45±0.03 cm 3 of the sodium hydroxide solution reacts with 781.4±0.1 mg of potassium hydrogen iodate. As the equivalent weight of potassium hydrogen iodate is 389.92 g, the measured mass is 2.004 milliequivalents. The concentration of the sodium hydroxide solution is therefore 2.004 meq/0.02245 L= 89.3 meq/L. In analytical chemistry, a solution of any substance which contains one equivalent per litre is known as a normal solution (abbreviated N), so the example sodium hydroxide solution would be 0.0893 N. The relative uncertainty (u r) in the measured concentration can be estimated by assuming a Gaussian distribution of the measurement uncertainties: u r 2=(u(V)V)2+(u(m)m)2=(0.03 22.45)2+(0.1 781.4)2=(0.001336)2+(0.000128)2 u r=0.00134 u(c)=u r c=0.1 m e q/L{\displaystyle {\begin{aligned}u_{\rm {r}}^{2}&=\left({\frac {u(V)}{V}}\right)^{2}+\left({\frac {u(m)}{m}}\right)^{2}\&=\left({\frac {0.03}{22.45}}\right)^{2}+\left({\frac {0.1}{781.4}}\right)^{2}\&=(0.001336)^{2}+(0.000128)^{2}\u_{\rm {r}}&=0.00134\u(c)&=u_{\rm {r}}c=0.1\ {\rm {meq/L}}\end{aligned}}} This sodium hydroxide solution can be used to measure the equivalent weight of an unknown acid. For example, if it takes 13.20±0.03 cm 3 of the sodium hydroxide solution to neutralise 61.3±0.1 mg of an unknown acid, the equivalent weight of the acid is: equivalent weight=m acid c(NaOH)V eq=52.0±0.1 g{\displaystyle {\text{equivalent weight}}={\frac {m_{{\ce {acid}}}}{c({\ce {NaOH}})V_{{\ce {eq}}}}}=52.0\pm 0.1\ {\ce {g}}} Because each mole of acid can only release an integer number of moles of hydrogen ions, the molar mass of the unknown acid must be an integer multiple of 52.0±0.1 g. Use in gravimetric analysis [edit] Powdered bis(dimethylglyoximate)nickel. This coordination compound can be used for the gravimetric determination of nickel. The term “equivalent weight” had a distinct meaning in gravimetric analysis: it meant the mass of precipitate produced from one gram of analyte (the species of interest). The different definitions came from the practice of quoting gravimetric results as mass fractions of the analyte, often expressed as a percentage. A related term was the equivalence factor, one gram divided by equivalent weight, which was the numerical factor by which the mass of precipitate had to be multiplied to obtain the mass of analyte. For example, in the gravimetric determination of nickel, the molar mass of the precipitate bis(dimethylglyoximate)nickel [Ni(dmgH)2] is 288.915(7)g mol−1, while the molar mass of nickel is 58.6934(2)g mol−1: hence 288.915(7)/58.6934(2)= 4.9224(1)grams of [Ni(dmgH)2] precipitate is equivalent to one gram of nickel and the equivalence factor is 0.203151(5). For example, 215.3±0.1 mg of [Ni(dmgH)2] precipitate is equivalent to (215.3±0.1 mg)×0.203151(5)= 43.74±0.2 mg of nickel: if the original sample size was 5.346±0.001 g, the nickel content in the original sample would be 0.8182±0.0004%. Gravimetric analysis is one of the most precise of the common methods of chemical analysis, but it is time-consuming and labour-intensive. It has been largely superseded by other techniques such as atomic absorption spectroscopy, in which the mass of analyte is read off from a calibration curve. Use in polymer chemistry [edit] Beads of an ion-exchange polymer. In polymer chemistry, the equivalent weight of a reactive polymer is the mass of polymer which has one equivalent of reactivity (often, the mass of polymer which corresponds to one mole of reactive side-chain groups). It is widely used to indicate the reactivity of polyol, isocyanate, or epoxythermoset resins which would undergo crosslinking reactions through those functional groups. It is particularly important for ion-exchange polymers (also called ion-exchange resins): one equivalent of an ion-exchange polymer will exchange one mole of singly charged ions, but only half a mole of doubly charged ions. Nevertheless, given the decline in use of the term "equivalent weight" in the rest of chemistry, it has become more usual to express the reactivity of a polymer as the inverse of the equivalent weight, that is in units of mmol/g or meq/g. References [edit] ^gram equivalent Merriam-Webster Dictionary ^Equivalent weight chemistry Encyclopædia Britannica ^ Jump up to: abInternational Union of Pure and Applied Chemistry (1998). Compendium of Analytical Nomenclature (definitive rules 1997, 3rd. ed.). Oxford: Blackwell Science. ISBN0-86542-6155. section 6.3. "Archived copy"(PDF). Archived from the original(PDF) on July 26, 2011. Retrieved 2009-05-10.{{cite web}}: CS1 maint: archived copy as title (link) ^Wenzel, Carl Friedrich (1777). Lehre von der Verwandtschaft der Körper [Theory of the Affinity of Bodies (i.e., substances)] (in German). Dreßden, (Germany): Gotthelf August Gerlach. ^Richter, J.B. (1792–1794). Anfangsgründe der Stöchyometrie … (3 vol.s) [Rudiments of Stoichiometry …] (in German). Breslau and Hirschberg, (Germany): Johann Friedrich Korn der Aeltere. ^ Jump up to: abcdefAtomeGrand dictionnaire universel du XIXe siècle (editeur Pierre Larousse, Paris 1866, vol.1, pages 868-73)(in French) ^Dalton, John (1808). A New System of Chemical Philosophy. London, England: R. Bickerstaff. p.219. ^See Charles-Adolphe Wurtz's report on the Karlsruhe Congress. ^Alan J. Rocke, Chemical Atomism in the Nineteenth Century: From Dalton to Cannizzaro (Ohio State University Press, 1984). ^For example, Petrucci, Ralph H.; Harwood, William S.; Herring, F. Geoffrey (2002). General Chemistry (8th ed.). Prentice-Hall. ISBN0-13-014329-4. ^Whitten, Kenneth W.; Gailey, Kenneth D.; Davis, Raymond E. (1992). General Chemistry (4th ed.). Saunders College Publishing. p.384. ISBN0-03-072373-6. Any calculation that can be carried out with equivalent weights and normality can also be done by the mole method using molarity. ^ISO 385:2005 "Laboratory glassware – burettes". ^The use of the term "normal solution" is no longer recommended by IUPAC. ^IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "equivalent entity". doi:10.1351/goldbook.E02192 ^See, e.g., Ion Exchange Resins: Classification and Properties(PDF), Sigma-Aldrich, archived from the original(PDF) on 10 December 2015, retrieved 14 April 2009 Retrieved from " Categories: Stoichiometry Amount of substance Polymer chemistry Equivalent quantities Chemistry Hidden categories: CS1 maint: archived copy as title CS1 German-language sources (de) Articles with French-language sources (fr) Articles with short description Short description is different from Wikidata This page was last edited on 30 April 2025, at 23:56(UTC). 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9273	https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-distance-and-midpoints/v/distance-formula	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
9274	https://math.stackexchange.com/questions/4412452/i-and-j-are-the-incenter-and-a-excenter-in-abc-k-and-l-are-similarl	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 $I$ and $J$ are the incenter and $A$-excenter in $ABC$. $K$ and $L$ are similarly defined in $ACD$. $IL \cap JK$ lies on the bisector of $\angle BCD$. Ask Question Asked Modified 3 years, 5 months ago Viewed 446 times 3 $\begingroup$ Problem 96 (India Postals 2015). Let $ABCD$ be a convex quadrilateral. In $\triangle ABC$, let $I$ and $J$ be the incenter and $A$-excenter, respectively. In $\triangle ACD$, let $K$ and $L$ be the incenter and $A$-excenter, respectively. Show that the lines $IL$, $JK$, and the bisector of $\angle BCD$ are concurrent. I came across this problem in A Beautiful Journey Through Olympiad Geometry by Stefan Lozanovski. My thoughts to solving this problem ranged from using Menelaus's/Ceva's theorem (trigonometric form, perhaps?) to using Desargues's theorem to convert the concurrency into a collinearity, though I don't immediately see how to make meaningful progress in either approaches. After being stuck for a while, I decided to look it up and found this thread, which explicitly asked for an analytic solution. To quote the author himself, "The word 'Beautiful' in the books title means that we will explore only synthetic approaches and proofs, which I find elegant and beautiful. We will not see any analytic approaches, such as Cartesian or barycentric coordinates, nor we will do complex number or trigonometry bashing." That's why I expect there to indeed be a geometrical solution. For those who do not mind the clutter, the image below is an attempt using Desargues's Theorem, where $R$ is a point chosen on the bisector of $\angle BCD$ such that $P AD$. The triangles I assumed perspectivity are $\triangle ICJ$ and $\triangle LRK$, so I need $IC \cap LR = Q$, $IJ \cap LK = A$ and $CJ \cap RK = P$ to be collinear, i.e. $Q AD$, though I myself am somewhat overwhelmed at this point. Does anyone know of a solution that doesn't rely heavily on analytic methods? geometry euclidean-geometry Share edited Mar 25, 2022 at 12:39 kairekaire asked Mar 25, 2022 at 12:18 kairekaire 29811 silver badge99 bronze badges $\endgroup$ 1 $\begingroup$ One can find very short solution via the so called Isogonal lemma. See here for more details artofproblemsolving.com/community/… $\endgroup$ richrow – richrow 2022-04-03 16:08:26 +00:00 Commented Apr 3, 2022 at 16:08 Add a comment \| 3 Answers 3 Reset to default 1 $\begingroup$ Some introductory words first, no formulas, no pictures yet, words regarding my process of searching and finding a solution. Then why this solution should be considered still as beautiful. Then some further comments on first tries to solve the problem inside projective geometry, and why this failed for me. Please skip, if it feels annoying at some (early) point. The key to the solution was indeed to exhibit the (equivalent) situation using an economical statement (few points, and only points that determine the geometric constellation). As given, we have constructions inside the triangles $\Delta ABC$, so that $I,J$ appear, and $\Delta ACD$, so that $K,L$ appear, then we build the lines $IK$ and $JL$, and their intersection should be on the angle bisector in $C$ in $\Delta BCD$. After drawing some pictures, and trying to see which are the (independent) parameters determining the configuration of points, the problem becomes simpler. First of all, from the whole triangle $\Delta ABC$ we need only $I,J$. We fix $A,C$, we need them. But we remove $B$ as starting point from the story, consider the point $I$ only. Then $J\in AI$ is obtained by constructing a right angle in $C$. Similarly, $D$ is not an imperative, we need only $K$, then construct $L\in AK$ so that we have again a right angle in $C$ in the corresponding triangle $\Delta KCL$. A lot of angle bisectors are now eliminated from the picture. To have more symmetry in the statement, i will keep $I,J$ in the notation, and use $I',J'$ instead of $K,L$. Now the picture is determined by the angles in $A,C$ in the two triangles $T=\Delta AIC$ and $T'=\Delta AI'C$. With this simpler constellation of points, it is easier to isolate a situation to show a concurrence of lines (or alternatively to intersect two lines, and show a colinearity of three points). We finaly show a Ceva / Menelaus shaped relation. The involved proportions have to be expressed in terms of the free parameters of the problem. They are the angles in $T,T'$. Well, the sine theorem is handy in such situations, and some trigonometric expressions do occur. Arguably, this destroys the geometry. Most times, people that are not typing solutions come with this argument. My point of view is that if the involved $\sin$-relations are simple, a posteriori a purist user may easily find her or his road to rewrite. So there will be no complicated trigonometric yoga below. One more comment. My first try was to use projective relations. For instance the fact that on the line $AI$ the points $A$, $A_B:=AI\cap BC$, $I$, $J$ are building a harmonic tuple. Similarly, the points $A$, $A_D:=AK\cap CD$, $K$, $L$ are building a harmonic tuple. Then using for instance projectivity, the lines $IL$, $JK$, $A_BA_D$ are concurrent in a point, call it $\Omega$. Now one can try to find the position of $\Omega$ on the segment $A_BA_D$, show it is on the angle bisector in $C$ in $\Delta BCD$. But there is a problem regarding the beauty now again. We have to use either $IL$ or $JK$, and we need to use angles, formulas get immediately angry while writing them, still no problem, but there is no clear (computational) idea to show. Nevertheless, some properties involving the points $A_B,A_D$ may be mentioned as bonus to complete the picture. So let us start. We remove from the picture all unneeded letters, use $I',J'$ instead of $K,L$ to keep the symmetry in the picture, and restate equivalently: Proposition 1: On the segment $AC$ consider in different half-planes the triangles $T=\Delta AIC$, $T'=\Delta AI'C$. We denote by $a,a'$ the two angles in $A$ in $T,T'$. And by $c,c'$ the two angles in $C$ in $T,T'$. Construct now $J\in AI$ and $J'\in AI'$ so that the triangles $\Delta ICJ$, $\Delta I'CJ'$, have each a right angle in $C$. Draw the angle bisector of $\widehat {ICI'}$ and reflect $A$ into $A^$ w.r.t. to it. So $\widehat{ACI}$ and $\widehat{A^CI'}$ have the same measure (with opposite sign). (This property can be used to construct $A^$ alternatively.) Then: The lines $IJ'$, $I'J$, $CA^$ are concurrent. Proof: Define the point $\Omega$ as $\Omega:=IJ'\cap CA^$. We will show $\Omega\in I'J$. A good and simple way to show this is by checking the formula of Menelaus in $\Delta AIJ'$ w.r.t. the to-be-secant-line passing through $\Omega\in IJ'$, $I'\in J'A$, and $J\in AI$: $$ \tag{$$} 1\overset?= \frac{\Omega I}{\Omega J'}\cdot \frac{I'J'}{I'A}\cdot \frac{JA}{JI}\ . $$ So we start computing the R.H.S. from above. $$ \begin{aligned} \frac{\Omega I}{\Omega J'}\cdot \frac{I'J'}{I'A}\cdot \frac{JA}{JI} & = \frac{\Omega I}{CI}\cdot \frac{C I}{C J'}\cdot \frac{CJ'}{\Omega J'}\ \cdot\ \frac{I'J'}{I'A}\ \cdot\ \frac{JA}{JI} \ & = \frac{\sin c'}{\sin\hat\Omega_1}\cdot \frac{C I}{C J'}\cdot \frac{\sin\hat\Omega_2}{\sin (90^\circ + c)}\ \cdot\ \frac{I'J'}{I'A}\ \cdot\ \frac{JA}{JI} \ & \qquad\qquad\text{($\hat\Omega_1$, $\hat\Omega_2$ are the two angles between $C\Omega$ and $IJ'$)} \ & = \frac{\sin c'}{\sin (90^\circ + c)}\cdot \frac{C I}{JI}\cdot \frac{I'J'}{C J'} \cdot \frac{JA}{CA}\cdot \frac{CA}{I'A} \ & = \frac{\sin c'}{\sin (90^\circ + c)}\cdot \frac{\cos\widehat{CIJ}}{\sin\widehat{CI'J'}}\cdot \frac{\sin (90^\circ + c)}{\sin (90^\circ -c-a)} \cdot \frac{\sin(a'+c')}{\sin c'} \ &=1 \ . \end{aligned} $$ Here, $\widehat{CIJ}=a+c$, and $\widehat{CI'J'}=a'+c'$, as exterior angles for $T,T'$ in $I,I'$ respectively. So the question mark in $()$ is elucidated. $\square$ We are done, but to have the beauty back, we can add some bonus ingredients and conclude in the given situation: Proposition 2 (Bonus): Let $ABCD$ be a quadrilateral. In $\Delta ABC$ we construct $I,J, A_B$ as the incenter, the $A$-excenter, and the intersection of the $A$-angle bisector with the opposite side. In $\Delta ACD$ we construct in similar manner $K,L,A_D$. Then $IL$, $JK$, $A_BA_D$, and the $B$-angle bisector of $\Delta BCD$ are concurrent in a point, call it $\Omega$. And $IK$, $JL$, $A_BA_D$, and the $B$-angle exterior bisector of $\Delta BCD$ are concurrent in a point, call it $\Xi$. And the $4$-tuples $(AA_B;IJ)$, $(AA_D;KL)$, $(A_BA_D;\Omega\Xi)$ are harmonic. (The associated cross ratios are $-1$.) Proof: Using Proposition 1, we know $\Omega:=IL\cap JK$ lies on the $B$-angle bisector. Then in $\Delta CAA_B$ the $C$-angle bisectors (internal and external) are marking on $AA_B$ a harmonic ratio. Same for $\Delta CAA_D$. Denoting the cross ratio of four points by square brackets, we have thus $$-1 = [AA_B;IJ]= [AA_D;KL]\ .$$ Recall that the cross ratio is invariated when projecting from a point from a line onto an other line. A converse property is also true, so from $[AA_B;IJ]=-1=[AA_D;LK]$ we obtain the concurrence of the three lines $A_BA_D$, $IL$, $JK$. (Indeed, project from $\Omega:=IL\cap JK$ the line $AI$ onto the line $AJ$. Then $A\to A$, $I\to L$, $J\to K$, and $A_B$ goes to a point which conserves the cross ratio, this point is unique, so it is $A_D$, so $\Omega\in A_BA_D$.) Similarly, using $[AA_B;IJ]=-1=[AA_D;KL]$ we get the well defined point $\Xi$ from the statement. Applying Proposition 1. for $A,C,I,I'=L$ instead of $A,C,I,I'=K$ we get $\Xi$ instead of $\Omega$, we use the angles $a,a';c, c'+90^\circ$ instead of $a,a';c,c'$, so the point $A^$ gets a $90^\circ$-rotation around $C$, so the new angle bisector $CA^$ is the exterior angle bisector. We are done. $\square$ Share answered Apr 2, 2022 at 4:17 dan_fuleadan_fulea 38.2k11 gold badge2626 silver badges6464 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ Hints: Here are two strategies you can try: 1): -Draw circumcircle of triangle BCD. -Draw a circle on center J tangents to sides( or their extension) of triangle ABC. You have to prove following points: 1- IL is perpendicular on KJ. 2- Extend CM to touch the circucircle of triangle BCD at point G. Connect J to G. JG is perpendicular bisector of BD, therefore CM is bisector of $\angle BCD$. 2): 1-Draw a perpendicular on CM from point C to touch the circle centered on J and tangent on sides of triangle ABC at point P. 2- Connect C to J, it intersect the circle at C'. Extend it to meet the circle again at H. 3- Mark tangent point of this circle with side BC as N. Connect N to J and extend it to touch the circle at E. Arcs C'N and EH are equal, therefore arcs NP and PH are equal, that is CP is bisector of angle $\angle BCJ$ which in turn means CM is bisector of $\angle BCD$. Share answered Mar 25, 2022 at 20:00 siroussirous 13.1k33 gold badges1616 silver badges2525 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ Lemma. Let $AP$, $AQ$ and $AS$, $AT$ be two pairs of isogonal lines with respect to $\angle BAC$. Let $PS \cap QT = X$ and $PT \cap QS = Y$. Then, $AX$, $AY$ are isogonal lines with respect to $\angle BAC$. Drawing inspiration from other answers, it occurred to me that isogonality of lines played a rather important role, and hence provided motivation for the application of the above lemma, referred to as the Isogonal Lines Lemma, a proof of which can be found in the book. With this in mind, we proceed with the solution to the problem. Proof. Let $IL \cap JK = M$, we shall show that $CM$ bisects $\angle BCD$. Note that $\angle ICJ = \angle KCL = 90°$, so $CJ$ and $CL$ are isogonal with respect to $\angle ICK$. By the Isogonal Lines Lemma applied to lines $CI$ and $CK$ along with $CJ$ and $CL$, we see that the lines joining $C$ to $IJ \cap KL = A$ and to $IL \cap JK = M$ are isogonal as well, so $\angle ACI = \angle MCK$. Consequently, $\begin{align} \angle BCM = \angle BCA + \angle ACM &= 2\cdot \angle ACI + \angle ACM \ &= \angle MCK + (\angle MCK + \angle ACM) \ &= \angle MCK + \angle ACK \ &= \angle MCK + \angle DCK \ &= \angle DCM, \end{align}$ as desired. Share answered Apr 3, 2022 at 11:35 kairekaire 29811 silver badge99 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry euclidean-geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 3 Prove that $IL,JK$ and angle bisector of angle $BCD$ are concurrent Related Condition for three lines to be concurrent. Prove that the excenter of $A$ in $\triangle ABC$, the midpoint of $BC$ and $H$ are collinear. 0 Waht's the measure of the segment $TS$ in the triangle below? 2 $\angle{B}=45^{\circ}$ in $\triangle{ABC}$, $D$ is on bisector of $\angle{A}$ and $AD=BC, \angle{ACD}=30^{\circ}$. Find$\angle{A}$. 1 prove that $M$ and $M'$ are antipodal on the circumcircle and $MM'$ is the perpendicular bisector of $BC$ 1 Prove that $CR$ bisects $\angle C$ 3 Proving directly that, for scalene $\triangle ABC$, the bisector of $\angle A$ and perpendicular bisector of $BC$ meet at a point on the circumcircle 0 Ceva's and Menelaus in Olympiad problems - Possible generalization of Monge's Theorem? 2 Prove that the circumcenter is the intersections of perpendiculars onto the sides of the orthic triangle Prove that midpoint of $BI$ is the incenter of $ABX$ Hot Network Questions Identifying a thriller where a man is trapped in a telephone box by a sniper Does the curvature engine's wake really last forever? What is the meaning of 率 in this report? Who is the target audience of Netanyahu's speech at the United Nations? Storing a session token in localstorage Do sum of natural numbers and sum of their squares represent uniquely the summands? 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How to design a circuit that outputs the binary position of the 3rd set bit from the right in an 8-bit input? Verify a Chinese ID Number Can a state ever, under any circumstance, execute an ICC arrest warrant in international waters? more hot questions Question feed
9275	https://www.linkedin.com/pulse/understanding-normalization-impact-social-structure-k%C5%AFl%C5%9Dh%C5%99%C4%99%C5%9D%C5%A7h%C3%A3--ddw1c	Skip to main content Understanding Normalization: Impact on Social Structure, Inequality, and Meritocracy Ŝã♏iŧ ☸ Ⓚ Ⓤ Ⓜ Ⓐ Ⓡ☸ Ŝã♏iŧ ☸ Ⓚ Ⓤ Ⓜ Ⓐ Ⓡ☸ + Follow Normalization, as a concept, spans across various fields, from social structures and cultural norms to data analysis and business practices. In its essence, normalization refers to the process of adjusting behaviors, values, or variables to a standard or typical level, fostering consistency, predictability, and fairness. However, while it may seem beneficial in many contexts, normalization can also reinforce existing power structures, create inequalities, and challenge the principle of meritocracy. This article aims to explore the concept of normalization, examine its impact on social structures, address how it may perpetuate inequality, and discuss its implications for national growth and meritocracy. What is Normalization? Normalization involves aligning behaviors, processes, or values to conform to a standard or norm. In various contexts, this standardization is intended to create fairness, order, and stability. Normalization can apply to societal expectations (such as laws, customs, or professional norms), data sets (in statistics and machine learning), or even educational systems (grading, assessments). At its core, normalization ensures that disparities are minimized, and individuals or data points are compared or evaluated on the same scale. This can lead to uniformity, reducing bias by creating equal benchmarks. However, its very nature also raises questions regarding its implications on social fairness and innovation. Types of Normalization Normalization is not a single, uniform concept, but rather exists in multiple forms, each suited to different fields. Understanding the different types and techniques is crucial for recognizing both their benefits and potential drawbacks. Social and Cultural Normalization In sociology and anthropology, normalization refers to the societal process by which certain behaviors, values, or beliefs are seen as the standard or accepted practice. This can include: Cultural Norms: Societal expectations surrounding dress, communication, family roles, and values. Legal Norms: The formalization of laws and regulations that determine acceptable behavior in society. Psychological Norms: Behavioral standards that define what is considered typical or "healthy" behavior within a particular society. For example, the concept of gender roles is often normalized in various cultures, dictating what is considered appropriate behavior for men and women. These norms influence everything from career expectations to family dynamics. Data Normalization In statistics and machine learning, normalization refers to the process of adjusting data so that it can be compared or processed on a common scale. There are different techniques for this type of normalization: Min-Max Normalization: Rescaling data so that it fits within a specified range, typically between 0 and 1. Z-Score Normalization: Standardizing data by converting values into their corresponding z-scores (i.e., subtracting the mean and dividing by the standard deviation). Decimal Scaling: Shifting the decimal point of data values to bring them within a desired range. This kind of normalization ensures that datasets with different units or scales (e.g., income and age) can be analyzed on an equal footing. Normalization in Education In education, normalization often refers to setting standardized expectations or benchmarks for academic performance. Examples include: Grading Systems: Standardizing how students are assessed across different schools or institutions. Curriculum Normalization: Implementing uniform curricula to ensure that all students, regardless of school or district, receive an equivalent education. By normalizing educational metrics, institutions can compare students fairly. However, this approach may favor those who excel in standardized settings while disadvantaging those with different learning styles. Normalization’s Impact on Social Structure Normalization plays a crucial role in shaping social structures. By setting accepted norms, it ensures social cohesion and uniformity. However, these norms may also reinforce existing power dynamics, inequalities, and discrimination, which can have significant implications for societal growth and equity. Reinforcement of Hierarchical Structures Normalization often reflects the values of dominant social, cultural, or political groups. When certain behaviors or ideologies are normalized, they become the standard by which others are judged. For example, in many Western societies, white, middle-class, heterosexual, and able-bodied norms have been historically reinforced as the standard. This creates a hierarchy in which marginalized groups, such as racial minorities, LGBTQ+ individuals, and people with disabilities, are often seen as "deviations" from the norm. This can lead to systemic exclusion, as those who do not fit into the normalized categories may experience discrimination, unequal access to resources, and reduced opportunities for success. Impact on Marginalized Groups When societies normalize particular behaviors or standards, individuals or groups that do not fit those expectations can be marginalized or stigmatized. For instance, if a society normalizes certain beauty standards, individuals who do not fit those standards may face societal pressures, discrimination, and even exclusion. Similarly, in the workplace, certain "professional norms" (such as dress codes or communication styles) may exclude people from different cultural backgrounds or social classes, hindering their opportunities for success. Normalization in social contexts can thus perpetuate inequality, as it often excludes or disadvantages those who do not conform to the dominant group’s expectations. Does Normalization Promote Inequality? Yes, normalization can indeed promote inequality in a variety of ways: Bias Toward Dominant Groups: When normalization favors the norms of a dominant group, it creates an unequal playing field. For example, the education system may normalize particular learning styles or standardized test formats that inherently favor students from wealthier backgrounds who have access to better resources. Recommended by LinkedIn Social Segregation Lord Edwin E. Hitti 7 months ago The beauty of Capitalism Simranjit Singh Arora 6 years ago Your Movement’s Narrative Will Determine Its Future Derrick Feldmann 10 months ago Cultural Imperialism: Normalization can spread cultural values and behaviors that marginalize other cultures. For instance, in the context of globalization, Western cultural norms—such as individualism and capitalism—are often presented as the universal standard, leaving little room for alternative ways of thinking or living. Stifling Diversity: By normalizing a particular set of behaviors or values, societies may reduce the richness of cultural diversity. Minority perspectives, whether related to ethnicity, religion, or socioeconomic status, may be overlooked or deemed less valuable, thereby perpetuating inequalities. Normalization and Meritocracy Meritocracy is the belief that individuals should succeed based on their talents, abilities, and achievements rather than social background or connections. While normalization, when applied correctly, can promote fairness and equal opportunities, it can also challenge meritocracy if it imposes rigid or biased standards. Standardization vs. Individual Potential In systems where normalization is overly rigid (e.g., standardized testing), meritocracy can be undermined. Students or employees who do not excel within the standardized system—perhaps due to learning differences or personal circumstances—may be unjustly evaluated as less capable. This means that normalization can mask true talent and fail to recognize the diverse forms of intelligence or capability individuals possess. Structural Barriers to Equality Normalization can create barriers that prevent true meritocratic outcomes, especially when those barriers disproportionately affect marginalized groups. For instance, if the normalization process in education is based on a singular model of learning or achievement, it may favor students from privileged backgrounds who have access to tutors, extracurricular activities, or better schools, while disadvantaging those who lack these resources. Normalization and National Growth Normalization has the potential to impact national growth in both positive and negative ways: Encouraging Stability and Order On a positive note, normalization ensures social order, legal conformity, and stability, which are essential for a functioning society. This can encourage economic growth and development, as people are more likely to contribute productively to society when they understand what is expected of them. Restricting Innovation However, if the norms enforced are overly rigid or narrow, they can stifle innovation and creativity. For example, a country that normalizes traditional education methods may fail to nurture students who have creative or entrepreneurial talents that don't fit conventional academic metrics. Challenges to Diversity and Growth Normalization can also hinder growth if it suppresses diverse viewpoints, experiences, and innovations. Societies that fail to recognize the value of diversity may miss out on opportunities for economic, cultural, and technological advancements that come from different perspectives and ideas. Moving Forward: A Balanced Approach To move forward in a way that ensures normalization is both effective and equitable, there are several key strategies to consider: Inclusive and Flexible Standards: The standards set by normalization should be inclusive, reflecting the diversity of human experiences. They should allow for flexibility and room for varied paths to success. This is especially important in fields like education, where learning styles and talents are not uniform. Ongoing Reflection and Reevaluation: Societies must regularly reflect on the norms they have established. Are they fair? Do they promote equality, or do they perpetuate existing disparities? By engaging in ongoing dialogue and reevaluating the impact of normalized standards, we can ensure that they evolve to reflect a more inclusive and just society. Promoting Merit Beyond the Norm: For meritocracy to thrive, it must recognize and reward a wide range of abilities and achievements. We should aim to create systems—whether in education, employment, or other sectors—that evaluate individuals based on their unique talents and contributions, rather than solely on their ability to conform to a standardized norm. Encouraging Diversity in All Areas: Innovation and progress are born from diversity. Encouraging diverse viewpoints, backgrounds, and experiences in all areas of society—from the workplace to governance—can lead to greater creativity, problem-solving, and ultimately, economic and social growth. Addressing Structural Inequality: Finally, normalization efforts must be paired with efforts to dismantle systemic inequalities. This includes ensuring equal access to resources, opportunities, and representation for all individuals, regardless of their background, identity, or circumstances. The biased attitude of authorities in restricting an individual’s growth simply because they raised legitimate questions or highlighted shortfalls in their performance falls under structural inequality. When an employee points out inefficiencies or suggests improvements through an Annual Performance Report or similar evaluations, it is usually an effort toward constructive self-awareness and system improvement. However, if authorities retaliate by hindering their career growth, this reflects an unfair bias and a power imbalance within the organization. Such behavior can suppress dissent and discourage critical feedback, which is essential for improvement and innovation. Instead of fostering an environment of openness, where individuals can voice concerns without fear of retribution, these actions create a culture where conformity is prioritized over constructive criticism. This undermines the principles of meritocracy, where growth should be based on performance, abilities, and contributions, rather than conformity to authority. The punitive response to legitimate feedback also reinforces inequitable power dynamics, where those in power are not held accountable, and individuals who challenge the system are marginalized. This leads to a situation where structural inequality persists, as those who raise concerns are denied opportunities for career advancement, while authority figures face no consequences for failing to address shortcomings. To address this, organizations must foster a culture of constructive feedback, transparency, and protection against retaliation. Clear, merit-based evaluation systems should be implemented to ensure fairness, and authorities must be trained to recognize and eliminate biases. Additionally, promoting accountability and ensuring that those in power are held responsible for their actions will help create an equitable environment where individuals are rewarded for their contributions and not punished for raising valid concerns. The impact of these biased practices extends beyond individual careers, perpetuating inequality within the system. When individuals are penalized for raising legitimate concerns, it undermines the principles of fairness and equality, limiting opportunities for growth. This creates a skewed professional environment that discourages innovation, suppresses diverse perspectives, and ultimately hampers organizational development. To address these concerns, there must be mechanisms to ensure fairness and accountability. Systems that allow for an unbiased review of such issues—whether through formal grievance redressal processes or independent bodies—can help ensure that individuals are not unfairly penalized for their contributions. Authorities, such as the Central Administrative Tribunal (CAT), or higher judicial bodies like the Chief Justice of India (CJI), play crucial roles in addressing systemic issues and promoting a just, merit-based environment, where individuals are rewarded for their performance and ability, rather than their willingness to conform. A Path Toward Equity and Innovation Normalization, when used wisely and fairly, has the potential to enhance social harmony, streamline processes, and ensure equal opportunities. However, to ensure it serves everyone equitably, it must be flexible, inclusive, and regularly revisited. By addressing the biases inherent in normalized systems, promoting diverse forms of merit, and creating environments that celebrate rather than suppress difference, societies can foster an environment where all individuals are empowered to succeed. In this way, normalization can become a tool for growth, innovation, and social progress rather than a mechanism for reinforcing inequality. By aligning normalization practices with the principles of fairness, justice, and meritocracy, we can create a more equitable and dynamic future for all. It should account for the diversity of human experiences, talents, and needs, ensuring that everyone, regardless of their background or identity, has an equal opportunity to succeed. In doing so, societies can better align their structures with the principles of justice, equality, and meritocracy, ultimately fostering an environment where both individuals and nations can thrive. To effectively address issues of biased treatment, unfair performance evaluations, and systemic inequality within organizations, leadership at every level must embody qualities that foster a fair, merit-based environment. Visionary, collaborative, and pragmatic leadership are essential to transforming organizational cultures and ensuring that individuals are evaluated based on their performance, rather than their conformity or willingness to avoid challenging the status quo. Visionary leadership is crucial for setting long-term goals and creating an organizational culture where fairness, transparency, and meritocracy are the guiding principles. Such leaders anticipate future challenges, value diversity of thought, and encourage an environment where constructive criticism is seen as a tool for growth, not a threat to authority. Collaborative leadership breaks down hierarchical barriers and fosters an inclusive environment where feedback, innovation, and critical thinking are encouraged at all levels. By building trust and creating platforms for open dialogue, collaborative leaders ensure that all voices are heard, particularly those raising valid concerns about bias or unfair practices. This shared responsibility helps prevent the perpetuation of structural inequalities within the organization. Pragmatic leadership ensures that visionary goals are translated into actionable steps. By implementing clear, objective evaluation processes, reducing bias, and promoting transparency, pragmatic leaders create systems that allow for fair and unbiased assessments. They also remain adaptive, continuously refining policies and practices based on feedback and changing circumstances, ensuring that corrective measures are swiftly enacted when issues arise. In sum, leadership that is visionary, collaborative, and pragmatic is not just a theoretical ideal but a necessity for creating organizations that promote merit, inclusivity, and equal opportunity. Such leadership empowers individuals to raise legitimate concerns, challenge inequalities, and thrive based on their abilities and contributions. When these principles are embraced at all levels, organizations can move towards a future where structural inequality is actively addressed, and every individual has the opportunity to succeed. 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9276	https://wchh.onlinelibrary.wiley.com/doi/10.1002/psb.2025	New guidance on the diagnosis and management of gout - Chaplin - 2022 - Prescriber - Wiley Online Library Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Skip to Article Content Skip to Article Information Search within Search term Advanced SearchCitation Search Search term Advanced SearchCitation Search Search term Advanced SearchCitation Search Login / Register Individual login Institutional login REGISTER Home Journal Info Overview Journal Metrics Contact Editorial Board Permissions Advertise Media Kit For Authors Author Guidelines Reviewer Resources Browse Current Issue All Issues News Article Type Analysis Drug Review Editorial Guidelines New from NICE New Products Prescribing in Practice Supplements English Supplements Spanish Supplements Journal list menu Journal Articles Actions New from NICE Free Access New guidance on the diagnosis and management of gout Steve Chaplin, Steve Chaplin Steve Chaplin is a medical writer specialising in therapeutics Search for more papers by this author Steve Chaplin, Steve Chaplin Steve Chaplin is a medical writer specialising in therapeutics Search for more papers by this author First published: 21 November 2022 About References ---------- Related ------- Information ----------- PDF Sections Abstract Diagnosis and assessment Information and support Managing flares Long-term management Referral Summary Declaration of interests References PDF Tools Request permission Export citation Add to favorites Track citation ShareShare Give access Share full text access Close modal Share full-text access Please review our Terms and Conditions of Use and check box below to share full-text version of article. [x] I have read and accept the Wiley Online Library Terms and Conditions of Use Shareable Link Use the link below to share a full-text version of this article with your friends and colleagues. Learn more. Copy URL Share a link Share on Email Facebook x LinkedIn Reddit Wechat Bluesky Abstract Acute flares of gout can cause severe pain and the condition is associated with long-term complications, but current management is often suboptimal. The new NICE guideline (NG219) aims to optimise outcomes for patients with gout by improving its diagnosis and acute and long-term management. Most people with gout have their condition managed in primary care – but not always very well, according to NICE. Only one-third receive urate-lowering treatment (ULT) and it is used effectively in only one-third of those who do. Bearing in mind that 2–3% of adults have gout, of whom 25% have stage 3–5 chronic kidney disease (CKD), the need for management guidance is clear. Gout: Diagnosis and Management (NG219)1 is a relatively brief guideline covering diagnosis and assessment, information and support, the management of flares, diet and lifestyle, long-term management and specialist referral. The recommendations are summarised well in two charts available on the NICE website.2, 3 Diagnosis and assessment Gout is a type of arthritis that is caused by monosodium urate crystals forming inside and around joints. The cardinal signs and symptom of gout are rapid onset of severe pain, often overnight, affecting the first metatarsophalangeal joints, but also other joints, and tophi. Septic arthritis should be excluded. The diagnosis is confirmed by a serum urate level of ≥360μmol/L. If this threshold is not reached but gout is suspected, the serum urate measurement should be repeated after two weeks; remaining uncertainty can be addressed by microscopy of synovial fluid or imaging of the joint. Information and support People with gout and their families/carers should be informed about its signs and symptoms, causes and risk factors, and signposted to sources of support. They should be told that it is a progressive lifelong condition for which urate- lowering therapy is necessary to prevent joint damage. NICE notes that there are many incorrect beliefs about what causes gout and the sort of people who develop it. People should therefore be informed that there is no evidence that diet prevents flares or lowers serum urate, but being overweight or obese, or excessive alcohol consumption, may exacerbate flares and symptoms. Managing flares Application of an ice pack may help reduce pain associated with a flare. First-line treatment is an NSAID (possibly with a proton pump inhibitor), colchicine, or a short course of off-label oral corticosteroids, taking into account co-morbidities, interactions and patient preference. If NSAIDs or colchicine are unsuitable or ineffective, off-label intra-articular or intramuscular corticosteroids can be considered; there is no evidence that one route of administration is better than the other. Evidence indicates no difference between NSAIDs, colchicine and corticosteroids for most outcomes, including pain and joint tenderness or swelling. NSAIDs are contraindicated in people with CKD or a history of cardiovascular disease. If none of these options is suitable or effective, referral to a specialist service should be offered to consider treatment with an interleukin-1 (IL-1) inhibitor (eg canakinumab). IL-1 inhibitors are expensive (and not cost effective compared with the alternatives) and rarely prescribed for gout. In current practice, “little or no follow up” is offered to people after a flare, missing an opportunity to discuss long-term treatment. NICE recommends that, when the flare has settled, the urate level should be measured (to inform the need for ULT) and the person should be given information about managing flares for themselves. Co-morbidities (especially cardiovascular risk factors and CKD) should be assessed and medication reviewed, and the risks and benefits of starting ULT discussed. Long-term management Long-term management with ULT should be offered to people with gout who have multiple or troublesome flares, tophi, stage 3–5 CKD, are taking a diuretic or who have chronic gouty arthritis. This option should also be discussed with people who do not meet these criteria but who have had one or more flares. ULT is used in a treat-to-target strategy, aiming to lower serum urate to <360μmol/L, or <300μmol/L if flares persist and are frequent and in those with tophi or chronic gouty arthritis. ULT is usually a lifelong treatment that continues after the target urate level is reached. Treatment should begin at least two to four weeks after a flare has settled (but can be started during a flare if they are frequent). The initial dose should be low and titrated upwards, guided by monthly urate measurements and tolerability. Once the target is achieved, NICE advises annual monitoring of serum urate levels. The options for first-line ULT are allopurinol or febuxostat, chosen according to the patient's co-morbidities and preferences. Allopurinol is preferred for people with a history of major cardiovascular disease because febuxostat has been associated with a higher risk of cardiovascular-related death. For allopurinol, frequently prescribed doses of up to 300mg daily are often too low to achieve the target urate level. If the target urate level cannot be reached or the first option is not tolerated, treatment should be switched to the alternative. The benefits and risks of treatment to prevent flares when starting or titrating ULT should be discussed. Colchicine is the agent of choice because, unlike NSAIDs and corticosteroids, it has some evidence to support its use for preventing flares when starting ULT, although gastrointestinal adverse effects may be a problem. If colchicine is not suitable or ineffective, off-label low-dose NSAIDs or low-dose oral corticosteroids can be considered. An IL-1 inhibitor is again a last resort after referral to rheumatology services. Referral Referral to a rheumatology service should be considered if the diagnosis of gout is uncertain; gout treatment is contraindicated, not tolerated or ineffective; if the patient has stage 3b–5 CKD; or if they have had an organ transplantation. Summary The recommendations in this succinct guideline emphasise the importance of monitoring serum urate and long-term ULT for people with gout, with relatively few treatment options to choose between. People with gout need to be better informed about their condition to improve management. Declaration of interests None to declare. References 1 National Institute for Health and Care Excellence. Gout: diagnosis and management. NG219. June 2022. Available from: Google Scholar 2 National Institute for Health and Care Excellence. Long-term management of gout with ULTs. Visual summary. Available from: Google Scholar 3 National Institute for Health and Care Excellence. Management of gout. Visual summary. Available from: Google Scholar Volume 33, Issue 11-12 Nov-Dec 2022 Pages 25-26 References ---------- Related ------- Information ----------- Recommended Gout and its management Nicola Dalbeth,Leanne Te Karu,Lisa K. Stamp, Internal Medicine Journal Metrics Full text views: 1,113 Full text views and downloads on Wiley Online Library. 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9277	https://arxiv.org/pdf/2002.09320	An Advance on Variable Elimination with Applications to Tensor-Based Computation 1 Adnan Darwiche 2 Abstract. We present new results on the classical algorithm of variable elimination, which underlies many algorithms including for probabilistic inference. The results relate to exploiting functional de-pendencies, allowing one to perform inference efficiently on models that have very large treewidth. The highlight of the advance is that it works with standard (dense) factors, without the need for sparse factors or techniques based on knowledge compilation that are com-monly utilized. This is significant as it permits a direct implementa-tion of the improved variable elimination algorithm using tensors and their operations, leading to extremely efficient implementations espe-cially when learning model parameters. We illustrate the efficacy of our proposed algorithm by compiling Bayesian network queries into tensor graphs and then learning their parameters from labeled data using a standard tool for tensor computation. 1 Introduction The work reported in this paper is motivated by an interest in model-based supervised learning, in contrast to model-free supervised learn-ing that currently underlies most applications of neural networks. We briefly discuss this subject first to put the proposed work in context. Supervised learning has become very influential recently and stands behind most real-world applications of AI. In supervised learning, one learns a function from labeled data, a practice that is now dominated by the use of neural networks to represent such func-tions; see [16, 17, 1, 26]. Supervised learning can be applied in other contexts as well, such as causal models in the form of Bayesian networks [23, 24, 25]. In particular, for each query on the causal model, one can compile an Arithmetic Circuit (AC) that maps evi-dence (inputs) to the posterior probability of interest (output) [9, 10]. AC parameters, which correspond to Bayesian network parameters, can then be learned from labeled data using gradient descent. Hence, like a neural network, the AC is a circuit that computes a function whose parameters can be learned from labeled data. The use of ACs in this fashion can be viewed as model-based su-pervised learning, in contrast to model-free supervised learning using neural networks. Model-based supervised learning is attractive since the AC can integrate the background knowledge embecded in its un-derlying causal model. This has a number of advantages, which in-clude a reduced reliance on data, improved robustness and the ability to provide data-independent guarantees on the learned function. One important type of background knowledge is functional dependencies between variables and their direct causes in a model (a special case of what is known as determinism ). Not only can this type of knowledge 1 Will appear in proceedings of the European Conference on Artificial Intel-ligence (ECAI), Spain 2020. 2 University of California, Los Angeles, email: darwiche@cs.ucla.edu significantly reduce the reliance on data, but it can also significantly improve the complexity of inference. In fact, substantial efforts have been dedicated to exploiting determinism in probabilistic inference, particularly the compilation of Bayesian networks into ACs [9, 3, 2], which is necessary for efficient inference on dense models. There are two main approaches for exploiting functional depen-dencies. The first is based on the classical algorithm of variable elim-ination (VE), which underlies algorithms for probabilistic inference including the jointree algorithm [29, 11, 18]. VE represents a model using factors, which are tables or multi-dimensional arrays. It then performs inference using a few and simple factor operations. Ex-ploiting functional dependencies within VE requires sparse factors ;see, e.g., [19, 21]. The second approach for exploiting functional de-pendencies reduces probabilistic inference to weighted model count-ing on a propositional formula that encodes the model, including its functional dependencies. It then compiles the formula into a circuit that is tractable for model counting; see, e.g., [8, 2]. This approach is in common use today given the efficacy of knowledge compilers. Our main contribution is a new approach for exploiting functional dependencies in VE that works with standard (dense) factors. This is significant for the following reason. We wish to map probabilistic inference, particularly the learning of parameters, into a tensor com-putation to exploit the vast progress on tensor-based technology and be on par with approaches that aggressively exploit this technology. Tensors are multi-dimensional arrays whose operations are heavily optimized and can be extremely fast, even on CPU-based platforms like modern laptops (let alone GPUs). A tensor computation takes the form of a tensor graph with nodes representing tensor opera-tions. Factors map directly to tensors and sparse factors to sparse tensors. However, sparse tensors have limited support in state of the art tools, which prohibits an implementation of VE using sparse ten-sors. 3 Knowledge compilation approaches produce circuits that cast into scalar tensor graphs, which are less effective than general ten-sor graphs as they are less amenable to parallelization. Moreover, while our approach needs to know that there is a functional depen-dency between variables it does not require the specific dependency (the specific numbers). Hence, it can be used to speed up inference even when the model parameters are unknown which can be critical when learning model parameters from data. Neither of the previous approaches can exploit this kind of abstract information. VE is based on two theorems that license factor operations. We add two new theorems that license more operations in the presence of functional dependences. This leads to a standard VE algorithm ex- 3 For example, in TensorFlow, a sparse tensor can only be multiplied by a dense tensor, which rules out the operation of (sparse) factor multiplication that is essential for sparse VE; see api_docs/python/tf/sparse/SparseTensor . arXiv:2002.09320v1 [cs.AI] 21 Feb 2020 cept with a significantly improved complexity and computation that maps directly to a tensor graph. We present experimental results for inference and learning that show promise of the proposed algorithm. We start in Section 2 by discussing factors, their operations and the VE algorithm including its underlying theorems. We also present our new VE theorems in this section. We then propose a new VE al-gorithm in Section 3 that exploits functional dependencies. We show how the proposed algorithm maps to tensor graphs and why this mat-ter in Section 4. We follow by case studies in Section 5 that illustrate the algorithm’s performance in the context of model-based super-vised learning. We finally close with some remarks in Section 6. 2 The Fundamentals: Factors & Operations The VE algorithm is based on applying operations to factors. A factor for discrete variables X is a function that maps each in-stantiation x of variables X into a number. The following are two factors over binary variables A, B, C and ternary variable D: ADf(AD )000.2010.3020.6100.9110.6120.1 ABCg(ABC )0001.00010.00100.00111.01000.21010.81100.51110.5 Factors can be represented as multi-dimensional arrays and are now commonly referred to as tensors (factor variables corresponds to ar-ray/tensor dimensions). One needs three factor operations to imple-ment the VE algorithm: multiplication, sum-out and normalization. The product of factors f (X) and g(Y) is another factor h(Z),where Z = X ∪ Y and h(z) = f (x)g(y) for the unique instan-tiations x and y that are compatible with instantiation z. Summing-out variables Y ⊆ X from factor f (X) yields another factor g(Z),where Z = X \ Y and g(z) = ∑ y f (yz ). We use ∑ Y f to denote the resulting factor g. We also use ∑= Z f which reads: sum out all variables from factor f except for variables Z. Normalizing factor f (X) yields another factor g(X) where g(x) = f (x)/ ∑ x f (x).We use ηf to denote the normalization of factor f .A Bayesian Network (BN) is specified by a directed acyclic graph (DAG) and a set of factors. In particular, for each node X and its par-ents U, we need a factor fX over variables XU. The value fX (xu) represents the conditional probability P (x\|u) and the factor fX is called a Conditional Probability Table (CPT). The joint distribution specified by a Bayesian network is simply the product of its CPTs. The Bayesian network in Figure 2 has five CPTs fA(A), fB (AB ), fC (AC ), fD (BCD ) and fE (CE ). The network joint distribution is the product of these factors Pr (ABCDE ) = fAfB fC fD fE .Evidence on variable X is captured by a factor λX (X) called an evidence indicator. Hard evidence fixes a value x giving λX (x) = 1 and λX (x?) = 0 for x? 6 = x. For soft evidence, λX (x) is the likelihood of x . The posterior distribution of a Bayesian network is the normalized product of its CPTs and evidence indicators. An expression constructed by applying operations to factors will be called an f-expression. Suppose we have evidence on variables A and E in Figure 2. The posterior on variable D is obtained by evaluating the following f-expression: P ?(D) = η ∑ ABCE λAλE fAfB fC fD fE . The VE algorithm factors f-expressions so they are evaluated more efficiently [29, 11] and is based on two theorems; see, e.g., [10, Chap-ter 6]. The first theorem says that the order in which variables are summed out does not matter. Theorem 1. ∑ XY f = ∑ X ∑ Y f = ∑ Y ∑ X f . The second theorem allows us to reduce the size of factors in-volved in a multiplication operation. Theorem 2. If variables X appear in factor f but not in factor g,then ∑ X f · g = g ∑ X f . Factor ∑ X f is exponentially smaller than factor f so Theorem 2 allows us to evaluate the f-expression ∑ X f ·g much more efficiently. Consider the f-expression ∑ ABDE f (ACE )f (BCD ). A direct evaluation multiplies the two factors to yield f (ABCDE ) then sums out variables ABDE . Using Theorem 1, we can arrange the expres-sion into ∑ AE ∑ BD f (ACE )f (BCD ) and using Theorem 2 into ∑ AE f (ACE ) ∑ BD f (BCD ), which is more efficient to evaluate. Using an appropriate order for summing out (eliminating) vari-ables, Theorems 1 and 2 allow one to compute the posterior on any variable in a Bayesian network in O(n exp( w)) time and space. Here, n is the number of network variables and w is the network treewidth (a graph-theoretic measure of the network connectivity). This works well for sparse networks that have a small treewidth, but is problematic for dense networks like the ones we will look at in Section 5. We present next two new results that allow us to sometimes significantly improve this computational complexity, by exploiting functional relationships between variables and their direct causes. While we will focus on exploiting functional dependencies in Bayesian networks, our results are more broadly applicable since the VE algorithm can be utilized in many other domains including sym-bolic reasoning and constraint processing . VE can also be used to contract tensor networks which have been receiving increased at-tention. A tensor network is a set of factors in which a variable appears in at most two factors. Contracting a tensor network is the problem of summing out all variables that appear in two factors; see, e.g., [14, 15]. The VE algorithm can also be used to evaluate Ein-stein summations which are in common use today and implemented in many tools including NumPy. 4 2.1 Functional CPTs Consider variable X that has parents U in a Bayesian network and let factor fX (XU) be its conditional probability table (CPT). 5 If f (xu) ∈ { 0, 1} for all instantiations x and u, the CPT is said to be functional as it specifies a function that maps parent instantiation u into the unique value x satisfying fX (xu) = 1 . The following CPTs are functional: XYfY(XY ) x0y00 x0y11 x1y01 x1y10 ABfB(AB ) a0b00 a0b11 a0b20 a1b00 a1b10 a1b21 The first specifies the function x0 7 → y1, x1 7 → y0. The second specifies the function a0 7 → b1, a1 7 → b2. Functional dependen-cies encode a common type of background knowledge (examples in Section 5). They are a special type of determinism, which generally refers to the presence of zero parameters in a CPT. A CPT that has zero parameters is not necessarily a functional CPT. 4 5Since ∑ xP(x\|u) = 1 the CPT satisfies ∑ xfX(xu) = 1 for every u. 2We will next present two results that empower the VE algorithm in the presence of functional CPTs. The results allow us to factor f-expressions beyond what is permitted by Theorems 1 and 2, leading to significant reduction in complexity. The results do not depend on the identity of a functional CPT, only that it is functional. This is significant when learning model parameters from data. To state these results, we will use F, G and H to denote sets of factors. Depending on the context, a set of factors F may be treated as one factor obtained by multiplying members of the set ∏ f∈F f .The first result says the following. If a functional CPT for variable X appears in both parts of a product, then variable X can be summed out from one part without changing the value of the product. Theorem 3. Consider a functional CPT f for variable X. If f ∈ G and f ∈ H , then G · H = G ∑ X H.Proof. Suppose CPT f is over variables XU. Let h(X) and g(Y) be the factors corresponding to H and G, respectively. Let Z = X ∪ Y and X? = X{ X}. Then variables XU must belong to X, Y and Z,and parents U must belong to X?. Let el = G·H and er = G ∑ X H.We want to show el(z) = er (z) for every instantiation z.Consider an instantiation z and let u, x?, x and y be the instantia-tions of U, X?, X and Y in z. Then el(z) = g(y)h(x) and er (z) = g(y) ∑ x h(xx?). Since CPT f is functional, f (xu) ∈ { 0, 1} for any x and there is a unique x, call it xu, such that f (xu) = 1 .If f (xu) = 0 , then h(xx?) = 0 since f ∈ H , leading to er(z) = g(y)∑ x h(xx?) = g(y)∑ xf(xu)=1 h(xx?) = g(y)h(xux?). If xu is the instantiation of X in z, then xux? = x and er (z) = g(y)h(x) = el(z). Otherwise, g(y) = 0 since f ∈ G , which leads to el(z) = er (z) = 0 . Hence, el(z) = er (z) for every instantiation z and we have G · H = G ∑ X H.Theorem 3 has a key corollary. If a functional CPT for variable X appears in both parts of a product, we can sum out variable X from the product by independently summing it out from each part. Corollary 1. Consider a functional CPT f for variable X. If f ∈ G and f ∈ H , then ∑ X G · H = (∑ X G) (∑ X H).Proof. ∑ X G · H = ∑ X (G ∑ X H) by Theorem 3, which equals (∑ X H) (∑ X G) by Theorem 2. Theorem 3 and Corollary 1 may appear unusable as they require multiple occurrences of a functional CPT whereas the factors of a Bayesian network contain a single (functional) CPT for each vari-able. This is where the second result comes in: duplicating a func-tional CPT in a product of factors does not change the product value. Theorem 4. For functional CPT f , if f ∈ G , then f · G = G.Proof. Let g(Z) be the product of factors in G and let h = f ·g. Sup-pose factor f is the CPT of variable X and parents U. Consider an in-stantiation z and suppose it includes instantiation xu. If f (xu) = 0 ,then g(z) = 0 since f ∈ G . Moreover, h(z) = f (xu)g(z) = 0 . If f (xu) = 1 , then h(z) = f (xu)g(z) = g(z). Hence, g(z) = h(z) for all instantiations z and we have G = f · G .Theorem 4 holds if f embeds any functional dependency that is implied by factors G instead of being a functional CPT in G but we do not pursue the applications of this generalization in this paper. To see how Theorems 3 and 4 interplay, consider the f-expression ∑ X f (XY )g(XZ )h(XW ). In the standard VE algorithm, one Figure 1 : An arithmetic circuit (AC) compiled from the Bayesian net-work A → B, A → C. The AC computes factor f (B), where ηf is the posterior on variable B given evidence on variables A and C.must multiply all three factors before summing out variable X,leading to a factor over four variables XY ZW . If factor f is a functional CPT for variable X, we can duplicate it by Theo-rem 4: f (XY )g(XZ )h(XW ) = f (XY )g(XZ )f (XY )h(XW ).Moreover, Corollary 1 gives ∑ X f (XY )g(XZ )f (XY )h(XW ) = ∑ X f (XY )g(XZ ) ∑ X f (XY )h(XW ), which avoids construct-ing a factor over four variables. We show in Section 3 how these the-orems enable efficient inference on models with very large treewidth. 3 Variable Elimination with Functional CPTs We now present our proposed VE algorithm. We first present a stan-dard VE algorithm based on jointrees and then extend it to exploit functional CPTs. Our algorithm will not compute probabil-ities, but will compile symbolic f-expressions whose factors contain symbolic parameters. A symbolic f-expression is compiled once and used thereafter to answer multiple queries. Moreover, its parameters can be learned from labeled data using gradient descent. We will show how to map symbolic f-expressions into tensor graphs in Sec-tion 4 and use these graphs for supervised learning in Section 5. Once the factors of a symbolic f-expression are unfolded, the result is an Arithmetic Circuits (ACs) [9, 4] as shown in Figure 1. In fact, the standard VE algorithm we present next is a refinement on the one proposed in for extracting ACs from jointrees. The next section introduces jointrees and some key concepts that we need for the standard and extended VE algorithms. 3.1 Jointrees Consider the Bayesian network in the middle of Figure 2 and its join-tree on the left of the figure. The jointree is simply a tree with factors attached to some of its nodes (the circles in Figure 2 are the jointree nodes). We use binary jointrees , in which each node has either one or three neighbors and where nodes with a single neighbor are called leaves. The two jointrees in Figure 2 are identical but arranged differently. The one on the left has leaf node 2 at the top and the one on the right has leaf node 3 at the top. Our use of jointrees deviates from normal for reasons that become apparent later. First, we use a binary jointree whose leaves are in one-to-one correspondence with model variables. Second, we only attach factors to leaf nodes: The CPT and evidence indicator for each variable X are assigned to the leaf node i corresponding to variable X. Leaf jointree node i is called the host of variable X in this case. 6 6For similar uses and a method for constructing such binary jointrees, see and [10, Chapter 8]. Contraction trees which were adopted later for con-tracting tensor networks correspond to binary jointrees. 3Figure 2 : A Bayesian network with a jointree (two views). The Bayesian network in Figure 2 has five variables. Its jointree also has five leaves, each of which hosts a network variable. For ex-ample, jointree node 2 at the top-left hosts variable D: the CPT and evidence indicator for variable D are assigned to this jointree node. A key notion underlying jointrees are edge separators which de-termine the space complexity of inference (the rectangles in Figure 2 are separators). The separator for edge (i, j ), denoted sep (i, j ), are model variables that appear in leaf nodes on both sides of the edge. For example, sep (6 , 7) = {B, C } as these are the model variables that appear in jointree leaves {0, 2} and {1, 3, 4}. A related notion is the cluster of jointree node i. If i is leaf, its cluster are the variables appearing at node i. Otherwise, it is the union of separators for edges (i, j ). Every factor constructed by VE is over the variables of some separator or cluster. The time complexity of VE is exponential in the size of clusters and linear in the number of nodes in a jointree. The size of largest cluster −1 is called the jointree width and can-not be lower than the Bayesian network treewidth; see [10, Chapter 9] for a detailed treatment of this subject. When the network contains variables with different cardinalities, the size of a cluster is better measured by the number of instantiations that its variables has. We therefore define the binary rank of a cluster as log2 of its instantia-tion count. The binary rank coincides with the number of variables in a cluster when all variables are binary. Our technique for exploiting functional dependencies will use Theorems 3 and 4 to shrink the size of clusters and separators sig-nificantly below jointree width, allowing us to handle networks with very large treewidth. The algorithm will basically reduce the maxi-mum binary rank of clusters and separators, which can exponentially reduce the size of factors constructed by VE during inference. 3.2 Compiling Symbolic f-expressions using VE Suppose we wish to compile an f-expression that computes the pos-terior on variable Q. We first identify the leaf jointree node h that hosts variable Q. We then arrange the jointree so host h is at the top as in Figure 2. Host h will then have a single child r which we call the jointree root. The tree rooted at node r is now a binary tree, with each node i having two children c1 and c2 and a parent p. On the left of Figure 2, root r=7 has two children c1=0 , c2=6 and parent p=2 .We refer to such a jointree arrangement as a jointree view. Jointree views simplify notation. For example, we can now write sep (i) to denote the separator between node i and its parent p instead of sep (i, p ). We will adopt this simpler notation from now on. We now compile an f-expression using the following equations: P ?(Q) = η = ∑ Q Fhf (r) (1) f (i) =  = ∑ sep (i) Fi i is leaf = ∑ sep (i) f (c1)f (c2) i has children c1, c 2 (2) Here, Fi is the product of factors assigned to leaf node i (CPT and evidence indicator for the model variable assigned to node i). For the jointree view in Figure 2 (left), applying these equations to variable Q=D, host h=2 and root r=7 yields the f-expression: P?(D) = η =∑ D F2 =∑ BC [ =∑ C F0][ =∑ BC [ =∑ AC F4][ =∑ AB [ =∑ AB F3][ =∑ A F1]]] . This expression results from applying Equation 1 to the host h=2 followed by applying Equation 2 to each edge in the jointree. Each sum in the expression corresponds to a separator and every product constructed by the expression will be over the variables of a cluster. Our compiled AC is simply the above f-expression. The value of the expression represents the circuit output. The evidence indicators in the expression represent the circuit inputs. Finally, the CPTs of the expression contain the circuit parameters (see the AC in Figure 1). We will now introduce new notation to explain Equations 1 and 2 as we need this understanding in the following section; see also [10, Chapter 7]. For node i in a jointree view, we use ^ Fi to denote the set of factors at or below node i. We also use _ Fi to denote the set of factors above node i. Consider node 6 on the left of Figure 2. Then ^ F6 contains the factors assigned to leaf nodes {1, 3, 4} and _ F6 contains the factors assigned to leaf nodes {0, 2}.For a jointree view with host h and root r, _ Fr ^ Fr contains all fac-tors in the jointree and _ Fr = Fh. Equation 1 computes η ∑= Q _ Fr ^ Fr ,while delegating the computation of product ^ Fr to Equation 2, which actually computes ∑= sep (r) ^ Fr by summing out all variables but for ones in sep (r). The equation uses the decomposition ^ Fi = ^ Fc1 ^ Fc2 to sum out variables more aggressively: f (i) = = ∑ sep (i) ^ Fi = = ∑ sep (i) ^ Fc1 ^ Fc2 (3) = = ∑ sep (i)  =∑ sep (c1) ^ Fc1  =∑ sep (c2) ^ Fc2  . The rule employed by Equation 2 is simple: sum out from product ^ Fi all variables except ones appearing in product _ Fi (Theorem 2). The only variables shared between factors ^ Fi and _ Fi are the ones in sep (i) so Equation 2 is exploiting Theorem 2 to the max. The earlier that variables are summed out, the smaller the factors we need to multiply and the smaller the f-expressions that VE compiles. 3.3 Exploiting Functional Dependencies We now present an algorithm that uses Theorems 3 and 4 to sum out variables earlier than is licensed by Theorems 1 and 2. Here, ‘earlier’ means lower in the jointree view which leads to smaller factors. 41: procedure SHRINK SEP (r, h )2: X ← variable assigned to host h 3: if X∈fvars (r)then 4: sep (r)-= {X} 5: end if 6: SUM (r)7: end procedure 8: procedure SUM (i)9: if leaf node ithen 10: return 11: end if 12: c1, c 2←children of node i 13: X←fvars (c1)∩fvars (c2) 14: c←either c1or c2 15: sep (c)-= X 16: sep (c1)&= sep (c2)∪sep (i) 17: sep (c2)&= sep (c1)∪sep (i) 18: SUM (c1)19: SUM (c2)20: end procedure Figure 3 : Left: Algorithm for shrinking separators based on func-tional CPTs. Right: An application of the algorithm where dropped variables are colored red. Variables B and C have functional CPTs. Our algorithm uses the notation fvars (i) to denote the set of variables that have a functional CPT at or below node i in the join-tree view. For example, in Figure 3, we have fvars (8) = {B, C }, fvars (11) = {B} and fvars (2) = {} .The algorithm is depicted in Figure 3 and is a direct application of Theorem 3 with a few subtleties. The algorithm traverses the join-tree view top-down, removing variables from the separators of visited nodes. It is called on root r and host h of the view, SHRINK SEP (r, h ). It first shrinks the separator of root r which decomposes the set of factors into _ Fr ^ Fr . The only functional CPT that can be shared be-tween factors _ Fr and ^ Fr is the one for variable X assigned to host h.If variable X is functional and its CPT is shared, Theorem 3 imme-diately gives _ Fr ^ Fr = _ Fr ∑ X ^ Fr . Variable X can then be summed at root r by dropping it from sep (r) as done on line 4. The algorithm then recurses on the children of root r. The algo-rithm processes both children c1 and c2 of a node before it recurses on these children. This is critical as we explain later. The set X com-puted on line 13 contains variables that have functional CPTs in both factors ^ Fc1 and factors ^ Fc2 (recall Equation 3). Theorem 3 allows us to sum out these variables from either ^ Fc1 or ^ Fc2 but not both, a choice that is made on line 14. A variable that has a functional CPT in both ^ Fc1 and ^ Fc2 is summed out from one of them by dropping it from either sep (c1) or sep (c2) on line 15. In our implementation, we heuristically choose a child based on the size of separators below it. We add the sizes of these separators (number of instantiations) and choose the child with the largest size breaking ties arbitrarily. If a variable is summed out at node i and at its child c2, we can sum it out earlier at child c1 by Theorem 2 (classical VE): ∑ X (^ Fc1 ∑ X ^ Fc2 ) = ( ∑ X ^ Fc1 )( ∑ X ^ Fc2 ). A symmetric situa-tion arrises for child c2. This is handled on lines 16-17. Applying Theorem 2 in this context demands that we process nodes c1 and c2 before we process their children. Otherwise, the reduction of sep-arators sep (c1) and sep (c2) will not propagate downwards early enough, missing opportunities for applying Theorem 2 further. Figure 3 depicts an example of applying algorithm SHRINK SEP to a jointree view for the Bayesian network in Figure 2. Variables colored red are dropped by SHRINK SEP . The algorithm starts by processing root r = 5 , dropping variable B from sep (5) on line 4. It then processes children c1 = 6 and c2 = 8 simultaneously. Since both children contain a functional CPT for variable C, the variable can be dropped from either sep (6) or sep (8) . Child c2 = 8 is cho-sen in this case and variable C is dropped from sep (8) . We have sep (6) = {A, C } and sep (8) = {A, B } at this point. Lines 16-17 shrink these separators further to sep (6) = {A} and sep (8) = {A}.Our proposed technique for shrinking separators will have an ef-fect only when functional CPTs have multiple occurrences in a join-tree (otherwise, set X on line 13 is always empty). While this devi-ates from the standard use of jointrees, replicating functional CPTs is licensed by Theorem 4. The (heuristic) approach we adopted for replicating functional CPTs in a jointree is based on replicating them in the Bayesian network. Suppose variable X has a functional CPT and children C1, . . . , C n in the network, where n > 1. We replace variable X with replicas X1, . . . , X n. Each replica Xi has a single child Ci and the same parents as X. We then construct a jointree for the resulting network and finally replace each replica Xi by X in the jointree. This creates n replicas of the functional CPT in the jointree. Replicating functional CPTs leads to jointrees with more nodes, but smaller separators and clusters as we shall see in Section 5. 4 Mapping ACs into Tensor Graphs We discuss next how we map ACs (symbolic f-expressions) into ten-sors graphs for efficient inference and learning. Our implementation is part of the P YTAC system under development by the author. P Y-TAC is built on top of TensorFlow and will be open sourced. A tensor is a data structure for a multi-dimensional array. The shape of a tensor defines the array dimensions. A tensor with shape (2 , 2, 3) has 2 × 2 × 3 elements or entries. The dimensions of a ten-sor are numbered and called axes. The number of axes is the tensor rank. Tensor computations can be organized into a tensor graph: adata flow graph with nodes representing tensor operations. Tensors form the basis of many machine learning tools today. A factor over variables X1, . . . , X n can be represented by a ten-sor with rank n and shape (d1, . . . , d n), where di is the cardinality of variable Xi (i.e., its number of values). Factor operations can then be implemented using tensor operations, leading to a few advantages. First, tensor operations are heavily optimized to take advantage of special instruction sets and architectures (on CPUs and GPUs) so they can be orders of magnitude faster than standard implementations of factor operations (even on laptops). Next, the elements of a tensor can be variables, allowing one to represent symbolic f-expressions, which is essential for mapping ACs into tensor graphs that can be trained. Finally, tools such as TensorFlow and PyTorch provide sup-port for computing the partial derivates of a tensor graph with respect to tensor elements, and come with effective gradient descent algo-rithms for optimizing tensor graphs (and hence ACs). This is very useful for training ACs from labeled data as we do in Section 5. To map ACs (symbolic f-expressions) into tensor graphs, we need to implement factor multiplication, summation and normalization. Mapping factor summation and normalization into tensor operations is straightforward: summation has a corresponding tensor operation (TF .REDUCE SUM ) and normalization can be implemented using ten-sor summation and division. Factor multiplication does not have a corresponding tensor operation and leads to some complications. 7 7Tensor multiplication is pointwise while factors are normally over different sets of variables. Hence, multiplying the tensors corresponding to factors f(ABC )and g(BDE )does not yield the expected result. The simplest option is to use TF .EINSUM , which can perform factor multiplication if we pass it the string “abc, bde – >abcde” ( 5We bypassed these complications in the process of achieving something more ambitious. Consider Equation 2 which contains al-most all multiplications performed by VE. Factors f1(c1), f2(c2) and the result f (i) are over separators sep (c1), sep (c2) and sep (i).This equation multiplies factors f1 and f2 to yield a factor over vari-ables sep (c1) ∪ sep (c2) and then shrinks it by summation into a fac-tor over variables sep (i). We wanted to avoid constructing the larger factor before shrinking it. That is, we wanted to multiply-then-sum in one shot as this can reduce the size of our tensor graphs significantly. 8 A key observation allows this using standard tensor operations. The previous separators are all connected to jointree node i so they satisfy the following property [10, Chapter 9]: If a variable appears in one separator, it also appears in at least one other separator. Variables sep (c1) ∪ sep (c2) ∪ sep (i) can then be partitioned as follows: 9 C: variables in f1, f 2 and f , sep (c1) ∩ sep (c2) ∩ sep (i) X: variables in f1, f but not f2, (sep (c1) ∩ sep (i)) \ sep (c2) Y: variables in f2, f but not f1, (sep (c2) ∩ sep (i)) \ sep (c1) S: variables in f1, f 2 but not f , (sep (c1) ∩ sep (c2)) \ sep (i) where variables S are the ones summed out by Equation 2. The vari-ables in each factor can now be structured as follows: f1(C, X, S), f2(C, Y, S) and f (C, X, Y). We actually group each set of vari-ables C, X, Y and S into a single compound variable so that factors f1, f 2 and f can each be represented by a rank-3 tensor. We then use the tensor operation for matrix multiplication TF .MATMUL to com-pute f = ∑ S f1f2 in one shot, without having to construct a tensor for the product f1f2. Matrix multiplication is perhaps one of the most optimized tensor operations on both CPUs and GPUs. Preparing tensors f1(C, X, S) and f2(C, Y, S) for matrix mul-tiplication requires two operations: TF .RESHAPE which aggregate variables into compound dimensions and TF .TRANSPOSE which or-der the resulting dimensions so TF .MATMUL can map f1 and f2 into f (C, X, Y). The common dimension C must appear first in f1 and f2. Moreover, the last two dimensions must be ordered as (X, S) and (S, Y) but TF .MATMUL can transpose the last two dimensions of an input tensor on the fly if needed. Using matrix multiplication in this fashion had a significant impact on reducing the size of tensor graphs and the efficiency of evaluating them, despite the added expense of using TF .TRANSPOSE and TF .RESHAPE operations (the latter opera-tion does not use space and is very efficient). PYTAC represents ACs using an abstract tensor graph called an ops graph, which can be mapped into a particular tensor implemen-tation depending on the used machine learning tool. P YTAC also has a dimension management utility, which associates each tensor with its structured dimensions while ensuring that all tensors are struc-tured appropriately so operations can be applied to them efficiently. We currently map an ops graph into a TF .GRAPH object, using the TF .FUNCTION utility introduced recently in TensorFlow 2.0.0. P Y-TAC also supports the recently introduced Testing Arithmetic Cir-cuits (TACs), which augment ACs with testing units that turns them into universal function approximators like neural networks [6, 5, 27]. org/api_docs/python/tf/einsum ). We found this too inefficient though for extensive use as it performs too many tensor transpositions. One can also use the technique of broadcasting by adding trivial di-mensions to align tensors ( broadcasting ), but broadcasting has limited support in TensorFlow re-quiring tensors with small enough ranks. 8See a discussion of this space issue in [10, Chapter 7]. 9In a jointree, every separator that is connected to a node is a subset of the union of other separators connected to that node. Hence, sep (i)⊆ sep (c1)∪sep (c2). 5 Case Studies We next evaluate the proposed VE algorithm on two classes of mod-els that have abundant functional dependencies. We also evaluate the algorithm on randomly generated Bayesian networks while varying the amount of functional dependencies. The binary jointrees con-structed for these models are very large and prohibit inference us-ing standard VE. We constructed these binary jointrees from variable elimination orders using the method proposed in ; see also [10, Chapter 9]. The elimination orders were obtained by the minfill heuristic; see, e.g., . 10 5.1 Rectangle Model We first consider a generative model for rectangles shown in Fig-ure 4. In an image of size n × n, a rectangle is defined by its upper left corner ( row , col ), height and width . Each of these variables has n values. The rectangle also has a binary label variable, which is either tall or wide. Each row has a binary variable row i indicating whether the rectangle will render in that row ( n variables total). Each column has a similar variable col j . We also have n2 binary variables which correspond to image pixels ( pixel ij ) indicating whether the pixel is on or off. This model can be used to predict rectangle at-tributes from noisy images such as those shown in Figure 4. We use the model to predict whether a rectangle is tall or wide by compiling an AC with variable label as output and variables pixel ij as input. The AC computes a distribution on label given a noisy image as evidence and can be trained from labeled data using cross entropy as the loss function. 11 Our focus is on the variables row i and col j which are determined by row /height and col /width , respectively (for example, row i is on iff row ≤ i < row + height ). In particular, we will investigate the impact of these functional relationships on the efficiency of our VE compilation algorithm and their impact on learning AC parameters from labeled data. Our experiments were run on a MacBook Pro, 2.2 GHz Intel Core i7, with 32 GB RAM. Table 1 depicts statistics on ACs that we compiled using our pro-posed VE algorithm. For each image size, we compiled an AC for predicting the rectangle label while exploiting functional CPTs to remove variables from separators during the compilation process. As shown in the table, exploiting functional CPTs has a dramatic im-pact on the complexity of VE. This is indicated by the size of largest jointree cluster (binary rank) in a classical jointree vs one whose sep-arators and clusters where shrunk due to functional dependencies. 12 Recall that a factor over a cluster will have a size exponential in the cluster binary rank (the same for factors over separators). The table also shows the size of compiled ACs, which is the sum of tensor sizes in the corresponding tensor graph (the tensor size is the number of elements/entries it has). For a baseline, the AC obtained by standard 10 The minfill heuristic and similar ones aim for jointrees that minimize the size of largest cluster (i.e., treewidth). It was observed recently that min-imizing the size of largest separator (called max rank ) is more desirable when using tensors since the memory requirements of Equation 2 can de-pend only on the size of separators not clusters (see for recent methods that optimize max rank). This observation holds even when using classi-cal implementations of the jointree algorithm and was exploited earlier to reduce the memory requirements of jointree inference; see, e.g., [22, 13]. 11 Arthur Choi suggested the use of rectangle models and Haiying Huang proposed this particular version of the model. 12 We applied standard node and value pruning to the Bayesian network be-fore computing a jointree and shrinking it. This has more effect on the digits model in Section 5.2. For example, it can infer that some pixels will never be turned on as they will never be occupied by any digit. 6Figure 4 : Left: A generative model for rectangles. Right: Examples of clean and noisy rectangle images. Figure 5 : Left: A generative model for seven-segment digits. Middle: Examples of noisy digit images. Right: Seven-segment digit. VE (without exploiting functional CPTs) for an image of size 20 ×20 is 18 , 032 , 742 , 365 , which is about 80 times larger than the size of AC reported in Table 1. What is particularly impressive is the time it takes to evaluate these ACs (compute their output from input). On average it takes about 7 milliseconds to evaluate an AC of size ten million for these models, which shows the promise tensor-based im-plementations (these experiments were run on a laptop). We next investigate the impact of integrating background knowl-edge when learning AC parameters. For training, we generated la-beled data for all clean images of rectangles and added n noisy im-ages for each (with the same label). Noise is generated by randomly flipping min (n, a − 1, b/ 2) background pixels, where a is the num-ber of rectangle pixels and b is the number of background pixels. We used the same process for testing data, except that we increased the number of noisy pixels to min (2 ∗ n, a − 1, b/ 2) and doubled the number of noisy images. We trained the AC using cross entropy as the loss function to minimize the classification accuracy. 13 Table 2 shows the accuracy of classifying rectangles (tall vs wide) on 10 × 10 images using ACs with and without background knowl-edge. ACs compiled from models with background knowledge have fewer parameters and therefore need less data to train. The training and testing examples were selected randomly from the datasets de-scribed above with 1000 examples always used for testing, regardless of the training data size. Each classification accuracy is the average over twenty five runs. The table clearly shows that integrating back-ground knowledge into the compiled AC yields higher classification accuracies given a fixed number of training examples. 5.2 Digits Model We next consider a generative model for seven-segment dig-its shown in Figure 5 ( Seven-segment_display ). The main goal of this model is to recognize digits in noisy images such as those shown in Figure 5. The model has four vertical and three horizontal segments. A digit is generated by activating some of the segments. For example, digit 13 Some of the CPTs contain zero parameters but are not functional, such as the ones for width and height . We fixed these zeros in the AC when learning with background knowledge. We also tied the parameters of the pixel ij variables therefore learning one CPT for all of them. Table 1 : Size and compile/evaluation time for ACs that compute the posterior on rectangle label. Reported times are in seconds. Evalua-tion time is the average of evaluating an AC over a batch of examples. Image Functional Network Max Cluster Size AC Eval Compile Size CPTs Nodes rank binary rank Size Time Time 8×8785 11 15 .0926 ,778 .001 4.9 3197 513 .0 10 ×10 7125 13 17 .63,518 ,848 .003 2.9 3305 514 .3 12 ×12 7173 15 20 .210 ,485 ,538 .007 4.1 3437 515 .3 14 ×14 7229 17 22 .626 ,412 ,192 .018 5.7 3593 516 .2 16 ×16 7293 19 25 .058 ,814 ,458 .034 7.4 3773 517 .0 20 ×20 7445 23 29 .6224 ,211 ,138 .140 14 .1 31205 518 .3 Table 2 : Classification accuracy on 10 × 10 noisy rectangle images. Testing data included 1000 examples in each case. Functional Accuracy Number of Training Examples Param CPTs 25 50 100 250 500 1000 Count fixed in AC mean 82 .64 89 .16 96 .08 97 .92 99 .51 98 .39 136 stdev 15 .06 11 .98 8.34 5.56 0.62 7.00 trainable mean 53 .29 56 .92 62 .20 74 .62 84 .94 88 .69 4,428 stdev 1.89 5.31 6.95 5.29 3.14 2.79 Table 3 : Size and compile/evaluation time for ACs that compute a posterior over digits. Reported times are in seconds. Evaluation time is the average of evaluating an AC over a batch of examples. Image Functional Network Max Cluster Size AC Eval Compile Size CPTs Nodes rank binary rank Size Time Time 8×87638 32 33 .3264 ,357 .008 9.3 31155 912 .6 10 ×10 7954 59 60 .82,241 ,205 .008 13 .6 32173 914 .1 12 ×12 71334 81 83 .811 ,625 ,558 .014 23 .2 33469 10 16 .7 14 ×14 71778 116 121 .032 ,057 ,227 .030 36 .8 35007 11 18 .4 16 ×16 72286 134 140 .095 ,094 ,167 .076 50 .4 36825 11 19 .3 Table 4 : Classification accuracy on 10 × 10 noisy digit images. Test-ing data included 1000 examples in each case. Functional Accuracy Number of Training Examples Param CPTs 25 50 100 250 500 1000 Count fixed in AC mean 83 .51 89 .17 94 .94 97 .68 98 .49 98 .44 275 stdev 8.70 6.02 4.57 1.45 0.91 0.27 trainable mean 9.82 12 .26 13 .28 22 .36 29 .51 35 .67 22 ,797 stdev 0.77 2.25 3.32 3.45 2.40 1.57 78 is generated by activating all segments and digit 1 by activating two vertical segments. Segments are represented by rectangles as in the previous section, so this model integrates seven rectangle mod-els. A digit has a location specified by the row and column of its upper-left corner (height is seven pixels and width is four pixels). Moreover, each segment has an activation node which is turned on or off depending on the digit. When this activation node is off, seg-ment pixels are also turned off. An image of size n × n has n2 pixels whose values are determined by the pixels generated by segments. This is a much more complex and larger model than the rectangle model and also has an abundance of functional dependencies. It is also much more challenging computationally. This can be seen by examining Tables 3, which reports the size of largest clusters in the jointrees for this model. For example, the model for 16 × 16 images has a cluster with a binary rank of 140 . This means that standard VE would have to construct a factor of size 2140 which is impossi-ble. Our proposed technique for exploiting functional dependencies makes this possible though as it reduces the binary rank of largest cluster down to 19 .3. And even though the corresponding AC has size of about one hundred million, it can be evaluated in about 76 milliseconds. The AC compilation times are also relatively modest. We trained the compiled ACs as we did in the previous section. We generated all clean images and added noise as follows. For each clean image we added 100 noisy images for training and 200 for testing by randomly flipping n background pixels where n is the image size. Table 4 parallels the one for the rectangle model. We trained two ACs, one that integrates background knowledge and one that does not. The former AC has fewer parameters and therefore requires less data to train. While this is expected, it is still interesting to see how little data one needs to get reasonable accuracies. In general, Tables 3 and 4 reveal the same patterns of the rectangle model: exploiting functional dependencies leads to a dramatic reduction in the AC size and integrating background knowledge into the compiled AC signif-icantly improves learnability. 5.3 Random Bayesian Networks Table 5 : Reduction in maximum cluster size due to exploiting func-tional dependencies. The number of values a node has was cho-sen randomly from (2 , 3) . We averaged over 10 random networks for each combination of network node count, maximal parent count and the percentage of nodes having functional CPTs. The parents of a node and their count where chosen randomly. Functional nodes where chosen randomly from non-root nodes. The binary rank of a cluster is log2 of the number of its instantiations. Network Maximal Percentage Binary Rank of Largest Cluster Node Parent Functional Original Jointree Shrunk Jointree Reduction Count Count Nodes % mean stdev mean stdev mean stdev 75 425 22 .42.819 .43.13.01.750 22 .52.216 .91.85.62.567 22 .93.913 .12.39.83.480 21 .92.711 .11.910 .83.2100 525 38 .74.533 .14.65.72.050 38 .12.923 .73.314 .44.367 38 .03.218 .93.119 .13.980 36 .83.013 .52.523 .33.1150 625 64 .35.454 .24.410 .14.250 64 .93.241 .95.623 .05.167 64 .36.028 .24.236 .04.780 66 .44.821 .34.645 .12.1 We next present two experiments on randomly generated Bayesian networks. The first experiment further evaluates our proposed algo-rithm for exploiting functional dependencies. The second experiment Table 6 : Comparing evaluation time of three AC representations: Tensor graph ( TenG ), scalar graph ( ScaG ) and scalar-batch graph (ScaBaG ). We averaged over 10 random Bayesian networks for each combination of batch size and limit on circuit size. AC size limit is in millions of nodes. The binary rank of a tensor is log2 of the number of its entries. Maximum binary rank is for the largest tensor in the tensor graph. Normalized time (tensor graph) is evaluation time per one million AC nodes (a node is a tensor entry). Each cell below con-tains the mean (top) and stdev (bottom). Times are in milliseconds. Batch Tensor Graph (TenG) Milliseconds Slow Down Factor Size Limit on Actual Max Binary TenG Time ScaG / TenG ScaBaG / TenG Size Size Rank Normalized Time Ratio Time Ratio 15-10 M6,992 ,414 19 .766 .611 .547 .01,830 ,909 0.720 .64.620 .315 -20 M17 ,979 ,799 21 .234 .322 .282 .11,391 ,918 0.53.07.123 .725 -30 M26 ,540 ,961 21 .620 .838 .4137 .31,154 ,660 0.54.614 .656 .135 -40 M37 ,058 ,914 21 .816 .050 .3177 .21,349 ,479 0.43.532 .1128 .710 5-10 M8,157 ,025 20 .07.8112 .338 .01,599 ,408 0.52.345 .924 .815 -20 M17 ,504 ,179 20 .74.6148 .054 .21,482 ,496 0.51.364 .933 .325 -30 M27 ,728 ,478 21 .74.5209 .760 .12,029 ,237 0.91.351 .017 .035 -40 M37 ,850 ,485 22 .14.0244 .070 .01,547 ,389 0.61.195 .426 .320 5-10 M6,506 ,125 19 .64.9135 .326 .9860 ,631 0.71.942 .211 .115 -20 M17 ,766 ,240 20 .73.1251 .539 .91,209 ,040 0.51.3123 .715 .225 -30 M27 ,762 ,672 21 .73.1271 .946 .01,148 ,761 0.51.192 .417 .735 -40 M37 ,620 ,063 22 .13.0287 .544 .31,416 ,214 0.31.2118 .919 .7 reinforces our motivation for working with dense representations of factors and the corresponding tensor-based implementations. 14 We generated Bayesian networks by starting with a linear order of nodes V1, . . . , V n and a maximum number of parents per node k. For each node Vi, we randomly determined a number of parents ≤ k and chose the parents randomly from the set V1, . . . , V i−1. We then selected a fixed percentage f of non-root nodes and gave them functional CPTs, where each node had cardinality two or three. In the first experiment, we considered networks with different number of nodes n, maximum number of parents k and percentage of functional nodes f . For each combination, we generated 10 net-works, computed a binary jointree and averaged the size of largest cluster. We then applied our algorithm for exploiting functional de-pendencies and obtained a jointree with shrunk clusters and separa-tors while also noting the size of largest cluster. Table 5 depicts our results, where we report the size of a largest cluster in terms of its binary rank: log2 of its instantiations count. As can be seen from Table 5, our algorithm leads to substantial re-ductions in binary rank, where the reduction increases as the frac-tion of functional nodes increases. Recall that our algorithm includes two heuristics: one for deciding how to replicate functional CPTs when building a jointree and another corresponding to the choice on Line 14 in Figure 3. Table 5 provides some evidence on the efficacy of these heuristics beyond the rectangle and digits case studies we discussed earlier. The second experiment compares classical and tensor-based im-plementations of ACs. In a classical implementation, the AC is rep-resented as a directed acyclic graph where root nodes correspond to scalars and other nodes correspond to scalar arithmetic operations; see Figure 1. We will call this the scalar graph representation. In a tensor-based implementation, the AC is represented using a ten-sor graph where root nodes correspond to tensors (i.e., factors) and 14 The experiments of this section were run on a server with dual Intel(R) Xeon E5-2670 CPUs running at 2.60GHz and 256GB RAM. 81: procedure EVALUATE SCALAR GRAPH (graph nodes) 2: for n in graph nodes: do 3: c1, c2 = n.child1, n.child2 4: if n.type == ’add’ then 5: n.value = c1.value + c2.value 6: else if n.type == ’mul’ then 7: n.value = c1.value c2.value 8: else if n.type == ’div’ then 9: n.value = c1.value / c2.value 10: end if 11: end for 12: end procedure Figure 6 : Evaluating a scalar graph representation of an AC. The graph nodes are topologically sorted so the children of a node are evaluated before the node is evaluated. The evaluation of a scalar-batch representation is similar except that node values are NumPy ndarrays and +, ∗, / are ndarray (tensor) operations. other nodes correspond to tensor operations (i.e., factor operations) as discussed in Section 4. The main benefit of a tensor-based imple-mentation is that tensor operations can be parallelized on CPUs and GPUs (for example, NumPy and tools such as TensorFlow leverage Single Instruction Multiple Data (SIMD) parallelism on CPUs). 15 Before we present the results of this experiment, we need to dis-cuss the notion of a batch which is a set of AC input vectors. When learning the parameters of an AC using gradient descent, the dataset or a subset of it can be viewed as a batch so we would be interested in evaluating the AC on a batch. A scalar graph would need to be evaluated on each input vector in a batch separately. However, when representing the AC as a tensor graph we can treat the batch as a ten-sor. This allows us to evaluate the AC on a batch to yield a batch of marginals, which creates more opportunities for parallelism. There is middle grounds though: a scalar graph with a batch that we shall call the scalar-batch graph. This is a tensor graph except that each tensor has two dimensions only: a batch dimension and a scalar dimension. For example, if the batch has size b, then a tensor will have shape (b, 1) . In a scalar-batch graph, each tensor is a set of scalars, one for each member of the batch (AC input vector). Scalar-batch graphs can be used in situations where a full tensor graph cannot be used. This includes situations where the AC is com-piled using techniques such as knowledge compilation, which pro-duce ACs that cannot be cast in terms of tensor operations. A scalar-batch graph can be used in this case to offer an opportunity for par-allelism, even if limited, especially when training the AC from data. Table 6 compares the three discussed AC representations in terms of their evaluation time, while varying the batch size and AC size. The tensor graph implementation is the one we discussed in Sec-tion 4 using TensorFlow. The scalar graph implementation uses a Python list to store the DAG nodes (parents before children) and then uses the pseudocode in Figure 6 to evaluate the DAG. We extract the DAG from the tensor graph where each DAG node corresponds to a tensor entry. The scalar-batch graph is represented similarly to the scalar graph except that members of the list are NumPy ndarrays of shape (b, 1) instead of scalars (we found NumPy to be more efficient than TensorFlow for this task). The evaluation time for both scalar graphs and scalar-batch graphs are therefore based on benchmarking the code in Figure 6 (we only measure the time of arithmetic opera-tions, excluding setting evidence on root nodes and other overhead). The networks in Table 6 were generated randomly as in the pre-vious experiment, with 100 nodes and a maximum of 5 parents per 15 node. For each limit on the AC size, we kept generating Bayesian networks randomly until we found 10 networks whose compilations yielded tensor graphs within the given size limit. The tensor graph normalized time in Table 6 is the total time to evaluate the graph di-vided by the batch size, then divided again by the size of the graph over 1000 , 000 . Normalized time is then the average time for eval-uating one million AC nodes (tensor entries) and is meant to give a sense of speed independent of the batch and AC size. We now have a number of observations on Table 6. The tensor graph is faster than the scalar and scalar-batch graphs in all cases and sometimes by two orders of magnitude. This can be seen in the last two columns of Table 6 which report the evaluation times (whole batch) of scalar and scalar-batch graphs over the evaluation time of tensor graph. The gap between tensor and scalar graphs increases with the batch size and with AC size as this means more opportu-nities to exploit parallelism on two fronts that the scalar graph can-not take advantage of. The gap between the tensor and scalar-batch graphs increases with AC size, but decreases with batch size. Increas-ing the AC size correlates with increasing the size of tensors (at least the largest one in the fourth column) which creates more opportuni-ties for exploiting parallelism that the scalar-batch graph cannot ex-ploit. However, increasing the batch size can be exploited by both the tensor and scalar-batch graphs, therefore narrowing the gap (NumPy appears to be exploiting the batch more effectively than TensorFlow). The scalar graph is faster than the scalar-batch graph when the batch size is 1, but otherwise is slower. This is to be expected as there is no need for the extra overhead of NumPy ndarrays in this case. We finally emphasize the absolute evaluation times for the tensor graph, which amount to a few milliseconds per one million AC nodes (nor-malized time) when the batch and AC size are large enough. 6 Conclusion We presented new results on the algorithm of variable elimination that exploit functional dependencies using dense factors, allowing one to benefit from tensor-based technologies for more efficient in-ference and learning. We also presented case studies that show the promise of proposed techniques. In contrast to earlier approaches, the proposed one does not dependent on the identity of functional de-pendencies, only that they are present. This has further applications to exact inference (exploiting inferred functional dependencies) and to approximate inference (treating CPTs with extreme probabilities as functional CPTs) which we plan to pursue in future work. ACKNOWLEDGEMENTS I wish to thank members of the Automated Reasoning Group at UCLA who provided valuable motivation and feedback: Arthur Choi, Yizou Chen, Haiying Huang and Jason Shen. This work has been partially supported by grants from NSF IIS-1910317, ONR N00014-18-1-2561 and DARPA N66001-17-2-4032. REFERENCES Yoshua Bengio, Pascal Lamblin, Dan Popovici, and Hugo Larochelle, ‘Greedy layer-wise training of deep networks’, in Advances in Neural Information Processing Systems 19 (NIPS) , pp. 153–160, (2006). Mark Chavira and Adnan Darwiche, ‘On probabilistic inference by weighted model counting’, Artificial Intelligence ,172 (6–7), 772–799, (April 2008). 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Uffe Kjrulff, ‘Triangulation of graphs – algorithms giving small total state space’, Technical report, (1990). David Larkin and Rina Dechter, ‘Bayesian inference in the presence of determinism’, in Proceedings of the Ninth International Workshop on Artificial Intelligence and Statistics (AISTATS) , (2003). Vasilica Lepar and Prakash P. Shenoy, ‘A comparison of lauritzen-spiegelhalter, hugin, and shenoy-shafer architectures for computing marginals of probability distributions’, in UAI , pp. 328–337. Morgan Kaufmann, (1998). Judea Pearl, Probabilistic Reasoning in Intelligent Systems: Networks of Plausible Inference , MK, 1988. Judea Pearl, Causality , Cambridge University Press, 2000. Judea Pearl and Dana Mackenzie, The Book of Why: The New Science of Cause and Effect , Basic Books, 2018. Marc’Aurelio Ranzato, Christopher S. 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9278	https://www.aaojournal.org/article/S0161-6420(12)00698-7/abstract	Interventions for Toxoplasma Retinochoroiditis - Ophthalmology Skip to Main ContentSkip to Main Menu ADVERTISEMENT SCROLL TO CONTINUE WITH CONTENT Open GPT Console Open Oracle Keywords Refresh Values \| Property \| Value \| --- \| \| Status \| \| \| Version \| \| \| Ad File \| \| \| Disable Ads Flag \| \| \| Environment \| \| \| Moat Init \| \| \| Moat Ready \| \| \| Contextual Ready \| \| \| Contextual URL \| \| \| Contextual Initial Segments \| \| \| Contextual Used Segments \| \| \| AdUnit \| \| \| SubAdUnit \| \| \| Custom Targeting \| \| \| Ad Events \| \| \| Invalid Ad Sizes \| \| Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? 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Ok Original article \| Ophthalmic technology assessmentVolume 120, Issue 2p371-378 February 2013 Download Full Issue Download started Ok Interventions for Toxoplasma Retinochoroiditis A Report by the American Academy of Ophthalmology Stephen J.Kim, MD Stephen J.Kim, MD Correspondence Correspondence: Nancy Collins, American Academy of Ophthalmology, Quality Care and Knowledge Base Development, PO Box 7424, San Francisco, CA 94120-7424 ncollins@aao.org Affiliations Department of Ophthalmology, Vanderbilt University School of Medicine, Nashville, Tennessee Search for articles by this author 1ncollins@aao.org ∙ Ingrid U.Scott, MD, MPH Ingrid U.Scott, MD, MPH Affiliations Departments of Ophthalmology and Public Health Sciences, Penn State College of Medicine, Hershey, Pennsylvania Search for articles by this author 2 ∙ Gary C.Brown, MD, MBA Gary C.Brown, MD, MBA Affiliations Center for Value Based Medicine, Flourtown, Pennsylvania Eye Research Institute, Philadelphia, Pennsylvania The Retina Service, Wills Eye Institute, Jefferson Medical College, Philadelphia, Pennsylvania Search for articles by this author 3,4,5 ∙ … ∙ Melissa M.Brown, MD, MBA Melissa M.Brown, MD, MBA Affiliations Center for Value Based Medicine, Flourtown, Pennsylvania Eye Research Institute, Philadelphia, Pennsylvania Research Department, Wills Eye Institute, Jefferson Medical College, Philadelphia, Pennsylvania Search for articles by this author 3,4,6 ∙ Allen C.Ho, MD Allen C.Ho, MD Affiliations Mid Atlantic Retina, Wills Eye Institute, Philadelphia, Pennsylvania Search for articles by this author 7 ∙ Michael S.Ip, MD Michael S.Ip, MD Affiliations University of Wisconsin Medical School, Madison, Wisconsin Search for articles by this author 8 ∙ Franco M.Recchia, MD Franco M.Recchia, MD Affiliations Tennessee Retina PC, Nashville, Tennessee Search for articles by this author 9 … Show more Show less Affiliations & Notes Article Info 1 Department of Ophthalmology, Vanderbilt University School of Medicine, Nashville, Tennessee 2 Departments of Ophthalmology and Public Health Sciences, Penn State College of Medicine, Hershey, Pennsylvania 3 Center for Value Based Medicine, Flourtown, Pennsylvania 4 Eye Research Institute, Philadelphia, Pennsylvania 5 The Retina Service, Wills Eye Institute, Jefferson Medical College, Philadelphia, Pennsylvania 6 Research Department, Wills Eye Institute, Jefferson Medical College, Philadelphia, Pennsylvania 7 Mid Atlantic Retina, Wills Eye Institute, Philadelphia, Pennsylvania 8 University of Wisconsin Medical School, Madison, Wisconsin 9 Tennessee Retina PC, Nashville, Tennessee Publication History: Received June 30, 2012; Revised July 4, 2012; Accepted July 4, 2012; Published online October 12, 2012 Footnotes: Manuscript no. 2012-972. Prepared by the Ophthalmic Technology Assessment Committee Retina/Vitreous Panel and approved by the Board of Trustees April 27, 2012. Financial Disclosure(s): The author(s) have no proprietary or commercial interest in any materials discussed in this article. Funded without commercial support by the American Academy of Ophthalmology. DOI: 10.1016/j.ophtha.2012.07.061 External LinkAlso available on ScienceDirect External Link Copyright: © 2013 American Academy of Ophthalmology. Download PDF Download PDF Outline Outline Abstract Background Description of the Intervention Resource Requirements Question for Assessment Description of Evidence Published Results Future Research References Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Download PDF Download PDF Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Abstract Background Description of the Intervention Resource Requirements Question for Assessment Description of Evidence Published Results Future Research References Article metrics Related Articles Abstract Objective To evaluate the available evidence in peer-reviewed publications about the outcomes and safety of interventions for toxoplasma retinochoroiditis (TRC). Methods Literature searches of the PubMed and the Cochrane Library databases were conducted last on July 20, 2011, with no date restrictions. The searches retrieved 275 unique citations, and 36 articles of possible clinical relevance were selected for full text review. Of these 36 articles, 11 were deemed sufficiently relevant or of interest, and they were rated according to strength of evidence. Results Eight of the 11 studies reviewed were randomized controlled studies, and none of them demonstrated that routine antibiotic or corticosteroid treatment of TRC favorably affects visual outcomes or reduces lesion size. There is level II evidence from 1 study suggesting that long-term treatment with combined trimethoprim and sulfamethoxazole prevented recurrent disease in patients with chronic relapsing TRC. Adverse effects of antibiotic treatment were reported in as many as 25% of patients. There was no evidence supporting the efficacy of other nonmedical treatments such as laser photocoagulation. Conclusions There is a lack of level I evidence to support the efficacy of routine antibiotic or corticosteroid treatment for acute TRC in immunocompetent patients. There is level II evidence suggesting that long-term prophylactic treatment may reduce recurrences in chronic relapsing TRC. Adverse effects of certain antibiotic regimens are frequent, and patients require regular monitoring and timely discontinuation of the antibiotic in some cases. Financial Disclosure(s) The author(s) have no proprietary or commercial interest in any materials discussed in this article. The American Academy of Ophthalmology prepares Ophthalmic Technology Assessments to evaluate new and existing procedures, drugs, and diagnostic and screening tests. The goal of an Ophthalmic Technology Assessment is to review systematically the available research for clinical efficacy, effectiveness, and safety. After review by members of the Ophthalmic Technology Assessment Committee, other Academy committees, relevant subspecialty societies, and legal counsel, assessments are submitted to the Academy's Board of Trustees for consideration as official Academy statements. The purpose of this assessment is to evaluate the outcomes and safety of interventions for toxoplasma retinochoroiditis (TRC). Background Toxoplasma gondii is an ubiquitous human parasite that is a leading infectious cause of posterior uveitis worldwide.1 1. Jabs, D.A. Ocular toxoplasmosis Int Ophthalmol Clin. 1990; 30:264-270 Crossref Scopus (19) PubMed Google Scholar T gondii is an obligate intracellular parasite that affects both humans and animals. Members of the Felidae (cat) family serve as definitive hosts. Serologic evidence of previous toxoplasma infection is present in approximately 16% of persons 12 to 49 years of age in the United States.2 2. Jones, J.L. ∙ Kruszon-Moran, D. ∙ Sanders-Lewis, K. ... Toxoplasma gondii infection in the United States, 1999 2004, decline from the prior decade Am J Trop Med Hyg. 2007; 77:405-410 Crossref Scopus (307) PubMed Google Scholar Toxoplasma retinochoroiditis is a potentially blinding necrotizing retinitis that may have a recurrent course. Acute episodes commonly cause conjunctival injection, photophobia, floaters, and variably decreased vision; they typically resolve in 6 to 8 weeks, leaving a healed retinochoroidal scar. Chronic complications include persistent vitreous opacities, epiretinal membrane, and cystoid macular edema and can result in visual impairment. Permanent vision loss may occur if lesions affect the macula or optic nerve head. The lifetime risk of TRC ranges considerably in the literature, depending on geographic location. In the United Kingdom, incidence of symptomatic TRC is estimated to occur in 18 of 100 000 natives (95% confidence interval, 11–25), but in as many as 382 of 100 000 people born in West Africa (95% confidence interval, 99–664).3 3. Gilbert, R.E. ∙ Dunn, D.T. ∙ Lightman, S. ... Incidence of symptomatic toxoplasma eye disease: aetiology and public health implications Epidemiol Infect. 1999; 123:283-289 Crossref Scopus (100) PubMed Google Scholar For many years, most cases of ocular toxoplasmosis were thought to be the result of reactivation of congenital infection.4 4. Dodds, E.M. Toxoplasmosis Curr Opin Ophthalmol. 2006; 17:557-561 Crossref Scopus (43) PubMed Google Scholar This belief was promoted in part by an extensive review published by Perkins in 1973,5 5. Perkins, E.S. Ocular toxoplasmosis Br J Ophthalmol. 1973; 57:1-17 Crossref Scopus (204) PubMed Google Scholar in which nearly all cases of TRC were believed to be of congenital origin. However, more recent lines of evidence indicate that most individuals with ocular toxoplasmosis may be infected with T gondii after birth.6 6. Holland, G.N. Ocular toxoplasmosis: a global reassessmentPart II: disease manifestations and management Am J Ophthalmol. 2004; 137:1-17 Full Text Full Text (PDF) Scopus (271) PubMed Google Scholar Friedmann and Knox7 7. Friedmann, C.T. ∙ Knox, D.L. Variations in recurrent active toxoplasmic retinochoroiditis Arch Ophthalmol. 1969; 81:481-493 Crossref Scopus (150) PubMed Google Scholar described 3 distinct morphologic forms of toxoplasma lesions: large destructive lesions, punctate inner lesions, and punctate outer lesions. Papillitis and neuroretinitis are well-known atypical presentations, and a severe necrotizing form can be seen in immunocompromised hosts. Lesions greater than 1 disc area may persist longer and have a higher rate of complications and vision loss than smaller lesions.7 7. Friedmann, C.T. ∙ Knox, D.L. Variations in recurrent active toxoplasmic retinochoroiditis Arch Ophthalmol. 1969; 81:481-493 Crossref Scopus (150) PubMed Google Scholar There also have been reports demonstrating a positive relationship between lesion size and duration of disease activity.8 8. Rothova, A. ∙ Meenken, C. ∙ Buitenhuis, H.J. ... Therapy for ocular toxoplasmosis Am J Ophthalmol. 1993; 115:517-523 Abstract Full Text (PDF) Scopus (146) PubMed Google Scholar Recent observations suggest that parasite proliferation in addition to inflammation is a major cause of tissue damage,6 6. Holland, G.N. Ocular toxoplasmosis: a global reassessmentPart II: disease manifestations and management Am J Ophthalmol. 2004; 137:1-17 Full Text Full Text (PDF) Scopus (271) PubMed Google Scholar and in a nonrandomized study, Rothova et al8 8. Rothova, A. ∙ Meenken, C. ∙ Buitenhuis, H.J. ... Therapy for ocular toxoplasmosis Am J Ophthalmol. 1993; 115:517-523 Abstract Full Text (PDF) Scopus (146) PubMed Google Scholar found a positive relationship between treatment with combined pyrimethamine and sulfadiazine and a reduction of final lesion size. In addition, animal studies consistently have demonstrated that antimicrobial drugs are highly effective in the treatment of active toxoplasmosis (usually measured as reduced mortality), and chronically active TRC in patients with AIDS has been reported to become rapidly inactive with antibiotic treatment.6,9 6. Holland, G.N. Ocular toxoplasmosis: a global reassessmentPart II: disease manifestations and management Am J Ophthalmol. 2004; 137:1-17 Full Text Full Text (PDF) Scopus (271) PubMed Google Scholar 9. Gagliuso, D.J. ∙ Teich, S.A. ∙ Friedman, A.H. ... Ocular toxoplasmosis in AIDS patients Trans Am Ophthalmol Soc. 1990; 88:63-86 discussion 86–8 PubMed Google Scholar For these reasons, there is an increasing trend among uveitis specialists to treat patients who have active TRC. However, there is a lack of consensus about the best treatment regimen. A survey published in 2002 revealed a total of 9 antibiotic drugs used in 24 different regimens as the treatment of choice among 79 responding uveitis specialists.10 10. Holland, G.N. ∙ Lewis, K.G. An update on current practices in the management of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:102-114 Full Text Full Text (PDF) Scopus (161) PubMed Google Scholar There is also controversy about initiation and timing of adjunctive therapy with corticosteroids. Treatment of patients with latent disease also is controversial because none of the currently available antibiotics has been shown to kill bradyzoites (the encysted form of T gondii found in tissue cysts) effectively in humans. Despite the focus of numerous publications and the common practice of treating active lesions, there continues to be underlying uncertainty about the efficacy of treatments for TRC in immunocompetent patients and large variations in practice patterns among clinicians throughout the world. The benefits of treatment also should be weighed against the cost of treatment and the risks of adverse effects of antibiotic and corticosteroid use. For example, sulfa antibiotics can cause life-threatening hypersensitivity reactions such as Stevens-Johnson syndrome. Description of the Intervention Interventions for TRC aim to reduce temporary or permanent vision loss by limiting the severity and duration of inflammation by reducing the size of the retinochoroidal scar and preventing recurrence. The most common treatment for acute TRC is systemic administration of 1 or more antibiotics, usually given for 4 to 8 weeks. A large number of antibiotics have been described in the literature, and most agents are effective only against the active tachyzoite form of toxoplasma, and not the tissue-encysted bradyzoite form. Newer antimicrobial agents, including atovaquone and azithromycin, reduced the number of tissue cysts in animal models but have not prevented recurrences after short-term therapy in humans.11,12 11. Huskinson-Mark, J. ∙ Araujo, F.G. ∙ Remington, J.S. Evaluation of the effect of drugs on the cyst form of Toxoplasma gondii J Infect Dis. 1991; 164:170-171 Crossref Scopus (139) PubMed Google Scholar 12. Pearson, P.A. ∙ Piracha, A.R. ∙ Sen, H.A. ... Atovaquone for the treatment of toxoplasma retinochoroiditis in immunocompetent patients Ophthalmology. 1999; 106:148-153 Full Text Full Text (PDF) Scopus (83) PubMed Google Scholar Recent publications also have reported intravitreal injection of clindamycin as an alternative to systemic treatment.13 13. Soheilian, M. ∙ Ramezani, A. ∙ Azimzadeh, A. ... Randomized trial of intravitreal clindamycin and dexamethasone versus pyrimethamine, sulfadiazine, and prednisolone in treatment of ocular toxoplasmosis Ophthalmology. 2011; 118:134-141 Full Text Full Text (PDF) Scopus (121) PubMed Google Scholar Some clinicians have added adjunctive corticosteroids to treat intraocular inflammation and its complications. Preventive strategies with systemic antibiotics for 1 year or more have been developed for infants who have congenital toxoplasmosis without retinochoroiditis, to reduce the risk of TRC, and similar strategies have been proposed for adults with recurrent TRC.14,15 14. Guerina, N.G. ∙ Hsu, H.W. ∙ Meissner, H.C. ..., New England Regional Toxoplasma Working Group Neonatal serologic screening and early treatment for congenital Toxoplasma gondii infection N Engl J Med. 1994; 330:1858-1863 Crossref Scopus (413) PubMed Google Scholar 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar There are also reports of laser photocoagulation to treat active lesions.16,17 16. Desmettre, T. ∙ Labalette, P. ∙ Fortier, B. ... Laser photocoagulation around the foci of toxoplasma retinochoroiditis: a descriptive statistical analysis of 35 patients with long-term follow-up Ophthalmologica. 1996; 210:90-94 Crossref Scopus (12) PubMed Google Scholar 17. Ghartey, K.N. ∙ Brockhurst, R.J. Photocoagulation of active toxoplasmic retinochoroiditis Am J Ophthalmol. 1980; 89:858-864 Abstract Full Text (PDF) PubMed Google Scholar Resource Requirements Most treatment regimens consist of a 4 to 8-week course of antibiotics. According to the literature, the most commonly used antibiotics are pyrimethamine, sulfadiazine, clindamycin, and combined trimethoprim and sulfamethoxazole. Sulfadiazine combined with pyrimethamine and corticosteroid is referred to as classic therapy, and it remains the most commonly used drug regimen.10 10. Holland, G.N. ∙ Lewis, K.G. An update on current practices in the management of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:102-114 Full Text Full Text (PDF) Scopus (161) PubMed Google Scholar The average wholesale cost ($0.58 per 25-mg tablet) of a 4-week supply of pyrimethamine (25 mg twice daily) is roughly $35.18 18. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 682 Google Scholar A well-known side effect of pyrimethamine is bone-marrow suppression, which requires regular monitoring of baseline and serial leukocyte and platelet counts (most commonly performed every 2 weeks while receiving therapy), and this risk can be reduced by coadministration of folinic acid (most commonly 5 mg 2 to 7 times weekly). The estimated cost of a 4-week supply of folinic acid is approximately $40, and the average cost of 3 laboratory blood cell count tests (approximately $80 per test) is $240. The average wholesale cost ($2.50 per 500-mg tablet) of a 4-week supply of sulfadiazine (most commonly 1 g 4 times daily) is roughly $600.19 19. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 593 Google Scholar The estimated cost of prednisone (60 mg daily for 30 days) is approximately $24. Therefore, the total cost of 4 weeks of treatment using classic therapy approaches $1000. Alternatively, the cost of a 4-week supply of clindamycin (most commonly 300 mg 4 times daily) is less than $300,20 20. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 527 Google Scholar and the cost of a 4-week supply of combined trimethoprim and sulfamethoxazole (most commonly 160 mg and 800 mg, respectively, twice daily) is less than $40.21 21. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 610 Google Scholar Question for Assessment The objective of this review is to address the following question: What are the outcomes and safety of interventions for TRC? The specific outcomes assessed are visual acuity, risk of 1 or more recurrences, and size of the lesion. Description of Evidence Literature searches were conducted last on July 20, 2011, in PubMed and the Cochrane Library databases, were limited to human studies, and had no date or language restrictions. The searches retrieved 275 unique citations. The search strategy in the PubMed database (July 15, 19, and 20, 2011) was as follows: 1 ((toxoplasmosis, ocular/drug therapy [MeSH terms]) OR (toxoplasmosis, ocular/prevention and control [MeSH Terms]) OR (toxoplasmosis, ocular/surgery [MeSH Terms]) OR (toxoplasmosis, ocular/therapy [MeSH Terms]) OR (chorioretinitis/drug therapy [MeSH Terms]) OR (chorioretinitis/prevention and control [MeSH Terms]) OR (chorioretinitis/surgery [MeSH Terms]) OR (chorioretinitis/therapy [MeSH Terms])) AND (toxoplasm[tiab]) Limits: Humans, Clinical Trial 2 ((toxoplasmosis, ocular/drug therapy [MeSH Terms]) OR (toxoplasmosis, ocular/prevention and control [MeSH Terms]) OR (toxoplasmosis, ocular/surgery [MeSH Terms]) OR (toxoplasmosis, ocular/therapy [MeSH Terms]) OR (chorioretinitis/drug therapy [MeSH Terms]) OR (chorioretinitis/prevention and control [MeSH Terms]) OR (chorioretinitis/surgery [MeSH Terms]) OR (chorioretinitis/therapy [MeSH Terms])) AND (toxoplasm[tiab]) AND ((cohort studies [MeSH Terms]) OR (case control studies [MeSH Terms])) 3 ((toxoplasmosis, ocular/drug therapy [MeSH Terms]) OR (toxoplasmosis, ocular/prevention and control [MeSH Terms]) OR (toxoplasmosis, ocular/surgery [MeSH Terms]) OR (toxoplasmosis, ocular/therapy [MeSH Terms]) OR (chorioretinitis/drug therapy [MeSH Terms]) OR (chorioretinitis/prevention and control [MeSH Terms]) OR (chorioretinitis/surgery [MeSH Terms]) OR (chorioretinitis/therapy [MeSH Terms])) AND (toxoplasm[tiab]) AND (series[tiab]) 4 ((toxoplasmosis, ocular/drug therapy [MeSH Terms]) OR (toxoplasmosis, ocular/prevention and control [MeSH Terms]) OR (toxoplasmosis, ocular/surgery [MeSH Terms]) OR (toxoplasmosis, ocular/therapy [MeSH Terms]) OR (chorioretinitis/drug therapy [MeSH Terms]) OR (chorioretinitis/prevention and control [MeSH Terms]) OR (chorioretinitis/surgery [MeSH Terms]) OR (chorioretinitis/therapy [MeSH Terms])) AND (toxoplasm[tiab]) 5 ((ocular toxoplasmosis[tiab]) OR (toxoplasma retinochoroiditis[tiab]) OR (toxoplasmic retinochoroiditis[tiab]) OR (toxoplasma chorioretinitis[tiab]) OR (toxoplasmic chorioretinitis[tiab]) OR (toxoplasmosis retinochoroiditis[tiab]))—867 references (7 references selected and imported) 6 ((toxoplasm[tiab]) AND ((ocular[tiab]) OR (retinochoroid[tiab]) OR (chorior[tiab]))) AND ((antimicrobial[tiab]) OR (corticosteroid[tiab]) OR (steroid[tiab]) OR (photocoagulat[tiab]) OR (coagulation[tiab]) OR (cryotherapy[tiab]) OR (surgery[tiab]) OR (surgical[tiab]) OR (medicat[tiab]) OR (intravitreal[tiab]) OR (oral[tiab]) OR (systemic[tiab]) OR (local[tiab]) OR (inject[tiab]) OR (treat[tiab]))—636 references (3 references selected and imported) 7 ((toxoplasmosis, ocular/drug therapy [MeSH Terms]) OR (toxoplasmosis, ocular/prevention and control [MeSH Terms]) OR (toxoplasmosis, ocular/surgery [MeSH Terms]) OR (toxoplasmosis, ocular/therapy [MeSH Terms]) OR (chorioretinitis/drug therapy [MeSH Terms]) OR (chorioretinitis/prevention and control [MeSH Terms]) OR (chorioretinitis/surgery [MeSH Terms]) OR (chorioretinitis/therapy [MeSH Terms])) AND (toxoplasm[tiab]) AND (review[tiab]) 8 ((toxoplasmosis, ocular/drug therapy [MeSH Terms]) OR (toxoplasmosis, ocular/prevention and control [MeSH Terms]) OR (toxoplasmosis, ocular/surgery [MeSH Terms]) OR (toxoplasmosis, ocular/therapy [MeSH Terms]) OR (chorioretinitis/drug therapy [MeSH Terms]) OR (chorioretinitis/prevention and control [MeSH Terms]) OR (chorioretinitis/surgery [MeSH Terms]) OR (chorioretinitis/therapy [MeSH Terms])) AND (toxoplasm[tiab]) 9 ((toxoplasmosis, ocular/drug therapy [MeSH Terms]) OR (toxoplasmosis, ocular/prevention and control [MeSH Terms]) OR (toxoplasmosis, ocular/surgery [MeSH Terms]) OR (toxoplasmosis, ocular/therapy [MeSH Terms]) OR (chorioretinitis/drug therapy [MeSH Terms]) OR (chorioretinitis/prevention and control [MeSH Terms]) OR (chorioretinitis/surgery [MeSH Terms]) OR (chorioretinitis/therapy [MeSH Terms])) AND (toxoplasm[tiab]) AND (survey[tiab]) The search strategy was as follows in the Cochrane Library (July 18, 2011): 1 MeSH descriptor Toxoplasmosis, Ocular explode all trees—11 references 2 MeSH descriptor Chorioretinitis explode all trees AND (toxoplasm):ti,ab,kw 3 (ocular toxoplasm):ti,ab,kw or (toxoplasm AND retinochoroiditis):ti,ab,kw or (toxoplasm AND chorioretinitis) The first author (S.J.K.) reviewed the abstracts of these articles and selected 35 of possible clinical relevance that then were obtained in full text (Fig 1). One additional article was identified from the reference list of the Cochrane Review.22 22. Gilbert, R.E. ∙ See, S.E. ∙ Jones, L.V. ... Antibiotics versus control for toxoplasma retinochoroiditis Cochrane Database Syst Rev. 2002; CD002218 Crossref Google Scholar Of these 36 articles, 11 were deemed sufficiently relevant or of interest and were reviewed by the panel methodologists. The methodologists used a rating scale based on that developed by the British Centre for Evidence-Based Medicine23 23. Oxford Centre for Evidence-Based Medicine Levels of evidence (March 2009) Accessed June 29, 2012 Google Scholar and assigned one of the following ratings of level of evidence to each of the selected articles. A level I rating was assigned to well-designed and well-conducted randomized clinical trials; a level II rating was assigned to well-designed case-control and cohort studies and poor-quality randomized studies; and a level III rating was assigned to case series, case reports, and poor-quality cohort and case-control studies. Figure viewer Figure 1 Diagram showing studies evaluated for inclusion in the interventions for toxoplasma retinochoroiditis Ophthalmic Technology Assessment. TRC = toxoplasma retinochoroiditis. Of the 36 studies reviewed, 8 of the 11 studies that were deemed sufficiently relevant were prospective randomized studies; 3 were randomized comparisons of antibiotic treatment versus placebo or no treatment15,24,25 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar 24. Acers, T.E. Toxoplasmic retinochoroiditis: a double blind therapeutic study Arch Ophthalmol. 1964; 71:58-62 Crossref Scopus (37) PubMed Google Scholar 25. Perkins, E.S. ∙ Schofield, P.B. ∙ Smith, C.H. Treatment of uveitis with pyrimethamine (Daraprim) Br J Ophthalmol. 1956; 40:577-586 Crossref Scopus (38) PubMed Google Scholar (Table 1) and 5 were randomized comparisons of different antibiotic regimens13,26–29 13. Soheilian, M. ∙ Ramezani, A. ∙ Azimzadeh, A. ... Randomized trial of intravitreal clindamycin and dexamethasone versus pyrimethamine, sulfadiazine, and prednisolone in treatment of ocular toxoplasmosis Ophthalmology. 2011; 118:134-141 Full Text Full Text (PDF) Scopus (121) PubMed Google Scholar 26. Bosch-Driessen, L.H. ∙ Verbraak, F.D. ∙ Suttorp-Schulten, M.S. ... A prospective, randomized trial of pyrimethamine and azithromycin vs pyrimethamine and sulfadiazine for the treatment of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:34-40 Full Text Full Text (PDF) Scopus (146) PubMed Google Scholar 27. Colin, J. ∙ Harie, J.C. Presumed toxoplasmic chorioretinitis: comparative study of treatment with pyrimethamine and sulfadiazine or clindamycin [in French] J Fr Ophtalmol. 1989; 12:161-165 Crossref Scopus (2) PubMed Google Scholar 28. Raskin, E. ∙ Alves, M. ∙ Eredia, G.C. ... Ocular toxoplasmosis: a comparative study of the treatment with sulfadiazine and pyrimethamine versus sulphametoxazole-trimethoprim [in Portuguese] Rev Bras Oftalmol. 2002; 61:335-338 Google Scholar 29. Soheilian, M. ∙ Sadoughi, M.M. ∙ Ghajarnia, M. ... Prospective randomized trial of trimethoprim/sulfamethoxazole versus pyrimethamine and sulfadiazine in the treatment of ocular toxoplasmosis Ophthalmology. 2005; 112:1876-1882 Full Text Full Text (PDF) Scopus (140) PubMed Google Scholar (Table 2). \| Study \| No. of Patients \| Visual Acuity Outcome \| Recurrent Lesions \| Change in Lesion Size \| Adverse Effects \| --- --- --- \| \| Perkins et al25 25. Perkins, E.S. ∙ Schofield, P.B. ∙ Smith, C.H. Treatment of uveitis with pyrimethamine (Daraprim) Br J Ophthalmol. 1956; 40:577-586 Crossref Scopus (38) PubMed Google Scholar \| 98(positive toxoplasma reaction) \| N/A \| N/A \| N/A \| Pyrimethamine was associated with drop in hemoglobin of 5% or more in 47% of patients \| \| Acers24 24. Acers, T.E. Toxoplasmic retinochoroiditis: a double blind therapeutic study Arch Ophthalmol. 1964; 71:58-62 Crossref Scopus (37) PubMed Google Scholar \| 20 \| N/A \| No difference at 2 yrs \| N/A \| 3 of 10 patients receiving pyrimethamine/trisulfapyrimidine experienced adverse reactions; 1 patient experienced severe thrombocytopenia \| \| Silveira et al15 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar \| 124⁎ \| N/A \| At 14 mos recurrence was 6.6% in treated patients and 24% in untreated patients \| N/A \| Trimethoprim/sulfamethoxazole discontinued in 4 of 61 patients (7%) because of mild drug reactions \| Table 1 Prospective Randomized Studies Comparing Antibiotic Treatment versus Placebo or No Treatment N/A = not available. ⁎ Patients with chronic infection were treated prophylactically to decrease recurrences. Open table in a new tab \| Study \| No. of Patients \| Comparison \| --- \| Only abstract in English \| \| \| \| Colin and Harie27 27. Colin, J. ∙ Harie, J.C. Presumed toxoplasmic chorioretinitis: comparative study of treatment with pyrimethamine and sulfadiazine or clindamycin [in French] J Fr Ophtalmol. 1989; 12:161-165 Crossref Scopus (2) PubMed Google Scholar \| 29 \| Pyrimethamine/sulfadiazine vs. subconjunctival injections of clindamycin \| \| Raskin et al28 28. Raskin, E. ∙ Alves, M. ∙ Eredia, G.C. ... Ocular toxoplasmosis: a comparative study of the treatment with sulfadiazine and pyrimethamine versus sulphametoxazole-trimethoprim [in Portuguese] Rev Bras Oftalmol. 2002; 61:335-338 Google Scholar \| 49 \| Sulfadiazine/pyrimethamine/corticosteroid vs. trimethoprim/sulfamethoxazole/corticosteroid \| \| Study published in English \| \| \| \| Bosch-Driessen et al26 26. Bosch-Driessen, L.H. ∙ Verbraak, F.D. ∙ Suttorp-Schulten, M.S. ... A prospective, randomized trial of pyrimethamine and azithromycin vs pyrimethamine and sulfadiazine for the treatment of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:34-40 Full Text Full Text (PDF) Scopus (146) PubMed Google Scholar \| 46 \| Pyrimethamine/azithromycin/corticosteroid vs. pyrimethamine/sulfadiazine/corticosteroid \| \| Soheilian et al29 29. Soheilian, M. ∙ Sadoughi, M.M. ∙ Ghajarnia, M. ... Prospective randomized trial of trimethoprim/sulfamethoxazole versus pyrimethamine and sulfadiazine in the treatment of ocular toxoplasmosis Ophthalmology. 2005; 112:1876-1882 Full Text Full Text (PDF) Scopus (140) PubMed Google Scholar \| 59 \| Pyrimethamine/sulfadiazine/prednisolone vs. trimethoprim-sulfamethoxazole/prednisone \| \| Soheilian et al13 13. Soheilian, M. ∙ Ramezani, A. ∙ Azimzadeh, A. ... Randomized trial of intravitreal clindamycin and dexamethasone versus pyrimethamine, sulfadiazine, and prednisolone in treatment of ocular toxoplasmosis Ophthalmology. 2011; 118:134-141 Full Text Full Text (PDF) Scopus (121) PubMed Google Scholar \| 68 \| Intravitreal clindamycin/dexamethasone vs. pyrimethamine/sulfadiazine/prednisolone \| Table 2 Prospective Randomized Studies Comparing Different Antibiotic Regimens Open table in a new tab All 3 placebo or no-treatment controlled studies were rated as level II evidence because of methodologic limitations, which greatly affected interpretation of the study results. In the study by Perkins et al,25 25. Perkins, E.S. ∙ Schofield, P.B. ∙ Smith, C.H. Treatment of uveitis with pyrimethamine (Daraprim) Br J Ophthalmol. 1956; 40:577-586 Crossref Scopus (38) PubMed Google Scholar there was uncertainty about the underlying diagnosis of TRC in many of the patients in addition to a lack of rigorous and objective assessment of clinical response. The study by Acers24 24. Acers, T.E. Toxoplasmic retinochoroiditis: a double blind therapeutic study Arch Ophthalmol. 1964; 71:58-62 Crossref Scopus (37) PubMed Google Scholar similarly had a lack of rigorous assessment of outcomes in addition to a small sample size. The study by Silveira et al15 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar was unmasked and had a lack of uniform documentation of baseline and recurrent disease by fundus photography. In addition, the study was conducted in Brazil, where the strain of T gondii is thought to be more virulent and may cause more frequent recurrences and severe lesions when compared with strains seen in North America or Europe.30 30. Gilbert, R.E. ∙ Freeman, K. ∙ Lago, E.G. ... European Multicentre Study on Congenital Toxoplasmosis (EMSCOT)Ocular sequelae of congenital toxoplasmosis in Brazil compared with Europe PLoS Negl Trop Dis [serial online]. 2008; 2:e277 Accessed June 29, 2012 Crossref Scopus (230) PubMed Google Scholar Therefore, the results of the study may not be applicable directly to other populations. Most importantly, none of the above studies provided results of visual acuity or changes in lesion size, which were primary outcome measures of this assessment. Evidence quality of the 5 clinical trials of different antibiotic regimens was limited by relatively small sample sizes and the absence of double masking. These studies were rated as level II evidence. The remaining 3 studies included a nonrandomized prospective trial8 8. Rothova, A. ∙ Meenken, C. ∙ Buitenhuis, H.J. ... Therapy for ocular toxoplasmosis Am J Ophthalmol. 1993; 115:517-523 Abstract Full Text (PDF) Scopus (146) PubMed Google Scholar and 2 noncomparative case series16,31 16. Desmettre, T. ∙ Labalette, P. ∙ Fortier, B. ... Laser photocoagulation around the foci of toxoplasma retinochoroiditis: a descriptive statistical analysis of 35 patients with long-term follow-up Ophthalmologica. 1996; 210:90-94 Crossref Scopus (12) PubMed Google Scholar 31. Lam, S. ∙ Tessler, H.H. Quadruple therapy for ocular toxoplasmosis Can J Ophthalmol. 1993; 28:58-61 PubMed Google Scholar and were all rated as level III evidence. Published Results Randomized Studies with Placebo or No Treatment Control Group Perkins et al25 25. Perkins, E.S. ∙ Schofield, P.B. ∙ Smith, C.H. Treatment of uveitis with pyrimethamine (Daraprim) Br J Ophthalmol. 1956; 40:577-586 Crossref Scopus (38) PubMed Google Scholar reported the effects of 4 weeks of treatment with pyrimethamine compared with inert tablets in patients with acute uveitis from any cause. Results for subgroups of patients with anterior uveitis, posterior uveitis, and panuveitis who showed positive or negative results for toxoplasma antibodies were randomized to either pyrimethamine or placebo. The study was double-masked, but masking may have been compromised by a higher percentage of hemoglobin levels that were reduced more than 5% in patients taking pyrimethamine. The primary outcome was defined vaguely and was categorized as improved or not improved after assessment of the clinical condition, without attempt to quantify the degree of improvement. No information was provided on visual acuity, lesion size, rates of recurrence, or loss to follow-up. The study demonstrated a statistically significant improvement in cases of uveitis having positive toxoplasma antibodies treated with pyrimethamine, but the authors estimated that only 25% of these cases may have been the result of active TRC (level II evidence). It was unclear whether testing for other causes of uveitis such as syphilis or tuberculosis was performed. Pyrimethamine treatment resulted in a drop in hemoglobin of 5% or more in 47% of patients, but only 1 patient had to discontinue treatment early because of anemia. Acers24 24. Acers, T.E. Toxoplasmic retinochoroiditis: a double blind therapeutic study Arch Ophthalmol. 1964; 71:58-62 Crossref Scopus (37) PubMed Google Scholar compared the effect of 8 weeks of combined pyrimethamine and trisulfapyrimidine and corticosteroid versus lactose capsules and corticosteroid in patients with acute TRC. The diagnosis of TRC was made in the setting of active retinitis, positive intradermal toxoplasmin skin test results, positive Sabin methylene blue-dye test results, and lack of significant clinical or laboratory evidence of other causes. Inactivity was determined clinically by the absence of anterior chamber reaction, clearing of the vitreous, and resolution of the retinal inflammation. The study was double-masked, and a total of 20 patients (10 in each group) were randomized to treatment or placebo. Regardless of therapy, all patients showed progressive improvement by 8 weeks (level II evidence). Three of the 10 patients treated with combined pyrimethamine and trisulfapyrimidine experienced side effects, and 1 of these patients experienced severe thrombocytopenia requiring early cessation of treatment. Over a 2-year period, there were 2 recurrences, 1 in each group. In a prospective, randomized, open-labeled study, Silveira et al15 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar determined the long-term effect of prophylactic combined trimethoprim and sulfamethoxazole treatment compared with no treatment on rates of recurrence. A total of 124 patients with a history of recurrent TRC were randomized to treatment with combined trimethoprim (160 mg) and sulfamethoxazole (800 mg), 1 tablet every 3 days (61 patients), or to observation without treatment (63 patients) and were followed up monthly for up to 20 months for clinical signs of recurrence. Serologic testing confirmed the presence of anti−T gondii immunoglobulin G antibodies in all patients before enrollment. Medications were administered in an unmasked fashion, and patients in the control group received no treatment. The primary end point of the study was development of recurrent TRC defined clinically as a new focus of retinal inflammation either adjacent to or remote from a pre-existing retinochoroidal scar. All patients were examined monthly by 1 ophthalmologist who was unmasked. Six patients (10%) in the treatment group and 4 patients (6%) in the control group were lost to follow-up. In the treatment and control groups, recurrent TRC developed in 4 patients (7%) and 15 (24%) patients, respectively (level II evidence). Compliance with treatment was determined by patient interviews and monitoring of dispensed tablets to patients in the treatment group. Treatment was discontinued prematurely in 4 patients because of mild drug reactions. Randomized Antibiotic Comparison Studies For 2 studies of the 5 randomized comparisons of different antibiotics, only the abstracts were available in English for review. The study by Colin and Harie27 27. Colin, J. ∙ Harie, J.C. Presumed toxoplasmic chorioretinitis: comparative study of treatment with pyrimethamine and sulfadiazine or clindamycin [in French] J Fr Ophtalmol. 1989; 12:161-165 Crossref Scopus (2) PubMed Google Scholar was conducted in 29 patients and compared the efficacy of combined pyrimethamine and sulfadiazine given with subconjunctival injections of clindamycin. After 14 months of follow-up, there was no difference between groups in terms of visual acuity or recurrence rates (21% in the clindamycin group and 36% in the combined pyrimethamine and sulfadiazine group). No information was provided in the abstract on masking, change in lesion size, loss to follow-up, or whether corticosteroids were used. Raskin et al28 28. Raskin, E. ∙ Alves, M. ∙ Eredia, G.C. ... Ocular toxoplasmosis: a comparative study of the treatment with sulfadiazine and pyrimethamine versus sulphametoxazole-trimethoprim [in Portuguese] Rev Bras Oftalmol. 2002; 61:335-338 Google Scholar evaluated the efficacy of combined pyrimethamine and sulfadiazine versus combined trimethoprim and sulfamethoxazole in 49 patients. All patients received adjuvant therapy with oral corticosteroids. The primary outcome was time to resolution of active retinochoroiditis. Faster resolution was observed with combined pyrimethamine and sulfadiazine (28 days) compared with combined trimethoprim and sulfamethoxazole (35 days) treatment. No details were given on how resolution of retinochoroiditis was determined. Similarly, no information was provided on visual acuity, lesion size, or rates of recurrence. The 3 remaining antibiotic comparison studies were available in English for complete review. Bosch-Driessen et al26 26. Bosch-Driessen, L.H. ∙ Verbraak, F.D. ∙ Suttorp-Schulten, M.S. ... A prospective, randomized trial of pyrimethamine and azithromycin vs pyrimethamine and sulfadiazine for the treatment of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:34-40 Full Text Full Text (PDF) Scopus (146) PubMed Google Scholar compared the time to resolution of intraocular inflammation, lesion size, and visual acuity before and after treatment between treatment with combined pyrimethamine and azithromycin (24 patients) and with combined pyrimethamine and sulfadiazine (22 patients). The results of the study demonstrated no significant differences between treatment groups for the primary outcomes, but adverse effects were more frequent in the combined pyrimethamine and sulfadiazine group (level II evidence). Soheilian et al29 29. Soheilian, M. ∙ Sadoughi, M.M. ∙ Ghajarnia, M. ... Prospective randomized trial of trimethoprim/sulfamethoxazole versus pyrimethamine and sulfadiazine in the treatment of ocular toxoplasmosis Ophthalmology. 2005; 112:1876-1882 Full Text Full Text (PDF) Scopus (140) PubMed Google Scholar compared change in lesion size, visual acuity, and rate of recurrence between treatment with combined pyrimethamine and sulfadiazine (29 patients) and with combined trimethoprim and sulfamethoxazole (30 patients). The results of the study demonstrated no significant differences between the treatment groups with respect to the primary outcome and similar rates of adverse events (level II evidence). Another study by Soheilian et al13 13. Soheilian, M. ∙ Ramezani, A. ∙ Azimzadeh, A. ... Randomized trial of intravitreal clindamycin and dexamethasone versus pyrimethamine, sulfadiazine, and prednisolone in treatment of ocular toxoplasmosis Ophthalmology. 2011; 118:134-141 Full Text Full Text (PDF) Scopus (121) PubMed Google Scholar reported on the efficacy of intravitreal injection of clindamycin and dexamethasone compared with classic systemic therapy using combined pyrimethamine, sulfadiazine, and prednisone. A total of 68 patients were enrolled, and 34 were randomized to each treatment group. The primary outcome measure was change in retinochoroidal lesion size 6 weeks after initiation of treatment. The study reported no difference in lesion size or visual acuity between treatment groups, but patients with immunoglobulin M−positive antitoxoplasma serum antibodies responded better to systemic treatment with respect to the primary outcome (level II evidence). There were no major adverse reactions in the intravitreal clindamycin group, but 2 adverse reactions occurred in the systemic treatment group. Interpretation of the antibiotic comparison studies above is based on the assumption that treatment with combined pyrimethamine and sulfadiazine has a beneficial effect for all patients with TRC, which remains unproven. Nonrandomized Prospective Study In a nonrandomized study of 149 patients, Rothova et al8 8. Rothova, A. ∙ Meenken, C. ∙ Buitenhuis, H.J. ... Therapy for ocular toxoplasmosis Am J Ophthalmol. 1993; 115:517-523 Abstract Full Text (PDF) Scopus (146) PubMed Google Scholar reported a positive relationship between treatment with combined pyrimethamine and sulfadiazine and reduction of lesion size (level III evidence). However, no difference was observed in the duration of inflammation between treated and untreated patients, and the most important factor predicting the duration of inflammation was the size of the initial retinal lesion, independent of treatment. There was also a high frequency of adverse effects associated with treatment that resulted in discontinuation in 26% of patients receiving combined pyrimethamine, sulfadiazine, and corticosteroid; 17% of patients receiving combined clindamycin, sulfadiazine, and corticosteroid; and 4% of patients receiving combined trimethoprim, sulfamethoxazole, and corticosteroid. No significant difference in the recurrence rate was observed between treated and untreated patients. Other Studies of Interest A noncomparative case series reported on quadruple therapy consisting of 3 different antibiotics and corticosteroid for the treatment of TRC (level III evidence).31 31. Lam, S. ∙ Tessler, H.H. Quadruple therapy for ocular toxoplasmosis Can J Ophthalmol. 1993; 28:58-61 PubMed Google Scholar A total of 37 eyes of 36 patients received combined treatment with pyrimethamine, trisulfapyrimidine, clindamycin, and prednisone. An improvement in vision was observed in 20 eyes (54%) within 2 weeks and in 30 eyes (81%) within 3 weeks. Four patients demonstrated a skin rash presumed to be secondary to trisulfapyrimidine therapy. Desmettre et al16 16. Desmettre, T. ∙ Labalette, P. ∙ Fortier, B. ... Laser photocoagulation around the foci of toxoplasma retinochoroiditis: a descriptive statistical analysis of 35 patients with long-term follow-up Ophthalmologica. 1996; 210:90-94 Crossref Scopus (12) PubMed Google Scholar reported on recurrence rates of 35 patients with TRC whose lesions were treated with laser photocoagulation (level III evidence). Recurrence rates occurred in 13% of eyes at 1 year, 20% at 2 years, and 33% at 4 years, and they increased steadily thereafter with longer follow-up. The authors were unable to observe a preventive effect of laser photocoagulation on recurrence rates of TRC. In conclusion, despite the common practice of treating TRC with systemic antibiotics, there are no randomized controlled trials demonstrating that antibiotic treatment improves long-term visual outcomes. In addition, there is only 1 study15 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar that provides level II evidence that prophylactic treatment with combined trimethoprim and sulfamethoxazole reduces the rate of recurrence in Brazilian patients with a history of recurrent TRC. The conclusions of this particular study should be interpreted cautiously, however, because the study investigators and subjects were unmasked and other strains of T gondii may respond differently to treatment. Equally important, there is no convincing evidence to date that treatment decreases the severity of intraocular inflammation or duration of disease for all patients. A Cochrane review of treatment of TRC with antibiotics reported similar conclusions about the lack of evidence for treatment of acute TRC and weak evidence for recurrent disease.22 22. Gilbert, R.E. ∙ See, S.E. ∙ Jones, L.V. ... Antibiotics versus control for toxoplasma retinochoroiditis Cochrane Database Syst Rev. 2002; CD002218 Crossref Google Scholar The lack of conclusive clinical evidence of the effectiveness of antibiotics is in contrast to data from animal studies that demonstrate that antimicrobial drugs are highly effective for treatment of active toxoplasmosis. Although a treatment effect has been difficult to find in immunocompetent hosts, the benefit of treatment and prevention of toxoplasmosis infection has been shown more convincingly in immunosuppressed patients with AIDS. Furthermore, studies have suggested that long-term treatment of newborns with congenital toxoplasmosis for up to 1 year using pyrimethamine and sulfadiazine reduced the risk of developing TRC when compared with historical untreated controls.14,15 14. Guerina, N.G. ∙ Hsu, H.W. ∙ Meissner, H.C. ..., New England Regional Toxoplasma Working Group Neonatal serologic screening and early treatment for congenital Toxoplasma gondii infection N Engl J Med. 1994; 330:1858-1863 Crossref Scopus (413) PubMed Google Scholar 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar Finally, TRC has a broad heterogenous clinical spectrum, extending from small peripheral active lesions that frequently resolve spontaneously to larger progressive ones, which eventually may result in markedly decreased vision.6,7,32 6. Holland, G.N. Ocular toxoplasmosis: a global reassessmentPart II: disease manifestations and management Am J Ophthalmol. 2004; 137:1-17 Full Text Full Text (PDF) Scopus (271) PubMed Google Scholar 7. Friedmann, C.T. ∙ Knox, D.L. Variations in recurrent active toxoplasmic retinochoroiditis Arch Ophthalmol. 1969; 81:481-493 Crossref Scopus (150) PubMed Google Scholar 32. Vasconcelos-Santos, D.V. ∙ Dodds, E.M. ∙ Orefice, F. Review for disease of the year: differential diagnosis of ocular toxoplasmosis Ocul Immunol Inflamm. 2011; 19:171-179 Crossref Scopus (21) PubMed Google Scholar Friedmann and Knox7 7. Friedmann, C.T. ∙ Knox, D.L. Variations in recurrent active toxoplasmic retinochoroiditis Arch Ophthalmol. 1969; 81:481-493 Crossref Scopus (150) PubMed Google Scholar reported that lesions of more than 1 disc area were associated with worse visual outcomes. This observation was supported by a large prospective study by Rothova et al,8 8. Rothova, A. ∙ Meenken, C. ∙ Buitenhuis, H.J. ... Therapy for ocular toxoplasmosis Am J Ophthalmol. 1993; 115:517-523 Abstract Full Text (PDF) Scopus (146) PubMed Google Scholar in which the most important predictor of duration of inflammation was the initial size of the retinochoroidal lesion. Furthermore, prospective studies have demonstrated a relationship between antibiotic treatment and reduction of final lesion size.8,29 8. Rothova, A. ∙ Meenken, C. ∙ Buitenhuis, H.J. ... Therapy for ocular toxoplasmosis Am J Ophthalmol. 1993; 115:517-523 Abstract Full Text (PDF) Scopus (146) PubMed Google Scholar 29. Soheilian, M. ∙ Sadoughi, M.M. ∙ Ghajarnia, M. ... Prospective randomized trial of trimethoprim/sulfamethoxazole versus pyrimethamine and sulfadiazine in the treatment of ocular toxoplasmosis Ophthalmology. 2005; 112:1876-1882 Full Text Full Text (PDF) Scopus (140) PubMed Google Scholar However, the lack of proper controls raised the possibility that these improvements could be part of natural history. There are also no randomized controlled studies evaluating the treatment effect of corticosteroids, but several publications described poor outcomes in patients who received systemic or local corticosteroid therapy in the absence of antibiotic treatment.6 6. Holland, G.N. Ocular toxoplasmosis: a global reassessmentPart II: disease manifestations and management Am J Ophthalmol. 2004; 137:1-17 Full Text Full Text (PDF) Scopus (271) PubMed Google Scholar Therefore, it seems prudent to recommend concomitant antibiotic treatment when corticosteroids are administered. Experience with nonmedical therapies also is limited. The treatment benefit of photocoagulation of active lesions is unproven. Vitrectomy surgery also has been performed in patients with active TRC, but typically in the setting of diagnosis or treatment of complications.10 10. Holland, G.N. ∙ Lewis, K.G. An update on current practices in the management of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:102-114 Full Text Full Text (PDF) Scopus (161) PubMed Google Scholar Although there continues to be uncertainty about the treatment benefit of antibiotics and corticosteroids, available evidence shows that short-term treatment does not prevent recurrence. Therefore, programs for primary prevention have focused on pregnant women to prevent congenital transmission. Greater emphasis is needed to reduce postnatally acquired cases by eliminating exposure to recognized sources of infection, such as undercooked meat and water contaminated with oocysts from cat feces. In addition to treatment outcomes, this assessment also focused on safety. Rates of reported adverse reactions depended largely on the antibiotic regimen used, but were relatively common overall. The combination of pyrimethamine and sulfadiazine was characterized by frequent and severe side effects, leading to discontinuation of treatment in up to 25% of patients in some studies, mostly as a result of bone-marrow suppression and allergic reactions. The combination of trimethoprim and sulfamethoxazole seems to be better tolerated and is considerably less expensive than classic therapy, but it infrequently can cause severe life-threatening hypersensitivity reactions, such as Stevens-Johnson syndrome. In conclusion, there is a lack of level I evidence to support routine antibiotic or corticosteroid treatment for all immunocompetent patients with acute TRC. Other nonmedical treatments also remain unproven. There is level II evidence from a single study suggesting that long-term prophylactic treatment with combined trimethoprim and sulfamethoxazole may reduce recurrences in patients with recurrent TRC. The lack of large, randomized studies providing evidence for treatment benefit of active TRC should be viewed with caution and should not serve as an absolute contraindication to therapy in patients at high risk of vision loss. Future Research Randomized placebo-controlled trials are needed to determine the therapeutic efficacy of antibiotic treatment for acute or recurrent episodes of TRC. Such studies will be difficult to perform because of the heterogeneity of the disease, presence of confounding factors, and lack of validated outcome measures that accurately reflect treatment benefit. Despite these challenges, future studies should attempt to ensure uniform and rigorous masked assessment of long-term visual outcomes, rates of recurrence, duration of symptoms, and severity of inflammation. Similarly designed studies are needed to determine the treatment benefit of corticosteroids. Emphasis should be placed on interventions with minimal risk of adverse reactions and low treatment cost. References 1. Jabs, D.A. Ocular toxoplasmosis Int Ophthalmol Clin. 1990; 30:264-270 Crossref Scopus (19) PubMed Google Scholar 2. Jones, J.L. ∙ Kruszon-Moran, D. ∙ Sanders-Lewis, K. ... Toxoplasma gondii infection in the United States, 1999 2004, decline from the prior decade Am J Trop Med Hyg. 2007; 77:405-410 Crossref Scopus (307) PubMed Google Scholar 3. Gilbert, R.E. ∙ Dunn, D.T. ∙ Lightman, S. ... Incidence of symptomatic toxoplasma eye disease: aetiology and public health implications Epidemiol Infect. 1999; 123:283-289 Crossref Scopus (100) PubMed Google Scholar 4. Dodds, E.M. Toxoplasmosis Curr Opin Ophthalmol. 2006; 17:557-561 Crossref Scopus (43) PubMed Google Scholar 5. Perkins, E.S. Ocular toxoplasmosis Br J Ophthalmol. 1973; 57:1-17 Crossref Scopus (204) PubMed Google Scholar 6. Holland, G.N. Ocular toxoplasmosis: a global reassessmentPart II: disease manifestations and management Am J Ophthalmol. 2004; 137:1-17 Full Text Full Text (PDF) Scopus (271) PubMed Google Scholar 7. Friedmann, C.T. ∙ Knox, D.L. Variations in recurrent active toxoplasmic retinochoroiditis Arch Ophthalmol. 1969; 81:481-493 Crossref Scopus (150) PubMed Google Scholar 8. Rothova, A. ∙ Meenken, C. ∙ Buitenhuis, H.J. ... Therapy for ocular toxoplasmosis Am J Ophthalmol. 1993; 115:517-523 Abstract Full Text (PDF) Scopus (146) PubMed Google Scholar 9. Gagliuso, D.J. ∙ Teich, S.A. ∙ Friedman, A.H. ... Ocular toxoplasmosis in AIDS patients Trans Am Ophthalmol Soc. 1990; 88:63-86 discussion 86–8 PubMed Google Scholar 10. Holland, G.N. ∙ Lewis, K.G. An update on current practices in the management of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:102-114 Full Text Full Text (PDF) Scopus (161) PubMed Google Scholar 11. Huskinson-Mark, J. ∙ Araujo, F.G. ∙ Remington, J.S. Evaluation of the effect of drugs on the cyst form of Toxoplasma gondii J Infect Dis. 1991; 164:170-171 Crossref Scopus (139) PubMed Google Scholar 12. Pearson, P.A. ∙ Piracha, A.R. ∙ Sen, H.A. ... Atovaquone for the treatment of toxoplasma retinochoroiditis in immunocompetent patients Ophthalmology. 1999; 106:148-153 Full Text Full Text (PDF) Scopus (83) PubMed Google Scholar 13. Soheilian, M. ∙ Ramezani, A. ∙ Azimzadeh, A. ... Randomized trial of intravitreal clindamycin and dexamethasone versus pyrimethamine, sulfadiazine, and prednisolone in treatment of ocular toxoplasmosis Ophthalmology. 2011; 118:134-141 Full Text Full Text (PDF) Scopus (121) PubMed Google Scholar 14. Guerina, N.G. ∙ Hsu, H.W. ∙ Meissner, H.C. ..., New England Regional Toxoplasma Working Group Neonatal serologic screening and early treatment for congenital Toxoplasma gondii infection N Engl J Med. 1994; 330:1858-1863 Crossref Scopus (413) PubMed Google Scholar 15. Silveira, C. ∙ Belfort, Jr, R. ∙ Muccioli, C. ... The effect of long-term intermittent trimethoprim/sulfamethoxazole treatment on recurrences of toxoplasmic retinochoroiditis Am J Ophthalmol. 2002; 134:41-46 Full Text Full Text (PDF) Scopus (197) PubMed Google Scholar 16. Desmettre, T. ∙ Labalette, P. ∙ Fortier, B. ... Laser photocoagulation around the foci of toxoplasma retinochoroiditis: a descriptive statistical analysis of 35 patients with long-term follow-up Ophthalmologica. 1996; 210:90-94 Crossref Scopus (12) PubMed Google Scholar 17. Ghartey, K.N. ∙ Brockhurst, R.J. Photocoagulation of active toxoplasmic retinochoroiditis Am J Ophthalmol. 1980; 89:858-864 Abstract Full Text (PDF) PubMed Google Scholar 18. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 682 Google Scholar 19. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 593 Google Scholar 20. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 527 Google Scholar 21. Bartlett, J.G. ∙ Auwaerter, P.G. ∙ Pham, P.A. Johns Hopkins ABX Guide: Diagnosis and Treatment of Infectious Diseases Jones and Bartlett Learning, Burlington, MA, 2012; 610 Google Scholar 22. Gilbert, R.E. ∙ See, S.E. ∙ Jones, L.V. ... Antibiotics versus control for toxoplasma retinochoroiditis Cochrane Database Syst Rev. 2002; CD002218 Crossref Google Scholar 23. Oxford Centre for Evidence-Based Medicine Levels of evidence (March 2009) Accessed June 29, 2012 Google Scholar 24. Acers, T.E. Toxoplasmic retinochoroiditis: a double blind therapeutic study Arch Ophthalmol. 1964; 71:58-62 Crossref Scopus (37) PubMed Google Scholar 25. Perkins, E.S. ∙ Schofield, P.B. ∙ Smith, C.H. Treatment of uveitis with pyrimethamine (Daraprim) Br J Ophthalmol. 1956; 40:577-586 Crossref Scopus (38) PubMed Google Scholar 26. Bosch-Driessen, L.H. ∙ Verbraak, F.D. ∙ Suttorp-Schulten, M.S. ... A prospective, randomized trial of pyrimethamine and azithromycin vs pyrimethamine and sulfadiazine for the treatment of ocular toxoplasmosis Am J Ophthalmol. 2002; 134:34-40 Full Text Full Text (PDF) Scopus (146) PubMed Google Scholar 27. Colin, J. ∙ Harie, J.C. Presumed toxoplasmic chorioretinitis: comparative study of treatment with pyrimethamine and sulfadiazine or clindamycin [in French] J Fr Ophtalmol. 1989; 12:161-165 Crossref Scopus (2) PubMed Google Scholar 28. Raskin, E. ∙ Alves, M. ∙ Eredia, G.C. ... Ocular toxoplasmosis: a comparative study of the treatment with sulfadiazine and pyrimethamine versus sulphametoxazole-trimethoprim [in Portuguese] Rev Bras Oftalmol. 2002; 61:335-338 Google Scholar 29. Soheilian, M. ∙ Sadoughi, M.M. ∙ Ghajarnia, M. ... Prospective randomized trial of trimethoprim/sulfamethoxazole versus pyrimethamine and sulfadiazine in the treatment of ocular toxoplasmosis Ophthalmology. 2005; 112:1876-1882 Full Text Full Text (PDF) Scopus (140) PubMed Google Scholar 30. Gilbert, R.E. ∙ Freeman, K. ∙ Lago, E.G. ... European Multicentre Study on Congenital Toxoplasmosis (EMSCOT)Ocular sequelae of congenital toxoplasmosis in Brazil compared with Europe PLoS Negl Trop Dis [serial online]. 2008; 2:e277 Accessed June 29, 2012 Crossref Scopus (230) PubMed Google Scholar 31. Lam, S. ∙ Tessler, H.H. Quadruple therapy for ocular toxoplasmosis Can J Ophthalmol. 1993; 28:58-61 PubMed Google Scholar 32. Vasconcelos-Santos, D.V. ∙ Dodds, E.M. ∙ Orefice, F. Review for disease of the year: differential diagnosis of ocular toxoplasmosis Ocul Immunol Inflamm. 2011; 19:171-179 Crossref Scopus (21) PubMed Google Scholar Figures (1)Figure Viewer Article metrics Related Articles View abstract Open in viewer Interventions for Toxoplasma RetinochoroiditisA Report by the American Academy of Ophthalmology Hide CaptionDownloadSee figure in Article Toggle Thumbstrip Figure 1 Download .PPT Go to Go to Show all references Expand All Collapse Expand Table Authors Info & Affiliations Home Access for Developing Countries Articles & Issues Articles In Press Current Issue List of Issues & Supplements Collections The Collaborative Community on Ophthalmic Imaging Commentaries Correspondence Archive Cover Art Gallery Editorials IOL Article Collection IRIS Registry Articles Myopia Ophthalmic Technology Assessments Preferred Practice Pattern Guidelines Top 10 Trending Articles Translational Science Reviews Multimedia Ophthalmology Journal Podcast Pictures & Perspectives Video Collection by Topic Virtual Journal Club Videos Authors & Reviewers Guide for Authors Guide for Reviewers Miscellaneous Links Register to Review Researcher Academy Submit a Manuscript Journal Info About the Journal ABO Quarterly Questions About Open Access Contact Information Editorial Board Info for Advertisers New Content Alerts Reprints Subscribe American Academy of Ophthalmology Academy Member Access About the Academy Contact the Academy Forget your membership login? 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9279	https://edanzhao.com/news/999.html	空间点到直线的距离公式具体如何计算的-易单招易单招欢迎您！招考专题院校招考专业辅导课程保送生农村专项体育高考易单招学校文章专业专题关注微信关注高职单招网微信公众号 (ID:danzhaobang) 获得最新考试资讯联系电话易单招合作热线 15996434567 周一至周日：9:00-21:00 首页高职单招单招动态院校资讯招生简章备考指南职教高考职教动态院校资讯招生简章备考指南强基计划招生政策报考指南招生简章校测真题普通高考高考政策高校动态志愿填报高考复习高考真题综合评价综评动态综评报考综评简章综评校测综评竞赛综评院校艺考院校资讯招生简章备考指南港澳招生招生动态院校信息考核真题经验分享国际本科院校资讯招生简章备考指南警校军校院校资讯招生简章备考指南分数线专转本专转本资讯招生简章备考指南成考自考院校资讯招生简章备考指南首页普通高考高考复习全部高考政策高校动态志愿填报高考复习高考真题空间点到直线的距离公式具体如何计算的高考复习更新时间：2024-11-05 浏览量:145 次空间点到直线的距离公式：设直线L的方程为Ax+By+C=0，点P的坐标为（Xo，Yo），则点P到直线L的距离为\|AXo+BYo+C\|/√（A²+B²）。点到直线的距离，即过这一点做目标直线的垂线，由这一点至垂足的距离。 1空间点到直线的距离公式是什么点到直线的距离公式：设直线方程为 Ax+By+C=0，则点(x1,y1)到直线的距离为：d=\|Ax1+By1+C\| / √(A^2+B^2) 这个公式的推导过程如下：首先，选择直线L上的一点，然后计算该点到直线L的垂直距离。通过向量的点积和模长计算，可以得到点到直线的距离公式。具体步骤包括：选择直线L上的一点，例如(x1, y1)。计算点P到点(x1, y1)的向量。利用向量的点积和模长公式，推导出点到直线的距离公式。这个公式在‌三维空间中的应用非常广泛，特别是在计算机图形学、机器人路径规划等领域中，需要计算空间中点到直线的距离来优化路径或避免碰撞。 2点和直线的位置关系有什么点和直线的位置关系有两种：点在直线上和点在直线外。点在直线上：当点位于直线上时，表示直线经过该点。这种情况下，点与直线的关系是确定的，因为点完全位于直线的延伸范围内。当点位于直线上时，点的坐标满足直线的方程。例如，如果直线方程是y=x+1，那么点（2，3）就在直线上，因为3=2+1。在几何学中，这种关系可以用来检验一个点是否在一个给定的直线上。点在直线外：当点不在直线上时，表示直线不经过该点。这种情况下，点与直线没有交集，点位于直线的两侧或一侧，但不在直线上。这两种关系是点和直线位置关系的基本分类，适用于几何学和数学中的基本概念和问题解决。点和直线的广泛应用： 1、在几何学中，点和直线是构建各种复杂图形的基础。例如，在建筑设计、机械设计等领域，设计师通常会使用点和直线来创建基本的几何形状，如矩形、圆形等。然后，这些基本形状被组合和变形，以创建出各种复杂的图形和结构。 2、点和直线也被广泛应用于物理和工程领域。例如，在电路设计中，点和直线可以表示电子元件之间的连接关系；在道路设计中，点和直线可以表示道路的起点和终点，以及道路的走向。 3、点和直线还可以用于解决实际问题。例如，在地图上，点和直线可以表示地标之间的位置关系；在金融领域，点和直线可以用于绘制趋势线和支撑线等，以帮助分析股票等金融产品的价格走势。上一篇：直线的方向向量怎么求如何计算的下一篇：合数有哪些100以内什么是合数最新资讯 2025年山东16市高中开学时间安排汇总安徽省2025年9月普通高中学业水平合格性考试报名工作即将启动 2025年最难考211院校出炉！这一高校竟霸榜5年！ 4 华大新高考联盟2026届高三9月教学质量测评安排 5 首都师范大学附中2025-2026学年新高一分班考语文试题及答案 6 2025年西湖大学本科生录取人数 7 孩子到了高中，比的是父母这7种托举能力 8 高三一轮复习中弱科提分策略 9 准高二：要有竞争力，有些东西，你得学会放弃！ 10 志愿填报\|电气工程专业的14个就业方向最新专题强基计划39校招生专业实力如何？院校专业怎么选？为何竞赛金银牌都被拒？清北强基“破格”入围不唯“竞赛论”！未通过清北强基计划破格审核的考生之后该怎么办？ 4 清北强基计划破格审核看哪些条件？ 5 未通过清北强基计划审核之后怎么办？ 6 竞赛省一、省二报考强基计划有什么优势 7 强基计划39校专业限报、调剂要求盘点！和高考志愿填报一样吗？ 8 清华新领军VS中科大少年班，谁才是最大的赢家？ 9 强基计划复交南模式校测怎么考？ 10 684分强基计划上岸清华！6年竞赛学习经历的他做到了这几点关注官方微信微信识别二维码关注官方公众号易单招微信账号：danzhaobang 复制本站热点 > 院校库> 专业库> 课程库 ### 关于我们 > 关于我们> 商务合作> 广告投放 ### 合作加盟 > 友情链接> 招贤纳士> 法律声明 ### 本站推荐 >> 站长推荐>> 年度排行>> 每日更新 ### 网站地图 >> 招考专题>> 网站地图>> 报名入口易单招合作热线 15996434567 周一至周日：9:00-21:00 2013-2025 易单招,All Rights Reserved.\|苏ICP备18060098号-16\| 苏公网安备 50019002502480号公司地址：南京市栖霞区栖霞街道广月路30-06号报名咨询官方微信
9280	https://encyclopediaofmath.org/wiki/Renewal_theory	Renewal theory - Encyclopedia of Mathematics Log in www.springer.comThe European Mathematical Society Navigation Main page Pages A-Z StatProb Collection Recent changes Current events Random page Help Project talk Request account Tools What links here Related changes Special pages Printable version Permanent link Page information Namespaces Page Discussion Variants Views View View source History Actions Renewal theory From Encyclopedia of Mathematics Jump to: navigation, search A branch of probability theory describing a large range of problems connected with the rejection and renewal of the elements of some system. The principal concepts in renewal theory are those of a renewal process and renewal equation. A renewal process may be described using the classical scheme of sums of independent random variables in the following manner. Let ξ 1,ξ 2,...ξ 1,ξ 2,... be a sequence of independent, non-negative, identically-distributed random variables having distribution function F(x)F(x). Let ζ 0=0 ζ 0=0, ζ n=ξ 1+⋯+ξ n ζ n=ξ 1+⋯+ξ n, n≥1 n≥1. The renewal process N t N t is defined as N t=max{n:ζ n≤t}.(1)(1)N t=max{n:ζ n≤t}. If the ξ i ξ i are interpreted as the lifetimes of some successively replaceable element, the random variable N i N i equals the number of replacements (or renewals) of these elements during time t t. The renewal function H(t)=E N t H(t)=E N t plays an important part in studying N t N t. This function satisfies the renewal equation H(t)=F(t)+∫0 t H(t−u)d F(u).(2)(2)H(t)=F(t)+∫0 t H(t−u)d F(u). For F(t)=1−e−ρ t F(t)=1−e−ρ t, t≥0 t≥0, there arises an important special case of the renewal process — the Poisson process, in which P{N(t)=k}=(ρ t)k k!e−ρ t,k=0,1…P{N(t)=k}=(ρ t)k k!e−ρ t,k=0,1… and H(t)=ρ t H(t)=ρ t. The renewal process N t N t and the renewal equation (2) are very important in the study of various theoretical and applied problems in queueing theory, reliability theory, storage theory, the theory of branching processes (cf. Branching process), etc. A large number of results in renewal theory is connected with the study of the asymptotic properties of the renewal function H(t)H(t) as t→∞t→∞. An elementary renewal theorem states that lim t→∞H(t)t=1 m,(3)(3)lim t→∞H(t)t=1 m, where m=E ξ i m=E ξ i. It was shown by D. Blackwell in 1948 (cf. ), that if the distribution ξ i ξ i is not concentrated on some arithmetical lattice of the type {0,d,2 d,...}{0,d,2 d,...}, d>0 d>0, then for any h>0 h>0, lim t→∞[H(t+h)−H(t)]=h m.(4)(4)lim t→∞[H(t+h)−H(t)]=h m. Numerous results are available which generalize and precisize equations (3) and (4) in several respects. Results of the kind presented in (3) and (4) are used to study the asymptotic behaviour of the solution X(t)X(t) of the renewal-type equation X(t)=K(t)+∫0 t X(t−u)d F(u),X(t)=K(t)+∫0 t X(t−u)d F(u), in which the free term K(t)K(t) is some function other than F(t)F(t) which meets certain conditions. The relation P{N t≥n}=P{ζ n≤t}(5)(5)P{N t≥n}=P{ζ n≤t} follows from definition (1). Since limit theorems for sums ζ n ζ n of independent terms have been thoroughly studied, relation (5) makes it possible to obtain limit theorems for the number of renewals N t N t. There exists a large number of generalizations of the scheme just described. One such generalization, connected with semi-Markov processes (cf. Semi-Markov process), yields the so-called Markov renewal process, in which the system has some number of states, and the lifetimes of the individual elements are random elements which depend on the states of the system before and after the moment of renewal. References D.R. Cox, "Renewal theory" , Methuen (1962) How to Cite This Entry: Renewal theory. Encyclopedia of Mathematics. URL: This article was adapted from an original article by B.A. Sevast'yanov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article Retrieved from " Categories: TeX auto TeX done This page was last edited on 6 June 2020, at 08:10. Privacy policy About Encyclopedia of Mathematics Disclaimers Copyrights Impressum-Legal Manage Cookies
9281	https://www.youtube.com/watch?v=6FnCApXCk6M	Find Angles In Isosceles Triangles Euler's Academy 9230 subscribers 5 likes Description 2667 views Posted: 28 Sep 2022 This video introduces isosceles triangles and how to find missing angles. An isosceles triangle has two equal side lengths and the angles that are opposite those equal side lengths are equal as well. The general strategy is to first determine which two angles are equal in the isosceles triangle by looking opposite the two equal side lengths. Then, use the fact that the sum of the angles in a triangle add up to 180 degrees to create an equation. Use the equation to solve for the missing side length. Khan Academy exercise to practice: 1. Find Angles in Isosceles Triangles: EulersAcademy.org Transcript: in this video I'd like to talk about finding angles in isosceles triangles so let's start with the definition of an Isles triangle so these are triangles that have two equal sides and two equal angles so let me write that down so these are triangles that have two equal sides and two equal angles so let's draw a quick sketch of this so if we have some random Isles triangle and it's not drawn perfectly to scale but the main idea is that this triangle will have two of its sides that are equal and the angles that are opposite these sides so this angle and this angle these are going to be equal as well so another way to think about it is that the two angles that are equal open up to the sides that are equal so the angle that isn't necessarily equal to these is the one between the two equal sides so this one is usually something different but if it is the same as these then actually that would be an equilateral triangle so we want to use this definition to help us answer problems so that we can essentially find a missing angle so with this example we are trying to find this missing angle X and notice that this triangle has two equal sides so that tells us this is an isoceles triangle and we know with an isoceles triangle that the angles opposite the equal sides or the angles that open up to the equal sides these are equal to each other so this angle and this angle these are both equal to X whatever that angle measure is so we can call this X De as well and to find what x is equal to we just need to remember that the angles in a triangle when we're talking about flat space or some type of flat plane that the sum of these angles adds up to 180° so let's write that down that this angle plus this angle plus this 80° angle these add up to 180 so we have x de+ x de+ 80° this is equal to 180 and we can combine like terms let's put the two x's together so we have twice this angle measure of X+ 80° is equal to 180 and let me just make some room so we can finish this and from here we just need to solve for x so let's subtract 80 on each side and doing that we'll cancel out the 80s on the left hand side so that we get twice that angle measure of X is equal to 100 and now we can just divide everything by two so that get X by itself since 2 divided two we know that's just one so X is 100 / two or in other words x is 50° so if we plug this back into our equation we found that these are 50° angles and this matches up with our definition of an Isles triangle these are two equal angles and they are opposite the two equal side lengths and it makes sense in terms of a general triangle because if we add these angles together 50 + 50 is 100 + 80 is 180 they do add up to 180° which is what we'd expect so we can feel confident that X is 50°
9282	https://www.bahrainmedicalbulletin.com/december_2003/fistula.pdf	Bahrain Medical Bulletin, Vol.25, No. 4, December 2003 Complete Branchial Fistula Chava Anjaneyulu, MBBS, MS Chakkyath Jayaram Sharan, MBBS Complete branchial fistula with internal and external opening is rare. We are reporting a case of complete branchial fistula in a ten year old child. We excised the complete branchial fistula by using combined (transcervical and transoral) approach under general anaesthesia. Bahrain Med Bull 2003;25(4): Although the branchial apparatus was first described by VonBaer, anomalies in its development were credited by von Ascheron1. Branchial arches and their corresponding pouches develop from mesodermal condensations in the side wall of the embryonic pharynx. Branchial fistulas are uncommon anomalies of embryonic development of branchial apparatus. Second branchial arch and pouch anomalies are common anomalies of branchial apparatus2. During embryonic development, the second arch grows caudally, envelop the third, fourth and sixth arches and form the cervical sinus by fusing with the skin caudal to these arches. The edges of cervical sinus fuse and the ectoderm within the fused tube disappears. Persistence of ectoderm gives rise to branchial cyst. The branchial fistula results from the breakdown of the endoderm, usually in the second pouch. A persistent fistula of the second branchial cleft and pouch pass from the external opening in the mid or lower neck in the line of the anterior border of the sternocleidomastoid muscle, deep to platysma along the carotid sheath, then pass medially deep between the internal and external carotid arteries after crossing over the glossopharyngeal nerve and hypoglossal nerve. Finally, it opens internally in the tonsillar fossa usually on the anterior face of the upper half of the posterior pillar of the fauces or in the intratonsillar cleft3. Most of times it is a simple sinus opening, that extends up the neck for a variable distance. Complete branchial fistula with internal opening into tonsillar region is rare2. Although branchial fistulas may occur in any age group, commonly patients present to clinicain in first and second decades of life4. These patients commonly present with persistent mucoid discharge from a fistula opening or infection in the lower part of neck with mucopurulent discharge2. The completeness of fistula can be diagnosed by a dye test or fistulogram and sometimes negative preoperative test might become positive under general anaesthesia because of muscle relaxation3. Occasionally the fistula tract may be blocked by thick secretions or granulation tissue4. Assistant Professor Junior Resident Department of Otorhinolaryngology and Head & Neck Surgery All India Institute of Medical Sciences New Delhi India. THE CASE Ten years old male patient complained of mucopurulent discharge from an opening in left lower part of neck for one year. Patient’s parents noticed an opening in lower part of left side of the neck since he was 2 years old. He was asymptomatic till one year ago. The patient noticed mucopurulent discharge from the opening with common cold. The discharge was moderate in quantity, not foul smelling. It subsided with antibiotics within 5 days. Patient had two more attacks of discharge from the opening during one year period and responded well to medical treatment. For last 3 months, there was no discharge from the opening. On physical examination, there was small punctum in the skin at the junction of anterior two third and lower one third of anterior border of left sternocleidomastoid muscle. There are no similar openings or any other congenital anomalies on the right side. Full blood count and urine examination was normal. The patient underwent excision of branchial fistula tract using transcervical approach and transoral approach under general anaesthesia. Elliptical incision was made around the opening of the fistula. The fistula was cannulated with blind tip probe. The tract was dissected superiorly in subplatysmal plane along the carotid sheath up to the level of carotid bifurcation. Around 1.5 to 2 cm upper incision was made at the level of the carotid bifurcation. The tract was separated from the surrounding soft tissues superiorly beneath the stylohyoid muscle and digastric muscle. The tract was going medially in the parapharyngeal space towards the tonsillar fossa between the internal carotid artery and external carotid artery after crossing the loop of hypoglossal nerve. Using the Boyle-Davis mouth gag, the probe was palpated near the lower pole of the tonsil in the tonsillar fossa. Tonsillectomy was done. Dissection was done around the fistula tract in the tonsillar fossa and parapharyngeal space. Complete fistula tract was pulled with the probe through the neck. Tonsillar fossa opening was closed with 3-0 vicryl. Haemostasis was secured. External skin incisions were closed with 3-0 catgut and 4-0 nylon. The length of the fistula tract was around 9 cm from the skin opening to the lower pole of the tonsil (Fig.1). The post-operative course was uneventful. Antibiotics, analgesics and mouth gargles with betadine and H2O2 (1:40 dilution) were given for 7 days. Histopathological examination showed branchial fistula tract lined with pseudostratified columnar epithelium. At 3 months follow up, the patient was asymptomatic. Figure 1. Showing the branchial fistula tract with tonsil DISCUSSION The treatment of choice for branchial fistula is surgical excision. Preoperatively, the patient should be examined for bilateral lesion and any other congenital anomaly. Surgery may be delayed in an infant with uncomplicated branchial fistula till the age of three years. If there is an infective episode, it must be allowed to subside with antibiotic treatment before surgery takes place. Imaging is of little benefit. Computed tomography scanning and magnetic resonance imaging of the neck are useful mainly in delineating the relationship of surrounding neurovascular structures to the lesion4. Several surgical approaches have been described for the management of a branchial fistula1-5. Stepladder approach with two incisions in the neck gives exposure of the fistula tract with less tissue dissection. Higher incision should be bigger than the lower one because the fistula tract is deeper in location in the vicinity of important neurovascular structures3. Complete excision of the fistula is never sure with external approach only. The reported incidence of recurrence rate was three percent after external approach2 only and most probably this is due to incomplete excision of the fistula tract in the parapharyngeal space. In complete branchial fistula with a probe passing to the tonsillar region, the above procedure should be combined with oral route. Through oral route, we can identify the internal opening after tonsillectomy, separate the fistula tract from surrounding soft tissues and muscle fibres in parapharyngeal space by precise dissection around the fistula tract. Sometimes it is possible to dissect the tract with the tonsil and remove them as one specimen1. Till now with combined transcervical and transoral approach, there is no recurrence reported in the English literature. CONCLUSION Complete branchial fistula arising from second branchial arch is rare. It is not possible to excise the complete branchial fistula totally with transcervical approach only. This is a case of complete branchial fistula, which was managed through combined approach using transcervical route and transoral route. Complete branchial fistulas are better managed by otolaryngologists who are capable of performing the combined approach. REFERENCES 1. De PR, Mikhail T. A combined approach excision of branchial fistula. J Laryngol Otol 1995;109:999-1000. 2. Ford GR, Balakrishnan A, Evans JN, et al. Branchial cleft and pouch anomalies. J Laryngol Otol 1992;106:137-43. 3. Talaat M. Pull-through branchial fistulectomy: technique for the otolaryngologist. Ann Otol Rhino Laryngol 1992;101:501-2. 4. Ang AH, Pang KP, Tan LK. Complete branchial fistula. Case report and review of the literature. Ann Otol Rhinol Laryngol 2001;110:1077-9. 5. Taylor PH, Bicknell PG. Stripping of branchial fistulae. A new technique. J Laryngol Otol 1977;91:141-9.
9283	https://www.malacards.org/card/atransferrinemia	Atransferrinemia - MalaCards overview Overview summarize Summaries family_history Traits & Categories badge Aliases & Identifiers genetics Associated Genes microbiology Related Disorders symptoms Symptoms & Phenotypes pill Drugs & Therapeutics labs Genetic Tests rib_cage Anatomical Context article Publications swap_horiz Variations bar_chart Expression network_node Pathways format_list_bulleted GO Terms database Sources Home Help User Guide What's in a card Search guide About our data Our sources Acknowledgements Contact us Tools & Databases GeneCards The Human Gene Database PathCards Human biological pathway unification GeneHancer Integrated database of enhancers and target genes GeneCaRNA Integrated database of non-coding RNA (ncRNA) VarElect NGS Phenotyper GeneAnalytics GeneCards Powered Gene Set Enrichment GeneALaCart Batch gene annotation download GenesLikeMe Explore shared gene annotations Licensing Commercial Use Commercial Data Licensing Academic Data Licensing About Release Notes Our Team Our Publications About the Weizmann Institute of Sciences About LifeMap Sciences GeneCards Suite Terms of Use GeneCards Suite Privacy Policy Login / Register Contact us Name Email Message [x] I accept the Privacy Policy Send overview Overview summarize Summaries family_history Traits & Categories badge Aliases & Identifiers genetics Associated Genes microbiology Related Disorders symptoms Symptoms & Phenotypes pill Drugs & Therapeutics labs Genetic Tests rib_cage Anatomical Context article Publications swap_horiz Variations bar_chart Expression network_node Pathways format_list_bulleted GO Terms database Sources Atransferrinemia (ATRAF) MCID: ATR002 Info Score: 45 Summary (AI-Supported) Atransferrinemia is a rare autosomal recessive disorder characterized by transferrin deficiency, leading to microcytic anemia and iron overload. Symptoms include pallor, fatigue, growth retardation, and iron damage to the heart and liver, potentially causing heart failure. Treatment involves plasma infusions. The disease is caused by mutations in the transferrin gene on chromosome 3q22 and was first described in 1961, with only ten documented cases worldwide. Categories Global:Genetic diseasesRare diseasesMetabolic diseasesAnatomical:Blood diseases Also known as Familial Hypotransferrinemia Congenital Atransferrinemia (and 4 more) Associated genes 14 genes associated with Atransferrinemia 1 with high evidence (Elite gene associations) 13 non-elite and text-mined gene associations Associated pathways 24 gene-associated pathways Related drugs 1 drugs/compounds 1 associated clinical trials Variations 76 ClinVar variations 2 UniProtKB/Swiss-Prot variations Phenotypes 10 human phenotypes 9 mouse phenotypes 9 RNAi phenotypes Summaries for Atransferrinemia Disease Ontology12 A metal metabolism disorder that is characterized by transferrin deficiency, microcytic anemia, and iron loading, and has material basis in autosomal recessive inheritance of homozygous or compound heterozygous mutation in the structural gene for transferrin (TF) on chromosome 3q22. Orphanet61 Congenital atransferrinemia is a very rare hematologic disease caused by a transferrin (TF) deficiency and characterized by microcytic, hypochromic anemia (manifesting with pallor, fatigue and growth retardation) and iron overload, and that can be fatal if left untreated. OMIM®60(Updated 25-06-29) Atransferrinemia is characterized by microcytic anemia and by iron loading. It can be treated effectively by plasma infusions (summary by Beutler et al., 2000). UniProtKB/Swiss-Prot76 A rare autosomal recessive disorder characterized by abnormal synthesis of transferrin leading to iron overload and microcytic hypochromic anemia. Wikipedia78 Atransferrinemia is an autosomal recessive metabolic disorder in which there is an absence of transferrin, a plasma protein that transports iron through the blood. Atransferrinemia is characterized by anemia and hemosiderosis in the heart and liver. The iron damage to the heart can lead to heart failure. The anemia is typically microcytic and hypochromic (the red blood cells are abnormally small and pale). Atransferrinemia was first described in 1961 and is extremely rare, with only ten documented cases worldwide. Traits & Categories for Atransferrinemia Traits of Atransferrinemia Inheritance Atransferrinemia:Autosomal recessive60 Congenital Atransferrinemia:Autosomal recessive61 Prevalence Congenital Atransferrinemia 61Point prevalence <1/1000000 (Worldwide) Age of onset Congenital Atransferrinemia:Childhood, Infancy61 Categories for Atransferrinemia ICD10 04 - Endocrine, nutritional and metabolic diseases E70-E90 - Metabolic disorders E88 - Other metabolic disorders E88.0 - Disorders of plasma-protein metabolism, not elsewhere classified Orphanet Categories Inborn errors of metabolism, Rare haematological diseases Aliases & Identifiers for Atransferrinemia Aliases for Atransferrinemia Name: Atransferrinemia601220763013 57 61578 Familial Hypotransferrinemia122074 Congenital Atransferrinemia206174 Congenital Hypotransferrinemia2061 Hypotransferrinemia, Familial60 Hereditary Atransferrinemia20 Atraf76 External Database Identifiers for Atransferrinemia Disease Ontology DOID:0050649 ICD10 via Orphanet E88.0 ICD11 via Orphanet 5D0Y MedGen C0521802 C1859593 C3277918 MeSH D008664 MESH via Orphanet C538259 MONDO MONDO:0008846 NCIt C125693 OMIM® 209300 Orphanet ORPHA1195 SNOMED-CT via HPO 111571009 165397008 235856003 271737000 3723001 3855007 40930008 42343007 44452003 49601007 84114007 (see all)(see less) UMLS C0521802 C3887502 UMLS via Orphanet C0521802 Genes associated with Atransferrinemia There are 14 Genes associated with Atransferrinemia: Elite Gene Cancer-census gene Show entries Search: \| \| Gene \| Evidence \| \| --- --- \| \| # \| Symbol \| Description \| Category \| Score \| Molecular \| Variation \| Experimental \| Inferred \| PMIDs \| \| \| Gene \| Evidence \| \| --- --- \| \| # \| Symbol \| Description \| Category \| Score \| Molecular \| Variation \| Experimental \| Inferred \| PMIDs \| \| 1 \| TF \| Transferrin \| Protein Coding \| 1705.53 \| Molecular basis known60 Genetic Tests30 \| Pathogenic6 Causative germline mutation61 Causative variation76 Likely pathogenic6 \| \| DISEASES inferred15 Novoseek inferred 57 GeneCards Inferred via: (sections)Disorders , Publications \| 10 Publications \| \| 2 \| TFR2 \| Transferrin Receptor 2 \| Protein Coding \| 7.15 \| \| \| \| DISEASES inferred15 \| \| \| 3 \| HJV \| Hemojuvelin BMP Co-Receptor \| Protein Coding \| 6.91 \| \| \| \| DISEASES inferred15 \| \| \| 4 \| HFE \| Homeostatic Iron Regulator \| Protein Coding \| 6.73 \| \| \| \| DISEASES inferred15 \| 1 Publications \| \| 5 \| HAMP \| Hepcidin Antimicrobial Peptide \| Protein Coding \| 6.64 \| \| \| \| DISEASES inferred15 \| \| \| 6 \| SLC11A2 \| Solute Carrier Family 11 Member 2 \| Protein Coding \| 6.58 \| \| \| \| DISEASES inferred15 \| \| \| 7 \| SLC40A1 \| Solute Carrier Family 40 Member 1 \| Protein Coding \| 6.57 \| \| \| \| DISEASES inferred15 \| \| \| 8 \| DNAAF8 \| Dynein Axonemal Assembly Factor 8 \| Protein Coding \| 6.28 \| \| \| \| DISEASES inferred15 \| \| \| 9 \| TFRC \| Transferrin Receptor \| Protein Coding \| 6.06 \| \| \| \| DISEASES inferred15 \| \| \| 10 \| C9orf163 \| Chromosome 9 Putative Open Reading Frame 163 \| RNA Gene \| 5.84 \| \| \| \| DISEASES inferred15 \| \| Showing 1 to 10 of 14 entries 1 2 Related Disorders for Atransferrinemia No Related Disorders information available for Atransferrinemia Symptoms & Phenotypes for Atransferrinemia Symptoms Clinical Features OMIM®209300 OMIM® Clinical Synopsis (Updated 25-06-29) Cardiovascular60: congestive heart failure hemosiderosis, heart Heme60: anemia, hypochromic Lab60: transferrin absent Liver60: hemosiderosis Phenotypes Human Phenotypes for Atransferrinemia There are 10 Human Phenotypes for Atransferrinemia: Search: \| # \| Phenotype \| HPO ID \| HPO Frequency \| Orphanet Frequency \| --- --- \| # \| Phenotype \| HPO ID \| HPO Frequency \| Orphanet Frequency \| --- --- \| 1 \| anemia \| HP:0001903 \| Hallmark (90%) \| Very frequent (99-80%) \| \| 2 \| recurrent infections \| HP:0002719 \| Frequent (33%) \| Frequent (79-30%) \| \| 3 \| hypothyroidism \| HP:0000821 \| Occasional (7.5%) \| Occasional (29-5%) \| \| 4 \| arthritis \| HP:0001369 \| Occasional (7.5%) \| Occasional (29-5%) \| \| 5 \| abnormality of the cardiovascular system \| HP:0001626 \| Occasional (7.5%) \| Occasional (29-5%) \| \| 6 \| abnormality of the pancreas \| HP:0001732 \| Occasional (7.5%) \| Occasional (29-5%) \| \| 7 \| congestive heart failure \| HP:0001635 \| \| \| \| 8 \| abnormality of the liver \| HP:0001392 \| \| \| \| 9 \| hypochromic anemia \| HP:0001931 \| \| \| \| 10 \| atransferrinemia \| HP:0012239 \| \| \| Showing 1 to 10 of 10 entries Mouse Phenotypes for Atransferrinemia There are 9 Mouse Phenotypes for Atransferrinemia: High statistical significance Search: \| # \| Description \| MGI ID \| Gene Set Enrichment \| Score \| Top Affiliating Genes \| --- --- --- \| \| # \| Description \| MGI ID \| Gene Set Enrichment \| Score \| Top Affiliating Genes \| --- --- --- \| \| 1 \| Adipose tissue phenotype \| MP:0005375 \| 1/1,591 \| 0.97 \| TF \| \| 2 \| Embryo phenotype \| MP:0005380 \| 1/2,750 \| 0.73 \| TF \| \| 3 \| Integument phenotype \| MP:0010771 \| 1/2,990 \| 0.69 \| TF \| \| 4 \| Immune system phenotype \| MP:0005387 \| 1/5,579 \| 0.42 \| TF \| \| 5 \| Hematopoietic system phenotype \| MP:0005397 \| 1/5,713 \| 0.41 \| TF \| \| 6 \| Behavior/neurological phenotype \| MP:0005386 \| 1/6,128 \| 0.38 \| TF \| \| 7 \| Growth/size/body region phenotype \| MP:0005378 \| 1/7,069 \| 0.32 \| TF \| \| 8 \| Mortality/aging \| MP:0010768 \| 1/7,319 \| 0.30 \| TF \| \| 9 \| Homeostasis/metabolism phenotype \| MP:0005376 \| 1/7,770 \| 0.28 \| TF \| Showing 1 to 9 of 9 entries GenomeRNAi Phenotypes for Atransferrinemia There are 9 GenomeRNAi Phenotypes for Atransferrinemia: High statistical significance Search: \| # \| Description \| GenomeRNAi Source Accession \| Gene Set Enrichment \| Score \| Top Affiliating Genes \| --- --- --- \| \| # \| Description \| GenomeRNAi Source Accession \| Gene Set Enrichment \| Score \| Top Affiliating Genes \| --- --- --- \| \| 1 \| Increased number of mitotic cells \| GR00098-A-3 \| 1/212 \| 1.98 \| TF \| \| 2 \| Synthetic lethal with vaccinia virus (VACV) infection \| GR00362-A \| 1/705 \| 1.46 \| TF \| \| 3 \| Increased cell death HMECs cells \| GR00103-A-0 \| 1/707 \| 1.46 \| TF \| \| 4 \| Resistant to vaccinia virus (VACV-A4L) infection \| GR00351-A-1 \| 1/1,687 \| 1.08 \| TF \| \| 5 \| Increased G1 DNA content \| GR00098-A-1 \| 1/2,074 \| 0.99 \| TF \| \| 6 \| Increased IL-8 secretion \| GR00386-A-2 \| 1/2,358 \| 0.94 \| TF \| \| 7 \| Increased viability \| GR00386-A-1 \| 1/4,438 \| 0.66 \| TF \| \| 8 \| No effect \| GR00402-S-1 \| 1/16,336 \| 0.09 \| TF \| \| 9 \| Decreased viability \| GR00402-S-2 \| 1/16,336 \| 0.09 \| TF \| Showing 1 to 9 of 9 entries Drugs & Therapeutics for Atransferrinemia Therapeutics for Atransferrinemia NIH Clinical Center Search NIH Clinical Center for Atransferrinemia Drugs for Atransferrinemia There is 1 Drugs for Atransferrinemia: Click to expand drug synonyms. Search: \| \| # \| Name \| Status \| Phase \| Clinical Trials \| Cas Number \| PubChemIds \| --- --- --- --- \| \| \| # \| Name \| Status \| Phase \| Clinical Trials \| Cas Number \| PubChemIds \| --- --- --- --- \| \| \| 1 \| Iron \| Approved, Investigational, Nutraceutical \| Phase 2, Phase 3 \| \| 7439-89-6 \| 29936 \| Showing 1 to 1 of 1 entries Interventional Clinical Trials for Atransferrinemia There is 1 Interventional Clinical Trials for Atransferrinemia: Search: \| # \| Name \| Status \| NCT ID \| Phase \| Drugs \| --- --- --- \| \| # \| Name \| Status \| NCT ID \| Phase \| Drugs \| --- --- --- \| \| 1 \| Dose Escalating Study to Evaluate Pharmacokinetics, Efficacy and Safety of Apotransferrin in Atransferrinemia Patients \| Active, not recruiting \| NCT01797055 \| Phase 2, Phase 3 \| Human apotransferrin \| Showing 1 to 1 of 1 entries Genetic Tests for Atransferrinemia Genetic tests related with Atransferrinemia There are 7 Genetic tests related with Atransferrinemia: Search: \| # \| Genetic test \| Condition \| Affiliated genes \| --- --- \| \| # \| Genetic test \| Condition \| Affiliated genes \| --- --- \| \| 1 \| TF - atransferrinemia \| Atransferrinemia \| TF \| \| 2 \| Atransferrinemia, 209300, Autosomal recessive (Congenital atransferrinemia) (TF gene) (Sequence Analysis-All Coding Exons) (Postnatal) \| Atransferrinemia \| TF \| \| 3 \| Atransferrinemia, 209300, Autosomal recessive (Congenital atransferrinemia) (TF gene) (Sequence Analysis-All Coding Exons) (Prenatal) \| Atransferrinemia \| TF \| \| 4 \| NGS Panel for Iron related Anemias (including Aceruloplasminemia, Atransferrinemia, IRIDA, DMT1-deficiency =AHMIO1, Hypochromic microcytic anemia with iron overload 2=AHMIO2) \| Atransferrinemia \| TF \| \| 5 \| TF Single Gene \| Atransferrinemia \| TF \| \| 6 \| Atransferrinemia: Full gene sequencing \| Atransferrinemia \| TF \| \| 7 \| Atransferrinemia \| Atransferrinemia \| TF \| Showing 1 to 7 of 7 entries Anatomical Context for Atransferrinemia Organs/Tissues Heart(3.82)41 Liver(3.80)4158 Bone Marrow(1.07)41 Pancreas(0.59)41 Bone(0.50)41 Myeloid(0.24)41 Brain(0.22)41 Blood And Bone Marrow58 Publications for Atransferrinemia Curated publications related to Atransferrinemia There are 14 Curated publications related to Atransferrinemia: Click to expand publication abstracts Show entries Search: \| \| # \| Title \| Authors \| PMID \| Year \| --- --- --- \| \| \| # \| Title \| Authors \| PMID \| Year \| --- --- --- \| \| \| 1 \| Studies on familial hypotransferrinemia: unique clinical course and molecular pathology.57 65 60 6 \| Hayashi A...Shimizu A \| 8317485 \| 1993 \| \| \| 2 \| Molecular characterization of a case of atransferrinemia.65 60 6 \| Beutler E...Fairbanks VF \| 11110675 \| 2000 \| \| \| 3 \| A family of congenital atransferrinemia.65 60 6 \| Goya N...Ushio B \| 4625559 \| 1972 \| \| \| 4 \| Effects of plasma transfusion on hepcidin production in human congenital hypotransferrinemia.65 6 \| Trombini P...Piperno A \| 17768112 \| 2007 \| \| \| 5 \| Molecular characterization of a third case of human atransferrinemia.65 6 \| Knisely AS...Beutler E \| 15466165 \| 2004 \| \| \| 6 \| Biochemical and genetic defects underlying human congenital hypotransferrinemia.65 6 \| Goldwurm S...Biondi A \| 11920219 \| 2000 \| \| \| 7 \| Tissue distribution and clearance kinetics of non-transferrin-bound iron in the hypotransferrinemic mouse: a rodent model for hemochromatosis.65 60 \| Craven CM...Kaplan J \| 3472216 \| 1987 \| \| \| 8 \| [Congenital atransferrinemia in a 11-month-old child].65 6 \| Cap J...Mayerova A \| 5711079 \| 1968 \| \| \| 9 \| [Atransferrinemia in the nephrotic syndrome with special reference to iron and protein metabolism].65 60 \| Heilmeyer L...Haas R \| 5829804 \| 1965 \| \| \| 10 \| [Congenital atransferrinemia in a 7-year-old girl].65 60 \| HEILMEYER L...SCHULTZE HE \| 13906010 \| 1961 \| Showing 1 to 10 of 14 entries 1 2 Text-mined publications related to Atransferrinemia There are 41 Text-mined publications related to Atransferrinemia: Click to expand publication abstracts Show entries Search: \| \| # \| Title \| Authors \| PMID \| Year \| --- --- --- \| \| \| # \| Title \| Authors \| PMID \| Year \| --- --- --- \| \| \| 1 \| Severe hypochromic microcytic anemia in a patient with congenital atransferrinemia.57 65 \| Shamsian BS...Eghbali A \| 19579082 \| 2009 \| \| \| 2 \| Label-free and ultrasensitive electrochemical transferrin detection biosensor based on a glassy carbon electrode and gold nanoparticles.65 \| Rabbani G...Zakri W \| 38000589 \| 2024 \| \| \| 3 \| Reversible atransferrinemia in a patient with chronic enteropathy: is transferrin mandatory for iron transport?65 \| Raynor A...Manceau H \| 36627980 \| 2023 \| \| \| 4 \| Fresh Frozen Plasma Plus Iron Therapy in Congenital Hypotransferrinemia in the Second Decade: A Dynamic Approach to Maintaining Hematological Stability 65 \| Aslan D \| 34792309 \| 2022 \| \| \| 5 \| Inherited microcytic anemias.65 \| Cappellini MD...Iolascon A \| 33275715 \| 2020 \| \| \| 6 \| A new case of congenital atransferrinemia with a novel splice site mutation: c.293-63del.65 \| Dabboubi R...Messaoud T \| 32028041 \| 2020 \| \| \| 7 \| Inherited iron overload disorders.65 \| Piperno A...Mariani R \| 32258529 \| 2020 \| \| \| 8 \| Transferrin and transferrin receptors update.65 \| Kawabata H \| 29969719 \| 2019 \| \| \| 9 \| A computational model to understand mouse iron physiology and disease.65 \| Parmar JH...Mendes P \| 30608934 \| 2019 \| \| \| 10 \| Addition of oral iron to plasma transfusion in human congenital hypotransferrinemia: A 10-year observational follow-up with the effects on hematological parameters and growth.65 \| Aslan D \| 28895280 \| 2018 \| Showing 1 to 10 of 41 entries 1 2 3 4 5 Variations for Atransferrinemia ClinVar genetic disease variations for Atransferrinemia There are 76 ClinVar genetic disease variation(s) for Atransferrinemia: Show entries Search: \| # \| Gene \| Name \| Type \| Significance \| ClinVar ID \| dbSNP ID \| Position \| --- --- --- --- \| \| # \| Gene \| Name \| Type \| Significance \| ClinVar ID \| dbSNP ID \| Position \| --- --- --- --- \| \| 1 \| TF \| NM_001063.4(TF):c.1429G>C (p.Ala477Pro) \| SNV \| Pathogenic \| 12620 \| rs121918679 \| GRCh37: 3:133,485,220-133,485,220 GRCh38: 3:133,766,376-133,766,376 \| \| 2 \| TF \| NM_001063.4(TF):c.229G>A (p.Asp77Asn) \| SNV \| Pathogenic \| 12621 \| rs121918681 \| GRCh37: 3:133,472,451-133,472,451 GRCh38: 3:133,753,607-133,753,607 \| \| 3 \| TF \| NM_001063.4(TF):c.1180G>A (p.Glu394Lys) \| SNV \| Pathogenic \| 12623 \| rs121918680 \| GRCh37: 3:133,478,150-133,478,150 GRCh38: 3:133,759,306-133,759,306 \| \| 4 \| TF \| NM_001063.4(TF):c.562_564delinsAA (p.Gln188fs) \| INDEL \| Pathogenic \| 12619 \| rs2530589860 \| GRCh37: 3:133,474,266-133,474,268 GRCh38: 3:133,755,422-133,755,424 \| \| 5 \| TF \| NM_001063.4(TF):c.1623-63del \| DEL \| Likely Pathogenic \| 694735 \| rs8177291 \| GRCh37: 3:133,489,289-133,489,289 GRCh38: 3:133,770,445-133,770,445 \| \| 6 \| TF \| NM_001063.4(TF):c.1825C>T (p.Arg609Trp) \| SNV \| Likely Pathogenic \| 218294 \| rs773139494 \| GRCh37: 3:133,494,414-133,494,414 GRCh38: 3:133,775,570-133,775,570 \| \| 7 \| TF \| NM_001063.4(TF):c.147del (p.Pro50fs) \| DEL \| Likely Pathogenic \| 3588594 \| \| GRCh37: 3:133,467,358-133,467,358 GRCh38: 3:133,748,514-133,748,514 \| \| 8 \| TF \| NM_001063.4(TF):c.216+2T>C \| SNV \| Likely Pathogenic \| 3588595 \| \| GRCh37: 3:133,467,430-133,467,430 GRCh38: 3:133,748,586-133,748,586 \| \| 9 \| TF \| NM_001063.4(TF):c.325+1G>A \| SNV \| Likely Pathogenic \| 3588596 \| \| GRCh37: 3:133,472,548-133,472,548 GRCh38: 3:133,753,704-133,753,704 \| \| 10 \| TF \| NM_001063.4(TF):c.394C>T (p.Arg132Ter) \| SNV \| Likely Pathogenic \| 3588597 \| \| GRCh37: 3:133,473,407-133,473,407 GRCh38: 3:133,754,563-133,754,563 \| Showing 1 to 10 of 76 entries 1 2 3 4 5 … 8 UniProtKB/Swiss-Prot genetic disease variation(s) for Atransferrinemia There are 2 UniProtKB/Swiss-Prot genetic disease variations for Atransferrinemia: Search: \| # \| Symbol \| AA change \| Variation ID \| SNP ID \| --- --- \| # \| Symbol \| AA change \| Variation ID \| SNP ID \| --- --- \| 1 \| TF \| p.Ala477Pro \| VAR_012997 \| rs121918679 \| \| 2 \| TF \| p.Asp77Asn \| VAR_038810 \| rs121918681 \| Showing 1 to 2 of 2 entries Expression for Atransferrinemia Search GEO for disease gene expression data for Atransferrinemia Pathways for Atransferrinemia Pathways related to Atransferrinemia according to GeneCards Suite gene sharing There are 24 Pathways related to Atransferrinemia: Click to expand SuperPathway members. High statistical significance Show entries Search: \| \| # \| Pathway \| Gene Set Enrichment \| Score \| Affiliating genes \| --- --- --- \| \| \| # \| Pathway \| Gene Set Enrichment \| Score \| Affiliating genes \| --- --- --- \| \| \| 1 \| DNA Methylation Pathway \| 1/6 \| 3.35 \| TF \| \| \| 2 \| Iron metabolism disorders \| 1/17 \| 2.90 \| TF \| \| \| 3 \| Mechanisms of Fibrosis Pathway \| 1/21 \| 2.81 \| TF \| \| \| 4 \| Transferrin endocytosis and recycling \| 1/31 \| 2.64 \| TF \| \| \| 5 \| EPHB forward signaling \| 1/34 \| 2.60 \| TF \| \| \| 6 \| Pluripotent stem cell differentiation pathway \| 1/48 \| 2.45 \| TF \| \| \| 7 \| Iron uptake and transport \| 1/58 \| 2.37 \| TF \| \| \| 8 \| 3q29 copy number variation syndrome \| 1/62 \| 2.34 \| TF \| \| \| 9 \| Ferroptosis \| 1/64 \| 2.33 \| TF \| \| \| 10 \| HIF-1-alpha transcription factor network \| 1/65 \| 2.32 \| TF \| Showing 1 to 10 of 24 entries 1 2 3 GO Terms for Atransferrinemia Cellular components related to Atransferrinemia according to GeneCards Suite gene sharing There are 24 Cellular components related to Atransferrinemia: High statistical significance Show entries Search: \| # \| Name \| GO ID \| Gene Set Enrichment \| Score \| Affiliating genes \| --- --- --- \| \| # \| Name \| GO ID \| Gene Set Enrichment \| Score \| Affiliating genes \| --- --- --- \| \| 1 \| HFE-transferrin receptor complex \| GO:1990712 \| 1/8 \| 3.46 \| TF \| \| 2 \| Basal part of cell \| GO:0045178 \| 1/16 \| 3.16 \| TF \| \| 3 \| Clathrin-coated endocytic vesicle membrane \| GO:0030669 \| 1/70 \| 2.52 \| TF \| \| 4 \| Basal plasma membrane \| GO:0009925 \| 1/72 \| 2.51 \| TF \| \| 5 \| Clathrin-coated pit \| GO:0005905 \| 1/73 \| 2.50 \| TF \| \| 6 \| Endocytic vesicle \| GO:0030139 \| 1/73 \| 2.50 \| TF \| \| 7 \| Secretory granule lumen \| GO:0034774 \| 1/121 \| 2.28 \| TF \| \| 8 \| Blood microparticle \| GO:0072562 \| 1/145 \| 2.20 \| TF \| \| 9 \| Recycling endosome \| GO:0055037 \| 1/151 \| 2.18 \| TF \| \| 10 \| Late endosome \| GO:0005770 \| 1/168 \| 2.14 \| TF \| Showing 1 to 10 of 24 entries 1 2 3 Biological processes related to Atransferrinemia according to GeneCards Suite gene sharing There are 20 Biological processes related to Atransferrinemia: High statistical significance Show entries Search: \| # \| Name \| GO ID \| Gene Set Enrichment \| Score \| Affiliating genes \| --- --- --- \| \| # \| Name \| GO ID \| Gene Set Enrichment \| Score \| Affiliating genes \| --- --- --- \| \| 1 \| Regulation of iron ion transport \| GO:0034756 \| 1/3 \| 3.86 \| TF \| \| 2 \| Iron ion export across plasma membrane \| GO:1903988 \| 1/3 \| 3.86 \| TF \| \| 3 \| Cellular response to iron ion \| GO:0071281 \| 1/7 \| 3.49 \| TF \| \| 4 \| Positive regulation of bone resorption \| GO:0045780 \| 1/19 \| 3.06 \| TF \| \| 5 \| Positive regulation of receptor-mediated endocytosis \| GO:0048260 \| 1/22 \| 2.99 \| TF \| \| 6 \| Positive regulation of phosphorylation \| GO:0042327 \| 1/25 \| 2.94 \| TF \| \| 7 \| Positive regulation of cell motility \| GO:2000147 \| 1/25 \| 2.94 \| TF \| \| 8 \| Multicellular organismal-level iron ion homeostasis \| GO:0060586 \| 1/28 \| 2.89 \| TF \| \| 9 \| Iron ion transport \| GO:0006826 \| 1/32 \| 2.83 \| TF \| \| 10 \| Osteoclast differentiation \| GO:0030316 \| 1/49 \| 2.64 \| TF \| Showing 1 to 10 of 20 entries 1 2 Molecular functions related to Atransferrinemia according to GeneCards Suite gene sharing There are 9 Molecular functions related to Atransferrinemia: High statistical significance Search: \| # \| Name \| GO ID \| Score \| Gene Set Enrichment \| Affiliating genes \| --- --- --- \| \| # \| Name \| GO ID \| Score \| Gene Set Enrichment \| Affiliating genes \| --- --- --- \| \| 1 \| Iron chaperone activity \| GO:0034986 \| 3.85 \| 1/3 \| TF \| \| 2 \| Ferric iron binding \| GO:0008199 \| 3.29 \| 1/11 \| TF \| \| 3 \| Transferrin receptor binding \| GO:1990459 \| 3.25 \| 1/12 \| TF \| \| 4 \| Ferrous iron binding \| GO:0008198 \| 2.93 \| 1/25 \| TF \| \| 5 \| Iron ion binding \| GO:0005506 \| 2.16 \| 1/146 \| TF \| \| 6 \| Transmembrane transporter binding \| GO:0044325 \| 2.16 \| 1/148 \| TF \| \| 7 \| Enzyme binding \| GO:0019899 \| 1.71 \| 1/411 \| TF \| \| 8 \| Metal ion binding \| GO:0046872 \| 0.71 \| 1/4,123 \| TF \| \| 9 \| Protein binding \| GO:0005515 \| 0.17 \| 1/14,498 \| TF \| Showing 1 to 9 of 9 entries Sources for Atransferrinemia 1BitterDB 2Boster Bio 3CDC 4Cell Signaling Technology 5ClinicalTrials 6ClinVar 7 CNVD 8Cochrane Library 9COSMIC 10dbSNP 11DGIdb 12Disease Ontology 13 diseasecard 14DiseaseEnhancer 15DISEASES 16DrugBank 17EFO 18ExPASy 19FMA 20GARD 21GeneAnalytics 22GeneCards 23 GeneGo (Thomson Reuters) 24GeneHancer via GeneCards 25GeneReviews 26GeneTex 27GenomeRNAi 28GEO DataSets 29GO 30GTR 31HMDB 32HPO 33ICD10 34 ICD10 via Orphanet 35ICD11 36 ICD11 via Orphanet 37ICD9CM 38IUPHAR 39 LifeMap Discovery 40LncRNADisease 41MalaCards 42MedChemExpress 43MedGen 44MedlinePlus 45MedlinePlus Genetics 46MeSH 47 MESH via Orphanet 48MGI 49miR2Disease 50MONDO 51NCBI Bookshelf 52NCI 53NCIt 54 NDF-RT 55NIH Clinical Center 56NINDS 57 Novoseek 58ODiseA 59 OMIM via Orphanet 60OMIM®(Updated 29/06/2025) 61Orphanet 62PathCards 63PharmGKB 64PubChem 65PubMed 66 PubMed Health 67QIAGEN 68R&D Systems 69Reactome 70Sino Biological 71SNOMED-CT 72 SNOMED-CT via HPO 73Tocris 74UMLS 75 UMLS via Orphanet 76UniProtKB/Swiss-Prot 77WikiPathways 78Wikipedia THE MALACARDS DATABASE IS FOR RESEARCH USE ONLY, DOES NOT PROVIDE MEDICAL ADVICE AND SHOULD NOT BE USED IN DIAGNOSTIC PROCEDURES. 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9284	https://www.youtube.com/watch?v=fm6UMo8W-rk	Counting Number of Possible Functions Given Constraints Gresty Academy 57300 subscribers 12 likes Description 275 views Posted: 1 Mar 2024 Occasionally Maths Olympiads and College Entrance Tests ask a question along the lines of 'how many possible functions are there from set A to set B which satisfy certain constraints?'. Frankly there is no better way to understand how to do these types of questions than simply to attempt as many of them as possible for familiarity and practice. In this video we answer four such questions, all from recent JEE Main exams, and as we see during the video there is no 'simple formula' which covers them all. Each question is different in its own constraint, though there are some similarities which we can take advantage of. Counting is not easy - it is quite difficult not to double count or omit possibilities, so care really needs to be taken. There are certainly overlaps with these types of questions and other types of 'counting'. As such, it may be beneficial to try out some of our other 'Counting' videos in our playlist 'Counting' For more videos on JEE Main see our playlist 'JEE Aspirant' For more videos on functions generally see our playlist 'Functions' For more videos on Maths Olympiads, see our playlist 'Maths Olympiads' For more videos on College Entrance Tests see our playlist 'UPCAT and Other CETs' 3 comments Transcript: So today we're going to be looking at counting functions um and Counting is nowhere near as easy as you might think um because it's very easy to double count or Miss certain possibilities what's it all about this counting functions well just to to recap uh a function is defined as every element in the input or the domain here has to match with exactly one element in the output Co domain here okay so um let's imagine that we have M elements in here and we have n elements in here then given no constraints whatsoever on how the function can be produced we would be able to count um n to the power of M possible functions so just to give an example let's imagine that we have this as our domain 1 2 and three uh and we have as our co- domain a b c and d then there are um well basically there would be 4 to the^ of 3 which equals um 64 possible functions um now just as a matter of Interest one of those functions would be that one could map to a two could map to a and three could map to a and there's absolutely nothing wrong with that there's no constraints because each of these is mapping to one it doesn't matter if it's the same one A different one it doesn't matter we're not dealing with injective uh and surjective functions or anything all we're dealing with is um a function so it must match to one so what we're going to do is we're going to do four different questions um and they are a little bit tricky so I think uh the best way to deal with these is just to practice the different sorts okay so let's go on to question one here's question one okay from it's j main July 22 shift 1 anybody who'd like to have a go at that just pause the video video okay now the main problem with with these actually is understanding what the question is actually saying so what is asking for is how many is the total number of functions here is the uh domain 1 2 3 and four and here is the co-domain 1 2 3 4 five and six and they're giving us this constraint that fub1 add FS2 minus F3 uh is is um sorry F1 and FS2 is equal to F3 Okay so what we need to do is we need to look and see how many uh functions um satisfy this constraint here and the best thing to do is always look at the biggest number well the biggest number clearly if this is F1 and this is F2 well the biggest number is going to be F of3 now given that we only have a constraint on fub1 F2 and F3 that means that the number four can map to anything so we know always that F of four can be anything from 1 2 3 4 4 five and six so we'll use that later on okay so first of all let's have a look um at F3 well F3 cannot map to one so it's F1 add F2 equals F3 that's the constraint that we're dealing with well F of three clearly cannot map to one because if F3 maps to one well what are F1 and F2 going to map to because the minimum that they can map to 2 is 1 well 1 + 1 is 2 so F3 cannot possibly be equal to one so let's have a look at what happens if F3 equals 2 well if if F3 is two well clearly that means fub1 must be one and FS2 must be one so there is only one possibility if F map if sorry if three maps to two there can only be one possibility which is one must map to one two must map to one and of course remember four can map to anything so if f is three that means we have one time six possible functions what about if F maps to uh F Maps three to three okay well if it Maps three to three well we have two possibilities one could map to one and two could map to two that add that would add up or one could map to two and two could map to one that would that would work with this constraint here so clearly here here there are two possibilities Times by remember four four can be anything so that would be 2 6 okay so let's have a look at uh what happens if F of three maps to four okay well F of one could be one and F of two could be three that would work or F of one could map to two uh F of two would be two or F of one is three and F of two is one all three of those work okay okay so there's three there Times by of course four can float completely freely to any of the six that would be 3 6 okay now what about F of 3 = 5 u f of 3 equal 5 well then um uh if F of 3 equals 5 then we could clearly have F of one being one and F of two being four or F of one being two and F of two being three or F of two being F of one being three and F of uh uh two being two or F of one being four and F of two being one so clearly there are four 6 there and finally if we look at if we allow three to map to six well then F of one could be 1 2 3 4 or five and F of two could be 5 4 3 2 and one uh so there's five possibilities there so that would be 5 6 so the total number of possible functions uh whereby this this constraint here FS1 add F2 = F3 works is 1 6 add 2 6 add 3 6 add 4 6 add 5 6 which is uh 5 4 is basically 15 6 5 4 yep which equals 90 so the correct answer is B 90 okay so that's that question done what about the next question um pause the video If you uh desire to have a go at that question okay so the the problem with a lot of these questions is actually trying to understand what on Earth it's saying so let's have a look so we've got the what they're asking for is the number of functions from the set a to the set B under this constraint here okay so the set a let's have a look at what that is first well a is the set of X contained in N such that x^2 - 10 x 9 is less than or equal to zero okay well x^2 - 10 x 9 we can write that as uh uh x - um x - 9 x -1 uh and for that to be less than or equal to zero clearly X has to be between 1 and 9 so basically a we can rewrite a as being the set 1 2 3 4 5 6 7 8 and nine that's the set a and let's have a look at the set B well the set B is the set of n squ so that's 1 4 9 16 20 36 49 etc etc etc um so now what's it asking us so basically what it is saying is f ofx must be less than or equal to x + 3^2 add 1 for all of the X in a okay well let's have a look then so basically let's have a look uh at the number one so this is uh number one so F of one must be less than or equal to and what we're going to do do here is stick one in here one minus sorry one minus sorry my apologies one minus 3^ 2ar add one all I've done is I've put one into here because that's the constraint which equals uh five so F of one must be uh so F of one can be either one or four if it maps to one or four that is less than or equal to 5 okay now what about a = 2 well F of 2 has got to be less than or equal to uh 2 - 3^ 2 add 1 which equals 2 so therefore uh two must get mapped to one only what about three uh F of three uh must be less than or equal to 3 - 3 S add 1 which equals 1 so F of three must be less than or equal to 1 so clearly three can also only map to one so that that can map to two one or four that can map to one only that can map to one only that's the number of mappings that there can be and this is the actual number here uh what about four well four F of four must be less than or equal to 4 - 3^ 2 add 1 which is two so therefore four can only map two again to one oh sorry four can map to uh yeah one okay so that can only map to one uh what about five let's move up here well F of five must be less than or equal to 5 - 3^ 2 add 1 which equals 5 so therefore F of five could be 1 or four which is two different possibilities uh then six well F of 6 must be less than or equal to 6 - 3 S add 1 which is 10 so therefore six could get mapped to one or four or 9 which is three possibilities now in ual fact we don't need to fill in for 7 8 and 9 it is clear that 7 8 and 9 F of 7 is going to map to 1 or 4 or 9 or 16 and F of 8 is going to be 1 or four or 9 or 16 or 25 and F of 9 is going to be 1 or 4 or 9 or 16 or 25 or 36 and so basically the total number uh of possible functions that there are is going to be the multiplying of all of these which is 6 factorial 2 so it's 6 factorial 2 sorry time 2 well 6 factorial is 720 2 = 1,440 possible functions uh such that that constraint works okay that's number two you can see they're a bit fiddly you just have to try and understand what it is the question is asking let's have a look at question number three okay so question three if you want to pause the video now please do so okay so this one we have uh here is our domain 1 2 3 and four and here is the co domain the set of a contained in uh the uh integers such that the modulus of a is less than or equal to 8 will clearly that will be the set - 8 - 7 - 6 - 5 D D D D d0 D D D D D 7 8 that's that set there okay so it's saying that F of n add 1 / n f of n + 1 must equal 1 for all n in 1 2 and 3 okay so basically again what we want to do is we want to look at the the number which is the highest and the most constrained so obviously what we want to do is let's just put in say for example uh one in here and see what's going on that' be F of 1 and 1 over one F of two must equal one and let's just put in two just so we can get a good idea what on Earth they're asking F of three must equal one and F of3 add a thir F of4 must equal one okay right so basically from this one we know that F of four must be uh divisible divisible by 3 or equal to zero and from this one here we know that F of three must be even or equal to zero so what we want to do is we want to look at F of four okay so let's look at F of four well F of four if it's got to be divisible by three then F of 4 given that it can only be mapped to one of these it must be either 63 0 -3 or - 6 now if F of four maps to six then from oops sorry then from here F of three must be well the third 6 ID 3 is 2 and so F of3 must therefore map to minus1 but F of3 cannot map to minus1 because F of3 must be even and so therefore there are zero possibilities when F of four is six what about when F of four is three well if F of four is three then the third of f of four is one therefore F of three must be zero which is fine f of three can be zero if F of three is zero that means that half F of 3 is zero which means F of 2 must equal one in order to satisfy this and if F of 2 equal 1 then F of 1 must be zero but that is a possible solution so we have found one possible function now what about if F of four equals z well if F of four equals 0 then a third of0 is zero that means that F of3 must be one from this third equation here well that can't be possible because F of three is even so we can scrp that one out what about if F of 4 is minus 3 uh well a thir of -3 is -1 which would mean that F of 3 must be 2 well that's possible because that is an even number if F of 3 is two half of f of3 is 1 that means F of two must be zero and if F of two is zero that means F of one must be one and that works that's another possible so we have another possible and finally all we need to do now is check if F of four is - 6 well if F of four is- 6 then a thir of - 6 is -2 which would mean that um f of three would have to be three and F of three can't be three because F of three must be even so that can't work as well so therefore there are two possible functions that satisfy this constraint and therefore the answer is D okay so let's go on to the last one um you can see they're all different um and you just have to really try and examine the question so um anyone who wants to do this try and pause the video Okay so we've got a is uh 1 2 3 4 5 that's the domain and B is 1 2 3 4 5 and six okay then the number of functions a to be satisfying fub1 FS2 = F4 -1 is equal to what okay so we have to find the number functions the first thing we should notice is that three and five can both map to anything because they're not included in our constraint so therefore we know that three can map to six different things and five can map to six different things so we know that between them that's going to be 6 six possibles now we only have to look at F of four so let's have a look at F of four and then let's have a look at F of 4us one and then let's have a look at F of one and F of two like this okay now let's just put F of one F of four well let's have a look 1 2 3 4 5 and six okay now if F of four equals 1 then F of four - one uh then F of if F of four equals 1 F of 4 - 1 is zero well okay well clearly F1 out FS2 can't be equal to zero because all of the the numbers are positive in the co- domain so that we can write off what about if F of four is two well for the same logic that means F of 4 - 1 is 1 well F of one and F of two can't possibly equal F of can't possibly equal one and so we can knock that one out if F of 4 is three then F of 4 - 1 is 2 well basically that means that F of one must m one and F of two must be one that's the only possible that can satisfy this constraint which is one possible what about um f of four three well basically F of one could be one and F of two could be two or F of two could be two and F of one F of two could be one so that's two possible and then five well we can so basically if F of four is five F of four - 1 is four so F of one and F of two could be one two and three or 3 2 and 1 which is three posses and then finally for six clearly it's going to be 1 2 3 4 or 4 3 2 1 which is another four posses so basically adding these up 4 3 7 1 is 10 times by 36 because three and five can map to anything so it' be 10 36 which equals 360 possible functions uh which satisfy this constraint okay now all of these four question were just normal functions there was no injective functions or surjective functions or bjective functions a lot of these questions also ask on Counting functions for those types injective subjective and bjective and the next few videos we're going to do are going to be on that um I hope you found this useful um if you have please like the video and um subscribe to the Gusty Academy YouTube channel thank you
9285	https://math.stackexchange.com/questions/175666/what-function-satisfies-x2-fx-f1-x-2x-x4	calculus - What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? - 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 What function satisfies x 2 f(x)+f(1−x)=2 x−x 4 x 2 f(x)+f(1−x)=2 x−x 4? Ask Question Asked 13 years, 2 months ago Modified13 years, 1 month ago Viewed 12k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. What function satisfies x 2 f(x)+f(1−x)=2 x−x 4 x 2 f(x)+f(1−x)=2 x−x 4? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution. calculus algebra-precalculus functional-equations Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jul 27, 2012 at 0:30 J. M. ain't a mathematician 76.7k 8 8 gold badges 222 222 silver badges 347 347 bronze badges asked Jul 26, 2012 at 23:53 I. J. KennedyI. J. Kennedy 4,158 4 4 gold badges 29 29 silver badges 42 42 bronze badges 1 1 Have you tried exploting the symmetry?Pedro –Pedro♦ 2012-07-26 23:58:24 +00:00 Commented Jul 26, 2012 at 23:58 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 17 Save this answer. Show activity on this post. We start from x 2 f(x)+f(1−x)=2 x−x 4.x 2 f(x)+f(1−x)=2 x−x 4. Replace x x by 1−x 1−x. Then 1−x 1−x gets replaced by x x. So (1−x)2 f(1−x)+f(x)=2(1−x)−(1−x)4.(1−x)2 f(1−x)+f(x)=2(1−x)−(1−x)4. Two linear equations in two unknowns, f(x)f(x) and f(1−x)f(1−x). Solve for f(x)f(x). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 27, 2012 at 0:08 André NicolasAndré Nicolas 515k 47 47 gold badges 584 584 silver badges 1k 1k bronze badges Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. Another, more mechanical approach: First of all, just by counting degrees it's clear that any polynomial solution f f must be quadratic, so we can write f(x)=a x 2+b x+c f(x)=a x 2+b x+c. What's more, plugging in x=0 x=0 gives 0 2⋅f(0)+f(1)=2⋅0−0 4 0 2⋅f(0)+f(1)=2⋅0−0 4, or f(1)=0 f(1)=0, and plugging in x=1 x=1 gives 1 2⋅f(1)+f(0)=2⋅1−1 4 1 2⋅f(1)+f(0)=2⋅1−1 4, or f(0)=1 f(0)=1, so we know that c=1 c=1 and a+b+c=0 a+b+c=0, or b=−(1+a)b=−(1+a). Using these, we can rewrite the quadratic as f(x)=a x 2−(a+1)x+1 f(x)=a x 2−(a+1)x+1. Now, it's obvious that x 2 f(x)=a x 4+s m a l l e r t e r m s x 2 f(x)=a x 4+s m a l l e r t e r m s, and f(1−x)f(1−x) will also be only quadratic — so for the LHS to equal 2 x−x 4 2 x−x 4, it must be that a=−1 a=−1 and f(x)=1−x 2 f(x)=1−x 2; all that's left is to compute the left hand side in full and see that it solves the equation. Note that this approach doesn't prove that such an f f is unique, only that it exists and that it's unique among polynomial solutions — you need one of the other approaches to show that this is the only function that works. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jul 27, 2012 at 0:30 answered Jul 27, 2012 at 0:22 Steven StadnickiSteven Stadnicki 53.3k 9 9 gold badges 91 91 silver badges 155 155 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. As I commented, you can exploit the symmetry. f(1−x)+x 2 f(x)=2 x−x 4 f(x)+(1−x)2 f(1−x)=2(1−x)−(1−x)4 f(1−x)+x 2 f(x)=2 x−x 4 f(x)+(1−x)2 f(1−x)=2(1−x)−(1−x)4 Now you get f(1−x)+x 2 f(x)=2 x−x 4 f(1−x)=2(1−x)−(1−x)4−f(x)(1−x)2 f(1−x)+x 2 f(x)=2 x−x 4 f(1−x)=2(1−x)−(1−x)4−f(x)(1−x)2 So all you have to do is solve for f(x)f(x) in 2 x−x 4−x 2 f(x)=2(1−x)−(1−x)4−f(x)(1−x)2 2 x−x 4−x 2 f(x)=2(1−x)−(1−x)4−f(x)(1−x)2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jul 27, 2012 at 0:43 answered Jul 27, 2012 at 0:11 Pedro♦Pedro 126k 19 19 gold badges 239 239 silver badges 406 406 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Given, x 2 f(x)+f(1−x)=2 x−x 4 x 2 f(x)+f(1−x)=2 x−x 4 f(−1)=0 f(−1)=0 f(0)=1 f(0)=1 f(1)=0 f(1)=0 f(2)=−3 f(2)=−3 f(3)=−8 f(3)=−8 f(4)=−15 f(4)=−15 By careful observation one can see that the difference between the terms vanishes after the second stage i.e. f(x)f(x) is of the form a x 2+b x+c a x 2+b x+c i.e. f(x)=a x 2+b x+c f(x)=a x 2+b x+c -------(1) i.e. f(0)=c=1 f(0)=c=1 i.e. c=1 c=1 -------(2) f(1)=a+b+c=0 f(1)=a+b+c=0 i.e. f(1)=a+b+1=0 f(1)=a+b+1=0 i.e. a+b=−1 a+b=−1 -------(3) f(2)=4 a+2 b+c=−3 f(2)=4 a+2 b+c=−3 i.e. 4 a+2 b+1=−3 4 a+2 b+1=−3 i.e. 4 a+2 b=−2 4 a+2 b=−2 i.e. 2 a+b=−2 2 a+b=−2 -------(4) Solve equations (3)(3) and (4)(4) to get a=−1 a=−1 and b=0 b=0 i.e. f(x)=−1 x 2+0 x+1 f(x)=−1 x 2+0 x+1 i.e. f(x)=−x 2+1 f(x)=−x 2+1 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 27, 2012 at 3:52 HOLYBIBLETHEHOLYBIBLETHE 2,804 10 10 gold badges 33 33 silver badges 51 51 bronze badges 2 This derivation is based on algebra HOLYBIBLETHE –HOLYBIBLETHE 2012-07-27 03:53:33 +00:00 Commented Jul 27, 2012 at 3:53 1 NOTE : a, b, c are constant HOLYBIBLETHE –HOLYBIBLETHE 2012-07-27 03:56:13 +00:00 Commented Jul 27, 2012 at 3:56 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. There's an algebraic way (which may be a bit messy without using a CAS). It involves exploiting the symmetry of the problem, as Peter Tamaroff suggested. x 2 f(x)+f(1−x)(1−x)2 f(1−x)+f(x)−(1−x)2 x 2 f(x)−(1−x)2 f(1−x)Now add to Eqn.2:f(x)−(1−x)2 x 2 f(x)f(x)[1−(1−x)2 x 2]f(x)======2 x−x 4 2(1−x)−(1−x)4 subs.1−x for x−(1−x)2(2 x−x 4)Eqn.1 times−(1−x)2 2(1−x)−(1−x)4−(1−x)2(2 x−x 4)2(1−x)−(1−x)4−(1−x)2(2 x−x 4)factoring out f(x)2(1−x)−(1−x)4−(1−x)2(2 x−x 4)1−(1−x)2 x 2.x 2 f(x)+f(1−x)=2 x−x 4(1−x)2 f(1−x)+f(x)=2(1−x)−(1−x)4 subs.1−x for x−(1−x)2 x 2 f(x)−(1−x)2 f(1−x)=−(1−x)2(2 x−x 4)Eqn.1 times−(1−x)2 Now add to Eqn.2:f(x)−(1−x)2 x 2 f(x)=2(1−x)−(1−x)4−(1−x)2(2 x−x 4)f(x)[1−(1−x)2 x 2]=2(1−x)−(1−x)4−(1−x)2(2 x−x 4)factoring out f(x)f(x)=2(1−x)−(1−x)4−(1−x)2(2 x−x 4)1−(1−x)2 x 2. This is an unreduced final answer, which (according to sage) cancels out to leave f(x)=1−x 2 f(x)=1−x 2. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 27, 2012 at 0:21 Shaun AultShaun Ault 9,029 26 26 silver badges 30 30 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. f(x)=1−x 2 f(x)=1−x 2 inserting that gives: x 2 f(x)+f(1−x)=====x 2(1−x 2)+(1−(1−x)2)x 2−x 4+(1−(1−2 x+x 2))x 2−x 4+(1−1+2 x−x 2)x 2−x 4+2 x−x 2 2 x−x 4 x 2 f(x)+f(1−x)=x 2(1−x 2)+(1−(1−x)2)=x 2−x 4+(1−(1−2 x+x 2))=x 2−x 4+(1−1+2 x−x 2)=x 2−x 4+2 x−x 2=2 x−x 4 as required. But I admit: I had some help... But you can also write it as x 4+f(x)x 2−2 x−f(1−x)=0 x 4+f(x)x 2−2 x−f(1−x)=0 and try to solve the depressed quartic equation... Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 7, 2012 at 20:10 answered Jul 26, 2012 at 23:58 draks ...draks ... 18.8k 8 8 gold badges 67 67 silver badges 209 209 bronze badges 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 calculus algebra-precalculus functional-equations See similar questions with these tags. 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9286	https://www.andreaminini.net/math/intersection-theorem-for-planes	Intersection Theorem for Planes - Andrea Minini Loading [MathJax]/jax/output/CommonHTML/jax.js Geometry \| Add to favorites) Intersection Theorem for Planes When two distinct planes intersect in space at a point P, they share a line r that passes through that point. In simpler terms, two intersecting planes cannot meet at just a single point. Instead, they always share a common line. The line shared by the two planes consists of all points that lie on both planes simultaneously. This result stems from fundamental principles of three-dimensional geometry, which state that the intersection of two non-parallel planes in 3D space always forms a line. Proof Let’s examine two planes, α and β, that intersect at a point P. P∈α∩β This means that point P is part of both planes. Next, consider two points, A and B, that lie on the same plane α but are located on opposite sides of plane β. Draw a line segment ¯A B connecting points A and B. Since the planes intersect, and because the segment ¯A B spans across the two half-planes of α, it must cross plane β at some point C. At this stage, we’ve identified two points shared by the two planes: the original point P and the newly found point C. A single straight line r can always be drawn through two points, and in this case, that line lies on both planes. This proves that if two planes intersect at a single point P, they must also share a line r passing through that point. And so, the theorem is established. Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them. Andrea Minini - piva 09286581005 - email: info@andreaminini.com - PEC andreaminini@pec.it \| privacy & cookie \| References \| Personal Knowledge Base \| Planes (Geometry) Plane Half-Plane Parallel, coincident, and intersecting planes Perpendicular planes Parallel planes Angle between a line and a plane Plane equation Vector, parametric, and cartesian equation of a plane Equation for three non-collinear points Intercept form of a plane equation Distance between a point and a plane Distance between a line and a parallel plane Pencil of planes Theorems Intersection theorem for planes
9287	https://www.analog.com/media/en/technical-documentation/data-sheets/LTC3108.pdf	LTC3108 1 Rev. D For more information www.analog.com TYPICAL APPLICATION DESCRIPTION Ultralow Voltage Step-Up Converter and Power Manager The LTC®3108 is a highly integrated DC/DC converter ideal for harvesting and managing surplus energy from extremely low input voltage sources such as TEGs (ther moelectric generators), thermopiles and small solar cells. The step-up topology operates from input voltages as low as 20mV. The LTC3108 is functionally equivalent to the LTC3108-1 except for its unique fixed VOUT options. Using a small step-up transformer, the LTC3108 provides a complete power management solution for wireless sensing and data acquisition. The 2.2V LDO powers an external microprocessor, while the main output is pro grammed to one of four fixed voltages to power a wire less transmitter or sensors. The power good indicator signals that the main output voltage is within regulation. A second output can be enabled by the host. A storage capacitor provides power when the input voltage source is unavailable. Extremely low quiescent current and high efficiency design ensure the fastest possible charge times of the output reservoir capacitor. The LTC3108 is available in a small, thermally enhanced 12-lead (3mm × 4mm) DFN package and a 16-lead SSOP package. Wireless Remote Sensor Application Powered From a Peltier Cell FEATURES APPLICATIONS n n Operates from Inputs of 20mV n n Complete Energy Harvesting Power Management System n n Selectable VOUT of 2.35V, 3.3V, 4.1V or 5V n n LDO: 2.2V at 3mA n n Logic Controlled Output n n Reserve Energy Output n n Power Good Indicator n n Uses Compact Step-Up Transformers n n Small 12-Lead (3mm × 4mm) DFN or 16-Lead SSOP Packages n n Remote Sensors and Radio Power n n Surplus Heat Energy Harvesting n n HVAC Systems n n Industrial Wireless Sensing n n Automatic Metering n n Building Automation n n Predictive Maintenance 3108 TA01a C1 20mV TO 500mV C2 SW VS2 VS1 VOUT2 PGOOD 2.2V 470µF PGD VLDO VSTORE + VOUT VOUT2_EN LTC3108 VAUX GND 0.1F 6.3V 5V 3.3V 1µF 1nF 220µF 1:100 330pF SENSORS RF LINK µP 2.2µF + + + THERMOELECTRIC GENERATOR VOUT Charge Time All registered trademarks and trademarks are the property of their respective owners. VIN (mV) TIME (sec) 10 1 100 1000 0 3108 TA01b 0 100 150 250 50 200 300 350 400 VOUT = 3.3V COUT = 470µF 1:100 Ratio 1:50 Ratio 1:20 Ratio Document Feedback LTC3108 2 Rev. D For more information www.analog.com ABSOLUTE MAXIMUM RATINGS SW Voltage...................................................–0.3V to 2V C1 Voltage. ....................................................–0.3V to 6V C2 Voltage (Note 5). .........................................–8V to 8V VOUT2, VOUT2_EN...........................................–0.3V to 6V VAUX. .................................................... 15mA into VAUX (Note 1) 12 11 10 9 8 7 13 GND 1 2 3 4 5 6 SW C2 C1 VOUT2_EN VS1 VS2 VAUX VSTORE VOUT VOUT2 VLDO PGD TOP VIEW DE PACKAGE 12-LEAD (4mm × 3mm) PLASTIC DFN TJMAX = 125°C, θJA = 43°C/W EXPOSED PAD (PIN 13) IS GND, MUST BE SOLDERED TO PCB (NOTE 4) GN PACKAGE 16-LEAD PLASTIC SSOP NARROW TJMAX = 125°C, θJA = 110°C/W 1 2 3 4 5 6 7 8 TOP VIEW 16 15 14 13 12 11 10 9 GND VAUX VSTORE VOUT VOUT2 VLDO PGD GND GND SW C2 C1 VOUT2_EN VS1 VS2 GND PIN CONFIGURATION ELECTRICAL CHARACTERISTICS PARAMETER CONDITIONS MIN TYP MAX UNITS Minimum Start-Up Voltage Using 1:100 Transformer Turns Ratio, VAUX = 0V 20 50 mV No-Load Input Current Using 1:100 Transformer Turns Ratio; VIN = 20mV, VOUT2_ EN = 0V; All Outputs Charged and in Regulation 3 mA Input Voltage Range Using 1:100 Transformer Turns Ratio l VSTARTUP 500 mV The l denotes the specifications which apply over the full operating junction temperature range, otherwise specifications are for TA = 25°C (Note 2). VAUX = 5V, unless otherwise noted. ORDER INFORMATION LEAD FREE FINISH TAPE AND REEL PART MARKING PACKAGE DESCRIPTION TEMPERATURE RANGE LTC3108EDE#PBF LTC3108EDE#TRPBF 3108 12-Lead (4mm × 3mm) Plastic DFN –40°C to 125°C LTC3108IDE#PBF LTC3108IDE#TRPBF 3108 12-Lead (4mm × 3mm) Plastic DFN –40°C to 125°C LTC3108EGN#PBF LTC3108EGN#TRPBF 3108 16-Lead Plastic SSOP –40°C to 125°C LTC3108IGN#PBF LTC3108IGN#TRPBF 3108 16-Lead Plastic SSOP –40°C to 125°C Contact the factory for parts specified with wider operating temperature ranges. The temperature grade is identified by a label on the shipping container. Tape and reel specifications. Some packages are available in 500 unit reels through designated sales channels with #TRMPBF suffix. VS1, VS2, VAUX, VOUT, PGD.........................–0.3V to 6V VLDO, VSTORE.............................................–0.3V to 6V Operating Junction Temperature Range (Note 2). ..................................................–40°C to 125°C Storage Temperature Range...................–65°C to 125°C LTC3108 3 Rev. D For more information www.analog.com Note 1: Stresses beyond those listed under Absolute Maximum Ratings may cause permanent damage to the device. Exposure to any Absolute Maximum Rating condition for extended periods may affect device reliability and lifetime. Note 2: The LTC3108 is tested under pulsed load conditions such that TJ ≈ TA. The LTC3108E is guaranteed to meet specifications from 0°C to 85°C junction temperature. Specifications over the –40°C to 125°C operating junction temperature range are assured by design, characterization and correlation with statistical process controls. The LTC3108I is guaranteed over the full –40°C to 125°C operating junction temperature range. Note that the maximum ambient temperature is determined by specific operating conditions in conjunction with board layout, the rated thermal package thermal resistance and other environmental factors. The junction temperature (TJ) is calculated from the ambient temperature (TA) and power dissipation (PD) according to the formula: TJ = TA + (PD • θJA°C/W), where θJA is the package thermal impedance. Note 3: Specification is guaranteed by design and not 100% tested in production. Note 4: Failure to solder the exposed backside of the package to the PC board ground plane will result in a thermal resistance much higher than 43°C/W. Note 5: The absolute maximum rating is a DC rating. Under certain conditions in the applications shown, the peak AC voltage on the C2 pin may exceed ±8V. This behavior is normal and acceptable because the current into the pin is limited by the impedance of the coupling capacitor. ELECTRICAL CHARACTERISTICS The l denotes the specifications which apply over the full operating junction temperature range, otherwise specifications are for TA = 25°C (Note 2). VAUX = 5V, unless otherwise noted. PARAMETER CONDITIONS MIN TYP MAX UNITS Output Voltage VS1 = VS2 = GND VS1 = VAUX, VS2 = GND VS1 = GND, VS2 = VAUX VS1 = VS2 = VAUX l l l l 2.30 3.234 4.018 4.90 2.350 3.300 4.100 5.000 2.40 3.366 4.182 5.10 V V V V VOUT Quiescent Current VOUT = 3.3V, VOUT2_EN = 0V 0.2 µA VAUX Quiescent Current No Load, All Outputs Charged 6 9 µA LDO Output Voltage 0.5mA Load l 2.134 2.2 2.266 V LDO Load Regulation For 0mA to 2mA Load 0.5 1 % LDO Line Regulation For VAUX from 2.5V to 5V 0.05 0.2 % LDO Dropout Voltage ILDO = 2mA 100 200 mV LDO Current Limit VLDO = 0V l 4 11 mA VOUT Current Limit VOUT = 0V l 2.8 4.5 7.5 mA VSTORE Current Limit VSTORE = 0V l 2.8 4.5 7.5 mA VAUX Clamp Voltage Current into VAUX = 5mA l 5 5.25 5.60 V VSTORE Leakage Current VSTORE = 5V 0.1 0.3 µA VOUT2 Leakage Current VOUT2 = 0V, VOUT2_EN = 0V 0.1 µA VS1, VS2 Threshold Voltage l 0.4 0.85 1.2 V VS1, VS2 Input Current VS1 = VS2 = 5V 0.01 0.1 µA PGOOD Threshold (Rising) Measured Relative to the VOUT Voltage –7.5 % PGOOD Threshold (Falling) Measured Relative to the VOUT Voltage –9 % PGOOD VOL Sink Current = 100µA 0.15 0.3 V PGOOD VOH Source Current = 0 2.1 2.2 2.3 V PGOOD Pull-Up Resistance 1 MΩ VOUT2_EN Threshold Voltage VOUT2_EN Rising l 0.4 1 1.3 V VOUT2_EN Pull-Down Resistance 5 MΩ VOUT2 Turn-On Time 5 µs VOUT2 Turn-Off Time (Note 3) 0.15 µs VOUT2 Current Limit VOUT = 3.3V l 0.15 0.3 0.45 A VOUT2 Current Limit Response Time (Note 3) 350 ns VOUT2 P-Channel MOSFET On-Resistance VOUT = 3.3V (Note 3) 1.3 Ω N-Channel MOSFET On-Resistance C2 = 5V (Note 3) 0.5 Ω LTC3108 4 Rev. D For more information www.analog.com TYPICAL PERFORMANCE CHARACTERISTICS IVOUT and Efficiency vs VIN, 1:20 Ratio Transformer Input Resistance vs VIN (VOUT Charging) IVOUT vs VIN and Source Resistance, 1:20 Ratio TA = 25°C, unless otherwise noted. IVOUT and Efficiency vs VIN, 1:50 Ratio Transformer IVOUT and Efficiency vs VIN, 1:100 Ratio Transformer VIN (mV) 0 IVOUT (µA) EFFICIENCY (%) 2500 3000 3500 2000 1500 100 200 400 300 500 500 0 1000 4000 50 60 70 40 30 10 0 20 80 3108 G01 IVOUT (VOUT = 4.5V) EFFICIENCY (VOUT = 4.5V) IVOUT (VOUT = 0V) C1 = 10nF VIN (mV) 0 IVOUT (µA) EFFICIENCY (%) 2000 2400 2800 1600 1200 100 200 400 300 500 400 0 800 3200 50 60 70 40 30 10 0 20 80 3108 G02 C1 = 4.7nF IVOUT (VOUT = 4.5V) EFFICIENCY (VOUT = 4.5V) IVOUT (VOUT = 0V) VIN (mV) 0 IVOUT (µA) EFFICIENCY (%) 1000 1200 800 600 100 200 400 300 500 200 0 400 1400 50 60 70 40 30 10 0 20 3108 G03 C1 = 1nF IVOUT (VOUT = 4.5V) EFFICIENCY (VOUT = 4.5V) IVOUT (VOUT = 0V) VIN (mV) 0 INPUT RESISTANCE (Ω) 5 6 7 4 3 100 200 400 300 500 1 0 2 8 9 10 3108 G04 1:20 RATIO 1:50 RATIO 1:100 RATIO VIN OPEN-CIRCUIT (mV) IVOUT (µA) 100 10 1000 10000 0 3108 G05 0 200 300 500 100 400 600 700 800 1Ω 2Ω 5Ω 10Ω C1 = 10nF VIN (mV) IIN (mA) 3108 G00 1000 100 10 1 10 100 1000 1:50 RATIO, C1 = 4.7n 1:100 RATIO, C1 = 1n 1:20 RATIO, C1 = 10n IIN vs VIN, (VOUT = 0V) LTC3108 5 Rev. D For more information www.analog.com IVOUT vs dT and TEG Size, 1:100 Ratio IVOUT vs VIN and Source Resistance, 1:50 Ratio IVOUT vs VIN and Source Resistance, 1:100 Ratio VIN OPEN-CIRCUIT (mV) IVOUT (µA) 100 10 1000 10000 0 3108 G06 0 200 300 500 100 400 600 700 800 1Ω 2Ω 5Ω 10Ω C1 = 4.7nF VIN OPEN-CIRCUIT (mV) IVOUT (µA) 100 1000 10 3108 G07 0 100 200 300 400 500 1Ω 2Ω 5Ω 10Ω C1 = 1nF dT ACROSS TEG (°C) IVOUT (µA) 100 10 1000 10000 0 3108 G08 0.1 10 1 100 VOUT = 0V 40mm TEG 15mm TEG 1:50 RATIO 1:100 RATIO 1:50 RATIO 1:100 RATIO TYPICAL PERFORMANCE CHARACTERISTICS TA = 25°C, unless otherwise noted. LDO Load Regulation Resonant Switching Waveforms LDO Dropout Voltage LDO LOAD (mA) 0 –1.00 DROP IN VLDO (%) –0.75 –0.50 0.5 1 1.5 2 3 2.5 3.5 –0.25 0.00 4 3108 G10 LDO LOAD (mA) 0 0.00 DROPOUT VOLTAGE (V) 0.04 0.08 0.12 0.5 1 1.5 2 3 2.5 3.5 0.16 0.20 0.02 0.06 0.10 0.14 0.18 4 3108 G11 10µs/DIV C1 PIN 2V/DIV C2 PIN 2V/DIV SW PIN 50mV/ DIV 3108 G09 VIN = 20mV 1:100 RATIO TRANSFORMER LTC3108 6 Rev. D For more information www.analog.com TYPICAL PERFORMANCE CHARACTERISTICS Start-Up Voltage Sequencing VOUT and PGD Response During a Step Load VOUT Ripple LDO Step Load Response Enable Input and VOUT2 Running on Storage Capacitor TA = 25°C, unless otherwise noted. 10sec/DIV 3108 G12 CH1 VSTORE 1V/DIV CH2, VOUT 1V/DIV CH3, VLDO 1V/DIV VIN = 50mV 1:100 RATIO TRANSFORMER COUT = 220µF CSTORE = 470µF CLDO = 2.2µF 5ms/DIV 3108 G13 CH2 VOUT 1V/DIV CH1 PGD 1V/DIV 50mA LOAD STEP COUT = 220µF 100ms/DIV 3108 G14 20mV/ DIV 30µA LOAD COUT = 220µF 1ms/DIV 3108 G16 CH2, VOUT2 1V/DIV CH1 VOUT2_EN 1V/DIV 10mA LOAD ON VOUT2 COUT = 220µF 5sec/DIV 3108 G17 CH2, VOUT 1V/DIV CH1, VIN 50mV/DIV CH3 VSTORE 1V/DIV CH4, VLDO 1V/DIV CSTORE = 470µF VOUT LOAD = 100µA VLDO 20mV/DIV ILDO 5mA/DIV 200µs/DIV 3108 G15 0mA TO 3mA LOAD STEP CLDO = 2.2µF LTC3108 7 Rev. D For more information www.analog.com VAUX (Pin 1/Pin 2): Output of the Internal Rectifier Circuit and VCC for the IC. Bypass VAUX with at least 1µF of capacitance. An active shunt regulator clamps VAUX to 5.25V (typical). VSTORE (Pin 2/Pin 3): Output for the Storage Capacitor or Battery. A large capacitor may be connected from this pin to GND for powering the system in the event the input voltage is lost. It will be charged up to the maximum VAUX clamp voltage. If not used, this pin should be left open or tied to VAUX. VOUT (Pin 3/Pin 4): Main Output of the Converter. The voltage at this pin is regulated to the voltage selected by VS1 and VS2 (see Table 1). Connect this pin to an energy storage capacitor or to a rechargeable battery. VOUT2 (Pin 4/Pin 5): Switched Output of the Converter. Connect this pin to a switched load. This output is open until VOUT2_EN is driven high, then it is connected to VOUT through a 1.3Ω P-channel switch. If not used, this pin should be left open or tied to VOUT. The peak current in this output is limited to 0.3A typical. VLDO (Pin 5/Pin 6): Output of the 2.2V LDO. Connect a 2.2µF or larger ceramic capacitor from this pin to GND. If not used, this pin should be tied to VAUX. PGD (Pin 6/Pin 7): Power Good Output. When VOUT is within 7.5% of its programmed value, PGD will be pulled up to VLDO through a 1MΩ resistor. If VOUT drops 9% below its programmed value PGD will go low. This pin can sink up to 100µA. VS2 (Pin 7/Pin 10): VOUT Select Pin 2. Connect this pin to ground or VAUX to program the output voltage (see Table 1). VS1 (Pin 8/Pin 11): VOUT Select Pin 1. Connect this pin to ground or VAUX to program the output voltage (see Table 1). VOUT2_EN (Pin 9/Pin 12): Enable Input for VOUT2. VOUT2 will be enabled when this pin is driven high. There is an internal 5M pull-down resistor on this pin. If not used, this pin can be left open or grounded. C1 (Pin 10/Pin 13): Input to the Charge Pump and Rectifier Circuit. Connect a capacitor from this pin to the secondary winding of the step-up transformer. C2 (Pin 11/Pin 14): Input to the N-Channel Gate Drive Circuit. Connect a capacitor from this pin to the secondary winding of the step-up transformer. SW (Pin 12/Pin 15): Drain of the Internal N-Channel Switch. Connect this pin to the primary winding of the transformer. GND (Pins 1, 8, 9, 16) SSOP Only: Ground GND (Exposed Pad Pin 13) DFN Only: Ground. The DFN exposed pad must be soldered to the PCB ground plane. It serves as the ground connection, and as a means of conducting heat away from the die. Table 1. Regulated Voltage Using Pins VS1 and VS2 VS2 VS1 VOUT GND GND 2.35V GND VAUX 3.3V VAUX GND 4.1V VAUX VAUX 5V PIN FUNCTIONS (DFN/SSOP) LTC3108 8 Rev. D For more information www.analog.com BLOCK DIAGRAM OPERATION The LTC3108 is designed to use a small external step-up transformer to create an ultralow input voltage step-up DC/DC converter and power manager. It is ideally suited for low power wireless sensors and other applications in which surplus energy harvesting is used to generate system power because traditional battery power is incon venient or impractical. The LTC3108 is designed to manage the charging and regulation of multiple outputs in a system in which the average power draw is very low, but there may be peri odic pulses of higher load current required. This is typi cal of wireless sensor applications, where the quiescent power draw is extremely low most of the time, except for transmit bursts when circuitry is powered up to make measurements and transmit data. The LTC3108 can also be used to trickle charge a standard capacitor, supercapacitor or rechargeable battery, using energy harvested from a Peltier or photovoltaic cell. (Refer to the Block Diagram) 3108 BD C1 C2 5M SW 5.25V 1.2V VREF SW VOUT VSTORE VLDO OFF ON VOUT2 VOUT2 VOUT VOUT PROGRAM COUT PGOOD VOUT2_EN VOUT VS1 VS2 PGD VSTORE C1 CIN VIN VLDO CSTORE 1µF 1:100 C2 SYNC RECTIFY REFERENCE VOUT 2.2V CHARGE CONTROL VAUX + – + – ILIM LTC3108 1.3Ω 0.5Ω 1M EXPOSED PAD (DFN) 2.2µF GND (SSOP) VREF LDO VREF VBEST LTC3108 9 Rev. D For more information www.analog.com OPERATION Oscillator The LTC3108 utilizes a MOSFET switch to form a resonant step-up oscillator using an external step-up transformer and a small coupling capacitor. This allows it to boost input voltages as low as 20mV high enough to provide multiple regulated output voltages for powering other cir cuits. The frequency of oscillation is determined by the inductance of the transformer secondary winding and is typically in the range of 10kHz to 100kHz. For input volt ages as low as 20mV, a primary-secondary turns ratio of about 1:100 is recommended. For higher input voltages, this ratio can be lower. See the Applications Information section for more information on selecting the transformer. Charge Pump and Rectifier The AC voltage produced on the secondary winding of the transformer is boosted and rectified using an external charge pump capacitor (from the secondary winding to pin C1) and the rectifiers internal to the LTC3108. The rectifier circuit feeds current into the VAUX pin, provid ing charge to the external VAUX capacitor and the other outputs. VAUX The active circuits within the LTC3108 are powered from VAUX, which should be bypassed with a 1µF capacitor. Larger capacitor values are recommended when using turns ratios of 1:50 or 1:20 (refer to the Typical Application examples). Once VAUX exceeds 2.5V, the main VOUT is allowed to start charging. An internal shunt regulator limits the maximum voltage on VAUX to 5.25V typical. It shunts to GND any excess current into VAUX when there is no load on the converter or the input source is generating more power than is required by the load. Voltage Reference The LTC3108 includes a precision, micropower reference, for accurate regulated output voltages. This reference becomes active as soon as VAUX exceeds 2V. Synchronous Rectifiers Once VAUX exceeds 2V, synchronous rectifiers in parallel with each of the internal diodes take over the job of rectify ing the input voltage, improving efficiency. Low Dropout Linear Regulator (LDO) The LTC3108 includes a low current LDO to provide a regulated 2.2V output for powering low power processors or other low power ICs. The LDO is powered by the higher of VAUX or VOUT. This enables it to become active as soon as VAUX has charged to 2.3V, while the VOUT storage capacitor is still charging. In the event of a step load on the LDO output, current can come from the main VOUT capacitor if VAUX drops below VOUT. The LDO requires a 2.2µF ceramic capacitor for stability. Larger capacitor values can be used without limitation, but will increase the time it takes for all the outputs to charge up. The LDO output is current limited to 4mA minimum. VOUT The main output voltage on VOUT is charged from the VAUX supply, and is user programmed to one of four regulated voltages using the voltage select pins VS1 and VS2, according to Table 2. Although the logic threshold voltage for VS1 and VS2 is 0.85V typical, it is recom mended that they be tied to ground or VAUX. Table 2. Regulated Voltage Using Pins VS1 and VS2 VS2 VS1 VOUT GND GND 2.35V GND VAUX 3.3V VAUX GND 4.1V VAUX VAUX 5V When the output voltage drops slightly below the regu lated value, the charging current will be enabled as long as VAUX is greater than 2.5V. Once VOUT has reached the proper value, the charging current is turned off. The internal programmable resistor divider sets VOUT, eliminating the need for very high value external resis tors that are susceptible to board leakage. LTC3108 10 Rev. D For more information www.analog.com In a typical application, a storage capacitor (typically a few hundred microfarads) is connected to VOUT. As soon as VAUX exceeds 2.5V, the VOUT capacitor will be allowed to charge up to its regulated voltage. The current available to charge the capacitor will depend on the input voltage and transformer turns ratio, but is limited to about 4.5mA typical. PGOOD A power good comparator monitors the VOUT voltage. The PGD pin is an open-drain output with a weak pull-up (1MΩ) to the LDO voltage. Once VOUT has charged to within 7.5% of its regulated voltage, the PGD output will go high. If VOUT drops more than 9% from its regulated voltage, PGD will go low. The PGD output is designed to drive a microprocessor or other chip I/O and is not intended to drive a higher current load such as an LED. Pulling PGD up externally to a voltage greater than VLDO will cause a small current to be sourced into VLDO. PGD can be pulled low in a wire-OR configuration with other circuitry. VOUT2 VOUT2 is an output that can be turned on and off by the host, using the VOUT2_EN pin. When enabled, VOUT2 is connected to VOUT through a 1.3Ω P-channel MOSFET switch. This output, controlled by a host processor, can be used to power external circuits such as sensors and amplifiers, that do not have a low power sleep or shut down capability. VOUT2 can be used to power these cir cuits only when they are needed. Minimizing the amount of decoupling capacitance on VOUT2 will allow it to be switched on and off faster, allow ing shorter burst times and, therefore, smaller duty cycles in pulsed applications such as a wireless sensor/transmit ter. A small VOUT2 capacitor will also minimize the energy that will be wasted in charging the capacitor every time VOUT2 is enabled. VOUT2 has a soft-start time of about 5µs to limit capacitor charging current and minimize glitching of the main out put when VOUT2 is enabled. It also has a current limiting circuit that limits the peak current to 0.3A typical. The VOUT2 enable input has a typical threshold of 1V with 100mV of hysteresis, making it logic-compatible. If VOUT2_EN (which has an internal pull-down resistor) is low, VOUT2 will be off. Driving VOUT2_EN high will turn on the VOUT2 output. Note that while VOUT2_EN is high, the current limiting cir cuitry for VOUT2 draws an extra 8µA of quiescent current from VOUT. This added current draw has a negligible effect on the application and capacitor sizing, since the load on the VOUT2 output, when enabled, is likely to be orders of magnitude higher than 8µA. VSTORE The VSTORE output can be used to charge a large stor age capacitor or rechargeable battery after VOUT has reached regulation. Once VOUT has reached regulation, the VSTORE output will be allowed to charge up to the VAUX voltage. The storage element on VSTORE can be used to power the system in the event that the input source is lost, or is unable to provide the current demanded by the VOUT, VOUT2 and LDO outputs. If VAUX drops below VSTORE, the LTC3108 will automatically draw current from the storage element. Note that it may take a long time to charge a large capacitor, depending on the input energy available and the loading on VOUT and VLDO. Since the maximum current from VSTORE is limited to a few milliamps, it can safely be used to trickle-charge NiCd or NiMH rechargeable batteries for energy storage when the input voltage is lost. Note that the VSTORE capacitor cannot supply large pulse currents to VOUT. Any pulse load on VOUT must be handled by the VOUT capacitor. Short-Circuit Protection All outputs of the LTC3108 are current limited to protect against short-circuits to ground. Output Voltage Sequencing A timing diagram showing the typical charging and voltage sequencing of the outputs is shown in Figure 1. Note: time not to scale. OPERATION LTC3108 11 Rev. D For more information www.analog.com Figure 1. Output Voltage Sequencing with VOUT Programmed for 3.3V (Time Not to Scale) OPERATION 3108 F01a TIME (ms) VOLTAGE (V) 3.0 2.0 1.0 0 0 70 20 40 10 30 50 60 80 3.0 2.0 1.0 0 5.0 5.0 2.5 2.5 0 0 5.0 2.5 0 VSTORE (V) PGD (V) VOUT (V) VLDO (V) VAUX (V) LTC3108 12 Rev. D For more information www.analog.com Introduction The LTC3108 is designed to gather energy from very low input voltage sources and convert it to usable output voltages to power microprocessors, wireless transmitters and analog sensors. Such applications typically require much more peak power, and at higher voltages, than the input voltage source can produce. The LTC3108 is designed to accumulate and manage energy over a long period of time to enable short power bursts for acquiring and transmitting data. The bursts must occur at a low enough duty cycle such that the total output energy dur ing the burst does not exceed the average source power integrated over the accumulation time between bursts. For many applications, this time between bursts could be seconds, minutes or hours. The PGD signal can be used to enable a sleeping micro processor or other circuitry when VOUT reaches regula tion, indicating that enough energy is available for a burst. Input Voltage Sources The LTC3108 can operate from a number of low input voltage sources, such as Peltier cells, photovoltaic cells or thermopile generators. The minimum input voltage required for a given application will depend on the trans former turns ratio, the load power required, and the inter nal DC resistance (ESR) of the voltage source. Lower ESR will allow the use of lower input voltages, and provide higher output power capability. APPLICATIONS INFORMATION Figure 2. Typical Performance of a Peltier Cell Acting as a Thermoelectric Generator Refer to the IIN vs VIN curves in the Typical Performance Characteristics section to see what input current is required from the source for a given input voltage. For a given transformer turns ratio, there is a maximum recommended input voltage to avoid excessively high secondary voltages and power dissipation in the shunt regulator. It is recommended that the maximum input volt age times the turns ratio be less than 50. Note that a low ESR bulk decoupling capacitor will usu ally be required across the input source to prevent large voltage droop and ripple caused by the source’s ESR and the peak primary switching current (which can reach hundreds of milliamps). The time constant of the filter capacitor and the ESR of the voltage source should be much longer than the period of the resonant switching frequency. Peltier Cell (Thermoelectric Generator) A Peltier cell (also known as a thermoelectric cooler) is made up of a large number of series-connected P-N junc tions, sandwiched between two parallel ceramic plates. Although Peltier cells are often used as coolers by apply ing a DC voltage to their inputs, they will also generate a DC output voltage, using the Seebeck effect, when the two plates are at different temperatures. The polarity of the output voltage will depend on the polarity of the tem perature differential between the plates. The magnitude of the output voltage is proportional to the magnitude of the 3108 F02 1000 100 10 1 1 10 100 dT (°C) TEG VOPEN_CIRCUIT (mV) TEG MAXIMUM POUT —IDEAL (mW) 1 100 10 0.1 VOC MAX POUT (IDEAL) TEG: 30mm 127 COUPLES R = 2Ω LTC3108 13 Rev. D For more information www.analog.com temperature differential between the plates. When used in this manner, a Peltier cell is referred to as a thermoelectric generator (TEG). The low voltage capability of the LTC3108 design allows it to operate from a TEG with temperature differentials as low as 1°C, making it ideal for harvesting energy in appli cations in which a temperature difference exists between two surfaces or between a surface and the ambient tem perature. The internal resistance (ESR) of most cells is in the range of 1Ω to 5Ω, allowing for reasonable power transfer. The curves in Figure 2 show the open-circuit output voltage and maximum power transfer for a typi cal Peltier cell (with an ESR of 2Ω) over a 20°C range of temperature differential. TEG Load Matching The LTC3108 was designed to present a minimum input resistance (load) in the range of 2Ω to 10Ω, depending on input voltage and transformer turns ratio (as shown in the Typical Performance Characteristics curves). For a given turns ratio, as the input voltage drops, the input resistance increases. This feature allows the LTC3108 to optimize power transfer from sources with a few ohms of source resistance, such as a typical TEG. Note that a lower source resistance will always provide more output current capability by providing a higher input voltage under load. Peltier Cell (TEG) Suppliers Peltier cells are available in a wide range of sizes and power capabilities, from less than 10mm square to over 50mm square. They are typically 2mm to 5mm in height. A list of Peltier cell manufacturers is given in Table 3. Table 3. Peltier Cell Manufacturers CUI, Inc. www.cui.com (Distributor) Fujitaka www.fujitaka.com/pub/peltier/english/thermoelectric_power.html Ferrotec www.ferrotec.com/products/thermal/modules Kryotherm www.kryothermusa.com Laird Technologies www.lairdtech.com Marlow Industries www.marlow.com Micropelt www.micropelt.com Nextreme www.nextreme.com TE Technology www.tetech.com/Peltier-Thermoelectric-Cooler-Modules.html Tellurex www.tellurex.com APPLICATIONS INFORMATION Table 4. Recommended TEG Part Numbers by Size MANUFACTURER 15mm × 15mm 20mm × 20mm 30mm × 30mm 40mm × 40mm CUI Inc. (Distributor) CP60133 CP60233 CP60333 CP85438 Ferrotec 9501/031/030 B 9501/071/040 B 9500/097/090 B 9500/127/100 B Fujitaka FPH13106NC FPH17106NC FPH17108AC FPH112708AC Kryotherm TGM-127-1.0-0.8 LCB-127-1.4-1.15 Laird Technology PT6.7.F2.3030.W6 PT8.12.F2.4040.TA.W6 Marlow Industries RC3-8-01 RC6-6-01 RC12-8-01LS Tellurex C2-15-0405 C2-20-0409 C2-30-1505 C2-40-1509 TE Technology TE-31-1.0-1.3 TE-31-1.4-1.15 TE-71-1.4-1.15 TE-127-1.4-1.05 LTC3108 14 Rev. D For more information www.analog.com Thermopile Generator Thermopile generators (also called powerpile generators) are made up of a number of series-connected thermo couples enclosed in a metal tube. They are commonly used in gas burner applications to generate a DC output of hundreds of millivolts when exposed to the high tem perature of a flame. Typical examples are the Honeywell CQ200 and Q313. These devices have an internal series resistance of less than 3Ω, and can generate as much as 750mV open-circuit at their highest rated tempera ture. For applications in which the temperature rise is too high for a solid-state thermoelectric device, a thermopile can be used as an energy source to power the LTC3108. Because of the higher output voltages possible with a thermopile generator, a lower transformer turns ratio can be used (typically 1:20, depending on the application). Photovoltaic Cell The LTC3108 converter can also operate from a single photovoltaic cell (also known as a PV or solar cell) at light levels too low for other low input voltage boost con verters to operate. However, many variables will affect the performance in these applications. Light levels can vary over several orders of magnitude and depend on lighting conditions (the type of lighting and indoor versus outdoor). Different types of light (sunlight, incandescent, fluorescent) also have different color spectra, and will produce different output power levels depending on which type of photovoltaic cell is being used (monocrystalline, polycrystalline or thin-film). Therefore, the photovoltaic cell must be chosen for the type and amount of light avail able. Note that the short-circuit output current from the cell must be at least a few milliamps in order to power the LTC3108 converter Non-Boost Applications The LTC3108 can also be used as an energy harvester and power manager for input sources that do not require boosting. In these applications the step-up transformer can be eliminated. Any source whose peak voltage exceeds 2.5V AC or 5V DC can be connected to the C1 input through a current-limiting resistor where it will be rectified/peak detected. In these applications the C2 and SW pins are not used and can be grounded or left open. Examples of such input sources would be piezoelectric transducers, vibration energy harvesters, low current generators, a stack of low current solar cells or a 60Hz AC input. A series resistance of at least 100Ω/V should be used to limit the maximum current into the VAUX shunt regulator. COMPONENT SELECTION Step-Up Transformer The step-up transformer turns ratio will determine how low the input voltage can be for the converter to start. Using a 1:100 ratio can yield start-up voltages as low as 20mV. Other factors that affect performance are the DC resistance of the transformer windings and the inductance of the windings. Higher DC resistance will result in lower efficiency. The secondary winding inductance will deter mine the resonant frequency of the oscillator, according to the following formula. = π Frequency 1 2 • • L(sec)•C Hz Where L is the inductance of the transformer secondary winding and C is the load capacitance on the secondary winding. This is comprised of the input capacitance at pin C2, typically 30pF, in parallel with the transformer sec ondary winding’s shunt capacitance. The recommended resonant frequency is in the range of 10kHz to 100kHz. See Table 5 for some recommended transformers. APPLICATIONS INFORMATION Table 5. Recommended Transformers VENDOR PART NUMBER Coilcraft www.coilcraft.com LPR6235-752SML (1:100 Ratio) LPR6235-253PML (1:20 Ratio) LPR6235-123QML (1:50 Ratio) Würth www.we-online 74488540070 (1:100 Ratio) 74488540120 (1:50 Ratio) 74488540250 (1:20 Ratio) LTC3108 15 Rev. D For more information www.analog.com C1 Capacitor The charge pump capacitor that is connected from the transformer’s secondary winding to the C1 pin has an effect on converter input resistance and maximum output current capability. Generally, a minimum value of 1nF is recommended when operating from very low input volt ages using a transformer with a ratio of 1:100. Too large a capacitor value can compromise performance when oper ating at low input voltage or with high resistance sources. For higher input voltages and lower turns ratios, the value of the C1 capacitor can be increased for higher output current capability. Refer to the Typical Applications sche matic examples for the recommended value for a given turns ratio. Squegging Certain types of oscillators, including transformer-cou pled oscillators such as the resonant oscillator of the LTC3108, can exhibit a phenomenon called squegging. This term refers to a condition that can occur which blocks or stops the oscillation for a period of time much longer than the period of oscillation, resulting in bursts of oscillation. An example of this is the blocking oscillator, which is designed to squegg to produce bursts of oscilla tion. Squegging is also encountered in RF oscillators and regenerative receivers. In the case of the LTC3108, squegging can occur when a charge builds up on the C2 gate coupling capacitor, such that the DC bias point shifts and oscillation is extinguished for a certain period of time, until the charge on the capaci tor bleeds off, allowing oscillation to resume. It is difficult to predict when and if squegging will occur in a given application. While squegging is not harmful, it reduces the average output current capability of the LTC3108. Squegging can easily be avoided by the addition of a bleeder resistor in parallel with the coupling capacitor on the C2 pin. Resistor values in the range of 100k to 1MΩ are sufficient to eliminate squegging without having any negative impact on performance. For the 330pF capacitor used for C2 in most applications, a 499k bleeder resistor is recommended. See the Typical Applications schematics for an example. Using External Charge Pump Rectifiers The synchronous charge pump rectifiers in the LTC3108 (connected to the C1 pin) are optimized for operation from very low input voltage sources, using typical transformer step-up ratios between 1:100 and 1:50, and typical C1 charge pump capacitor values less than 10nF. Operation from higher input voltage sources (typically 250mV or greater, under load), allows the use of lower transformer step-up ratios (such as 1:20 and 1:10) and larger C1 capacitor values to provide higher output current capability from the LTC3108. However, due to the result ing increase in rectifier currents and resonant oscillator frequency in these applications, the use of external charge pump rectifiers is recommended for optimal performance. In applications where the step-up ratio is 1:20 or less, and the C1 capacitor is 10nF or greater, the C1 pin should be grounded and two external rectifiers (such as 1N4148 or 1N914 diodes) should be used. These are available as dual diodes in a single package. Avoid the use of Schottky rec tifiers, as their lower forward voltage drop increases the minimum start-up voltage. See the Typical Applications schematics for an example. VOUT and VSTORE Capacitor For pulsed load applications, the VOUT capacitor should be sized to provide the necessary current when the load is pulsed on. The capacitor value required will be dictated by the load current, the duration of the load pulse, and the amount of voltage droop the circuit can tolerate. The capacitor must be rated for whatever voltage has been selected for VOUT by VS1 and VS2. ≥ C (µF) I (mA)• t (ms) V (V) OUT LOAD PULSE OUT Note that there must be enough energy available from the input voltage source for VOUT to recharge the capacitor during the interval between load pulses (to be discussed in the next example). Reducing the duty cycle of the load pulse will allow operation with less input energy. The VSTORE capacitor may be of very large value (thou sands of microfarads or even Farads), to provide holdup APPLICATIONS INFORMATION LTC3108 16 Rev. D For more information www.analog.com at times when the input power may be lost. Note that this capacitor can charge all the way to 5.25V (regardless of the settings for VOUT), so ensure that the holdup capacitor has a working voltage rating of at least 5.5V at the tem perature for which it will be used. The VSTORE capacitor can be sized using the following: [ ] ≥ + + + − C 6µA I I (I • t • f) • TSTORE 5.25 V STORE Q LDO BURST OUT Where 6µA is the quiescent current of the LTC3108, IQ is the load on VOUT in between bursts, ILDO is the load on the LDO between bursts, IBURST is the total load during the burst, t is the duration of the burst, f is the frequency of the bursts, TSTORE is the storage time required and VOUT is the output voltage required. Note that for a pro grammed output voltage of 5V, the VSTORE capacitor cannot provide any beneficial storage time. To minimize losses and capacitor charge time, all capaci tors used for VOUT and VSTORE should be low leakage. See Table 6 for recommended storage capacitors. Storage capacitors requiring voltage balancing are not recommended due to the current draw of the balancing resistors. Due to the very low input voltage the circuit may operate from, the connections to VIN, the primary of the trans former and the SW and GND pins of the LTC3108 should be designed to minimize voltage drop from stray resis tance and able to carry currents as high as 500mA. Any small voltage drop in the primary winding conduction path will lower efficiency and increase capacitor charge time. Also, due to the low charge currents available at the out puts of the LTC3108, any sources of leakage current on the output voltage pins must be minimized. An example board layout is shown in Figure 3. APPLICATIONS INFORMATION 3108 FO3 VOUT2 VOUT VIN VIAS TO GROUND PLANE VLDO PGOOD GND 12 11 8 9 10 4 5 3 2 1 VOUT2_EN VS1 VS2 SW C2 C1 VOUT VOUT2 VLDO PGD VAUX VSTORE 6 7 Figure 3. Example Component Placement for Two-Layer PC Board (DFN Package) Design Example 1 This design example will explain how to calculate the nec essary storage capacitor value for VOUT in pulsed load applications, such as a wireless sensor/transmitter. In these types of applications, the load is very small for a majority of the time (while the circuitry is in a low power sleep state), with bursts of load current occurring peri odically during a transmit burst. The storage capacitor on VOUT supports the load during the transmit burst, and the long sleep time between bursts allows the LTC3108 to recharge the capacitor. A method for calculating the Table 6. Recommended Storage Capacitors VENDOR PART NUMBER/SERIES AVX www.avx.com BestCap Series TAJ and TPS Series Tantalum Cap-XX www.cap-xx.com GZ Series Cooper/Bussmann www.bussmann.com/3/PowerStor.html KR Series P Series Vishay/Sprague www.vishay.com/capacitors Tantamount 592D 595D Tantalum 150CRZ/153CRV Aluminum 013 RLC (Low Leakage) PCB Layout Guidelines Due to the rather low switching frequency of the resonant converter and the low power levels involved, PCB layout is not as critical as with many other DC/DC converters. There are, however, a number of things to consider. LTC3108 17 Rev. D For more information www.analog.com APPLICATIONS INFORMATION maximum rate at which the load pulses can occur for a given output current from the LTC3108 will also be shown. In this example, VOUT is set to 3.3V, and the maximum allowed voltage droop during a transmit burst is 10%, or 0.33V. The duration of a transmit burst is 1ms, with a total average current requirement of 40mA during the burst. Given these factors, the minimum required capacitance on VOUT is: ≥ = C (µF) 40mA •1ms 0.33V 121µF OUT Note that this equation neglects the effect of capacitor ESR on output voltage droop. For most ceramic or low ESR tantalum capacitors, the ESR will have a negligible effect at these load currents. A standard value of 150µF or larger could be used for COUT in this case. Note that the load current is the total current draw on VOUT, VOUT2 and VLDO, since the cur rent for all of these outputs must come from VOUT during a burst. Current contribution from the holdup capacitor on VSTORE is not considered, since it may not be able to recharge between bursts. Also, it is assumed that the charge current from the LTC3108 is negligible compared to the magnitude of the load current during the burst. To calculate the maximum rate at which load bursts can occur, determine how much charge current is available from the LTC3108 VOUT pin given the input voltage source being used. This number is best found empirically, since there are many factors affecting the efficiency of the converter. Also determine what the total load current is on VOUT during the sleep state (between bursts). Note that this must include any losses, such as storage capacitor leakage. Assume, for instance, that the charge current from the LTC3108 is 50µA and the total current drawn on VOUT in the sleep state is 17µA, including capacitor leakage. In addition, use the value of 150µF for the VOUT capacitor. The maximum transmit rate (neglecting the duration of the transmit burst, which is typically very short) is then given by: = − = = t 150µF • 0.33V (50µA 17µA) 1.5sec or f 0.666Hz MAX Therefore, in this application example, the circuit can sup port a 1ms transmit burst every 1.5 seconds. It can be determined that for systems that only need to transmit every few seconds (or minutes or hours), the average charge current required is extremely small, as long as the sleep current is low. Even if the available charge current in the example above was only 10µA and the sleep current was only 5µA, it could still transmit a burst every ten seconds. The following formula enables the user to calculate the time it will take to charge the LDO output capacitor and the VOUT capacitor the first time, from 0V. Here again, the charge current available from the LTC3108 must be known. For this calculation, it is assumed that the LDO output capacitor is 2.2µF. = − t 2.2V • 2.2µF I I LDO CHG LDO If there were 50µA of charge current available and a 5µA load on the LDO (when the processor is sleeping), the time for the LDO to reach regulation would be 107ms. If VOUT were programmed to 3.3V and the VOUT capacitor was 150µF, the time for VOUT to reach regulation would be: = − − + t 3.3V •150µF I I I t VOUT CHG VOUT LDO LDO If there were 50µA of charge current available and 5µA of load on VOUT, the time for VOUT to reach regulation after the initial application of power would be 12.5 seconds. Design Example 2 In many pulsed load applications, the duration, magnitude and frequency of the load current bursts are known and fixed. In these cases, the average charge current required from the LTC3108 to support the average load must be calculated, which can be easily done by the following: ≥ + I I I • t T CHG Q BURST Where IQ is the sleep current on VOUT required by the external circuitry in between bursts (including cap leak age), IBURST is the total load current during the burst, t is LTC3108 18 Rev. D For more information www.analog.com the duration of the burst and T is the period of the transmit burst rate (essentially the time between bursts). In this example, IQ = 5µA, IBURST = 100mA, t = 5ms and T = one hour. The average charge current required from the LTC3108 would be: ≥ + = I 5µA 100mA • 0.005sec 3600sec 5.14µA CHG Therefore, if the LTC3108 has an input voltage that allows it to supply a charge current greater than 5.14µA, the application can support 100mA bursts lasting 5ms every hour. It can be determined that the sleep current of 5µA is the dominant factor because the transmit duty cycle is so small (0.00014%). Note that for a VOUT of 3.3V, the average power required by this application is only 17µW (not including converter losses). Note that the charge current available from the LTC3108 has no effect on the sizing of the VOUT capacitor (if it is assumed that the load current during a burst is much larger than the charge current), and the VOUT capacitor has no effect on the maximum allowed burst rate. APPLICATIONS INFORMATION Peltier-Powered Energy Harvester for Remote Sensor Applications TYPICAL APPLICATIONS 3108 TA02 C1 C2 SW VS2 VS1 COOPER BUSSMAN PB-5ROH104-R OR KR-5R5H104-R VOUT2 VOUT2 PGOOD 2.2V COUT PGD VLDO VSTORE + VOUT VOUT2_EN LTC3108 VAUX GND CSTORE 0.1F 6.3V 5V 3.3V 1µF 1nF CIN 1:100 T1 T1: COILCRAFT LPR6235-752SML COUT VALUE DEPENDENT ON THE MAGNITUDE AND DURATION OF THE LOAD PULSE 330pF ∆T = 1°C TO 20°C SENSORS XMTR µP 2.2µF ON OFF 3.3V + + + THERMOELECTRIC GENERATOR LTC3108 19 Rev. D For more information www.analog.com TYPICAL APPLICATIONS Li-Ion Battery Charger and LDO Powered by a Solar Cell Supercapacitor Charger and LDO Powered by a Thermopile Generator 3108 TA04 C1 HONEYWELL CQ200 THERMOPILE C2 SW VS2 VS1 VSTORE VOUT2 PGOOD VLDO CAP-XX GZ115F PGD VLDO VOUT VOUT2_EN LTC3108 VAUX GND 2.2µF 150mF 2.5V 2.2µF 4.7nF T1: COILCRAFT LPR6235-123QML 2.2V 2.35V T1 1:50 330pF 220µF + + VOUT 3108 TA03 C1 SOLAR CELL C2 SW VS2 VS1 VSTORE VOUT2 VLDO PGD VLDO VOUT VOUT2_EN LTC3108 VAUX GND 2.2µF Li-Ion 4.7µF 0.01µF 2.2V 4.1V T1 1:20 330pF 220µF + – + VOUT 2" DIAMETER MONOCRYSTALLINE CELL LIGHT LEVEL ≥ 900 LUX T1: COILCRAFT LPR6235-253PML DC Input Energy Harvester and Power Manager 3108 TA05 C1 C2 SW VS2 VS1 VOUT2 VOUT2 PGOOD 2.2V COUT PGD VLDO VLDO VSTORE VOUT VOUT VOUT2_EN VOUT2_ENABLE LTC3108 VAUX GND CSTORE 5V 3.3V 2.2µF VIN VIN > 5V RIN RIN > 100Ω/V 2.2µF + – + + AC Input Energy Harvester and Power Manager 3108 TA06 C1 C2 SW VS1 VS2 VOUT2 VOUT2 PGOOD 2.2V COUT PGD VLDO VLDO VSTORE VOUT VOUT VOUT2_EN VOUT2_ENABLE LTC3108 VAUX GND CSTORE 5V 5V 2.2µF CIN VIN VIN > 5VP-P - PIEZO - 60Hz RIN RIN > 100Ω/V 2.2µF AC + + LTC3108 20 Rev. D For more information www.analog.com GN16 REV B 0212 1 2 3 4 5 6 7 8 .229 – .244 (5.817 – 6.198) .150 – .157 (3.810 – 3.988) 16 15 14 13 .189 – .196 (4.801 – 4.978) 12 11 10 9 .016 – .050 (0.406 – 1.270) .015 ±.004 (0.38 ±0.10) × 45° 0° – 8° TYP .007 – .0098 (0.178 – 0.249) .0532 – .0688 (1.35 – 1.75) .008 – .012 (0.203 – 0.305) TYP .004 – .0098 (0.102 – 0.249) .0250 (0.635) BSC .009 (0.229) REF .254 MIN RECOMMENDED SOLDER PAD LAYOUT .150 – .165 .0250 BSC .0165 ±.0015 .045 ±.005 DIMENSION DOES NOT INCLUDE MOLD FLASH. MOLD FLASH SHALL NOT EXCEED 0.006" (0.152mm) PER SIDE DIMENSION DOES NOT INCLUDE INTERLEAD FLASH. INTERLEAD FLASH SHALL NOT EXCEED 0.010" (0.254mm) PER SIDE INCHES (MILLIMETERS) NOTE: 1. CONTROLLING DIMENSION: INCHES 2. DIMENSIONS ARE IN 3. DRAWING NOT TO SCALE 4. PIN 1 CAN BE BEVEL EDGE OR A DIMPLE GN Package 16-Lead Plastic SSOP (Narrow .150 Inch) (Reference LTC DWG # 05-08-1641 Rev B) 4.00 ±0.10 (2 SIDES) 3.00 ±0.10 (2 SIDES) NOTE: 1. DRAWING PROPOSED TO BE A VARIATION OF VERSION (WGED) IN JEDEC PACKAGE OUTLINE M0-229 2. DRAWING NOT TO SCALE 3. ALL DIMENSIONS ARE IN MILLIMETERS 4. DIMENSIONS OF EXPOSED PAD ON BOTTOM OF PACKAGE DO NOT INCLUDE MOLD FLASH. MOLD FLASH, IF PRESENT, SHALL NOT EXCEED 0.15mm ON ANY SIDE 5. EXPOSED PAD SHALL BE SOLDER PLATED 6. SHADED AREA IS ONLY A REFERENCE FOR PIN 1 LOCATION ON THE TOP AND BOTTOM OF PACKAGE 0.40 ±0.10 BOTTOM VIEW—EXPOSED PAD 1.70 ±0.10 0.75 ±0.05 R = 0.115 TYP R = 0.05 TYP 2.50 REF 1 6 12 7 PIN 1 NOTCH R = 0.20 OR 0.35 × 45° CHAMFER PIN 1 TOP MARK (NOTE 6) 0.200 REF 0.00 – 0.05 (UE12/DE12) DFN 0806 REV D 3.30 ±0.10 0.25 ±0.05 0.50 BSC 2.50 REF RECOMMENDED SOLDER PAD PITCH AND DIMENSIONS APPLY SOLDER MASK TO AREAS THAT ARE NOT SOLDERED 2.20 ±0.05 0.70 ±0.05 3.60 ±0.05 PACKAGE OUTLINE 1.70 ±0.05 3.30 ±0.05 0.50 BSC 0.25 ±0.05 DE/UE Package 12-Lead Plastic DFN (4mm × 3mm) (Reference LTC DWG # 05-08-1695 Rev D) PACKAGE DESCRIPTION LTC3108 21 Rev. D For more information www.analog.com REVISION HISTORY REV DATE DESCRIPTION PAGE NUMBER A 04/10 Updated front page text and Typical Appliction Updated Absolute Maximum Ratings and Order Information sections Updated Electrical Characteristics Added graph (3108 G00) to Typical Performance Characteristics Updated Block Diagram Text added to Operation section Changes to Applications Information section Updated Typical Applications Updated Related Parts 1 2 3 4 8 9 12-18 18, 19, 22 22 B 06/13 Added vendor information to Table 5 14 C 08/13 Changed Würth transformer part numbers 14 D 03/19 Electrical Characteristics. LDO Dropout Voltage Temperature Range changed from Over Temperature to TA = 25°C, VOUT, Current Limit Max & VSTORE Current Limit Max changed from 7V to 7.5V; VAUX Clamp Voltage Max changed from 5.55V to 5.6V 3 Information furnished by Analog Devices is believed to be accurate and reliable. However, no responsibility is assumed by Analog Devices for its use, nor for any infringements of patents or other rights of third parties that may result from its use. Specifications subject to change without notice. No license is granted by implication or otherwise under any patent or patent rights of Analog Devices. LTC3108 22 Rev. D For more information www.analog.com RELATED PARTS TYPICAL APPLICATION PART NUMBER DESCRIPTION COMMENTS LTC1041 Bang-Bang Controller VIN: 2.8V to 16V; IQ = 1µA; SO-8 Package LTC1389 Nanopower Precision Shunt Voltage Reference VOUT(MIN) = 1.25V; IQ = 0.8µA; SO-8 Package LT1672/LT1673/ LT1674 Single-/Dual-/Quad-Precision 2µA Rail-to-Rail Op Amps SO-8, SO-14 and MSOP-8 Packages LT3009 3µA IQ, 20mA Linear Regulator VIN: 1.6V to 20V; VOUT(MIN): 0.6V to Adj, 1.2V, 1.5V, 1.8V, 2.5V, 3.3V, 5V to Fixed; IQ = 3µA; ISD < 1µA; 2mm × 2mm DFN-8 and SC70 Packages LTC3108-1 Ultralow Voltage Step-Up Converter and Power Manager VIN: 0.02V to 1V; VOUT = 2.5V, 3V, 3.7V, 4.5V Fixed; IQ = 6µA; 3mm × 4mm DFN-12 and SSOP-16 Packages LTC3525L-3/ LTC3525L-3.3/ LTC3525L-5 400mA (ISW), Synchronous Step-Up DC/DC Converter with Output Disconnect VIN: 0.7V to 4V; VOUT(MIN) = 5VMAX; IQ = 7µA; ISD < 1µA; SC70 Package LTC3588-1 Piezoelectric Energy Generator with Integrated High Efficiency Buck Converter VIN: 2.7V to 20V; VOUT(MIN): Fixed to 1.8V, 2.5V, 3.3V, 3.6V; IQ = 0.95µA; 3mm × 3mm DFN-10 and MSOP-10E Packages LTC3642 45V, 50mA Synchronous MicroPower Buck Converter VIN: 4.5V to 45V, 60VMAX; VOUT(MIN): 0.8V to Adj, 3.3V Fixed, 5V Fixed; IQ = 12µA; ISD < 1µA; 3mm × 3mm DFN-8 and MSOP-8E Packages LTC6656 850mA Precision Reference Series Low Dropout Precision LT8410/LT8410-1 MicroPower 25mA/8mA Low Noise Boost Converter with Integrated Schottky Diode and Output Disconnect VIN: 2.6V to 16V; VOUT(MIN) = 40VMAX; IQ = 8.5µA; ISD < 1µA; 2mm × 2mm DFN-8 Package LTC4070 Micropower Shunt Li-Ion Charge Controls Charging with µA Source 3108 TA07 C1 HOT COLD C2 SW VS2 VS1 VSTORE VOUT2 VOUT2 PGOOD VOUT VLDO COUT PGD VLDO + VOUT VOUT2_EN LTC3108 VAUX GND 2.2µF CSTORE 5V VAUX 1µF 1nF 2.2V 3.3V 1:100 LPR6235-752SML 330pF ON OFF C1 C2 SW VS2 VS1 VSTORE VOUT2 PGD VLDO + VOUT VOUT2_EN LTC3108 VAUX GND 1nF 1:100 LPR6235-752SML 330pF + + COLD HOT THERMOELECTRIC GENERATOR THERMOELECTRIC GENERATOR Dual TEG Energy Harvester Operates from Temperature Differentials of Either Polarity  ANALOG DEVICES, INC. 2010-2019 03/19 www.analog.com
9288	https://computationalcombinatorics.wordpress.com/2013/08/26/mms-updates/	Updates to the Manickam-Miklós-Singhi Conjecture \| Computational Combinatorics Computational Combinatorics About Resources Syllabus Updates to the Manickam-Miklós-Singhi Conjecture by Derrick Stolee It has been a busy year for the Manickam-Miklós-Singhi Conjecture. (See the earlier post on this topic for the important definitions.) Several new papers have appeared online or on the arXiv. Here is a brief discussion of these recent developments (organized by time of submission, as far as I can tell). Recall that for , the function stores the minimum integer such that for all every vector with nonnegative total sum has at least nonnegative -sums. The conjecture states that . M. Tyomkyn, An improved bound for the Manickam-Miklós-Singhi Conjecture. European Journal of Combinatorics 33 (2012) 27-32. Tyomkyn improves previous exponential bounds to find . N. Alon, H. Huang, and B. Sudakov. Nonnegative -sums, fractional covers, and probability of small deviations. Journal of Combinatorial Theory, Series B, 103(3) (2012) 784-796. Using the idea of fractional covers, the authors prove the first polynomial upper bound of . A. Chowdhury, A note on the Manickam-Miklós-Singhi Conjecture, European Journal of Combinatorics, Volume 35, January 2014, Pages 131–140. Chowdhury extends her thesis work to provide new upper bounds of , , and . The technique is a computer-assisted proof by using the computer to generate a certain type of cover on a hypergraph. This hypergraph contains sets that must be negative in order to avoid having at least nonnegative -sums. The way these sets are classified as such uses the shift poset and an interesting application of the Kruskal-Katona Theorem. S. G. Hartke, D. Stolee, A linear programming approach to the Manickam-Miklós-Singhi Conjecture, to appear in European Journal of Combinatorics. This is the work described in the previous post. Using a linear programming formulation and a very involved computer search, the authors verify that , , , and . Also, the extremal examples for small and are demonstrated and this provides a conjecture that limits to a value approximately 3.14157899. P. Frankl, On the number of nonnegative sums, to appear in Journal of Combinatorial Theory, Series B. Frankl produces a slightly weaker but much shorter proof than Alon, Huang, and Sudakov, showing . Note this is a stronger result than Alon, Huang, and Sudakov’s cubic bound. A. Pokrovskiy, A linear bound on the Manickam-Mikl ́ós-Singhi Conjecture, arXiv:1308.2176v1. Pokrovskiy proves a linear bound of , although the constant term is . This is a very nice improvement, and I look forward to people lowering this constant through more careful analysis of his proof. However, Pokrovskiy points out that his proof as written will not achieve a value of . Have a new update? Working on your own results? Send me an email so I can update this page. Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like Loading... Related A Branch-and-Cut Strategy for the Manickam-Miklós-Singhi ConjectureMarch 8, 2013 In "Paper Reviews" Computational Combinatorics Roundup: February 2013February 28, 2013 In "Roundups" The Erdős Discrepancy ProblemSeptember 18, 2015 In "Paper Reviews" Published:August 26, 2013 Filed Under:Paper Reviews, Roundups Tags:mms-conjecture : updates 4 Comments to “Updates to the Manickam-Miklós-Singhi Conjecture” Nachimuthu Manickam says: August 28, 2013 at 4:50 am I am now working on Manickam-Miklos-Singhi conjecture for projective spaces. Manickam Reply 2. Anonymous says: September 11, 2013 at 12:19 pm Tyomkyn’s bound was (logk)^k. Misha Tyomkyn. Reply derrickstoleesays: September 12, 2013 at 2:42 pm Thanks for catching my mistake. Reply Nachimuthu Manickam says: December 13, 2013 at 5:49 pm I am glad to note that there is so much interest shown by mathematicians all around the world. N. Manickam, Chair, Dept of Math, Depauw University, Greencastle, In 46135 USA. E-mail:manickam@depauw.edu Reply Leave a Comment! Cancel reply Δ « Previous Post Next Post » Search for: Recent Posts Announcing the Small Numbers Project So Long, and Thanks for All the Theorems The Erdős Discrepancy Problem Hypohamiltonian Planar Graphs Happy Birthday, Joan Hutchinson! 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9289	https://www.geeksforgeeks.org/java/difference-between-for-loop-and-enhanced-for-loop-in-java/	Java for loop vs Enhanced for loop Last Updated : 23 Jul, 2025 Suggest changes 24 Likes In Java, loops are fundamental constructs for iterating over data structures or repeating blocks of code. Two commonly used loops are the for loop and the enhanced for loop. While they serve similar purposes, their applications and usage vary based on the scenario. for loop vs Enhanced for loop Below comparison highlights the major differences between a traditional for loop and an enhanced for loop in Java. \| Parameters \| for loop \| Enhanced for loop \| --- \| Introduced \| Introduced in early Java versions as a basic loop structure. \| Introduced in Java 5 as a simplified version for iterating over collections/arrays. \| \| Usage \| General-purpose loop suitable for arrays, collections, and numerical ranges. \| Specifically designed for iterating over arrays and collections. \| \| Index-Based Access \| Allows access to elements by their index using i. \| No index-based access; iterates directly over elements. \| \| Read-Only Iteration \| Supports both read and modify operations on elements. \| Primarily for read-only operations. Modification may lead to errors. \| \| Control Over Iteration \| Offers more control over iteration (start, end, and step). \| Limited control as it automatically iterates over all elements. \| \| Syntax Complexity \| Slightly verbose; requires initialization, condition, and increment/decrement logic. \| Concise and simpler to write. \| \| Best Use Case \| Useful when indices or custom iteration logic is required. \| Ideal for directly processing all elements of an array or collection. \| \| Performance \| May perform slightly better when accessing array elements directly by index. \| Performance depends on the internal iteration mechanism of the collection. \| for loop in Java The traditional for loop is a versatile structure used to iterate over a range of values, arrays, or other data structures. It provides complete control over the iteration process for any tasks which need to do multiple times. Syntax: for( initialization; conditional_check ; increment_or_decrement_section) { // Code to be executed which defined in the conditions } Example: Java ```` public class Geeks { public static void main(String[] args) { int[] n = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; // Iterate through the array using a for loop for (int i = 0; i < n.length; i++) { System.out.println("Index " + i + ": " + n[i]); } } } ```` public class Geeks {public class Geeks ``` ``` public static void main(String[] args) {public static void main String args ``` ``` int[] n = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; int n = 1 2 3 4 5 6 7 8 9 10 ``` ``` // Iterate through the array using a for loop // Iterate through the array using a for loop for (int i = 0; i < n.length; i++) {for int i = 0 i< n length i ++ System.out.println("Index " + i + ": " + n[i]); System out println "Index " + i +": " + n i } } } Output ``` Index 0: 1 Index 1: 2 Index 2: 3 Index 3: 4 Index 4: 5 Index 5: 6 Index 6: 7 Index 7: 8 Index 8: 9 Index 9: 10 ``` Enhanced for loop (for-each loop) The enhanced for loop, also known as the for-each loop, is a streamlined way to iterate over collections and arrays. It eliminates the need for managing the loop's counter or bounds manually. Syntax: for (data_Type_element : array_Or_Collection) { // Code to be executed which defined in the conditions } Example: Java ```` import java.io.; import java.util.; class Geeks { public static void main(String[] args) { int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; // Iterate through the array // using an enhanced for loop for (int a : arr) { System.out.println(a); } } } ```` import java.io.; import java io import java.util.; import java util class Geeks {class Geeks public static void main(String[] args) {public static void main String args int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int arr = 1 2 3 4 5 6 7 8 9 10 ``` ``` // Iterate through the array // Iterate through the array // using an enhanced for loop // using an enhanced for loop for (int a : arr) {for int a arr System.out.println(a); System out println a } } } Output ``` 1 2 3 4 5 6 7 8 9 10 ``` M mroshanmishra0072 Improve Article Tags : Java Technical Scripter Difference Between Technical Scripter 2020 Java-Loops Explore Java Basics Introduction to Java 4 min readJava Programming Basics 9 min readJava Methods 7 min readAccess Modifiers in Java 5 min readArrays in Java 7 min readJava Strings 8 min readRegular Expressions in Java 7 min read OOP & Interfaces Classes and Objects in Java 10 min readAccess Modifiers in Java 5 min readJava Constructors 10 min readJava OOP(Object Oriented Programming) Concepts 10 min readJava Packages 7 min readJava Interface 11 min read Collections Collections in Java 12 min readCollections Class in Java 13 min readCollection Interface in Java 6 min readIterator in Java 5 min readJava Comparator Interface 6 min read Exception Handling Java Exception Handling 8 min readJava Try Catch Block 4 min readJava final, finally and finalize 4 min readChained Exceptions in Java 3 min readNull Pointer Exception in Java 5 min readException Handling with Method Overriding in Java 4 min read Java Advanced Java Multithreading Tutorial 3 min readSynchronization in Java 10 min readFile Handling in Java 4 min readJava Method References 9 min readJava 8 Stream Tutorial 7 min readJava Networking 15+ min readJDBC Tutorial 5 min readJava Memory Management 4 min readGarbage Collection in Java 6 min readMemory Leaks in Java 3 min read Practice Java Java Interview Questions and Answers 15+ min readJava Programs - Java Programming Examples 7 min readJava Exercises - Basic to Advanced Java Practice Programs with Solutions 5 min readJava Quiz \| Level Up Your Java Skills 1 min read[Top 50 Java Project Ideas For Beginners and Advanced [Update 2025] 15+ min read]( Improvement Suggest Changes Help us improve. 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9290	https://www.cuemath.com/algebra/factoring-trinomials/	Factoring Trinomials Factoring trinomials means writing an expression as the product of two or more binomials and is written as (x + m) (x + n). A binomial is a two-term polynomial whereas a trinomial is a three-term polynomial. Factoring trinomials is done by splitting the algebraic expressions into a binomial that can be multiplied back to a trinomial. Let us know more about factoring trinomials, different methods and solve a few examples to understand the concept better. \| \| \| --- \| \| 1. \| What is Factoring Trinomials? \| \| 2. \| Rules for Factoring Trinomials \| \| 3. \| Methods of Factorizing Trinomials \| \| 4. \| Factoring Trinomial Formula \| \| 5. \| FAQs on Factoring Trinomials \| What is Factoring Trinomials? Factoring trinomials is converting an algebraic expression from a trinomial expression to a binomial expression. A trinomial is a polynomial with three terms with the general expression as ax2 + bx + c, where a and b are coefficients and c is a constant. There are three simple steps to remember while factoring trinomials: Identify the values of b (middle term) and c (last term). Find two numbers that add to b and multiply to c. Use these numbers to factor the expression to obtain the factored terms. Two integers such as r and s are considered to factor a trinomial, whose sum is b and whose product is ac. We can rewrite the trinomial as ax2 + rx + sx + c and then use grouping and the distributive property to factor the polynomial. After the trinomial is undergone the process of factoring, the expression becomes a binomial in the form (x + r) (x + s). Here is an image to understand this better. Rules for Factoring Trinomials For factoring a trinomial there are points or rules to remember. These rules are based on mathematical signs such as (+) and (-) that play an important role while factoring trinomials and make it simple in factoring trinomials. The rules are as follows: If all terms of the trinomial are positive, then all terms of the binomials will be positive. If the last term of the trinomial is negative but the middle term and the first term are positive, then one term of the binomial will be negative and the other will be positive. (The greater factor will be positive and the smaller will be negative). If the middle term and the last term of the trinomial are negative and the first term is positive, then the sign for one binomial will be positive and the other will be negative. (The greater factor will be negative and the smaller will be positive). If the last term and the first term of the trinomial are positive but the middle term is negative, then both signs of the binomials will be negative. Look for common factors for the trinomial ax2 + bx + c, where a is 1. First factor the common factor then factor the rest of the expression. If ax2 is negative in a trinomial, you can factor −1 out of the whole trinomial first. Methods of Factorizing Trinomials Factoring a trinomial means expanding an equation into the product of two or more binomials. It is written as (x + m) (x + n). A trinomial can be factorized in many ways. Let's discuss each case. Quadratic Trinomial in One Variable The general form of quadratic trinomial formula in one variable is ax2 + bx + c, where a, b, c are constant terms and neither a, b, or c is zero. For the value of a, b, c, if b2 - 4ac > 0, then we can always factorize a quadratic trinomial. It means that ax2 + bx + c = a(x + h)(x + k), where h and k are real numbers. Now let's learn how to factorize a quadratic trinomial with an example. Example: Factorize: 3x2 - 4x - 4 Solution: Step 1:- First multiply the coefficient of x2 and the constant term. 3 × -4 = -12 Step 2:- Break the middle term -4x such that on multiplying the resulting numbers, we get the result -12 (obtained from the first step). -4x = -6x + 2x -6 × 2 = -12 Step 3:- Rewrite the main equation by applying the change in the middle term. 3x2 - 4x - 4 = 3x2 - 6x + 2x - 4 Step 4:- Combine the first two terms and the last two terms, simplify the equation and take out any common numbers or expressions. 3x2 - 6x + 2x - 4 = 3x (x - 2) + 2(x - 2) Step 5:- Again take (x - 2) common from both the terms. 3x (x - 2) + 2(x - 2) = (x - 2) (3x + 2) Therefore, (x - 2) and (3x + 2) are the factors of 3x2 - 4x - 4. Quadratic Trinomial in Two Variable There is no specific way to solve a quadratic trinomial in two variables. Let's take an example. Example: Factorize: x2 + 3xy + 2y2 Solution Step 1: These types of trinomials also follow the same rule as above, i.e., we need to break the middle term. x2 + 3xy + 2y2 = x2 + 2xy + xy + 2y2 Step 2: Simplify the equation and take out common numbers of expressions. x2 + 2xy + xy + 2y2 = x (x + 2y) + y (x + 2y) Step 3: Again take (x + 2y) common from both the terms. x (x + 2y) + y (x + 2y) = (x + y) (x + 2y) Therefore, (x + y) and (x + 2y) are the factors of x2 + 3xy + 2y2 If Trinomial is an Identity Let's see some algebraic identities that are mentioned in the table below: \| \| \| --- \| \| Identity \| Expanded Form \| \| (x + y)2 \| x2 + 2xy + y2 \| \| (x - y)2 \| x2 - 2xy + y2 \| \| (x2 - y2) \| (x + y) (x - y) \| Example: Factorize: 9x2 + 12xy + 4y2 Solution: Step 1: Identify which identity can be applied in the expression. We can apply (x + y)2 = x2 + 2xy + y2 Step 2: Rearrange the expression so that it can appear in the form of the above identity. 9x2 + 12xy + 4y2 = (3x)2 + 2 × 3x × 2y + (2y)2 Step 3: Once the expression is arranged in the form of the identity, write its factors. (3x)2 + 2 × 3x × 2y + (2y)2 = (3x + 2y)2 = (3x + 2y) (3x + 2y) Therefore, (3x + 2y) is the factor of 9x2 + 12xy + 4y2. Leading coefficient of 1 Let us look at an example. Example: Factorize x2 + 7x + 12 Solution: Step 1: Compare the given equation with the standard form to obtain the coefficients. ax2 + bx + c is the standard form, comparing the equation x2 + 7x + 12 we get a = 1, b = 7, and c = 12 Step 2: Find the paired factors of c i.e 12 such that their sum is equal to b i.e 7. The pair factor of 12 are (1, 12), (2, 6), and (3, 4). Therefore, the suitable pair is 3 and 4. Step 3: Add each number to x separately. (x + 3) (x + 4) Therefore, (x + 3) (x + 4) are the factors for x2 + 7x + 12. Factorizing with GCF When the trinomial needs to be factorized where the leading coefficient is not equal to 1, the concept of GCF(Greatest Common Factor) is applied. Let us see the steps: Write the trinomial in descending order, from highest to lowest power. Find the GCF by factorization. Find the product of the leading coefficient 'a' and the constant 'c.' Find the factors of the product 'a' and 'c'. Pick a pair that sums up to get the number instead of 'b'. Rewrite the original equation by replacing the term “bx” with the chosen factors. Factor the equation by grouping. Negative Terms In some situations, a is negative, as in −ax2 + bx + c. To make the factoring of the trinomial simpler, we factor out -1 from ax2 as the first step and factor the rest of the expression. Let us look at an example. Example: Factorize -4x2 - 8x - 3. Solution: Step 1: Factor out -1 from the expression which changes the signs of the entire expression. -1 (4x2 + 8x + 3) Step 2: Multiply the first term and the constant term. 4 × 3 = 12 Step 3: Break the middle term 8x such that on multiplying the resulting numbers, we get the result 12 (obtained from the previous step) 8x = 6x + 2x 6 × 2 = 12. Step 4: Rewrite the middle term and group them. -1 (4x2 + 6x + 2x + 3) -1 ([4x2 + 2x] + [6x + 3]) Step 5: Factor the grouped terms. -1 (2x[2x + 1] + 3[2x + 1]) Step 6: Write it as a binomial. -1 [(2x + 1)(2x + 3)] (-2x -1)(2x -3) Therefore, (-2x -1) (2x -3) are the factors of -4x2 - 8x - 3. Factoring Trinomial Formula A trinomial can be a perfect square or a non-perfect square. We have two formulas to factorize a perfect square trinomial. But for factorizing a non-perfect square trinomial, we do not have any specific formula, instead, we have a process. The factoring trinomials formulas of perfect square trinomials are: a2 + 2ab + b2 = (a + b)2 a2 - 2ab + b2 = (a - b)2 For applying either of these formulas, the trinomial should be one of the forms a2 + 2ab + b2 (or) a2 - 2ab + b2. The process of factoring a non-perfect trinomial ax2 + bx + c is: Step 1: Find ac and identify b. Step 2: Find two numbers whose product is ac and whose sum is b. Step 3: Split the middle term as the sum of two terms using the numbers from step - 2. Step 4: Factor by grouping. To factorize a trinomial of the form ax2 + bx + c, we can use any of the below-mentioned formulas: a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) a2 - 2ab + b2 = (a - b)2 = (a - b) (a - b) a2 - b2 = (a + b) (a - b) a3 + b3 = (a + b) (a2 - ab + b2) a3 - b3 = (a - b) (a2 + ab + b2) Related Topics Listed below are a few interesting topics that are related to factoring trinomials, take a look. Perfect Square Trinomial Quadratic Equation Binomial Distribution Read More Download FREE Study Materials Factoring Trinomials Worksheets Factoring Trinomials Worksheets Worksheets on Factoring Trinomials Examples on Factoring Trinomials Example 1: Help Ben factroize x2 + 12x + 8. Solution: x2 + 6x + 8 = x2 + 4x + 2x + 4 × 2 = x2 + 4x + 2x + 8 = x(x + 4) + 2(x + 4) = (x + 2) (x + 4). Therefore, the factors of x2 + 12x + 8 is (x + 2) (x + 4). 2. Example 2: Jen needs to find the factors for 3x3 - 3x2 - 90x, let us help her. Solution: 3x3 - 3x2 - 90x Factor out 3 from the expression as it is the common factor. 3x (x2 - x - 30) Let us factor the expression within the bracket. 3x (x2 – 6x + 5x – 30) 3x [(x2 – 6x) + (5x – 30)] 3x [x(x - 6) + 5(x - 6)] 3x (x + 5)(x - 6). Therefore, the factors of 3x3 - 3x2 - 90x is 3x (x + 5)(x - 6). 3. Example 3: Factor the trinomial 2x2 - x - 3. Solution: We use the factoring trinomial formula of non-perfect trinomials to factor the given trinomial. Comparing 2x2 - x - 3 with ax2 + bx + c, we get a = 2, b = -1, and c = -3. Here ac = 2(-3) = -6 and b = -1. Two numbers whose product is -6 and whose sum is -1 are -3 and 2. We will split the middle term -x as -3x + 2x and then we factor by grouping the terms. 2x2 - x - 3 = 2x2 - 3x + 2x - 3 = x (2x - 3) + 1 (2x - 3) = (2x - 3)(x + 1) Therefore, the factors of 2x2 - x - 3 is (2x - 3)(x + 1). View Answer > Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Practice Questions on Factoring Trinomials Check Answer > FAQs on Factoring Trinomials What is Factoring Trinomials? Factoring trinomials is the process of finding factors for a given trinomial expression. These factors are expressed in the form of binomials that are the sum and product of the terms in a trinomial. The general form of a trinomial is ax2 + bx + c which is converted to a binomial in the form of (x + m)(x + n). How Do You Factor Trinomials? A trinomial can be factored in the form x2 + bx + c. First, we need to find two integers (r and s) whose product sums up to c and through addition, it sums up to b. Once we find the two numbers, we rewrite the trinomial as x2 + rx + sx + c and use the grouping and distributive property of the factor to find out the factors of the expression. The factors will be (x + r) (x + s). What is the Formula to Factor a Trinomial? The factoring trinomials formulas of perfect square trinomials are: a2 + 2ab + b2 = (a + b)2 a2 - 2ab + b2 = (a - b)2 For applying either of these formulas, the trinomial should be one of the forms a2 + 2ab + b2 (or) a2 - 2ab + b2. What are the Basic Rules to Factoring Trinomials? The rules or points to remember while factoring a trinomial are: If all terms of the trinomial are positive, then all terms of the binomials will be positive. If the last term of the trinomial is negative but the middle term and the first term are positive, then one term of the binomial will be negative and the other will be positive. (The greater factor will be positive and the smaller will be negative). If the middle term and the last term of the trinomial are negative and the first term is positive, then the sign for one binomial will be positive and the other will be negative. (The greater factor will be negative and the smaller will be positive). If the last term and the first term of the trinomial are positive but the middle term is negative, then both signs of the binomials will be negative. Look for common factors for the trinomial ax2 + bx + c, where a is 1. First factor the common factor then factor the rest of the expression. If ax2 is negative in a trinomial, you can factor −1 out of the whole trinomial first. What are the Algebraic Identities Used for Factoring Trinomials? The three basic algebraic identities used in factorizing trinomials are: (x + y)2 = x2 + 2xy + y2 (x - y)2 = x2 - 2xy + y2 (x2 - y2) = (x + y) (x - y) How do you Factor a Perfect Square Trinomial? A perfect square trinomial has three terms which may be in the form of (ax)2+ 2abx + b2= (ax + b)2 (or) (ax)2−2abx + b2 = (ax−b)2.The steps to be followed to factor a perfect square polynomial are as follows. Verify whether the given perfect square trinomial is of the form a2+ 2ab + b2 (or) a2−2ab + b2. Check if the middle term is twice the product of the first and the third term. Also, check the sign of the middle term. If the middle term is positive, then compare the perfect square trinomial with a2 + 2ab + b2 and if the middle term is negative, then compare the perfect square trinomial with a2 - 2ab + b2. If the middle term is positive, then the factors are (a+b) (a+b) and if the middle term is negative, then the factors are (a-b) (a-b). Math worksheets and visual curriculum FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team MATH WORKSHEETS Kindergarten Worksheets 1st Grade Worksheets 2nd Grade Worksheets 3rd Grade Worksheets 4th Grade Worksheets 5th Grade Worksheets 6th Grade Worksheets 7th Grade Worksheets 8th Grade Worksheets 9th Grade Worksheets 10th Grade Worksheets Terms and ConditionsPrivacy Policy
9291	https://z.c2.com/cybords/wiki.cgi?PositionalNotation	Your connection is not private Attackers might be trying to steal your information from z.c2.com (for example, passwords, messages, or credit cards). 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9292	https://vsp.pnnl.gov/help/Vsample/Design_Stratified_Pro_Test.htm	VSP Help Click here to see this page in full context Stratified Sampling for Estimating a Proportion Background Information The primary purpose of this sampling design is to estimate the proportion for the entire site, i.e., for all strata combined. Preexisting information is used to divide the site into non-overlapping strata that are expected to be more homogeneous internally than for the entire site (all strata combined). There are two related steps to calculating the number of samples needed to estimate the proportion for the site: Total number of samples Number of samples in each stratum Equation Used to Determine Total Number of Samples The equation depends on the specific method: Method 1: Minimize Variance of Sample Proportion for Fixed Cost The total number of samples is computed to maximize the precision of the estimated population proportion for a pre-specified fixed total cost, (C-c_0), of collecting and measuring samples. Note that the calculation is for the total number of samples, i.e., for combined strata, rather than individual strata. The formula used to calculate the total number of samples is: $$n = \frac{(C-c_0)\displaystyle\sum_{h=1}^{L}\frac{W_h\sqrt{P_h(1-P_h)}}{\sqrt{c_h}}}{\displaystyle\sum_{h=1}^{L}W_h\sqrt{P_h(1-P_h)}\sqrt{c_h}}$$ where \| (L) \| is the number of strata, (h=1,2,...,L), \| \| (P_h) \| is the estimated proportion of measurements in stratum (h), \| \| (W_h = N_h/N) \| is the weight associated with stratum (h), \| \| (N_h) \| is the total number of possible sampling locations (units) in stratum (h), \| \| (N) \| is the total number of possible units in all strata combined, (N = \displaystyle\sum_{h=1}^{L}N_h), \| \| (C) \| is the total sampling budget, (C = c_0 + \displaystyle\sum_{h=1}^{L}c_hn_h), \| \| (c_0) \| is the fixed overhead cost, \| \| (c_h) \| is the cost of collecting and measuring a sample in stratum (h), and \| \| (n_h) \| is the number of samples collected in stratum (h). \| Method 2: Minimize Cost for Required Variance of Sample Proportion The total number of samples is computed to achieve the pre-specified precision of the estimated population proportion for specified stratum costs, but no restriction on total costs. Note that the calculation is for the total number of samples, i.e., for combined strata, rather than individual strata. The formula used to calculate the total number of samples is: $$n = \frac{\Bigg(\displaystyle\sum_{h=1}^{L}W_h\sqrt{P_h(1-P_h)}\sqrt{c_h}\Bigg)\displaystyle\sum_{h=1}^{L}\frac{W_h\sqrt{P_h(1-P_h)}}{\sqrt{c_h}}}{V+\frac{1}{N}\displaystyle\sum_{h=1}^{L}W_hP_h(1-P_h)}$$ where \| (L) \| is the number of strata, (h=1,2,...,L), \| \| (P_h) \| is the estimated proportion of measurements in stratum (h), \| \| (W_h = N_h/N) \| is the weight associated with stratum (h), \| \| (N_h) \| is the total number of possible sampling locations (units) in stratum (h), \| \| (N) \| is the total number of possible units in all strata combined, (N = \displaystyle\sum_{h=1}^{L}N_h), \| \| (V) \| is the pre-specified variance or precision, and \| \| (c_h) \| is the cost of collecting and measuring a sample in stratum (h). \| Method 3: Predetermined Number The user supplies the total number. VSP provides no assurance that this user-supplied number is adequate for any particular design goal. Equation Used to Determine Allocation of Samples to Strata The total number of samples is allocated to the individual strata on an optimal basis using the formula: $$n_h = n\frac{N_h\sqrt{P_h(1-P_h)}/\sqrt{c_h}}{\displaystyle\sum_{h=1}^{L}N_h\sqrt{P_h(1-P_h)}/\sqrt{c_h}}$$ where \| (n_h) \| is the number of samples allocated to stratum (h), \| \| (L) \| is the number of strata, \| \| (N_h) \| is the total number of units in stratum (h), \| \| (P_h) \| is the proportion of measurements in stratum (h), \| \| (c_h) \| is the cost per population unit in stratum (h), \| \| (n) \| is the total number of units sampled in all strata, (n = \displaystyle\sum_{h=1}^{L}n_h). \| Statistical Assumptions The assumptions associated with the formulas for computing the number of samples are: The estimated stratum standard deviation, (P_h), are reasonable and representative of the stratum populations being sampled. The sampling locations are selected using simple random sampling. The stratum costs, (C_h), and the fixed costs (C_0), are accurate. The first and third assumptions will be assessed in a post data collection analysis. The second assumption is valid because simple random sampling is used. Reference: Cochran, W.G. 1977. Sampling Techniques, 3rd edition. John Wiley & Sons, New York. The Stratified dialog contains the following controls: Total Number of Samples Method Minimize Variance of Sample Proportion for Fixed Cost Method: Specify Total Budget Minimize Cost for Required Variance of Sample Proportion Method: Specify Required Variance Predetermined Number Method: Specify Total Number of Samples Number of Strata
9293	https://www.youtube.com/watch?v=A7L--_R9tgM	Visual Proof of Cos²θ + Sin²θ = 1 GCSE Mathematics Zone 37800 subscribers 20 likes Description 615 views Posted: 3 Jan 2024 Explore the fascinating concept of Trigonometry with this visually engaging video that provides concrete proof of the identity Cos^2θ + Sin^2θ = 1. Dive into the world of angles, triangles, and circles as we demonstrate this fundamental relationship using interactive diagrams and clear explanations. Whether you're a student or a math enthusiast, this video will enhance your understanding of this essential trigonometric identity. Trigonometryvisualproof Visualizingcos²θ+sin²θ=1 Explainingcos²θ+sin²θ=1withvisuals 3 comments Transcript: hello many people don't know where this trigonometric identity comes from but in this video I'm going to show you exactly where this identic comes from so to prove that cos s Theta + sin 2 Theta = 1 You Begin by a circle okay a circle on the X or Y plane like this all right so this is the X AIS and this is the Y AIS okay now this is the center the center o now any distance from the center o to the circumference is R okay now if we drop a line from the circumference to the X plane you're are going to form a 90° here all right so now you can see that uh we have formed a right angle triangle here now this point here is XA 1A y because uh this circle is on the X or Y plane okay this point is X Comm y now the distance from here from the center to here where we've dropped this line from is X distance okay then even this one will be y distance then you have an angle here let's say an angle Theta so for you to see nicely I'm going to draw this triangle here okay here all right okay we have this so we have this x then the Y this y then the radius R and the angle Theta all right now by Pythagoras Theorem if you want to find R it means you're supposed to add the square of X and the square of Y so meaning for you to find R you supposed to say uh X2 + y^ 2 is = r squ and this is the first equation and this is Pythagoras okay now because you want to prove that cos thet sin thet isal 1 now let's write uh X and Y in terms of cosine and uh sign so to do that first of all let's write so so this is very helpful so so the first letters stand for this is s cos and tan tangent all right so we write X and Y in terms of cosine and S so let's begin cosine cosine it means here cosine is adjacent over hypotenuse okay [Music] um cosine is adjacent over hypoten so we're going to say because we are referring to this angle here we're going to say cos Theta is equal to adjacent over hypoten adjacent is the side that is attached to the angle apart from the the hypotenuse okay so cosine is adjacent over hypotenuse adjacent over hypotenuse so in this case the adjacent is X so in case you want you may you may get confused to say how do you pick the adjacent the adjacent is the side that is attached to the angle other than the hypotenuse so the side that is attached to the angle is X so we have X over hypotenuse is the radius R all right so which means that cos Theta is = to x / r x/ r now let's solve for x okay so that we come and replace here so to solve for x we multiply both sides of this equation by by R so even here by R all right then we're going to have r cos Theta is equal to this and this R cancels and then and then you remain with X so which means that X isal to R cos Theta all right so this is equation number two uh where we we have written X in terms of cosine Theta right let's also write uh Y in terms of uh sign okay so because we're talking about sign so s sin Theta is equal to Here sign is opposite over hypoten so we are going to have opposite over hypoten which is equal to the opposite we talking about the side that is opposite to this angle which is y so we have y over um hypotenuse is R meaning we are having sin Theta is = y / R again let's solve for y so we're going to multiply by y both sides we're going to to multiply by R sorry both sides then we're going to have R sin Theta is equal to this RoR and this RoR cancels then we have y so this this implies that you're going to have y is equal to R sin Theta now this is equation number three at this stage substitute substitute equation 2 and three [Music] into equation one all right but remember equation one is this one so equation one is X2 + y^ 2 is = to [Music] r² so now we are going to replace uh x with r cos Theta because we said we found that xal R cos Theta so we're going to have R cos Theta and everything squared because X is squared here plus again we found that Y is R sin Theta R sin Theta and everything s is equal r s so we are going to [Music] have R squ cos s thet please don't make a mistake to square the angle is the co that is being squared plus r 2 sin 2 Theta is equal to r² now at this stage you can see that R r² is common in in this term so we're going to factor it out so we have r s cos s theta plus sin 2 Theta is = to r² okay so now at this stage divide throughout by r s okay so we're going to have this soide r s even here r s r s r s cancels you have one then here r s r s this also cancels and it's one okay so what is going to remain is this cos² theta plus sin 2 Theta is equal to this is 1 r² R 2 is [Music] 1 as required as required thank you very much for watching if you're new to this Channel please consider subscribing hit the like button and share
9294	https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_21?srsltid=AfmBOoql_Nec4NkNaBq6VwoAzgD8Z6zgtfz5vCvWEmam6iUWUgmLYP4O	Art of Problem Solving 2024 AMC 12A Problems/Problem 21 - 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 2024 AMC 12A Problems/Problem 21 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2024 AMC 12A Problems/Problem 21 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 (lazy + quick) 5 Solution 4 (transform) 6 Solution 5 (transform) 7 See also Problem Suppose that and the sequence satisfies the recurrence relation for all What is the greatest integer less than or equal to Solution 1 Multiply both sides of the recurrence to find that . Let . Then the previous relation becomes We can rewrite this relation for values of until and use telescoping to derive an explicit formula: Summing the equations yields: Now we can substitute back into our equation: Thus the sum becomes We know that , and we also know that , so the requested sum is equivalent to . All that remains is to calculate , and we know that this value lies between and (see the note below for a proof). Thus, so and thus the answer is . ~eevee9406 Note:. It is obvious that the sum is greater than 1 (since it contains as one of its terms). If you forget this and have to derive this on the exam, here is how: and it is clear that . ~eevee9406 Solution 2 According to the equation given, Suppose , , then , where , then we obtain that Hence, Notice that, so and thus the answer is . ~reda_mandymath Solution 3 (lazy + quick) We'll first try to isolate in terms of . Now, as with many, many of these large summation problems, if we just evaluate the first few values in the series, a pattern should emerge quickly. Here it works out well since our product on the LHS cancels out. Here it becomes glaringly obvious that . So, . We proceed with the same summation strategy as Solution 1 and get our answer of . Note: You only have find the answer's units digit from the answer choices; that's for each sum, giving choice B. ~nm1728 Solution 4 (transform) Set ~luckuso Solution 5 (transform) According to the equation given, The rest continues similar to Solution 1 or 2 ~luckuso See also 2024 AMC 12A (Problems • Answer Key • Resources) Preceded by Problem 20Followed by Problem 22 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
9295	http://www.math.clemson.edu/~sgao/papers/BGM93.pdf	Explicit Factorization of x2k + 1 over Fp with Prime p ≡3 mod 4 Ian F. Blake Department of Electrical and Computer Engineering University of Waterloo Waterloo, Ontario, N2L 3G1, Canada E-mail: ifblake@claude2.uwaterloo.ca Shuhong Gao, Ronald C. Mullin Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario, N2L 3G1, Canada E-mail: sgao@violet.uwaterloo.ca, rcmullin@watmath.uwaterloo.ca Revised August 28, 1992 Abstract. In this note we give a complete explicit factorization of x2k + 1 into irreducible polynomials over Fp for a prime p ≡3 mod 4. As a result we can construct an irreducible polynomial over Fp of degree of any power of 2. Some interesting properties of the coeﬃcients of the irreducibles are noted. We also mention that our results may be useful in applying the Fast Fourier Transform over ﬁnite ﬁelds. Key words: ﬁnite ﬁeld, irreducible polynomial, primitive root of unity, Fast Fourier Transform (FFT). Contact author: Shuhong Gao, Department of Combinatorics and Optimization, University of Waterloo, Waterloo, Ontario, N2L 3G1, Canada. 1 Introduction Let p be a prime with p ≡3 mod 4. We consider the problem of completely factoring x2k + 1 over Fp. As the roots of x2k + 1 are primitive 2k+1th roots of unity (in some extension ﬁeld of Fp), this problem is equivalent to constructing all the minimal poly-nomials over Fp of primitive 2k+1th roots of unity for an arbitrary integer k ≥1. Since the degree of any irreducible factor of x2k + 1 is known to be a power of 2, it is also related to the problem of constructing an irreducible polynomial of degree 2e for any given integer e. The last problem is considered by Lenstra and Shoup . Let 2a be the highest power of 2 in p + 1. Then 2a+1 is the highest power of 2 in p2 −1. Let α ∈Fp2 be of order 2a+1. Then it is known that x2e −α is irreducible over Fp2 for any integer e ≥0. Thus (x2e −α)(x2e −αp) is irreducible over Fp of degree 2e+1 and its roots are primitive 2a+e+1th roots of unity. Both Lenstra and Shoup give simple ways to construct an element of order 2a+1 in Fp2. Since p ≡3 mod 4, −1 is a quadratic nonresidue in Fp and therefore x2 + 1 is irreducible over Fp. So Fp2 = Fp(i) where i = √−1. Let f be the map f : Fp2 →Fp2 deﬁned by f(x) = (1 + x)(p−1)/2. We deﬁne f [k] recursively: f 0 = x, f k+1 = f(f k), k ≥0. Lenstra [4, page 344] points out that, for every k with 2 ≤k ≤a + 1, f k−2 ∈Fp2 has multiplicative order 2k, in particular f a−1 has order 2a+1. (Actually f(x) is one of the square roots of x−1 if xp+1 = 1.) Shoup [7, page 439] suggests taking a−1 successive square roots of i, then the resulting element is of order 2a+1. Taking square roots in Fp2 can be done using the following formula √α = iα(p+1)/4, if α(p−1)/2 = −1, (1 + α(p−1)/2)(p−1)/2α(p+1)/4, otherwise, which holds for any quadratic residue α in Fp(i). Both Lenstra’s and Shoup’s methods construct explicitly one element in Fp2 of order 2k for any k ≤a + 1 by taking square roots. Note that if α is of order 2k then both √α and −√α have order 2k+1 provided k > 0. One may modify their methods 2 to ﬁnd all the elements of order 2k recursively by starting at i = √−1. In order to factor x2k+1 + 1, one then needs to compute the minimal polynomials over Fp of all the elements constructed. We shall show that all the irreducible factors of x2k+1 can be obtained by computing directly their coeﬃcients. As a consequence, we ﬁnd some interesting properties of the coeﬃcients. All the operations will be in Fp. Main Results We assume that p is a prime such that 2a\|(p + 1), 2a+1 ∤(p + 1) with a ≥2. Then 2a+1 is the highest power in p2 −1. Theorem 1 Let H1 = {0}. Recursively deﬁne Hk = {±(u + 1 2 )(p+1)/4 : u ∈Hk−1} for k = 2, 3, · · · , a −1 and Ha = {±(u −1 2 )(p+1)/4 : u ∈Ha−1}. Then, for 1 ≤k ≤a −1, Hk has cardinality 2k−1, x2k + 1 = Y u∈Hk (x2 −2ux + 1), (1) and for any integer e ≥0, x2a+e + 1 = Y u∈Ha (x2e+1 −2ux2e −1). (2) All the factors in the above products are irreducible over Fp. Proof. First note that Fp2 contains all the 2a+1th roots of unity, since 2a+1\|(p2 −1). Note that since 22 ∤(p −1), for 1 ≤k ≤a, every primitive 2k+1th root of unity is of degree 2 over Fp. We prove (1) and (2) by induction on k. 3 For k = 1, note that p ≡3 mod 4, −1 is a quadratic nonresidue in Fp. Hence x2 +1 is irreducible over Fp. Therefore (1) is true for k = 1. Assume that (1) is true for k with 1 ≤k < a. For k + 1, we prove that (1) is true if k + 1 < a and (2) with e = 0 is true if k + 1 = a. Substituting the x in (1) by x2 yields x2k+1 + 1 = Y u∈Hk (x4 −2ux2 + 1). and for a complete factorization it is required to factor x4 −2ux2 + 1 (3) for any u ∈Hk. Let β be a root of (3). Then β is of order 2k+2. As k + 2 ≤a + 1, β is of degree 2 over Fp. The minimal polynomial of β is of the form x2 −2rx + s, (4) where r, s ∈Fp. As β is a root of both (3) and (4), we have β2 + s = 2rβ, (5) and β4 = 2uβ2 −1. (6) Squaring (5) gives β4 = (4r2 −2s)β2 −s2. (7) From (6) and (7) we have (4r2 −2s)β2 −s2 = 2uβ2 −1. As β2 ̸∈Fp (since β2 has order 2k+1 and 2k+1 ∤(p −1)), we must have 4r2 −2s = 2u and s2 = 1. So s = ±1, (8) 4 and r = ± r u + s 2 = ±(u + s 2 )(p+1)/4. (9) The last equation follows from the fact that if w is a quadratic residue in Fp then w(p+1)/4 is a square root of w. We prove that s must be 1 if k < a −1, and −1 if k = a −1. Case 1 k < a −1. Then k + 1 ≤a −1 and k + 3 ≤a + 1. Suppose s = −1 in (8) and (9). Then, from (4), x2 −2rx −1 is irreducible and its roots are primitive 2k+2th roots of unity. Hence the roots of x4 −2rx2 −1 are primitive 2k+3th roots of unity. As k + 3 ≤a + 1, x4 −2rx2 −1 has two irreducible factors of degree 2, and assume x2 −2¯ rx + ¯ s is one of them. Then, by a similar argument leading to (8) and (9), we ﬁnd that ¯ s2 = −1 (10) and 4¯ r2 −2¯ s = 2r (11) have at least one solution (¯ r, ¯ s) ∈Fp × Fp. This is impossible , as −1 is a quadratic nonresidue in Fp. Therefore s = 1 in (8) and (9). Since (3) has irreducible factors of degree 2, for every u ∈Hk, (u + 1)/2 must be a quadratic residue in Fp. Let u1 = ((u + 1)/2)(p+1)/4. Then x4 −2ux2 + 1 = (x2 −2u1x + 1)(x2 −2(−u1)x + 1). So (1) is true for k + 1. Case 2 k = a −1. In this case, k + 2 = a + 1, k + 3 = a + 2 > a + 1. Suppose s = 1 in (8) and (9). Then both x2 −2rx + 1 and x2 + 2rx + 1 are irreducible and have roots being primitive 2a+1th roots of unity. Thus the roots of x4 −2rx2 + 1 (12) 5 and x4 + 2rx2 + 1 (13) are primitive 2a+2th roots of unity. Since p has order 4 modulo 2a+2, a primitive 2a+2th root of unity is of degree 4 over Fp. So (12) and (13) must be irreducible over Fp. It is easy to see that if (r + 1)/2 = ¯ r2 for some ¯ r ∈Fp, then x2 −2¯ rx + 1 divides (12); if (r −1)/2 = ˜ r2 for some ˜ r ∈Fp, then x2 −2˜ rx −1 divides (12). So for (12) to be irreducible, both (r + 1)/2 and (r −1)/2 must be quadratic nonresidues. Similarly, for (13) to be irreducible, both of (−r + 1)/2 and (−r −1)/2 must also be quadratic nonresidues. This is impossible, since −1 is a quadratic nonresidue in Fp and one of (r + 1)/2 and −(r + 1)/2 is a quadratic nonresidue in Fp. Therefore s = −1 in (8) and (9). Hence, for each u ∈Hk, (u −1)/2 is a quadratic residue in Fp. Let u1 = ((u −1)/2)(p+1)/4. Then x4 −2ux2 + 1 = (x2 −2u1x −1)(x2 −2(−u1)x −1). So (2) is true for e = 0. This proves by induction that (1) and (2) with e = 0 hold. As the factors in (1) and (2) (with e = 0) are minimal polynomials of roots of unity, they are all irreducible over Fp. For e > 0, (2) obviously holds as it is true for e = 0. We just need to prove that every factor in (2) is irreducible over Fp. For any u ∈Ha, we have proved that x2 −2ux −1 is irreducible over Fp. Let α1, α2 be its two roots. We know that α1, α2 ∈Fp2 and have order 2a+1. By Theorem 3.75 [5, page 124], x2e −α1 and x2e −α2 are irreducible over Fp2 for any integer e ≥1. Hence (x2e −α1)(x2e −α2) = x2e+1 −2ux2e −1 is irreducible over Fp. This completes the whole proof. 2 6 Note that when p ≡−1 mod 8 (so a > 2), 1/2 is a quadratic residue in Fp. From the above proof we see that if k < a−1 then, for every u ∈Hk, (u+1)/2 is a quadratic residue in Fp,thus u + 1 is a quadratic residue. Observe that the irreducibility of x2 −2ux + 1 = (x −u)2 −(u2 −1) implies that u2 −1 = (u −1)(u + 1) is a quadratic nonresidue. So u −1 is a quadratic nonresidue. Similarly, for u ∈Ha−1, u −1 is a quadratic residue and u+1 is a quadratic nonresidue. For u ∈Ha, we can only say that u2 + 1 is a quadratic nonresidue due to the irreducibility of x2 −2ux −1. In summary, we have Corollary 2 If p ≡−1 mod 8 (hence a > 2), then (a) for each 1 ≤k < a −1 and u ∈Hk, u + 1 is a quadratic residue in Fp and u −1 is a quadratic nonresidue in Fp; (b) for each u ∈Ha−1, u −1 is a quadratic residue in Fp and u + 1 is a quadratic nonresidue in Fp; (c) for each u ∈Ha, u2 + 1 is a quadratic nonresidue in Fp. This solves, in a theoretical sense, a problem arising from primality testing [3, (11.6)(a)] and [2, section 5], as remarked by Lenstra [4, page 344]. Corollary 3 For 1 ≤k ≤a, let u ∈Hk. Deﬁne v = (1 −u2)(p+1)/4, if k < a, (−1 −u2)(p+1)/4, if k = a. Then u + iv ∈Fp2 = Fp(i) is a 2k+1th primitive root of unity where i = √−1. Proof. For u ∈Hk with k < a, we know from Corollary 2 that 1 −u2 is a quadratic residue in Fp. So v = (1 −u2)(p+1)/4 is a square root of 1 −u2, that is, v2 = 1 −u2. Hence u + iv is a root of x2 −2ux + 1. By Theorem 1, u + iv is a 2k+1th primitive root of unity. For u ∈Ha, the proof is similar. 2 7 As x2t −1 = (x −1) Qt−1 i=0(x2i + 1), the following corollary is an immediate conse-quence of Theorem 1. Corollary 4 For any integer t ≥1, the following factorization over Fp is complete: (a) if t < a + 1, then x2t −1 = (x −1)(x + 1) t−1 Y i=1 Y u∈Hi (x2 −2ux + 1); (b) if t ≥a + 1, then x2t −1 = (x −1)(x + 1) Y u∈Hi 1≤i≤a−1 (x2 −2ux + 1) Y u∈Ha 0≤r≤t−a−1 (x2r+1 −2ux2r −1). Remark We mention a possible application of the preceding results in applying the Fast Fourier Transform (FFT) over ﬁnite ﬁelds [6, Chapter IX] and [1, Chapter 7]. The FFT is widely used in many areas including computing the convolution of data, digital signal processing and computing products of polynomials or integers. In , to apply the FFT over ﬁnite ﬁelds one chooses an appropriately large N = 2e and a prime p of the form Nk + 1. If an Nth root of unity ω in Fp is given, then the FFT evaluates a polynomial in Fp[x] of degree at most N at the N points 1, ω, ω2, . . . , ωN−1 in time O(N log N). The problem here is that, when an integer e and a prime p = 2ek + 1 are given, there is currently no deterministic polynomial time (in log p and e) algorithm to construct a 2eth primitive root of unity in Fp. It is suggested in to apply the FFT over the ring Zm of integers modulo m where m = 2N/2 + 1 (which is not necessarily a prime). One advantage of Zm is that 2 is known to be a primitive Nth root of unity in Zm. Since the number m is exponential in N, the computation in Zm may be expensive for large N. In the following we show that such problems do not exist if one operates the FFT over Fp2. 8 Let e ≥1 be a positive integer and N = 2e. Let p be any prime of the form 2Nk−1. Deﬁne u = ue inductively: u1 = 0 and uk = (1 + uk−1 2 )(p+1)/4, k = 2, 3, . . . , e. Let v = (1 −u2)(p+1)/4. Then, by Theorem 1 and Corollary 3, ω = u + iv ∈Fp2 = Fp(i) is a 2eth primitive root of unity where i = √−1. Here the number of Fp-operations needed to get u + iv is O(e log p). So one can compute a 2eth primitive root of unity in Fp2 quickly for any given integer e and prime p of the form 2Nk −1. Also, for ﬁxed N = 2e, the generalized prime number theorem implies that the number of primes 2Nk −1 ≤N 2 is approximately N/(2e). This means that primes of the required form exist in reasonable abundance and their sizes can be bounded by N 2. So the problems encountered in and are avoided when the FFT is applied over Fp2. Conclusion We have given a direct way to compute the coeﬃcients of the irreducible factors of x2k + 1 over Fp for a prime p ≡3 mod 4 and for any given integer k. From the coeﬃcients of these irreducible factors, one can produce many quadratic residues and quadratic nonresidues in Fp. It was also noticed that our results may be useful in applying the Fast Fourier Transform over ﬁnite ﬁelds. References Aho, A. V., Hopcroft, J. E., Ullman, J. D.: The Design and Analysis of Computer Algorithms. Addison-Wesley, Reading, MA, 1974. 9 Borho, W., J. Buhl, H. Hoﬀmann, S. Mertens, E. Nebgen and R. Reckow: Große Primzahlen und Befreundete Zahlen: ¨ Uber den Lucas-Test und Thabit-Regeln. Mitt. Math. Ges. Hamburg 11, 232–256 (1983). Cohen, H., Lenstra, H. W., Jr.: Primality testing and Jacobi sums. Math. Comp. 42, 297–330 (1984). Lenstra, H. W.: Finding isomorphisms between ﬁnite ﬁelds. Math. Comp. 56, 329–347 (1991). Lidl, R., Niederreiter H.: Finite Fields. Addison-Wesley, Reading, MA, 1983. Lipson, J. D.: Elements of Algebra and Algebraic Computing. Ben-jamin/Cummings, 1981. Shoup, V.: New algorithms for ﬁnding irreducible polynomials over ﬁnite ﬁelds. Math. Comp. 54, 435–447 (1990). 10
9296	https://brainly.com/question/45351795	[FREE] Consider a circle with its center lying on the focus of the parabola y^2=2px such that it touches the - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +38,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +37k Ace exams faster, with practice that adapts to you Practice Worksheets +7,1k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Consider a circle with its center lying on the focus of the parabola y 2=2 p x such that it touches the directrix of the parabola. Then the point of intersection of the circle and parabola can be: A. (−p/2,p) B. (p/2,p) C. (p/2,−p) D. (p/2,−p/2) 1 See answer Explain with Learning Companion NEW Asked by tacojordan5093 • 12/29/2023 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 2183652 people 2M 0.0 0 Upload your school material for a more relevant answer The intersection point of the circle and parabola, where the center of the circle lies on the focus of the parabola and it touches the directrix, is at (p/2, p), representing option B. Explanation The student is asking about the intersection points of a circle and a parabola where the center of the circle is at the focus of the parabola, and it touches the directrix of the parabola. Given the parabola y2 = 2px, the focus is at (p/2, 0) and the directrix is the line x = -p/2. The circle with its center at the focus will have a radius p/2 to touch the directrix. For the circle and the parabola to intersect, we must solve the system of equations of both shapes. The equation of the circle with center (p/2, 0) and radius p/2 is (x - p/2)2 + y2 = (p/2)2. Substituting the equation of the parabola into the circle's equation, we can solve for x and y. After solving, we determine that the intersection point is at (p/2, p), which corresponds to option B. The point (p/2, p) lies on both the parabola and the circle, implying that this is the point of intersection for the two shapes. Answered by ArloHugo •17.6K answers•2.2M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 2183652 people 2M 0.0 0 Celestial Mechanics - Jeremy Tatum Optics - Sander Konijnenberg, Aurèle J.L. Adam, & H. Paul Urbach Physics for K-12 - Sunil Singh Upload your school material for a more relevant answer The intersection point of the circle and the parabola, where the center of the circle is at the focus of the parabola and it touches the directrix, is at (2 p,p), representing option B. This conclusion is drawn by analyzing the properties of both the parabola and the circle mathematically. Ultimately, we find that this point satisfies both equations involved in the problem. Explanation To determine the intersection point of the circle and the parabola, we first need to analyze the given parabola defined by the equation y 2=2 p x. Step 1: Understanding the Parabola The vertex of the parabola is at the origin (0,0). The focus of the parabola is at the point (2 p,0) and the directrix is at the line x=−2 p. Step 2: Circle Touching the Directrix Since the circle is centered at the focus (2 p,0) and touches the directrix, its radius must equal the distance from the center to the directrix. The distance from (2 p,0) to the line x=−2 p is: Distance=2 p−(−2 p)=p. Therefore, the radius of the circle is p. The equation of the circle can be described as: (x−2 p)2+y 2=p 2. Step 3: Solving for Intersections We now need to find the intersection points of this circle with the parabola. Substitute the expression for y 2 from the parabola into the circle's equation: (x−2 p)2+2 2 p x=p 2. Simplifying this equation will yield the points of intersection. After substituting and solving through these equations, we can find the potential intersection points. By substituting different x-values into the parabola's equation and checking for both equations to hold, we focus on checking the provided answer choices. The calculated points of intersection based on analysis show that one likely point is (2 p,p). Conclusion The viable intersection point that satisfies both the circle and parabola conditions and allows for the geometric properties outlined leads us to conclude that the point of intersection of the circle and parabola is at (2 p,p), corresponding to option B: (2 p,p). Examples & Evidence Consider the parabola described; you can graph it along with the circle to visualize how they intersect. By calculating specific coordinates, you can derive intersection points and verify their positions on both graphs. For instance, calculating for specific p-values may clarify this further. The evidence supporting the derived intersection point comes from the properties of conic sections, particularly the geometric definitions of parabolas and circles, as well as the symmetry inherent in these figures. Thanks 0 0.0 (0 votes) Advertisement tacojordan5093 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer Normal at the point P to the parabola , intersects the circle with SP as diameter at Q also. If PQ = 2 units (given that point S is focus of the given parabola ), then Abscissa of point P is 2 3 4 Community Answer let pq be a focal chord of the parabola 2 x py 4 . show that the circle with diameter pq is tangent to the directrix of the parabola Community Answer Consider the parabola with focus at (0, 2 3 ) and directrix at y=− 2 1 . (a) Determine the standard form equation of the parabola. (b) What are the points where the latus rectum intersects the parabola? (c) What is the standard form equation of the circle with the latus rectum of the parabola as the diameter of the circle? Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Using the Pythagorean theorem, find the length of a leg of a right triangle if the other leg is 8 feet long and the hypotenuse is 10 feet long. A. 41 f t B. 12.81 ft. C. 6 ft. D. 36 ft. Hence, for 0⩽u⩽360 and 0⩽v⩽36 0∘, solve the simultaneous equations 2 sin u+cos v=2 sin u−cos v=−2 1. [2 2 x−3]={3 x−2}. The distance from the center of a round table top to the edge of the table top is 4 ft. What's the area of the table top? A. 25.14 sq. ft. B. 50.24 sq. ft. C. 38 sq. ft. D. 16 sq. ft. Solve the following system of equations. {−3 x−4 y=−10−6 x−7 y=−13 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
9297	https://discourse.mc-stan.org/t/treating-a-missing-count-data-as-a-parameter-and-putting-truncated-binomial-prior-for-it/33098	Treating a missing count data as a parameter and putting truncated binomial prior for it Dear Stan users and developers, I am a new Stan user transitioning from JAGS. In my model, I want to treat a few missing count data points as parameters and put truncated binomial priors on them (some example code is shown below). The “p” parameter in the truncated binomial will be shared in some other places in the model system, and the “N” and truncated boundary “upper” are known data. It seems that Stan does not support this kind of integer parameter specification (i.e., Y_miss) since in the “parameter” section, only “real” type is supported (the code below will result in the error “No distribution ‘binomial’ was found with the correct signature.”). Does anyone have similar experience or have any suggestions for this issue when using Stan? Thanks very much for any discussions. (In JAGS, this kind of prior can be specified and this model can work.) data { int<lower=1> N; int<lower=1, upper=N> upper; // other data are left out } parameter { real Y_miss; // other parameters are left out } model { Y_miss ~ binomial(N, p) T[0, upper]; //I want to put truncated binomial prior for this Y_miss parameter // likelihood function depending on Y_miss (left out) } Sincerely, Terrence Welcome! Could you say a bit more about the other aspects of your model, or share the full code? It’s possible that your specific model might have some simplified form for the missing data that others will be able to identify. At the very least, you could create a model for Y and use that to infer the probabilities of each value of Y_miss. You could then marginalize over those probabilities to account for the uncertainty in the true value. This would be pretty easy and efficient if the rest of your model is straight-forward and upper is modest. It would get inefficient if, for example, upper = 10000. upper upper = 10000 You should be able to find a handful of topics here on marginalizing unobserved discrete variables. The Stan User Guide’s discussion of finite mixture models is also helpful. 5 Finite Mixtures \| Stan User’s Guide Thanks very much for your reply! I will go to check that Guide later. Let me provide some more details about the example data and model code. I have many observations of multinomial counts (3 categories), e.g., row 1: (10, 10, 10), row 2: (12, 8, 10), …, row S: (9, 9, 12). My primary target is to estimate the 3 probability parameters corresponding to the 3 multinomial categories. The first S multivariate observations are fully observed and can be fed into the common multinomial likelihood (with N=30, p=p[1:3]). Due to some special reasons, I also collect 5 extra observations with (NA, NA, NA), with informative missingness. The informative missingness means, I know the missing value for category 1 should be smaller than or equal to 10, and the missing value for category 2 is affirmed to be smaller than or equal to 8. Due to this informative missingness, the missing observations should contribute in the model. The strategy I planned for this is, treating the missing category 1 value as a parameter called Y_miss and putting prior, Truncated Binomial(N=30, p=p)T[0,10] on it. And then, conditional on the category 1 value (Y_miss), the category 2 count’s corresponding random variable will follow Binomial(N-Y_miss, p=p/(p+p)). For also incorporating the information that “category 2 value should be <=8”, we can add the cumulative version of the binomial distribution function above into the likelihood function to help enhance the estimation of p[1:3] (see the sample code below). data { int S; int<lower=0> Y[S,3]; } parameters { real<lower=0, upper=1> p; real Y_miss; //the type here is problematic, but I do not know how to deal with this given the request below } model { // some priors can be put for p[1:3], e.g., Dirichlet for (s in 1:S){ //for those S rows of complete data target += multinomial_lupmf(Y[s,]\|to_vector(p)); } for (i in 1:5){ //for those 5 rows of informative NA's Y_miss[i] ~ binomial(30, p) T[0, 10]; //I want to put truncated binomial prior for this Y_miss parameter target += binomial_lcdf(8 \| 30-Y_miss[i], p/(p+p)) //this is my planned likelihood function to handle this informative missing problem, not 100% sure this is appropriate } I hope the explanation and the example code clarifies my inquiry better. This model looks a little overparameterized by involving those missing value parameters, but they are nuisance and ancillary ones. Thanks for any comments on the Stan implementation of this idea or any critiques on my model thoughts for this problem. Sincerely, Terrence Hi Terrence, I would interpret the observation model you describe as “censoring” rather than truncation. The “informative missingness” tells you that the true number of counts in some bin is below some value (i.e., left-censored data). This is for instance analogous to survival analysis, in which right-censoring occurs (we know persons’ or items’ survival up to point in time, after which we stopped observing them). Truncation is a different observation process, i.e., events below/under some threshold do not appear in the data at all. So, the likelihood for these informatively missing counts in your model should be a cumulative probability of that count being under the thresholds you mention. As explained by @simonbrauer, integrating over all possible counts by means of the HMC itself is not possible in Stan as integer parameters are not allowed. So integration by marginalising over all possible counts is the way to go. Now, I’m not sure that using the binomial distribution here is correct, as counts in any of two of the three categories are correlated with one another. The binomial only applies when if missingness is informative in only one of categories. So, you need a cumulative probability function for a multinomial distribution. I don’t what this is from the top of my head, but imagine it’s manageable. Maybe someone else here or the interweb itself knows. [EDIT: typo correction] A quick search for multinomial cumulative distribution function yielded this 1981 (pay walled from the train so haven’t read): The multinomial CDF came up over in the Julia forums a while back. There are a handful of methods available but some are way over my head. Frey wrote some R code for one approach that I think could be re-written in Stan–though it might not be efficient enough to be practically useful in this case. Lebrun describes a more efficient method, as well as a nice summary of other methods and their limitations. With at most a \times b = 80 possible combinations of censoring (a <8 counts in one category and b<10 counts in another), a brute force integration by marginalising may not even be so bad, @simonbrauer? That does seem likely. Given the current model has no additional predictors of the outcome, the probabilities will be the same for all unobserved cases. So @TerrenceT could calculate those probabilities once rather than 5 times. Related topics \| Topic \| \| Replies \| Views \| Activity \| --- --- \| Truncated model for neg_binomial_2 Modeling \| 20 \| 1707 \| June 9, 2017 \| \| Estimating the binomial rate when number of trials is uncertain Modeling \| 12 \| 2645 \| June 27, 2018 \| \| Assigning bernoulli prior to missing entries in covariate Modeling \| 2 \| 1066 \| August 15, 2019 \| \| Marginalising out many discrete latent variables / modelling augmented count data Modeling techniques \| 12 \| 1487 \| July 17, 2018 \| \| A question on "stan cannot deal with discrete parameters" Developers \| 19 \| 6759 \| February 28, 2018 \| Powered by Discourse, best viewed with JavaScript enabled
9298	https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/14%3A_Differentiation_of_Functions_of_Several_Variables/14.02%3A_Limits_and_Continuity	Published Time: 2016-07-11T18:47:27Z 14.2: Limits and Continuity - Mathematics LibreTexts Processing math: 21% 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 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Home 2. Bookshelves 3. Calculus 4. Calculus (OpenStax) 5. 14: Differentiation of Functions of Several Variables 6. 14.2: Limits and Continuity Expand/collapse global location 14.2: Limits and Continuity Last updated Feb 6, 2025 Save as PDF 14.1E: Exercises for Section 14.1 14.2E: Exercises for Section 14.2 Page ID 2601 Gilbert Strang & Edwin “Jed” Herman OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Limit of a Function of Two Variables 1. Definition: δ Disks 2. Definition: limit of a function of two variables 3. Theorem 14.2.1: Limit Laws for Functions of Two Variables 4. Example 14.2.1: Finding the Limit of a Function of Two Variables 1. Solution 5. Exercise 14.2.1:/14:_Differentiation_of_Functions_of_Several_Variables/14.02:_Limits_and_Continuity#Exercise_.5C(.5CPageIndex.7B1.7D.5C):) 6. Example 14.2.2: Limits That Fail to Exist/14:_Differentiation_of_Functions_of_Several_Variables/14.02:_Limits_and_Continuity#Example_.5C(.5CPageIndex.7B2.7D.5C):_Limits_That_Fail_to_Exist) 1. Solution/14:_Differentiation_of_Functions_of_Several_Variables/14.02:_Limits_and_Continuity#Solution_2) 7. Exercise 14.2.2:/14:_Differentiation_of_Functions_of_Several_Variables/14.02:_Limits_and_Continuity#Exercise_.5C(.5CPageIndex.7B2.7D.5C):) Interior Points and Boundary Points Definition: interior and boundary points Definition: Open and closed sets Definition: connected sets and Regions Definition Example 14.2.3: Limit of a Function at a Boundary Point Solution Exercise 14.2.3 Continuity of Functions of Two Variables Definition: continuous Functions Example 14.2.4: Demonstrating Continuity for a Function of Two Variables Solution Exercise 14.2.4 Theorem 14.2.2: The Sum of Continuous Functions is Continuous Theorem 14.2.3: The Product of Continuous Functions is Continuous Theorem 14.2.4: The Composition of Continuous Functions is Continuous Example 14.2.5: More Examples of Continuity of a Function of Two Variables Solution Exercise 14.2.5 Functions of Three or More Variables Definition: δ-balls Example \PageIndex{6}: Finding the Limit of a Function of Three Variables Solution Exercise \PageIndex{6} Key Concepts Glossary Learning Objectives Calculate the limit of a function of two variables. Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach. State the conditions for continuity of a function of two variables. Verify the continuity of a function of two variables at a point. Calculate the limit of a function of three or more variables and verify the continuity of the function at a point. We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don’t occur with functions of one variable. Limit of a Function of Two Variables Recall from Section 2.5 that the definition of a limit of a function of one variable: Let f(x) be defined for all x≠a in an open interval containing a. Let L be a real number. Then \lim_{x→a}f(x)=L \nonumber if for every ε>0, there exists a δ>0, such that if 0<\|x−a\|<δ for all x in the domain of f, then \|f(x)−L\|<ε. \nonumber Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables. Definition: \delta Disks Consider a point (a,b)∈\mathbb{R}^2. A δdisk centered at point (a,b) is defined to be an open disk of radius δ centered at point (a,b) —that is, {(x,y)∈\mathbb{R}^2∣(x−a)^2+(y−b)^2<δ^2} \nonumber as shown in Figure \PageIndex{1}. Figure \PageIndex{1}: A δ disk centered around the point (2,1). The idea of a δ disk appears in the definition of the limit of a function of two variables. If δ is small, then all the points (x,y) in the δ disk are close to (a,b). This is completely analogous to x being close to a in the definition of a limit of a function of one variable. In one dimension, we express this restriction as a−δ<x<a+δ. \nonumber In more than one dimension, we use a δ disk. Definition: limit of a function of two variables Let f be a function of two variables, x and y. The limit of f(x,y) as (x,y) approaches (a,b) is L, written \lim_{(x,y)→(a,b)}f(x,y)=L \nonumber if for each ε>0 there exists a small enough δ>0 such that for all points (x,y) in a δ disk around (a,b), except possibly for (a,b) itself, the value of f(x,y) is no more than ε away from L (Figure \PageIndex{2}). Using symbols, we write the following: For any ε>0, there exists a number δ>0 such that \|f(x,y)−L\|<ε \nonumber whenever 0<\sqrt{(x−a)^2+(y−b)^2}<δ. \nonumber Figure \PageIndex{2}: The limit of a function involving two variables requires that f(x,y) be within ε of L whenever (x,y) is within δ of (a,b). The smaller the value of ε, the smaller the value of δ. Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in the section titled The Limit Laws in single-variable calculus. Theorem \PageIndex{1}: Limit Laws for Functions of Two Variables Let f(x,y) and g(x,y) be defined for all (x,y)≠(a,b) in a neighborhood around (a,b), and assume the neighborhood is contained completely inside the domain of f. Assume that L and M are real numbers such that \lim_{(x,y)→(a,b)}f(x,y)=L \nonumber and \lim_{(x,y)→(a,b)}g(x,y)=M, \nonumber and let c be a constant. Then each of the following statements holds: Constant Law: \lim_{(x,y)→(a,b)}c=c \nonumber Identity Laws: \lim_{(x,y)→(a,b)}x=a \nonumber \lim_{(x,y)→(a,b)}y=b \nonumber Sum Law: \lim_{(x,y)→(a,b)}(f(x,y)+g(x,y))=L+M \nonumber Difference Law: \lim_{(x,y)→(a,b)}(f(x,y)−g(x,y))=L−M \nonumber Constant Multiple Law: \lim_{(x,y)→(a,b)}(cf(x,y))=cL \nonumber Product Law: \lim_{(x,y)→(a,b)}(f(x,y)g(x,y))=LM \nonumber Quotient Law: \lim_{(x,y)→(a,b)}\dfrac{f(x,y)}{g(x,y)}=\dfrac{L}{M} \text{ for } M≠0 \nonumber Power Law: \lim_{(x,y)→(a,b)}(f(x,y))^n=L^n \nonumber for any positive integer n. Root Law: \lim_{(x,y)→(a,b)}\sqrt[n]{f(x,y)}=\sqrt[n]{L} \nonumber for all L if n is odd and positive, and for L≥0 if n is even and positive. The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions. Example \PageIndex{1}: Finding the Limit of a Function of Two Variables Find each of the following limits: \displaystyle \lim_{(x,y)→(2,−1)}(x^2−2xy+3y^2−4x+3y−6) \displaystyle \lim_{(x,y)→(2,−1)}\dfrac{2x+3y}{4x−3y} Solution a. First use the sum and difference laws to separate the terms: \begin{align} \lim_{(x,y)→(2,−1)}(x^2−2xy+3y^2−4x+3y−6)\ = \left(\lim_{(x,y)→(2,−1)}x^2 \right)− \left(\lim_{(x,y)→(2,−1)}2xy \right)+ \left(\lim_{(x,y)→(2,−1)}3y^2 \right)−\left(\lim_{(x,y)→(2,−1)}4x\right) \ + \left(\lim_{(x,y)→(2,−1)}3y \right)−\left(\lim_{(x,y)→(2,−1)}6\right). \end{align} Next, use the constant multiple law on the second, third, fourth, and fifth limits: \begin{align} =(\lim_{(x,y)→(2,−1)}x^2)−2(\lim_{(x,y)→(2,−1)}xy)+3(\lim_{(x,y)→(2,−1)}y^2)−4(\lim_{(x,y)→(2,−1)}x) \[4pt] +3(\lim_{(x,y)→(2,−1)}y)−\lim_{(x,y)→(2,−1)}6.\end{align} Now, use the power law on the first and third limits, and the product law on the second limit: \begin{align} \left(\lim_{(x,y)→(2,−1)}x\right)^2−2\left(\lim_{(x,y)→(2,−1)}x\right) \left(\lim_{(x,y)→(2,−1)}y\right)+3\left(\lim_{(x,y)→(2,−1)}y\right)^2 \ −4\left(\lim_{(x,y)→(2,−1)}x\right)+3\left(\lim_{(x,y)→(2,−1)}y\right)−\lim_{(x,y)→(2,−1)}6. \end{align} Last, use the identity laws on the first six limits and the constant law on the last limit: \begin{align} \lim_{(x,y)→(2,−1)}(x^2−2xy+3y^2−4x+3y−6) = (2)^2−2(2)(−1)+3(−1)^2−4(2)+3(−1)−6 \[4pt] =−6. \end{align} b. Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law, \begin{align} \lim_{(x,y)→(2,−1)}(4x−3y) =\lim_{(x,y)→(2,−1)}4x−\lim_{(x,y)→(2,−1)}3y \[4pt] =4(\lim_{(x,y)→(2,−1)}x)−3(\lim_{(x,y)→(2,−1)}y) \[4pt] =4(2)−3(−1)=11. \end{align} Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law: \begin{align} \lim_{(x,y)→(2,−1)}(2x+3y) =\lim_{(x,y)→(2,−1)}2x+\lim_{(x,y)→(2,−1)}3y \[4pt] =2(\lim_{(x,y)→(2,−1)}x)+3(\lim_{(x,y)→(2,−1)}y) \[4pt] =2(2)+3(−1)=1. \end{align} Therefore, according to the quotient law we have \begin{align} \lim_{(x,y)→(2,−1)}\dfrac{2x+3y}{4x−3y} =\dfrac{\displaystyle \lim_{(x,y)→(2,−1)}(2x+3y)}{\displaystyle \lim_{(x,y)→(2,−1)}(4x−3y)} \[4pt] =\dfrac{1}{11}. \end{align} Exercise \PageIndex{1}: Evaluate the following limit: \lim_{(x,y)→(5,−2)}\sqrt{\dfrac{x^2−y}{y^2+x−1}}. \nonumber Hint Use the limit laws. Answer \displaystyle \lim_{(x,y)→(5,−2)}\sqrt{\dfrac{x^2−y}{y^2+x−1}}=\dfrac{3}{2} \nonumber Since we are taking the limit of a function of two variables, the point (a,b) is in \mathbb{R}^2, and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward (a,b). If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken. Example \PageIndex{2}: Limits That Fail to Exist Show that neither of the following limits exist: \displaystyle \lim_{(x,y)→(0,0)}\dfrac{2xy}{3x^2+y^2} \displaystyle \lim_{(x,y)→(0,0)}\dfrac{4xy^2}{x^2+3y^4} Solution a. The domain of the function f(x,y)=\dfrac{2xy}{3x^2+y^2} consists of all points in the xy-plane except for the point (0,0) (Figure \PageIndex{3}). To show that the limit does not exist as (x,y) approaches (0,0), we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point (0,0). First, consider the line y=0 in the xy-plane. Substituting y=0 into f(x,y) gives f(x,0)=\dfrac{2x(0)}{3x^2+0^2}=0 \nonumber for any value of x. Therefore the value of f remains constant for any point on the x-axis, and as y approaches zero, the function remains fixed at zero. Next, consider the line y=x. Substituting y=x into f(x,y) gives f(x,x)=\dfrac{2x(x)}{3x^2+x^2}=\dfrac{2x^2}{4x^2}=\tfrac{1}{2}. \nonumber This is true for any point on the line y=x. If we let x approach zero while staying on this line, the value of the function remains fixed at \tfrac{1}{2}, regardless of how small x is. Choose a value for ε that is less than 1/2—say, 1/4. Then, no matter how small a δ disk we draw around (0,0), the values of f(x,y) for points inside that δ disk will include both 0 and \tfrac{1}{2}. Therefore, the definition of limit at a point is never satisfied and the limit fails to exist. Figure \PageIndex{3}: Graph of the function f(x,y)=\dfrac{2xy}{3x^2+y^2}. Along the line y=0, the function is equal to zero; along the line y=x, the function is equal to \tfrac{1}{2}. b. In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the x-axis (i.e., y=0), then the function remains fixed at zero. The same is true for the y-axis. Suppose we approach the origin along a straight line of slope k. The equation of this line is y=kx. Then the limit becomes \begin{align} \lim_{(x,y)→(0,0)}\dfrac{4xy^2}{x^2+3y^4} = \lim_{(x,y)→(0,0)}\dfrac{4x(kx)^2}{x^2+3(kx)^4} \ = \lim_{(x,y)→(0,0)}\dfrac{4k^2x^3}{x^2+3k^4x^4} \ =\lim_{(x,y)→(0,0)}\dfrac{4k^2x}{1+3k^4x^2} \ = \dfrac{\displaystyle \lim_{(x,y)→(0,0)}(4k^2x)}{\displaystyle \lim_{(x,y)→(0,0)}(1+3k^4x^2)} \ = 0. \end{align} regardless of the value of k. It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation x=y^2. Substituting y^2 in place of x in f(x,y) gives \begin{align}\lim_{(x,y)→(0,0)}\dfrac{4xy^2}{x^2+3y^4} = \lim_{(x,y)→(0,0)}\dfrac{4(y^2)y^2}{(y^2)^2+3y^4} \ = \lim_{(x,y)→(0,0)}\dfrac{4y^4}{y^4+3y^4} \ = \lim_{(x,y)→(0,0)}1 \ = 1. \end{align} By the same logic in part a, it is impossible to find a δ disk around the origin that satisfies the definition of the limit for any value of ε<1. Therefore, \displaystyle \lim_{(x,y)→(0,0)}\dfrac{4xy^2}{x^2+3y^4} \nonumber does notexist. Exercise \PageIndex{2}: Show that \lim_{(x,y)→(2,1)}\dfrac{(x−2)(y−1)}{(x−2)^2+(y−1)^2} \nonumber does not exist. Hint Pick a line with slope k passing through point (2,1). Answer If y=k(x−2)+1, then \lim_{(x,y)→(2,1)}\dfrac{(x−2)(y−1)}{(x−2)^2+(y−1)^2}=\dfrac{k}{1+k^2}. Since the answer depends on k, the limit fails to exist. Interior Points and Boundary Points To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology. Definition: interior and boundary points Let S be a subset of \mathbb{R}^2 (Figure \PageIndex{4}). A point P_0 is called an interior point of S if there is a δ disk centered around P_0 contained completely in S. A point P_0 is called a boundary point of S if every δ disk centered around P_0 contains points both inside and outside S. Figure \PageIndex{4}: In the set S shown, (−1,1) is an interior point and (2,3) is a boundary point. Definition: Open and closed sets Let S be a subset of \mathbb{R}^2 (Figure \PageIndex{4}). S is called an open set if every point of S is an interior point. S is called a closed set if it contains all its boundary points. An example of an open set is a δ disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a δ disk but not the other half, then the set is neither open nor closed. Definition: connected sets and Regions Let S be a subset of \mathbb{R}^2 (Figure \PageIndex{4}). An open set S is a connected set if it cannot be represented as the union of two or more disjoint, nonempty open subsets. A set S is a region if it is open, connected, and nonempty. The definition of a limit of a function of two variables requires the δ disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the δdisk is not contained inside the domain. By definition, some of the points of the δdisk are inside the domain and some are outside. Therefore, we need only consider points that are inside both the δ disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point. Definition Let f be a function of two variables, x and y, and suppose (a,b) is on the boundary of the domain of f. Then, the limit of f(x,y) as (x,y) approaches (a,b) is L, written \lim_{(x,y)→(a,b)}f(x,y)=L, \nonumber if for any ε>0, there exists a number δ>0 such that for any point (x,y) inside the domain of f and within a suitably small distance positive δ of (a,b), the value of f(x,y) is no more than ε away from L (Figure \PageIndex{2}). Using symbols, we can write: For any ε>0, there exists a number δ>0 such that \|f(x,y)−L\|<ε\, \text{whenever}\, 0<\sqrt{(x−a)^2+(y−b)^2}<δ. \nonumber Example \PageIndex{3}: Limit of a Function at a Boundary Point Prove \lim_{(x,y)→(4,3)}\sqrt{25−x^2−y^2}=0. \nonumber Solution The domain of the function f(x,y)=\sqrt{25−x^2−y^2} is \big{(x,y)∈\mathbb{R}^2∣x^2+y^2≤25\big}, which is a circle of radius 5 centered at the origin, along with its interior as shown in Figure \PageIndex{5}. Figure \PageIndex{5}: Domain of the function f(x,y)=\sqrt{25−x^2−y^2}. We can use the limit laws, which apply to limits at the boundary of domains as well as interior points: \begin{align} \lim_{(x,y)→(4,3)}\sqrt{25−x^2−y^2} =\sqrt{\lim_{(x,y)→(4,3)}(25−x^2−y^2)} \ = \sqrt{\lim_{(x,y)→(4,3)}25−\lim_{(x,y)→(4,3)}x^2−\lim_{(x,y)→(4,3)}y^2} \ =\sqrt{25−4^2−3^2} \ = 0 \end{align} See the following graph. Figure \PageIndex{6}: Graph of the function f(x,y)=\sqrt{25−x^2−y^2}. Exercise \PageIndex{3} Evaluate the following limit: \lim_{(x,y)→(5,−2)}\sqrt{29−x^2−y^2}. \nonumber Hint Determine the domain of f(x,y)=\sqrt{29−x^2−y^2}. Answer \lim_{(x,y)→(5,−2)}\sqrt{29−x^2−y^2} = \sqrt{29−5^2−(-2)^2} = \sqrt{29−25−4} = \sqrt{0} = 0\nonumber Continuity of Functions of Two Variables In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for f(x) to be continuous at point x=a f(a) exists. \displaystyle \lim_{x→a}f(x) exists. \displaystyle \lim_{x→a}f(x)=f(a). These three conditions are necessary for continuity of a function of two variables as well. Definition: continuous Functions A function f(x,y) is continuous at a point (a,b) in its domain if the following conditions are satisfied: f(a,b) exists. \displaystyle \lim_{(x,y)→(a,b)}f(x,y) exists. \displaystyle \lim_{(x,y)→(a,b)}f(x,y)=f(a,b). Example \PageIndex{4}: Demonstrating Continuity for a Function of Two Variables Show that the function f(x,y)=\dfrac{3x+2y}{x+y+1} \nonumber is continuous at point (5,−3). Solution There are three conditions to be satisfied, per the definition of continuity. In this example, a=5 and b=−3. f(a,b) exists. This is true because the domain of the function f consists of those ordered pairs for which the denominator is nonzero (i.e., x+y+1≠0). Point (5,−3) satisfies this condition. Furthermore, f(a,b)=f(5,−3)=\dfrac{3(5)+2(−3)}{5+(−3)+1}=\dfrac{15−6}{2+1}=3. \nonumber \displaystyle \lim_{(x,y)→(a,b)}f(x,y) exists. This is also true: \begin{align} \lim_{(x,y)→(a,b)}f(x,y) =\lim_{(x,y)→(5,−3)}\dfrac{3x+2y}{x+y+1} \ =\dfrac{\displaystyle \lim_{(x,y)→(5,−3)}(3x+2y)}{\displaystyle \lim_{(x,y)→(5,−3)}(x+y+1)} \ = \dfrac{15−6}{5−3+1} \ = 3. \end{align} \displaystyle \lim_{(x,y)→(a,b)}f(x,y)=f(a,b). This is true because we have just shown that both sides of this equation equal three. Exercise \PageIndex{4} Show that the function f(x,y)=\sqrt{26−2x^2−y^2}\nonumber is continuous at point (2,−3). Hint Use the three-part definition of continuity. Answer 1. The domain of f contains the ordered pair (2,−3) because f(a,b)=f(2,−3)=\sqrt{16−2(2)^2−(−3)^2}=3 2. \displaystyle \lim_{(x,y)→(a,b)}f(x,y)=3 3. \displaystyle \lim_{(x,y)→(a,b)}f(x,y)=f(a,b)=3 Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point (x_0,y_0) in its domain if for every ε>0 there exists a δ>0 such that, whenever \sqrt{(x−x_0)^2+(y−y_0)^2}<δ it is true, \|f(x,y)−f(a,b)\|<ε. This definition can be combined with the formal definition (that is, the epsilon–delta definition) of continuity of a function of one variable to prove the following theorems: Theorem \PageIndex{2}: The Sum of Continuous Functions is Continuous If f(x,y) is continuous at (x_0,y_0), and g(x,y) is continuous at (x_0,y_0), then f(x,y)+g(x,y) is continuous at (x_0,y_0). Theorem \PageIndex{3}: The Product of Continuous Functions is Continuous If g(x) is continuous at x_0 and h(y) is continuous at y_0, then f(x,y)=g(x)h(y) is continuous at (x_0,y_0). Theorem \PageIndex{4}: The Composition of Continuous Functions is Continuous Let g be a function of two variables from a domain D⊆\mathbb{R}^2 to a range R⊆R. Suppose g is continuous at some point (x_0,y_0)∈D and define z_0=g(x_0,y_0). Let f be a function that maps R to R such that z_0 is in the domain of f. Last, assume f is continuous at z_0. Then f∘g is continuous at (x_0,y_0) as shown in Figure \PageIndex{7}. Figure \PageIndex{7}: The composition of two continuous functions is continuous. Let’s now use the previous theorems to show continuity of functions in the following examples. Example \PageIndex{5}: More Examples of Continuity of a Function of Two Variables Show that the functions f(x,y)=4x^3y^2 and g(x,y)=\cos(4x^3y^2) are continuous everywhere. Solution The polynomials g(x)=4x^3 and h(y)=y^2 are continuous at every real number, and therefore by the product of continuous functions theorem, f(x,y)=4x^3y^2 is continuous at every point (x,y) in the xy-plane. Since f(x,y)=4x^3y^2 is continuous at every point (x,y) in the xy-plane and g(x)=\cos x is continuous at every real number x, the continuity of the composition of functions tells us that g(x,y)=\cos(4x^3y^2) is continuous at every point (x,y) in the xy-plane. Exercise \PageIndex{5} Show that the functions f(x,y)=2x^2y^3+3 and g(x,y)=(2x^2y^3+3)^4 are continuous everywhere. Hint Use the continuity of the sum, product, and composition of two functions. Answer The polynomials g(x)=2x^2 and h(y)=y^3 are continuous at every real number; therefore, by the product of continuous functions theorem, f(x,y)=2x^2y^3 is continuous at every point (x,y) in the xy-plane. Furthermore, any constant function is continuous everywhere, so g(x,y)=3 is continuous at every point (x,y) in the xy-plane. Therefore, f(x,y)=2x^2y^3+3 is continuous at every point (x,y) in the xy-plane. Last, h(x)=x^4 is continuous at every real number x, so by the continuity of composite functions theorem g(x,y)=(2x^2y^3+3)^4 is continuous at every point (x,y) in the xy-plane. Functions of Three or More Variables The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function f(x,y,z) that gives the temperature at a physical location (x,y,z) in three dimensions. Or perhaps a function g(x,y,z,t) can indicate air pressure at a location (x,y,z) at time t. How can we take a limit at a point in \mathbb{R}^3? What does it mean to be continuous at a point in four dimensions? The answers to these questions rely on extending the concept of a δ disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables. Definition: δ-balls Let (x_0,y_0,z_0) be a point in \mathbb{R}^3. Then, a δ-ball in three dimensions consists of all points in \mathbb{R}^3 lying at a distance of less than δ from (x_0,y_0,z_0) —that is, \big{(x,y,z)∈\mathbb{R}^3∣\sqrt{(x−x_0)^2+(y−y_0)^2+(z−z_0)^2}<δ\big}. \nonumber To define a δ-ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point P=(w_0,x_0,y_0,z_0) in \mathbb{R}^4, a δ ball around P can be described by \big{(w,x,y,z)∈\mathbb{R}^4∣\sqrt{(w−w_0)^2+(x−x_0)^2+(y−y_0)^2+(z−z_0)^2}<δ\big}. \nonumber To show that a limit of a function of three variables exists at a point (x_0,y_0,z_0), it suffices to show that for any point in a δ ball centered at (x_0,y_0,z_0), the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well. Example \PageIndex{6}: Finding the Limit of a Function of Three Variables Find \lim_{(x,y,z)→(4,1,−3)}\dfrac{x^2y−3z}{2x+5y−z}. \nonumber Solution Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, the identity law, and the constant law, \begin{align}\lim_{(x,y,z)→(4,1,−3)}(2x+5y−z) =2(\lim_{(x,y,z)→(4,1,−3)}x)+5(\lim_{(x,y,z)→(4,1,−3)}y)−(\lim_{(x,y,z)→(4,1,−3)}z) \ = 2(4)+5(1)−(−3) \ = 16. \end{align} Since this is nonzero, we next find the limit of the numerator. Using the product law, power law, difference law, constant multiple law, and identity law, \begin{align} \lim_{(x,y,z)→(4,1,−3)}(x^2y−3z) =(\lim_{(x,y,z)→(4,1,−3)}x)^2(\lim_{(x,y,z)→(4,1,−3)}y)−3\lim_{(x,y,z)→(4,1,−3)}z \ =(4^2)(1)−3(−3) \ = 16+9 \ = 25 \end{align} Last, applying the quotient law: \lim_{(x,y,z)→(4,1,−3)}\dfrac{x^2y−3z}{2x+5y−z}=\dfrac{\displaystyle \lim_{(x,y,z)→(4,1,−3)}(x^2y−3z)}{\displaystyle \lim_{(x,y,z)→(4,1,−3)}(2x+5y−z)}=\dfrac{25}{16} \nonumber Exercise \PageIndex{6} Find \lim_{(x,y,z)→(4,−1,3)}\sqrt{13−x^2−2y^2+z^2} \nonumber Hint Use the limit laws and the continuity of the composition of functions. Answer \lim_{(x,y,z)→(4,−1,3)}\sqrt{13−x^2−2y^2+z^2}=2 \nonumber Key Concepts To study limits and continuity for functions of two variables, we use a δ disk centered around a given point. A function of several variables has a limit if for any point in a δ ball centered at a point P, the value of the function at that point is arbitrarily close to a fixed value (the limit value). The limit laws established for a function of one variable have natural extensions to functions of more than one variable. A function of two variables is continuous at a point if the limit exists at that point, the function exists at that point, and the limit and function are equal at that point. Glossary boundary pointa point P_0 of R is a boundary point if every δ disk centered around P_0 contains points both inside and outside Rclosed seta set S that contains all its boundary pointsconnected setan open set S that cannot be represented as the union of two or more disjoint, nonempty open subsets δdiskan open disk of radius δ centered at point (a,b)δballall points in \mathbb{R}^3 lying at a distance of less than δ from (x_0,y_0,z_0)interior pointa point P_0 of \mathbb{R} is a boundary point if there is a δ disk centered around P_0 contained completely in \mathbb{R}open seta set S that contains none of its boundary pointsregionan open, connected, nonempty subset of \mathbb{R}^2 This page titled 14.2: Limits and Continuity is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform. 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9299	https://brightchamps.com/en-us/math/math-formulas/section-formula-in-mathematics	Our Programs Learn More About us Login Table Of Contents List of Formulas for the Section Formula Section Formula for Internal Division Section Formula for External Division Application of Section Formula in Geometry Importance of the Section Formula Tips and Tricks for Memorizing the Section Formula Common Mistakes and How to Avoid Them While Using the Section Formula Examples of Problems Using the Section Formula FAQs on the Section Formula Glossary for Section Formula Summarize this article: 142 Learners Last updated on August 10, 2025 Section Formula in Mathematics In geometry, the section formula is used to find the coordinates of a point dividing a line segment in a given ratio. This concept is essential for understanding the division of lines in coordinate geometry. In this topic, we will learn how to apply the section formula. Section Formula in MathematicsforUSStudents List of Formulas for the Section Formula The section formula helps in finding the coordinates of a point dividing a line segment internally or externally. Let’s learn the formulas to calculate these coordinates. Section Formula for Internal Division The section formula for internal division finds the coordinates of a point dividing a line segment internally in the ratio (m:n). If the coordinates of the endpoints are A(x1, y1) and B(x2, y2), the coordinates x, y of the dividing point are given by: x = frac{mx2 + nx1}{m + n} y = frac{my2 + ny1}{m + n} Section Formula for External Division The section formula for external division finds the coordinates of a point dividing a line segment externally in the ratio (m:n). If the coordinates of the endpoints are A(x1, y1)) and B(x2, y2), the coordinates x, y of the dividing point are given by: x = frac{mx2 - nx1}{m - n} y = frac{my2 - ny1}{m - n} Application of Section Formula in Geometry The section formula is crucial in geometry for solving problems related to dividing line segments, finding centroids, and working with coordinates. It provides a foundation for more advanced topics in coordinate geometry. Importance of the Section Formula The section formula is an essential tool in mathematics and real-life applications involving coordinate systems. Here are some key points about its importance: It helps in dividing line segments in a specific ratio, aiding in construction and design tasks. Understanding the section formula allows students to grasp advanced concepts in geometry and vector analysis. It is used to find specific points such as centroids and in calculations involving distances and midpoints. Tips and Tricks for Memorizing the Section Formula Students often find the section formula tricky, but with some tips and tricks, it can be mastered. Remember that the formula involves ratios, and the point divides the line segment either internally or externally. Visual aids like diagrams can help in understanding and memorizing the formula. Practice with real-life scenarios, such as dividing lengths in projects or maps, to reinforce understanding. Common Mistakes and How to Avoid Them While Using the Section Formula Students often make errors when applying the section formula. Here are some common mistakes and how to avoid them. Mistake 1 Confusing Internal and External Division Students may confuse the formulas for internal and external division. To avoid this, remember that internal division uses addition in the denominator, while external division uses subtraction. Mistake 2 Incorrectly Applying Ratios Errors can occur if the ratio is applied incorrectly. Always double-check which part of the line segment the ratio applies to and whether it is internal or external. Mistake 3 Mixing Up Coordinates When using the section formula, students might mix up the coordinates of the endpoints. To avoid this, label and organize the coordinates clearly before applying the formula. Mistake 4 Forgetting to Simplify Sometimes students forget to simplify the fractions in the formula. To avoid this, simplify the expressions in the formula after substituting the values. Mistake 5 Ignoring Negative Signs Students may ignore negative signs, especially in external division. Pay close attention to the signs when calculating the coordinates to ensure accuracy. Hey! Examples of Problems Using the Section Formula Problem 1 Find the point dividing the line segment joining (2, 3) and (10, 7) internally in the ratio 3:2. Okay, lets begin The point is (6, 5) Explanation Using the section formula for internal division: x = frac{3 × 10 + 2 × 2}{3 + 2} = frac{30 + 4}{5} = frac{34}{5} = 6.8 y = \frac{3 × 7 + 2 × 3}{3 + 2} = frac{21 + 6}{5} = frac{27}{5} = 5.4 Thus, rounding to the nearest integer, the point is (6, 5). Well explained 👍 Problem 2 Find the coordinates of the point dividing the line segment joining (-1, 4) and (3, 8) externally in the ratio 1:3. Okay, lets begin The point is (7, 10) Explanation Using the section formula for external division: x = frac{1 × 3 - 3 × (-1)}{1 - 3} = frac{3 + 3}{-2} = -3 y = frac{1 × 8 - 3 × 4}{1 - 3} = frac{8 - 12}{-2} = 2 Thus, the coordinates of the point are (7, 10). Well explained 👍 Problem 3 Determine the point dividing the line segment from (5, -2) to (15, 4) internally in the ratio 2:3. Okay, lets begin The point is (11, 2) Explanation Using the section formula for internal division: x = frac{2 × 15 + 3 × 5}{2 + 3} = frac{30 + 15}{5} = frac{45}{5} = 9 y = frac{2 × 4 + 3 × (-2)}{2 + 3} = frac{8 - 6}{5} = frac{2}{5} = 0.4 Thus, rounding to the nearest integer, the point is (11, 2). Well explained 👍 Problem 4 Find the point dividing the segment joining (0, 0) and (6, 8) externally in the ratio 2:1. Okay, lets begin The point is (12, 16) Explanation Using the section formula for external division: x = frac{2 × 6 - 1 × 0}{2 - 1} = frac{12}{1} = 12 y = \frac{2 × 8 - 1 × 0}{2 - 1} = frac{16}{1} = 16 Thus, the coordinates of the point are (12, 16). Well explained 👍 FAQs on the Section Formula 1.What is the section formula for internal division? The section formula for internal division is: x = frac{mx2 + nx1}{m + n} and y = frac{my2 + ny1}{m + n}, where A(x1, y1) and B(x2, y2) are the endpoints. 2.What is the section formula for external division? The section formula for external division is: x = frac{mx2 - nx1}{m - n} and y = frac{my2 - ny1}{m - n}, where A(x1, y1)) and (B(x2, y2) are the endpoints. 3.How is the section formula used in geometry? The section formula is used to find the coordinates of points dividing lines in specific ratios, crucial for solving problems in coordinate geometry. 4.Can the section formula be used for three-dimensional coordinates? Yes, the section formula can be extended to three dimensions with an additional term for the z-coordinate. 5.What is a practical application of the section formula? One practical application is in navigation, where it helps find intermediate points on a map or route. Glossary for Section Formula Section Formula: A mathematical formula used to determine the coordinates of a point dividing a line segment in a given ratio. Internal Division: Dividing a line segment between two points within the segment. External Division: Dividing a line segment beyond its endpoints. Ratio: A relationship between two numbers indicating how many times the first number contains the second. Coordinate Geometry: A branch of geometry where the position of points is defined using coordinates. Jaskaran Singh Saluja About the Author Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles. Fun Fact : He loves to play the quiz with kids through algebra to make kids love it. 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