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Hdu4612 connectivity, calculate the tree diameter, and add one side to obtain the least Bridge A Free Trial That Lets You Build Big! Start building with 50+ products and up to 12 months usage for Elastic Compute Service • Sales Support 1 on 1 presale consultation • After-Sales Support 24/7 Technical Support 6 Free Tickets per Quarter Faster Response • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.	{"url":"https://topic.alibabacloud.com/a/hdu4612-connectivity-calculate-the-tree-diameter-and-add-one-side-to-obtain-the-least-bridge_8_8_31711512.html","timestamp":"2024-11-04T23:48:22Z","content_type":"text/html","content_length":"80973","record_id":"<urn:uuid:90bd4b12-40ad-4771-a292-4149448ca5f3>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00828.warc.gz"}
Data Interpretation Quant Quiz For Sbi PO \| Sbi Clerk \| LIC \| RRB \| FCI \| CWC \| RBI & Other Exams - GovernmentAdda Home Data Interpretation Data Interpretation Data Interpretation Quant Quiz For Sbi PO \| Sbi Clerk \| LIC... Data Interpretation Quant Quiz For Sbi PO \| Sbi Clerk \| LIC \| RRB \| FCI \| CWC \| RBI & Other Exams Directions (01-05): In the Bar-chart, total members enrolled in different years from 1990 to 1994 in two gymnasium A and B. Based on this Bar chart solve the following questions- 1. In the year 1995, 30% increase in total number of members enrolled in 1994 of both gymnasium, then find the total no. of members enrolled in 1995? (a) 282 (c) 292 (e) none of these 2. The ratio between total members of both gymnasium in 1991 to total members in 1994 of both gymnasium is- (a) 22:27 (b) 21:11 (c) 11:21 (d) 25:13 (e) 27:22 3. The number of members of gymnasium A in 1991 is what % of the no. of members of gymnasium B in 1994. (a) 60% (b) 55% (c) 58% (d) 62% (e) none of these 4. The average of members enrolled in gymnasium A from 1991 to 1994 together is what percent more then the The average of members enrolled in gymnasium B in 1993 and 1994 together?(Rounded off to 2 decimal places) (a) 10.51% (b) 20.51% (c) 15.51% (d) 17.51% (e) none of these 5. Total member enrolled in gymnasium B in 1993 and 1994 together is what % more then members enrolled in gymnasium A in 1990 and 1994 together? (a) 60% (b) 65% (c) 62.5% (d) 61.5% (e) none of these Direction (6-10): Study the following table and answer the questions that follow. Given line graph shows the number of students appeared from state A and state B in an examination. Q6. Number of students appeared from state B in 2009 is about what percent of total students appeared from state A? Q7. What is the difference between the total number of students from state A in 2004 and 2005 together and those of state B in 2008 and 2009 together? Q8. What is the ratio of number of students appeared in examination from state B in 2004 ,2006 and 2008 to the number of students appeared from state A in 2005,2007 and 2009? Q9. If in 2010 the number of students appeared from state A is increase by 10%and those from state B increased by 15% as compared to the number of students from respective states in year 2009,then what is the ratio of number of students from state A and state B in 2010? Q10. What is the difference between average number of students from state A and state B? Directions (11-15): There are five companies and we have given the no. of employee working in different companies. In the table we have also given the percentage of male and female employees of HR and Marketing department. 11. If 60% of the employees of company T in HR department have MBA degree and 40% of the employees of the same company in the Marketing dept. have MBA degree, then how many employees have MBA degree in company T in both dept. together. (a) 98 (b) 108 (c) 106 (d) 92 (e) 66 12. The ratio of female employee of company Q in HR dept. to male employee of company R in Marketing dept. ? (a) 4:13 (b) 5:22 (c) 22:5 (e) none of these 13. Total number of HR employees of company P is what % more then the total no. of marketing employee in company T? (a) 236.76% (b) 226.67% (c) 276.76% (d) 246.67% (e) none of these 14. The ratio of male employee in HR dept. of company P and R together to female employee of Marketing department in company S and T together? (a) 187:27 (b) 43:188 (c) 188:43 (d) 27:187 (e) none of these 15. Difference between female employees of HR dept. in all companies together (excluding company S) and the female employees of Marketing dept. in all companies together (excluding company Q)? (a) 139 (b) 129 (c) 135 (d) 141 (e) none of these LEAVE A REPLY Cancel reply Government Scheme GK GS GA	{"url":"https://governmentadda.com/data-interpretation-quant-quiz-for-sbi-po-sbi-clerk-railway-nabard-rbi-other-exams-3/","timestamp":"2024-11-02T11:22:20Z","content_type":"text/html","content_length":"211918","record_id":"<urn:uuid:6346601f-aa4b-4d4b-81e6-524e2602e1b0>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00226.warc.gz"}
Algebra and Geometry Seminar Welcome to the Algebra and Geometry Seminar at Iowa State University, organized by Jonas Hartwig, Jason McCullough, and Tathagata Basak. During Fall 2024, the seminar runs on Thursdays at 3:10pm–4:00pm in Carver 401. Grad students are especially encouraged to attend. Topics include: • associative, commutative, and Lie algebras; • algebraic, differential, and hyperbolic geometry; • representation theory and tensor categories; • mathematics of quantum theory and fundamental physics. August 29, 2024 (Meet and Greet) The first meeting is a meet and greet where we introduce ourselves and share our interests. September 5, 2024 Introduction to DG Algebras and Minimal Models Jason McCullough (Iowa State University) Free resolutions are an essential technique in commutative algebra to study modules. Often free resolutions carry an algebra structure, turning it into a differential graded (DG) algebra. I will introduce DG algebras, survey known results, and define minimal models — i.e. how one creates a DG-algebra resolution even when the minimal free resolution carries no such structure. September 12, 2024 Introduction to Homotopy Lie Algebras Jason McCullough (Iowa State University) To any local ring, one can associate a graded Lie algebra, called the homotopy Lie algebra. Useful properties of the ring can be encoded by properties of the Lie algebra (such as when the ring is a complete intersection, Golod, Koszul, etc.). I will introduce homotopy Lie algebras, compute a few examples, and highlight a few results in the literature. September 19, 2024 Weight Modules and the Restricted Yoneda Embedding Dylan Fillmore (Iowa State University) Given an algebra $A$ and an appropriately chosen subcategory $\mathcal{B}$ of ${}_A\mathsf{Mod}$, the weight spaces of an $A$–module are given by the restricted Yoneda embedding from ${}_A\ mathsf{Mod}$ to the functor category $[\mathcal{B}^{\rm op}, \mathsf{Vect}]$. We discuss the existence of a left adjoint to the restricted Yoneda embedding, and give a sufficient condition for this adjunction to restrict to an equivalence. September 26, 2024 (No talk) October 3 and October 10, 2024 Hyperbolic Lattices from Strongly Regular Graphs Tathagata Basak (Iowa State University) Given a strongly regular graph, we'll define lattice of signature $(n,1)$. The construction is analogous to definition of root lattices, given a simply laced dynkin diagram. In many examples these lattices have interesting hyperbolic reflection group. For instance, the 275 vertex McLaughlin graph produces a signature $(21, 1)$ lattice on which the McLaughlin group naturally acts and whose reflection group contain 275 reflections that braid or commute according to the McLaughlin graph. A variant of this construction works for certain regular bipartite graphs and a variant of the construction produces hyperbolic Hermitian lattices over rings of integers of some quadratic number fields. In particular, we will talk about two examples over the ring of Eisenstein integers. One reflection group, in $U(4, 1)$, is related to the fundamental group of moduli space of cubic surfaces. Another, in $U(13, 1)$, is conjecturally related to the monster simple group. October 17, 2024 (No talk) October 24, 2024 From Interpolation Problems to Matroids Paolo Mantero (University of Arkansas) Interpolation problems are long-standing problems at the intersection of Algebraic Geometry, Commutative Algebra, Linear Algebra and Numerical Analysis, aiming at understanding the set of all polynomial equations passing through a given finite set X of points with given multiplicities. In this talk we discuss the problem for matroidal configurations, i.e. sets of points arising from the strong combinatorial structure of a matroid. Starting from the special case of uniform matroids, we will discover how an interplay of commutative algebra and combinatorics allows us to solve the interpolation problem for any matroidal configuration. It is the widest class of points for which the interpolation problem is solved. Along the way, we will touch on several open problems and conjectures. The talk is based on two joint projects with Vinh Nguyen. Last updated: Back to top ⇧	{"url":"http://jthartwig.net/ag/index.html","timestamp":"2024-11-10T00:00:07Z","content_type":"text/html","content_length":"7313","record_id":"<urn:uuid:03ff3d50-e5a9-4806-93ad-c907a236fecd>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00347.warc.gz"}
Fracture mechanics studies of failures of lead zirconate titanate ceramics under mechanical and/or electrical loadings - Rare & Special e-Zone Lead zirconate titanate (PZT) ferroelectric ceramics are widely applied in actuators, sensors, controlling devices and smart structures. Service conditions often involve combined mechanical-electrical loadings, which may deteriorate material properties and thus reduce reliability. Essentially, PZT ceramics are brittle and crack easily at all scales from domains to devices. Most industrial applications require the reliability of this type of piezoelectric materials to be well ensured. Thus fracture mechanics studies become an essential step for a better understanding of fracture behaviors of the materials. Even though the response of ferroelectrics on applied fields is essentially nonlinear, linear analysis is still fundamental and important. We started from the linear electro-elastic theoretical analysis through Stroh's formalism and coordinate transformation, to investigate the crack driving force as a function of applied mechanical and electrical loadings for different crack orientations with respect to the poling direction. We found that the energy release rate for the propagation of a slit crack is related to the slit geometry and the ratio of dielectric constants. For an insulating crack, the energy release rate is smaller when the crack plane is closer to the poling direction, indicating that the material will be easily fractured when the poling direction is perpendicular to the crack plane. For a conducting crack, however, the energy release rate behaves differently in different loading conditions. But the tendency to crack is most likely along the crack plane parallel to the poling direction, along which the electric field is applied. To explore the non-linearity around the crack tip in the piezoelectric material under the mixed mechanical and electric loadings, we studied the interaction of a piezoelectric screw dislocation with an insulating cavity and presented the general solution in series, using the image dislocation approach with consideration of the electric field in the cavity. When the cavity is reduced into a crack, three cases appear, which correspond to three different electrical boundary conditions along the crack faces, depending on the ratio of α/β, where α is the ratio of the minor semi-axis to the major semi-axis of the ellipse and β is the ratio of the dielectric constant of the cavity to the dielectric constant of the piezoelectric material. The crack is electrically impermeable when α/β→∞, while the crack becomes electrically permeable as α/β→0. Since the minimum of dielectric constant has a finite nonzero value and a real crack has also a nonzero width, the ratio of α/β will generally have a finite nonzero value, resulting in a semi-impermeable crack. Furthermore, the difference in the electric boundary conditions lead to great differences in the image force acting on the dislocation, and in the intensity factors and the energy release rate induced by the dislocation. Experimentally we investigated the fracture behavior of conductive cracks in PZT-4 piezoelectric ceramics by using compact tension specimens under electrical and/or mechanical loading. Finite element calculations were conducted to obtain the energy release rate, the stress intensity factor and the intensity factor of the electric field strength for the specimens. The experimental results show that the critical energy release rate under either pure electrical or pure mechanical loading is a constant, independent of the ligament length. However, the critical energy release rate under combined electrical and mechanical loading depends on the weighting of the electrical load in comparison with the mechanical load. We normalized the critical stress intensity factor by the critical stress intensity factor under pure mechanical loading and normalized the critical intensity factor of electric field strength by the critical intensity factor of electric field strength under pure electric loading. Then, a quadratic function describes the relationship between the normalized critical stress intensity factor and the normalized critical intensity factor of the electric field strength, which can serve as a failure criterion of conductive cracks in piezoelectric ceramics under combined electrical and mechanical loading. While a pure electric field can fracture poled ferroelectric ceramics, and electric fracture toughness exists for PZT ceramics, we then further studied how the electric fields effect depoled lead zirconate titanate ceramics with conductive cracks. With the same methodologies used for poled ferroelectric ceramics, we found the principle of fracture mechanics can also be used to measure the electrical fracture toughness for depoled PZT ceramics. The electrical fracture toughness G = 263±35 N / m is about 9 times higher than the mechanical fracture toughness G = 30.4±3.9 N / m . The high electrical fracture toughness arises from the greater energy dissipation around the conductive crack tip under pure electric loading, which is impossible in the brittle depoled ceramics under pure mechanical loading. The energy release rate for combined mechanical and electric loadings, however, depends on the weighting of the electrical load in comparison with the mechanical load. When we normalized the critical intensity factor under combined loading by the critical stress intensity factor under purely mechanical loading and the critical electric intensity factor under combined loading by the critical electric intensity factor under purely electrical loading, we obtained a quadratic function to represent their relationship, which can be regarded as a failure criterion for depoled ferroelectric ceramics with conductive cracks. We conducted a theoretical nonlinear analysis and computer simulations to understand the fracture behavior of a conductive crack in a dielectric material. A polarization saturation-free zone (SFZ) model is proposed to establish a failure criterion for conductive cracks in dielectric Theceramics under electric and/or mechanical loads. The SFZ model treats dielectric ceramics as mechanically brittle and electrically ductile material and allows the local intensity factors of electric field strength and electric displacement, as well as local stress and strain intensity factors, to have finite nonzero values. At failure we apply the Griffith criterion to balance the critical value of local energy release rate at the crack tip with the specific surface energy. Failure occurs when the local energy release rate exceeds the critical value. The experimental results in Chapter 5 agree with the predictions from the proposed SFZ model. The computer simulations are based on the results of the nonlinear analysis, which gives a polarization saturation zone, and a domain-switching model. In computations, we ignore the transformation strain in the depoled ceramics and use an electric dipole to represent the local net saturation polarization in an element such that a polydomain structure is simulated with a grid of points where each polarization is fixed at each grid point. Considering the interactions among the electric dipoles and the crack, we calculate the total electric field, which acts on a linear dielectric background medium. The entire material response is then described by the behavior of the background medium under the integrated loading from the electric dipoles and applied. To verify the algorithm, we simulated P~E curves under a remotely uniform loading for a finite medium with a single edge crack and obtained very satisfactory results, thereby ensuring the accuracy of the algorithm. We took the local energy release rate as the fracture criterion. Domain switching in the polarization saturation zone shields the conductive crack tip from the applied loads. That is why electrical fracture toughness is much higher than the mechanical fracture toughness. The simulation results illustrate that this combined polarization saturation zone and domain-switching model explain our experimental observations and facilitate the establishment of a failure criterion for conductive cracks in piezoelectric ceramics under combined mechanical and electrical loading.	{"url":"https://lbezone.hkust.edu.hk/rse/?p=11145","timestamp":"2024-11-14T14:23:00Z","content_type":"text/html","content_length":"64665","record_id":"<urn:uuid:db8dd382-8276-4a02-b3ba-ad5bd0a6c5f4>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00523.warc.gz"}
Back to Basics: S-parameters Figure 1: S-matrices for one-, two-, and three-port RF networks Suppose you have an optical lens of some sort onto which you shine a light with a known photonic output. While most of the incident light passes through the lens, some fraction of the light is reflected and some is absorbed (the behavior is also dependent on the wavelength of the incident light). You'd like to characterize that lens: Exactly how much light was reflected? How much passed through? What is it about the lens that prevented all of the light from passing through? Now, let's apply that same sort of thinking to a two-port network. When we apply a signal to the network's input, the same thing will happen as it did with the lens: Some fraction of the signal will propagate through the network but some is reflected, or scattered, back through the port it entered through. Some of the signal will be dissipated as heat, or even as electromagnetic radiation. As with the lens, we want to know what's happening with our two-port network. What is causing the scattering of the signal? Bear in mind that the “signal” is an electromagnetic wave propagating through the medium of the network, just as the light passing through the lens is also an electromagnetic wave. The answer lies in the application of a mathematical construct called a scattering matrix (or S-matrix), which quantifies how energy propagates through a multi-port network. The S-matrix is what allows us to accurately describe the properties of incredibly complicated networks as simple "black boxes". The S-matrix for an N-port network contains N2 coefficients (S-parameters), each one representing a possible input-output path. By “multi-port network,” we could be referring to, for example, a cable or a microstrip line. How would that interface affect a 5-Gbps USB 3.0 signal? This is the sort of practical application in which S-parameters shine. S-parameters are a “frequency-domain” description of the electrical behavior of a network. They are complex numbers, and are expressed in terms of both magnitude and phase. That's because both the magnitude and phase of the input signal are changed by the network due to losses, reflections, and propagation time. Quite often we refer to the magnitude only, because how much gain or loss occurs is typically of the most interest. However, the phase information is extremely important and should not be ignored. S-parameters are defined for a given set of frequencies and port impedance (typically 50 ohms), and vary as a function of frequency for any real-world network. There are also mixed-mode S-parameters. Say you have a differential lane in your circuit and you want to quantify the losses in a differential signal. S-parameters for the differential lane can be transformed into differential S-parameters, with which you consider differential signal characteristics alongside of common signal characteristics. In their most basic sense, S-parameters refer to the ratio of "voltage out versus voltage in." S-parameters come in a matrix, with the number of rows and columns equal to the number of ports. For the S-parameter subscripts "ij", j is the port that is excited (the input port), and "i" is the output port. Thus, S11 refers to the ratio of signal that reflects from port one for a signal incident on port one, as a function of frequency. S-parameters along the diagonal of the S-matrix are referred to as reflection coefficients because they only refer to what happens at a single port, while off-diagonal S-parameters are referred to as transmission coefficients, because they refer to what happens from one port to another. Figure 1 shows the S-matrices for one, two and three-port In time-domain reflectometer measurements (TDRs), a fast rising step edge is sent into the DUT and the reflected signal measured. In addition, we can look at the signal that is transmitted through the DUT. This is the time-domain transmitted signal (TDT). The signal incident into the DUT can be thought of as being composed of a series of sine waves, each with a different frequency, amplitude, and phase. Each sine wave component will interact with the DUT independently. When a sine wave reflects from the DUT, the amplitude and phase may change a different amount for each frequency. This variation gives rise to the particular reflected pattern. Figure 2: There are time-domain measurements and there are S-parameters; each emphasize different aspects of a transmission line's Likewise, the transmitted signal will have each incident frequency component with a different magnitude and phase. There is no difference in the information content between the time-domain view of the TDR or TDT signal, and the frequency-domain view. Using Fourier transform techniques, the time-domain response can be mathematically transformed into the frequency-domain response and back again without changing or losing any information. While these two domains tell the same story, they emphasize different parts of the story (Figure 2). TDR measurements are more sensitive to the instantaneous impedance profile of the interconnect, while frequency-domain measurements are more sensitive to the total input impedance looking into the DUT. To distinguish these two domains, we also use different words to refer to them. Time-domain measurements are TDR and TDT responses, while frequency-domain reflection and transmission responses are referred to as S (scattering) parameters. S11 is the reflected signal and S21 is the transmitted signal. They are also often termed the return loss and the insertion loss. It’s worth noting, however, that TDR and TDT typically refer to the raw measurements from the TDR instrument. From these measurements, a transform from the S-parameters yields other responses, such as impedance, step response, impulse response, single-bit responses, and more. The raw TDR and TDT measurements are typically less accurate because they are often uncalibrated (unless you’re using a Teledyne LeCroy SPARQ network analyzer; see below). Depending on the question asked, the answer may be obtained faster from one domain or the other. If the question concerns the characteristic impedance of the uniform transmission line, the time-domain display of the information will get us the answer faster. If the question is about the bandwidth of the model, the frequency-domain display of the information will get us the answer faster. Having said that, the time-consuming TDR calibration procedure is prone to errors that can yield incorrect S-parameters. Teledyne LeCroy’s SPARQ signal-integrity network analyzers avoid this pitfall by performing automatic calibrations with an internally connected calibration kit.	{"url":"https://blog.teledynelecroy.com/2014/05/back-to-basics-s-parameters.html","timestamp":"2024-11-02T04:15:07Z","content_type":"application/xhtml+xml","content_length":"92783","record_id":"<urn:uuid:edcf52c1-26d9-4140-8368-42131d99e160>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00273.warc.gz"}
CORMSIS Seminar by Henri Bonnel Event 11:00 - 12:00 30 September 2013 Room 8031 (8c), Building 54 For more information regarding this event, please email Prof. Joerg Fliege at J.Fliege@soton.ac.uk . Event details You are invited to a CORMSIS seminar on Monday 30th Sep at 11:00, which will be presented by Henri Bonnel, a guest of Joerg Fliege, from University of New Caledonia. Tea/coffee will be available before the seminar. Post-Pareto Analysis for Multiobjective Stochastic Problems The solution set (Pareto or efficient set) of a multiobjective optimization problem is often very large (infinite and even unbounded). The grand coalition of a cooperative game can be written as a multiobjective optimal control problem. Assuming that this game is supervised by a decision maker (DM), the DM can use his own (scalar) objective for choosing a solution. Of course this solution must satisfy all the players of the grand coalition, hence must be a Pareto solution. Another interest for the study of the problem of optimizing a scalar function over a Pareto set is that it may be possible to avoid the generation of all the Pareto set. For multiobjective mathematical programming problems (finite dimensional optimization) there are many contributions in this field (see e.g. [5] for an extensive bibliography). Some recent results for the case of the multiobjective control problems can be found in [1,4]. Generalization of this problem for the semivectorial bilevel problems has been studied in [3]. My talk deals with a different setting: mulitobjective stochastic optimization, and it is based on the paper [1]. Thus I will consider the problem of minimizing the expectation of a real valued random function over the weakly Pareto or Pareto set associated with a Stochastic Multi-Objective Optimization Problem, whose objectives are expectations of random functions. Assuming that the closed form of these expectations is difficult to obtain, the Sample Average Approximation method is applied in order to approach this problem. I will show that the Hausdorff-Pompeiu distance between the Sample Average Approximation of size N weakly Pareto sets and the true weakly Pareto set converges to zero almost surely as the sample size N goes to infinity, assuming that our Stochastic Multi-Objective Optimization Problem is strictly convex. Also every cluster point of any sequence of Sample Average Approximation optimal solutions is almost surely a true optimal solution. To handle also the non-convex case, it is assumed that the real objective to be minimized over the Pareto set depends on the expectations of the objectives of the Stochastic Optimization Problem, i.e. it is considered the problem of optimizing a scalar function over the Pareto outcome space of the Stochastic Optimization Problem. Then, without any convexity hypothesis, some similar results hold for the Pareto sets in the outcome spaces. Finally I will show that the sequence of Sample Average Approximation optimal values converges almost surely to the true optimal value as the sample size goes to infinity. [1] Henri Bonnel. Post-Pareto Analysis for Multiobjective Parabolic Control Systems. Ann. Acad. Rom. Sci. Ser. Math. Appl., 5: 13-34, 2013. [2] H. Bonnel and J. Collonge. Stochastic Optimization over a Pareto Set Associated with a Stochastic Multi-objective Optimization Problem. Journal of Optimization Theory and Applications, (online first, DOI 10.1007/s10957-013-0376-8) 2013. [3] H. Bonnel and J. Morgan. Semivectorial Bilevel Convex Optimal Control Problems: An Existence Result. SIAM Journal on Control and Optimization, 50, (6): 3224-3241, 2012. [4] H. Bonnel and Y. Kaya. Optimization Over the Efficient Set of Multi-objective Convex Optimal Control Problems. Journal of Optimization Theory and Applications, 147, (1): 93-112, 2010. [5] Y. Yamamoto. Optimization over the efficient set : an overview. J. Global Optim., 22: 285-317, 2002. Speaker information Henri Bonnel,University of New Caledonia	{"url":"https://www.southampton.ac.uk/cormsis/news/events/2013/0930.page","timestamp":"2024-11-09T23:55:58Z","content_type":"application/xhtml+xml","content_length":"41237","record_id":"<urn:uuid:9595f696-46ae-4422-a8d3-e3621bf43efc>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00481.warc.gz"}
Is Momentum a Vector? Understanding Momentum in Physics Momentum is a fundamental concept in physics that describes the motion of objects. Whether momentum is a vector quantity or not is a crucial question to understand its application in various physical situations. In this article, we will delve into the nature of momentum, discuss its vector properties, and explore how to calculate it in different scenarios. Understanding the vector nature of momentum is essential for properly applying momentum principles in your studies and real-world applications. We’ll define momentum, explore its vector characteristics, and provide examples of calculating momentum in different situations. By the end of this article, you’ll have a firm grasp of the vector nature of momentum and how to effectively apply it in your physics problem-solving. Momentum is a physical quantity that represents the amount of motion an object possesses. It is defined as the product of an object’s mass and its velocity. As a vector quantity, momentum has both a magnitude and a direction, which is crucial for understanding its behavior in various physical scenarios. Introduction to Momentum Momentum is a fundamental concept in physics that describes the motion of an object. It is a physical quantity that represents the amount of motion an object possesses. Momentum is defined as the product of an object’s mass and its velocity. In physics, quantities can be either scalar or vector. Scalar quantities have only a magnitude, while vector quantities have both magnitude and direction. Understanding whether momentum is a scalar or vector quantity is essential for correctly applying momentum principles. Scalar quantities, such as mass and temperature, have a single numerical value that represents their magnitude. They do not have a specific direction associated with them. On the other hand, vector quantities, like velocity and force, have both a magnitude and a direction. Determining whether momentum is a scalar or vector quantity is crucial for understanding how it behaves in different physical situations. Scalar Quantities Vector Quantities Have only a magnitude Have both magnitude and direction Examples: mass, time, temperature Examples: velocity, force, momentum By understanding the vector nature of momentum, you can accurately apply momentum principles in various physical scenarios, such as analyzing collisions, predicting the motion of objects, and solving complex physics problems. This knowledge is essential for a comprehensive understanding of the fundamental laws of physics governing the motion of objects. Defining Momentum Momentum (p) is defined as the product of an object’s mass (m) and its velocity (v). The momentum formula is: p = m × v This means that the momentum of an object is directly proportional to its mass and velocity. The greater the mass and velocity of an object, the greater its momentum. Momentum is a vector quantity, meaning it has both a magnitude and a direction. To calculate momentum, you simply need to multiply the object’s mass by its velocity. For example, if an object has a mass of 5 kilograms and a velocity of 10 meters per second, its momentum would p = 5 kg × 10 m/s = 50 kg·m/s Understanding the momentum formula and how to calculate momentum in different scenarios is essential for applying momentum principles in physics problems and real-world situations. Vector Nature of Momentum As a vector quantity, momentum has both a magnitude and a direction. The direction of momentum is the same as the direction of the object’s velocity. This means that if an object is moving in a certain direction, its momentum will also be in that same direction. The magnitude of momentum is determined by the object’s mass and velocity, as defined by the momentum formula. Understanding the vector properties of momentum is crucial for correctly analyzing and applying momentum principles in various physical situations. Recognizing the momentum direction is essential when dealing with the motion of objects, as it helps you accurately predict the path and behavior of the object in question. Characteristic Description Vector Quantity Momentum has both magnitude and direction, making it a vector quantity. Momentum Direction The direction of momentum is the same as the direction of the object’s velocity. Magnitude The magnitude of momentum is determined by the object’s mass and velocity, as defined by the momentum formula. Understanding the vector nature of momentum is essential for correctly applying momentum principles in various physical situations, such as analyzing the motion of objects, predicting their behavior, and solving momentum-related problems. Calculating Momentum in Different Scenarios Mastering the art of calculating momentum is a crucial skill in the realm of physics. Whether you’re dealing with a fast-moving sports car or a slowly drifting satellite, understanding how to apply the momentum formula can unlock a deeper comprehension of various physical phenomena. For instance, if you know an object’s mass and velocity, you can easily determine its momentum by simply multiplying these two values together. Conversely, if you’re given an object’s momentum and either its mass or velocity, you can rearrange the formula to solve for the missing component. Applying these principles in different scenarios is essential for problem-solving in physics. From analyzing the forces at play in a collision to predicting the trajectory of a projectile, the ability to calculate momentum is a valuable asset. By mastering this skill, you’ll be better equipped to navigate the complexities of the physical world and uncover the underlying principles that govern the motion of objects.	{"url":"https://likesgainer.com/is-momentum-a-vector-understanding-momentum-in-physics/","timestamp":"2024-11-13T11:37:22Z","content_type":"text/html","content_length":"121832","record_id":"<urn:uuid:1b2a708d-ba0c-4ad2-9fb4-d1082e6c23e5>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00752.warc.gz"}
Free Random Number Generator FreeCalculator.net’s sole focus is to provide fast, comprehensive, convenient, free online calculators in a plethora of areas. Currently, we have over 100 calculators to help you “do the math” quickly in areas such as finance, fitness, health, math, and others, and we are still developing more. Our goal is to become the one-stop, go-to site for people who need to make quick calculations. Additionally, we believe the internet should be a source of free information. Therefore, all of our tools and services are completely free, with no registration required. We coded and developed each calculator individually and put each one through strict, comprehensive testing. However, please inform us if you notice even the slightest error – your input is extremely valuable to us. While most calculators on FreeCalculator.net are designed to be universally applicable for worldwide usage, some are for specific countries only.	{"url":"https://freecalculator.net/random-number-generator","timestamp":"2024-11-08T23:32:17Z","content_type":"text/html","content_length":"49224","record_id":"<urn:uuid:46d53aeb-1112-49e5-8027-0982b7c36656>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00047.warc.gz"}
Kindergarten Math Worksheets for Counting, Numbers, Shapes, Comparing and More • Cupcakes & CurriculumKindergarten Math Worksheets for Counting, Numbers, Shapes, Comparing and More This BUNDLED kindergarten math worksheets resource includes printable and digital math worksheets that give your kindergarteners practice with ALL kindergarten math skills like addition and subtraction, geometry, word problems, comparing and classifying objects and more. These Common Core aligned math worksheets are perfect for morning work, assessment, homework, review, fast-finisher activities, exit tickets, and math centers. These kindergarten math worksheets are provided in TWO formats to best fit your classroom needs – Print and Go and JPEG. Included in this BUNDLED kindergarten math worksheets resource K.NBT.1 – Compose/Decompose Numbers • Match the number to the correct ten frame (2 pages) • Count the base ten cubes and write how many tens and ones; write the number (2 pages) • Match the number to the correct base ten blocks (2 pages) • Fill in the missing number (2 pages) • Fill in the ten frames to match each equation (2 pages) • Draw base ten blocks to match the equation (2 pages) • Fill in the missing numbers (2 pages) • Color in the ten frames to match the words and write the sum (2 pages) KCC1 – Counting to 100 by 1s and 10s • Trace numbers (11 pages) • Write numbers (4 pages) • Fill in missing numbers from the 100 chart (3 pages) • Fill in the missing numbers 1-20 (1 page) • Connect the dots to count (11 pages) KCC2 – Counting Forward • Count up starting with the first number (6 pages) • Count up on a number line (6 pages) • Count up using scoops of ice cream (4 pages) KCC3 – Recognize and Write Numbers 1-20 • Count the objects and write the number (3 pages) • Count the dots in the ten frame (3 pages) • Match the number to the correct ten frame (3 pages) • Count the dots on the domino then write the number (3 pages) KCC4 – Numbers and Quantities • Draw the number of circles specified (2 pages) • Circle how many objects are in each row (2 pages) • Fill the 5 frame with the correct number (1 page) • Fill the 10 frames with the correct number (4 pages) • Cut and paste the dice under the correct number (1 page) • Cut and paste the dominos under the correct number (1 page) • Cut and paste the ten frames under the correct number (1 page) KCC5 – Counting to Ask HOW MANY? • Draw the number of circles specified (1 page) • Count and write how many (3 pages) • Match the number to its quantity (3 pages) • Cut and paste the numbers next to the quantity (3 pages) KCC6 – Comparing Quantities • Count the gumballs in each gumball machine. Write the number and show on a ten frame. Circle the machine with more gumballs (5 pages) • Color the sets with numbers more or less than the given picture (6 pages) • Write GREATER or LESS in each blank (2 pages) • Match the boxes that have equal quantities (2 pages) KCC7 – Comparing Numbers • Circle the number in the bubbles that are more or less (4 pages) • Circle the number in each box that is greater or less (2 pages) • Cut and paste numbers and order them from least to greatest (1 page) • Cut and paste symbols to compare the numbers (3 pages) K.MD.1 – Describing Objects • Match the picture to the descriptive word (2 pages) • Identifying given pictures as heavy or light (1 page) • Identifying given pictures as tall or short (1 page) • Writing the number of given cubes long a pencil is (2 pages) • Circling the correct descriptive word to describe a picture (2 pages) • Using a word bank, coming up with a descriptive word for a picture (2 pages) K.MD.2 – Comparing Objects • Identifying 2 things as taller/shorter or lighter/heavier (2 pages)• • Coloring the object that holds more (1 page) • Coloring the object that holds less (1 page) • Coloring the object that is taller (1 page) • Coloring the object that is shorter (1 page) • Coloring the object that is heavier (1 page) • Coloring the object that is lighter (1 page) • Coloring pencils that are longer or shorter than the given example (2 pages) K.MD.3 – Classifying Objects • Classifying shapes by coloring and counting (2 pages) • Classifying circles by coloring and counting (3 pages) • Classifying objects into two categories and counting; cut & paste (5 pages) K.OA.1 – Addition and Subtraction Within 10 • Use pictures to help you add (2 pages) • Use pictures to help you write the equation (1 page) • Use pictures to help you subtract (4 pages) • Add the numbers on the die (2 pages) • Add the numbers on the give frame (2 pages) • Match the picture to the equation (1 page) K.OA.2 – Addition and Subtraction Word Problems • Solve an addition word problem by showing a picture, number line, equation, and tracing a sentence. (10 pages) • Solve a subtraction word problem by showing a picture, number line, equation, and tracing a sentence. (10 pages) K.OA.3 – Decomposing Numbers • Show 3 ways to make 3 (1 page) • Show 4 ways to make 4 (1 page) • Show 5 ways to make 5 (1 page) • Show 5 ways to make 6 (1 page) • Show 5 ways to make 7 (1 page) • Show 5 ways to make 8 (1 page) • Show 5 ways to make 9 (1 page) • Make each number bond true (2 pages) • Complete each part-part-whole table (2 pages) • Fill in the blanks to make each equation (3 pages) K.OA.4 – Making Ten • Color cubes and write an equation to make ten (1 page) • Write an equation based on a picture of cubes to make ten (1 page) • Color in a ten frame to match an equation that equals ten (2 pages) • Complete the part-part-whole (1 page) • Complete the part-part-whole with the sum and 1 addend given (1 page) • Write an equation that matches the picture of the ten frame (2 pages) • Complete the number bond (1 page) • Complete the number bond with the sum and 1 addend given (1 page K.OA.5 – Fluently Add and Subtract Within 5 • Solve vertical addition equations (2 pages) • Solve horizontal addition equations (2 pages) • Match the sum to the equation (2 pages) • Complete the number bond (2 pages) • Complete the part-part-whole (2 pages) • Match the difference to the equation (2 pages) • Solve vertical subtraction equations (2 pages) • Solve horizontal subtraction equations (2 pages) K.G.1 – Shapes and Positional Words • Match the description to the picture (3 pages) • Circle the picture that matches the description (4 pages) • Color the shape that makes the description true (3 pages) K.G.2 – Naming Shapes • Match the shape to the name (2 pages) • Trace the word and draw the shape (2 pages) • Write the shape using a word bank (4 pages) • Color the correct shape (2 pages) • Cut and paste the shapes into the correct column (2 pages) K.G.3 – 2D & 3D Shapes • Color 2D shapes and 3D shapes (2 pages) • Circle the correct shape in each row (2 pages) • Cut and paste the shape into the correct column (1 page) • Finish the sentence with FLAT or SOLID (1 page) • Circle the correct object in each row (2 pages) • Match the shape to the real world object (2 pages) K.G.4 – Analyze and Compare Shapes • Match description to shape (1 page) • Finish each sentence with numbers to describe shapes (1 page) • Draw to shapes and write how they are same/different (3 pages) • Identify number of vertices (1 page) • Cut and paste into correct categories (4 pages) K.G.5 – Model Shapes • Complete dot to dot and name shapes using word bank (2 pages) • Trace and draw shapes (1 page) • Finish drawing shapes and name shapes using word bank (3 pages) • Construct shapes and name attributes (2 pages) • Trace shapes and use ruler to draw 5 more (2 pages) K.G.6 – Compose Simple Shapes to Form Larger Shapes • Use smaller shapes to make new shape -trace and draw (2 pages) • Use tangrams to draw a picture (1 page) • Make given pictures with tangrams and count shapes used in each (3 pages) • Cut smaller shapes and paste onto larger shapes (4 pages) Copyright © Cupcakes & Curriculum Permission to copy for single classroom use only. 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Great Squares Investigate how this pattern of squares continues. You could measure lengths, areas and angles. I was doodling the other day and drew a little square like this:- I supposed that the side was one (something) long. Well I wondered what would happen if I drew the four lines a bit longer, in fact twice as long so that the extra bits stuck out. This was quite a nice little design, I thought, and then I noticed that it looked at though the ends of these lines looked as though they could make a square. So I drew one! I've used a different colour to show this new square. Now, mathematical patterns usually go on repeating themselves so I used that idea to pretend that this new green square was my first one and so I drew the extra bits again, so that the lines were twice as long as the square. So I went on! A new square appeared, now red! Extend that one. . . I really liked what was happening here! and so on. . . . I had to stop there because it had up to the size of the paper. I suggest that you print these pages out so far and have a good look at the way that the pattern has grown and see what things you notice in this last picture. #set var="roll-text" value="Send us your findings if you get this far" --></p> <p>I found myself looking at a square and the extra lines and just one more square :-</p> <img alt="" src="/content/99/11/bbprob1/nov-dra3.jpg" /><br /> <br clear="all" /> <br /> <p>But I wanted it bigger and I thought about putting in some extra lines that again were just extensions of lines already there!</p> <img alt="" src="/content/99/11/bbprob1/ nov-sqrpa4.jpg" /><br /> <br clear="all" /> <br /> <p>Perhaps it would be good to print this out, if you have not done so already and explore the shapes, areas, lengths, angles etc.</p> <p>It would be really good to cut along the lines and see what happens. <!-- #set var="roll-text" value= "You can scan, photocopy, draw - then cut, rearrange and paste. Messy!" --><!-- #set var="roll-text" value When I looked at this shape I could imagine that I was looking through a square window at a pattern of squares but could only see the one square in the middle. So I printed out these:- Then I printed out the bigger square . . . . This one I traced onto a "see through" sheet and placed it in different places over the grid of smaller squares. This was great fun and led to some interesting conversations. Have a GO! I think that this is one of the most exciting shape investigations that I've put on the NRICH site, so let's have a lot of workings sent to Cambridge U.K. from all over the world so that we can show how people from different countries are thinking and working in their maths. Pester your teachers to collect some results and send them off. Student Solutions There are lots of answers to this problem, depending on what questions you choose to ask. Have a go yourself, and if you discover anything interesting, us to tell us what you've done! Please don't worry that your solution is not "complete" - we'd like to hear about anything you have tried. Teachers - you might like to send in a summary of your children's work. Teachers' Resources Why do this problem? is good for pupils working with squares and triangles. It will increase their familiarity with the properties of these shapes and is a good introduction to 'thinking outside the box'. Possible approach Working with the groups of pupils I've found it good to slowly present the first few stages of the growth of the squares. In this way I made sure that they understood each step. Key questions What do you notice? How do you know ...? (Following something they've said or recorded.) Possible extension This sheet gives some ideas for how to take the activity further. For more extension work Go to Extending Great Squares for ideas for this pupil. Possible support Some pupils will need help with accurate drawing, Squared paper is useful. Some pupils will benefit from using a computer drawing program.	{"url":"https://nrich.maths.org/problems/great-squares","timestamp":"2024-11-12T02:56:10Z","content_type":"text/html","content_length":"57166","record_id":"<urn:uuid:e0551728-a946-47de-95fc-ac078129ae01>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00065.warc.gz"}
University Digital Conservancy :: Browsing by Subject "Heterogeneity" Browsing by Subject "Heterogeneity" Now showing 1 - 9 of 9 • Results Per Page • Sort Options • Essays in Inequality and Heterogeneity Recent trends in both developed and developing economies show increasing inequality in income and wealth. Technological change is reshaping the nature of work for many, as automation, offshoring and other practices are adopted by firms around the globe. These changes to the type of jobs workers have are linked to changes in wages and labor earnings, in particular the adoption of new (worker-replacing) technologies has been linked to decreases in wages and increases in income inequality. Simultaneously, the trend towards higher inequality has sparked questions about the desirability (optimality) of inequality and whether governments should use the tools at their disposal to try to curb these trends. My dissertation contributes to the discussion on these topics in two distinct ways. The first two chapters deal with the effects of technological change in the nature of occupations, and its effects for wage inequality, while the third chapter deals with the implications of fiscal policy (particularly capital income and wealth taxation) in the face of wealth inequality caused by differences in the rate of return across individuals. The first part of my dissertation develops a new theory of how the specific tasks carried out by workers are determined, providing a flexible framework in which to study the implications for workers of automation, offshoring, skill-biased technological change among others. I use this framework along with U.S. occupational data to study the recent adoption of automation and its effects on the wage structure. The final chapter shows how the determinants of inequality matter for determining the optimal policy in the face of inequality. In the presence of rate of return heterogeneity wealth taxes dominate capital income taxes. Relative to capital income taxes, wealth taxes benefit the individuals who are more productive, increasing the allocative efficiency in the economy, in turn leading to potentially large welfare gains despite increases in inequality. • Essays in Inequality and Public Economics This dissertation consists of three chapters which contribute to quantitative and theoretical understanding of inequality and associated public policies. The first essay studies how different should income taxation be across singles and couples. I answer this question using a general equilibrium overlapping generations model that incorporates single and married households, intensive and extensive margins of labor supply, human capital accumulation, and uninsurable idiosyncratic labor productivity risk. The degree of tax progressivity is allowed to vary with marital status. I parameterize the model to match the U.S. economy and find that couples should be taxed less progressively than singles. Relative to the actual U.S. tax system, the optimal reform reduces progressivity for couples and increases it for singles. The key determinants of optimal policy for couples relative to singles include the detrimental effects of joint taxation and progressivity on labor supply and human capital accumulation of married secondary earners, the degree of assortative mating, and within-household insurance through responses of spousal labor supply. I conclude that explicitly modeling couples and accounting for the extensive margin of labor supply and human capital accumulation is qualitatively and quantitatively important for the optimal policy design. In the second essay, I develop a framework for assessing the welfare effects of labor income tax changes on married couples. I build a static model of couples' labor supply that features both intensive and extensive margins and derive a tractable expression that delivers a transparent understanding of how labor supply responses, policy parameters, and income distribution affect the reform-induced welfare gains. Using this formula, I conduct a comparative welfare analysis of four tax reforms implemented in the United States over the last four decades, namely the Tax Reform Act of 1986, the Omnibus Budget Reconciliation Act of 1993, the Economic Growth and Tax Relief Reconciliation Act of 2001, and the Tax Cuts and Jobs Act of 2017. I find that these reforms created welfare gains ranging from -0.16% to 0.62% of aggregate labor income. A sizable part of the gains is generated by the labor force participation responses of women. Despite three reforms resulting in aggregate welfare gains, I show that each reform created winners and losers. Furthermore, I uncover two patterns in the relationship between welfare gains and couples' labor income. In particular, the reforms of 1986 and 2017 display a monotonically increasing relationship, while the other two reforms demonstrate a U-shaped pattern. Finally, I characterize the bias in welfare gains resulting from the assumption about a linear tax function. I consider a reform that changes tax progressivity and show that the linearization bias is given by the ratio between the tax progressivity parameter and the inverse elasticity of taxable income. Quantitatively, it means that linearization overestimates the welfare effects of the U.S. tax reforms by 3.6-18.1%. The third essay studies the policies that are aimed at mitigating COVID-19 transmission. Most economic papers that explore the effects of COVID-19 assume that recovered individuals have a fully protected immunity. In 2020, there was no definite answer to whether people who recover from COVID-19 could be reinfected with the severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2). In the absence of a clear answer about the risk of reinfection, it is instructive to consider the possible scenarios. To study the epidemiological dynamics with the possibility of reinfection, I use a Susceptible-Exposed-Infectious-Resistant-Susceptible model with the time-varying transmission rate. I consider three different ways of modeling reinfection. The crucial feature of this study is that I explore both the difference between the reinfection and no-reinfection scenarios and how the mitigation measures affect this difference. The principal results are the following. First, the dynamics of the reinfection and no-reinfection scenarios are indistinguishable before the infection peak. Second, the mitigation measures delay not only the infection peak, but also the moment when the difference between the reinfection and no-reinfection scenarios becomes prominent. These results are robust to various modeling assumptions. • Essays in macro and labor economics (2013-06) Wiczer, David Geoffrey The first chapter studies the rate of long-term unemployment, which spiked during the Great Recession. To help explain this, I exploit the systematic and counter-cyclical differences in unemployment duration across occupations. This heterogeneity extends the tail of the unemployment duration distribution, which is necessary to account for the observed level of long-term unemployment and its increase since 2007. This chapter introduces a model in which unemployment duration and occupation are linked; it measures the effects of occupation-specific shocks and skills on unemployment duration. Here, a worker will be paid more for human capital in his old occupation but a bad shock may make those jobs scarce. Still, their human capital partly ``attaches'' them to their prior occupation, even when searching there implies a longer expected duration. Hence, unemployment duration rises and becomes more dispersed across occupations. Redistributive shocks and business cycles, as in the Great Recession, exacerbate this effect. For quantitative discipline, the model matches data on the wage premium to occupational experience and the co-movement of occupations' productivity. The distribution of duration is then endogenous. For comparison's sake, if a standard model with homogeneous job seekers matches the job finding rate, then it also determines expected duration and understates it. That standard model implies just over half of the long-term unemployment in 1976-2007 and almost no rise in the recent recession. But, with heterogeneity by occupation, this chapter nearly matches long-term unemployment in the period 1976-2007 and 70% of its rise during the Great Recession. The second chapter studies the link between wage growth and the match of a worker's occupation and skills. The notion here is that if human capital accumulation depends on match quality, poor matches can have long-lasting effects on lifetime earnings. I build a model that incorporates such a mechanism, in which human capital accumulation is affected by imperfect information about one's self. This informational friction leads to matches in which a worker accumulates human capital more slowly and has weaker earnings growth. To get direct evidence, the chapter pieces together two sets of data on the skills used by an occupation and the skills a worker is particularly good at. Data on occupations describes occupations by the intensity with which they use many dimensions of workers' knowledge, skills and abilities. To pair, we have data on tests taken by respondents in a panel that tracks occupations and earnings. The test designers created a mapping between their tests and the occupational descriptors, which allows us to create two measures. The first measure of match quality is just the dot product between the dimensions of workers' skills and utilization rate of these skills by occupations. The second measure mismatch relative to an optimal matching computed using the Gale-Shapley algorithm for stable pairs. In both, worse matches have significantly slower returns to occupational tenure. With the most conservative estimate, plus or minus one standard deviation of mismatch affects the return to occupational tenure by 1% per • Heterogeneous protein distribution during rapid and equilibrium freezing (2013-04) Twomey, Alan Michael Interactions between proteins and ice were studied in situ using FTIR and confocal Raman microspectroscopy under equilibrium and non-equilibrium conditions over a range of temperatures. During quasi-equilibrium freezing of aqueous solutions of dimethyl sulfoxide (DMSO) and bovine serum albumin, preferential exclusion of albumin and/or DMSO was observed. It was hypothesized that the albumin may be adsorbed onto the ice interface or entrapped in the ice phase. To investigate protein-ice interactions during freezing under non-equilibrium conditions, confocal Raman microspectroscopy was used to map the distribution of albumin and the cryoprotective agent trehalose. Microheterogeneity was found in the composition of the freeze-concentrated liquid phase that indicated that albumin was preferentially distributed near or at the boundary of the ice phase. The observed microheterogeneity did not occur under all freezing protocols, which suggests that the technique developed here could be used to develop freezing protocols that would reduce harmful protein-ice interactions. • Innovative Statistical Methods for Meta-analyses with Between-study Heterogeneity To assess the benefits and harms of medical interventions, meta-analysis plays an important role in combining results from multiple studies. While the notion of combining independent results is motivated by similarities between studies, a pooled estimate may be insufficient in the presence of between-study heterogeneity in a meta-analysis. The sources of between-study heterogeneity come from studies being: 1) different and unrelated (possibly due to a mixture of non-replicable study findings); 2) different but similar (i.e., drawn from the same distribution); or 3) susceptible to modeling using covariates. In the first, studies do not replicate each other, and meta-analysis is not considered an option. In the second, a random-effects model may be used to reflect the similarity of studies, and in the third, a meta-regression analysis is suggested. To differentiate the first from the others, it is essential to develop a statistical framework establishing whether multiple studies give sufficiently similar results, i.e., replicate each other, before undertaking a meta-analysis. However, traditional meta-analysis approaches cannot effectively distinguish whether the between-study difference is from non-replicability or unknown study-specific characteristics. No rigorous statistical methods exist to characterize the non-replicability of multiple studies in a meta-analysis. In Chapter 2, we introduce a new measure, the externally standardized residuals from a leave-m-studies-out procedure, to quantify replicability. We also explore its asymptotic properties and use extensive simulations and three real-data studies to illustrate this measure's performance. We also provide the R package "repMeta" to implement the proposed approach. The remainder of this dissertation concerns scenarios when substantial heterogeneity still exists among replicable studies in a meta-analysis. Such heterogeneity may or may not decrease by incorporating available covariates in a meta-analysis, given that the sources of effect heterogeneity are commonly unknown and unmeasured. A proxy for those unknown and unmeasured factors may still be available in a meta-analysis, namely the baseline risk. Chapter 3 proposes using the bivariate generalized linear mixed-effects model (BGLMM) to 1) account for the potential correlation of the baseline risk with the treatment effect measure, and 2) obtain estimated effects conditioning on the baseline risk. We demonstrate a strong negative correlation between study effects and the baseline risk, and the conditional effects notably vary with baseline risks. Chapter 4 reinforces the suggestion that a meta-analysis should model the heterogeneity in effect measures with respect to baseline risks and study conditions. It finds that two commonly-used binary effect measures, the odds ratio (OR) and risk ratio (RR), have a similar dependence on the baseline risk in 20,198 meta-analyses from the Cochrane Database of Systematic Reviews, a leading source of healthcare evidence. This empirical evidence contrasts with a false argument that OR does not vary with study conditions. We illustrate that understanding effect heterogeneity is essential to patient-centered practice in an actual meta-analysis of the interventions addressing the chronic hepatitis B virus infection. • Mechanical Heterogeneity and Mechanoadaptation in Cerebral Aneurysms (2022-12) Shih, Elizabeth Cerebral aneurysms are abnormal dilations of blood vessels in the brain found in 2% of the population. While rupture is rare, it is fatal or will most likely cause neurological deficits. The prevalence of unexpected ruptures suggests that the current predictive measurements to evaluate rupture risk are incomprehensive and require more investigation. To understand progression and stabilization versus rupture, we adopt a biomechanical approach to investigate how cellular mechanism influence tissue-scale mechanics. In my first aim, I mechanically characterize the local heterogeneity in acquired human cerebral aneurysm and arterial specimens using the Generalized Anisotropic Inverse Mechanics method. I find that both ruptured and unruptured aneurysms are considerably weaker and more heterogeneous than normal arteries, suggesting that maladaptive remodeling results in complex mechanical properties arising from initially ordered structures. From these changes, stress concentrations at boundaries between stiff and weak regions and diverse cell microenvironments are all likely to influence stabilization versus rupture. After identifying that aneurysms contain a wide range of microenvironment stiffnesses, I investigate how local extracellular stiffnesses influence the mechanically dominant and mechanosensitive vascular smooth muscle cells using cellular microbiaxial stretching. First, I examine the common assumptions used in inverse calculations of cell tractions and find that a crucial filtering term must be scaled accordingly to cell substrate mechanical properties to ensure accurate calculations. When this term is adjusted across different microenvironment/substrate groups, I find that healthy smooth muscle cells are remarkably robust across a wide range of substrate moduli. Lastly, I develop a continuum model to capture the physical forces exerted on single cells during aneurysm progression, in which cell density begins to decrease and cells are only able to remodel their immediate surroundings. The model introduces a strain factor for vascular smooth muscle cells, which combines the homogeneous rule-of-mixtures approach with an Eshelby-based strain factor to describe a single inclusion in an infinite matrix. This model will be incorporated into future growth and remodeling laws to describe aneurysm progression. Taken together, the results of this work elucidate the complex tissue and cell mechanics that govern aneurysm development, stabilization, and rupture. This provides a basis to eventually identify new metrics for risk evaluation and improve future predictive models for clinical translation, ultimately aiding aneurysm diagnoses and treatment plans. • Queueing Analysis of Computer Systems (2022-08) Abdul Jaleel, Jazeem Heterogeneity and Performance Interference are two characteristics of modern large scale computer systems. The policies developed for simple homogeneous computer systems do not scale well when accounting for Heterogeneity and Performance Interference. In this thesis, we develop novel “power-of-d” policies that better address such systems. In the first part of this thesis, we approach the challenges of load balancing in large scale heterogeneous server systems. We sequentially develop effective load balancing models and introduce a framework to help characterize the different load balancing policies based on their querying and assignment rules. We compare the performance of these novel optimizable policies with the conventional policies present in literature for different parameter settings. Our policy framework allows us to develop complex load balancing policies --- i.e., allowing for probabilistic querying and/or job assignment not following simple rules such as Join-the-Idle-Queue, Join-the-Shortest-Queue, Join-the-Shortest-Expected-Wait-Queue and Join-the-Shortest-Expected-Delay-Queue--- which is a novel addition to the literature. Furthermore, our work makes it possible to do a comprehensive numerical study of policies that consider each queried server's queue length and speed information for job assignment. Prior to our work, conducting such a study required simulations which was computationally infeasible. Our performance comparisons for different mixes of querying and assignment rules allows us to identify the trade-off between policy simplicity and performance. In the last chapter of this thesis, we address a new area of interest which is load balancing models for systems where servers undergo performance interference. We first develop simple novel “power-of-d” policies for such systems that consider queried server's idleness and/or interference state information for job assignment. Using analytical proofs and numerical experiments we analyze the performance of these policies and identify system parameter regions that favors different heuristics. The analysis further motivates us to develop a more general and complex optimizable policy that has better performance under all parameter settings. • Statistical methods for meta-analysis Meta-analysis has become a widely-used tool to combine findings from independent studies in various research areas. This thesis deals with several important statistical issues in systematic reviews and meta-analyses, such as assessing heterogeneity in the presence of outliers, quantifying publication bias, and simultaneously synthesizing multiple treatments and factors. The first part of this thesis focuses on univariate meta-analysis. We propose alternative measures to robustly describe between-study heterogeneity, which are shown to be less affected by outliers compared with traditional measures. Publication bias is another issue that can seriously affect the validity and generalizability of meta-analysis conclusions. We present the first work to empirically evaluate the performance of seven commonly-used publication bias tests based on a large collection of actual meta-analyses in the Cochrane Library. Our findings may guide researchers in properly assessing publication bias and interpreting test results for future systematic reviews. Moreover, instead of just testing for publication bias, we further consider quantifying it and propose an intuitive publication bias measure, called the skewness of standardized deviates, which effectively describes the asymmetry of the collected studies’ results. The measure’s theoretical properties are studied, and we show that it can also serve as a powerful test statistic. The second part of this thesis introduces novel ideas in multivariate meta-analysis. In medical sciences, a disease condition is typically associated with multiple risk and protective factors. Although many studies report results of multiple factors, nearly all meta-analyses separately synthesize the association between each factor and the disease condition of interest. We propose a new concept, multivariate meta-analysis of multiple factors, to synthesize all available factors simultaneously using a Bayesian hierarchical model. By borrowing information across factors, the multivariate method can improve statistical efficiency and reduce biases compared with separate analyses. In addition to synthesizing multiple factors, network meta-analysis has recently attracted much attention in evidence-based medicine because it simultaneously combines both direct and indirect evidence to compare multiple treatments and thus facilitates better decision making. First, we empirically compare two network meta-analysis models, contrast- and arm-based, with respect to their sensitivity to treatment exclusions. The arm-based method is shown to be more robust to such exclusions, mostly because it can use single-arm studies while the contrast-based method cannot. Then, focusing on the currently popular contrast-based method, we theoretically explore the key factors that make network meta-analysis outperform traditional pairwise meta-analyses. We prove that evidence cycles in the treatment network play critical roles in network meta-analysis. Specifically, network meta-analysis produces posterior distributions identical to separate pairwise meta-analyses for all treatment comparisons when a treatment network does not contain cycles. This equivalence is illustrated using simulations and a case study. • Three essays on Public Economics and heterogeneity. (2009-08) Schneider, Anderson Luis This thesis presents a collection of essays about Public Economics and individual heterogeneity. The essays are motivated by two different subjects. The first subject refers to the relation between economic outcomes and majority voting in a democratic regime. More specifically, the outcome regarding redistributive labor income taxes is analyzed when heterogeneous individuals vote, once and for all, over an infinite sequence of taxes. The second subject refers to time consistency problems in Public Economics. Issues about optimal fiscal policy are considered in an environment where different individuals hold distinct information about (sequential) action performed by the government. This friction prevents the standard punishment mechanism that enforces good policy outcomes or, alternatively, inhibits the occurrence of time consistency problems. The best equilibrium outcome is then analyzed in this new situation. Chapters 2 focus on the first subject. Moreover, the chapter explores the relationship between changes in labor income inequality and movements in labor taxes over the last decades in the US. In order to do so, this relation is modeled through a political economy channel by developing a median voter result over sequence of taxes. We consider an infinite horizon economy in which agents are heterogeneous with respect to both initial wealth and labor skills. We study indirect preferences over redistributive fiscal policies - sequences of affine taxes on and capital income - that can be supported as a competitive equilibrium. The paper assumes balanced growth preferences and full commitment. The first result is the following: if initial capital holdings are an affine function of skills, then the best fiscal policy for the agent with the median labor skill is preferred to any other policy by at least half of the individuals in the economy. The second result provides the characterization of the most preferred tax sequence by the median agent: marginal taxes on labor depend directly on the absolute value of the distance between the median and the mean value of the skill distribution. We extend the above results to an economy in which the distribution of skills evolves stochastically over time. A temporary increase in inequality could imply either higher or lower labor taxes, depending on the sign of the correlation between inequality and aggregate labor. The calibrated model does a good job on fitting both the increasing trend and the levels of labor taxes in the last decades, and also on matching some short run co-movements. Chapter 3 generalizes the median voter theorem developed in chapter 2 to a situation where there is no commitment or, alternatively, voting is sequential over time. More specifically, the same equilibrium definition as in Bernheim and Slavov (2008) is adopted. Chapter 4 deals with optimal fiscal policy when the government takes actions sequentially over time and cannot commit to a pre-specified plan of actions. These features potentially generate what is known in the literature as time consistency problems. Although these problems play an important role in public policy, game theoretical models in macroeconomics seem to indicate the opposite. Due to the complexity of this kind of models, it is commonly assumed that information is complete and perfect. In turn, this assumption becomes the key element that allows agents to coordinate perfectly to punish the government if it does not do what private agents want. As a result, a wide range of feasible payoffs can be sustained as equilibrium, including the best payoff under commitment. Since this approach is widely used for normative purposes a natural question emerges: are the above results robust to small variations in information? This paper analyzes an investment taxation problem in an economy with incomplete information. Specifically, we study an environment with the following main characteristics: 1) the aggregate productivity (fundamental) is stochastic, 2) only the government observes it and; 3) every agent privately receives a noisy signal about the fundamental. The first characteristic implies that the best policy (tax on investment) with commitment is state contingent. The second and third characteristics make the information incomplete. In particular, agents have different information sets, and therefore different beliefs, about the true state of the economy. As a result, independently of the accuracy of the signal, incomplete information reduces the set of equilibrium payoffs. First, we show that any policy that depends solely on the fundamental cannot be an equilibrium. Second, the best equilibrium policy is independent of the fundamental. Finally, for any discount factor strictly smaller than one and for any size of the noise, the best equilibrium is inefficient.	{"url":"https://conservancy.umn.edu/browse/subject?value=Heterogeneity","timestamp":"2024-11-05T19:41:52Z","content_type":"text/html","content_length":"494171","record_id":"<urn:uuid:4289179b-d899-4c5f-9d6c-8ba1942129ab>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00662.warc.gz"}
Calibrate new allometric equations — est_params Calibrate new allometric equations This function calibrates new allometric equations from sampling previous ones. New allometric equations are calibrated for each species and location by resampling the original compiled equations; equations with a larger sample size, and/or higher taxonomic rank, and climatic similarity with the species and location in question are given a higher weight in this process. species = NULL, new_eqtable = NULL, wna = 0.1, w95 = 500, nres = 10000 a character vector, containing the genus (e.g. "Quercus") of each tree. a numeric vector of length 2 with longitude and latitude (if all trees were measured in the same location) or a matrix with 2 numerical columns giving the coordinates of each tree. a character vector (same length as genus), containing the species (e.g. "rubra") of each tree. Default is NULL, when no species identification is available. Optional. An equation table created with the new_equations() function. Default is the compiled allodb equation table. a numeric vector, this parameter is used in the weight_allom() function to determine the dbh-related and sample-size related weights attributed to equations without a specified dbh range or sample size, respectively. Default is 0.1. a numeric vector, this parameter is used in the weight_allom() function to determine the value at which the sample-size-related weight reaches 95% of its maximum value (max=1). Default is 500. number of resampled values. Default is "1e4". An object of class "data.frame" of fitted coefficients (columns) of the non-linear least-square regression: $$AGB = a * dbh ^ b + e, \space \mathit{with} \space e ~ N(0, sigma^2)$$ # calibrate new allometries for all Lauraceae species lauraceae <- subset(scbi_stem1, Family == "Lauraceae") genus = lauraceae$genus, species = lauraceae$species, coords = c(-78.2, 38.9) #> # A tibble: 2 × 7 #> genus species long lat a b sigma #> <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> #> 1 Lindera benzoin -78.2 38.9 0.0834 2.52 354. #> 2 Sassafras albidum -78.2 38.9 0.0817 2.52 339.	{"url":"https://docs.ropensci.org/allodb/reference/est_params.html","timestamp":"2024-11-14T02:07:23Z","content_type":"text/html","content_length":"16646","record_id":"<urn:uuid:ca099f35-cbca-4560-8809-94cceaaed73e>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00736.warc.gz"}
1 Unite de Recherche en Genetique Moleculaire et en Hematologie, TNSERM U.91 CN RS U A 607, Hopital Henri Mondor , 94010 Creteil, France. 2 INSERM U 152, lnstitut Cochin de Genetique Moleculaire (ICGM), Hopital Cochin, 27 Rue du Faubourg Saint Jacques, 75014 Paris, France. 3 Institut Curie, 27 Rue d'Ulm, 75005 Paris, France. Most acutely transforming leukemogenic or sarcomagenic retroviruses have transduced in their genomes altered cellular genes named oncogenes [1]. These viruses are usually defective in replication as a result of the deletion of structural genes, and they require a helper virus for propagation. Isolation of new acutely transforming retroviruses, comprehensive studies of their physiopathological processes, and molecular analysis of their genomes remain, therefore, powerful means for the identification of key genes involved in the regulation of cell growth, differentiation, or development. The purpose of this paper is to summarize the data that we have accumulated over the past few years on a recently isolated acute leukemogenic murine retrovirus named myeloproliferative leukemia virus (MPL V). Isolation of the MPLV We isolated the MPL V in 1985 at the Curie Institute (Orsay, France) during a research program designed to evaluate the in vivo transforming properties of different Friend helper viruses (F MuL V). While several clonal F-MuL V isolates have the capacity to induce a rapid erythroblastosis in newborninoculated NIH Swiss or BALB/c mice [2], DBA/2 mice were found to be resistant to this early erythroleukemia [3]. Nevertheless, they developed various types of hematopoietic malignancies after a latent interval of712 months [4-6]. In general, these leukemias ( either myelogenous, lymphoid, or erythroid) were associated with a more or less severe anemia. Out of 238 DBA/2 mice inoculated at birth with F -MuL V clone 57 [7], one mouse developed, after 7 months of infection, an hepatosplenomegaly unusually accompanied with a polycythemia. Cellfree extract prepared from the original leukemic spleen or supernatant medium from an in vitro permanent cell line derived from the leukemic spleen cells caused an explosive leukemia upon inoculation into adult mice of most strains, including C 57 Bl strains. The disease was characterized by hepa tosplenomegaly, polycythemia, pronounced myelemia but no thymus or lymph node involvement, and death within 1- 3 months. Spleen and liver were extensively infiltrated with maturing precursor cells belonging to the granulocytic, erythroblastic and megakaryocytic lineages. Typically, the blood of severely diseased animals was also massively invaded by morphologically normal polymorphonuclears, erythroblasts, and platelets. Several hematopoietic lineages were obviously involved in this disease, hence our name for the virus isolate, "myeloproliferative leukemia virus" [8]. Genetic Analysis of MPL V Isolate Virologic studies of MPL V by Penciolelli et al. [9] demonstrated that this highly leukemogenic virus isolate contained two dissociable retroviral genomes: one was the parental replication-competent FMuL V 57, and the second was a new replication-defective component now designated as MPL V. A comparison of viral RN A species expressed in F -MuLV alone or F-MuLV + MPLV-producing cells by Northern blot analysis showed that MPLV was 0.8 kb shorter than F-MuLV and that a deletion had probably occurred in the MPL V env gene. This was further confirmed by the establishment of the MPL V restriction endonuclease map which was compared with that of F-MuLV [10]. From their data, these investigators concluded that the MPLVdefective genome (a) was derived from F-MuLV, (b) had conserved the F -M uL V gag and pol regions, and (c) was deleted and rearranged in its env region [9]. Although MPLV does not transform fibroblasts in culture, its isolation free of replicating F-MuLV in nonproducer cells was feasible since the MPL V titer in the original isolate was approximately equivalent to that of F-MuLV. By the technique of limiting dilution and singlecell cloning, nonproducer cells containing MPL V were derived from Mus dunni fibroblasts [9]. Supernatant medium from these nonproducer cells did not cause any disease in inoculated mice demonstrating the defectiveness of MPL V. However , when superinfected with a variety of replicating helper viruses, supernatants reproduced the same acute myeloproliferative syndrome as caused by the original isolate. These experiments provided circumstantial evidence that the helper-dependent MPL V genome contained the genetic information necessary for the observed pathological processes. Genomic Composition of MPLV In an attempt to define the origin and nature of the genetic sequences contained in the MPL V -rearranged env region, Souyri et al. [11] derived cDNA probes which were nonhomologous to sequences contained in F-MuLV. Two probes were found to be MPL V specific, in that they hybridized to RNA of MPLV -containing nonproducer cells but did not hybridize to RNA ofecotropic MuLVs norto RNA of amphotropic or xenotropic murine viruses. This indicated that, in contrast to Friend spleen focus-forming viruses (SFFV), MPL V did not result from a recombination between F-MuLV and a portion of the env gene of murine xenotropic virus [ 12, 13] . A full-length biologically active MPL V provirus was molecularly cloned from a genomic library of a nonproducer Mus dunni clone [11 ]. Sequence analysis revealed that the MPL V env gene contains a large open reading frame which could code for a polypeptide of 284 amino acids. This protein would contain 64 amino acids derived from the amino terminus of the F-MuL V gp70, including the signal peptide, 36 amino acids from a central region of the F-MuLV env gene, and 184 amino acids that are specific to MPL V (Fig. 1). A hydrophobicity plot of the amino acids sequence revealed that, in addition to the 34 hydrophobic amino acids of the gp 70 signal peptide, the MPL V -specific domain contained a stretch of 22 uncharged amino acids. Thus, the putative MPL V env product presents the features of a transmembrane protein comprising an extracellular domain of 143 amino acids, a single transmembrane domain of 22 amino acids, and a cytoplasmic domain of 119 amino acids without consensus sequence for kinase activity [14]. Computer analysis of the deduced amino acid sequence revealed that the MPL V -specific sequence did not correspond to any known genes. Fig. I. Schematic representation of the putative MPLV env product. The ENV-mpl fusion protein consists of three fragments: part 1 derives from the NH2-terminal region of the F-MuLV envelope gene (gp70) and contains a signal peptide; part 2 derives from a central part of the F- MuLV gp 70; part 3 corresponds to a transduced cellular sequence, most probably truncated at its N-terminus. The product encoded by this rearranged gene has the features of a single mernbrane-spanning domain (TM). The extra cellular domain of v-mpl possesses the amino acid sequence WSAWS highly conserved in the hematopoietin receptor superfamily, while the cytoplasmic domain does not contain consensus sequence for know catalytic activity MPLV Has Transduced a Novel Oncogene Since nonviral sequences found in the genome of acutely transforming retroviruses derive from cellular genes and are conserved phylogenetically, we looked for the presence of MPLV -specific sequences in genomic DNA from different mammals. Under stringent hybridization conditions, discrete bands were revealed in DNAs from mouse, rat, mink, dog, cow, and human. In addition, MPLV specific probes recognized a 3.0-kb mRNA in spleen and bone marrow from adult mice and in fetal liver cells, but not in nonhematopoietic tissues [11 ]. Thus, taking into consideration the biological properties of MPL V, the cellular origin of the sequence contained in its env gene, the conservation in the genome of mammals and the expression in normal hematopoietic tissues, we concluded that MPL V had transduced a novel oncogene which was designated as v-mpl. By in situ hybridization and genetic analysis studies, chromosomal localization of the c-mpl proto-oncogene was assigned to mouse chromosome 4 (Vigon et al., unpublished data) and to human chromosome 1 p34 [15]. Leukemogenic Properties of MPLV In vivo studies by Wendling and coworkers have indicated that MPL V induced a rapid suppression of growth factor requirements for in vitro colony formation of a large spectrum of committed as well as multipotential progenitor cells [16, 17]. The primary manifestation of viral infection was a switch to erythropoietin (EPO) independence of the colony forming unit-erythroid (CFU-E) population which was complete in the spleen after 6 days of infection. A possible stimulating effect of EPO present or secreted in the culture medium was ruled out by the addition of neutralizing antiEPO antibodies to the culture system. The effects of MPL V infection on the early and primitive erythroid progenitor cells (BFU-E) was assessed in methylcellulose serum-free cultures. It was found that well hemoglobinized pure and mixed erythroid colonies developed without the addition of interleukin-3 or EPO. Moreover, while a majority of colonies contained erythroblasts mixed with megakaryocytes, about 12% revealed three or more lineages of differentiation [16]. Further in vivo studies have documented that MPL V infection also induced the spontaneous colony formation of myeloid progenitors, i.e., granulocyte macrophage colony-forming cell (GMCFC) granulocyte (G)-CFC, mega karyocyte (Meg)-CFC, and mixed CFC, probably as a result of direct infection of these progenitors and not as a consequence of a paracrine secretion of soluble colony stimulating factors by the accessory cells [17]. These observations supported the conclusion that MPL V acts on various progenitors, inducing their proliferation and terminal differentiation independently of signals normally provided by colony stimulating factors, interleukins, EPO, or any conditioned medium. However, formal proof that MPL V can transform hematopoietic target cells in the absence of coinfection with a replicating MuL V was not provided by these experiments. We addressed this question by producing helper-free MPL V stocks using the packaging psi-CRE cell line that produces a high titer of infectious, nonreplicating particles but does not yield helper virus [18]. When adult ICFW mice were intravenously given helper-free preparations of MPL V, more than 90% of the mice were healthy 2 months after inoculation. Nevertheless, we observed that MPL V induced a mild but transient spleen enlargement with the appearance of colonies well visible on the spleen surface on days 5, 10, and 15 after inoculation. Histologically, colonies were composed of erythroblasts, or erythroblasts, granulocytes, and megakaryocytes clustered together in the splenic red pulp. On day 25 and thereafter, these colonies disappeared, leaving spleens with a normal aspect. In contrast, when helperfree preparations of MPL V were injected into mice pretreated with the aplastic drug 5- fluorouracil ( 5- FU, 150 mg/kg body weigth, 4 days before virus inoculation), all animals developed atypical MPL V syndrome and died from overt leukemia within 2 months (Wendling et al., unpublished data). Together these data indicate that (a) the MPL V component is primarily responsible for the myeloproliferative effects of the viral complex, (b) expression of MPL V in erythroid and myeloid progenitors abolishes their growth factor requirement for in vitro colony formation, and (c) induction of leukemia occurs in 5FU -pretreated mice, suggesting that stable infection of cycling primitive progenitors is critical for leukemia development. In Vitro Transformation Properties of MPLV An area of current research in our laboratory is related to the ability of a helperfree preparation of MPL V to transform hematopoietic cells in vitro. A 2-h incubation of bone marrow cells enriched in highly dividing primitive progenitors by treatment of mice with 5- FU was sufficient to induce autonomous colony formation of about 30% of the colonyforming cells present in the preparation. Cytologically, half of these spontaneous colonies were composed of either granulocytes, megakaryocytes, or erythrocytes, while the remainders were mixed colonies of which about 20% contained three or more lineages of differentiation. Upon replating, the multilineage colonies produced secondary and tertiary mixed colonies, suggesting self-renewal [11]. The question of whether or not transformation of hematopoietic progenitors would lead to the generation of immortalized cell lines was then investigated. When marrow cells were cultured in liquid medium, it was observed that rapidly dividing nonadherent cell populations were produced in MPL V -infected cultures. After 10 to 12 days, these nonadherent populations could be transferred into fresh flasks devoid of stromal feeder layers. Cells continued to proliferate and generated permanent suspension cultures containing polymorphonuclears, megakaryocytes and erythroblasts. Upon continuous passages, the majority of the cell lines evolved towards a more restricted phenotype which remained stable over several months. Diverse immortalized megakaryocytic, myelomonocytic, erythroblastic, or mastocytic cell lines retaining the ability to differentiate could easily be obtained. Since these permanent cell lines evolved from a multipotential to a more restricted phenotype, we investigated whether they were polyclonal or monoclonal by studying proviral-cell DNAjunctions. Cultures were polyclonal 5 days after initiation. However, after 3 weeks and at a time where all cultures displayed a multipotential phenotype, one or a few major proliferating clones were detected in each cell line. Interestingly, the same clones were still found after 3 months of continuous passages when the cell lines appeared to be restricted in their differentiation potential [11]. Thus, it seems likely that MPLV induces the clonal outgrowth of a single or few transformed, probably multipotentiaI, stem cells (clonal selection), the full differentiation capabilities of which being lost along with continuous culturing (clonal evolution). The obtaining of immortalized in vitro cell lines raised the question of whether cells were tumorigenic. To approach this problem, 2 x 106 cells were subcutaneously grafted into either syngeneic or nude mice. Upon repeated assays, none of the cell lines developed tumor nodules at the site of inoculation when cells from cultures less than 4 months old were grafted. After prolonged passages (more than 7 months), 60% of the cell lines produced hematopoietic subcutaneous tumoral nodules, suggesting that additional genetic events must have occurred to reach a full malignant state. Summary and Current Knowledge The myeloproliferative leukemia virus isolate consists of two distinct viral components: a replicating F-MuLV and a helper-dependent MPL V. MPL V accounts for the rapid in vivo and in vitro transformation of a broad spectrum of multipotential, myeloid, and erythroid progenitors which acquire growth factor independent proliferation and differentiation. By sequence analysis of a biologically active clone, MPL V has been shown to be an env recombinant virus containing sequences derived from the F-MuL V env gene and additional nonviral cellular sequences. These nonviral sequences are conserved in various mammals and are expressed in hemopoietic tissues from normal mice. MPLV was thus generated by transduction of an oncogene (v-mpl) in the envelope region of an F -MuLV genome. v-mpl does not correspond to any known gene, but the putative MPL V env fusion product has the features of a transmembrane protein with the Nterminal signal sequence of the F-MuLV gp70 directing the polypeptide across the membrane and a single transmembrane domain. Interestingly, the extracellular domain of v-mpl possesses, 13 amino acids upstream to the membranespanning domain, the amino acid sequence WSXWS, highly conserved in all cy to kine receptors that make up the hematopoietin receptor superfamily [19]. In addition, a significant number of conserved amino acids were found when the extracellular domain ofv-mplwas aligned with that of the IL-2ß, IL-3, IL-4, IL-6, IL- 7, GM-CSF, G-CSF, and EPO receptors [11]. Since the N-terminal part of the fusion protein consists of F -M uL V derived sequences, it is not yet known whether the c-mpl proto-oncogene product would contain the highly conserved cysteine residues characteristically found in the ligand-binding domain of each of these receptors [ 19] .Nevertheless, with regard to the general features of v-mpr, it is tempting to speculate that MPL V has transduced a truncated form of a putative cytokine receptor. Cloning of the protooncogene cDNA is currently underway in our laboratory to allow further comparison. A major focus of future research will be to understand the mechanism by which this viral oncogene can short-circuit the growth-regulatory signals delivered by the binding of various hematopoietic growth factors to their specific receptors. This requires further studies on the mechanism of signal transduction by MPL V and by other receptors of the same family. We thank Martine Charon, Laurence Cocault and Paule Varlet who provided excellent technical assistance. This work was supported by grants from the Institut National de la Recherche Medicale (INSERM), the Centre National de la Recherche Scientifique (CNRS), the Association pour la Recherche contre la Cancer (ARC), the Fondation pour la Recherche Medicale, and the Ministere de la Recherche et de la 1. Reddy EP, Salka AM, Curran T (eds) (1988) The oncogene handbook. Elsevier, Amsterdam 2. Troxler DH, Scolnick EM (1978) Rapid leukemia induced by cloned Friend strain of replicating murine type-C virus. virology 85:17-27 3. Ruscetti S, Davis L, Fields J, Oliff AI (1981) Friend murine leukemia virusinduced leukemia is associated with the formation of mink cell focus-inducing viruses and is blocked in mice expressing endogenous mink cell focus-inducing xenotropic viral envelope genes. J Exp Med 154:907-920 4. Wendling F, Fichelson S, Heard JM, Gisselbrecht S, Varet B, Tambourin P (1983) Induction of myeloid leukemias in mice by biologically cloned ecotropic F-MuL V. In: Scolnick EM, Levine AJ (eds) Tumor viruses and differentiation. Liss, New York, pp 357- 362 5. Shibuya T, Mak TW (1982) Host control of susceptibility to erythroleukemia and to the types of leukemia induced by Friend murine leukemia virus: initial and late stages. Cell 31 :483~493 6. Chesebro B, Portis JL, Wehrly K, Nishio J (1983) Effect of murine host genotype on MCF virus expression, latency and leukemia types of leukemias induced by Friend murine helper virus. Virology 128: 7. Oliff AJ, Hager G L, Chang EH, Scolnick EM, Chan HW, Lowy DR (1980) Transfection of molecularly cloned Friend murine leukemia virus DNA yields a highly leukemogenic helper-independent type C virus. J Virol 33 :475-486 8. Wendling F, Varlet P, Charon M, Tambourin P (1986) MPLV: a retrovirus complex inducing an acute myeloproliferative leukemic disorder in adult mice. Virology 149:242-246 9. Penciolelli JF, Wendling F, Robert Lezenes J, Barque JP, Tambourin P, Gisselbrecht S (1987) Genetic analysis of myeloproliferative leukemia virus, a novel acute leukemogenic replication-defective retrovirus. J Virol 61: 579- 10. Chattopadhyay SK, Oliff Al, Linemeyer MR, Lander MR, Lowy DR (1981) Genomes of murine leukemia viruses isolated from wild mice. J ViroI39:777-791 11. Souyri M, Vigon I, Penciolelli JF, Heard JM, Tambourin P, Wendling F (1990) A putative truncated cytokine receptor gene transduced by the myeloproliferative leukemia virus immortalizes hematopoietic progenitors. Cell 63:1137-1147 12. Ruscetti S, Linemeyer D, Feild J, Troxler D, Scolnick EM (1979) Characterization of a protein found in cells infected with spleen focus-forming virus that shares immunological cross-reactivity with the gp70 found in mink cell focus-inducing virus particles. J Virol 30: 787- 798 13. Clark SP, Mak TW (1983) Complete nucleotide sequence of an infectious clone of Friend spleen focus-forming provirus: gp55 is an envelope fusion glycoprotein. Proc Natl Acad Sci USA 80:5037-5041 14. Hanks SK, Quinn AM, Hunter T (1988) The protein kinase family conserved features and deduced phylogeny of catalytic domains. Science 241 :42-52 15. Le Conniat M, Souyri M, Vigon I, Wendling F, Tambourin P, Berger R (1989) The human homolog of the myeloproliferative leukemia virus maps to chromosome band 1p34. Hum Genet 83:194-196 16. Wendling F, Penciolelli JF, Charon M, Tambourin P (1989) Factor-independent erythropoietic progenitor cells in leukemia induced by the myeloproliferative leukemia virus. Blood 73:1161-1167 17. Wendling F, Vigon I, Souyri M, Tambourin P (1989) Myeloid progenitor cells transformed by the myeloproliferative leukemia virus proliferate and differentiate in vitro without the addition of growth factors. Leukemia 3 :475-480 18. Danos 0, Mulligan R (1988) Safe and efficient generation of recombinant retroviruses with amphotropic and ecotropic host ranges. Proc Natl Acad Sci USA 85:6460-6464 19. Cosman D, Lyman SC, ldzerda RL, Beckmann MP, Parks LS, Goodwin RG, March CJ (1990) Anew cytokine receptor superfamily. TIBS 15:265-270	{"url":"http://www.science-connections.com/trends/human_leukemia/158.htm.htm","timestamp":"2024-11-02T12:34:39Z","content_type":"text/html","content_length":"29262","record_id":"<urn:uuid:2292445c-919f-49e8-a45c-8c8e065714ae>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00521.warc.gz"}
Information for Families We’d like to introduce you to the Illustrative Mathematics curriculum. This problem-based curriculum makes rigorous elementary school mathematics accessible to all learners. What is a problem-based curriculum? In a problem-based curriculum, students spend most of their time in class working on carefully crafted and sequenced problems. Teachers help students understand the problems, ask questions to push their thinking, and orchestrate discussions to be sure that the mathematical takeaways are clear. Learners gain a rich and lasting understanding of mathematical concepts and procedures and experience applying this knowledge to new situations. Students frequently collaborate with their classmates—they talk about math, listen to each other’s ideas, justify their thinking, and critique the reasoning of others. They gain experience communicating their ideas both verbally and in writing, developing skills that will serve them well throughout their lives. This kind of instruction may look different from what you experienced in your own math education. Current research says that students need to be able to think flexibly in order to use mathematical skills in their lives (and also on the types of tests they will encounter throughout their schooling). Flexible thinking relies on understanding concepts and making connections between them. Over time, students gain the skills and the confidence to independently solve problems that they've never seen before. What supports are in the materials to help my student succeed? • Warm-ups: Each lesson begins with a warm-up routine that is an invitation to the mathematics of the lesson. The same routines are used throughout the entire curriculum, and students become very familiar with the structure of the routines. During warm-up routines, all students are encouraged to share their developing ideas, ask questions, and respond to the reasoning of others. • Activity and Lesson Syntheses: Each activity and lesson includes a synthesis that provides an opportunity for students to discuss key mathematical ideas of the activity/lesson and incorporate their new insights into their big-picture understanding. • Section Summaries: Each section is followed by a section summary that describes the key mathematical ideas discussed in the section. The summaries include visuals and worked examples of problems when relevant. Students can use the section summaries to review the topics covered in the section. • Representations: There are a limited number of representations thoughtfully introduced in the curriculum and students are encouraged to use the representations that make sense to them. These representations help students develop an understanding of the content as well as solve problems. • Family Support Materials: Included in each unit, is an overview of the unit's math content and questions to ask or problems to work on with your student. What can my student do to be successful in this course? Learning how to learn in a problem-based classroom can be a challenge for students at first. Over time, students gain independence as learners when they share their rough drafts of ideas, compare their existing ideas to new things they are learning, and revise their thinking. Many students and families tell us that while this was challenging at first, becoming more active learners in math helped them build skills to take responsibility for their learning in other settings. Here are some ideas for encouraging your student: • If you’re not sure how to get started on a problem, that’s okay! What can you try? Could you draw a picture or diagram? Could you make a guess? Could you describe an answer that’s definitely • If you’re feeling stuck, write down what you notice and what you wonder, or a question you have, and then share that when it’s time to work with others or discuss. • Your job when working on problems in this class is to come up with ideas and share them. You don’t have to be right or confident at first, but sharing your thinking will help everyone learn. If that feels hard or scary, it’s okay to say, “This is just an idea . . .” or “I’m not really sure but I think . . .” • Whether you’re feeling stuck or feeling confident with the material, listen to your classmates and ask them about their ideas. One way that learning happens is by comparing your ideas to other people’s ideas. We are excited to be able to support your student in their journey toward knowing, using, and enjoying mathematics.	{"url":"https://im.kendallhunt.com/k5/families/grade-2/information-for-families.html","timestamp":"2024-11-12T14:57:14Z","content_type":"text/html","content_length":"76167","record_id":"<urn:uuid:0cd96db8-d543-40f2-8c24-0880162e20a4>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00645.warc.gz"}
Talk:Density of a set From Encyclopedia of Mathematics "Theorem 4 Let $\mu$ be a locally finite Radon measure on $\mathbb R^n$. If the $n$-dimensional density $\theta^n (\mu, x)$ exists for $\lambda$-a.e. $x$, then the measure $\mu$ is given by the formula (1) where $f = \theta^n (\mu, \cdot)$." — Really? But what happens if $\mu$ is an atom at a point? --Boris Tsirelson 19:21, 3 August 2012 (CEST) Indeed, the correct assumption is that the density must exist for $\mu$-a.e. $x$. Thanks for pointing it out. The same error appears in Theorem 5.Camillo 08:24, 4 August 2012 (CEST) How to Cite This Entry: Density of a set. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Density_of_a_set&oldid=27346	{"url":"https://encyclopediaofmath.org/index.php?title=Talk:Density_of_a_set&oldid=27346","timestamp":"2024-11-13T02:52:06Z","content_type":"text/html","content_length":"14286","record_id":"<urn:uuid:82861f54-7a5b-4977-a16b-582a16d63c36>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00338.warc.gz"}
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August 2017 Monthly Archives: August 2017 Here’s a link to my final project, creating three artifacts to teach Periodicity in Chemistry: Login: etecstudente0101 Password: etecstudent Many concepts are difficult for students to understand without some form of visualization to aid the description. This week I have integrated T-GEM with a water conductivity PhET Simulation to create a lesson activity which can address misconceptions. A regular misconception that students have is that pure water conducts electricity. By the end of this lesson students should be able to communicate what makes water conductive. Step 1: Introduction -Students form groups of 2-3, each group will have a computer, paper, and writing utensil -Review as a class: 1 – What is electricity? What is flowing? 2 – How do we make electricity? 3 – What is concentration (with respect to dissolved solids)? 4 – How could we make things more or less concentrated? Step 2: Generate Have students as a group generate ideas around the following key questions: 1 – Is water conductive? 2 – How does electricity move in water? 3 – Would dissolving solids in the water change its properties? Each group will write their predictions down on a piece of paper. Step 3: Evaluate Have groups use PhET Simulation “Sugar and Solutions” to investigate water conductivity. Ask students to test if their predictions are correct. Step 4: Modify Ask students to create a situation where their new knowledge of water conductivity could be useful. (The use of Makey-Makeys can be used so students can make an apparatus which physically uses water conductivity to control a computer) Step 5: Reflect Have students revise their original ideas and together formulate the main points of water conductivity. Friedrichsen, P. M., & Pallant, A. (2007). French fries, dialysis tubing & computer models: Teaching diffusion & osmosis through inquiry & modeling. The American Biology Teacher, 69(2), 22-27. Khan, S. (2010). New Pedagogies on Teaching Science with Computer Simulations. J Sci Educ Technol, 20(3), 215–232. http://doi.org/10.1007/s10956-010-9247-2 Srinivasan, S., Perez, L. C., Palmer,R., Brooks,D., Wilson,K., & Fowler. D. (2006). Reality versus simulation. Journal of Science Education and Technology, 15 (2), 137-141. http:// Coding and computational thinking is rapidly making it’s way into mathematics education. The ability to break a problem down into pieces, and use variables and computations to complete an action is an ideal way to teach students how to unite scientific thinking, mathematical practice, and digital creation. CodeCombat is a great tool which I have been using with middle-school classes to teach computational thinking and coding. The program is set up like a video game where a student must use one of several programming languages to instruct their hero how to navigate mazes and defeat enemies. The programming is text-based and is a great way to introduce students who have up till now only seen visual-based programming. The free version is playable for between 1 and 2 hours with a class. There are many license options available if you wish to proceed further with students. Traditionally, educational professionals believed that knowledge in math or science must be constructed by first learning the simple mechanical and fact-based aspects before being able to integrate these fundamentals into real-world problems. While it may make sense to construct a building by first focusing on fundamental pieces such as a foundation and framing, this method may be too simplified to apply to students who are embodied in a world with a plethora of problems to be solved, some of which they may never have experienced before. Carraher et al (1985) looked at math skills in the practical world and discovered that youth with very little formal education developed successful strategies to deal with real life mathematical problems in a market. The youth could successfully solve 95% of problems in the informal market setting while only being able to successfully solve 73% of the problems given to them in a formal test setting. “It seems quite possible that children might have difficulty with routines learned at school and yet at the same time be able to solve the mathematical problems for which these routines were devised in other more effective ways” (Carraher et al, 1985). Thus, as educators it can be useful to use real-life problems in the world to help students gain more applicable and effective knowledge. Two ways in which students can use real-life experiences to guide their learning is through networked communities such as GLOBE and Exploratorium. In the GLOBE project, scientists are linked with teachers and students to gather data from around the world (Butler & MacGregor, 2003). Students are taught data collection techniques and can visually display their and other’s collected data to analyse and interpret. An example of such is looking at the carbon cycle in different biomes; students collect topsoil data from their region and compare it with data from other students in different parts of the world. With this program, students can directly participate in global knowledge generation on a global scale. Further, Exploratorium presents a virtual museum which allows students to interact and learn with interactive tools, hands-on activities, apps, blogs, and videos to learn about science. “Many innovative educational applications, tools, and experiences are being specifically designed to capture the interests and attention of learners to support everyday learning” (Hsi, 2008). Such tools allow students to generate knowledge in and out of the classroom as the line between formal and informal education becomes blurred. The goal from informal learning is to create a passion for life-long learning in students. If students can self-motivate, knowledge construction can become limitless. Butler, D.M., & MacGregor, I.D. (2003). GLOBE: Science and education. Journal of Geoscience Education, 51(1), 9-20. Carraher, T. N., Carraher, D. W., & Schliemann, A. D. (1985). Mathematics in the streets and in schools. British journal of developmental psychology, 3(1), 21-29. Hsi, S. (2008). Information technologies for informal learning in museums and out-of-school settings. International handbook of information technology in primary and secondary education, 20(9), Embodied learning revolves around the notion that it is not just the brain participating in learning activities, rather the whole body participates, interacts, and manufactures new concepts (Winn, 2003). Winn works to defeat the notion that the mind and body are separate entities, rather the interactions we have with our environment is key to the learning process. Removal of environmental learning experiences works to limit student’s ability to adapt with the environment and therefore form a more complete learning experience. Winn (2003) encourages educators to create a “framework that integrates three concepts, embodiment, embeddedness and adaptation.” This embodied learning can be extended to artificial environments where students actively engage and become “coupled” (Winn, 2003). A person in a coupled learning environment actively engages and interacts through problem solving and discovery learning. The use of mobile technology can allow students to become coupled learners with artificial environments. “They also have the potential to establish participatory narratives that can aid learners in developing a contextual understanding of what are all too often presented as decontextualized scientific facts, concepts, or principles” (Barab & Dede, 2007). Mobile technologies can allow students to become fully immersed in virtual scenarios where they must participate in scientific processes, or partially immersed where they use mobile technology to aid a real-life investigation. Barab & Dede (2007) highlight the use of game-based technologies to target academic content learning in more embodied and integrated formats. Zurina & Williams (2011) helped my understanding of embodied learning in the classroom by analyzing how children may gesticulate to solve problems they are working on. Gesticulation is required by these children as they work integrated with their bodies and environment in the learning process. Without realizing it, as I explored this topic I realise that I model embodied behaviour to my students through instruction. For example, when analysing linear and polynomial graphs, I teach students to use the left arm rule to determine if the leading coefficient is positive or negative (one holds their left arm up to recreate slope of the graph; if it is easy the slope is positive, if it is difficult to contort your arm in such a way the slope must be negative). Are there other embodied actions which help teachers reach their students? Are you performing gestures which aid in learning without knowing it? Barab, S., & Dede, C. (2007). Games and immersive participatory simulations for science education: an emerging type of curricula. Journal of Science Education and Technology, 16(1), 1-3. Winn, W. (2003). Learning in artificial environments: Embodiment, embeddedness, and dynamic adaptation. Technology, Instruction, Cognition and Learning, 1(1), 87-114. Full-text document retrieved on January 17, 2013. Retreived from: http://www.hitl.washington.edu/people/tfurness/courses/inde543/READINGS-03/WINN/winnpaper2.pdf Zurina, H., & Williams, J. (2011). Gesturing for oneself. Educational Studies in Mathematics, 77(2-3), 175-188. TELES present an engaging and relevant way to encourage student learning. In Module B, we took a look at four models of TELEs which help to bridge the gap between real-world science and the textbook-based model many science teachers have become reliant on. These models are T-GEM, Anchored Instruction, SKI/WISE, and LfU. Some of the major similarities between these models are collaborative approaches, solving real-world problems, working with real-world data sets, and scaffolding new observations with student preconceptions. Such approaches strive to break the cycle of textbook and fact-based learning which do little for generating a realistic view of science, motivating students, and developing critical thinking skills. The exploration of these four models has greatly increased my confidence in building an effective science classroom. These topics are a refreshing way to integrate high level cognitive skills into a science classroom and will greatly aid my ability to design and run science classes which can remain interesting, relevant, and applicable to students. While working through this module I still struggle to think of creative ways to include these four models into my mathematics instruction. Apart from data analysis and the use of math in multi-disciplinary problems, I remain lost in how to change my mathematics instruction in similar constructive ways as in science. I find the access to online Math instructional aides to be limited in ability and scope. As the new BC Ministry of Education rolls out to all the grades, learning activities such as the four which we looked at in this module will become valuable assets to fulfill curricular and core competencies. Cognition and Technology Group at Vanderbilt (1992). The Jasper Experiment: An exploration of issues in learning and instructional design. Educational Technology, Research and Development, 40(1), Edelson, D.C. (2001). Learning-for-use: A framework for the design of technology-supported inquiry activities. Journal of Research in Science Teaching,38(3), 355-385. http://ezproxy.library.ubc.ca/ login?url=http://dx.doi.org/ 10.1002/1098-2736(200103)38:3<355::aid-tea1010>3.0.CO;2-M Khan, S. (2007). Model-based inquiries in chemistry. Science Education, 91(6), 877-905. Williams, M. & Linn, M. C.(2002) WISE Inquiry in Fifth Grade Biology. Research in Science Education, 32(4), 415-436. I utilize a simulation in my senior business classes that I think could also be useful in a mathematics classroom in the lower grades. Junior Achievement Titan is a business simulation where your company creates and markets a fictional product. You are given a lot of control over how many parameters you want your students to be able to control depending on their level of understanding. The resource is free to use and a representative from Junior Achievement will spend about an hour with you on the phone walking you through how the simulation works and strategies to implement in your classroom. I hope this helps! Assignment #2, The Design of TELEs encouraged us to shared our videos. I produced a guide to PBL that finally gives a more research grounded basis for our STEM program. Very satisfying! This video was designed to highly one aspect of the guide: The Role of Technology in STEM. I hope it provides concrete examples of how we can use the affordances of technology to transform how we I’ve enjoyed learning from you all. It’s been awesome to be in an environment for sharing so many good ideas. Enjoy the rest of the summer! Model-based learning is a theory that allows students to learn from building, critiquing and changing our ways of thinking on how the world works (Khan 2007). One of the big ideas from BC’s grade 6 math curricula is: Properties of objects and shapes can be described, measured, and compared using volume, area, perimeter, and angles. I’ve decided to use a T-GEM model to support student inquiry while using an information visualization technology from a website called Illuminations. In particular, to examine the challenging concept of how the angles of shapes add up when manipulated. Triona and Klahr argued that computer simulations can be as productive a learning tool as hands-on equipment, given the same curriculum and educational setting (As cited in Finkelstein et al., 2005). T-GEM Model for understanding how angles in shapes work: Introduction: Students will be introduced to pictures of different environments: downtown of cities, houses, construction, buildings, cars et. What shapes do you see? Patterns? Can you determine the measurement of each angle? The sum of all angles within these shapes? Generate: Using the following link below, students will choose a polygon and reshape it by dragging the vertices to different locations. The students will see that when the figure changes shape, the angle measures will automatically update. Are there any patterns? What relationships do you see with triangles, quadrilaterals, pentagons, hexagons? Does the sum of all angles change or remain the same when they are manipulated? They will record their observations down. Evaluate: Find a formula that relates the number of sides (n) to the sum of the interior angle measures. Why are we learning this? Can you think of any real-world examples when you would need to know the different angles within shapes? On the applet, play the animated clip in the lower right corner. Is there a different result for different shapes? Compare your results with a different group. Did you find the same formula? Modify: Can you now look at shapes and determine the sum of all angles? Are there instances where you are not sure? Do shapes must be a certain size? Reflection: Students will revisit the simulations again and create their own shapes. They will test their peers by asking them questions such as, “What will happen to sum of all angles when I move vertices upward?” They will reflect on their findings by posting a response to a collaborative tool called Padlet. Finkelstein, N. D., Adams, W. K., Keller, C. J., Kohl, P. B., Perkins, K. K., Podolefsky, N. S., … & LeMaster, R. (2005). When learning about the real world is better done virtually: A study of substituting computer simulations for laboratory equipment. Physical Review Special Topics-Physics Education Research, 1(1), 010103. Khan, S. (2007). Model‐based inquiries in chemistry. Science Education, 91(6), 877-905.	{"url":"https://blogs.ubc.ca/stem2017/2017/08/","timestamp":"2024-11-08T12:18:01Z","content_type":"text/html","content_length":"66851","record_id":"<urn:uuid:0ea823fb-890c-4f36-ad44-8efae4c72a8d>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00558.warc.gz"}
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If you are planning for Industry Certifications, then this course brings considerable preparatory material helping you achieve your objective This is not another Solar PV course but in fact this course brings abundance of material which shall take your knowledge and understanding to a level which shall be sufficient to understand all professional theory behind solar photovoltaics and associated equipment like charge controllers, inverters and batteries, and will cover various topics related to solar energy, PV cells and arrays including sun, solar system, solar radiation, solar energy, PV modules, PV arrays/systems, charge controllers, inverters and batteries, which are necessary to grasp the professional understanding for designing and installing of solar PV systems. Moreover, this course brings to you detailed System Planning and Designing comprising of Customer Requirements, Economics of Solar PV Systems, Payback Period/NPV/IRR, Understanding Project Scope and System Specifications, RFP, User Load Analysis, Critical Load Analysis, Site Survey and Planning, Shading Analysis, Calculating inter-row distances, PV Array Orientation & Optimization, Mechanical Design, Mounting Structures Types and Materials and detailed Sizing of On-Grid/Off-Grid System(s) in Microsoft-Excel Sheets. Additionally, this course shall help the learner to go through some common PV-related questions and learn how to calculate and solve the problems. All theory is explained with calculations and reasoning. These questions will certainly enhance the confidence of the learner. Efforts have been made that all such topics and questions are covered and comprehensive learning takes place in a couple of hours through theory and calculations. Course Content • INTRODUCTION TO SOLAR PV SYSTEMS □ History of Solar Energy □ PV Applications and Industry □ PV Industry Stakeholders □ Solar Energy Technologies • The Sun • Solar Radiation and Solar Light • Extraterrestrial Solar Radiation and Solar Constant • Terrestrial Solar Radiation and Air Mass • Peak Sun Value and Hours • Solar Irradiance and Irradiation • Solar Radiation Data and Maps • Measuring Solar Radiation and Sunshine • Earth Orbit and Rotation • Solar Time and Equation of Time Graph • Sun Path Charts and Solar Window • Photovoltaic Module Azimuth and Tilt Angles • Solar and Magnetic Declination • Power and Energy Basic Electrical Equations • Atom, Semiconductors and Band-Gap • Silicon Element Structure and Doping • Photovoltaic Effect and Solar Cell Working Principle • Structure, Materials and Fabrication of Solar Cell • Current-Voltage-Power Curves of Solar PV Modules • Temperature Coefficient and Calculating Voltages • Efficiency and Fill Factor • Module Series and Parallel Connections • Bypass Diodes • PV Labels • Test Conditions • PV International Standards • Charge Controller Functions • Types of Charge Contoller • Voltage Settings of Charge Controller • Selection Parameters • Installation of a Charge Controller • AC, DC and Quality of Power -1 • AC, DC and Quality of Power -2 • Switching, Power Conditioning and Control -1 • Switching, Power Conditioning and Control -2 • Types of Solar Inverters -1 • Types of Solar Inverters -2 • Earthing & Grounding • Selection Parameters of Inverter -1 • Selection Parameters of Inverter -2 • Selection Parameters of Inverter -3 • Selection Parameters of Inverter -4 • Characteristics and Parameters of a Battery • Battery of Batteries -I • Battery of Batteries -II • Calculating Battery Bank Size • Selection Parameters of a Battery • Operation and Installation of a Battery • Battery Standards • Calculate Voltage Drop in a Cable • Calculate System Power based upon Solar Irradiance, Module rated Power and Efficiency • Finding No. of Strings and PV Modules Combination given System Data • Calculating Voltage and Current Output at Standard Test Conditions given System Data • Calculate AC Output of a Solar PV System given System Losses and Inverter Efficiency • Finding least amount of PV Modules given System Data • Understanding purpose of Bypass Diodes • Find best solution for having multiple orientations of PV Modules • Calculate MPPT voltage of the system at Standard Test Conditions • Find best Voltage for charging a 12V Lead Acid battery • Find characteristics of a Wire that lead to more Voltage drop • Calculate PV Size given Area and PV Module Efficiency • Calculate minimum number of PV Modules to be installed in series given System data • What kind of ground faults do most inverters detect • Calculate total System size given PV Modules data and number of Strings • Calculate Voltage of PV Cells given other combination • Calculate No of Batteries required given Load Watt-Hours and Volts • Calculate No. of Batteries required given Load data • Which one affect the production of Current most in PV Modules • Calculate Energy Units consumed by a Load • Understanding Air Mass • Calculate power produced by the PV Array given PV Module area and efficiency • Calculate size of Breaker at output of an Inverter • Understand factors required to calculate Inter-row spacing of PV Modules in an Array • Troubleshooting a PV System consisting of one source circuit not producing power • Find optimimed PV Tilt Angle during the Summer season for an Off-Grid PV System • Calculate Maximum No of Modules to be connected in Series given System data • Calculate the Energy Units consumed per day by a residential load • Find worst Orientation Design for the installation of the PV system • What can be the maximum array size given Charge Controller data • Finding Tilt Angle of the Array given Location and Latitude • Calculate Energy produces by a PV System in one year given System data • Calculate Maximum No. of PV Modules to be installed in series given System data • Troubleshooting an Off-Grid PV system which often Shuts down during Winter • Calculate annual System Output given System data • Introduction to System Design Stages • Customer Requirements and Challenges • System Selection and Economics of Solar PV System • I - Selection of Type of System • II - Economics of Solar PV System • III - PayBack NPV IRR • IV - Net Metering State Incentives • V - Payback Period-NPV-IRR Examples • VI - Levelized Cost of Energy • Project Scope of Work and Agreement • Estimating User Load Requirements • System Specifications • Request for Proposal or Tender Document • Site Survey and Data Collection • Types of Shadows • Shading Analysis • I - Introduction to Shading Analysis • II - Calculate Inter-Row Distances and area of PV Array • III - Inter-Row Distance Graph, Window of Obstruction, Energy Yield • PV Array Location Considerations • PV Array Orientation and Energy Yield Optimization • Optimizing String Connections • Existing Electrical and Interconnection Equipment • Factors for PV Array Mounting Design • PV Array Mounting Locations • Mounting Structure Material and Types • Roofing Structure and Basics • Mechanical Design and Techniques • Applicable Codes for Installation • Introduction to Solar PV System Sizing • PV System Configuration Concepts and Inverter Selection • Load and Critical Design Analysis • Equipment Key Design Concepts and Parameters • System Losses • System Sizing Calculations • I - On-Grid System Sizing Part-1 • II - On-Grid System Sizing Part-2 • III - Off-Grid System Sizing • IV - Finding-PSH-Temp-Data • V - PV System Sizing Excel Sheets UNDERSTANDING CORE CONCEPTS OF SOLAR PV SYSTEM AS PER NEC 690 • Introduction & Definitions: NEC 690.1 & 690.2 • Maximum Voltage and Calculations: NEC 690.7 • Maximum Current and Circuit Sizing: NEC 690.8 • Rapid Shutdown System: NEC 690.12 • Over-Current Protection (OCPD): NEC 690.9 WHO IS THIS COURSE FOR: • Beginners, engineers, technicians or others who are interested to learn and understand solar PV science and technology on sound and professional footings • Engineers, technicians or professionals who are already in this field, and quickly want to refresh their knowledge on sound basis • Those who may decide in future to go for industry certifications Course outline 14 modules 18 - 24 hours to complete 17:31 hours of video lectures Welcome • 1 assignments Orientation Materials This is a self-study on demand course. This course is self-paced, so you don’t need to be logged in at any specific time. You can get started immediately after you enroll and the course materials will remain in your account with minimum guaranteed access for 12 months (1 year) after enrollment. • Understanding Fundamentals of a Solar Photovoltaic System in 60 minutes (01:12:11 hours) Module 1 • 12 assignments 1) Photovoltaics 2) History of Solar Energy 3) PV Applications 4) PV Industry Stakeholders 5) Solar Energy Technologies • Photovoltaics (03:54 minutes) • Advantages & Disadvantages of Solar Photovoltaics • History of Solar Energy (03:04 minutes) • History and Timeline of Solar Cells (.pdf) • PV Applications and Industry (03:35 minutes) • Trends in the PV Industry: Report by IEA 2020 (.pdf) • Solar Industry Update 2020 by NREL (.pdf) • Future of Solar Photovoltaics (.pdf) • Photovoltaic Industry Stakeholders (07:54 minutes) • Solar Energy Industry Association • Top Solar Associations • Solar Technologies (Thermal Energy Collectors) (04:47 minutes) Module 2 • 25 assignments • The Sun □ Solar Radiation and Solar Light □ Extraterrestrial Solar Radiation and Solar Constant □ Terrestrial Solar Radiation and Air Mass □ Peak Sun Value and Hours □ Solar Irradiance and Irradiation □ Solar Radiation Data and Maps □ Measuring Solar Radiation and Sunshine □ Earth Orbit and Rotation □ Solar Time and Equation of Time Graph □ Sun Path Charts and Solar Window □ Photovoltaic Module Azimuth and Tilt Angles □ Solar and Magnetic Declination □ Power and Energy Basic Electrical Equations • The Sun (03:16 minutes) • Solar Radiation and Solar Light (02:11 minutes) • Extraterrestrial Solar Radiation and Solar Constant (01:57 minutes) • Terrestrial Solar Radiation and Air Mass (04:26 minutes) • Introduction to Solar Radiation • Peak Sun Value and Hours (03:06 minutes) • Solar Irradiance and Irradiation (01:20 minutes) • Solar Radiation Data and Maps (01:47 minutes) • Global Solar Atlas • National Solar Radiation Database (NSRDB) by NREL • Measuring Solar Radiation (07:05 minutes) • Solar Radiation Measurements (.pdf) • Review of Devices used to Measure Solar Radiation (.pdf) • Best Practics Handbook for use of Solar Resource Data for Solar Energy Applications (.pdf) • Earth Orbit and Rotation (03:47 minutes) • Terrestrial Coordinates (.pdf) • SPACEPEDIA: A Resource of Solar System • Solar Time and Equation of Time (04:10 minutes) • Finding Sunrise/Sunset/Noon and Solar Position - Online Tool • Sun Path and Charts (02:24 minutes) • Sun Path Chart - Online Tool • Photovoltaic Module Azimuth and Tilt Angles (04:11 minutes) • Solar Magnetic Declination and Tilt Angles (02:29 minutes) • Tilt & Azimuth Angle: What Angle Should I Tilt My Solar Panels? • Power Energy Equations (03:48 minutes) Module 3 • 14 assignments • Atom, Semiconductors and Band-Gap □ Silicon Element Structure and Doping □ Photovoltaic Effect and Solar Cell Working Principle □ Structure, Materials and Fabrication of Solar Cell □ Current-Voltage-Power Curves of Solar PV Modules □ Temperature Coefficient and Calculating Voltages □ Efficiency and Fill Factor □ Module Series and Parallel Connections □ Bypass Diodes □ PV Labels □ Test Conditions □ PV International Standards • Atom, Semiconductors and Band-Gap (08:03 minutes) • Silicon Element Structure and Doping (04:25 minutes) • Photovoltaic Effect and Solar Cell Working Principle (06:15 minutes) • Structure Materials and Fabrication (06:29 minutes) • IV-Curves of Solar Module (09:01 minutes) • Temperature Coefficient and Calculating Voltages (17:08 minutes) • Maximum Power Point Tracking Algorithms (.pdf) • Results of I-V Curves and Visual Inspection of PV Modules Deployed at TEP Solar Test Yard (.pdf) • Efficiency and Fill Factor (06:16 minutes) • Module Series and Parallel Connections (06:56 minutes) • Bypass Diodes (03:30 minutes) • PV Labels (01:39 minutes) • Standard Test Conditions (02:52 minutes) • PV International Standards (05:23 minutes) Module 4 • 9 assignments • Charge Controller Functions □ Types of Charge Contoller □ Voltage Settings of Charge Controller □ Selection Parameters □ Installation of a Charge Controller • Introduction to Charge Controllers (02:16 minutes) • Charge Controller Functions (08:32 minutes) • Types of Charge Contollers (08:03 minutes) • Voltage Settings of a Charge Controller (05:38 minutes) • Selection Parameters of a Charge Controller (05:05 minutes) • How to select a solar charge controller • Choosing the Right Solar Charge Controller/Regulator • Solar Charge Controller Sizing and How to Choose One • Installation of a Charge Controller (03:32 minutes) Module 5 • 15 assignments • AC, DC and Quality of Power -1 □ AC, DC and Quality of Power -2 □ Switching, Power Conditioning and Control -1 □ Switching, Power Conditioning and Control -2 □ Types of Solar Inverters -1 □ Types of Solar Inverters -2 □ Earthing & Grounding □ Selection Parameters of Inverter -1 □ Selection Parameters of Inverter -2 □ Selection Parameters of Inverter -3 □ Selection Parameters of Inverter -4 • Introduction to Inverters (05:06 minutes) • AC, DC and Quality of Power - I (09:04 minutes) • AC, DC and Quality of Power - II (09:02 minutes) • Switching, Power Conditioning and Control - I (06:54 minutes) • Switching, Power Conditioning and Control - II (09:34 minutes) • Types of Solar Inverters - I (08:10 minutes) • Types of Solar Inverters - II (07:15 minutes) • Earthing & Grounding (03:35 minutes) • Guidelines for Designing Grounding Systems for Solar PV Installations • Selection Parameters of Inverter - I (08:17 minutes) • Selection Parameters of Inverter - II (07:35 minutes) • Selection Parameters of Inverter - III (08:08 minutes) • Selection Parameters of Inverter - IV (08:55 minutes) • Choosing the Right Solar Inverter • Solar Inverter Comparison Chart Module 6 • 8 assignments • Characteristics and Parameters of a Battery □ Battery of Batteries -I □ Battery of Batteries -II □ Calculating Battery Bank Size □ Selection Parameters of a Battery □ Operation and Installation of a Battery □ Battery Standards • Introduction to Batteries (19:15 minutes) • Characteristics and Parameters of a Battery (20:53 minutes) • Types of Batteries - I (12:13 minutes) • Types of Batteries - II (13:53 minutes) • Calculating Battery Bank Size (19:51 minutes) • Selection Parameters of a Battery (03:58 minutes) • Operation and Installation of a Battery (08:10 minutes) • Battery Standards (05:11 minutes) Module 7 • 35 assignments APPLIED TECHNICAL CALCULATION QUESTIONS: All Theory and Solutions Explained • Calculate Voltage Drop in a Cable □ Calculate System Power based upon Solar Irradiance, Module rated Power and Efficiency □ Finding No. of Strings and PV Modules Combination given System Data □ Calculating Voltage and Current Output at Standard Test Conditions given System Data □ Calculate AC Output of a Solar PV System given System Losses and Inverter Efficiency □ Finding least amount of PV Modules given System Data □ Understanding purpose of Bypass Diodes □ Find best solution for having multiple orientations of PV Modules □ Calculate MPPT voltage of the system at Standard Test Conditions □ Find best Voltage for charging a 12V Lead Acid battery □ Find characteristics of a Wire that lead to more Voltage drop □ Calculate PV Size given Area and PV Module Efficiency □ Calculate minimum number of PV Modules to be installed in series given System data □ What kind of ground faults do most inverters detect □ Calculate total System size given PV Modules data and number of Strings □ Calculate Voltage of PV Cells given other combination □ Calculate No of Batteries required given Load Watt-Hours and Volts □ Calculate No. of Batteries required given Load data □ Which one affect the production of Current most in PV Modules □ Calculate Energy Units consumed by a Load □ Understanding Air Mass □ Calculate power produced by the PV Array given PV Module area and efficiency □ Calculate size of Breaker at output of an Inverter □ Understand factors required to calculate Inter-row spacing of PV Modules in an Array □ Troubleshooting a PV System consisting of one source circuit not producing power □ Find optimimed PV Tilt Angle during the Summer season for an Off-Grid PV System □ Calculate Maximum No of Modules to be connected in Series given System data □ Calculate the Energy Units consumed per day by a residential load □ Find worst Orientation Design for the installation of the PV system □ What can be the maximum array size given Charge Controller data □ Finding Tilt Angle of the Array given Location and Latitude □ Calculate Energy produces by a PV System in one year given System data □ Calculate Maximum No. of PV Modules to be installed in series given System data □ Troubleshooting an Off-Grid PV system which often Shuts down during Winter □ Calculate annual System Output given System data • Calculate Voltage Drop in a Cable (03:10 minutes) • Calculate System Power based upon Solar Irradiance, Module rated Power and Efficiency (03:40 minutes) • Finding No. of Strings and PV Modules Combination given System Data (04:37 minutes) • Calculating Voltage and Current Output at Standard Test Conditions given System Data (08:40 minutes) • Calculate AC Output of a Solar PV System given System Losses and Inverter Efficiency (11:30 minutes) • Finding least amount of PV Modules given System Data (06:37 minutes) • Purpose of Bypass Diodes (07:34 minutes) • Best solution for having multiple orientations of PV Modules (09:52 minutes) • Calculate MPPT voltage of the system at Standard Test Conditions (04:22 minutes) • Best Voltage for charging a 12V Lead Acid battery (06:39 minutes) • Characteristics of a Wire that lead to more Voltage drop (02:35 minutes) • Calculate PV Size given Area and PV Module Efficiency (03:31 minutes) • Calculate minimum number of PV Modules to be installed in series given System data (08:22 minutes) • What kind of ground faults do most inverters detect (06:47 minutes) • Calculate total System size given PV Modules data and number of Strings (02:23 minutes) • Calculate Voltage of PV Cells given other combination (03:15 minutes) • Calculate No of Batteries required given Load Watt-Hours and Volts (06:09 minutes) • Calculate No. of Batteries required given Load data (04:48 minutes) • Which one affect the production of Current most in PV Modules (06:12 minutes) • Calculate Energy Units consumed by a Load (03:59 minutes) • What is Air Mass ? (05:54 minutes) • Calculate power produced by the PV Array given PV Module area and efficiency (05:54 minutes) • Calculate size of Breaker at output of an Inverter (04:11 minutes) • Factors required to calculate Inter-row spacing of PV Modules in an Array (07:23 minutes) • Troubleshooting a PV System consisting of one source circuit not producing power (08:05 minutes) • Optimum PV Tilt Angle during the Summer season for an Off-Grid PV System (03:59 minutes) • Calculate Maximum No of Modules to be connected in Series given System data (21:25 minutes) • Calculate the Energy Units consumed per day by a residental load (09:04 minutes) • Worst Orientation Design for the installation of the PV system (19:01 minutes) • What can be the maximum array size given Charge Controller data (03:03 minutes) • Finding Tilt Angle of the Array given Location and Latitude (06:39 minutes) • Calculate Energy produces by a PV System in one year given System data (12:18 minutes) • Calculate Maximum No. of PV Modules to be installed in series given System data (05:16 minutes) • Troubleshooting an Off-Grid PV system which often Shuts down during Winter (05:02 minutes) • Calculate annual System Output given System data (05:30 minutes) Module 8 • 13 assignments • System Selection and Economics of Solar PV System • I - Selection of Type of System • II - Economics of Solar PV System • III - PayBack NPV IRR • IV - Net Metering State Incentives • VI - Payback Period-NPV-IRR Examples • VI - Levelized Cost of Energy • Project Scope of Work and Agreement • Estimating User Load Requirements • System Specifications • Request for Proposal or Tender Document • Introduction to System Design Stages (02:30 minutes) • Customer Requirements and Challenges (02:34 minutes) • Economics of Solar PV System - I (Selection of Type of System) (07:31 minutes) • Economics of Solar PV System - II (What is Economics of Solar PV System) (04:42 minutes) • Economics of Solar PV System - III (Understanding PayBack, NPV, IRR) (25:20 minutes) • Economics of Solar PV System - IV (Net Metering State Incentives) (04:56 minutes) • Economics of Solar PV System - V (Payback Period-NPV-IRR Examples) (23:55 minutes) • Economics of Solar PV System - V (Payback Period-NPV-IRR Excel Sheet) (.xlsx) • Economics of Solar PV System - VI (Levelized Cost of Energy) (06:08 minutes) • Project Scope of Work and Agreement (05:49 minutes) • Estimating User Load Requirements (10:34 minutes) • System Specifications (06:37 minutes) • Request for Proposal or Tender Document (06:50 minutes) Module 9 • 10 assignments • Types of Shadows □ Shading Analysis □ I - Introduction to Shading Analysis □ II - Calculate Inter-Row Distances and area of PV Array □ III - Inter-Row Distance Graph-Window of Obstruction-Energy Yield □ PV Array Location Considerations □ PV Array Orientation and Energy Yield Optimization □ Optimizing String Connection □ Existing Electrical and Interconnection Equipment • Site Survey and Data Collection (12:04 minutes) • Types of Shadows (03:10 minutes) • Shading Analysis - I (Introduction) (12:32 minutes) • Shading Analysis - II (Calculate Inter-Row Distances and area of PV Array) (15:20 minutes) • Shading Analysis - III (Inter-Row Distance Graph, Window of Obstruction, Energy Yield) (07:18 minutes) • PV Array Location Considerations (04:10 minutes) • PV Array Orientation and Energy Yield Optimization (09:04 minutes) • Optimizing String Connections (10:10 minutes) • Existing Electrical and Interconnection Equipment (03:46 minutes) • PV System Sizing Excel Sheets - Attached (.xlsx) Module 10 • 1 assignments • Factors for PV Array Mounting Design □ PV Array Mounting Locations □ Mounting Structure Material and Types □ Roofing Structure and Basics □ Mechanical Design and Techniques □ Applicable Codes for Installation • PV Array Mounting Components, Considerations and Mechanical Design (.pdf) Module 11 • 11 assignments • PV System Configuration Concepts and Inverter Selection □ Load and Critical Design Analysis □ Equipment Key Design Concepts and Parameters □ System Losses □ System Sizing Calculations □ I - On-Grid System Sizing Part-1 □ II - On-Grid System Sizing Part-2 □ III - Off-Grid System Sizing □ IV - Finding-PSH-Temp-Data □ V - PV System Sizing Excel Sheets • Introduction to Solar PV System Sizing (02:42 minutes) • PV System Configuration Concepts and Inverter Selection (13:01 minutes) • Load and Critical Design Analysis (18:39 minutes) • Equipment Key Design Concepts and Parameters (08:26 minutes) • System Losses (07:16 minutes) • System Sizing Calculations - I (On-Grid System Sizing Part-1) (29:04 minutes) • System Sizing Calculations - II (On-Grid System Sizing Part-2) (28:34 minutes) • System Sizing Calculations - III (Off-Grid System Sizing) (22:37 minutes) • System Sizing Calculations - IV (Resources for PSH and Temperature Data ) (05:49 minutes) • Intro to attached PV System Sizing Excel Sheets (04:24 minutes) • PV System Sizing Excel Sheets (.xlsx) Module 12 • 7 assignments UNDERSTANDING CORE CONCEPTS OF SOLAR PV SYSTEM AS PER NEC 690 This Module shall take you to learn basic and core concepts of Voltage, Currents, Rapid Shutdown System and OCPD • Introduction & Definitions: NEC 690.1 & 690.2 □ Maximum Voltage and Calculations: NEC 690.7 □ Maximum Current and Circuit Sizing: NEC 690.8 □ Rapid Shutdown System: NEC 690.12 □ Over-Current Protection (OCPD): NEC 690.9 • Introduction and Definitions: NEC 690.1 & 690.2 (10:15 minutes) • Maximum Voltage and Calculation Methods: NEC 690.7 (06:16 minutes) • Maximum Current and Circuit Sizing: NEC 690.8 (10:53 minutes) • Circuit Sizing Two Methods: NEC 690.8 (.pdf) • Conductor Sizing NEC Table 310.16 (.png) • Rapid Shuwdown System: NEC 690.12 (04:34 minutes) • Over-Current Protection (OCPD) (08:37 minutes) Conclusion • 4 assignments Feedback and Additional Resources This is our last module but you still have access to the all of course materials for 12 months (1 year), so keep working and you'll be able to complete the course at your own pace. After your year of access expires you can optionally extend access with a HeatSpring Membership. Enjoy the course and keep in touch! • 1 Year of Access to Course Materials • Feedback: 2-minute Exit Survey • Consider Joining as a HeatSpring Member • Certificate of Completion: Request a Certificate Director, Innovasys Engineering Asif Khokher is seasoned Professional Engineer and certified NABCEP PVIP having 24+ Years of experience as Consultant/Owner's Engineer and Project Manager in various domains like Front End Engineering & Design, Tender Design, Detailed Design & System Integration, Design Review, Approvals, Inspection/ Commissioning in the field of Instrumentation & Control Systems, SCADA... How does this course work? You can begin this online course instantly upon enrollment. This 7 module course is delivered entirely online. This is a self-study, self-paced course and you can set your own schedule to complete the materials. You can begin the lecture videos and other course materials as soon as you enroll. After successfully completing the course, you will be able to generate a certificate of completion. How long do I have access to the materials? Students get unlimited access to the course materials as soon as they enroll and for one year (365 days) after enrollment. Rewatch videos and review assignments as many times as you want. View updates the instructor makes to the course as the industry advances. Return to your course anytime with online access from anywhere in the world. After the one year of access expires, access can be extended by joining as a HeatSpring member. A single membership extends access to course materials for all past enrollments. Is there a certificate of completion? Yes, when you complete this course you are eligible for a certificate of completion from HeatSpring. You can download your certificate as soon as you have completed all of the course requirements. Students can easily share their verified certificates on their LinkedIn profiles using our LinkedIn integration. Can I register multiple people?	{"url":"https://www.heatspring.com/courses/learn-essential-theory-planning-and-design-of-solar-pv-system-with-calculations","timestamp":"2024-11-14T18:08:20Z","content_type":"text/html","content_length":"253032","record_id":"<urn:uuid:f8a6921e-df1c-4417-b12d-530a3d195e34>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00034.warc.gz"}
5th May Mathematicians Of The Day 5th May On this day in 1777, Leonhard Euler was the first to use the symbol i for $\sqrt{-1}$ in the paper De Formulis Differentialibus Angularibus maxime irrationalibus quas tamen per logarithmos et arcus circulares integrare licet [Of the most irrational Angular Differential Formulas, which, however, may be integrated by means of logarithms and circular arcs], presented to the St Petersburg Academy:- Quoniam mihi quidem alia adhuc via non patet istud praestandi nisi per imaginaria procedendo, formulam $\sqrt{-1}$ littera i in posterum designabo, ita ut sit $i\:i = -1$ ideoque $\Large\frac 1 i ormalsize= -i$. [Since there is still no other way clear to me of accomplishing this except by an imaginary procedure, I will denote the formula $\sqrt{-1}$ by the letter i in the future, so that $i\:i = -1$ and therefore $\Large\frac 1 i ormalsize= -i$.] See THIS LINK. A (personally designed) stamp of one of today's mathematicians is at THIS LINK. I speak twelve languages: English is the bestest.	{"url":"https://mathshistory.st-andrews.ac.uk/OfTheDay/oftheday-05-05/","timestamp":"2024-11-11T03:44:07Z","content_type":"text/html","content_length":"32270","record_id":"<urn:uuid:12ab24cc-3d13-49c0-99fe-1b0470687ffb>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00572.warc.gz"}
Go to the source code of this file. subroutine dpbcon (UPLO, N, KD, AB, LDAB, ANORM, RCOND, WORK, IWORK, INFO) Function/Subroutine Documentation subroutine dpbcon ( character UPLO, integer N, integer KD, double precision, dimension( ldab, * ) AB, integer LDAB, double precision ANORM, double precision RCOND, double precision, dimension( * ) WORK, integer, dimension( * ) IWORK, integer INFO Download DPBCON + dependencies [TGZ] [ZIP] [TXT] DPBCON estimates the reciprocal of the condition number (in the 1-norm) of a real symmetric positive definite band matrix using the Cholesky factorization A = U*TU or A = LLT computed by DPBTRF. An estimate is obtained for norm(inv(A)), and the reciprocal of the condition number is computed as RCOND = 1 / (ANORM norm(inv(A))). UPLO is CHARACTER1 [in] UPLO = 'U': Upper triangular factor stored in AB; = 'L': Lower triangular factor stored in AB. N is INTEGER [in] N The order of the matrix A. N >= 0. KD is INTEGER [in] KD The number of superdiagonals of the matrix A if UPLO = 'U', or the number of subdiagonals if UPLO = 'L'. KD >= 0. AB is DOUBLE PRECISION array, dimension (LDAB,N) The triangular factor U or L from the Cholesky factorization A = UTU or A = LLT of the band matrix A, stored in the [in] AB first KD+1 rows of the array. The j-th column of U or L is stored in the j-th column of the array AB as follows: if UPLO ='U', AB(kd+1+i-j,j) = U(i,j) for max(1,j-kd)<=i<=j; if UPLO ='L', AB(1+i-j,j) = L(i,j) for j<=i<=min(n,j+kd). LDAB is INTEGER [in] LDAB The leading dimension of the array AB. LDAB >= KD+1. ANORM is DOUBLE PRECISION [in] ANORM The 1-norm (or infinity-norm) of the symmetric band matrix A. RCOND is DOUBLE PRECISION The reciprocal of the condition number of the matrix A, [out] RCOND computed as RCOND = 1/(ANORM AINVNM), where AINVNM is an estimate of the 1-norm of inv(A) computed in this routine. [out] WORK WORK is DOUBLE PRECISION array, dimension (3*N) [out] IWORK IWORK is INTEGER array, dimension (N) INFO is INTEGER [out] INFO = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value Univ. of Tennessee Univ. of California Berkeley Univ. of Colorado Denver NAG Ltd. November 2011 Definition at line 132 of file dpbcon.f.	{"url":"https://netlib.org/lapack/explore-html-3.4.2/d6/da0/dpbcon_8f.html","timestamp":"2024-11-06T04:56:12Z","content_type":"application/xhtml+xml","content_length":"13538","record_id":"<urn:uuid:fa80b154-3db9-4f73-bb8f-a977f5b7cde4>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00579.warc.gz"}
Classifications of internal storage Up to this point, you have learned some of the general functions of the cpu, the physical characteristics of memory, and how data is stored in the internal storage section. Now, we will explain yet another way to classify internal (primary or main) storage. This is by the different kinds of memories used within the cpu: read-only memory, random-access memory, programmable read-only memory, and erasable programmable read-only memory. READ-ONLY MEMORY (ROM) In most computers, it is useful to have often used instructions, such as those used to bootstrap (initial system load) the computer or other specialized programs, permanently stored inside the computer. Memory that enables us to do this without the programs and data being lost (even when the computer is powered down) is called read-only memory. Only the computer manufacturer can provide these programs in ROM and once done, they cannot be changed. Consequently, you cannot put any of your own data or programs in ROM. Many complex functions such as routines to extract square roots, translators for programming languages, and operating systems can be placed in ROM memory. Since these instructions are hard wired (permanent), they can be performed quickly and accurately. Another advantage of ROM is that your computer facility can order programs tailored for its needs and have them permanently installed in ROM by the manufacturer. Such programs are called microprograms or Another kind of memory used inside computers is called random-access memory (RAM) or read/write memory. RAM memory is rather like a blackboard on which you can scribble down notes, read them, and rub them out when you are finished with them. In the computer, RAM is the working memory. Data can be read (retrieved) from or written (stored) into RAM just by giving the computer the address of the location where the data is stored or is to be stored. When the data is no longer needed, you can simply write over it. This allows you to use the storage again for something else. Core, semiconductor, and bubble storage all have random access capabilities. An alternative to ROM is programmable read only memory (PROM) that can be purchased already programmed by the manufacturer or in a blank state. By using a blank PROM, you can enter any program into the memory. However, once the PROM has been written into, it can never be altered or changed. Thus you have the advantage of ROM with the additional flexibility to program the memory to meet a unique need. The main disadvantage of PROM is that if a mistake is made and entered into PROM, it cannot be corrected or erased. Also, a special device is needed to "burn" the program into PROM. The erasable programmable read-only memory (EPROM) was developed to overcome the drawback of PROM. EPROMs can also be purchased blank from the manufacturer and programmed locally at your command/ activity. Again, this requires special equipment. The big difference with EPROM is that it can be erased if and when the need arises. Data and programs can be retrieved over and over again without destroying the contents of the EPROM. They will stay there quite safely until you want to reprogram it by first erasing the EPROM with a burst of ultra-violet light. This is to your advantage, because if a mistake is made while programming the EPROM, it is not considered fatal. The EPROM can be erased and corrected. Also, it allows you the flexibility to change programs to include improvements or modifications in the future. Q.17 In what type of memory are often used instructions and programs permanently stored inside the computer? Q.18 Who provides the programs stored in ROM? Q.19 Can programs in ROM be changed? Q.20 What is another name for random-access memory (RAM)? Q.21 How is data read from or written into RAM? Q.22 In what two states can programmable read-only memory (PROM) be purchased? Q.23 What is the main disadvantage of PROM? Q.24 What does EPROM stand for? Q.25 How is EPROM erased? The last kind of memory we will briefly introduce here is called secondary storage or auxiliary storage. This is memory outside the main body of the computer (cpu) where we store programs and data for future use. When the computer is ready to use these programs and data, they are read into internal storage. Secondary (auxiliary) storage media extends the storage capabilities of the computer system. We need it for two reasons. First, because the computer's internal storage is limited in size, it cannot always hold all the data we need. Second, in secondary storage, data and programs do not disappear when power is turned off. Secondary storage is nonvolatile. This means information is lost only if you, the user, intentionally erase it. The three types of secondary storage we most commonly use are magnetic disk, tape, and drum. The popularity of disk storage devices is largely because of their direct-access capabilities. Most every system (micro, mini, and mainframe) will have disk capability. Magnetic disks resemble phonograph records (round platters), coated with a magnetizable recording material (iron oxide), but their similarities end there. Magnetic disks come in many different sizes and storage capacities. They range from 3 inches to 4 feet in diameter and can store from 2.5 million to 600 million characters (bytes) of data. They can be portable in that they are removable, or they can be permanently mounted in the storage devices called disk drive units or disk drives. They can be made of rigid metal (hard disks) or flexible plastic (floppy disks or diskettes) as shown in figure 2-6. Figure 2-6. - Various types and sizes of magnetic disk storage. Music is stored on a phonograph record in a continuous groove that spirals into the center of the record. But there are no grooves on a magnetic disk. Instead, data is stored on all disks in a number of invisible concentric circles called tracks. Each track has a designated number beginning with track 000 at the outer edge of the disk. The numbering continues sequentially toward the center to track 199, 800, or whatever the highest track number is. No track ever touches another (fig. 2-7). The number of tracks can vary from 35 to 77 on a floppy disk surface and from 200 to over 800 on hard disk surfaces. Figure 2-7. - Location of tracks on the disk's recording surface. Data is written as tiny magnetic bits (or spots) on the disk surface. Eight-bit codes are generally used to represent data. Each code represents a different number, letter, or special character. In chapter 4, you'll learn how the codes are formed. When data is read from the disk, the data on the disk remains unchanged. When data is written on the disk, it replaces any data previously stored on the same area of the disk. Characters are stored on a single track as strings of magnetized bits (0's and 1's) as shown in figure 2-8. The 1 bits indicate magnetized spots or ON bits. The 0 bits represent unmagnetized portions of the track or OFF bits. Although the tracks get smaller as they get closer to the center of the disk platter, each track can hold the same amount of data because the data density is greater on tracks near the center. Figure 2-8. - A string of bits written to disk on a single track. A track can hold one or more records. A record is a set of related data treated as a unit. The records on a track are separated by gaps in which no data is recorded, and each of the records is preceded by a disk address. This address indicates the unique position of the record on the track and is used to directly access the record. Figure 2-9 shows a track on which five records have been recorded. Because of the gaps and addresses, the amount of data we can store on a track is reduced as the number of records per track is increased. Records on disk can be blocked (grouped together). Only one disk address is needed per block, and as a result, fewer gaps occur. We can use the blocking technique to increase the amount of data we can store on one track. Figure 2-9. - Data records as they are written to disk on a single track.	{"url":"https://www.tpub.com/neets/book22/92b.htm","timestamp":"2024-11-12T13:18:40Z","content_type":"text/html","content_length":"23166","record_id":"<urn:uuid:717037c8-e434-43ee-9670-a870ed7958f0>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00796.warc.gz"}
Convert micrometers to yards ( um to yd ) Last Updated: 2024-11-03 01:44:37 , Total Usage: 1123874 Converting micrometers to yards is an exercise in translating a very small unit of measurement in the metric system to a larger, more traditional unit in the imperial system. This type of conversion is crucial in fields like textile manufacturing, microscopy, and material science, where it's often necessary to relate micro-scale measurements to more familiar macro-scale units. Historical or Origin Micrometers (µm): A micrometer, or micron, is a unit of length in the metric system, equal to one-millionth of a meter. It's widely used in scientific and engineering fields for measuring tiny dimensions, such as the thickness of materials or the size of biological cells. Yards (yd): The yard is a unit of length in the imperial system, primarily used in the United States and the United Kingdom. Historically, it was based on the length of a man's belt or girdle, but it has been standardized to exactly 0.9144 meters. Calculation Formula The formula to convert micrometers to yards is: \[ \text{Yards} = \text{Micrometers} \times \text{Conversion Factor} \] The conversion factor from micrometers to yards is approximately $1.09361 \times 10^{-6}$, as there are 1,093.61 yards in a meter, and a micrometer is one-millionth of a meter. Example Calculation For example, if you want to convert 10,000 micrometers to yards, you would use the formula: \[ \text{Yards} = 10,000 \times 1.09361 \times 10^{-6} \approx 0.0109361 \text{ yd} \] Why It's Needed and Use Cases This conversion is important in contexts where micro-scale measurements need to be understood or communicated in terms more familiar to those used to the imperial system. For instance, in textile manufacturing, the diameter of fibers might be measured in micrometers, but fabric lengths are often discussed in yards. Common Questions (FAQ) • Why do we convert between metric and imperial units? Converting between these systems is necessary because different countries and industries use different measurement systems. • How accurate is this conversion? The conversion is very accurate due to the precise definitions of a micrometer and a yard. However, the level of precision needed can depend on the specific • Is this conversion common in everyday life? While not common in everyday activities for most people, this conversion is quite relevant in certain professional and scientific fields. In summary, converting micrometers to yards bridges the gap between the microscopic world measured in metric units and the more macroscopic scales often conceptualized in imperial units. This conversion is crucial in various technical and scientific applications, facilitating a better understanding and communication across different measurement systems.	{"url":"https://calculator.fans/en/tool/um-to-yd-convertor.html","timestamp":"2024-11-06T18:53:16Z","content_type":"text/html","content_length":"12840","record_id":"<urn:uuid:7f20f789-d882-4c73-a2f5-7571489b8e2a>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00244.warc.gz"}
Realization-obstruction exact sequences for Clifford system extensions For every action φ ∈ Hom(G, Aut[k](K)) of a group G on a commutative ring K we introduce two abelian monoids. The monoid Cliff[k](φ) consists of equivalence classes of strongly G-graded algebras of type φ up to G-graded Clifford system extensions of K-central algebras. The monoid C[k](ϕ) consists of equivariance classes of homomorphisms of type φ from G to the Picard groups of K-central algebras (generalized collective characters). Furthermore, for every such φ there is an exact sequence of abelian monoids 0→H2(GKϕ∗)→Cliffk(ϕ)→Ck(ϕ)→H3(GKϕ∗). This sequence describes the obstruction to realizing a generalized collective character of type φ, that is it determines if such a character is associated to some strongly G-graded k-algebra. The rightmost homomorphism is often surjective, terminating the above sequence. When φ is a Galois action, then the well-known restriction-obstruction sequence of Brauer groups is an image of an exact sequence of sub-monoids appearing in the above Bibliographical note Publisher Copyright: © 2022, The Hebrew University of Jerusalem. ASJC Scopus subject areas Dive into the research topics of 'Realization-obstruction exact sequences for Clifford system extensions'. Together they form a unique fingerprint.	{"url":"https://cris.haifa.ac.il/en/publications/realization-obstruction-exact-sequences-for-clifford-system-exten","timestamp":"2024-11-15T02:23:24Z","content_type":"text/html","content_length":"52877","record_id":"<urn:uuid:d35c0fbc-e6b8-4788-a471-362954d9ca8a>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00141.warc.gz"}
Abstract: I will report on recent progress on free quantum groups (with Bichon and Collins), notably on the construction of a 4-th example. The first 3 examples, due to Wang, correspond in the large N limit to the semicircular, circular and free Poisson laws. The 4-th one, that we construct, is related to a more complicated law, that I will describe in detail.	{"url":"https://www.math.ucla.edu/~shlyakht/berkeley-07/conference/abstracts.php","timestamp":"2024-11-13T01:33:00Z","content_type":"text/html","content_length":"19483","record_id":"<urn:uuid:35c011d5-9696-4798-b2d4-07bc9deaa871>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00820.warc.gz"}
On the regularity of area minimizing currents at boundaries with arbitrary multiplicity Inserted: 9 oct 2024 Year: 2024 In this paper, we consider an area-minimizing integral $m$-current $T$, within a submanifold $\Sigma \in C^{3,\kappa}$ of $\mathbb{R}^{m+n}$, with arbitrary boundary multiplicity $\partial T = Q[\![\ Gamma]\!]$, where $\Gamma\subset\Sigma$ of class $C^{3,\kappa}$. We prove that the set of density $Q/2$ boundary singular points of $T$ is $\mathcal{H}^{m-3}$-rectifiable. This result generalizes Allard's boundary regularity theorem to a higher multiplicity setting. In particular, if $\Gamma$ is a closed manifold which lies at the boundary of a uniformly convex set $\Omega$ and $\Sigma=\mathbb{R}^{m+n}$ the whole boundary singular set is $\mathcal{H}^{m-3}$ As a structural consequence of our regularity theory, we show that the boundary regular set, without any assumptions on the density, is open and dense in $\Gamma$.	{"url":"https://cvgmt.sns.it/paper/6820/","timestamp":"2024-11-09T18:51:29Z","content_type":"text/html","content_length":"8610","record_id":"<urn:uuid:b0fb1299-1d10-4264-93ec-beddb0a9707a>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00041.warc.gz"}
Generalized Legendre Conjecture Generalized Legendre Conjecture: Is there a prime between n^s and (n+1)^s for s &lt 2? Equipped with the functions isPrime(n) and nextPrime(n), we can now easily test hypotheses and conjectures about primes. (A conjecture is a statement we believe to be true but have not proved; when someone eventually proves a conjecture, it becomes a theorem.) Here are some interesting conjectures all of them apparently true: [Click Legendre's conjecture. For each integer n > N[2] = 0, there is a prime p between n^2 and (n+1)^2. The n^5/3 conjecture. For each integer n > N[5/3] = 0, there is a prime p between n^5/3 and (n+1)^5/3. The n^8/5 conjecture. For each integer n > N[8/5] = 0, there is a prime p between n^8/5 and (n+1)^8/5. The n^3/2 conjecture. For each integer n > N[3/2] = 1051, there is a prime p between n^3/2 and (n+1)^3/2. The n^4/3 conjecture. For each integer n > N[4/3] = 6776941, there is a prime p between n^4/3 and (n+1)^4/3. The n^5/4 conjecture. For each integer n > N[5/4] ≥ 50904310155, there is a prime p between n^5/4 and (n+1)^5/4. The n^6/5 conjecture. For each integer n > N[6/5] ≥ 833954771945899, there is a prime p between n^6/5 and (n+1)^6/5. The above statements are verifiable up to at least 18-digit primes. (The existing knowledge of maximum prime gaps up to low 19-digit numbers simplifies the verification; beyond that, the verification becomes less practical.) These statements suggest the following generalization: The generalized Legendre conjecture. (A) There exist infinitely many pairs (s, N[s]), 1 < s ≤ 2, such that for each integer n > N[s] there is a prime between n^s and (n+1)^s. (Weak formulation.) (B) For each s > 1, there exists an integer N[s] such that for each integer n > N[s] there is a prime between n^s and (n+1)^s. (Strong formulation.) Discussion. Some of the pairs (s, N[s]) are the above special cases: (2,0), (5/3, 0), (8/5, 0), (3/2, 1051), (4/3, 6776941). N[s] is a function of s; namely, N[s] denotes the greatest counterexample for the n^s conjecture: if we proceed from small to large n, the last interval [n^s, (n+1)^s] containing no primes happens to occur at n = N[s]; and for each n > N[s] there is at least one prime between n^s and (n+1)^s. We readily see that s > 1 is indeed a necessary condition. (A short explanation for a younger reader: should we have s ≤ 1, our intervals [n^s, (n+1)^s] would be way too narrow to contain a prime or any integer at all, in most cases). Moreover, is it plausible that s > s[min] > 1 should also be satisfied, with a certain lower bound s[min]> 1, in order for N[s] to exist? If so, part (A) would still be true, but part (B) would be invalidated for some s very close to 1. Questions to explore further: (1) Is N[s] a monotonic function of s? (No, it is not there are counterexamples.) (2) What the lower bound s[min] might be, for the n^s conjecture to still make sense? (A tongue-in-cheek guess: less than 1.1. A more serious answer: s > 1 is likely enough; no additional s[min] is needed. Here is a hint.) (3) How is the n^s conjecture related to other known conjectures and theorems about the distribution of primes? (The n^s conjecture follows from the Cramer-Granville conjecture when n is large Here is a partial computational verification of special cases of the generalized Legendre conjecture. Copyright © 2011, Alexei Kourbatov, JavaScripter.net.	{"url":"http://javascripter.net/math/primes/generalizedlegendreconjecture.htm","timestamp":"2024-11-08T17:54:14Z","content_type":"text/html","content_length":"17521","record_id":"<urn:uuid:0f9dfd22-5fb2-430e-b50f-528f67e34a2e>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00400.warc.gz"}
operator % abstract method Euclidean modulo of this number by other. Returns the remainder of the Euclidean division. The Euclidean division of two integers a and b yields two integers q and r such that a == b * q + r and 0 <= r < b.abs(). The Euclidean division is only defined for integers, but can be easily extended to work with doubles. In that case, q is still an integer, but r may have a non-integer value that still satisfies 0 <= r < \|b\|. The sign of the returned value r is always positive. See remainder for the remainder of the truncating division. The result is an int, as described by int.%, if both this number and other are integers, otherwise the result is a double. print(5 % 3); // 2 print(-5 % 3); // 1 print(5 % -3); // 2 print(-5 % -3); // 1 double operator %(num other);	{"url":"https://main-api.flutter-io.cn/flutter/dart-core/double/operator_modulo.html","timestamp":"2024-11-04T00:52:31Z","content_type":"text/html","content_length":"9169","record_id":"<urn:uuid:f54957a6-434d-4511-b2df-7f9ba8fdaf05>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00260.warc.gz"}
Ordering Rational Numbers Comparing and ordering rational numbers is a complex skill with many steps, often complicated by the number of ways the problem can be presented. These tips and ideas will help scaffold this skill and make it more hands-on and engaging for your students. Comparing and ordering rational numbers is predominantly a 6th grade skill in both Common Core and TEKS. It will also support students locating and plotting rational numbers on the coordinate plane. 1. 6.NS.7 Understand ordering and absolute value of rational numbers. 2. 6.2D Order a set of rational numbers arising from mathematical and real-world contexts (Readiness Standard) Released Staar Question Looking at a test question isn’t a way to “teach to the test,” but rather, a way for me to make sure that I am teaching the complexity of the standard. By looking at the test question above, it helps me to understand that: • Students will need to order a set of 5 (or possibly more) rational numbers • These rational numbers will vary in type (fractions, mixed numbers, improper fractions, whole numbers, and decimals) so I can assume that test questions can also include percentages • It also helps me recognize that I need to make sure students pay attention to the order in which the question is asking □ Greatest to least – which can also include words like “descending” □ Least to greatest – which can also include words like “ascending” Convert to the Same Form (fraction, decimal, percent) Before you can jump into ordering rational numbers, you will need to spend a few days on converting between fractions, decimals, and percentages. But I wouldn’t start by ordering a set of numbers that include fractions, decimals, and percentages. Start by scaffolding! Order a set of decimals, then a set of percentages, and lastly a set of fractions before introducing a mixed bag of rational numbers. For decimals, I suggest asking students to line up the decimal point and compare the digits going from left to right. Have students add zeros to numbers after the decimal point so no student confuses a number with more digits as greater. For percentages, I would use the same method. Remind students where the decimal point is in a percentage if it is not visible. For fractions, this is where I don’t have strong opinions. Perhaps because I have not seen one method to be more successful than others. Butterfly method, finding a common denominator, converting fractions into a decimal, reasoning through it – I think each student will gravitate toward their favorite based on their past experiences. Once students have mastered ordering from the same type of rational numbers, have students order a small set of 3 rational numbers by converting them to the same form. In my experience, students (and me, the teacher) like to convert most rational numbers to decimals. I think this is because they are the easiest to compare and then order. Use a Number Line When negative numbers are included, a number line is an absolute must. The number line gives us context! A quick sketch of a number line with a zero in the middle immediately serves as a reminder that the numbers with the least value will be the furthest left and the numbers with the greatest value will be the furthest right! In fact, a number line was a requirement for work shown when we ordered numbers in class. A dry erase marker and a desk allowed for lots of space for converting between the fractions, decimals, and percentages and placing them on the number line. Quick Hits • Have a half day or need an extension activity? Have students make flashcards of benchmark fractions: ¼, ½, ¾, ⅕, ⅖, ⅗, ⅘. Once my students could ace this in a flyswatter challenge, I would add ⅛, ⅜, ⅝, ⅞. If students had these benchmark fractions memorized, it lightened the mental load for ordering rational numbers. • Make ordering numbers hands on! This can be done by using these cards. Print a set, laminate, and then have a living number line using the sticky side of painter’s tape (see the idea here). • Get students moving! Give each student a card with a rational number. Tell them to get into groups of 4 and stand in order from least to greatest. Then tell them to find a new group with 3 people and stand in order from greatest to least. Go around the room and listen to their reasoning. You can do this as many times as you want using a different number of students and in a different order. You can finish the activity by having all the students stand in ascending or descending order. How do you teach comparing and ordering rational numbers? 1 Comment 1. EA says There are some great ideas in here, but the percent trick just fuels misunderstandings about what a percent represents and its value. In addition, this will not work for the 8th grade standard where students are asked to order a mixed representation of rational numbers.	{"url":"http://bayareabikesapp.com/index-175.html","timestamp":"2024-11-02T16:53:54Z","content_type":"text/html","content_length":"177148","record_id":"<urn:uuid:3e14b509-7e51-42a4-b249-84f4236b04d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00680.warc.gz"}
Estimando a Natureza, as Posições Horizontais e Verticais de Fontes 3D Usando a Deconvolução de Euler We present a new method that drastically reduces the number of the source location estimates in 3D Euler deconvolution to one only. Our method is grounded on the analytical estimators of the base level and of the horizontal and vertical source positions in 3D Euler deconvolution as a function of the x- and y-coordinates of the observations. By assuming any tentative structural index (defining the geometry of the sources), our method automatically locates on the maps of the horizontal coordinate estimates plateaus, indicating consistent estimates that are very close to the true corresponding coordinates. These plateaus are located at the neighborhood of the highest values of the anomaly and show a contrasting behavior with those estimates which form inclined planes at the anomaly borders. The plateaus are automatically located on the maps of the horizontal coordinate estimates by fitting a first-degree polynomial to these estimates in a moving-window scheme spanning all estimates. The positions where the angular coefficient estimates are closest to zero identify the plateaus of the horizontal coordinate estimates. The sample means of these horizontal coordinate estimates are the best horizontal location estimates. After mapping each plateau, our method takes as the best structural index the one which yields the minimum correlation between the magnetic and the estimated base level over each plateau. By using the estimated structural index for each plateau, our approach extracts the vertical coordinate estimates over the corresponding plateau. The sample mean of these estimates are the best depth location estimates in our method. When applied to synthetic data, our method yielded good performances even when the study area does not cover entirely the anomalies. A test on real data over intrusions in the Goiás Alkaline Province, Brazil, retrieved sphere-like sources suggesting three-dimensional bodies.	{"url":"https://www.pinga-lab.org/thesis/melo-msc.html","timestamp":"2024-11-03T00:38:50Z","content_type":"text/html","content_length":"9163","record_id":"<urn:uuid:c2692107-eb03-4a63-966d-fe2388873b78>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00592.warc.gz"}
9.2: Watch these air parcels move and change. Last updated Page ID $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$ $ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$ $ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$ $ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vectorC}[1]{\textbf{#1}} $ $ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $ $ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $ $ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $ $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\ evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\ newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y} $ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real} {\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec} [3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array} {r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\ wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\ newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var} {\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left\|#1\right\|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\ bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\ widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$ The water vapor image from the GOES 13 satellite, above, indicates different air masses over the United States. As we know from Lesson 7, the water vapor image actually shows the top of a column of water vapor that strongly absorbs in the water vapor channel wavelengths, but it is not a bad assumption to think that there is a solid column of moister air underneath the water vapor layer that is emitting and is observed by the satellite. In a single snapshot, it is not possible to see what happens to the air parcels over time. But if we look at a loop, then we can see the air parcels moving and changing shape as they move. Water vapor in the atmosphere over North America showing the behavior of different air parcels as they interact. Credit: NOAA Watch a Loop Visit this website to see a loop. Pick any air parcel with more water vapor in the first frame and then watch it evolve over time. What does it do? Maybe it moves; it spins; it stretches; it shears; it grows. Maybe it does only a few of these things; maybe it does them all. We can break each air parcel’s complex behavior down into a few basic types of flows and then mathematically describe them. We will just describe these basic motions here and show how they lead to An air parcel Credit: W. Brune Assume that we have an air parcel as in the figure above. We focus on motion in the two horizontal directions to aid in the visualization (and because most motion in the atmosphere is horizontal) but the concepts apply to the vertical direction as well. If the air parcel is moving and does not change its orientation, shape, or size, then it is only undergoing translation (see figure below). Air parcel undergoing translation in x-direction (top) and at 45^o (bottom). Black arrows are the wind field; green arrows are the motion of the air parcel. The orientation, shape, and size of the air parcel does not change as it is translated. Credit: W. Brune The air parcel can do more than just translate. It can undergo changes relative to translation, and its total motion will then be a combination of translation and relative motion. Let’s suppose that different parts of the air parcel have slightly different velocities. This situation is depicted in the figure below. Air parcel with relative motion for two different points in the air parcel separated by dx in the x direction and dy in the y direction and with different velocities at each point. Credit: W. Brune, after R. Najjar If we consider very small differences dx and dy, then we can write u and v at point (x[o] + dx,y[o] + dy) as a Taylor series expansion in two dimensions: \[u\left(x_{o}+d x, y_{o}+d y\right) \approx u\left(x_{o}, y_{o}\right)+\frac{\partial u}{\partial x} d x+\frac{\partial u}{\partial y} d y\] \[v\left(x_{o}+d x, y_{o}+d y\right) \approx v\left(x_{o}, y_{o}\right)+\frac{\partial v}{\partial x} d x+\frac{\partial v}{\partial y} d y\] We see that u(x[o],y[o]) and v(x[o],y[o]) are the translation, and the relative motion is expressed as gradients of u in the x and y directions and gradients of v in the x and y directions. There are four gradients represented by the four partial derivatives. Each can be either positive or negative for each partial derivative. $\frac{\partial u}{\partial x}$ is the following change in velocity in the $x$ direction: $\frac{\partial v}{\partial y}$ is the following change in velocity in the $y$ direction: $\frac{\partial u}{\partial v}$ is the following change in velocity in the $y$ direction: \frac{\partial u}{\partial v} \text { is the following change in velocity in the } y \text { direction: } \frac{\partial v}{\partial x} \text { is the following change in velocity in the } x \text { direction: } Note that a partial derivative is positive if a positive value is becoming more positive or a negative value is becoming less negative. Similarly, a negative partial derivative occurs when a positive value is becoming less positive or a negative value is becoming more negative. Be sure that you have this figured out before you go on. Watch this video (2:38) for further explanation: Partials Velocity Distance Click here for transcript of the Partials Velocity Distance video. I want to make sure that you understand the partial derivatives of the u and v velocity with respect to x and y because we will soon be using these terms a lot. Let's start with the partial derivative u with respect to x. Consider a constantly increasing x so that the change in x is positive. As x increases, u becomes initially less negative, hence a positive change; then becomes positive, another positive change; and then becomes more positive, another positive change. Since the change in u and the change in x are both always positive, the partial derivative is positive, greater than 0. Look at the case where a partial derivative is less than 0, or negative. As x increases, u becomes less positive hence, a negative change. Then becomes negative, another negative change, then becomes more negative, another negative change. Since the change in u is always negative with a positive change in x, the partial derivative is always negative. Same logic applies to the partial derivative of v with respect to y. Up is positive for y, so you should look at how v changes as y becomes more positive. Look at the case of the change in u with respect to y. It does not matter that u is in the x direction perpendicular to y because we are interested in how u changes as a function of y. Let's look at what happens as y becomes more positive. On the left, u becomes less negative, a positive change in u, then positive, and more positive. Thus the partial derivative is a positive change in u over a positive change in y and therefore is positive, or greater than 0. The change in u with respect to y is always positive in this case. Using the same logic on the right, we see that the change in u with respect to y is always negative. And because a change in y is positive, the partial derivative is negative, or less than 0. The same logic applies to the partial derivative of v with respect to x. To the right is positive for x. So you can determine how v changes as x becomes more positive to see whether the partial derivative is positive or negative.	{"url":"https://geo.libretexts.org/Bookshelves/Meteorology_and_Climate_Science/Book%3A_Fundamentals_of_Atmospheric_Science_(Brune)/09%3A_Kinematics/9.02%3A_Watch_these_air_parcels_move_and_change.","timestamp":"2024-11-13T19:16:04Z","content_type":"text/html","content_length":"135923","record_id":"<urn:uuid:b5cbb4b0-05e5-495d-aa2b-16954dd17027>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00068.warc.gz"}
Question about equipment Gentlemen experts, I wanted to understand some of the logic associated with equipment. Equipment is capable of changing the player's statistics, it can be a static value or in percentage terms. I understand it this way: if both options are indicated, then a static value is added first, then a percentage of the total available statistics amount. It means what kind of equipment the player will wear first. For example, there is "Helmet" (defense 20) (defense + 5%) and there is "Armor" (defense 40) (defense + 5%). If the player first puts on "Armor", and then "Helmet", then the final defence will be greater than if first "Helmet" and then "Armor". Tell me, do I understand everything correctly? And I still don't understand how "Stat Bonus Range" works in terms of percentage change in statistics. P.S. So far, I cannot verify this myself, so I am asking about it here. Edited by Yuri 5 answers to this question • 0 I'm pretty sure the %boost is calculated over your base stats. So it doesn't matter which one goes first • 0 9 minutes ago, Beefy Kasplant said: I'm pretty sure the %boost is calculated over your base stats. So it doesn't matter which one goes first That is, if the "Helmet" has a defence +20, then the percentages are counted from 20? Or do you mean basic character stats? Edited by Yuri • 0 All armor base is added up first, then the % bonuses are applied I believe. So if you're wearing 40, 20Â = 60 Defense * %bonuses = 60 * 1.10 = 66 Requires some testing before I'm 100% sure of this but this is how I believe it works. • 0 10 minutes ago, AisenArvalis said: All armor base is added up first, then the % bonuses are applied I believe. So if you're wearing 40, 20Â = 60 Defense * %bonuses = 60 * 1.10 = 66 Requires some testing before I'm 100% sure of this but this is how I believe it works. What do you think about the "Stat Bonus Range"? How does this parameter fit with statistics percentages? • 0 I seem to be starting to understand a little. When the character puts on a new item, all percentages are recalculated and added to all available statistics.	{"url":"https://www.ascensiongamedev.com/topic/5949-question-about-equipment/","timestamp":"2024-11-14T12:36:13Z","content_type":"text/html","content_length":"118364","record_id":"<urn:uuid:cc2a5d95-28e9-4d33-9141-1cbc1922ade2>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00069.warc.gz"}
Area of Trapezoids -7th Grade Math Learn How to Find the Area of Trapezoid. Definition of Trapezoid: Trapezoid is a four sided flat shape with straight sides that has a pair of opposite sides parallel To find the area of Trapezoid, we have to add the two bases and then divide by 2 and then Multiply with Height of Trapezoid. Formula for Area of Trapezoid =((Base1+ Base2)/2) X Height How to Find Area of Trapezoid Shape -7th Grade Math Tutorial No comments:	{"url":"http://www.broandsismathclub.com/2015/08/area-of-trapezoids-7th-grade-math.html","timestamp":"2024-11-06T14:36:21Z","content_type":"text/html","content_length":"54728","record_id":"<urn:uuid:35c63ac6-f728-462e-8ba2-42dabb6ca1ff>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00416.warc.gz"}
Science stencils For preparation of daily chemical laboratory experiment and preparation of SOP, those are basic glassware drawings including beaker, flask, etc. A stencil for labquest spectrophotometer with different solvents. Direct Logic: Ladder Logic. Use these in Keynote to animate blocks of code. This stencil is #2 of 8 Direct Logic: Ladder Logic. Use these in Keynote to animate blocks of code. This stencil is #1 of 8 Direct Logic: Ladder Logic. Use these in Keynote to animate blocks of code. This stencil is #4 of 8 Direct Logic: Ladder Logic. Use these in Keynote to animate blocks of code. This stencil is #8 of 8. Direct Logic: Ladder Logic. Use these in Keynote to animate blocks of code. This stencil is #6 of 8 Direct Logic: Ladder Logic. Use these in Keynote to animate blocks of code. This stencil is #7 of 8 Continuous Process Improvement (CPI) tools for integrating into quality plans to include Define, Measure, Analyze, Improve, and Control (DMAIC) scientific model. Includes Bell Shaped Curve for plotting, Project Charter Matrix, Fishbone Diagram showing root causes, SIPOC Chart and Process Control Plan. Soon to come: Weighted...more	{"url":"https://www.graffletopia.com/categories/science?page=8&sort=name&view=thumbnails","timestamp":"2024-11-10T04:42:12Z","content_type":"text/html","content_length":"16260","record_id":"<urn:uuid:0abf993f-3bec-4e30-9a39-24af33bea58b>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00690.warc.gz"}
Mathematics for BCA Mathematics for BCA is specially designed for students of BCA, who are at the threshold of a completely new domain. Keeping this in mind, the book has been planned with utmost care in the exposition of concepts, choice of illustrative examples, and also in sequencing of topics. The language is simple yet precise. A large number of worked out problems have been included to familiarize the students with the techniques to solving them, and to instill confidence. The topics are interdependent and must be studied in the same order as given in the book.	{"url":"https://www.vikaspublishing.com/books/engineering/core-engineering/mathematics-bca/9788125949589/","timestamp":"2024-11-04T10:55:07Z","content_type":"text/html","content_length":"44606","record_id":"<urn:uuid:cfbaa5dd-22ab-4840-8d82-cc728ad16fd9>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00246.warc.gz"}
Permutations and Combinations-Topics in IB Mathematics Permutation and Combination Permutations and Combinations ‘Permutations and Combinations‘ is the next post of my series Online Maths Tutoring. It is very useful and interesting as a topic. It’s also very useful in solving problems of Probability. Our IB Maths Tutors say that to understand Permutations and Combinations, we first need to understand Factorial. Definition of Factorial- If we multiply n consecutive natural numbers together, then the product is called factorial of n. Its shown by n! or by for example : Some Properties of Factorials (i) Factorials can only be calculated for positive integers at this level. We use gamma functions to define non-integer factorial that’s not required at this level (ii) Factorial of a number can be written as a product of that number with the factorial of its predecessor (iii) you can watch this video for the explanation. (iv) If we want to simplify a “permutations and combinations” expression that has factorials in the numerator as well as in the denominator, we make all the factorials equal to the smallest Exponent of Prime Number p in n! Let’s assume that p is a prime number and n is a positive integer, then exponent of p in n! is denoted by E[p ](n!) We can’t use this result to find the exponent of composite numbers. Fundamental Principle of Counting Almost all IB Online Tutors, teach the first exercise of Permutations and Combinations that is based on the Fundamental Principle of Counting. We can learn it in two steps. Principle of Addition If there are x different ways to do a work and y different ways to two another work and both the works are independent of each other then there are (x+y) ways to do either first OR second work If we can choose a man in a team by 6 different ways and a woman by 4 different ways then we can choose either a man or a woman by 6+4=10 different ways. The principle of Multiplication If there are x different ways to do a work and y different ways to do another work and both the works are independent of each other then there are (x.y) ways to do both first AND second works. If we can choose a man in a team by 6 different ways and a woman by 4 different ways then we can choose a man and a woman by 6*4=24 different ways. Definition of Permutation The process of making different arrangements of objects, letters and words etc by changing their position is known as permutation A, B, and C are four books then we can arrange them BY 6 DIFFERENT WAYS ABC, ACB BCA, BAC CAB CBA. so we can say that there are 6 different permutations of this arrangement. Number of Permutations of n different objects taken all at a time If we want to arrange n objects at n different places then the total number of ways of doing this or the total number of permutations = =n! here P represents permutations Number of Permutations of n different objects taken r at a time If we want to arrange n objects at r different places then the total number of ways of doing this or the total number of permutations = = here represent permutations of n objects taken r at a time. Number of Permutations of n objects when all objects are not different If we have n objects in total out of which p are of one type, q are of another type, r are of any other type, remaining objects are all different from each other, the total number of ways of arranging them= Number of Permutations of n different objects taken all at a time when repetition of objects is allowed If we want to arrange n objects at n different places and we are free to repeat objects as many times as we wish, then the total number of ways of doing this or the total number of permutations = Number of Permutations of n different objects taken r at a time when repetition of objects is allowed If we want to arrange n objects at r different places(taking r at a time) and we are free to repeat objects as many times as we wish, then the total number of ways of doing this or the total number of permutations= Circular Permutations- When we talk about arrangements of objects, it usually means linear arrangements. But if we wish, we can also arrange objects in a loop. Like we can ask our guests to sit around a round dining table. These types of arrangements are called circular permutations. If we want to arrange n objects in a circle, then the total number of ways/circular permutations=(n-1)! this case works when there is some difference between clock-wise and anti-clockwise orders IF there is no distinction between clock-wise and anti-clockwise orders, the total number of permutations=(n-1)!/2 Take help from Maths Tutors for free in case of any difficulty Restricted Permutations There may be following cases of restricted permutation (a) Number of arrangements of ‘n’ objects, taken ‘r’ at a time, when a particular object is to be always included = (b) Number of arrangements of ‘n’ objects, taken ‘r’ at a time, when a particular object is fixed: = (c) The number of arrangements of ‘n’ objects, taken ‘r’ at a time, when a particular object is never taken: = ^n-1 P[r.] (d) The number of arrangements of ‘n’ objects, taken ‘r’ at a time, when ‘m’ specific objects always come with each-other = (e) The number of arrangements of ‘n’ things, taken all at a time, when ‘m’ specific objects always come with each other= In my next post, I will discuss in detail about combinations and will share a large worksheet based on P & C. In the meantime you can download and solve these questions. Also, Check the below-given post on Permutation and Combination Click to getting FREE Online Maths Tutoring Whatsapp at +919911262206 or fill the form to get 1 hr free Class	{"url":"https://ibelitetutor.com/blog/permutations-and-combinations/","timestamp":"2024-11-06T02:22:47Z","content_type":"text/html","content_length":"67849","record_id":"<urn:uuid:e98295d6-10c0-4a89-9f0b-34f1df22bbd3>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00714.warc.gz"}
Physics – The Quantum Pontiff Tommaso Toffoli’s 3-input 3-output logic gate, central to the theory of reversible and quantum computing, recently featured on a custom cake made for his 70’th birthday. Nowadays scientists’ birthday celebrations often take the form of informal mini-conferences, festschrifts without the schrift. I have had the honor of attending several this year, including ones for John Preskill’s and Wojciech Zurek’s 60’th birthdays and a joint 80’th birthday conference for Myriam Sarachik and Daniel Greenberger, both physics professors at City College of New York. At that party I learned that Greenberger and Sarachick have known each other since high school. Neither has any immediate plans for retirement. Resolution of Toom's rule paradox A few days ago our Ghost Pontiff Dave Bacon wondered how Toom’s noisy but highly fault-tolerant 2-state classical cellular automaton can get away with violating the Gibbs phase rule, according to which a finite-dimensional locally interacting system, at generic points in its phase diagram, can have only only one thermodynamically stable phase. The Gibbs rule is well illustrated by the low-temperature ferromagnetic phases of the classical Ising model in two or more dimensions: both phases are stable at zero magnetic field, but an arbitrarily small field breaks the degeneracy between their free energies, making one phase metastable with respect to nucleation and growth of islands of the other. In the Toom model, by contrast, the two analogous phases are absolutely stable over a finite area of the phase diagram, despite biased noise that would seem to favor one phase over the other. Of course Toom’s rule is not microscopically reversible, so it is not bound by laws of equilibrium thermodynamics. Nevertheless, as Dave points out, the distribution of histories of any locally interacting d-dimensional system, whether microscopically reversible or not, can be viewed as an equilibrium Gibbs distribution of a d+1 dimensional system, whose local Hamiltonian is chosen so that the d dimensional system’s transition probabilities are given by Boltzmann exponentials of interaction energies between consecutive time slices. So it might seem, looking at it from the d+1 dimensional viewpoint, that the Toom model ought to obey the Gibbs phase rule too. The resolution of this paradox, described in my 1985 paper with Geoff Grinstein, lies in the fact that the d to d+1 dimensional mapping is not surjective. Rather it is subject to the normalization constraint that for every configuration X(t) at time t, the sum over configurations X(t+1) at time t+1 of transition probabilities P(X(t+1)\|X(t)) is exactly 1. This in turn forces the d+1 dimensional free energy to be identically zero, regardless of how the d dimensional system’s transition probabilities are varied. The Toom model is able to evade the Gibbs phase rule because • being irreversible, its d dimensional free energy is ill-defined, and • the normalization constraint allows two phases to have exactly equal d+1 dimensional free energy despite noise locally favoring one phase or the other. Just outside the Toom model’s bistable region is a region of metastability (roughly within the dashed lines in the above phase diagram) which can be given an interesting interpretation in terms of the d+1 dimensional free energy. According to this interpretation, a metastable phase’s free energy is no longer zero, but rather -ln(1-Γ)≈Γ, where Γ is the nucleation rate for transitions leading out of the metastable phase. This reflects the fact that the transition probabilities no longer sum to one, if one excludes transitions causing breakdown of the metastable phase. Such transitions, whether the underlying d-dimensional model is reversible (e.g. Ising) or not (e.g. Toom), involve critical fluctuations forming an island of the favored phase just big enough to avoid being collapsed by surface tension. Such critical fluctuations occur at a rate Γ≈ exp(-const/s^(d-1)) where s>0 is the distance in parameter space from the bistable region (or in the Ising example, the bistable line). This expression, from classical homogeneous nucleation theory, makes the d+1 dimensional free energy a smooth but non-analytic function of s, identically zero wherever a phase is stable, but lifting off very smoothly from zero as one enters the region of metastability. Below, we compare the d and d+1 dimensional free energies of the Ising model with the d+1 dimensional free energy of the Toom model on sections through the bistable line or region of the phase We have been speaking so far only of classical models. In the world of quantum phase transitions another kind of d to d+1 dimensional mapping is much more familiar, the quantum Monte Carlo method, nicely described in these lecture notes, whereby a d dimensional zero-temperature quantum system is mapped to a d+1 dimensional finite-temperature classical Monte Carlo problem. Here the extra dimension, representing imaginary time, is used to perform a path integral, and unlike the classical-to-classical mapping considered above, the mapping is bijective, so that features of the d+1 dimensional classical system can be directly identified with corresponding ones of the d dimensional quantum one. A Paradox of Toom's Rule? Science is slow. You can do things like continue a conversation with yourself (and a few commenters) that started in 2005. Which is what I’m now going to do 🙂 The below is probably a trivial observation for one of the cardinals, but I find it kind of interesting. Let’s begin by recalling the setup. Toom’s rule is a cellular automata rule for a two dimensional cellular automata on a square grid. Put +1 and -1’s on the vertices of a square grid, and then use the following update rule at each step: “Update the value with the majority vote of your own state, the state of your neighbor to the north, and the state of your neighbor to the east.” A few steps of the rule are shown here with +1 as white and -1 as blue: As you can see Toom’s rule “shrinks” islands of “different” states (taking away such different cells from the north and east sides of such an island.) It is this property which gives Toom’s rule some cool properties in the presence of noise. So now consider Toom’s rule, but with noise. Replace Toom’s update rule with the rule followed by, for each and every cell a noise process. For example this noise could be to put the cell into state +1 with p percent probability and -1 with q percent probability. Suppose now you are trying to store information in the cellular automata. You start out at time zero, say, in the all +1 state. Then let Toom’s rule with noise run. If p=q and these values are below a threshold, then if you start in the +1 state you will remain in a state with majority +1 with a probability that goes to one exponentially as a function of the system size. Similarly if you start in -1. The cool thing about Toom’s rule is that this works not just for p=q, but also for some values of p not equal to q (See here for a picture of the phase diagram.) That is there are two stable states in this model, even for biased noise. Contrast Toom’s rule with a two dimensional Ising model which is in the process of equilibriating to temperature T. If this model has no external field applied, then like Toom’s rule there is a phase where the mostly +1 and the mostly -1 states are stable and coexist. These are from zero temperature (no dynamics) to a threshold temperature T, the critical temperature of the Ising model. But, unlike in Toom’s rule, if you now add an external field, which corresponds to a dynamics where there is now a greater probability of flipping the cell values to a particular value (p not equal to q above), then the Ising model no longer has two stable phases. In fact there is a general argument that if you look at a phase diagram as a function of a bunch of parameters (say temperature and applied magnetic field strength in this case), then the places where two stable regimes can coexist has to be a surface with one less dimension than your parameter space. This is known as Gibbs’ phase rule. Toom’s rule violates this. It’s an example of a nonequilibrium system. So here is what is puzzling me. Consider a three dimensional cubic lattice with +1,-1 spins on its vertices. Define an energy function that is a sum over terms that act on the spins on locations (i,j,k), (i+1,j,k), (i,j+1,k), (i,j,k+1) such that E = 0 if the spin at (i,j,k+1) is in the correct state for Toom’s rule applied to spins (i,j,k), (i+1,j,k), and (i,j+1,k) and is J otherwise. In other words the terms enforce that the ground state locally obey’s Toom’s rule, if we imagine rolling out Toom’s rule into the time dimension (here the z direction). At zero temperature, the ground state of this system will be two-fold degenerate corresponding to the all +1 and all -1 state. At finite temperature this model well behave as a symmetric noise Toom’s rule model (see here for why.) So even at finite temperature this will preserve information, like the d>2 Ising model and Toom’s CA rule. But since this behaves like Toom’s rule, it seems to me that if you add an external field, then this system is in a bit of paradox. On the one hand, we know from Gibb’s phase rule, that this should not be able to exhibit two stable phases over a range of external fields. On the other hand, this thing is just Toom’s rule, laid out spatially. So it would seem that one could apply the arguments about why Toom’s rule is robust at finite field. But these contradict each other. So which is it? 4 Pages Walk up to a physicist at a party (we could add a conditional about the amount of beer consumed by the physicist at this point, but that would be redundant, it is a party after all), and say to him or her “4 pages.” I’ll bet you that 99 percent of the time the physicist’s immediate response will be the three words “Physical Review Letters.” PRL, a journal of the American Physical Society, is one of the top journals to publish in as a physicist, signaling to the mating masses whether you are OK and qualified to be hired as faculty at (insert your college name here). I jest! (As an aside, am I the only one who reads what APS stands for and wonders why I have to see the doctor to try out for high school tennis?) In my past life, before I passed away as Pontiff, I was quite proud of the PRLs I’d been lucky enough to have helped with, including one that has some cool integrals, and another that welcomes my niece into the world. Wait, wht?!? Yes, in “Coherence-Preserving Quantum Bits” the acknowledgement include a reference to my brother’s newborn daughter. Certainly I know of no other paper where such acknowledgements to a beloved family member is given. The other interesting bit about that paper is that we (okay probably you can mostly blame me) originally entitled it “Supercoherent Quantum Bits.” PRL, however, has a policy about new words coined by authors, and, while we almost made it to the end without the referee or editor noticing, they made us change the title because “Supercoherent Quantum Bits” would be a new word. Who would have thought that being a PRL editor meant you had to be a defender of the lexicon? (Good thing Ben didn’t include qubits in his title.) Which brings me to the subject of this post. This is a cool paper. It shows that a very nice quantum error correcting code due to Bravyi and Haah admits a transversal (all at once now, comrades!) controlled-controlled-phase gate, and that this, combined with another transversal gate (everyone’s fav the Hadamard) and fault-tolerant quantum error correction is universal for quantum computation. This shows a way to not have to use state distillation for quantum error correction to perform fault-tolerant quantum computing, which is exciting for those of us who hope to push the quantum computing threshold through the roof with resources available to even a third world quantum computing company. What does this have to do with PRL? Well this paper has four pages. I don’t know if it is going to be submitted or has already been accepted at PRL, but it has that marker that sets off my PRL radar, bing bing bing! And now here is an interesting thing I found in this paper. The awesome amazing very cool code in this paper is defined via its stabilizer I I I I I I IXXXXXXXX; I I I I I I I ZZZZZZZZ, I I IXXXXI I I IXXXX; I I I ZZZZ I I I I ZZZZ, IXXI IXXI IXXI IXX; I ZZ I I ZZ I I ZZ I I ZZ, XIXIXIXIXIXIXIX; Z I Z I Z I Z I Z I Z I Z I Z, This takes up a whopping 4 lines of the article. Whereas the disclaimer, in the acknowledgements reads The U.S. Government is authorized to reproduce and distribute reprints for Governmental pur- poses notwithstanding any copyright annotation thereon. Disclaimer: The views and conclusions contained herein are those of the authors and should not be interpreted as necessarily representing the official policies or endorse- ments, either expressed or implied, of IARPA, DoI/NBC, or the U.S. Government. Now I’m not some come-of-age tea party enthusiast who yells at the government like a coyote howls at the moon (I went to Berkeley damnit, as did my parents before me.) But really, have we come to a point where the god-damn disclaimer on an important paper is longer than the actual definition of the code that makes the paper so amazing? Before I became a ghost pontiff, I had to raise money from many different three, four, and five letter agencies. I’ve got nothing but respect for the people who worked the jobs that help supply funding for large research areas like quantum computing. In fact I personally think we probably need even more people to execute on the civic duty of getting funding to the most interesting and most trans-form-ative long and short term research projects. But really? A disclaimer longer than the code which the paper is about? Disclaiming, what exactly? Erghhh. Non-chaotic irregularity In principle, barring the intervention of chance, identical causes lead to identical effects. And except in chaotic systems, similar causes lead to similar effects. Borges’ story “Pierre Menard” exemplifies an extreme version of this idea: an early 20’th century writer studies Cervantes’ life and times so thoroughly that he is able to recreate several chapters of “Don Quixote” without mistakes and without consulting the original. Meanwhile, back at the ShopRite parking lot in Croton on Hudson, NY, they’d installed half a dozen identical red and white parking signs, presumably all from the same print run, and all posted in similar environments, except for two in a sunnier location. The irregular patterns of cracks that formed as the signs weathered were so similar that at first I thought the cracks had also been printed, but then I noticed small differences. The sharp corners on letters like S and E, apparently points of high stress, usually triggered near-identical cracks in each sign, but not always, and in the sunnier signs many additional fine cracks formed. Another example of reproducibly irregular dynamics was provided over 30 years ago by Ahlers and Walden’s experiments on convective turbulence, where a container of normal liquid helium, heated from below, exhibited nearly the same sequence of temperature fluctuations in several runs of the experiment. Test your intuition The name of this post was shamelessly stolen from Gil Kalai’s popular series Test Your Intuition. But today’s post will be testing our physics intuition, rather than our mathematical intuition. Although this is a quantum blog, we’ll look at the behavior of a classical fluid. The question is: what happens when you soak a washcloth with water and then ring it out… in zero gravity? Think about it for a few minutes before watching the result of the actual experiment below. Throwing cold water on the Quantum Internet There has been a lot of loose talk lately about a coming “Quantum Internet”. I was asked about it recently by a journalist and gave him this curmudgeonly answer, hoping to redirect some of the naive …First let me remark that “quantum internet” has become a bit of a buzzword that can lead to an inaccurate idea of the likely role of quantum information in a future global information infrastructure. Although quantum concepts such as qubit and entanglement have revolutionized our understanding of the nature of information, I believe that the Internet will remain almost entirely classical, with quantum communication and computation being used for a few special purposes, where the unique capabilities of quantum information are needed. For other tasks, where coherent control of the quantum state is not needed, classical processing will suffice and will remain cheaper, faster, and more reliable for the foreseeable future. Of course there is a chance that quantum computers and communications links may some day become so easy to build that they are widely used for general purpose computing and communication, but I think it highly unlikely. Would the quantum internet replace the classical one, or would the two somehow coexist? As remarked above, I think the two would coexist, with the Internet remaining mostly classical. Quantum communication will be used for special purposes, like sharing cryptographic keys, and quantum computing will be used in those few situations where it gives a significant speed advantage (factoring large numbers, some kinds of search, and the simulation of quantum systems), or for the processing of inherently quantum signals (say from a physics experiment). Would quantum search engines require qubits transmitted between the user’s computer and the web searcher’s host? Or would they simply use a quantum computer performing the search on the host machine, which could then return its findings classically? It’s not clear that quantum techniques would help search engines, either in transmitting the data to the search engine, or in performing the search itself. Grover’s algorithm (where coherent quantum searching gives a quadratic speedup over classical searching) is less applicable to the large databases on which search engines operate, than to problems like the traveling salesman problem, where the search takes place not over a physical database, but over an exponentially large space of virtual possibilities determined by a small amount of physical data. On the other hand, quantum techniques could play an important supporting role, not only for search engines but other Internet applications, by helping authenticate and encrypt classical communications, thereby making the Internet more secure. And as I said earlier, dedicated quantum computers could be used for certain classically-hard problems like factoring, searches over virtual spaces, simulating quantum systems, and processing quantum data. When we talk about quantum channels do we mean a quantum communication link down which qubits can be sent and which prevents them decohering, or are these channels always an entangled link? … A quantum channel of the sort you describe is needed, both to transmit quantum signals and to share entanglement. After entanglement has been shared, if one has a quantum memory, it can be stored and used later in combination with a classical channel to transmit qubits. This technique is called quantum teleportation (despite this name, for which I am to blame, quantum teleportation cannot be used for transporting material objects). But could we ever hope for quantum communication in which no wires are needed – but entanglement handles everything? The most common misconception about entanglement is that it can be used to communicate—transmit information from a sender to a receiver—perhaps even instantaneously. In fact it cannot communicate at all, except when assisted by a classical or quantum channel, neither of which communicate faster than the speed of light. So a future Internet will need wires, radio links, optical fibers, or other kinds of communications links, mostly classical, but also including a few quantum channels. How soon before the quantum internet could arrive? I don’t think there will ever be an all-quantum or mostly-quantum internet. Quantum cryptographic systems are already in use in a few places, and I think can fairly be said to have proven potential for improving cybersecurity. Within a few decades I think there will be practical large-scale quantum computers, which will be used to solve some problems intractable on any present or foreseeable classical computer, but they will not replace classical computers for most problems. I think the Internet as a whole will continue to consist mostly of classical computers, communications links, and data storage devices. Given that the existing classical Internet is not going away, what sort of global quantum infrastructure can we expect, and what would it be used for? Quantum cryptographic key distribution, the most mature quantum information application, is already deployed over short distances today (typically < 100 km). Planned experiments between ground stations and satellites in low earth orbit promise to increase this range several fold. The next and more important stage, which depends on further progress in quantum memory and error correction, will probably be the development of a network of quantum repeaters, allowing entanglement to be generated between any two nodes in the network, and, more importantly, stockpiled and stored until needed. Aside from its benefits for cybersecurity (allowing quantum-generated cryptographic keys to be shared between any two nodes without having to trust the intermediate nodes) such a globe-spanning quantum repeater network will have important scientific applications, for example allowing coherent quantum measurements to be made on astronomical signals over intercontinental distances. Still later, one can expect full scale quantum computers to be developed and attached to the repeater network. We would then finally have achieved a capacity for fully general processing of quantum information, both locally and globally—an expensive, low-bandwidth quantum internet if you will—to be used in conjunction with the cheap high-bandwidth classical Internet when the unique capabilities of quantum information processing are needed. Quantum Frontiers As a postdoc at Caltech, I would often have lunch with John Preskill. About once per week, we would play a game. During the short walk back, I would think of a question to which I didn’t know the answer. Then with maybe 100 meters to go, I would ask John that question. He would have to answer the question via a 20 minute impromptu lecture given right away, as soon as we walked into the Now, these were not easy questions. At least, not to your average person, or even your average physicist. For example, “John, why do neutrinos have a small but nonzero mass?” Perhaps any high-energy theorist worth their salt would know the answer to that question, but it simply isn’t part of the training for most physicists, especially those in quantum information science. Every single time, John would give a clear, concise and logically well-organized answer to the question at hand. He never skimped on equations when they were called for, but he would often analyze these problems using simple symmetry arguments and dimensional analysis—undergraduate physics! At the end of each lecture, you really felt like you understood the answer to the question that was asked, which only moments ago seemed like it might be impossible to answer. But the point of this post is not to praise John. Insead, I’m writing it so that I can set high expectations for John’s new blog, called Quantum Frontiers. Yes, that’s right, John Preskill has a blog now, and I hope that he’ll exceed these high expectations with content of similar or higher quality to what I witnessed in those after-lunch lectures. (John, if you’re reading this, no pressure.) And John won’t be the only one blogging. It seems that the entire Caltech IQIM will “bring you firsthand accounts of the groundbreaking research taking place inside the labs of IQIM, and to answer your questions about our past, present and future work on some of the most fascinating questions at the frontiers of quantum science.” This sounds pretty exciting, and it’s definitely a welcome addition to the (underrepresented?) quantum blogosphere. What increases when a self-organizing system organizes itself? Logical depth to the rescue. (An earlier version of this post appeared in the latest newsletter of the American Physical Society’s special interest group on Quantum Information.) One of the most grandly pessimistic ideas from the 19^th century is that of “heat death” according to which a closed system, or one coupled to a single heat bath at thermal equilibrium, eventually inevitably settles into an uninteresting state devoid of life or macroscopic motion. Conversely, in an idea dating back to Darwin and Spencer, nonequilibrium boundary conditions are thought to have caused or allowed the biosphere to self-organize over geological time. Such godless creation, the bright flip side of the godless hell of heat death, nowadays seems to worry creationists more than Darwin’s initially more inflammatory idea that people are descended from apes. They have fought back, using superficially scientific arguments, in their masterful peanut butter video. Much simpler kinds of complexity generation occur in toy models with well-defined dynamics, such as this one-dimensional reversible cellular automaton. Started from a simple initial condition at the left edge (periodic, but with a symmetry-breaking defect) it generates a deterministic wake-like history of growing size and complexity. (The automaton obeys a second order transition rule, with a site’s future differing from its past iff exactly two of its first and second neighbors in the current time slice, not counting the site itself, are black and the other two are white.) Fig 2. Time → But just what is it that increases when a self-organizing system organizes itself? Such organized complexity is not a thermodynamic potential like entropy or free energy. To see this, consider transitions between a flask of sterile nutrient solution and the bacterial culture it would become if inoculated by a single seed bacterium. Without the seed bacterium, the transition from sterile nutrient to bacterial culture is allowed by the Second Law, but prohibited by a putative “slow growth law”, which prohibits organized complexity from increasing quickly, except with low probability. Fig. 3 The same example shows that organized complexity is not an extensive quantity like free energy. The free energy of a flask of sterile nutrient would be little altered by adding a single seed bacterium, but its organized complexity must have been greatly increased by this small change. The subsequent growth of many bacteria is accompanied by a macroscopic drop in free energy, but little change in organized complexity. The relation between universal computer programs and their outputs has long been viewed as a formal analog of the relation between theory and phenomenology in science, with the various programs generating a particular output x being analogous to alternative explanations of the phenomenon x. This analogy draws its authority from the ability of universal computers to execute all formal deductive processes and their presumed ability to simulate all processes of physical causation. In algorithmic information theory the Kolmogorov complexity, of a bit string x as defined as the size in bits of its minimal program x, the smallest (and lexicographically first, in case of ties) program causing a standard universal computer U to produce exactly x as output and then halt. x = min{p: U(p)=x} Because of the ability of universal machines to simulate one another, a string’s Kolmogorov complexity is machine-independent up to a machine-dependent additive constant, and similarly is equal to within an additive constant to the string’s algorithmic entropy H[U](x), the negative log of the probability that U would output exactly x and halt if its program were supplied by coin tossing. Bit strings whose minimal programs are no smaller than the string itself are called incompressible, or algorithmically random, because they lack internal structure or correlations that would allow them to be specified more concisely than by a verbatim listing. Minimal programs themselves are incompressible to within O(1), since otherwise their minimality would be undercut by a still shorter program. By contrast to minimal programs, any program p that is significantly compressible is intrinsically implausible as an explanation for its output, because it contains internal redundancy that could be removed by deriving it from the more economical hypothesis p*. In terms of Occam’s razor, a program that is compressible by s bits deprecated as an explanation of its output because it suffers from s bits worth of ad-hoc assumptions. Though closely related[1] to statistical entropy, Kolmogorov complexity itself is not a good measure of organized complexity because it assigns high complexity to typical random strings generated by coin tossing, which intuitively are trivial and unorganized. Accordingly many authors have considered modified versions of Kolmogorov complexity—also measured in entropic units like bits—hoping thereby to quantify the nontrivial part of a string’s information content, as opposed to its mere randomness. A recent example is Scott Aaronson’s notion of complextropy, defined roughly as the number of bits in the smallest program for a universal computer to efficiently generate a probability distribution relative to which x cannot efficiently be recognized as atypical. However, I believe that entropic measures of complexity are generally unsatisfactory for formalizing the kind of complexity found in intuitively complex objects found in nature or gradually produced from simple initial conditions by simple dynamical processes, and that a better approach is to characterize an object’s complexity by the amount of number-crunching (i.e. computation time, measured in machine cycles, or more generally other dynamic computational resources such as time, memory, and parallelism) required to produce the object from a near-minimal-sized description. This approach, which I have called logical depth, is motivated by a common feature of intuitively organized objects found in nature: the internal evidence they contain of a nontrivial causal history. If one accepts that an object’s minimal program represents its most plausible explanation, then the minimal program’s run time represents the number of steps in its most plausible history. To make depth stable with respect to small variations in x or U, it is necessary also to consider programs other than the minimal one, appropriately weighted according to their compressibility, resulting in the following two-parameter definition. • An object x is called d-deep with s bits significance iff every program for U to compute x in time <d is compressible by at least s bits. This formalizes the idea that every hypothesis for x to have originated more quickly than in time d contains s bits worth of ad-hoc assumptions. Dynamic and static resources, in the form of the parameters d and s, play complementary roles in this definition: d as the quantifier and s as the certifier of the object’s nontriviality. Invoking the two parameters in this way not only stabilizes depth with respect to small variations of x and U, but also makes it possible to prove that depth obeys a slow growth law, without which any mathematically definition of organized complexity would seem problematic. • A fast deterministic process cannot convert shallow objects to deep ones, and a fast stochastic process can only do so with low probability. (For details see Bennett88.) Logical depth addresses many infelicities and problems associated with entropic measures of complexity. • It does not impose an arbitrary rate of exchange between the independent variables of strength of evidence and degree of nontriviality of what the evidence points to, nor an arbitrary maximum complexity that an object can have, relative to its size. Just as a microscopic fossil can validate an arbitrarily long evolutionary process, so a small fragment of a large system, one that has evolved over a long time to a deep state, can contain evidence of entire depth of the large system, which may be more than exponential in the size of the fragment. • It helps explain the increase of complexity at early times and its decrease at late times by providing different mechanisms for these processes. In figure 2, for example, depth increases steadily at first because it reflects the duration of the system’s actual history so far. At late times, when the cellular automaton has run for a generic time comparable to its Poincare recurrence time, the state becomes shallow again, not because the actual history was uneventful, but because evidence of that history has become degraded to the point of statistical insignificance, allowing the final state to be generated quickly from a near-incompressible program that short-circuits the system’s actual history. • It helps explain while some systems, despite being far from thermal equilibrium, never self-organize. For example in figure 1, the gaseous sun, unlike the solid earth, appears to lack means of remembering many details about its distant past. Thus while it contains evidence of its age (e.g. in its hydrogen/helium ratio) almost all evidence of particular details of its past, e.g. the locations of sunspots, are probably obliterated fairly quickly by the sun’s hot, turbulent dynamics. On the other hand, systems with less disruptive dynamics, like our earth, could continue increasing in depth for as long as their nonequilibrium boundary conditions persisted, up to an exponential maximum imposed by Poincare recurrence. • Finally, depth is robust with respect to transformations that greatly alter an object’s size and Kolmogorov complexity, and many other entropic quantities, provided the transformation leaves intact significant evidence of a nontrivial history. Even a small sample of the biosphere, such as a single DNA molecule, contains such evidence. Mathematically speaking, the depth of a string x is not much altered by replicating it (like the bacteria in the flask), padding it with zeros or random digits, or passing it though a noisy channel (although the latter treatment decreases the significance parameter s). If the whole history of the earth were derandomized, by substituting deterministic pseudorandom choices for all its stochastic accidents, the complex objects in this substitute world would have very little Kolmogorov complexity, yet their depth would be about the same as if they had resulted from a stochastic evolution. The remaining infelicities of logical depth as a complexity measure are those afflicting computational complexity and algorithmic entropy theories generally. • Lack of tight lower bounds: because of open P=PSPACE question one cannot exhibit a system that provably generates depth more than polynomial in the space used. • Semicomputability: deep objects can be proved deep (with exponential effort) but shallow ones can’t be proved shallow. The semicomputability of depth, like that of Kolmogorov complexity, is an unavoidable consequence of the unsolvability of the halting problem. The following observations can be made partially mitigating these infelicities. • Using the theory of cryptographically strong pseudorandom functions one can argue (if such functions exist) that deep objects can be produced efficiently, in time polynomial and space polylogarithmic in their depth, and indeed that they are produced efficiently by some physical processes. • Semicomputability does not render a complexity measure entirely useless. Even though a particular string cannot be proved shallow, and requires an exponential amount of effort to prove it deep, the depth-producing properties of stochastic processes can be established, assuming the existence of cryptographically strong pseudorandom functions. This parallels the fact that while no particular string can be proved to be algorithmically random (incompressible), it can be proved that the statistically random process of coin tossing produces algorithmically random strings with high probability. Granting that a logically deep object is one plausibly requiring a lot of computation to produce, one can consider various related notions: • An object y is deep relative to x if all near-minimal sized programs for computing y from x are slow-running. Two shallow objects may be deep relative to one another, for example a random string and its XOR with a deep string. • An object can be called cryptic if it is computationally difficult to obtain a near- minimal sized program for the object from the object itself, in other words if any near-minimal sized program for x is deep relative to x. One-way functions, if they exist, can be used to define cryptic objects; for example, in a computationally secure but information theoretically insecure cryptosystem, plaintexts should be cryptic relative to their ciphertexts. • An object can be called ambitious if, when presented to a universal computer as input, it causes the computer to embark on a long but terminating computation. Such objects, though they determine a long computation, do not contain evidence of it actually having been done. Indeed they may be shallow and even algorithmically random. • An object can be called wise if it is deep and a large and interesting family of other deep objects are shallow relative to it. Efficient oracles for hard problems, such as the characteristic function of an NP-complete set, or the characteristic set K of the halting problem, are examples of wise objects. Interestingly, Chaitin’s omega is an exponentially more compact oracle for the halting problem than K is, but it is so inefficient to use that it is shallow and indeed algorithmically random. Further details about these notions can be found in Bennett88. K.W. Regan in Dick Lipton’s blog discusses the logical depth of Bill Gasarch’s recently discovered solutions to the 17-17 and 18×18 four-coloring problem I close with some comments on the relation between organized complexity and thermal disequilibrium, which since the 19^th century has been viewed as an important, perhaps essential, prerequisite for self-organization. Broadly speaking, locally interacting systems at thermal equilibrium obey the Gibbs phase rule, and its generalization in which the set of independent parameters is enlarged to include not only intensive variables like temperature, pressure and magnetic field, but also all parameters of the system’s Hamiltonian, such as local coupling constants. A consequence of the Gibbs phase rule is that for generic values of the independent parameters, i.e. at a generic point in the system’s phase diagram, only one phase is thermodynamically stable. This means that if a system’s independent parameters are set to generic values, and the system is allowed to come to equilibrium, its structure will be that of this unique stable Gibbs phase, with spatially uniform properties and typically short-range correlations. Thus for generic parameter values, when a system is allowed to relax to thermal equilibrium, it entirely forgets its initial condition and history and exists in a state whose structure can be adequately approximated by stochastically sampling the distribution of microstates characteristic of that stable Gibbs phase. Dissipative systems—those whose dynamics is not microscopically reversible or whose boundary conditions prevent them from ever attaining thermal equilibrium—are exempt from the Gibbs phase rule for reasons discussed in BG85, and so are capable, other conditions being favorable, of producing structures of unbounded depth and complexity in the long time limit. For further discussion and a comparison of logical depth with other proposed measures of organized complexity, see B90. [1] An elementary result of algorithmic information theory is that for any probability ensemble of bit strings (representing e.g. physical microstates), the ensemble’s Shannon entropy differs from the expectation of its members’ algorithmic entropy by at most of the number of bits required to describe a good approximation to the ensemble. Why the laity hope Einstein was wrong. Although reputable news sources pointed out that most scientists think some more mundane explanation will be found for the too-early arrival of CERN-generated neutrinos in Gran Sasso, recently confirmed by a second round of experiments with much briefer pulse durations to exclude the most likely sources of systematic error, the take-home message for most non-scientists seems to have been “Einstein was wrong. Things can go faster than light.” Scientists trying to explain their skepticism often end up sounding closed-minded and arrogant. People say, “Why don’t you take evidence of faster-than-light travel at face value, rather than saying it must be wrong because it disagrees with Einstein.” The macho desire not to be bound by an arbitrary speed limit doubtless also helps explain why warp drives are such a staple of science fiction. At a recent dinner party, as my wife silently reminded me that a lecture on time dilation and Fitzgerald contraction would be inappropriate, the best I could come up with was an analogy to another branch of physics where where lay peoples’ intuition accords better with that of specialists: I told them, without giving them any reason to believe me, that Einstein showed that faster-than-light travel would be about as far-reaching and disruptive in its consequences as an engine that required no fuel. That was too crude an analogy. Certainly a fuelless engine, if it could be built, would be more disruptive in its practical consequences, whereas faster-than-light neutrinos could be accommodated, without creating any paradoxes of time travel, if there were a preferred reference frame within which neutrinos traveling through rock could go faster than light, while other particles, including neutrinos traveling though empty space, would behave in the usual Lorentz-invariant fashion supported by innumerable experiments and astronomical observations. But it is wrong to blame mere populist distrust of authority for this disconnect between lay and expert opinion. Rather the fault lies with a failure of science education, leaving the public with a good intuition for Galilean relativity, but little understanding of how it has been superseded by special relativity. So maybe, after dinner is over and my audience is no longer captive, I should retell the old story of cosmic ray-generated muons, who see the onrushing earth as having an atmosphere only a few feet thick, while terrestrial observers see the muons’ lifetime as having been extended manyfold by time dilation. It is this difference in appreciation of special relativity that accounts for the fact that for most people, faster-than-light travel seems far more plausible than time travel, whereas for experts, time travel, via closed timelike curves of general relativistic origin, is more plausible than faster-than-light travel in flat spacetime.	{"url":"https://dabacon.org/pontiff/category/physics/","timestamp":"2024-11-13T23:00:12Z","content_type":"text/html","content_length":"146392","record_id":"<urn:uuid:f2317baa-4c1b-4bcd-b1c2-56c77d412595>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00495.warc.gz"}
R - Gas Constant (SI units) `"R (Gas Constant)" = 8.31446261815324 " J/Mol" * "K"` The Gas Constant, R, from the Ideal Gas Law is 8.31446261815324 Joules / (moles • Kelvin). The gas constant (also known as the molar, universal, or ideal gas constant, denoted by the symbol R or R) is a physical constant which is featured in many fundamental equations in the physical sciences. The Gas Constant (R) is equivalent to the Boltzmann constant, but expressed in units of energy (i.e. the pressure-volume product) per temperature increment per mole (rather than energy per temperature increment per particle). The constant is also a combination of the constants from Boyle's law, Charles's law, Avogadro's law, and Gay-Lussac's law. The Gas Constant (R) appears in many formulas including the following: • Some descriptive text in the description of the Gas constant comes from Wikipedia: wikipedia/wiki/Gas_constant Enhance your vCalc experience with a free account Sign Up Now! Sorry, JavaScript must be enabled. Change your browser options, then try again.	{"url":"https://www.vcalc.com/wiki/ideal-gas-constant","timestamp":"2024-11-01T22:29:50Z","content_type":"text/html","content_length":"86515","record_id":"<urn:uuid:2b8a0db3-627e-4ebe-9f2e-56ced948b968>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00811.warc.gz"}
Calculate Time Differences in Excel & Google Sheets Last updated on February 6, 2023 Download Example Workbook Download the example workbook This tutorial will demonstrate how to calculate time differences in Excel & Google Sheets. Time Difference To calculate time differences in Excel simply subtract two times: Note if you time difference returns a negative number you’ll see this: To fix the output you can use the ABS Function to calculate the absolute value: Hours, Minutes, and Seconds Between Times To display only the hours, minutes, and seconds between times you can use the HOUR, MINUTE, and SECOND Functions: Hours Between Times To calculate the number of hours between times, subtract the times and multiply by 24 (24 hours in a day). To calculate only the number of full hours use the INT Function: Minutes Between Times To calculate the number of minutes between times do the same except multiply by 1440 (24 hours * 60 minutes): Seconds Between Times Multiply by 86,400 to calculate the number of seconds between times: Display Time Difference You can output a time difference as a string of text using the TEXT Function: Calculate Time Differences in Google Sheets All of the above examples work exactly the same in Google Sheets as Excel.	{"url":"https://www.automateexcel.com/formulas/time-difference/","timestamp":"2024-11-05T11:48:50Z","content_type":"text/html","content_length":"146053","record_id":"<urn:uuid:5611240d-4954-4922-9043-061913ab2c9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00172.warc.gz"}
Integer Exponents Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen Section 1.1 : Integer Exponents We will start off this chapter by looking at integer exponents. In fact, we will initially assume that the exponents are positive as well. We will look at zero and negative exponents in a bit. Let’s first recall the definition of exponentiation with positive integer exponents. If $a$ is any number and $n$ is a positive integer then, \[{a^n} = \underbrace {a \cdot a \cdot a \cdot \cdots \cdot a}_{n{\mbox{ times}}}\] So, for example, \[{3^5} = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 243\] We should also use this opportunity to remind ourselves about parenthesis and conventions that we have in regard to exponentiation and parenthesis. This will be particularly important when dealing with negative numbers. Consider the following two cases. \[{\left( { - 2} \right)^4}\hspace{0.5in}{\mbox{and}}\hspace{0.5in} - {2^4}\] These will have different values once we evaluate them. When performing exponentiation remember that it is only the quantity that is immediately to the left of the exponent that gets the power. In the first case there is a parenthesis immediately to the left so that means that everything in the parenthesis gets the power. So, in this case we get, \[{\left( { - 2} \right)^4} = \left( { - 2} \right)\left( { - 2} \right)\left( { - 2} \right)\left( { - 2} \right) = 16\] In the second case however, the 2 is immediately to the left of the exponent and so it is only the 2 that gets the power. The minus sign will stay out in front and will NOT get the power. In this case we have the following, \[ - {2^4} = - \left( {{2^4}} \right) = - \left( {2 \cdot 2 \cdot 2 \cdot 2} \right) = - \left( {16} \right) = - 16\] We put in some extra parenthesis to help illustrate this case. In general, they aren’t included and we would write instead, \[ - {2^4} = - 2 \cdot 2 \cdot 2 \cdot 2 = - 16\] The point of this discussion is to make sure that you pay attention to parenthesis. They are important and ignoring parenthesis or putting in a set of parenthesis where they don’t belong can completely change the answer to a problem. Be careful. Also, this warning about parenthesis is not just intended for exponents. We will need to be careful with parenthesis throughout this course. Now, let’s take care of zero exponents and negative integer exponents. In the case of zero exponents we have, \[{a^0} = 1\hspace{0.5in}{\mbox{provided }}a \ne 0\] Notice that it is required that $a$ not be zero. This is important since ${0^0}$ is not defined. Here is a quick example of this property. \[{\left( { - 1268} \right)^0} = 1\] We have the following definition for negative exponents. If $a$ is any non-zero number and $n$ is a positive integer (yes, positive) then, \[{a^{ - n}} = \frac{1}{{{a^n}}}\] Can you see why we required that $a$ not be zero? Remember that division by zero is not defined and if we had allowed $a$ to be zero we would have gotten division by zero. Here are a couple of quick examples for this definition, \[{5^{ - 2}} = \frac{1}{{{5^2}}} = \frac{1}{{25}}\hspace{0.5in}\hspace{0.25in}{\left( { - 4} \right)^{ - 3}} = \frac{1}{{{{\left( { - 4} \right)}^3}}} = \frac{1}{{ - 64}} = - \frac{1}{{64}}\] Here are some of the main properties of integer exponents. Accompanying each property will be a quick example to illustrate its use. We will be looking at more complicated examples after the 1. ${a^n}{a^m} = {a^{n + m}}$ Example : ${a^{ - 9}}{a^4} = {a^{ - 9 + 4}} = {a^{ - 5}}$ 2. ${\left( {{a^n}} \right)^m} = {a^{nm}}$ Example : ${\left( {{a^7}} \right)^3} = {a^{\left( 7 \right)\left( 3 \right)}} = {a^{21}}$ 3. $\displaystyle \frac{{{a^n}}}{{{a^m}}} = \left\{ {\begin{array}{l}{{a^{n - m}}}\\{\ Example : \(\begin{align}\frac{{{a^4}}}{{{a^{11}}}} & = {a^{4 - 11}} = {a^{ - 7}}\\ \frac{{{a^4}}}{{{a^ displaystyle \frac{1}{{{a^{m - n}}}}}\end{array}\,\,\,,\,\,\,\,a \ne 0} \right.$ {11}}}} & = \frac{1}{{{a^{11 - 4}}}} = \frac{1}{{{a^7}}} = {a^{ - 7}}\end{align}\) 4. ${\left( {ab} \right)^n} = {a^n}{b^n}$ Example : ${\left( {ab} \right)^{ - 4}} = {a^{ - 4}}{b^{ - 4}}$ 5. ${\displaystyle \left( {\frac{a}{b}} \right)^n} = \displaystyle \frac{{{a^n}}}{{{b^ Example : \({\left( {\displaystyle \frac{a}{b}} \right)^8} = \displaystyle \frac{{{a^8}}}{{{b^8}}}$ n}}},\,\,\,\,b \ne 0\) 6. $\displaystyle {\left( {\frac{a}{b}} \right)^{ - n}} = {\left( {\frac{b}{a}} \right) Example : \({\left( {\displaystyle \frac{a}{b}} \right)^{ - 10}} = {\left( {\displaystyle \frac{b}{a}} \right) ^n} = \frac{{{b^n}}}{{{a^n}}}$ ^{10}} = \displaystyle \frac{{{b^{10}}}}{{{a^{10}}}}\) 7. ${\left( {ab} \right)^{ - n}} = \displaystyle \frac{1}{{{{\left( {ab} \right)}^n}}} Example : \({\left( {ab} \right)^{ - 20}} = \displaystyle \frac{1}{{{{\left( {ab} \right)}^{20}}}}$ 8. $\displaystyle \frac{1}{{{a^{ - n}}}} = {a^n}$ Example : $\displaystyle \frac{1}{{{a^{ - 2}}}} = {a^2}$ 9. $\displaystyle \frac{{{a^{ - n}}}}{{{b^{ - m}}}} = \frac{{{b^m}}}{{{a^n}}}$ Example : $\displaystyle \frac{{{a^{ - 6}}}}{{{b^{ - 17}}}} = \frac{{{b^{17}}}}{{{a^6}}}$ 10. ${\left( {{a^n}{b^m}} \right)^k} = {a^{nk}}{b^{mk}}$ Example : ${\left( {{a^4}{b^{ - 9}}} \right)^3} = {a^{\left( 4 \right)\left( 3 \right)}}{b^{\left( { - 9} \ right)\left( 3 \right)}} = {a^{12}}{b^{ - 27}}$ 11. $\displaystyle {\left( {\frac{{{a^n}}}{{{b^m}}}} \right)^k} = \frac{{{a^{nk}}}}{{{b Example : \({\left( {\displaystyle \frac{{{a^6}}}{{{b^5}}}} \right)^2} = \displaystyle \frac{{{a^{\left( 6 \ ^{mk}}}}$ right)\left( 2 \right)}}}}{{{b^{\left( 5 \right)\left( 2 \right)}}}} = \frac{{{a^{12}}}}{{{b^{10}}}}\) Notice that there are two possible forms for the third property. Which form you use is usually dependent upon the form you want the answer to be in. Note as well that many of these properties were given with only two terms/factors but they can be extended out to as many terms/factors as we need. For example, property 4 can be extended as follows. \[{\left( {abcd} \right)^n} = {a^n}{b^n}{c^n}{d^n}\] We only used four factors here, but hopefully you get the point. Property 4 (and most of the other properties) can be extended out to meet the number of factors that we have in a given problem. There are several common mistakes that students make with these properties the first time they see them. Let’s take a look at a couple of them. Consider the following case. \[\begin{align}& { \mbox{Correct : }} & & {\mbox{ }}a{b^{ - 2}} = a\frac{1}{{{b^2}}} = \frac{a}{{{b^2}}}\\ & {\mbox{Incorrect : }} & & {\mbox{ }}a{b^{ - 2}} \ne \frac{1}{{a{b^2}}}\end{align}\] In this case only the $b$ gets the exponent since it is immediately off to the left of the exponent and so only this term moves to the denominator. Do NOT carry the $a$ down to the denominator with the $b$. Contrast this with the following case. \[{\left( {ab} \right)^{ - 2}} = \frac{1}{{{{\left( {ab} \right)}^2}}}\] In this case the exponent is on the set of parenthesis and so we can just use property 7 on it and so both the $a$ and the $b$ move down to the denominator. Again, note the importance of parenthesis and how they can change an answer! Here is another common mistake. \[\begin{align} & {\mbox{Correct : }} & & \frac{1}{{3{a^{ - 5}}}} = \frac{1}{3}\frac{1}{{{a^{ - 5}}}} = \frac{1}{3}{a^5}\\ & {\mbox{Incorrect : }} & & \frac{1}{{3{a^{ - 5}}}} \ne 3{a^5}\end{align} In this case the exponent is only on the $a$ and so to use property 8 on this we would have to break up the fraction as shown and then use property 8 only on the second term. To bring the 3 up with the $a$ we would have needed the following. \[\frac{1}{{{{\left( {3a} \right)}^{ - 5}}}} = {\left( {3a} \right)^5}\] Once again, notice this common mistake comes down to being careful with parenthesis. This will be a constant refrain throughout these notes. We must always be careful with parenthesis. Misusing them can lead to incorrect answers. Let’s take a look at some more complicated examples now. Example 1 Simplify each of the following and write the answers with only positive exponents. 1. ${\left( {4{x^{ - 4}}{y^5}} \right)^3}$ 2. ${\left( { - 10{z^2}{y^{ - 4}}} \right)^2}{\left( {{z^3}y} \right)^{ - 5}}$ 3. $\displaystyle \frac{{{n^{ - 2}}m}}{{7{m^{ - 4}}{n^{ - 3}}}}$ 4. $\displaystyle \frac{{5{x^{ - 1}}{y^{ - 4}}}}{{{{\left( {3{y^5}} \right)}^{ - 2}}{x^9}}}$ 5. ${\left( {\displaystyle \frac{{{z^{ - 5}}}}{{{z^{ - 2}}{x^{ - 1}}}}} \right)^6}$ 6. ${\left( {\displaystyle \frac{{24{a^3}{b^{ - 8}}}}{{6{a^{ - 5}}b}}} \right)^{ - 2}}$ Show All Solutions Hide All Solutions Show Discussion Note that when we say “simplify” in the problem statement we mean that we will need to use all the properties that we can to get the answer into the required form. Also, a “simplified” answer will have as few terms as possible and each term should have no more than a single exponent on it. There are many different paths that we can take to get to the final answer for each of these. In the end the answer will be the same regardless of the path that you used to get the answer. All that this means for you is that as long as you used the properties you can take the path that you find the easiest. The path that others find to be the easiest may not be the path that you find to be the easiest. That is okay. Also, we won’t put quite as much detail in using some of these properties as we did in the examples given with each property. For instance, we won’t show the actual multiplications anymore, we will just give the result of the multiplication. ${\left( {4{x^{ - 4}}{y^5}} \right)^3}$ Show Solution For this one we will use property 10 first. \[{\left( {4{x^{ - 4}}{y^5}} \right)^3} = {4^3}{x^{ - 12}}{y^{15}}\] Don’t forget to put the exponent on the constant in this problem. That is one of the more common mistakes that students make with these simplification problems. At this point we need to evaluate the first term and eliminate the negative exponent on the second term. The evaluation of the first term isn’t too bad and all we need to do to eliminate the negative exponent on the second term is use the definition we gave for negative exponents. \[{\left( {4{x^{ - 4}}{y^5}} \right)^3} = 64\left( {\frac{1}{{{x^{12}}}}} \right){y^{15}} = \frac{{64{y^{15}}}}{{{x^{12}}}}\] We further simplified our answer by combining everything up into a single fraction. This should always be done. The middle step in this part is usually skipped. All the definition of negative exponents tells us to do is move the term to the denominator and drop the minus sign in the exponent. So, from this point on, that is what we will do without writing in the middle step. ${\left( { - 10{z^2}{y^{ - 4}}} \right)^2}{\left( {{z^3}y} \right)^{ - 5}}$ Show Solution In this case we will first use property 10 on both terms and then we will combine the terms using property 1. Finally, we will eliminate the negative exponents using the definition of negative \[{\left( { - 10{z^2}{y^{ - 4}}} \right)^2}{\left( {{z^3}y} \right)^{ - 5}} = {\left( { - 10} \right)^2}{z^4}{y^{ - 8}}{z^{ - 15}}{y^{ - 5}} = 100{z^{ - 11}}{y^{ - 13}} = \frac{{100}}{{{z^{11}}{y^ There are a couple of things to be careful with in this problem. First, when using the property 10 on the first term, make sure that you square the “-10” and not just the 10 (i.e. don’t forget the minus sign…). Second, in the final step, the 100 stays in the numerator since there is no negative exponent on it. The exponent of “-11” is only on the $z$ and so only the $z$ moves to the $\displaystyle \frac{{{n^{ - 2}}m}}{{7{m^{ - 4}}{n^{ - 3}}}}$ Show Solution This one isn’t too bad. We will use the definition of negative exponents to move all terms with negative exponents in them to the denominator. Also, property 8 simply says that if there is a term with a negative exponent in the denominator then we will just move it to the numerator and drop the minus sign. So, let’s take care of the negative exponents first. \[\frac{{{n^{ - 2}}m}}{{7{m^{ - 4}}{n^{ - 3}}}} = \frac{{{m^4}{n^3}m}}{{7{n^2}}}\] Now simplify. We will use property 1 to combine the $m$’s in the numerator. We will use property 3 to combine the $n$’s and since we are looking for positive exponents we will use the first form of this property since that will put a positive exponent up in the numerator. \[\frac{{{n^{ - 2}}m}}{{7{m^{ - 4}}{n^{ - 3}}}} = \frac{{{m^5}n}}{7}\] Again, the 7 will stay in the denominator since there isn’t a negative exponent on it. It will NOT move up to the numerator with the $m$. Do not get excited if all the terms move up to the numerator or if all the terms move down to the denominator. That will happen on occasion. $\displaystyle \frac{{5{x^{ - 1}}{y^{ - 4}}}}{{{{\left( {3{y^5}} \right)}^{ - 2}}{x^9}}}$ Show Solution This example is similar to the previous one except there is a little more going on with this one. The first step will be to again, get rid of the negative exponents as we did in the previous example. Any terms in the numerator with negative exponents will get moved to the denominator and we’ll drop the minus sign in the exponent. Likewise, any terms in the denominator with negative exponents will move to the numerator and we’ll drop the minus sign in the exponent. Notice this time, unlike the previous part, there is a term with a set of parenthesis in the denominator. Because of the parenthesis that whole term, including the 3, will move to the numerator. Here is the work for this part. \[\frac{{5{x^{ - 1}}{y^{ - 4}}}}{{{{\left( {3{y^5}} \right)}^{ - 2}}{x^9}}} = \frac{{5{{\left( {3{y^5}} \right)}^2}}}{{x{y^4}{x^9}}} = \frac{{5\left( 9 \right){y^{10}}}}{{x{y^4}{x^9}}} = \frac{{45{y^ ${\left( {\displaystyle \frac{{{z^{ - 5}}}}{{{z^{ - 2}}{x^{ - 1}}}}} \right)^6}$ Show Solution There are several first steps that we can take with this one. The first step that we’re pretty much always going to take with these kinds of problems is to first simplify the fraction inside the parenthesis as much as possible. After we do that we will use property 5 to deal with the exponent that is on the parenthesis. \[{\left( {\frac{{{z^{ - 5}}}}{{{z^{ - 2}}{x^{ - 1}}}}} \right)^6} = {\left( {\frac{{{z^2}{x^1}}}{{{z^5}}}} \right)^6} = {\left( {\frac{x}{{{z^3}}}} \right)^6} = \frac{{{x^6}}}{{{z^{18}}}}\] In this case we used the second form of property 3 to simplify the $z$’s since this put a positive exponent in the denominator. Also note that we almost never write an exponent of “1”. When we have exponents of 1 we will drop them. ${\left( {\displaystyle \frac{{24{a^3}{b^{ - 8}}}}{{6{a^{ - 5}}b}}} \right)^{ - 2}}$ Show Solution This one is very similar to the previous part. The main difference is negative on the outer exponent. We will deal with that once we’ve simplified the fraction inside the parenthesis. \[{\left( {\frac{{24{a^3}{b^{ - 8}}}}{{6{a^{ - 5}}b}}} \right)^{ - 2}} = {\left( {\frac{{4{a^3}{a^5}}}{{{b^8}b}}} \right)^{ - 2}} = {\left( {\frac{{4{a^8}}}{{{b^9}}}} \right)^{ - 2}}\] Now at this point we can use property 6 to deal with the exponent on the parenthesis. Doing this gives us, \[{\left( {\frac{{24{a^3}{b^{ - 8}}}}{{6{a^{ - 5}}b}}} \right)^{ - 2}} = {\left( {\frac{{{b^9}}}{{4{a^8}}}} \right)^2} = \frac{{{b^{18}}}}{{16{a^{16}}}}\] Before leaving this section we need to talk briefly about the requirement of positive only exponents in the above set of examples. This was done only so there would be a consistent final answer. In many cases negative exponents are okay and in some cases they are required. In fact, if you are on a track that will take you into calculus there are a fair number of problems in a calculus class in which negative exponents are the preferred, if not required, form.	{"url":"https://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx","timestamp":"2024-11-11T06:37:56Z","content_type":"text/html","content_length":"90480","record_id":"<urn:uuid:0e08cc62-29fc-4264-8f38-4c7cc06816b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00774.warc.gz"}
Section Composer 4D BRIDGE SERIES - Features LARSA 4D Section Composer Tool LARSA Section Composer is a graphical companion tool for modeling arbitrary sections for use in LARSA 4D. Section Composer supports standard, parametric, nonprismatic, and custom shape sections. It is able to compute section properties in real time. Parametric Definition Sections in Section Composer are defined parametrically, meaning points are normally entered as equations of a few parameters, such width (b), depth (d), and thickness (t). Sections defined this way can be reused and resized as needed without re-computing the coordinates of control points. By simply changing a parameter, coordinates are immediately updated. Each point on a shape's perimeter is defined using mathematical equations. Equations in terms of section parameters, such as for creating the points on a rectangle (d,b), (-d,b), (-d,-b), (d,-b), make it possible to alter section dimensions without modifying each point, and to apply nonprismatic variation according to any user-enterable formula, such as d + x/100 for a variation that starts at d and increases on a linear 1:100 slope. Automatic Computation of Properties Section Composer can be used to model cross-sections with holes, composite parts, and built-up parts with any arbitrary shape. Properties including area, moment of inertia, radius of gyration, and torsion constant are all computed by Section Composer for any shape. Nonpristmatic Variation Accurate modeling of bridges requires the use of nonprismatic sections, sections whose dimensions vary along the length of the member. LARSA Section Composer makes it easy to define nonprismatic variation in sections by applying a formula to a parametric section definition. Formulas give the value of a parameter as a function of the position along the length of a span. Linear, parabolic, sinusoidal, and other types of functions can be attached to parameters, like depth, to control nonprismatic variation.	{"url":"https://www.larsa4d.com/products/features/4dbridge-section-composer.aspx","timestamp":"2024-11-14T11:16:42Z","content_type":"text/html","content_length":"19114","record_id":"<urn:uuid:ac65d330-9111-48f9-88f7-ed4a0c47ad5d>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00663.warc.gz"}
A Gentle Introduction to Monte Carlo Sampling for Probability Monte Carlo methods are a class of techniques for randomly sampling a probability distribution. There are many problem domains where describing or estimating the probability distribution is relatively straightforward, but calculating a desired quantity is intractable. This may be due to many reasons, such as the stochastic nature of the domain or an exponential number of random variables. Instead, a desired quantity can be approximated by using random sampling, referred to as Monte Carlo methods. These methods were initially used around the time that the first computers were created and remain pervasive through all fields of science and engineering, including artificial intelligence and machine learning. In this post, you will discover Monte Carlo methods for sampling probability distributions. After reading this post, you will know: • Often, we cannot calculate a desired quantity in probability, but we can define the probability distributions for the random variables directly or indirectly. • Monte Carlo sampling a class of methods for randomly sampling from a probability distribution. • Monte Carlo sampling provides the foundation for many machine learning methods such as resampling, hyperparameter tuning, and ensemble learning. Kick-start your project with my new book Probability for Machine Learning, including step-by-step tutorials and the Python To finish reading, please visit source site	{"url":"https://www.deeplearningdaily.com/a-gentle-introduction-to-monte-carlo-sampling-for-probability/","timestamp":"2024-11-08T11:53:23Z","content_type":"text/html","content_length":"154055","record_id":"<urn:uuid:74b40500-66ac-4bbd-9ad0-df6c3eca65e8>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00163.warc.gz"}
• The print method print.evpost avoids printing a long list by printing only the original function call. • The default value of inc_cens in kgaps_post() is now inc_cens = TRUE. • In the (extremely rare) cases where grimshaw_gp_mle() errors or returns an estimate for which the observation information is singular, a fallback function is used, which maximises the log-likelihood using stats::optim() • In the generalised Pareto example in the introductory vignette, it is now noted that for the Gulf of Mexico data a threshold set at the 95% threshold results in only a small number (16) of threshold excesses. • In the GP section of the introductory vignette a link is given to the binomial-GP analysis in the Posterior Predictive Extreme Value Inference vignette. • In the introductory vignette: corrected references to plots as “on the left” when in fact they were below, and corrected “random example” to “random sample”. • The microbenchmark results have been reinstated in the “Faster simulation using revdbayes” vignette. • Activated 3rd edition of the testthat package • LF line endings used in inst/include/revdbayes.h and inst/include/revdbayes_RcppExports.h to avoid CRAN NOTE. • The format of the data supplied to rpost() and rpost_rcpp() is checked and an error is thrown if it is not appropriate. • In rpost() and rpost_rcpp() an error is thrown if the input threshold thresh is lower than the smallest observation in data. This is only relevant when model = "gp", model = "bingp" or model = • The summary method for class “evpost” is now set up according to Section 8.1 of the R FAQ at (https://cran.r-project.org/doc/FAQ/R-FAQ.html). • A bug in grimshaw_gp_mle has been fixed, so that now solutions with K greater than 1 are discarded. (Many thanks to Leo Belzile.) • In grimshaw_gp_mle using the starting value equal to the upper bound can result in early termination of the Newton-Raphson search. A starting value away from the upper bound is now used (lines 282 and 519 of frequentist.R). (Many thanks to Jeremy Rohmer for sending me a dataset that triggered this problem.) • In set_prior() if prior = "norm" or prior = "loglognorm" then an explicit error is thrown if cov is not supplied. (Many thanks to Leo Belzile.) • The mathematics in the reference manual has been tidied. • The list returned from set_prior now contains default values for all the required arguments of a given in-built prior, if these haven’t been specified by the user. This simplifies the evaluation of prior densities using C++. • The GEV functions dgev, pgev, qgev, rgev and the GP functions dgp, pgp, qgp, rgp have been rewritten to conform with the vectorised style of the standard functions for distributions, e.g. those found at ?Normal. This makes these functions more flexible, but also means that the user take care when calling them with vectors arguments or different lengths. • The documentation for rpost has been corrected: previously it stated that the default for use_noy is use_noy = FALSE, when in fact it is use_noy = TRUE. • Bug fixed in plot.evpost : previously, in the d = 2 case, providing the graphical parameter col produced an error because col = 8 was hard-coded in a call to points. Now the extra argument points_par enables the user to provide a list of arguments to points. • All the (R, not C++) prior functions described in the documentation of set_prior are now exported. This means that they can now be used in the function posterior in the evdbayes package. • Unnecessary dependence on package devtools via Suggests is removed. • Bugs fixed in the (R) prior functions gp_norm, gev_norm and gev_loglognorm. The effect of the bug was negligible unless the prior variances are not chosen to be large. • In a call to rpost or rpost_rcpp with model = "os" the user may provide data in the form of a vector of block maxima. In this instance the output is equivalent to a call to these functions with model = "gev" with the same data. • A new vignette (Posterior Predictive Extreme Value Inference using the revdbayes Package) provides an overview of most of the new features. Run browseVignettes(“revdbayes”) to access. • S3 predict() method for class ‘evpost’ performs predictive inference about the largest observation observed in N years, returning an object of class evpred. • S3 plot() for the evpred object returned by predict.evpost. • S3 pp_check() method for class ‘evpost’ performs posterior predictive checks using the bayesplot package. • Interface to the bayesplot package added in the S3 plot.evpost method. • model = bingp can now be supplied to rpost() to add inferences about the probability of threshold exceedance to inferences about threshold excesses based on the Generalised Pareto (GP) model. set_bin_prior() can be used to set a prior for this probability. • rprior_quant(): to simulate from the prior distribution for GEV parameters proposed in Coles and Tawn (1996) [A Bayesian analysis of extreme rainfall data. Appl. Statist., 45, 463-478], based on independent gamma priors for differences between quantiles. • prior_prob(): to simulate from the prior distribution for GEV parameters based on Crowder (1992), in which independent beta priors are specified for ratios of probabilities (which is equivalent to a Dirichlet prior on differences between these probabilities).	{"url":"https://cran.mirror.garr.it/CRAN/web/packages/revdbayes/news/news.html","timestamp":"2024-11-03T15:17:29Z","content_type":"application/xhtml+xml","content_length":"20172","record_id":"<urn:uuid:aed95a4e-1a0b-4a94-8bf1-0f5893ba102a>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00775.warc.gz"}
Chapter 5: Exercises Short-Answer Questions, Exercises, and Problems ➢ Name and describe four cost behavior patterns. ➢ Describe two methods of determining the fixed and variable components of mixed costs. ➢ What is meant by the term break-even point? ➢ What are two ways in which the break-even point can be expressed? ➢ What is the relevant range? ➢ What is the formula for calculating the break-even point in sales revenue? ➢ What formula is used to solve for the break-even point in units? ➢ How can the break-even formula be altered to calculate the number of units that must be sold to achieve a desired level of income? ➢ Why might a business wish to lower its break-even point? How would it go about lowering the break-even point? ➢ What effect would you expect the mechanization and automation of production processes to have on the break-even point? ➢ Real world question Assume your college is considering hiring a lecturer to teach a special class in communication skills. Identify at least two costs that college administrators might consider in deciding whether to hire the lecturer and add the class. ➢ Real world question Two enterprising students are considering renting space and opening a class video recording service. They would hire camera operators to record large introductory classes. The students taking the classes would be charged a fee to rent and view the video on their laptops or smart phones. Identify as many costs of this business as you can and indicate which would be variable and which would be fixed. Exercise A Name and match the types of cost behavior with the appropriate diagram below:Exercise B Research Inc., performs laboratory tests. Use the high-low method to determine the fixed and variable components of a mixed cost, given the following observations: Volume (number of tests) Total cost 4,800 $6,000 19,200 9,600 Exercise C Compute the break-even point in sales dollars if fixed costs are $200,000 and the total contribution margin is 20% of revenue. Exercise D Barney Company makes and sells stuffed animals. One product, Michael Bears, sells for $28 per bear. Michael Bears have fixed costs of $100,000 per month and a variable cost of $12 per bear. How many Michael Bears must be produced and sold each month to break even? Exercise E Peter Garcia Meza is considering buying a company if it will break even or earn net income on revenues of $80,000 per month. The company that Peter is considering sells each unit it produces for $5. Use the following cost data to compute the variable cost per unit and the fixed cost for the period. Calculate the break-even point in sales dollars. Should Peter buy this company? Volume (units) Cost 8,000 $70,000 68,000 190,000 Exercise F Never Late Delivery currently delivers packages for $9 each. The variable cost is $3 per package, and fixed costs are $60,000 per month. Compute the break-even point in both sales dollars and units under each of the following independent assumptions. Comment on why the break-even points are different. 1. The costs and selling price are as just given. 2. Fixed costs are increased to $75,000. 3. Selling price is increased by 10%. (Fixed costs are $60,000.) 4. Variable cost is increased to $4.50 per unit. (Fixed costs are $60,000 and selling price is $9.) Exercise G Best Eastern Motel is a regional motel chain. Its rooms rent for $100 per night, on average. The variable cost is $40 a room per night. Fixed costs are $5,000,000 per year. The company currently rents 200,000 units per year, with each unit defined as one room for one night. Should this company undertake an advertising campaign resulting in a $500,000 increase in fixed costs per year, no change in variable cost per unit, and a 10% increase in revenue (resulting from an increase in the number of rooms rented)? What is the margin of safety before and after the campaign? Exercise H Fall-For-Fun Company sells three products. Last year’s sales were $600,000 for parachutes, $800,000 for hang gliders, and $200,000 for bungee jumping harnesses. Variable costs were: parachutes, $400,000; hang gliders, $700,000; and bungee jumping harnesses, $100,000. Fixed costs were $240,000. Find (a) the break-even point in sales dollars and (b) the margin of safety. Exercise I Early Horizons Day Care Center has fixed costs of $300,000 per year and variable costs of $10 per child per day. If it charges $25 a child per day, what will be its break-even point expressed in dollars of revenue? How much revenue would be required for Early Horizons Day Care to earn $100,000 net income per year? Problem A Assume the local franchise of Togorio Sandwich Company assigns you the task of estimating total maintenance cost on its delivery vehicles. This cost is a mixed cost. You receive the following data from past months: Month Units Costs March 8,000 $14,000 April 10,000 14,960 May 9,000 15,200 June 11,000 15,920 July 10,000 15,920 August 13,000 16,880 September 14,000 18,080 October 18,000 19,280 November 20,000 21,200 1. Using the high-low method, determine the total amount of fixed costs and the amount of variable cost per unit. Draw the cost line. 2. Prepare a scatter diagram, plot the actual costs, and visually fit a linear cost line to the points. Estimate the amount of total fixed costs and the amount of variable cost per unit. Problem B 1. Using the preceding graph, label the relevant range, total costs, fixed costs, break-even point, and profit and loss areas. 2. At 8,000 units, what are the variable costs, fixed costs, sales, and contribution margin amounts in dollars? 3. At 8,000 units, is there net income or loss? How much? Problem C The management of Bootleg Company wants to know the break-even point for its new line hiking boots under each of the following independent assumptions. The selling price is $50 pair of boots unless otherwise stated. (Each pair of boots is one unit.) 1. Fixed costs are $300,000; variable cost is $30 per unit. 2. Fixed costs are $300,000; variable cost is $20 per unit. 3. Fixed costs are $250,000; variable cost is $20 per unit. 4. Fixed costs are $250,000; selling price is $40; and variable cost is $30 per unit. Compute the break-even point in units and sales dollars for each of the four independent case. Problem D Refer to the previous problem. Bootleg Company’s sales are $1,100,000. Determine the margin (safety in dollars for cases (a) through (d). Problem E Using the data in the Bootleg Company problem (a through d), determine the level of sales dollars required achieve a net income of $125,000. Problem F Bikes Unlimited, Inc., sells three types of bicycles. It has fixed costs of $258,000 per month. The sales and variable costs of these products for April follow: Racing Mountain Touring Sales $1,00,000 $1,500,000 $2,500,000 Variable costs 700,000 900,000 1,250,000 Compute the break-even point in sales dollars. Problem G a. Assume that fixed costs of Celtics Company are $180,000 per year, variable cost is $12 per unit, and selling price is $30 per unit. Determine the break-even point in sales dollars. 1. Hawks Corporation breaks even when its sales amount to $89,600,000. In 2010, its sales were $14,400,000, and its variable costs amounted to $5,760,000. Determine the amount of its fixed costs. 2. The sales of Niners Corporation last year amounted to $20,000,000, its variable costs were $6,000,000, and its fixed costs were $4,000,000. At what level of sales dollars would the Niners Corporation break even? 3. What would have been the net income of the Niners Corporation in part (c), if sales volume had been 10% higher but selling prices had remained unchanged? 4. What would have been the net income of the Niners Corporation in part (c), if variable costs had been 10% lower? 5. What would have been the net income of the Niners Corporation in part (c), if fixed costs had been 10% lower? 6. Determine the break-even point in sales dollars for the Niners Corporation on the basis of the data given in (e) and then in (f). Answer each of the preceding questions. Problem H After graduating from college, M. J. Orth started a company that produced cookbooks. After three years, Orth decided to analyze how well the company was doing. He discovered the company has fixed costs of $1,200,000 per year, variable cost of $14.40 per cookbook (on average), and a selling price of $26.90 per cookbook (on average). How many units (that is, cookbooks) must be sold to break even? How many units will the company have to sell to earn $48,000? Problem I The operating results for two companies follow: Sierra Olympias Sales (20,000) units $1,920,000 $1,920,000 Variable costs 480,000 1,056,000 Contribution margin 1,440,000 864,000 Fixed costs 960,000 384,00 Net income 480,000 480,000 1. Prepare a cost-volume-profit chart for Sierra Company, indicating the break-even point, the contribution margin, and the areas of income and losses. 2. Compute the break-even point of both companies in sales dollars and units. 3. Assume that without changes in selling price, the sales of each company decline by 10%. Prepare income statements similar to the preceding statements for both companies. Problem J Soundoff, Inc., a leading manufacturer of electronic equipment, decided to analyze the profitability of its new portable compact disc (CD) players. On the CD player line, the company incurred $2,520,000 of fixed costs per month while selling 20,000 units at $600 each. Variable cost was $240 per unit. Recently, a new machine used in the production of CD players has become available; it is more efficient than the machine currently being used. The new machine would reduce the company’s variable costs by 20%, and leasing it would increase fixed costs by $96,000 per year. 1. Compute the break-even point in units assuming use of the old machine. 2. Compute the break-even point in units assuming use of the new machine. 3. Assuming that total sales remain at $12,000,000 and that the new machine is leased, compute the expected net income. 4. Should the new machine be leased? Why? Problem K Surething CD Company reports sales of $720,000, variable costs of $432,000, and fixed costs of $108,000. If the company spends $72,000 on a sales promotion campaign, it estimates that sales will be increased by $270,000. Determine whether the sales promotion campaign should be undertaken. Provide calculations. Alternate problems Alternate problem A Hear Right Company has identified certain variable and fixed costs in its production of hearing aids. Management wants you to divide one of its mixed costs into its fixed and variable portions. Here are the data for this cost: Month Units Costs January 20,800 $57,600 February 20,000 54,000 March 22,000 58,500 April 25,600 57,600 May 28,400 58,500 June 30,000 62,100 July 32,800 63,900 August 35,600 68,400 September 37,600 72,000 October 40,000 77,400 1. Using the high-low method, determine the total amount of fixed costs and the amount of variable cost per unit. Draw the cost line. 2. Prepare a scatter diagram, plot the actual costs, and visually fit a linear cost line to the points. Estimate the amount of total fixed costs and the variable cost per unit. Alternate problem B 1. Using the preceding graph, label the relevant range, total costs, fixed costs, break-even point, and profit and loss areas. 2. At 18,000 units, what would sales revenue, total costs, fixed and variable costs be? 3. At 18,000 units, would there be a profit or loss? How much? Alternate problem C Jefferson Company has a plant capacity of 100,000 units, at which level variable costs are $720,000. Fixed costs are expected to be $432,000. Each unit of product sells for $12. 1. Determine the company’s break-even point in sales dollars and units. 2. At what level of sales dollars would the company earn net income of $144,000? 3. If the selling price were raised to $14.40 per unit, at what level of sales dollars would the company earn $144,000? Alternate problem D a. Determine the break-even point in sales dollars and units for Cowboys Company that has fixed costs of $63,000, variable cost of $24.50 per unit, and a selling price of $35.00 per unit. 1. Wildcats Company breaks even when sales are $280,000. In March, sales were $670,000, and variable costs were $536,000. Compute the amount of fixed costs. 2. Hoosiers Company had sales in June of $84,000; variable costs of $46,200; and fixed costs of $50,400. At what level of sales, in dollars, would the company break even? 3. What would the break-even point in sales dollars have been in (c) if variable costs had been 10% higher? 4. What would the break-even point in sales dollars have been in (c) if fixed costs had been 10% higher? 5. Compute the break-even point in sales dollars for Hoosiers Company in (c) under the assumptions of (d) and (e) together. Answer each of the preceding questions. Alternate problem E See Right Company makes contact lenses. The company has a plant capacity of 200,000 units. Variable costs are $4,000,000 at 100% capacity. Fixed costs are $2,000,000 per year, but this is true only between 50,000 and 200,000 units. 1. Prepare a cost-volume-profit chart for See Right Company assuming it sells its product for $40 each. Indicate on the chart the relevant range, break-even point, and the areas of net income and 2. Compute the break-even point in units. 3. How many units would have to be sold to earn $200,000 per year? Alternate problem F Mr Feelds Cookies has fixed costs of $360,000 per year. It sells three types of cookies. The cost and revenue data for these products follow: Cream cake Goo fill Sweet tooth Sales $64,000 $95,0000 $131,000 Variable costs 38,400 55,100 66,000 Compute the break-even point in sales dollars. Beyond the numbers—Critical thinking Business decision case A Quality Furniture Company is operating at almost 100% of capacity. The company expects sales to increase by 25% in 2011. To satisfy the demand for its product, the company is considering two alternatives: The first alternative would increase fixed costs by 15% but not affect variable costs. The second alternative would not affect fixed costs but increase variable costs to 60% of the selling price of the company’s product. This is Quality Furniture Company’s condensed income statement for 2010: Sales $3,600,000 Variable $1,620,000 Fixed 330,000 1,950,000 Income before taxes $1,650,000 1. Determine the break-even point in sales dollars for 2011 under each of the alternatives. 2. Determine projected income for 2011 under each of the alternatives. 3. Which alternative would you recommend? Why? Business decision case B When the Weidkamp Company’s plant is completely idle, fixed costs amount to $720,000. When the plant operates at levels of 50% of capacity or less, its fixed costs are $840,000; at levels more than 50% of capacity, its fixed costs are $1,200,000. The company’s variable costs at full capacity (100,000 units) amount to $1,800,000. 1. Assuming that the company’s product sells for $60 per unit, what is the company’s break-even point in sales dollars? 2. Using only the data given, at what level of sales would it be more economical to close the factory than to operate it? In other words, at what level would operating losses approximate the losses incurred if the factory closed down completely? 3. Assume that Weidkamp Company is operating at 50% of its capacity and decides to reduce the selling price from $60 per unit to $36 per unit to increase sales. At what percentage of capacity must the company operate to break even at the reduced sales price? Business decision case C Monroe Company has recently been awarded a contract to sell 25,000 units of its product to the federal government. Monroe manufactures the components of the product rather than purchasing them. When the news of the contract was released to the public, President Mary Monroe, received a call from the president of the McLean Corporation, Carl Cahn. Cahn offered to sell Monroe 25,000 units of a needed component, Part J, for $15.00 each. After receiving the offer, Monroe calls you into her office and asks you to recommend whether to accept or reject Cahn’s offer. You go to the company’s records and obtain the following information concerning the production of Part J. Costs at current production level (200,000 units) Direct labor $1,248,000 Direct materials 576,000 Manufacturing overhead 600,000 Total cost $2,424,000 You calculate the unit cost of Part J to be $12.12 or ($2,424,000/200,000). But you suspect that this unit cost may not hold true at all production levels. To find out, you consult the production manager. She tells you that to meet the increased production needs, equipment would have to be rented and the production workers would work some overtime. She estimates the machine rental to be $60,000 and the total overtime premiums to be $108,000. She provides you with the following information: Costs at current production level (225,000 units) Direct labor $1,404,000 Direct materials 648,000 Manufacturing overhead (including equipmental rental and overtime premiums) Total cost $2,880,000 The production manager advises you to reject Cahn’s offer, since the unit cost of Part J would be only $12.80 or ($2,880,000/225,000 units) with the additional costs of equipment rental and overtime premiums. This amount still is less than the $15.00 that Cahn would charge. Undecided, you return to your office to consider the matter further. 1. Using the high-low method, compute the variable cost portion of manufacturing overhead. (Remember that the costs of equipment rental and overtime premiums are included in manufacturing overhead. Subtract these amounts before performing the calculation). 2. Compute the total costs to manufacture the additional units of Part J. (Note: include overtime premiums as a part of direct labor.) 3. Compute the unit cost to manufacture the additional units of Part J. 4. Write a report recommending that Monroe accept or reject Cahn’s offer. Business decision case D Refer to the “A broader perspective: Major television networks are finding it harder to break even” discussion of cost-volume-profit analysis for television networks. Write a memo to your instructor describing how the networks can reduce their break-even points. Group project E In teams of two or three students, develop a cost-volume-profit equation for a new business that you might start. Examples of such businesses are a portable espresso bar, a pizza stand, a campus movie theater, a package delivery service, a campus-to-airport limousine service, and a T-shirt printing business. Your equation should be in the form: Profits = (Price per unit X Volume) – (Variable cost per unit X Volume) – Fixed costs per period. Pick a period of time, say one month, and project the unit price, volume, unit variable cost, and fixed costs for the period. From this information, you will be able to estimate the profits—or losses—for the period. Select one spokesperson for your team to tell the class about your proposed business and its profits or losses. Good luck, and have fun. Group project F Refer to “A broader perspective: Even colleges use CVP” discussion of how cost-volume-profit analysis is used by colleges. In teams of two or three students, write a memo to your instructor defining step costs and explain why the step costs identified in the case are classified as such. Also include in your memo how the school might lower its break-even point. Group project G In teams of two or three students, address the following questions: • Why would a company consider increasing automation and decreasing the use of labor if the result would be an increase in the break-even point? • Would an increase in automation increase fixed costs over the short-run, long-run, or both? Write a memo to your instructor that addresses both questions. Be sure to explain your answers. Using the Internet—A view of the real world Visit the website for Intel Corporation, a high technology manufacturing company. Go to the company’s most recent financial statements and review the consolidated statement of income. What additional information, if any, would you need to perform cost-volume-profit analysis? Why is this information excluded from Intel’s income statement? Visit the website for Wal-Mart Corporation, a retail company. Go to the company’s most recent financial statements and review the statement of income. What additional information, if any, would you need to perform cost-volume-profit analysis? Why is this information excluded from Wal-Mart Corporation’s income statement? level (225,000 units)Direct labor$1,404,000Direct materials648,000 Manufacturing overhead (including equipmental rental and overtime premiums) 828,000Total cost$2,880,000 The production manager advises you to reject Cahn’s offer, since the unit cost of Part J would be only $12.80 or ($2,880,000/225,000 units) with the additional costs of equipment rental and overtime premiums. This amount still is less than the $15.00 that Cahn would charge. Undecided, you return to your office to consider the matter further. 1. Using the high-low method, compute the variable cost portion of manufacturing overhead. (Remember that the costs of equipment rental and overtime premiums are included in manufacturing overhead. Subtract these amounts before performing the calculation). 2. Compute the total costs to manufacture the additional units of Part J. (Note: include overtime premiums as a part of direct labor.) 3. Compute the unit cost to manufacture the additional units of Part J. 4. Write a report recommending that Monroe accept or reject Cahn’s offer.	{"url":"https://courses.lumenlearning.com/suny-managacct/chapter/unit-3-exercises/","timestamp":"2024-11-11T01:50:48Z","content_type":"text/html","content_length":"72252","record_id":"<urn:uuid:b9013066-4621-478e-b700-4afac471edae>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00860.warc.gz"}
Missing tests As public inputs for proofs verified by the universal batch verifier are padded with zero, Finding ref↗ would have been caught by a test case where a proof in the batch has no public inputs. However, the code-sampling test cases did not cover this edge case (extract from circuits/src/batch_verify/universal/types.rs): /// Samples a [`UniversalBatchVerifierInput`] for `config`. pub fn sample<R>(config: &UniversalBatchVerifierConfig, rng: &mut R) -> Self R: RngCore + ?Sized, let num_public_inputs = config.max_num_public_inputs as usize; let length = rng.gen_range(1..=num_public_inputs); let (proofs_and_inputs, vk) = sample_proofs_inputs_vk(length, 1, rng); let (proof, inputs) = proofs_and_inputs[0].clone(); assert!(length > 0); assert!(length + 1 == vk.s.len()); assert!(length == inputs.0.len()); Self { vk, proof, inputs } As can be seen here, only cases with at least one public input are tested. We recommend to change rng.gen_range(1..=num_public_inputs); to rng.gen_range(0..=num_public_inputs); to also cover the case of no public inputs. Testing only with randomly generated inputs generated as above may miss special edge-case values with low density such as a public input being zero (a random field element is extremely unlikely to be zero) or parameters occurring more than once. We thus recommend to also include test cases with such human-defined edge cases. Testing for all public inputs being zero was added in commits , , and .	{"url":"https://reports.zellic.io/publications/universal-proof-aggregator/sections/missing-tests-missing-tests","timestamp":"2024-11-12T17:17:42Z","content_type":"text/html","content_length":"68387","record_id":"<urn:uuid:a901ef24-266b-4c97-88e7-262ca887d8fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00556.warc.gz"}
describe pi as the ratio of the circumference of a circle to its diameter The circumference of a circle is the distance around that circle. But what is the formula to find the circumference? In this tutorial, you'll learn the formulas for the circumference of a circle. Take a look!	{"url":"https://virtualnerd.com/texasteks/texas-digits-grade-7/use-geometry-to-describe-or-solve-problems-involving-proportional-relationships/describe-%C3%8F%E2%82%AC-as-the-ratio-of-the","timestamp":"2024-11-13T12:37:40Z","content_type":"text/html","content_length":"16253","record_id":"<urn:uuid:c227b982-f097-442f-8428-e0b8a1083d5e>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00868.warc.gz"}
I originally put this on AxiomWiki created by Bill Page but since that has disappeared I moved it here. Bertfried Fauser wrote: Does AXIOM have monads (like Haskell)? Sorry not much time to investigate myself. Such things can be implemented very efficiently and in an encapsulated way by using monads (in Haskell). Formal definition If C is a category, a monad on C consists of a functor $T \colon C \to C$ together with two natural transformations: $\eta \colon 1_{C} \to T$ (where $1_C$ denotes the identity functor on C) and $\mu \colon T^{2} \to T$ (where $T^2$ is the functor $T \circ T$ from C to C). These are required to fulfill the following conditions (sometimes called coherence conditions): \mu \circ T\mu = & \mu \circ \mu T & \textrm{(as natural transformations } T^{3} \to T); \\ \mu \circ T \eta = & \mu \circ \eta T = 1_{T} & \textrm{(as natural transformations } T \to T; \\ & & \textrm{here } 1_T \textrm{ denotes the identity transformation from T to T).} We can rewrite these conditions using following commutative diagrams: & T^3\ar[r]^{T\mu}\ar[d]_{\mu T}&T^2\ar[d]^{\mu}\\ & T^2\ar[r]_{\mu}&T } & & \ar@{=}[dr]T\ar[r]^{\eta T}\ar[d]_{T\eta}&T^2\ar[d]^{\mu}\\ & T^2\ar[r]_{\mu}&T Ref: http://en.wikipedia.org/wiki/Monad_%28category_theory%29 On Fri, Nov 4, 2011 at 5:04 AM, Martin Baker wrote: As far as I can see there is already an SPAD category called Monad but its not really what we are talking about here? I have been thinking about what it would take to implement a monad in SPAD, I have been doing this by taking 'List' as being an example of a monad and then trying to generalize it. First test for List being monad (monad for a monoid) with: $\mu$ = 'concat', $\eta$ = 'list', and $T$ = [ ]. Here T and $\eta$ are effectively the same but I've used a different form 'list' and [ ] to distinguish between them. They are the same, in this case, because we are working in terms of components. It would be good to also be able to work in terms of functors and natural transformations, Haskell would do this because of automatic currying but with SPAD I think So I will check that the identities for monads are satisfied. First something to work with: T2a := [[A],[B]] T2b := [[C],[D]] then test associative square: concat[concat T2a,concat T2b] = concat(concat[T2a,T2b]) then test unit triangles: concat(list[A]) = [A] concat[list A] = [A] So this seems to have the properties needed, I would like to build a general monad with these properties. Haskell uses monads extensively so I thought that might provide an example of how it could be coded? Ref: http://en.wikipedia.org/wiki/Monad_%28functional_programming%29 In Haskell the monads in the prelude is in a different form as its purpose is to allow impure code to be wrapped in pure code. I have modified the Monad definition in the prelude to have a more category theoretic form as follows:: class Monad m where unit :: a -> m a -- usually called 'return' in Haskell mult :: m (m a) -> m a -- usually called 'join' in Haskell How could we write this as an SPAD category? I've tried to do a direct translation from a Haskell typeclass to an SPAD category (except I have changed 'm' to 'T' to give a more mathematical )abbrev category MONADC MonadCat MonadCat(A:Type,T: Type) : Category == with unit :A -> T A mult :T ( T A) -> T A It compiles but it doesn't look very useful? Everything is just a general type so there isn't any proper type checking and I would like the category to have some internal type structure so that we can relate it to the external functors and natural transformations. There are two axiom-categories that we need to define: - T - 'MonadCat' We can think of T as being an endofunctor on 'MonadCat' but also we want to turn things around by defining 'MonadCat' in terms of T, that is we start with a base 'type' and extend it by repeatedly applying T to it. So we want to define T as an endofunctor: T:\textrm{MonadCat}-> \textrm{MonadCat} But we also want to define T as a axiom-category which can be extended to domains which can then define our monad, for instance, if we want to define a list-monad (monad for a monoid) then we start by defining a list-t like this:: TList(A:Type) : Exports == Implementation where Rep := List A we can then define the list monad as:: MonadList(A:Type,T:TList Type) : Exports == Implementation where Rep := Union(T A,T %) Here is the code to represent the T ''endofunctor'' for a list which we want to generate a list monad. )abbrev domain TLIST TList TList(A:Type) : Exports == Implementation where Exports == CoercibleTo(OutputForm) with a:(a:List A) -> % toList:(a:%) -> List A baseType:(a:%) -> Type Implementation ==> Type add Rep := PrimitiveArray A a(a:List A):% == construct(a)$(PrimitiveArray A) toList(a:%):List A == entries(a::Rep)$(PrimitiveArray A) baseType(a:%):Type == A coerce(n: %):OutputForm == e := entries(n) if #e=1 then if A has CoercibleTo(OutputForm) then return (first e)::OutputForm return e::OutputForm We then use this to 'generate' a list monad: )abbrev domain MONADL MonadList MonadList(A:Type) : Exports == Implementation where Exports == CoercibleTo(OutputForm) with t:(inpu:TList(%)) -> % t:(inpu:List(%)) -> % unit:(inpu:TList(A)) -> % unit:(inpu:List(A)) -> % unit:(inpu:A) -> % mult:(inpu:%) -> % Implementation ==> Type add Rep := Union(leaf:TList(A),branch:TList(%)) ++ holds type of T A, T T A or T T T A... t(inpu:TList(%)):% == t(inpu:List(%)):% == unit(inpu:TList(A)):% == unit(inpu:List(A)):% == unit(inpu:A):% == mult(inpu:%):% == if inpu case leaf then error "cannot apply mult here" lst:List % := toList(inpu.branch) lst2:List List % := [toList(a.branch) for a in lst] lst3 := concat(lst2) coerce(n: %):OutputForm == if n case leaf e := toList(n.leaf) if #e=1 then if A has CoercibleTo(OutputForm) then return (first e)::OutputForm return e::OutputForm lst:List % := toList(n.branch) lst2 := [a::%::OutputForm for a in lst] if #lst2 = 1 then return hconcat("T "::OutputForm,(first lst2)::OutputForm) return hconcat("T "::OutputForm,lst2::OutputForm) It might be better expressed if we had dependant variables but I think we can do it this way: T2a := t[t[unit[A::Symbol]],t[unit[B::Symbol]]] T2b := t[t[unit[C::Symbol]],t[unit[D::Symbol]]] Going one way round the associativity square gives: mult(t[mult T2a,mult T2b]) Going the other way round the associativity square gives the same Also the identity triangle: From MartinBaker Sat Nov 5 03:02:52 -0700 2011 From: Martin Baker Date: Sat, 05 Nov 2011 03:02:52 -0700 Subject: further thoughts Message-ID: <20111105030252-0700@axiom-wiki.newsynthesis.org> I find myself wanting a way to define types inductively like this:: X := Union(A,T X) so X is an infinite sequence of types: A \/ T A \/ T T A \/ ... A:Type -- base type T:X -> X -- endofunctor but this is a circular definition, in order to break this circle perhaps we could start by defining T as:: T: Type->Type and then constraining it to T:X -> X in concrete implementations. Can anyone suggest a variation of the following that would be valid in SPAD: )abbrev category MONADC MonadCat MonadCat(A:Type,T: Type->Type) : Category == with X ==> Union(A,T X) unit :X -> T X mult :T ( T X) -> T X So if we take a concrete instance of List then:: X := Union(A,List A,List List A, List List List A ...) I get the impression that there is a hack in the List implementation that allows this? That is, when checking list types, the compiler does not look too deeply into the nested types? Perhaps if lists were implemented as monads such hacks would not be needed? The above is done purely in terms of types/categories which would be the ideal way to implement monads but perhaps there is a less powerful but more easily implemented way to implement monads in terms of domains. In this case representations can be implemented inductively (co-data types), in category theory terms perhaps this is like working in terms of components? So List is a monad with mult and unit defined as follows:: List(S: Type) is a domain constructor Abbreviation for List is LIST This constructor is exposed in this frame. ------------------------------- Operations -------------------------------- concat : List(%) -> % -- mult list : S -> % -- unit From MartinBaker Sat Nov 5 09:05:46 -0700 2011 From: Martin Baker Date: Sat, 05 Nov 2011 09:05:46 -0700 Subject: From reply on FriCAS forum by Bertfried Fauser Message-ID: <20111105090546-0700@axiom-wiki.newsynthesis.org> (MJB)I would like to keep all the relevant information together so I have taken the liberty of copying this reply to here - marked (BF) and added some comments myself (MJB). If this page could be worked up into a full tutorial then perhaps this information could be incorporated into the main text. (BF)First I tried to find where MONAD is defined in FriCAS, unfortunately the command (2) -> )show Monad Monad is a category constructor Abbreviation for Monad is MONAD This constructor is exposed in this frame. ------------------------------- Operations -------------------------------- ?? : (%,%) -> % ?=? : (%,%) -> Boolean ?^? : (%,PositiveInteger) -> % coerce : % -> OutputForm hash : % -> SingleInteger latex : % -> String ?~=? : (%,%) -> Boolean leftPower : (%,PositiveInteger) -> % rightPower : (%,PositiveInteger) -> % Does not say in which file the source code is (so greping is needed), it seems to be defined and used (in algebra) only in `naalgc.spad.pamphlet'. There it says: ++ Monad is the class of all multiplicative monads, i.e. sets ++ with a binary operation. and trouble starts. This is not what I understand of an monad, though they are related to monoids, but would better be understood as endofunctors in a category (or as a type constructor) Math: A monad (triple, standard construction) (T,mu,eta) is a triple consisting of an endo_functor_ T : C --> C on a category C, a multiplication m : T T ---> T and a unit eta : C --> TC such that eta is the unit, and mu is associative (defined by some diagrams) Let X be an object in C, then the natural transformations mu and eta have components mu_X and eta_X, which you seem to address (and the MONAD thing in AXIOM), this may be instantiated as a monoid (see the book by Arbib_Michael-Manes_E Arrows_structures_and_functors. _The_categorical_imperative-Academic_Press(1975)) which has lots of examples (power set, list, ...) A monad as in Haskel operates on the category (including function types) (MJB) I get the impression that the nomenclature for monads has changed over the years. Since what is currently in the Axiom code does not conform to modern conventions I suggest its name is changed to something less confusing. > First test for List being monad (monad for a monoid) with: > \mu = concat > \eta = list > T = [] (BF) ?? Monad =\= monoid. Math: A monoid is a set X with two operations mu : X x X --> X and unit E s.t. mu (E x X) = X = mu(X x E). (BF)I don't see why this should come with an `fmap' function (functoriality of T) (MJB) I have removed it in the page above. (BF) ListMonad(Set): This is actually a monad related to power sets, but lists also have order, while sets have not. Let X be a set, eta(X)= [X] is the singleton list containing this set, eta : Set --> T Set, where T Set is lists of sets. Now T T Set is lists of lists of sets, eg, [[X,Y],[Y],[X],[Z,W]] :: T T Set ; then the `multiplication' mu : T T --> T is `flatening the lists' e.g. forgetting the inner list's brackets -> [ X,Y,Y,X,Z,W] Associativity of m guarantees that if you have lists of lists of lists of sets (type T T T Set) then either of the two ways to forget the inner brackets (multiplying the monad functor T) gives the same result. The left/right unit of this multiplication is the empty list mu : [[],[X]] --> [X] The difference is that this construction is much more like a template class in C++, as it can be instantiated on any (suitable) category C. And this is what the list constructor actually does... > So I will check that the identities for monads are satisfied Yes, for lists that follows from the above definition... > )abbrev category MONADC MonadCat > MonadCat(A:Type, B:Type,T: Type) : Category == with > fmap :(A -> B) -> (T A -> T B) > unit :A -> T A > mult :T ( T A) -> T A > @ Indeed, A and B should be objects in the base category, so they have the same `type' (above A, B :: Set) . The A-->B is in Hom(A,B) = Set(A,B), the class of all functions from set A to set B, fmap allows to apply f inside the container T, and mult is actually used to flatten the container as types of type T T Set (say) appear under composing maps, as in unit o unit: A --> T T A > It compiles but it doesn't look very useful? It just defined what a monad is (provided you make sure in an implementation that your functor fulfils the correct associativity and unit laws. AXIOM checks in MONADWU (Monad with unit, ?) recip: % -> Union(%,"failed") ++ recip(a) returns an element, which is both a left and a right ++ inverse of \spad{a}, ++ or \spad{"failed"} if such an element doesn't exist or cannot ++ be determined (see unitsKnown). if such a unit can be found (has been given?) > We can think of T as being an endofunctor on MonadCat but also we want > to turn things around by defining MonadCat in terms of T, that is we > start with a base 'type' and extend it by repeatedly applying T to it. T (as a monad) is an endofunctor on `some' category, not on MonadCAT which is more or less a collection of monads (functor category of monads and monad law preserving natural transformations) > So we want to define T as an endofunctor: > \item T:(MonadCat)-> MonadCat But not like this. Think of T as a container which produces a new type out of an old Set --> Powerset Power set or Manes monad Set --> List of sets List monad Int --> List of ints List monad Measurable Spaces --> Measures on Measurable Spaces Giry monad etc (Manes has lots of examples) > TList(A:Type) : Exports == Implementation where > ... > Rep := List A Yes, much more like this. I cannot comment more on the later parts of your message. Note that monad could be much more complicated functors than 'list' T X = ( X x O)^I where you think of I as inputs, O as outputs and X a a state. But you could have trees, binary trees, labelled trees, ..... (MJB) If this page were worked into a full tutorial then I would like to see it restructured so that there is more theory at the start and then there are several worked examples like those here in addition to the list example. From MartinBaker Sat Nov 5 09:56:01 -0700 2011 From: Martin Baker Date: Sat, 05 Nov 2011 09:56:01 -0700 Subject: Algebras and Monads Message-ID: <20111105095601-0700@axiom-wiki.newsynthesis.org> I think it would be good if this page could explain why monads are important to include in a CAS, for me (MJB), this has something to do with the link between algebras and monads. I will make an attempt at explaining this and perhaps those more knowledgeable than me could correct any errors I make, A monad gives an algebraic theory. (here we are using the word 'theory' as in: 'group theory') An algebra for a monad is a model for that theory. (model theory is the study of mathematical structures using tools from mathematical logic.) An algebra takes an expression like this:: 3(9+4) - 5 and returns a value. This can be thought of as a monad where the endofunctor 'T' is the combination of the signatures of the operations. An 'expression' can be constructed inductively from 'terms'. For instance, in the simple case of natural numbers, we could define a expression like this:: <Expression> ::= <Num> , <Expression> + <Expression> <Num> ::= zero() , succ(<Num>) This abstract syntax of a context-free language defines a free algebra, we can add the laws of addition to define the natural numbers. So a term can contain a number, or other terms, or other terms containing other terms, and so on. We could think of this as a hierarchy of types: - N - T N - T T N - ... - N is a numerical value. - T N is a term containg a numerical value. - T T N is a term containing a term containing a numerical value. - ... So 'T' is an endofunctor which raises the level of the term in the expression. In order to construct and evaluate these expressions we need two natural transformations: - \eta:N -> T N - \mu:T T N -> T N - \eta wraps an element in a term. - \mu takes a term containing other terms and returns a term. \eta is sometimes called the 'unit' although it will wrap any number, not just '1'. In fact it will take any term and raise its level. \mu is sometimes called the 'multiplication' although this does not refer to the operation in the algebra being modelled so its not just multiplication or even just any binary operation, this could represent an operation of any 'arity'. It is possible to extend the standard category theory to 'multicategories', these are the same as categories except that instead of single arrow we can have 'arrows' with multiple inputs. This allows us to split up the function signature into individual functions. So if we want to add 2 to 3 to find the result, they both have type 'N' so we must first we must first make them terms: - \eta 2 - \eta 3 We can then construct (\eta 2 + \eta 3) which has type of T T N We can then apply \mu which evaluates this to give a term: \eta 5 We donâ€™t have a mechanism to unwrap this from a term and return a number. This is because pure F-algebras and inductive data types are 'initial' or 'constructive' that is they build these constructions but they are not designed to read out the data. Of course, in a practical program, it is easy to add a function to read out the value. However when we are modelling these structures from a theoretical point of view, then there might be advantages to working with a pure F-algebra or F-coalgebra. \section{Coalgebras and Coinductive types} Algebras are initial and we can construct them but not necessarily read them out. Coalgebras are the reverse, that is they are final and we can read them out but not construct them. Coalgebras and coinductive types are useful for: - Represent free algebras. - Infinite data types such as streams, infinite lists and non-well-founded sets. - Dynamical systems with a hidden, black-box state space, finite automita. - Bisimulations. - Objects with 'hidden' (indirectly available via methods) data. There are similarities between coinductive types and object oriented programming in that it is usually considerd good practice in OOP that the data (state space) is kept hidden from direct view and so data is only read using the objects methods (functions). A general CAS, such as FriCAS, is very rich and does not restrict users to F-algebras or F-coalgebras, or even have these concepts. A general CAS allows any function signature (although multiple outputs may need to be wrapped in a Record or a List) since, in most cases we need to construct and observe data. However in order to model these structures it would be useful to have the concept of a pure F-algebra or F-coalgebra. Applications might include: - converting between \lambda-calculus/combinators and a finite automata model of computation. - converting between functional programming and object oriented programming. - modelling algebraic, and co-algebraic structures. - modelling monads and comonads. We can convert between algebras and coalgebras with catamorphism (generalization of the concept of 'fold') anamorphism \section{Object Oriented Programming} (just checking if Ralf has read this far) So far we have: - Algebras: which allow us to construct instances of structures but not read them out. - Coalgebras: which allow us to read out structures but not construct instances of them. In a CAS it would be useful to be able to do both. Object Oriented Programming (OOP) is closer to coalgebra in that its structure is fixed when the programmer defines the class and can't be changed at runtime. Only the values of elements in the structure can be written and read at runtime. So how can OOP represent an algebra? Well the object does not represent the whole algebra, it represents a single element of the algebra. However, although an object contains only a single element of the algebra it knows about the structural framework of the whole algebra, so it knows how to link to other elements to form the whole algebra. A set of operator symbols, each operator has an 'arity' and all operators act only on the elements of the algebra. \section{F-Algebra and F-Coalgebra} F-Algebras and F-Coalgebras are the same as \Omega-(co)algebras except we denote all the operator signatures by a single functor. If I may misuse Axiom/FriCAS notation (why stop now? I hear you say) then I will notate this as: Poly := () } n0 times /\ (%) } n1 times /\ (%,%) } n2 times /\ (%,%,%) } n3 times So we can use this to define a free, pure algebra and a pure coalgebra as follows: UniversalAlgebra() : Metacategory == with Poly % -> % UniversalCoalgebra() : Metacategory == with % -> Poly % These represent free algebras and a particular instance can be formed by choosing values for the 'components': n0,n1,n2... F-algebras and F-coalgebras are themselves categories. They can also be defined over an arbitrary category '%' where 'Poly % -> %' is an endofunctor from % to itself. On their own pure algebra or coalgebra would not be much use, they don't have a way to interact with other algebras, but perhaps we could add that later. \section{Monads and T-Algebras} A T-Algebra is an F-Algebra, as defined above, together with a set of equations (axioms) built from the F-Algebra operations. So we have the endofunctor but what are the two natural transformations to construct a monad from it? Let T be a (finitary) equational theory, consisting of finitely many operational symbols, each of some arity and a set of equations between terms built from these operations. How do we specify operations (n-ary functions) in a T-algebra? We can use lambda calculus, or equivalently combinators. A requirement is that we need to be able express free algebras, as well as closed algebras, and algebras in-between. That is, some terms will simplify and some won't. \section{Algebras and Coalgebras from Adjoints} Algebraic constructs A method where one algebraic entity is constructed from another. A (co)algebra is defined 'over' another algebra and ultimately we get back to sets. For instance: - Complex numbers might be defined over real numbers. - Natural numbers might be defined over sets. - Monoid over sets: algebra=lists. - small category over graph. This relationship between a (co)algebra and its elements is an adjunction: F (or U) =forgetful algebra elements G (or F) =free Every adjunction gives rise to an monad: Given an adjunction F -\| G then GF=T for a monad where: - \eta:1c -> GF - \epsilon:FG -> 1d This relates to the natural transformations for monads as follows:: - unit for monad: \eta:1 -> T = \eta for adjunction - multiplication for monad:\mu:T T -> T = G \epsilon F Every monad gives rise to a class of adjunctions This generates a whole category of monads for each algebra with extremes (canonical forms being) being: - Eilenberg-Moore : In this case the monad is the terminal monad that is it represents the algebra - Kleisli : In this case the monad is the initial solution so it is the algebra of free algebras. In the Eilenberg-Moore case the algebra is a category with objects represented by a pair: (X,Ex: TX->X) X is the object of the elements T is the signature of the F-algebra (see below) this is the free functor applied to the forgetful functor (T=FU) \section{(Co)Algebra Example 1 - Natural Numbers} Natural numbers give the simplest algebra, that is, they are initial in the category of algebras. It can be represented as an algebra or a coalgebra as follows: - Algebra - is represented as an infinite set with a binary operation \verb{+:(%,%)-> %} - Coalgebra - is represented a two functions: \verb{zero:()-> %} \verb{succ:(%)-> %} Using the coalgebra representation of natural numbers would not be very practical as it would not be realistic to support the level of function nesting to represent large numbers. But from a theoretical point of view it might be useful to have natural numbers which are not defined over sets and if large literal values are not required it might have some use? \section{(Co)Algebra Example 2 - Groups} A group is traditionally defined as an algebra, that is as operations+equations defined over a set: - nullary operation: \verb{e:()-> %} - unary operation: \verb{inv:(%)-> %} - binary operation: \verb{:(%,%)-> %} equations defined in terms of arbitrary elements of the set:: - \forall x,y,z. x(yz)=(xy)z - \forall x. xinv(x)=e - \forall x. ex=x - \forall x. inv(x)x=e - \forall x. x*e=x Not all these equations are needed (there is some redundant information here) but there is no canonical form. There are other problems with defining an algebra in an algebraic way such as: - its defined over a set and we may want to define groups over other categories (known as group objects). - this type of definition may not scale up well, especially if we define multiplication for a specific group using Cayley table. Question - how does this relate to permutation groups? A permutation group looks to me like an endofunctor (viewing a group as a kind of category with just one object) and monad is related to monoid. Group defined as a co-algebra: Associativity:\mu\cdotp(id x \mu) = \mu\cdotp(\mu x id) Identity :\mu\cdotp(\eta x id) = id = \mu\cdotp(id x \eta) Inverse: \mu\cdotp(id x \delta)\cdotp\Delta = \eta\cdotp\epsilon = \mu\cdotp(\delta x id)\cdotp\Delta $\cdotp$ = function composition x = Cartesian product $\mu$ =multiplication $\eta$ = $\Delta$ = diagonalise $\delta$ = $\epsilon$ =	{"url":"http://www.euclideanspace.com/prog/scratchpad/spad/topics/category/monad/index.htm","timestamp":"2024-11-10T12:19:39Z","content_type":"text/html","content_length":"45022","record_id":"<urn:uuid:730e9636-77db-48e8-ab98-0352d14bea2d>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00401.warc.gz"}
S&P 500 Elliott Wave Technical Analysis – 14th September, 2012 by Lara \| Sep 16, 2012 \| S&P 500, S&P 500 + DJIA \| 3 comments Last analysis expected more upwards movement. The target at 1,469 has been exceeded by 5.51 points. Again, a careful analysis of structure is required to indicate whether or not this upwards trend is over. Today, I have one daily and one hourly charts for you. Click on the charts below to enlarge. Upwards movement is unlikely to be over at this stage. The structure within the final upwards wave is incomplete. At 1,513 wave v pink would reach 1.618 the length of wave iii pink. Also at 1,513 wave C blue would reach 1.618 the length of wave A blue. At primary degree the structure unfolding is an expanded flat correction. Within this structure the maximum common length of primary wave B would be about 1,464 which is 138% the length of primary wave A. Price is right at this point today. Within wave (Z) black wave A blue was a leading contracting diagonal. Wave B blue was a brief zigzag. Wave C blue is exhibiting alternation with wave A blue and unfolding as an impulse. Within wave C blue wave i pink lasted 4 sessions and wave iii pink lasted 13 sessions. I would expect wave v pink to last between 4 and 13 sessions. If wave v pink is to have a Fibonacci duration it may end after 1 more session, lasting a Fibonacci 8 sessions. The next Fibonacci number in the sequence would be 13, this would be met in another 6 sessions. When the parallel channel about wave (Z) black is clearly breached by downwards movement then we shall have confirmation of a trend change. Keep drawing the parallel channel about wave C blue. Draw the first trend line from the highs of i to iii pink, place a parallel copy upon the low of wave ii pink. Wave v pink may find resistance here at the upper edge of the channel, and it looks like we are seeing a small overshoot. Within wave (v) green wave v orange is yet to unfold. Ratios within wave iii orange are: wave 3 purple is 1.41 points longer than 4.236 the length of wave 1 purple, and wave 5 purple is 1.07 points longer than 2.618 the length of wave 1 purple and 1.95 points short of 0.618 the length of wave 3 purple. Wave (iii) green is 3.04 points longer than 2.618 the length of wave (i) green. With a Fibonacci ratio already exhibited within wave v pink I am not expecting to see one between wave (v) green and either of waves (i) or (iii) green. Within wave (v) green there is no Fibonacci ratio between waves iii and i orange. If wave iv orange has ended and price does not move it lower then at 1,503 wave v orange would reach equality in length with wave iii orange. When markets open on Monday then any further downwards extension of wave iv orange may not move into wave i orange price territory. This wave count is invalidated with movement below 1,437.76. The green parallel channel drawn here is a best fit. We may see wave v orange end about the upper edge of it. When this channel is clearly breached with downwards movement we shall have an early indication of a trend change. 3 Comments Karen on September 16, 2012 at 10:14 pm wave iii orange looks like my Major [3] top 1474.51 … looking for [4] 1419.15 then [5] = 1.62* [1] => 1603 Wei Zhu on September 16, 2012 at 3:44 pm You said the Maximum length of wave b is 1464, but your latest target is 1503, how to explain it? Thanks. Lara on September 17, 2012 at 1:23 am Yes I did. It is the maximum common length, it is not a point beyond which the count is invalidated. However, I am looking at the monthly charts and thinking it’s time now to swap over to the alternate. The expanding triangle scenario is looking pretty good. I’ll be updating this tomorrow.	{"url":"https://elliottwavestockmarket.com/2012/09/16/sp-500-elliott-wave-technical-analysis-14th-september-2012/","timestamp":"2024-11-09T03:19:52Z","content_type":"text/html","content_length":"42216","record_id":"<urn:uuid:5b1c274d-1593-4377-93ad-410d906d8bce>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00535.warc.gz"}
Quantification and Minimization of Crosstalk Sensitivity in Networks Dionysios Barmpoutis, Richard M Murray arXiv, q-bio.MN Crosstalk is defined as the set of unwanted interactions among the different entities of a network. Crosstalk is present in various degrees in every system where information is transmitted through a means that is accessible by all the individual units of the network. Using concepts from graph theory, we introduce a quantifiable measure for sensitivity to crosstalk, and analytically derive the structure of the networks in which it is minimized. It is shown that networks with an inhomogeneous degree distribution are more robust to crosstalk than corresponding homogeneous networks. We provide a method to construct the graph with the minimum possible sensitivity to crosstalk, given its order and size. Finally, for networks with a fixed degree sequence, we present an algorithm to find the optimal interconnection structure among their vertices.	{"url":"https://murray.cds.caltech.edu/index.php?title=Quantification_and_Minimization_of_Crosstalk_Sensitivity_in_Networks&oldid=19765","timestamp":"2024-11-07T03:29:28Z","content_type":"text/html","content_length":"18655","record_id":"<urn:uuid:d4e2d797-d57c-4f0d-be87-568e17a771a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00415.warc.gz"}
What’s Ag in Physics? What is A-g in Physics? If you think about doing it it’s simple. The simple idea is the fact that the curvature of a curved floor (the item ) can be really just a function of the amount of lines that could be drawn on such an surface and the area of the curvature (the quantity). The outside is that the actual object, but forms and also the curves are created by the notion of applying equations which paraphrase website mla are essential for solving troubles. In this situation, I’m discussing mathematics while in the sort of physics. In other words, a way to clarify how character functions. To know what is going on this, consider that the wonder of what is just really a square foot. This is among those questions with out understanding what is happening which individuals ask. At the most basic levelwe are able to think of this asking what’s the design of the square having a little bit of additional square»surplus» paraphraseexample.com on one aspect. In a world of differences among a flat and a curved world, the that I could shift. Inside this case, it can also differ from truly being a G to some G-squared block. It’s the location in the curve that has shifted, maybe not the shape of this curve. This really is really pretty much everything that is happening inside the rest of the item. Even the block that is G-squared is an instance of what’s called a kind of the G-space. It’s similar to two dimensional diagram that refers to a 3 dimensional system. We can adjust the way in which it behaves When we keep changing the job of the object and its own orientation. Thus, in this circumstance, it truly is the geometry, or the»shape» of this thing, and that is shifting. What is G is linked to what is called a differential equation. In this case, I am referring to the role of gravity. What’s essential is a way and the way it has an effect on objects. By way of example, once an item in some point goes away from your brute force, http://somvweb.som.umaryland.edu/absolutenm/templates/?a=3245 it will experience a downward flow. That which we need is ways to decrease this down flow to your line. What is really a G in Physics is similar to this»proper» definition of that which is G in the equation of movement. We now can modify if we adjust the position of the item and its orientation. Thus it really is the position of the thing that’s currently shifting. This example might show that there’s difference between what’s going on in a tangible thing and what is happening at a equation. It can also show that when we’re to comprehend something just like the cube, we must understand what’s going on from the equation that is physiological. There is a Connection between Them Both. What’s just a G in design may additionally mean»geometric size». These variables’ values describe how large an object is, or how big it’s in relation. The function of those variables could be considered being a formula that describes the object’s form. What is a G in Physics could be the manner that gravity functions. It’s the way that mass operates. A amazing many problems might be solved this way. Certainly one of the things which makes problems much easier to resolve is the fact that the things that we’re talking about can be clarified. It will not sound right to feel about a sphere because»No Thing», but alternatively to consider about the sphere for a chemical that exists. We can describe some thing that it behaves, gives us further insight into it. Добавить комментарий Отменить ответ Для отправки комментария вам необходимо авторизоваться.	{"url":"https://filterdom.com/whats-ag-in-physics/","timestamp":"2024-11-10T11:51:32Z","content_type":"text/html","content_length":"80974","record_id":"<urn:uuid:12136e4f-993a-4cc2-b333-649486e4aa9e>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00361.warc.gz"}
Crosspolarisation in beam-waveguide feeds Earlier studies have shown that the theory of the propagation of Gaussian beams takes into account the higher-order crosspolarized modes generated by offset reflectors used in beam waveguides. The paper demonstrates that by assuming an idealized Gaussian radiation pattern for the feedhorn and ignoring the finite sizes of the reflectors, the Gaussian mode analysis enables the residual crosspolarization transmitted by a beam waveguide to be predicted by relatively simple formulas. The discussion is limited to beam waveguide designs of paraboloid and ellipsoid configuration. The computed patterns radiated from the beam waveguide on a spherical surface corresponding to the Cassegrain subreflector position are plotted for the paraboloid and ellipsoid designs. Good agreement is found between Gaussian mode theory and numerical approach. It is concluded that the crosspolarization intrinsic to the reflection of radiation by offset curved surfaces is the dominant factor in the overall performance of a beam waveguide. Electronics Letters Pub Date: September 1976 □ Antenna Radiation Patterns; □ Beam Waveguides; □ Horn Antennas; □ Polarization Characteristics; □ Wave Excitation; □ Antenna Feeds; □ Ellipsoids; □ Microwave Transmission; □ Parabolic Reflectors; □ Subreflectors; □ Wave Reflection; □ Communications and Radar	{"url":"https://ui.adsabs.harvard.edu/abs/1976ElL....12..529H/abstract","timestamp":"2024-11-08T22:47:13Z","content_type":"text/html","content_length":"38002","record_id":"<urn:uuid:884e579e-2562-432b-9c3b-1ff62344284c>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00250.warc.gz"}
Statistics \| Bright top of page Central tendency and spread S.1.1 I can find the mode, median and range from a list of data Year 7 support Year 7 core Year 8 support Year 10 foundation S.1.2 I can calculate the mean from a list of data Year 7 support Year 7 core Year 8 support Year 10 foundation S.1.3 I can find the mode, median, mean and range from a list of data Year 7 extension Year 8 core Year 10 crossover S.1.4 I can identify the appropriate average to use in a given situation Year 7 support Year 7 core Year 7 extension Year 10 foundation Year 10 crossover S.1.5 I can interpret the mode, median, mean and range of two sets of data and make comparisons Year 7 core Year 7 extension Year 8 support Year 8 core Year 8 extension Year 10 foundation S.1.6 I can find the data based on information given on the averages and range Year 7 extension Year 8 core Year 8 extension Year 10 crossover S.1.7 I can adjust the mean when data is added or taken away from the set Year 7 extension Year 8 core Year 8 extension Year 10 crossover S.1.8 I can find the mode, range, median and mean from a discrete frequency table Year 7 extension Year 8 core Year 8 extension Year 10 foundation Year 10 crossover S.1.9 I can find the modal class, class in which the median lies and estimated mean from a grouped frequency table Year 8 extension Year 10 crossover S.1.10 I can compare distributions of grouped, discrete or continuous data using mean, mode, median and range No resources here yet - I'm working on it! Year 8 extension Year 10 crossover S.1.11 I can find the mode, range and median from a stem and leaf diagram Year 8 core Year 8 extension Year 10 foundation Year 10 crossover S.1.12 I can interpret and calculate quartiles and interquartile range No resources here yet - I'm working on it! Year 10 higher Year 10 extension S.1.13 I can find the interquartile range from a stem and leaf diagram No resources here yet - I'm working on it! Year 10 higher Year 10 extension S.2.1 I can read information from and complete a discrete frequency table S.2.2 I can read information from and complete a discrete or grouped frequency table No resources here yet - I'm working on it! Year 7 core Year 7 extension Year 8 support S.2.3 I can complete frequency trees from given information Year 7 core Year 7 extension Year 8 support Year 8 core Year 10 foundation S.2.4 I can solve complex frequency tree problems including fractions, percentages and ratios No resources here yet - I'm working on it! Year 8 extension Year 10 crossover S.3.1 I can read information from a two way table No resources here yet - I'm working on it! S.3.2 I can read, complete and interpret a two way table Year 7 core Year 7 extension Year 8 support Year 8 core S.4.1 I can read and complete a pictogram S.4.2 I can draw bars charts from a frequency table including dual/composite No resources here yet - I'm working on it! Year 7 support Yaer 7 core S.4.3 I can interpret bar charts and use them to solve problems Year 7 support Yaer 7 core Year 7 extension Year 8 support S.4.4 I can draw a stem and leaf diagram, including back to back Year 7 support Year 7 core Year 7 extension S.4.5 I can construct pie charts Year 8 core Year 8 extension S.4.6 I can read and interpret pie charts No resources here yet - I'm working on it! Year 8 core Year 8 extension S.4.7 I can identify misleading chart features Year 7 support Year 7 core Year 7 extension S.5.1 I can construct a histogram with unequal class widths Year 10 higher Year 10 extension S.5.2 I can interpret a histograms with unequal class widths (i.e finding the frequency) Year 10 higher Year 10 extension S.5.3 I can estimate from a histogram No resources here yet - I'm working on it! S.6.1 I can complete and interpret scatter graphs, including correlation and a line of best fit Year 7 core Year 8 support S.6.2 I can complete and interpret scatter graphs, including correlation, line of best fit and interpolation/extrapolation Year 7 extension Year 8 core Year 8 extension S.7.1 I can construct a cumulative frequency diagram Year 10 higher Year 10 extension S.7.2 I can construct and complete box plots No resources here yet - I'm working on it! Year 10 higher Year 10 extension S.7.3 I can interpret cumulative frequency diagrams No resources here yet - I'm working on it! Year 10 higher Year 10 extension S.7.4 I can interpret box plots No resources here yet - I'm working on it! Year 10 higher Year 10 extension S.7.5 I can make comparisons between two distributions using box plots Year 10 higher Year 10 extension S.8.1 I can apply statistics to describe a population No resources here yet - I'm working on it! Year 9 core Year 9 extension Year 10 higher Year 10 extension S.8.2 I can apply statistics to a capture and recapture problem No resources here yet - I'm working on it! Year 9 core Year 9 extension Year 10 higher Year 10 extension S.9.1 I can interpret line graphs for time series data No resources here yet - I'm working on it! Year 11 higher Year 11 extension bottom of page	{"url":"https://www.bright-maths.co.uk/statistics","timestamp":"2024-11-03T16:14:06Z","content_type":"text/html","content_length":"1051070","record_id":"<urn:uuid:57d6a775-eac8-4d69-9578-800e1be9580a>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00773.warc.gz"}
Nachi Avraham-Re'em - Talks From Poisson Suspensions to Infinitely Divisible Stationary Processes Spatial Poisson suspensions of Polish groups actions The Poisson suspension of Polish groups actions α-stable stationary processes The Poisson Suspension in Probability and Ergodic Theory • HUJI. Logic Seminar. January 2023. An introduction to the model theory of measure preserving group actions Stable Processes: from probability to non-singular ergodic theory 1st talk: An introduction to the ergodic theory of orbit equivalence classification of group actions 2nd talk: The orbital classification of the shift Symmetric stable processes indexed by amenable groups - ergodicity, mixing and spectral representation Ergodic theory of stable random fields indexed by amenable groups • HUJI. Dynamics & Probability Seminar. January 2021. The orbit equivalence class of markov subshifts of finite type • HUJI. Graduate Students Seminar. December 2020. Kakutani dichotomy and 0-1 laws in markov measures • Weizman. Students Probability Day VII. May 2019. The orbital equivalence class of the shift • HUJI. Dynamics Lunch Seminar. January 2019. Sethuraman-Varadhan's proof of the Central Limit Theorem for non-Homogeneous markov chains	{"url":"https://www.nachi.co.il/talks","timestamp":"2024-11-10T22:03:18Z","content_type":"text/html","content_length":"87393","record_id":"<urn:uuid:7cde1389-ff7e-4ccb-9f6a-9ce7803cd995>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00665.warc.gz"}
Explain Delta Modulation in detail with suitable diagram. Delta Modulation In PCM the signaling rate and transmission channel bandwidth are quite large since it transmits all the bits which are used to code a sample. To overcome this problem, Delta modulation is used. Working Principle Delta modulation transmits only one bit per sample. Here, the present sample value is compared with the previous sample value and this result whether the amplitude is increased or decreased is Input signal x(t) is approximated to step signal by the delta modulator.This step size is kept fixed. The difference between the input signal x(t) and staircase approximated signal is confined to two levels, i.e., +Δ and -Δ. Now, if the difference is positive, then approximated signal is increased by one step, i.e., ‘Δ’. If the difference is negative, then approximated signal is reduced by ‘Δ’ . When the step is reduced, ‘0’ is transmitted and if the step is increased, ‘1’ is transmitted. Hence, for each sample, only one binary bit is transmitted. Fig.1 shows the analog signal x(t) and its staircase approximated signal by the delta modulator. Fig.1. Delta Modulation Waveform Mathematical Expressions The error between the sampled value of x(t) and last approximated sample is given as: Where e( nT[s]) = error at present sample x(nT[s]) = sampled signal of x(t) If we assume u(nT[s]) as the present sample approximation of staircase output, then Let us define a quantity b( nT[s]) in such a way that, This means that depending on the sign of error e( nT[s]) , the sign of step size Δ is decided. In other words we can write Also if b (nT[s]) =+Δ then a binary ‘1’ is transmitted and if b (nT[s]) =-Δ then a binary ‘0’ is transmitted Here T[s ]= sampling interval. Fig. 2 (a) shows the transmitter . It is also known as Delta modulator. Fig.2 (a) Delta Modulation Transmitter It consists of a 1-bit quantizer and a delay circuit along with two summer circuits. The summer in the accumulator adds quantizer output (±Δ) with the previous sample approximation . This gives present sample approximation. i.e., The previous sample approximation u[(n-1)T[s ]] is restored by delaying one sample period T[s ]. The samples input signal x(nT[s ]) and staircase approximated signal xˆ(nT[s ]) are subtracted to get error signal e(nT[s ]). Thus, depending on the sign of e(nT[s ]), one bit quantizer generates an output of +Δ or -Δ . If the step size is +Δ, then binary ‘1’ is transmitted and if it is -Δ, then binary ‘0’ is transmitted . At the receiver end also known as delta demodulator, as shown in fig.2 (b) , it comprises of a low pass filter(LPF), a summer, and a delay circuit. The predictor circuit is eliminated here and hence no assumed input is given to the demodulator. Fig.2 (b) Delta Modulation Receiver The accumulator generates the staircase approximated signal output and is delayed by one sampling period T[s]. It is then added to the input signal. If the input is binary ‘1’ then it adds +Δ step to the previous output (which is delayed). If the input is binary ‘0’ then one step ‘Δ’ is subtracted from the delayed signal. Also, the low pass filter smoothens the staircase signal to reconstruct the original message signal x(t) .	{"url":"https://electronicspost.com/explain-delta-modulation-in-detail-with-suitable-diagram/","timestamp":"2024-11-11T16:25:22Z","content_type":"text/html","content_length":"63133","record_id":"<urn:uuid:f5602eb9-196f-41b1-9fa6-860019b703e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00372.warc.gz"}
Linear Supertypes 1. final def !=(arg0: Any): Boolean Definition Classes AnyRef → Any 2. final def ##(): Int Definition Classes AnyRef → Any 3. def +(other: String): String Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to any2stringadd[TupleCodec[A, B]] performed by method any2stringadd in scala.Predef. Definition Classes 4. def ->[B](y: B): (TupleCodec[A, B], B) Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ArrowAssoc[TupleCodec[A, B]] performed by method ArrowAssoc in scala.Predef. Definition Classes 5. def :+:[B](left: Codec[B]): CoproductCodecBuilder[:+:[B, :+:[(A, B), CNil]], ::[Codec[B], ::[Codec[(A, B)], HNil]], :+:[B, :+:[(A, B), CNil]]] Supports creation of a coproduct codec. 6. def ::[B](codecB: Codec[B]): Codec[::[B, ::[(A, B), HNil]]] When called on a Codec[A] where A is not a subytpe of HList, creates a new codec that encodes/decodes an HList of B :: A :: HNil. When called on a Codec[A] where A is not a subytpe of HList, creates a new codec that encodes/decodes an HList of B :: A :: HNil. For example, uint8 :: utf8 has type Codec[Int :: String :: HNil]. uint8 :: utf8 }}} Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithHListSupport[(A, B)] performed by method ValueCodecEnrichedWithHListSupport in scodec. Definition Classes 7. def :~>:[B](codecB: Codec[B])(implicit ev: =:=[Unit, B]): Codec[::[(A, B), HNil]] When called on a Codec[A], returns a new codec that encodes/decodes B :: A :: HNil. When called on a Codec[A], returns a new codec that encodes/decodes B :: A :: HNil. HList equivalent of ~>. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithHListSupport[(A, B)] performed by method ValueCodecEnrichedWithHListSupport in scodec. Definition Classes 8. final def <~[B](codecB: Codec[B])(implicit ev: =:=[Unit, B]): Codec[(A, B)] Assuming B is Unit, creates a Codec[A] that: encodes the A followed by a unit; decodes an A followed by a unit and discards the decoded unit. Assuming B is Unit, creates a Codec[A] that: encodes the A followed by a unit; decodes an A followed by a unit and discards the decoded unit. Operator alias of dropRight. Definition Classes 9. final def ==(arg0: Any): Boolean Definition Classes AnyRef → Any 10. def >>:~[L <: HList](f: ((A, B)) ⇒ Codec[L]): Codec[::[(A, B), L]] Creates a new codec that encodes/decodes an HList type of A :: L given a function A => Codec[L]. Creates a new codec that encodes/decodes an HList type of A :: L given a function A => Codec[L]. This allows later parts of an HList codec to be dependent on earlier values. Operator alias for Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithHListSupport[(A, B)] performed by method ValueCodecEnrichedWithHListSupport in scodec. Definition Classes 11. final def >>~[B](f: ((A, B)) ⇒ Codec[B]): Codec[((A, B), B)] Returns a new codec that encodes/decodes a value of type (A, B) where the codec of B is dependent on A. Returns a new codec that encodes/decodes a value of type (A, B) where the codec of B is dependent on A. Operator alias for flatZip. Definition Classes 13. def as[B](implicit as: Transformer[(A, B), B]): Codec[B] Transforms using implicitly available evidence that such a transformation is possible. Transforms using implicitly available evidence that such a transformation is possible. Typical transformations include converting: □ an F[L] for some L <: HList to/from an F[CC] for some case class CC, where the types in the case class are aligned with the types in L □ an F[C] for some C <: Coproduct to/from an F[SC] for some sealed class SC, where the component types in the coproduct are the leaf subtypes of the sealed class. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. Definition Classes 14. def asDecoder: Decoder[(A, B)] Gets this as a Decoder. 15. def asEncoder: Encoder[(A, B)] Gets this as an Encoder. 16. final def asInstanceOf[T0]: T0 17. def clone(): AnyRef Definition Classes @throws( ... ) 18. final def compact: Codec[(A, B)] Converts this codec to a new codec that compacts the encoded bit vector before returning it. Converts this codec to a new codec that compacts the encoded bit vector before returning it. Definition Classes Codec → GenCodec → Encoder 19. final def complete: Codec[(A, B)] Converts this codec to a new codec that fails decoding if there are remaining bits. Converts this codec to a new codec that fails decoding if there are remaining bits. Definition Classes Codec → GenCodec → Decoder 20. def contramap[C](f: (C) ⇒ (A, B)): GenCodec[C, (A, B)] Converts this GenCodec to a GenCodec[C, B] using the supplied C => A. Converts this GenCodec to a GenCodec[C, B] using the supplied C => A. Definition Classes GenCodec → Encoder Attempts to decode a value of type A from the specified bit vector. Attempts to decode a value of type A from the specified bit vector. error if value could not be decoded or the remaining bits and the decoded value Definition Classes TupleCodec → Decoder 22. def decodeOnly[AA >: (A, B)]: Codec[AA] Converts this to a codec that fails encoding with an error. Converts this to a codec that fails encoding with an error. Definition Classes Codec → Decoder 23. final def downcast[B <: (A, B)](implicit arg0: Manifest[B]): Codec[B] Safely lifts this codec to a codec of a subtype. Safely lifts this codec to a codec of a subtype. When a supertype of B that is not a supertype of A is decoded, an decoding error is returned. Definition Classes 24. final def dropLeft[B](codecB: Codec[B])(implicit ev: =:=[Unit, (A, B)]): Codec[B] Assuming A is Unit, creates a Codec[B] that: encodes the unit followed by a B; decodes a unit followed by a B and discards the decoded unit. Assuming A is Unit, creates a Codec[B] that: encodes the unit followed by a B; decodes a unit followed by a B and discards the decoded unit. Definition Classes 25. final def dropRight[B](codecB: Codec[B])(implicit ev: =:=[Unit, B]): Codec[(A, B)] Assuming B is Unit, creates a Codec[A] that: encodes the A followed by a unit; decodes an A followed by a unit and discards the decoded unit. Assuming B is Unit, creates a Codec[A] that: encodes the A followed by a unit; decodes an A followed by a unit and discards the decoded unit. Definition Classes 26. def econtramap[C](f: (C) ⇒ Attempt[(A, B)]): GenCodec[C, (A, B)] Converts this GenCodec to a GenCodec[C, B] using the supplied C => Attempt[A]. Converts this GenCodec to a GenCodec[C, B] using the supplied C => Attempt[A]. Definition Classes GenCodec → Encoder 27. def emap[C](f: ((A, B)) ⇒ Attempt[C]): GenCodec[(A, B), C] Converts this GenCodec to a GenCodec[A, C] using the supplied B => Attempt[C]. Converts this GenCodec to a GenCodec[A, C] using the supplied B => Attempt[C]. Definition Classes GenCodec → Decoder Attempts to encode the specified value in to a bit vector. Attempts to encode the specified value in to a bit vector. Definition Classes TupleCodec → Encoder 29. def encodeOnly: Codec[(A, B)] Converts this to a codec that fails decoding with an error. Converts this to a codec that fails decoding with an error. Definition Classes 30. def ensuring(cond: (TupleCodec[A, B]) ⇒ Boolean, msg: ⇒ Any): TupleCodec[A, B] Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to Ensuring[TupleCodec[A, B]] performed by method Ensuring in scala.Predef. Definition Classes 31. def ensuring(cond: (TupleCodec[A, B]) ⇒ Boolean): TupleCodec[A, B] Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to Ensuring[TupleCodec[A, B]] performed by method Ensuring in scala.Predef. Definition Classes 32. def ensuring(cond: Boolean, msg: ⇒ Any): TupleCodec[A, B] Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to Ensuring[TupleCodec[A, B]] performed by method Ensuring in scala.Predef. Definition Classes 33. def ensuring(cond: Boolean): TupleCodec[A, B] Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to Ensuring[TupleCodec[A, B]] performed by method Ensuring in scala.Predef. Definition Classes 34. final def eq(arg0: AnyRef): Boolean 35. def equals(arg0: Any): Boolean Definition Classes AnyRef → Any 36. final def exmap[B](f: ((A, B)) ⇒ Attempt[B], g: (B) ⇒ Attempt[(A, B)]): Codec[B] Transforms using two functions, A => Attempt[B] and B => Attempt[A]. Transforms using two functions, A => Attempt[B] and B => Attempt[A]. Definition Classes 37. def finalize(): Unit Definition Classes @throws( classOf[java.lang.Throwable] ) 38. def flatMap[B](f: ((A, B)) ⇒ Decoder[B]): Decoder[B] Converts this decoder to a Decoder[B] using the supplied A => Decoder[B]. Converts this decoder to a Decoder[B] using the supplied A => Decoder[B]. Definition Classes 39. def flatPrepend[L <: HList](f: ((A, B)) ⇒ Codec[L]): Codec[::[(A, B), L]] Creates a new codec that encodes/decodes an HList type of A :: L given a function A => Codec[L]. Creates a new codec that encodes/decodes an HList type of A :: L given a function A => Codec[L]. This allows later parts of an HList codec to be dependent on earlier values. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithHListSupport[(A, B)] performed by method ValueCodecEnrichedWithHListSupport in scodec. Definition Classes 40. final def flatZip[B](f: ((A, B)) ⇒ Codec[B]): Codec[((A, B), B)] Returns a new codec that encodes/decodes a value of type (A, B) where the codec of B is dependent on A. Returns a new codec that encodes/decodes a value of type (A, B) where the codec of B is dependent on A. Definition Classes 41. def flatZipHList[B](f: ((A, B)) ⇒ Codec[B]): Codec[::[(A, B), ::[B, HNil]]] Creates a new codec that encodes/decodes an HList type of A :: B :: HNil given a function A => Codec[B]. Creates a new codec that encodes/decodes an HList type of A :: B :: HNil given a function A => Codec[B]. If B is an HList type, consider using flatPrepend instead, which avoids nested HLists. This is the direct HList equivalent of flatZip. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithHListSupport[(A, B)] performed by method ValueCodecEnrichedWithHListSupport in scodec. Definition Classes 42. final def flattenLeftPairs(implicit f: FlattenLeftPairs[(A, B)]): Codec[Out] Converts this codec to an HList based codec by flattening all left nested pairs. Converts this codec to an HList based codec by flattening all left nested pairs. For example, flattenLeftPairs on a Codec[(((A, B), C), D)] results in a Codec[A :: B :: C :: D :: HNil]. This is particularly useful when combined with ~, ~>, and <~. Definition Classes 43. def formatted(fmtstr: String): String Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to StringFormat[TupleCodec[A, B]] performed by method StringFormat in scala.Predef. Definition Classes 44. final def fuse[AA <: (A, B), BB >: (A, B)](implicit ev: =:=[BB, AA]): Codec[BB] Converts this generalized codec in to a non-generalized codec assuming A and B are the same type. Converts this generalized codec in to a non-generalized codec assuming A and B are the same type. Definition Classes 45. final def getClass(): Class[_] Definition Classes AnyRef → Any 46. def hashCode(): Int Definition Classes AnyRef → Any 47. final def hlist: Codec[::[(A, B), HNil]] Lifts this codec in to a codec of a singleton hlist. Lifts this codec in to a codec of a singleton hlist. Definition Classes 48. final def isInstanceOf[T0]: Boolean 49. def map[C](f: ((A, B)) ⇒ C): GenCodec[(A, B), C] Converts this GenCodec to a GenCodec[A, C] using the supplied B => C. Converts this GenCodec to a GenCodec[A, C] using the supplied B => C. Definition Classes GenCodec → Decoder 50. final def narrow[B](f: ((A, B)) ⇒ Attempt[B], g: (B) ⇒ (A, B)): Codec[B] Transforms using two functions, A => Attempt[B] and B => A. Transforms using two functions, A => Attempt[B] and B => A. The supplied functions form an injection from B to A. Hence, this method converts from a larger to a smaller type. Hence, the name narrow. Definition Classes 51. final def ne(arg0: AnyRef): Boolean 52. final def notify(): Unit 53. final def notifyAll(): Unit 54. final def pairedWith[B](codecB: Codec[B]): Codec[((A, B), B)] Creates a Codec[(A, B)] that first encodes/decodes an A followed by a B. Creates a Codec[(A, B)] that first encodes/decodes an A followed by a B. Definition Classes 55. def pcontramap[C](f: (C) ⇒ Option[(A, B)]): GenCodec[C, (A, B)] Converts this GenCodec to a GenCodec[C, B] using the supplied partial function from C to A. Converts this GenCodec to a GenCodec[C, B] using the supplied partial function from C to A. The encoding will fail for any C that f maps to None. Definition Classes GenCodec → Encoder 56. def polyxmap[B](p: Poly, q: Poly)(implicit aToB: Aux[p.type, ::[(A, B), HNil], B], bToA: Aux[q.type, ::[B, HNil], (A, B)]): Codec[B] Polymorphic function version of xmap. Polymorphic function version of xmap. When called on a Codec[A] where A is not a subytpe of HList, returns a new codec that's the result of xmapping with p and q, using p to convert from A to B and using q to convert from B to A. polymorphic function that converts from A to B polymorphic function that converts from B to A Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithGenericSupport[(A, B)] performed by method ValueCodecEnrichedWithGenericSupport in scodec. Definition Classes 57. def polyxmap1[B](p: Poly)(implicit aToB: Aux[p.type, ::[(A, B), HNil], B], bToA: Aux[p.type, ::[B, HNil], (A, B)]): Codec[B] Polymorphic function version of xmap that uses a single polymorphic function in both directions. Polymorphic function version of xmap that uses a single polymorphic function in both directions. When called on a Codec[A] where A is not a subytpe of HList, returns a new codec that's the result of xmapping with p for both forward and reverse directions. polymorphic function that converts from A to B and from B to A Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithGenericSupport[(A, B)] performed by method ValueCodecEnrichedWithGenericSupport in scodec. Definition Classes 58. def selectEncoder[A](implicit inj: Inject[Coproduct with (A, B), A]): Encoder[A] When called on a Encoder[C] where C is a coproduct containing type A, converts to an Encoder[A]. When called on a Encoder[C] where C is a coproduct containing type A, converts to an Encoder[A]. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to EnrichedCoproductEncoder[Coproduct with (A, B)] performed by method EnrichedCoproductEncoder in scodec. Definition Classes Provides a bound on the size of successfully encoded values. Provides a bound on the size of successfully encoded values. Definition Classes TupleCodec → Encoder 60. final def synchronized[T0](arg0: ⇒ T0): T0 61. def toField[K]: Codec[FieldType[K, (A, B)]] Lifts this codec to a codec of a shapeless field -- allowing it to be used in records and unions. Lifts this codec to a codec of a shapeless field -- allowing it to be used in records and unions. Definition Classes 62. def toFieldWithContext[K <: Symbol](k: K): Codec[FieldType[K, (A, B)]] Lifts this codec to a codec of a shapeless field -- allowing it to be used in records and unions. Lifts this codec to a codec of a shapeless field -- allowing it to be used in records and unions. The specified key is pushed in to the context of any errors that are returned from the resulting Definition Classes 63. def toString(): String 64. final def unit(zero: (A, B)): Codec[Unit] Converts this to a Codec[Unit] that encodes using the specified zero value and decodes a unit value when this codec decodes an A successfully. Converts this to a Codec[Unit] that encodes using the specified zero value and decodes a unit value when this codec decodes an A successfully. Definition Classes 65. final def upcast[B >: (A, B)](implicit m: Manifest[(A, B)]): Codec[B] Safely lifts this codec to a codec of a supertype. Safely lifts this codec to a codec of a supertype. When a subtype of B that is not a subtype of A is passed to encode, an encoding error is returned. Definition Classes 66. final def wait(): Unit Definition Classes @throws( ... ) 67. final def wait(arg0: Long, arg1: Int): Unit Definition Classes @throws( ... ) 68. final def wait(arg0: Long): Unit Definition Classes @throws( ... ) 69. final def widen[B](f: ((A, B)) ⇒ B, g: (B) ⇒ Attempt[(A, B)]): Codec[B] Transforms using two functions, A => B and B => Attempt[A]. Transforms using two functions, A => B and B => Attempt[A]. The supplied functions form an injection from A to B. Hence, this method converts from a smaller to a larger type. Hence, the name widen. Definition Classes 70. def widenAs[X](to: (A, B) ⇒ X, from: (X) ⇒ Option[(A, B)]): Codec[X] 71. def widenOpt[B](f: ((A, B)) ⇒ B, g: (B) ⇒ Option[(A, B)]): Codec[B] Transforms using two functions, A => B and B => Option[A]. Transforms using two functions, A => B and B => Option[A]. Particularly useful when combined with case class apply/unapply. E.g., widenOpt(fa, Foo.apply, Foo.unapply). Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. Definition Classes 72. final def withContext(context: String): Codec[(A, B)] Creates a new codec that is functionally equivalent to this codec but pushes the specified context string in to any errors returned from encode or decode. Creates a new codec that is functionally equivalent to this codec but pushes the specified context string in to any errors returned from encode or decode. Definition Classes 73. final def withToString(str: String): Codec[(A, B)] Creates a new codec that is functionally equivalent to this codec but returns the specified string from toString. Creates a new codec that is functionally equivalent to this codec but returns the specified string from toString. Definition Classes 74. final def xmap[B](f: ((A, B)) ⇒ B, g: (B) ⇒ (A, B)): Codec[B] Transforms using the isomorphism described by two functions, A => B and B => A. Transforms using the isomorphism described by two functions, A => B and B => A. Definition Classes 75. final def ~[B](codecB: Codec[B]): Codec[((A, B), B)] Creates a Codec[(A, B)] that first encodes/decodes an A followed by a B. Creates a Codec[(A, B)] that first encodes/decodes an A followed by a B. Operator alias for pairedWith. Definition Classes 76. final def ~>[B](codecB: Codec[B])(implicit ev: =:=[Unit, (A, B)]): Codec[B] Assuming A is Unit, creates a Codec[B] that: encodes the unit followed by a B; decodes a unit followed by a B and discards the decoded unit. Assuming A is Unit, creates a Codec[B] that: encodes the unit followed by a B; decodes a unit followed by a B and discards the decoded unit. Operator alias of dropLeft. Definition Classes 78. def →[B](y: B): (TupleCodec[A, B], B) Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ArrowAssoc[TupleCodec[A, B]] performed by method ArrowAssoc in scala.Predef. Definition Classes Transforms using two functions, A => Attempt[B] and B => Attempt[A]. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. This implicitly inherited member is shadowed by one or more members in this class. To access this member you can use a type ascription: (tupleCodec: TransformSyntax[Codec, (A, B)]).exmap(f, g) Definition Classes Transforms using two functions, A => Attempt[B] and B => A. The supplied functions form an injection from B to A. Hence, this method converts from a larger to a smaller type. Hence, the name narrow. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. This implicitly inherited member is shadowed by one or more members in this class. To access this member you can use a type ascription: (tupleCodec: TransformSyntax[Codec, (A, B)]).narrow(f, g) Definition Classes Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to Tuple2CodecSupport[(A, B)] performed by method Tuple2CodecSupport in scodec. This implicitly inherited member is ambiguous. One or more implicitly inherited members have similar signatures, so calling this member may produce an ambiguous implicit conversion compiler To access this member you can use a type ascription: (tupleCodec: Tuple2CodecSupport[(A, B)]).self Definition Classes Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to EnrichedCoproductEncoder[Coproduct with (A, B)] performed by method EnrichedCoproductEncoder in scodec. This implicitly inherited member is ambiguous. One or more implicitly inherited members have similar signatures, so calling this member may produce an ambiguous implicit conversion compiler To access this member you can use a type ascription: (tupleCodec: EnrichedCoproductEncoder[Coproduct with (A, B)]).self Definition Classes Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithGenericSupport[(A, B)] performed by method ValueCodecEnrichedWithGenericSupport in scodec. This implicitly inherited member is ambiguous. One or more implicitly inherited members have similar signatures, so calling this member may produce an ambiguous implicit conversion compiler To access this member you can use a type ascription: (tupleCodec: ValueCodecEnrichedWithGenericSupport[(A, B)]).self Definition Classes Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueCodecEnrichedWithHListSupport[(A, B)] performed by method ValueCodecEnrichedWithHListSupport in scodec. This implicitly inherited member is ambiguous. One or more implicitly inherited members have similar signatures, so calling this member may produce an ambiguous implicit conversion compiler To access this member you can use a type ascription: (tupleCodec: ValueCodecEnrichedWithHListSupport[(A, B)]).self Definition Classes Supports TransformSyntax. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. This implicitly inherited member is ambiguous. One or more implicitly inherited members have similar signatures, so calling this member may produce an ambiguous implicit conversion compiler To access this member you can use a type ascription: (tupleCodec: TransformSyntax[Codec, (A, B)]).self Definition Classes Transforms using two functions, A => B and B => Attempt[A]. The supplied functions form an injection from A to B. Hence, this method converts from a smaller to a larger type. Hence, the name widen. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. This implicitly inherited member is shadowed by one or more members in this class. To access this member you can use a type ascription: (tupleCodec: TransformSyntax[Codec, (A, B)]).widen(f, g) Definition Classes Transforms using the isomorphism described by two functions, A => B and B => A. Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. This implicitly inherited member is shadowed by one or more members in this class. To access this member you can use a type ascription: (tupleCodec: TransformSyntax[Codec, (A, B)]).xmap(f, g) Definition Classes Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to ValueEnrichedWithTuplingSupport[TupleCodec[A, B]] performed by method ValueEnrichedWithTuplingSupport in scodec.codecs. This implicitly inherited member is shadowed by one or more members in this class. To access this member you can use a type ascription: (tupleCodec: ValueEnrichedWithTuplingSupport[TupleCodec[A, B]]).~(b) Definition Classes Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to Tuple2CodecSupport[(A, B)] performed by method Tuple2CodecSupport in scodec. This implicitly inherited member is shadowed by one or more members in this class. To access this member you can use a type ascription: (tupleCodec: Tuple2CodecSupport[(A, B)]).~~(B) Definition Classes Transforms using two functions, A => B and B => Option[A]. Particularly useful when combined with case class apply/unapply. E.g., pxmap(fa, Foo.apply, Foo.unapply). Implicit information This member is added by an implicit conversion from TupleCodec[A, B] to TransformSyntax[Codec, (A, B)] performed by method TransformSyntax in scodec. Definition Classes (Since version 1.7.0) Use widenOpt instead	{"url":"http://scodec.org/api/scodec-core/1.7.1/scodec/codecs/TupleCodec.html","timestamp":"2024-11-11T13:13:46Z","content_type":"text/html","content_length":"198077","record_id":"<urn:uuid:52257bbe-0aef-49a9-9baa-41e0279ae134>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00283.warc.gz"}
Trig Homework Help Download File ===== https://urlin.us/2tr Trig Homework Help Download File ===== https://urlin.us/2trFnP Trig Homework Help Trying to keep up with trig can be hard at first -- suddenly you're learning new terms like sine, cosine, and tangent, and having to figure out more triangles than you ever cared about. Fortunately it's just like any other math -- follow a set of rules, understand why it works how it does, and you'll be fine. Check out the lessons below if you need a refresher on trigonometry topics. Learn how to graph trigonometric functions, check homework answers, or let us help you study for your next trig quiz. Whatever you're working on, your online tutor will walk you step-by-step through the problem and the solution. Watch how it works. Our online classroom is equipped with all the tools you need for homework success. Upload a problem set, solve trigonometric equations on the interactive whiteboard, and chat with your tutor until your trig question is answered. We do not have monthly fees or minimum payments, and there are no hidden costs. Instead, the price is unique for every work order you submit. For tutoring and homework help, the price depends on many factors that include the length of the session, level of work difficulty, level of expertise of the tutor, and amount of time available before the deadline. You will be given a price up front and there is no obligation for you to pay. Homework library items have individual set prices. The area of mathematics called trigonometry focuses on triangles, with particular emphasis on the relationships among the sides and angles of a triangle. The six trigonometric functions are defined in this branch of mathematics, and the applications of those trig functions carry fundamental importance in science. The cyclical (periodic) nature of sine waves, for example, model everything from atomic vibrations to light waves. A basic course in trigonometry will cover the following topics: Students can find many books on trigonometry from Google Books and Amazon.com. There are also quite a few trigonometry tutorials available from such places as Clark University, United Arab Emirates University, 1728 Software Systems, Nipissing University, and Clemson University. Students can also follow the Journal of Trigonometry. Our tutors are just as dedicated to your success in class as you are, so they are available around the clock to assist you with questions, homework, exam preparation and any Trigonometry related assignments you need extra help completing. Because our college Trigonometry tutors are fully remote, seeking their help is easy. Rather than spend valuable time trying to find a local Trigonometry tutor you can trust, just call on our tutors whenever you need them without any conflicting schedules getting in the way. 7.Hello im looking for some help with this proof tree, im supposed to use the rules i have attached.I would...I would love to get a written solution.I did the other proof tree that you can see in the attaached files. So its supposed to look like thatThank You. View More These formulae are used to solve trigonometric functions and equation.Sometimes used to calculate the value for trigonometric value of higher angle.sin2A=2sinAcosA cos2A=A -A tan 2A =2tan A 1-A Trigonometric function relates sides of triangle with the angle of that triangle, they often called as circular function also. The six trigonometry functions are sine, cosine, tangent, cotangent, secant, and cosecant. We use trigonometric ratio in order to find the missing side of the triangle if we have one side say adjacent then by using tangent ratio we can find the opposite side and by using cosine ratio we can find the hypotenuse. Trigonometric ratio is the ratio of the sides of the right-angled triangle we named the three sides as the opposite, adjacent, and the hypotenuse, a ratio of these three sides with each other are known as a trigonometric ratio. How it works:Identify which concepts are covered on your quadratic functions homework.Find videos on those topics within this chapter.Watch fun v	{"url":"https://www.ercanaydin.com/group/mysite-231-group/discussion/f731be5e-7299-4725-8159-ee7012b91e48","timestamp":"2024-11-02T07:51:23Z","content_type":"text/html","content_length":"1050489","record_id":"<urn:uuid:dd8b329c-9fae-44c4-91ee-38f778cd5286>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00292.warc.gz"}
Heat Converters - Online Unit Conversion Tools Heat Converters Convert between different heat units using our heat converters. Effortlessly perform unit conversions related to heat and thermal energy with our online tools "Heat converters" is a specialized category crafted to facilitate the conversion of measurements related to heat and thermal properties. Understanding and managing heat is essential in various fields such as engineering, physics, chemistry, materials science, and environmental science. This category offers a comprehensive array of tools to enable accurate conversions between different units commonly used in heat-related calculations and analyses. Within "Heat converters," you'll find several sub-categories, each focusing on specific aspects of heat and thermal properties: 1. Heat Converters: This sub-category encompasses a broad range of conversions related to heat energy, including conversions between units such as joules, calories, British thermal units (BTU), kilowatt-hours, etc. Heat energy is fundamental in understanding thermal processes and energy transfer. 2. Fuel Efficiency - Mass Converter: Here, users can convert fuel efficiency measurements based on mass between units such as liters per 100 kilometers, miles per gallon (MPG), kilometers per liter, etc. Fuel efficiency is crucial in assessing the performance of vehicles and machinery in terms of energy consumption. 3. Fuel Efficiency - Volume Converter: This section focuses on converting fuel efficiency measurements based on volume between units such as liters per 100 kilometers, miles per gallon (MPG), gallons per mile, etc. Fuel efficiency measures how effectively fuel is utilized to produce mechanical work or heat. 4. Temperature Interval Converter: Enables the conversion of temperature interval measurements between units such as degrees Celsius, degrees Fahrenheit, kelvin, etc. Temperature interval conversions are essential for understanding temperature changes and differences. 5. Thermal Expansion Converter: In this sub-category, users can convert thermal expansion measurements between units such as meters per degree Celsius, inches per degree Fahrenheit, etc. Thermal expansion quantifies the increase in size or volume of a material with temperature. 6. Thermal Resistance Converter: Facilitates the conversion of thermal resistance measurements between units such as kelvin per watt, degrees Celsius per watt, etc. Thermal resistance quantifies the resistance of a material or structure to heat flow. 7. Thermal Conductivity Converter: This section focuses on converting thermal conductivity measurements between units such as watts per meter-kelvin, BTU per hour-foot-degree Fahrenheit, etc. Thermal conductivity quantifies the ability of a material to conduct heat. 8. Specific Heat Capacity Converter: Enables the conversion of specific heat capacity measurements between units such as joules per kilogram-kelvin, calories per gram-degree Celsius, etc. Specific heat capacity quantifies the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin. 9. Heat Density Converter: Converts heat density measurements between units such as joules per cubic meter, calories per cubic centimeter, BTU per cubic inch, etc. Heat density quantifies the amount of heat energy stored or transferred per unit volume of a substance or material. 10. Heat Flux Density Converter: This sub-category focuses on converting heat flux density measurements between units such as watts per square meter, BTU per square foot-second, etc. Heat flux density quantifies the rate of heat transfer per unit area. 11. Heat Transfer Coefficient Converter: Enables the conversion of heat transfer coefficient measurements between units such as watts per square meter-kelvin, BTU per hour-square foot-degree Fahrenheit, etc. Heat transfer coefficient quantifies the rate of heat transfer between two surfaces or mediums. By providing these specialized conversion tools, "Heat converters" empower users to work effectively with heat-related measurements, ensuring accuracy and consistency in various applications such as thermal engineering, HVAC (heating, ventilation, and air conditioning) systems, energy efficiency assessments, and more. Whether you're designing heat exchangers, analyzing thermal properties of materials, or optimizing energy usage, these tools offer valuable support for professionals, researchers, and students in the field of heat transfer and thermal sciences.	{"url":"https://toolsfairy.com/heat-converters","timestamp":"2024-11-02T14:12:52Z","content_type":"text/html","content_length":"24448","record_id":"<urn:uuid:7555498d-9894-45e5-8ade-7dce311c7172>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00894.warc.gz"}
PPT - Works of the Heart PowerPoint Presentation, free download - ID:580896 1. Works of the Heart Carve your name on Hearts and not on marble. - Charles H. Spurgeon TA-4-1a 3. Coordinate Graphs Do you think taller people have wider arm spans?: TK-3-1 4. Overview of Investigation 4 • Goals: • to implement the process of statistical investigation to answer questions • to review the process of measuring length, time, and distance • to analyze data by using coordinate graphs to explore relationship among variables • to explore intervals for scaling the vertical axis (y- axis) and the horizontal axis (x-axis) • Investigation 4 : Coordinate Graphs • Materials (p. 41j) • Student Handbook Pages (pp. 42 - 52) • Teaching the Investigation (pp. 52a – 52g) TK-3-2 5. Coordinate Graphs Is there really any relationship between a person’s height and his or her arm span?: TD-4-1 6. Coordinate Graphs How might we organize and display the data in a graph to help us answer this question?: TD-4-2 7. Relating Height to Arm Span Does the data in the double bar graph indicate any relationship between height and arm span? TD-4-3 8. Relating Height to Arm Span Does the data in the back-to-back stem plot indicate any relationship between height and arm span? TD-4-4 9. Problem 4.1 • Think about this question. If you know the measurement of a person’s arm span, do you know anything about his or her height? • To help you answer this question, you will need to collect some data. With you class, collect the height and arm span of each person in your class. Make a coordinate graph of your data. Then, use your graph to answer the question above. 10. Problem 4.1 follow-up • Draw a diagonal line through the points on the graph where the measures for arm span and height are the same. 1.How any of your classmates’ data are on this line? What is true about arm span compared to height for the points on this line? 2. What is true about arm span compared to height for the points below the line you drew? 3. What is true about arm span compared to height for the points above the line you drew? 11. Problem 4.2 • Study the graph on page 46, this graph was made using the data from Problem 3.1 • Look back at the data on page 31. On Labsheet 4.2, locate and label with initials the points for the first five students in the table. • If you know how long it takes a particular student to travel to school, can you know anything about the student’s distance from school? Use the graph to help you answer the question. Write a justification for your answer. 12. 4.2 Follow-up • Locat the points at (17, 4.50) and (60,4.50) on the coordiante graph on Labsheet 4.2. What can you tell about the students these points represent? • Locate the points (30, 2.00), (30, 3.00) and (30, 4.75). What can you tell about the students these points represent? • What would the graph look like if the same scale were used for both axes? 13. Overview of Investigation 5 • Goals: • to understand the mean as a number that “evens out” or “balances” a distribution • to create distributions with a designated mean • to find the mean of a set of data • to use the mean to help describe a set of data • to reason with a model that clarifies the development of the algorithm for finding the mean • to distinguish between the mean, median, and mode as ways to describe what is typical about a set of data • Investigation 5: What Do We Mean by Mean? • Materials (p. 52h) • Student Handbook Pages (pp. 53 - 67) • Teaching the Investigation (pp. 67a – 67l) TF-4-2 14. Terms • Mode “the value that occurs most frequently.” • Range “the spread of data values from the lowest value to the highest value.” • Median “the value that divides the data in half.” (half of the values are below the median, and half the values are above the median) • Mean “the average of the values of the data.” TF—4-1 15. Evening Things Out-Inv. 5 The purpose of the Census Bureau is to count the number of people living in the United States in order to compute the number of representatives each state will have in the United States House of Representatives. The census focuses on counting the people who live in households rather than “how many people are in a family.” TG-4-1 16. Ollie 2 people Ruth 4 people Yarnell 3 people Paul 6 people Gary 3 people Brenda 6 people Evening Things Out-Inv. 5.1 Six students in a middle school class determined the number of people in their households using the United States census guidelines. Their data is as follows: How could we determine the average number of people in these six households? TG-4-2 17. 5.1 Follow-up • The students had an idea for finding the average number of people in the households. They decided to rearrange the cubes to try to “even out” the number of cubes in each tower. Try this on your own and see what you find for the average number of people in the households, and then read on to see what the students did. (page 55-57) 18. Problem 5.2 A. Make a set of cube towers to show the size of each household. B. Make a line plot of the data. C. How many people are there in the six households altogether? Describe how you determined your answer. D. What is the mean number of people in the six households? Describe how you determined your answer. 19. Problem 5.2 Follow-up • How does the mean for this set of six students compare to the mean for the six students in Problem 5.1? • How does the median for this set of six students compare to the median for the six students in Problem 5.1? 20. The grateful heart is not only the greatest virtue but the parent of all others. - Cicero TK-4-1	{"url":"https://www.slideserve.com/orsin/works-of-the-heart","timestamp":"2024-11-13T19:37:25Z","content_type":"text/html","content_length":"92213","record_id":"<urn:uuid:f22ba67c-a767-401c-affc-d570b713ce8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00128.warc.gz"}
Nuclear induction lineshape: Non-Markovian diffusion with boundaries Data files Jan 19, 2024 version files 2.60 GB The dynamics of viscoelastic fluids are governed by a memory function, which is critical yet computationally intensive to determine, particularly when diffusion is restricted by boundaries. We introduce a computational method that effectively captures the memory effects by analyzing the time-correlation function of the pressure tensor, an indicator of viscosity, through the analytic continuation of the Stokes-Einstein equation to the Laplace domain. This equation is integrated with molecular dynamics (MD) simulations to obtain necessary parameters. Our method computes NMR lineshapes by employing a generalized diffusion coefficient that incorporates the influences of temperature and confinement geometry. This approach establishes a direct link between the memory function and thermal transport parameters, enabling precise computation of NMR signals for non-Markovian fluids in confined geometries. README: Nuclear Induction Lineshape: Non-Markovian Diffusion with Boundaries The simulation was executed on the Hoffman2 cluster at UCLA. The "submit_job.sh" script was used for running a batch of simulations. A sample of LAMMPS code for viscosity evaluation can be found in each folder, "in.visc_Bulk" and "in.visc_SiO2" for bulk and restricted simulations, respectively. For data analysis in Jupyter Lab, we provide a Python code in the "Viscosity Analysis.ipynb" notebook. Description of the data and file structure The various folders are marked to indicate different boundary conditions. The "Xegas.zip" folder contains the dataset for simulating Xe gas particles in bulk. Folders named A11-A75 hold datasets for cylindrical boundary conditions, with the tube radii in Angstroms represented by two digits in each folder's name. Each folder contains a file with the LAMMPS code named "in.visc_SiO2", a file titled "submit_job.sh" for batch submission of simulations to the HPC facility, several log files documenting code performance, and 90 data files. The data file names have a specific structure: the first part indicates the simulation phase and the monitored parameter, the second part denotes the temperature, and the third part specifies the random seed used for the simulation. Each simulation is conducted with 10 different random seeds across 9 temperatures: 240, 260, 280, 300, 320, 340, 360, 380, and 400 K. For instance, a file named "viscg_280_1.dat" represents the results of a viscosity simulation for Xe gas particles at 280 K using the first set of random distributions. These data are generated using a "ave/correlate" fix within the LAMMPS code. The resulting output includes rows displaying the timestep in the first column, followed by the simulated parameters in subsequent rows. Specifically, the components of the pressure tensor correlations, denoted as "v_pxy," "v_pxz," and "v_pyz," are presented in the last three columns. The Jupyter notebook contains a function demonstrating how to import and work with such a file. The Lammps code has the following sections: 1. Variable assignments: This section involves defining crucial variables, such as temperature (T), simulation dimensions (L), the number of particles (npart), and sampling intervals. 2. Unit conversion to SI: Here, unit conversions are encoded to ensure consistency with the International System of Units (SI). 3. Molecular Dynamics simulation setup: This part involves configuring the molecular dynamics simulation by specifying boundary conditions, defining the simulation region, setting coupling constants, and determining the time steps. 4. NVT canonical ensemble fix: A fix is defined to maintain the NVT canonical ensemble during the simulation. 5. Equilibration (Prerun): Prior to data collection, a prerun of 1e6 steps is executed to allow the system to reach an equilibrium state. 6. Correlation term computation: This section of the code focuses on calculating the correlation terms for the pressure tensor based on the Kubo formulation. The necessary summations are performed. 7. Main simulation run: The code proceeds to run for the desired number of steps, during which preliminary viscosity coefficient data is collected and printed for analysis. The simulation was performed for various temperatures in the range of 200-400 K for xenon particles in gas phase for both bulk and restricted diffusion. MD simulations were conducted to calculate the shear viscosity of gaseous xenon (Xe), as defined by Eq. (7). We utilized the Lennard-Jones (LJ) pair potential, expressed as U(r) = 4ϵ[(σ/r)^12 − (σ/r) ^6], where interactions between xenon atoms were characterized by ϵ = 1.77 kJ/mol, the depth of the potential well, and σ = 4.1 Å, the distance at which the potential energy becomes zero. Our simulations of bulk fluid were conducted for isotropic diffusion by placing 2000 xenon atoms within a box defined by periodic boundary conditions. Throughout the simulations, we maintained a consistent particle count, volume, and temperature, adhering to the canonical ensemble (NVT ensemble). Each set of simulations was repeated for ten different random seeds for the initial positions and velocities of the particles to ensure robust statistical sampling and accuracy of the results. In the simulations of restricted diffusion (i.e., diffusion limited by the nanotube geometry), nanotubes of a fixed length and various diameters were employed, and the number of particles was adjusted to maintain a constant particle density. For the simulations, we used xenon (Xe) particles. The interactions between the Xe particles and the cylindrical boundary were modeled using the Lennard-Jones potential, with parameters ϵ = 0.3 kJ/ mol and σ = 4.295 Å, representing a silica tube. To determine the viscosity coefficient, we integrated all three components of the pressure tensor, referred to as C[αβ](τ). Although the C[xy] component showed significantly higher values than the C[yz] and C[xz] components, given the tube’s orientation along the z-axis, we opted to integrate all three components together for a thorough	{"url":"https://datadryad.org:443/stash/dataset/doi:10.5061/dryad.m905qfv7n","timestamp":"2024-11-13T12:03:30Z","content_type":"text/html","content_length":"49886","record_id":"<urn:uuid:d3d0a1d9-e3f7-42ba-a062-a6dde69354fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00440.warc.gz"}
Empty Multiplication Table Worksheet Math, specifically multiplication, creates the foundation of numerous academic self-controls and real-world applications. Yet, for lots of learners, understanding multiplication can position a difficulty. To address this hurdle, educators and moms and dads have actually embraced an effective device: Empty Multiplication Table Worksheet. 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Urging constant method, providing assistance, and producing a positive knowing atmosphere are beneficial actions.	{"url":"https://crown-darts.com/en/empty-multiplication-table-worksheet.html","timestamp":"2024-11-13T21:35:11Z","content_type":"text/html","content_length":"29160","record_id":"<urn:uuid:9a16b62e-eab8-43b7-be44-48c4af162d63>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00309.warc.gz"}
Approximation schemes for degree-restricted MST and Red-Blue Separation Problem We develop a quasi-polynomial time approximation scheme for the Euclidean version of the Degree-restricted MST by adapting techniques used previously for approximating TSP. Given n points in the plane, d = 2 or 3, and ∈ > 0, the scheme finds an approximation with cost within 1 + ∈ of the lowest cost spanning tree with the property that all nodes have degree at most d. We also develop a polynomial time approximation scheme for the Euclidean version of the Red-Blue Separation Problem. Original language English (US) Title of host publication Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) Editors Jos C. M. Baeten, Jan Karel Lenstra, Joachim Parrow, Gerhard J. Woeginger Publisher Springer Verlag Pages 176-188 Number of pages 13 ISBN (Print) 3540404937, 9783540404934 State Published - 2003 Publication series Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) Volume 2719 ISSN (Print) 0302-9743 ISSN (Electronic) 1611-3349 All Science Journal Classification (ASJC) codes • Theoretical Computer Science • General Computer Science Dive into the research topics of 'Approximation schemes for degree-restricted MST and Red-Blue Separation Problem'. Together they form a unique fingerprint.	{"url":"https://collaborate.princeton.edu/en/publications/approximation-schemes-for-degree-restricted-mst-and-red-blue-sepa-3","timestamp":"2024-11-05T07:53:40Z","content_type":"text/html","content_length":"50579","record_id":"<urn:uuid:afc6a78d-b07e-4115-a321-2bc3e80426f7>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00398.warc.gz"}
EViews Help: align Align placement of multiple graphs. You must specify three numbers (each separated by a comma) in parentheses in the following order: the first number n is the number of columns in which to place the graphs, the second number h is the horizontal space between graphs, and the third number v is the vertical space between graphs. Spacing is specified in virtual inches. aligns MYGRAPH with graphs placed in three columns, horizontal spacing of 1.5 virtual inches, and vertical spacing of 1 virtual inch. var var1.ls 1 4 m1 gdp freeze(impgra) var1.impulse(m,24) gdp @ gdp m1 estimates a VAR, freezes the impulse response functions as multiple graphs, and realigns the graphs. By default, the graphs are stacked in one column, and the realignment places the graphs in two For a detailed discussion of customizing graphs, see “Graphing Data”	{"url":"https://help.eviews.com/content/graphcmd-align.html","timestamp":"2024-11-12T06:12:24Z","content_type":"application/xhtml+xml","content_length":"8556","record_id":"<urn:uuid:9549bacd-b13a-4046-ac0f-437ac2e08ee5>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00097.warc.gz"}
Casting Spells Problem C Casting Spells Casting spells is the least understood technique of dealing with real life. Actually, people find it quite hard to distinguish between real spells like “abrahellehhelleh” (used in the battles and taught at the mage universities) and screams like “rachelhellabracadabra” (used by uneducated witches for shouting at cats). Finally, the research conducted at the Unheard University showed how one can measure the power of a word (be it a real spell or a scream). It appeared that it is connected with the mages’ ability to pronounce words backwards. (Actually, some singers were burned at the stake for exactly the same ability, as it was perceived as demonic possession.) Namely, the power of a word is the length of the maximum subword of the form $ww^ Rww^ R$ (where $w$ is an arbitrary sequence of characters and $w^ R$ is $w$ written backwards). If no such subword exists, then the power of the word is $0$. For example, the power of abrahellehhelleh is $12$ as it contains hellehhelleh and the power of rachelhellabracadabra is $0$. Note that the power of a word is always a multiple of $4$. The input is a single line containing a word of length $3 \cdot 10^5$, consisting of (large or small) letters of the English alphabet. You should output one integer $k$, the power of the word. Sample Input 1 Sample Output 1 abrahellehhelleh 12 Sample Input 2 Sample Output 2 rachelhellabracadabra 0	{"url":"https://kth.kattis.com/courses/DD2458/popup17/assignments/csgzmn/problems/castingspells","timestamp":"2024-11-05T09:35:16Z","content_type":"text/html","content_length":"26569","record_id":"<urn:uuid:b37263b9-0365-416a-ac06-08c14d04f5db>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00234.warc.gz"}
Operational historian system retrieving summary data values and source data values based on alignment between a summarization cycle duration and a query cycle duration. A retrieval service process executing on a historian device utilizes a summarization cycle duration, including start and/or end times thereof, and a query cycle duration, including start and/or end times thereof, to determine whether to retrieve, via a communications network, source tag data and/or summary tag data from memory storage devices. Aspects of the present disclosure generally relate to the fields of networked computerized industrial control, automation systems, networked computerized systems utilized to monitor, log, and display relevant manufacturing/production events and associated data, and supervisory level control and manufacturing systems. More particularly, aspects relate to systems and methods for retrieving, via a communications network, source tag data and summary tag data from memory storage devices. Industry increasingly depends upon highly automated data acquisition and control systems to ensure that industrial processes are run efficiently, safely, and reliably while lowering their overall production costs. Data acquisition begins when a number of sensors measure aspects of an industrial process and periodically report their measurements back to a data collection and control system. Such measurements come in a wide variety of forms. By way of example, the measurements produced by a sensor/recorder include: temperature, pressure, pH, and mass/volume flow of material, as well as a tallied inventory of packages waiting in a shipping line and/or a photograph of a room in a factory. Storing, retrieving, and analyzing gathered process data is an important part of running an efficient process. Conventional systems and methods utilize retrieval services that are cyclical and the number of retrieved data values depends on the number of cycles. Data queries using the retrieval services may have large time intervals and/or cycle durations. For example, even if a query result contains only a few hundred rows, the retrieval service has to process millions of source data values. The conventional retrieval services are slow and overburden communications network bandwidth. Aspects of the disclosure improve operation of networked computerized industrial control, automation systems, networked computerized systems utilized to monitor, log, and display relevant manufacturing/production events and associated data, as well as supervisory level control and manufacturing systems by reducing extra and unnecessary utilization of processor resources and network bandwidth while satisfying query parameters. Aspects of the disclosure further utilize a summarization cycle duration, including start and/or end times thereof, and a query cycle duration, including start and/or end times thereof, to determine whether to retrieve, via a communications network, source tag data and/or summary tag data from memory storage devices. In an aspect, an operational historian system includes at least one processor and processor-executable instructions stored on at least one computer-readable storage medium. When executed by the processor, the processor-executable instructions implement a replication component and a retrieval component. The replication component is configured to generate a summary data value from source data values stored in a source database. The source data values are indicative of a physical property of a component within a continuous process. The summary data value comprises a statistical representation of the source data values for a summarization cycle duration. Moreover, the replication component is configured to store the summary tag data value in a summary database. The retrieval component is configured to receive a data query, which has a query cycle duration, from a client computing device via a communications network. The retrieval component is also configured to retrieve the summary data value from the summary database when the summarization cycle duration is less than or equal to the query cycle duration. In another aspect, a computer-implemented method includes a retrieval service, which is executing on historian of a distributed historization system, receiving a data query for source data values from a remote computing device. The data query has a query cycle duration and the source data values correspond to a physical property of a component in an industrial process. The method further includes the executing retrieval service retrieving summary tags from a metadata server executing on the historian. The summary tags each have a summary cycle duration and correspond to the source data values. The executing retrieval service further retrieves summary data values from a summary database of the distributed historization system when the query cycle duration includes a whole summary cycle duration. Furthermore, the executing retrieval service retrieves source data values from a source database of the distributed historization system when the query cycle duration includes a partial summary cycle duration. In yet another aspect, a distributed historization system includes a historian processor and a historian memory storage device that stores source data, summary data, and processor-executable instructions for execution by the historian processor to implement a summarization retrieval module. When executed by the historian processor, the processor-executable instructions are configured for receiving a query, by the summarization retrieval module via a communications network, from a client device. The received query has a query cycle duration. The processor-executable instructions are further configured for causing the summarization retrieval module to retrieve the summary data when the query cycle duration includes a whole summary cycle duration of the summary data and when both a start time and an end time of the query aligns with a start time and an end time of a summary cycle of the summary data. Moreover, the processor-executable instructions are configured for causing the summarization retrieval module to retrieve the source data when the query cycle duration includes a partial summary cycle and when the start time of the query is misaligned with the start time of the summary cycles and when the end time of the query is misaligned with the end time of the summary cycles. The processor-executable instructions are also configured for causing the summarization retrieval module to merge the retrieved summary data and the retrieved source data into a query result and transmit the query result from the historian memory storage device to the client device via the communications network. Other objects and features will be in part apparent and in part pointed out hereinafter. FIGS. 1A and 1B are block diagrams illustrating an exemplary system within which aspects of the disclosure may be incorporated. FIG. 2 is a block diagram illustrating an exemplary historian architecture according to an embodiment. FIGS. 3A to 3D illustrate an exemplary summarization retrieval process according to an embodiment. FIGS. 4 and 5 are diagrams illustrating exemplary selections of data values by the exemplary summarization retrieval process of FIGS. 3A to 3D. FIGS. 6 and 7 are diagrams illustrating exemplary selections of data values by the exemplary summarization retrieval process of FIGS. 3A to 3D using an integral retrieval mode. Corresponding reference characters indicate corresponding parts throughout the drawings. Operational historian systems include replication components/services for replicating data from one historian to one or more other historians. Utilization of these replication services creates a tiered relationship between the historians. For example, data values from a fill sensor may indicate a level of fluid within a tank at 1-second intervals, resulting in 86,400 data values for each 24-hour period. These data values may be stored as a fill sensor tag on a tier one (T1) historian that is geographically near the fill sensor but may need to be accessed by a client device that is geographically remote from the first historian, for instance. The historian system may utilize a tier two (T2) historian that is geographically nearer to the client device to provide data values to the client device. However, transferring all 86,400 data values, which is only for a single sensor in a continuous process that may utilize hundreds of thousands of sensors, from the T1 historian to the T2 historian would burden communications network bandwidth and other resources. To alleviate this burden, the replication components/services of the historian system may instead transfer summary tags (e.g., summary data values), which include statistical information about the data values, from the T1 historian to the T2 historian. U.S. Pat. No. 8,676,756, entitled Replicating Time-Series Data Values for Retrieved Supervisory Control and Manufacturing Parameter Values in a Multi-Tiered Historian Server Environment, provides additional details regarding tiered historians and is incorporated herein by reference in its entirety. The summary tags enable T1 historians to provide low resolution (e.g., 30-minute, 1-day, etc.) summary descriptions based upon a stream of high resolution (e.g., 1-second, 1-minute, etc.) data values received by the T1 historian. The T1 historian initially receives data for summary tags as a stream of non-summary data points for a particular tag. In accordance with a specified summary T2 tag, the T1 historian converts the streaming data for a cycle (e.g., time period) into a summary of the data received for the tag during the cycle. For example, the T1 historian analyzes and stores statistical information about the non-summary tag value at specified intervals, such as every 15 minutes. The summary tag is thereafter transmitted by the T1 historian to the T2 historian via a communications network. Exemplary types of summary tags include, but are not limited to, analog summary tags and state summary tags. Analog summary replication includes a T1 historian providing summary statistics for analog tags to a T2 historian. Analog summary tags include summary statistics derived from analog data acquired during a designated summary cycle. For example, analog summary statistics include the following attributes: First, FirstDateTime, Integral, IntegralOfSquares, Last, LastDateTime, MaxDateTime, Maximum, MinDateTime, Minimum, StartDateTime, TimeGood, and ValueCount. Result attributes for analog summaries calculated based on those attributes include: PercentGood, First, FirstDateTime, Last, LastDateTime, Minimum, MinDateTime, Maximum, MaxDateTime, Average, StdDev, Integral, and ValueCount. State summary replication includes a T1 historian summarizing discrete state values for a tag during a specified summary cycle. State summary tags facilitate analyzing discrete process variables, such as a machine state (e.g., running, starting, stopping, standby, off, etc.). For example, state summary statistics include the following attributes: MaxContained, MinContained, PartialEnd, PartialStart, StartDateTime, State, StateEntryCount, and TotalContained. Result attributes for state summaries calculated based on those attributes include: StateCount, ContainedStateCount, StateTimeMin, StateTimeMinContained, StateTimeMax, StateTimeMaxContained, StateTimeAvg, StateTimeAvgContained, StateTimeTotal, StateTimeTotalContained, StateTimePercent, and StateTimePercentContained. The contained designation refers to states that begin (e.g., enter) and end (e.g., exit) within a period of interest (e.g., a shift). Thus, a state that begins and/or ends outside a period of interest is not contained within the period of interest. As an example, co-pending, co-owned U.S. patent application Ser. No. 14/970,062, entitled Historical Summarization in a Process Control Environment, filed Dec. 15, 2015, discloses summarizing history associated with a historized reference or tag and is incorporated herein by reference in its entirety. The replication services/components of T1 historians also support simple (e.g., full data) replication, which retains full data resolution. In an embodiment, simple replication involves a straightforward copying of the tag data from a T1 historian to a T2 historian. When a tag is configured on a T1 historian for simple replication, all data values stored at the T1 historian for that tag are replicated to a T2 historian. Analog, discrete, and string data tags can be configured for simple replication, in an exemplary embodiment. Having provided a high-level summary of illustrative aspects of the exemplary T1/T2 historian replication arrangement, attention is directed to the figures and their associated written descriptions. It is noted that the following description is based on illustrative embodiments of the disclosure and should not be taken as limiting the disclosure with regard to alternative embodiments that are not explicitly described herein. FIGS. 1A and 1B illustrate an exemplary tiered historian system, generally indicated at 100, within which an embodiment of the disclosure may be incorporated. Referring further to FIG. 1A, the tiered historian system includes T1 historians 102, a communications network 104, a T2 historian 106, and a client device 108. Each of the T1 historians 102 receives and stores, on a memory storage device, data values from a continuous process associated therewith and transmits the data values (e.g., by simple replication and/or summary replication) to the T2 historian 106 via the communications network 104 for access by the client device 108. For example, the T1 historians 102 and T2 historian 106 may each be a single server computing device and/or a collection of networked server computing devices (e.g., a cloud). The T1 historians 102 and T2 historian 106 may also comprise a single computing device (e.g., machine) in accordance with one or more embodiments. In an embodiment, client device 108 includes any computing device capable of executing processor-executable instructions and providing a graphical user interface (GUI) including, but not limited to, personal computers, laptops, workstations, tablets, smartphones, mobile devices, and the like. As an example, co-pending, co-owned U.S. patent application Ser. No. 14/704,661, entitled Distributed Historization System, filed May 5, 2015, discloses a unified approach for historizing to the cloud and is incorporated herein by reference in its entirety. The communications network 104 is capable of facilitating the exchange of data among various components of system 100, including T1 historians 102 and T2 historian 106. The communications network 104 in the embodiment of FIGS. 1A and 1B includes a wide area network (WAN) that is connectable to other telecommunications networks, including other WANs or portions of the Internet or an intranet, including local area networks (LANs). The communications network 104 may be any telecommunications network that facilitates the exchange of data, such as those that operate according to the IEEE 802.3 (e.g., Ethernet) and/or the IEEE 802.11 (e.g., Wi-Fi) protocols, for example. In another embodiment, communications network 104 is any medium that allows data to be physically transferred through serial or parallel communication channels (e.g., copper wire, optical fiber, computer bus, wireless communication channel, etc.). In an embodiment, communications network 104 comprises at least in part a process control network. In another embodiment, communications network 104 comprises at least in part a SCADA system. In yet another embodiment, communications network 104 comprises at least in part an enterprise manufacturing intelligence (EMI)/operational intelligence (OI) system. Referring further to FIG. 1B, the system 100 also includes an exemplary plant, such as a fluid processing system 110. As illustrated, the fluid processing system 110 includes process controllers 112, tanks 114, valves 116, sensors 118, and a pump 120. In system 100, T1 historians 102, T2 historians 106, client device 108, process controllers 112, the tanks 114, the valves 116, sensors 118, and the pump 120 are communicatively coupled via communications network 104. Still referring to FIG. 1B, the fluid processing system 110 is adapted for changing or refining raw materials to create end products. It will be apparent to one skilled in the art that aspects of the present disclosure are capable of optimizing processes and processing systems other than fluid processing system 110 and that system 110 is presented for illustration purposes only. Additional exemplary processes include, but are not limited to, those in the chemical, oil and gas, food and beverage, pharmaceutical, water treatment, and electrical power industries. For example, processes may include conveyers, power distribution systems, and/or processes or operations that cannot be interrupted. In an embodiment, process controllers 112 provide an interface or gateway between components of fluid processing system 110 (e.g., valves 116, sensors 118, pump 120) and other components of system 100 (e.g., T1 historians 102, T2 historian 106, client device 108). In another embodiment, components of fluid processing system 110 communicate directly with T1 historians 102, T2 historian 106, and/or client device 108 via communications network 104. In yet another embodiment, process controllers 112 transmit data to and receive data from T1 historians 102, T2 historian 106, client device 108, valves 612, sensors 614, and/or pump 616 for controlling and/or monitoring various aspects of fluid processing system 110. The process controllers 112 of FIG. 1B are adapted to control and/or monitor aspects of fluid processing system 110. In an embodiment, processor controllers 112 are programmable logic controllers (PLC) that control and collect data from aspects of fluid processing system 110. FIG. 2 illustrates an exemplary architecture of an embodiment in which T1 historians 102 and T2 historian 106 comprise a single machine. In the illustrated embodiment, the historian includes a retrieval component 202, a metadata server 204, a summary database 206, a storage component 207, a replication component 208, and a source database 210. The retrieval component 202 of the exemplary embodiment is adapted to receive queries from client device 108, locate the requested data, perform necessary processing, and return the results to client device 108. In one form, retrieval component 202 creates new tag lists for multiple tag queries that may be a mix of original tags and summarization tags based on the available information. In an embodiment, retrieval component 202 is provided as processor-executable instructions that comprise a procedure, a function, a routine, a method, and/or a subprogram of the historian. Further details regarding retrieval component 202 are provided herein. The metadata server 204 is adapted to store and provide to retrieval component 202 metadata about which source tags stored in source database 210 correspond to a particular summary tag stored in summary database 206. In an embodiment, metadata server 204 is provided as processor-executable instructions that comprise a procedure, a function, a routine, a method, and/or a subprogram of the historian. Additional details regarding metadata server 204 are provided herein and in U.S. patent application Ser. No. 14/833,906, entitled Storing and Identifying Metadata through Extended Properties in a Historization System, which is incorporated herein by reference in its entirety. The replication component 208 is adapted to replicate data values from the source database 210. In one embodiment, replication component 208 provides summary replication by analyzing and producing summary statistics for data values stored in source database 210 as further explained herein. In another embodiment, replication component 208 provides data replication on a schedule having a fixed cycle duration. In yet another embodiment, replication component 208 provides data replication on a custom schedule having any duration. For example, the cycle duration may be stored in tag metadata on metadata server 204 (e.g., StorageRate). In one form, replication component 208 is provided as processor-executable instructions that comprise a procedure, a function, a routine, a method, and/or a subprogram of the historian. FIGS. 3A and 3B illustrate an exemplary summarization retrieval process in accordance with an aspect of the disclosure. Referring further to FIG. 3A, the process begins when retrieval component 202 receives a query for data from client device 108. The query has a cycle duration defined by a start time and an end time. At step 302, retrieval component 202 retrieves the summary tags corresponding to the queried source tags from metadata server 204. Retrieval component 202 then determines, at step 304, whether any of the retrieved summary tags have a cycle duration less than or equal to the query cycle duration. In an embodiment, retrieval component 202 utilizes a threshold when determining whether any of the retrieved summary tags have a cycle duration equal to the query cycle duration. For example, the retrieval component 202 may consider a summary tag cycle duration that is greater than the query cycle duration to be equal to the query cycle duration as long as the summary tag cycle duration is within a certain threshold (e.g., five percent) of the query cycle duration. When no summary tags exist that have a cycle duration less than or equal to the query cycle duration, retrieval component 202 retrieves the full source data from source database 210 using the source tags at step 306. In other words, the replication component 208 performs simple replication and the full tag data is directly copied from source database 210 to retrieval component 202. At step 308, retrieval component 202 processes the full source data for the query cycle and returns it to the client device 108, ending the process. When retrieval component 202 determines at step 304 that at least one of the retrieved summary tags has a cycle duration less than or equal to the query cycle duration, the process continues to step 310. At step 310, retrieval component 202 determines whether a start and/or end time of the query cycle is aligned with a start and/or end time of the summarization cycle. When the query cycle is misaligned with the summarization cycle, retrieval component 202 attempts to retrieve, at step 312, summary data from summary database 206 using the summarization tags with a shorter cycle duration for the misaligned period. In an embodiment, retrieval component 202 utilizes the exemplary Subroutine 1 illustrated in FIG. 3C. For instance, if the summarization cycle is 2:00-2:30, 2:30-3:00, etc. and the query cycle begins at 2:15, retrieval component 202 checks to determine if any summary cycles exist with an interval less than or equal to 15 minutes for the 2:15-2:30 period. If such summary cycles exist, retrieval component 202 will utilize those summary values rather than retrieving the full source data from source database 210. But if no such summary cycles exist, retrieval component 202 retrieves the full source data from source database 210 using the source tags for the misaligned period. As will be understood by one having skill in the art, Subroutine 1 illustrated in FIG. 3C may call itself such that there may be one or more levels of recursion. Referring further to FIG. 3A, when retrieval component 202 determines the query cycle is aligned with the summarization cycle at step 310 and/or after the retrieval component 202 retrieves data for the misaligned period at step 312, the process continues to step 314. At step 314, retrieval component 202 retrieves summary data from summary database 206 using the summary tags for the aligned period(s). Retrieval component 202 processes the retrieved summary data at step 316. Retrieval component 202 also processes the retrieved source data at step 316 for any misaligned periods. Referring further to FIG. 3B, retrieval component 202 detects gaps in the summary data at step 318. In an embodiment, replication component 208 guarantees that each summarization cycle has at least one data value. If a particular summarization cycle does not include at least one data value (i.e., has a gap), the value will need to be backfilled. At step 320, retrieval component 202 determines whether a gap exists in summary data for one or more summarization cycles. When retrieval component 202 determines that no summary data gaps exist, retrieval component 202 replaces metadata to source tag at step 322. When retrieval component 202 determines, at step 320, that a summary data gap exists for at least one summarization cycle, retrieval component 202 takes at least one of three actions. In an embodiment, retrieval component 202 attempts to retrieve, at step 324, summary data from summary database 206 using the summarization tags with a shorter cycle duration for the data gap period. In accordance with an aspect of the disclosure, retrieval component utilizes the exemplary Subroutine 2 illustrated in FIG. 3D. As illustrated in FIG. 3D, Subroutine 2 may call itself and/or Subroutine 1 such that there may be one or more levels of recursion. If no such summary cycles exist, retrieval component 202 retrieves the full source data from source database 210 using the source tags for the data gap period. After retrieving data at step 324, retrieval component 202 replaces metadata to source tag at step 322. In an additional or alternative embodiment, when retrieval component 202 determines, at step 320, that a summary data gap exists for at least one summarization cycle, retrieval component 202 utilizes, at step 326, a previous data value (e.g., last known summary value, last known source value, etc.) for that summarization cycle. In another additional or alternative embodiment, when retrieval component 202 determines, at step 320, that a summary data gap exists for at least one summarization cycle, retrieval component 202 ignores the data gap, as shown at step 328, and continues to step 322. After replacing metadata to source tag at step 322, retrieval component 202 merges data from summary tags and source tags at step 330 before returning the merged data to the client device 108 to end the process. In an additional or alternative embodiment, the number of summary data values is larger than the number of source data values for a query cycle having a short duration. For example, the query cycle duration may be 1 minute, 10 seconds and the duration of the summary cycles may be 1 minute. The selected tag may represent values for an infrequent action and thus may not have any data values for a 1-month period. In such a situation, summarization retrieval may need to process more data values using the summary data values than if the source data values are used. In this embodiment, retrieval component 202 may analyze, using a ValueCount field for example, the number of summary data values in the summarization cycle and determine whether to use the summary data values in summary database 206 or the source data values in source database 210. The following example is provided to help explain the process illustrated in FIGS. 3A and 3B and in no way limits the scope of the disclosure. As an example, a query may include client device 108 requesting a fill level of a fluid stored in tank 114-A during a query cycle of 30-minute intervals from 7:45 AM to 5:15 PM on a particular day. The source database 210 stores source tags for data values representing the fill level of tank 114-A for every 1-second interval during the particular day. The summary database 206 stores summary tags for the source tags stored in source database 210. For example, replication component 208 may have summarized the source data using a “best fit” approach on 30-minute intervals. In other words, the summary database 206 stores summary tags for data values representing the “best fit” of the source data for 30-minute intervals from 12:00:00 AM to 11:59:59 PM for the particular day. Upon receiving the query from client device 108, retrieval component 202 retrieves (step 302) the summary tags from metadata server 204 and compares (step 304) the summarization cycle duration to the query cycle duration. In the example, the summarization cycle duration is 30 minutes and the query cycle duration is also 30 minutes. Because the summarization cycle duration is equal to the query cycle duration, retrieval component 202 determines (step 310) whether the query cycle is aligned with the summarization cycles. Here, the query cycle begins at 7:45 AM, but the summarization cycle is on a schedule of 7:00-7:30, 7:30-8:00, 8:00-8:30, etc. so the beginning of the query cycle is misaligned with the summarization cycles. Moreover, the query cycle ends at 5:15 PM so it is misaligned with the summarization cycles that run from 4:30-5:00, 5:00-5:30, 5:30-6:00, etc. In an embodiment, the 7:45 AM to 8:00 AM and 5:00 PM to 5:15 PM periods may be referred to as partial cycles. Due to these misalignments, retrieval component 202 retrieves (step 312) the source data from source database 210 using the source tags for the partial cycles, namely, misaligned periods of 7:45 AM to 8:00 AM and 5:00 PM to 5:15 PM. The retrieval component 202 then retrieves (step 314) the summary data from summary database 206 for the aligned periods (e.g., 8:00-8:30, 8:30-9:00 . . . 4:00-4:30, 4:30-5:00), which may be referred to as full or whole cycles in one or more embodiments. When retrieval component 202 determines (step 320) that no data gaps exist for any of the summarization cycles between 8:00 AM and 5:00 PM, it replaces (step 322) the metadata to source tag, merges (step 330) the summary data and source data, and returns the merged data to client device 108. But, for example, sensor 118-A may have become disconnected or experienced an outage at 1:10 PM. Thus, there will be a gap in the summary data for the 1:00 PM to 1:30 PM summarization cycle. When retrieval component 202 detects this data gap (steps 318 and 320), it will retrieve the source data from source database 210 using the source tags for the gap period of 1:00 PM to 1:30 PM. Additionally or alternatively, retrieval component 202 may assign an uncertainty value to the summarization cycle from 1:00 PM to 1:30 PM or ignore the data gap, as further described herein. The retrieval component 202 then replaces (step 322) the metadata to source tag, merges (step 330) the summary data and source data, and returns the merged data to client device 108. Summarization cycles may be different than query cycles. For example, the duration of summarization cycles may be different than the duration of query cycles or the start and/or end times of the summarization cycles may be different from the start and/or end times of the query cycles. In an embodiment, aspects of system 100 may constrain client device 108 to query cycle durations and start and/or end times that match the summarization cycle durations and start and/or end times. Further to this example, summarization cycles may have 24-hour periods and client device 108 is constrained to query cycles of multiples of 24 hours (e.g., 24 hours, 48 hours, etc.) to match the summarization cycles. When a query cycle is aligned with summarization cycles retrieval component 202 will retrieve and process summary data from summary database 206, and when the query cycle is misaligned with the summarization cycles retrieval component 202 will retrieve and process the full source data from source database 210. In another embodiment, aspects of system 100 provide an approximation when query cycles and summarization cycles are misaligned. Referring to FIG. 4, the duration of a query cycle may be greater than the duration of a summarization cycle for a “best fit” retrieval mode. Under a best fit retrieval mode, the total time for the query is divided into even cycles (e.g., sub-periods) and then at least four values are returned for each cycle: first value in the cycle (“left”), last value in the cycle (“right”), minimum value in the cycle, and maximum value in the cycle. As illustrated in FIG. 4, a best fit summarization by replication component X resulted in four values (Left1, Right1, Min1, Max1) for the R1 summarization cycle and four values (Left2, Right2, Min2, Max2) for the R2 summarization cycle. The query cycle is misaligned with the summarization cycles and thus retrieval component 202 will select the best fit points from those points that are within the query cycle. As illustrated by the square formed by dashed lines, retrieval component 202 selects Max1 as the first value in the query cycle, Left2 as the last value in the query cycle, Min1 as the minimum value in the query cycle, and Left2 as the maximum value in the query cycle. Left2 is selected as the maximum value because the value of Left2 is greater than the maximum value of the R1 summarization cycle (i.e., Max1) and the maximum value of the R2 summarization cycle (i.e., Max2) is outside of the query cycle. Also as illustrated, this embodiment may sacrifice some accuracy by using the summarization data because a full data value (shown as LostMax) may exist within the duration of the query cycle and be the true maximum value within the query cycle. However, because the LostMax value was not within the R1 summarization period and was not the maximum value within the R2 summarization period, it was not included in the summary data available to retrieval component 202. In yet another embodiment, retrieval component 202 selects appropriate query cycles that will be aligned with summarization cycles. For example, retrieval component 202 may alter the time parameters of the query cycle to return a result that is aligned with summarization cycles. As illustrated in FIG. 5, the query cycle may be large enough to encompass two full summarization cycles, but the start and/or end times of each may not align. For Query Cycle 1, retrieval component 202 uses Summarization Cycle 2 and Summarization Cycle 3 and ignores the partial portions of Summarization Cycle 1 and Summarization Cycle 4. In this manner, Query Cycle 1 becomes aligned with the summarization cycles. For Query Cycle 2, retrieval component 202 uses Summarization Cycle 4 and Summarization Cycle 5 and ignores the partial portion of Summarization Cycle 6. In one form, retrieval component 202 utilizes the approach illustrated by FIG. 4 and the approach illustrated by FIG. 5, compares the accuracy of results for each, and uses the one having greater accuracy. The retrieval component 202 supports best fit retrieval modes, integral retrieval modes, time-weighted average retrieval modes, minimum retrieval modes, maximum retrieval modes, and like retrieval modes having cycle duration as a parameter. The retrieval component 202 supports retrieval modes that are delta in nature (e.g., best fit, minimum, maximum) and pure cyclic in nature (e.g., integral, time-weighted average). Best fit, minimum, and maximum retrieval modes do not require retrieval component to perform any calculations. In one form, retrieval component 202 uses an integral retrieval mode for query cycles misaligned with summary cycles in accordance with the approach illustrated by FIG. 4. The retrieval component 202 integrates the summarization cycles within the query cycle using the following formula: $R i = ∑ k = m n I k * t k ′ t k$ where m is the index of the first summarization cycle intersecting the query cycle, n is the index of the last summarization cycle intersecting the query cycle, i is the index of the query cycle, R[i ]is the integral of the i-th query cycle, I[k ]is the integral of the k-th summarization cycle that intersects the i-th query cycle, t′[k ]is the time of intersection between the k-th summarization cycle and the i-th query cycle, t[k ]is the total time of intersection between the k-th summarization cycle and the i-th query cycle, and T[i ]is the total time of the i-th query cycle. In the example illustrated in FIG. 6 with 1-day summarization cycles and 2.5-day query cycles, the integral for the first query cycle is R1=I2+I3+(I4*0.5) because the fourth day summarization cycle is intersected by the query cycle at mid-day. In this example, retrieval component 202 assumes that distribution of the 14 point was constant during the entire fourth day summarization cycle. In another form, retrieval component 202 uses an integral retrieval mode for query cycles misaligned with summary cycles in accordance with the approach illustrated by FIG. 5 (e.g., cycle duration of result will be different than the query cycle duration received from client device 108). The retrieval component 202 integrates summarization cycles that are completely within a particular query cycle (e.g., whole cycles) using the following formula: $R i = ∑ k = m n I k$ where m is the index of the first summarization cycle inside the query cycle, n is the index of the last summarization cycle inside the query cycle, i is the index of the query cycle, R[i ]is the integral of the i-th query cycle, I[k ]is the integral of the k-th summarization cycle that intersects the i-th query cycle, t[k ]is the time of intersection between the k-th summarization cycle and the i-th query cycle, T[i ]is the total time of the i-th query cycle, and the integral divisor is taken from the source tag. In the example illustrated in FIG. 7, with 1-day summarization cycles and 2.5-day query cycles, the integral for the first query cycle is R1=I2+I3 because only the two days (e.g., summarization cycles) that fit entirely within the query cycle are used. The retrieval component 202 can also calculate, based on integral value, a time-weighted average based on stair-step interpolation by the formula: $TWA = I T$ where I is the integral value of the cycle (e.g., calculated by one of the above methods) and T is the cycle time. Embodiments of the present disclosure may comprise a special purpose computer including a variety of computer hardware, as described in greater detail below. Embodiments within the scope of the present disclosure also include computer-readable media for carrying or having computer-executable instructions or data structures stored thereon. Such computer-readable media can be any available media that can be accessed by a special purpose computer. By way of example, and not limitation, computer-readable storage media include both volatile and nonvolatile, removable and non-removable media implemented in any method or technology for storage of information such as computer readable instructions, data structures, program modules or other data. Computer storage media are non-transitory and include, but are not limited to, random access memory (RAM), read only memory (ROM), electrically erasable programmable ROM (EEPROM), compact disk ROM (CD-ROM), digital versatile disks (DVD), or other optical disk storage, solid state drives (SSDs), magnetic cassettes, magnetic tape, magnetic disk storage, or other magnetic storage devices, or any other medium that can be used to carry or store desired program code means in the form of computer-executable instructions or data structures and that can be accessed by a general purpose or special purpose computer. When information is transferred or provided over a network or another communications connection (either hardwired, wireless, or a combination of hardwired or wireless) to a computer, the computer properly views the connection as a computer-readable medium. Thus, any such connection is properly termed a computer-readable medium. Combinations of the above should also be included within the scope of computer-readable media. Computer-executable instructions comprise, for example, instructions and data which cause a general purpose computer, special purpose computer, or special purpose processing device to perform a certain function or group of functions. The following discussion is intended to provide a brief, general description of a suitable computing environment in which aspects of the disclosure may be implemented. Although not required, aspects of the disclosure will be described in the general context of computer-executable instructions, such as program modules, being executed by computers in network environments. Generally, program modules include routines, programs, objects, components, data structures, etc. that perform particular tasks or implement particular abstract data types. Computer-executable instructions, associated data structures, and program modules represent examples of the program code means for executing steps of the methods disclosed herein. The particular sequence of such executable instructions or associated data structures represent examples of corresponding acts for implementing the functions described in such steps. Those skilled in the art will appreciate that aspects of the disclosure may be practiced in network computing environments with many types of computer system configurations, including personal computers, hand-held devices, multi-processor systems, microprocessor-based or programmable consumer electronics, network PCs, minicomputers, mainframe computers, and the like. Aspects of the disclosure may also be practiced in distributed computing environments where tasks are performed by local and remote processing devices that are linked (either by hardwired links, wireless links, or by a combination of hardwired or wireless links) through a communications network. In a distributed computing environment, program modules may be located in both local and remote memory storage An exemplary system for implementing aspects of the disclosure includes a special purpose computing device in the form of a conventional computer, including a processing unit, a system memory, and a system bus that couples various system components including the system memory to the processing unit. The system bus may be any of several types of bus structures including a memory bus or memory controller, a peripheral bus, and a local bus using any of a variety of bus architectures. The system memory includes nonvolatile and volatile memory types. A basic input/output system (BIOS), containing the basic routines that help transfer information between elements within the computer, such as during start-up, may be stored in ROM. Further, the computer may include any device (e.g., computer, laptop, tablet, PDA, cell phone, mobile phone, a smart television, and the like) that is capable of receiving or transmitting an IP address wirelessly to or from the internet. The computer may also include a magnetic hard disk drive for reading from and writing to a magnetic hard disk, a magnetic disk drive for reading from or writing to a removable magnetic disk, and an optical disk drive for reading from or writing to removable optical disk such as a CD-ROM or other optical media. The magnetic hard disk drive, magnetic disk drive, and optical disk drive are connected to the system bus by a hard disk drive interface, a magnetic disk drive-interface, and an optical drive interface, respectively. The drives and their associated computer-readable media provide nonvolatile storage of computer-executable instructions, data structures, program modules, and other data for the computer. Although the exemplary environment described herein employs a magnetic hard disk, a removable magnetic disk, and a removable optical disk, other types of computer readable media for storing data can be used, including magnetic cassettes, flash memory cards, digital video disks, Bernoulli cartridges, RAMs, ROMs, SSDs, and the like. Communication media typically embody computer readable instructions, data structures, program modules or other data in a modulated data signal such as a carrier wave or other transport mechanism and includes any information delivery media. One or more aspects of the disclosure may be embodied in computer-executable instructions (i.e., software), routines, or functions stored in system memory or nonvolatile memory as application programs, program modules, and/or program data. The software may alternatively be stored remotely, such as on a remote computer with remote application programs. Generally, program modules include routines, programs, objects, components, data structures, etc. that perform particular tasks or implement particular abstract data types when executed by a processor in a computer or other device. The computer executable instructions may be stored on one or more tangible, non-transitory computer readable media (e.g., hard disk, optical disk, removable storage media, solid state memory, RAM, etc.) and executed by one or more processors or other devices. As will be appreciated by one of skill in the art, the functionality of the program modules may be combined or distributed as desired in various embodiments. In addition, the functionality may be embodied in whole or in part in firmware or hardware equivalents such as integrated circuits, application specific integrated circuits, field programmable gate arrays (FPGA), and the like. The computer may operate in a networked environment using logical connections to one or more remote computers. The remote computers may each be another personal computer, a tablet, a PDA, a server, a router, a network PC, a peer device, or other common network node, and typically include many or all of the elements described above relative to the computer. The logical connections include a local area network (LAN) and a wide area network (WAN) that are presented here by way of example and not limitation. Such networking environments are commonplace in office-wide or enterprise-wide computer networks, intranets and the Internet. When used in a LAN networking environment, the computer is connected to the local network through a network interface or adapter. When used in a WAN networking environment, the computer may include a modem, a wireless link, or other means for establishing communications over the wide area network, such as the Internet. The modem, which may be internal or external, is connected to the system bus via the serial port interface. In a networked environment, program modules depicted relative to the computer, or portions thereof, may be stored in the remote memory storage device. It will be appreciated that the network connections shown are exemplary and other means of establishing communications over wide area network may be used. Preferably, computer-executable instructions are stored in a memory, such as the hard disk drive, and executed by the computer. Advantageously, the computer processor has the capability to perform all operations (e.g., execute computer-executable instructions) in real-time. The order of execution or performance of the operations in embodiments illustrated and described herein is not essential, unless otherwise specified. That is, the operations may be performed in any order, unless otherwise specified, and embodiments may include additional or fewer operations than those disclosed herein. For example, it is contemplated that executing or performing a particular operation before, contemporaneously with, or after another operation is within the scope of aspects of the disclosure. Embodiments may be implemented with computer-executable instructions. The computer-executable instructions may be organized into one or more computer-executable components or modules. Aspects of the disclosure may be implemented with any number and organization of such components or modules. For example, aspects of the disclosure are not limited to the specific computer-executable instructions or the specific components or modules illustrated in the figures and described herein. Other embodiments may include different computer-executable instructions or components having more or less functionality than illustrated and described herein. When introducing elements of aspects of the disclosure or the embodiments thereof, the articles “a”, “an”, “the” and “said” are intended to mean that there are one or more of the elements. The terms “comprising”, “including”, and “having” are intended to be inclusive and mean that there may be additional elements other than the listed elements. Having described aspects of the disclosure in detail, it will be apparent that modifications and variations are possible without departing from the scope of aspects of the disclosure as defined in the appended claims. As various changes could be made in the above constructions, products, and methods without departing from the scope of aspects of the disclosure, it is intended that all matter contained in the above description and shown in the accompanying drawings shall be interpreted as illustrative and not in a limiting sense. 21. A system for improving the bandwidth usage of a communications network comprising: a historian; one or more non-transitory computer readable mediums; and one or more processors; wherein the one or more non-transitory computer readable mediums comprise: a summary database; one or more summary tags; a source database; one or more source values; and computer executable instructions; wherein the one or more summary tags comprises a statistical representation of the one or more source data values for a summarization cycle duration; wherein the computer executable instructions cause the processor to: generate the one or more summary tags from the source data values stored in the source database and store the one or more summary tags in the summary database; receive a data query comprising a query cycle duration from a data query source; compare the query cycle duration with the summarization cycle duration; and return to the data query source, based on the comparison, one or more of: the one or more source data values, the one or more summary tags, or a combination of the one or more source data values and one or more summary 22. The system of claim 21, wherein the query cycle duration comprises a query start time and a query end time; wherein the one or more summary tags each comprise a summary start time and a summary end time; and wherein at least one of the one or more source data values are returned when the query start time is misaligned with the summary start time, and/or when the query end time is misaligned with the summary end time. 23. The system of claim 21, wherein the query cycle duration comprises a query start time and a query end time; wherein the one or more summary data values each comprise a summary start time and a summary end time; and wherein at least one of the one or more summary tags are returned when both the query start time and query end time is aligned with the summary start time and the summary end time, respectively. 24. The system of claim 21, wherein the query cycle duration comprises a query start time and a query end time; wherein the one or more summary tags each comprise a summary start time and a summary end time; and wherein the combination of the one or more source data values and the one or more summary data values are returned when: the query start time is misaligned with at least one of each summary start time, and/or when the query end time is misaligned with at least one of each summary end time; and at least one of each of the one or more summary tags comprise a summary start time and a summary end time within the query cycle duration. 25. The system of claim 21, wherein the query cycle duration comprises a query start time and a query end time; wherein the one or more summary data values each comprise a summary start time and a summary end time; wherein the one or more source data values are returned when the query start time is misaligned with the summary start time, and/or when the query end time is misaligned with the summary end time; and wherein the one or more summary tags are returned when both the query start time and the query end time is aligned with the summary start time and the summary end time, respectively. 26. The system of claim 25, wherein the computer executable instructions further cause the processor to: retrieve at least one of the one or more summary tags that comprise a summary start time and summary end time within the query cycle duration; retrieve at least one of the one or more source data values that have a source time greater than or equal to the query start time and less than or equal to the summary start time within the query cycle duration; retrieve the one or more source data values that have a source time less than or equal to the query end time and greater than or equal to the summary end time within the query cycle duration; and return the combination of all the retrieved one or more summary tags and the one or more source data values to the query source. 27. The system of claim 25, wherein the computer executable instructions further cause the processor to: replace the query start time and the query end time with an aligned query start time and aligned query end time, respectively; return the one or more summary data values to the query source; wherein the aligned query start time aligns with the summary start time of at least one summary data value; and wherein the aligned query end time aligns with the summary end time of at least one summary data value. 28. The system of claim 25, wherein the summary start time and end time define a time frame for each summary tag; wherein the computer executable instructions further cause the processor to: identify if the one or more summary data values comprises a time frame that the query start time falls therebetween; identify if the one or more summary data values comprises a time frame that the query end time falls therebetween; and return the identified one or more summary tags to the query source. 29. A system for improving bandwidth comprising: a historian system comprising one or more processors and one or more non-transitory computer readable mediums, a metadata server, a retrieval component, a summary database, and a source database; wherein the retrieval component is configured and arranged to receive a query, locate a requested data from the query, perform data processing, and return a result; wherein the query comprises a query cycle duration; wherein the source database comprises one or more source data values and one or more source tags; wherein the summary database comprises one or more summary tags; wherein the metadata server is configured and arranged to store and provide to the retrieval component metadata about which of a plurality of source tags stored in the source database correspond to a plurality of summary tags stored in the summary database, respectively; wherein the retrieval component is configured to determine if at least one of the one or more summary tags has a summarization cycle duration that includes a start and/or and time within the query cycle duration, wherein the retrieval component is configured to determine whether a start and/or end time of the query cycle duration is aligned with a start and/or end time of the summarization cycle duration; wherein when the query cycle duration includes a start and/or end time that is misaligned with the summarization cycle, the retrieval component retrieves the summarization tags from the summary database that include a summarization cycle duration that falls between the misaligned start and/or end time; and wherein the one or more summary tags comprise a statistical representation of the one or more source data values for the summarization cycle duration. 30. The system of claim 29, wherein the retrieval component determines how many of the one or more summary tags have a summarization cycle within the query cycle; and wherein the retrieval component determines whether to return at least one of the one or more summary tags, at least one of the one or more source data values, or a combination of at least one of the one or more summary tags and at least one of the one or more source data values. 31. The system of claim 29, wherein the one or more non-transitory computer readable mediums further comprise instructions that when executed by the one or more processors implement: a replication component; wherein the replication component is configured and arranged to replicate source data values in the source database and transfer the replicated source data values to the retrieval component; and/or wherein the replication component is configured and arranged to replicate the summary tags and transfer the replicated summary tags to the retrieval component. 32. The system of claim 29, wherein when the query cycle duration includes a start and/or end time that is misaligned with the summarization cycle, the retrieval component retrieves from the source database one or more source data values that include a source tag that is outside the summarization cycle duration and within the query cycle duration. 33. The system of claim 32, the one or more non-transitory computer readable mediums further comprise instructions that when executed by the one or more processors implement: a replication component; a first historian; and a second historian; wherein the replication component is configured and arranged to replicate source data values in the source database and transfer the replicated source data values from the first historian to the second historian; and/or wherein the replication component is configured and arranged to transfer the summary tags from the first historian to the second historian; and/or wherein the replication component is configured and arranged to replicate source data values in the source database and transfer the replicated source data values to the retrieval component; and/ or wherein the replication component is configured and arranged to replicate the summary tags and transfer the replicated summary tags to the retrieval component. 34. The system of claim 33, wherein the replication component is configured and arranged to generate the statistical representation; and wherein the replication component is configured to store the summary tag in the summary database. 35. The system of claim 34, wherein the retrieval component determines how many of the one or more summary tags have a summarization cycle within the query cycle; and wherein the retrieval component determines whether to return at least one of the one or more summary tags, at least one of the one or more source data values, or a combination of at least one of the one or more summary tags and at least one of the one or more source data values. 36. The system of claim 35, wherein the first historian and the second historian comprise a single computer comprising at least one of the one or more one or more processors and at least one of the one or more non-transitory computer readable mediums. 37. The system of claim 36, wherein the metadata server, the retrieval component, the summary database, the replication component, and the source database are implemented on at least one of the first historian and/or second 38. A method for improving bandwidth comprising: providing: a historian; one or more non-transitory computer readable mediums; and one or more processors; wherein the one or more non-transitory computer readable mediums comprise: a summary database; one or more summary tags; a source database; one or more source values; and computer executable instructions; wherein the one or more summary tags comprises a statistical representation of the one or more source data values for a summarization cycle duration; and wherein the computer executable instructions cause the processor to perform the steps of: generating the one or more summary tags from the source data values stored in the source database, storing the one or more summary tags in the summary database; receiving a data query comprising a query cycle duration from a data query source; comparing the query cycle duration with the summarization cycle duration; and returning to the data query source, based on the comparison, one or more of: the one or more source data values, the one or more summary tags, or a combination of the one or more source data values and one or more summary tags. 39. The system of claim 38, wherein the query cycle duration comprises a query start time and a query end time; wherein the one or more summary data values each comprise a summary start time and a summary end time; wherein the one or more source data values are returned when the query start time is misaligned with the summary start time, and/or when the query end time is misaligned with the summary end time; and wherein the one or more summary tags are returned when both the query start time and the query end time is aligned with the summary start time and the summary end time, respectively. 40. The system of claim 39, wherein the computer executable instructions further cause the processor to perform the steps of: retrieving at least one of the one or more summary tags that comprise a summary start time and summary end time within the query cycle duration; retrieving at least one of the one or more source data values that have a source time greater than or equal to the query start time and less than or equal to the summary start time within the query cycle duration; retrieving the one or more source data values that have a source time less than or equal to the query end time and greater than or equal to the summary end time within the query cycle duration; and returning the combination of all the retrieved one or more summary tags and the one or more source data values to the query Patent History Publication number : 20200142378 : Jan 2, 2020 Publication Date : May 7, 2020 Patent Grant number 11526142 Inventors Alexander Vasilyevich Bolotskikh (Ladera Ranch, CA), Vinay T. Kamath (Rancho Santa Margarita, CA), Yevgeny Naryzhny (Foothill Ranch, CA), Abhijit Manushree (Laguna Niguel, CA) Application Number : 16/732,956 International Classification: G05B 19/042 (20060101); G05B 19/418 (20060101); G06F 16/25 (20060101); G06F 16/2458 (20060101);	{"url":"https://patents.justia.com/patent/20200142378","timestamp":"2024-11-09T07:13:35Z","content_type":"text/html","content_length":"126541","record_id":"<urn:uuid:ba549300-d7d3-45b9-b6e3-326bcc5caec1>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00711.warc.gz"}
Search trees and core.logic Last week David Nolen (the author of core.logic) was visiting Hacker School so I decided to poke around inside core.logic. I made a PR that adds fair conjunction, user-configurable search and a parallel solver. First, a little background. From a high-level point of view, a constraint solver does three things: • specifies a search space in the form of a set of constraints • turns that search space into a search tree • searches the resulting tree for non-failed leaves Currently core.logic (and cKanren before it) complects all three of these. My patch partly decomplects the latter from the first two, allowing different search algorithms to be specified independently of the problem specification. Let's look at how core.logic works. I'm going to gloss over a lot of implementation details in order to make the core ideas clearer. The search tree in core.logic is representated as a lazy stream of the non-failed leaves of the tree. This stream can be: • nil - the empty stream • (Choice. head tail) - a cons cell Disjunction of two goals produces a new goal which contains the search trees of the two goals as adjacent branches. In core.logic, this is implemented by combining their streams with mplus. A naive implementation might look like this: (defn mplus [stream1 stream2] (nil? stream1) stream2 (choice? stream1) (Choice. (.head stream1) (mplus (.tail stream1) stream2)))) This amounts to a depth-first search of the leaves of the tree. Unfortunately, search trees in core.logic can be infinitely deep so a depth-first search can get stuck. If the first branch has an infinite subtree we will never see results from the second branch. ;; simple non-terminating goal (def forevero (fresh [] (run* [q] [(== q 1)])) ;; with depth-first search blocks immediately, returning (...) ;; with breadth-first search blocks after the first result, returning (1 ...) We can perform breadth-first search by adding a new stream type: • (fn [] stream) - a thunk representing a branch in the search tree And then interleaving results from each branch: (defn mplus [stream1 stream2] (fn? stream1) (fn [] (mplus stream2 (stream1))))) This is how core.logic implements fair disjunction (fair in the sense that all branches of conde will be explored equally). However, we still have a problem with fair conjunction. Conjunction is performed in core.logic by running the second goal starting at each of the leaves of the tree of the first goal. In terms of the stream representation, this looks like: (defn bind [stream goal] (nil? stream) nil ;; failure (choice? stream) (Choice. (bind (.head stream) goal) (bind (.tail stream) goal)) (fn? stream) (fn [] (bind (stream) goal)))) This gives rise to similar behaviour as the naive version of mplus: (run* [q] (!= q q))) ;; with unfair conjunction blocks immediately, returning (...) ;; with fair conjunction the second branch causes failure, returning () I suspect the reason that core.logic didn't yet have fair conjunction is entirely due to this stream representation, which complects all three stages of constraint solving and hides the underlying search tree. Since shackles is based on gecode it has the advantage of a much clearer theoretical framework (I strongly recommend this paper, not just for the insight into gecode but as a shining example of how mathematical intuition can be used to guide software design). The first step in introducing fair conjunction to core.logic is to explicitly represent the search tree. The types are similar: • nil - the empty tree • (Result. state) - a leaf • (Choice. left right) - a branch • (Thunk. state goal) - a thunk containing the current state and a sub-goal Defining mplus is now trivial since it is no longer responsible for interleaving results: (defn mplus [tree1 tree2] (Choice. tree1 tree2)) And we now have two variants of bind: (defn bind-unfair [tree goal] (nil? goal) nil ;; failure (result? tree) (goal (.state tree)) ;; success, start the second tree here (choice? tree) (Choice. (bind-unfair (.left tree) goal) (bind-unfair (.right tree) goal)) (thunk? tree) (Thunk. (.state tree) (bind-unfair ((.goal tree) state) goal)))) (defn bind-fair [tree goal] (nil? goal) nil ;; failure (result? tree) (goal (.state tree)) ;; success, start the second tree here (choice? tree) (Choice. (bind-fair (.left tree) goal) (bind-fair (.right tree) goal)) (thunk? tree) (Thunk. (.state tree) (bind-fair (goal state) (.goal tree))))) ;; interleave! The crucial difference here is that bind-fair takes advantage of the continuation-like thunk to interleave both goals, allowing each to do one thunk's worth of work before switching to the next. (We keep bind-unfair around because it tends to be faster in practice - when you know what order your goals will be run in you can use domain knowledge to specify the most optimal order. However, making program evaluation dependent on goal ordering is less declarative and there are also some problems that cannot be specified without fair conjunction. It's nice to have both.) Now that we explicity represent the tree we can use different search algorithms. My patch defaults to lazy, breadth-first search (to maintain the previous semantics) but it also supplies a variety of others including a parallel depth-first search using fork-join. I still need to write a few more tests and sign the clojure contributor agreement before this can be considered for merging. I also have a pesky performance regression in lazy searches - this branch sometimes does more work than the original when only finding the first solution. I'm not sure yet whether this is down to a lack of laziness somewhere or maybe just a result of a slightly different search order. Either way, it needs to be fixed. After this change, core.logic still complects the specification of the search space and the generation of the search tree (eg we have to choose between bind-unfair and bind-fair in the problem specification). At some point I would like to either fix that in core.logic or finish work on shackles. For now though, I'm going back to working on droplet.	{"url":"https://www.scattered-thoughts.net/writing/search-trees-and-core-dot-logic/","timestamp":"2024-11-03T04:02:50Z","content_type":"text/html","content_length":"20856","record_id":"<urn:uuid:8a46a6ba-e4d8-42f9-a124-a43a17a39f9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00214.warc.gz"}
=COSH formula \| Calculate the hyperbolic cosine of a number. Formulas / COSH Calculate the hyperbolic cosine of a number. =COSH(number) • COSH(0) The COSH function returns the hyperbolic cosine of an argument. In this example, the argument is 0, which returns the x-coordinate 1. Therefore, it will evaluate to 1. • COSH(a) The COSH function returns the hyperbolic cosine of an argument. In this example, the argument is a, which is a variable or a number. This formula will evaluate to the hyperbolic cosine of a. The COSH function is used to calculate the hyperbolic cosine of a number, which is the cosine of the hyperbolic angle of that number. COSH requires real number arguments to be provided. • The COSH function returns the x-component of the hyperbolic angle, which is a ray from the origin of the coordinate system that passes through a point on the hyperbola. • The hyperbolic angle area is equal to one-half the hyperbolic angle. Frequently Asked Questions What is the COSH function? The COSH function calculates the hyperbolic cosine for a number. The hyperbolic cosine of a number is the cosine of a number expressed as a hyperbolic angle. What is the argument of the COSH function? The argument for COSH is number. The number argument is any real number. What is required for the COSH function? The number argument is required for the COSH function. What are some examples of usage for the COSH function? The COSH function can be used to calculate the hyperbolic cosine of a number, expressed as a hyperbolic angle. It can also be used to compare angles in hyperbolic space. What are the benefits of using the COSH function? • It can provide insight into the relationships between hyperbolic angles. • It can help to determine trigonometric relationships in hyperbolic space. • It can help to visualize the pattern of the hyperbolic angle.	{"url":"https://sourcetable.com/formula/cosh","timestamp":"2024-11-11T03:29:47Z","content_type":"text/html","content_length":"58575","record_id":"<urn:uuid:881c8554-a585-4f58-a466-b9e124a6c8f3>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00191.warc.gz"}
Point interactions in two and three dimensions as models of small scatterers \| Request PDF We use cookies to offer you a better experience, personalize content, tailor advertising, provide social media features, and better understand the use of our services. To learn more or modify/prevent the use of cookies, see our Cookie Policy and Privacy Policy.	{"url":"https://archive.ymsc.tsinghua.edu.cn/pacm_download/436/9940-Pavel_Exner_449.pdf","timestamp":"2024-11-14T22:03:23Z","content_type":"text/html","content_length":"501217","record_id":"<urn:uuid:db1439dd-70c5-4fb9-96a7-f1c5906e8881>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00794.warc.gz"}
Effective Aperture Calculator In the field of telecommunications and electromagnetic wave physics, the Effective Aperture is a critical concept that describes an antenna's efficiency in capturing power from an incident electromagnetic wave. In simpler terms, it's the measure of an antenna's ability to extract power from the signal it's receiving. Effective Aperture Calculator Effective Aperture (A[e]) = m^2 Example Formula The Effective Aperture can be calculated using the following formula: A[e] = Gλ^2 / 4π 1. A[e]: This is the Effective Aperture. 2. G: This is the antenna gain. 3. λ: This is the wavelength of the incoming signal, which is related to the frequency of the signal. The wavelength can be found using the formula λ = c/f, where c is the speed of light, and f is the frequency. 4. π: This is the mathematical constant Pi. Who wrote/refined the formula The formula for calculating the Effective Aperture is a well-established principle in the field of telecommunications and has been refined over the years by many scientists and engineers. It is not attributed to a specific individual but is a cumulative result of the collective effort in the field. Real Life Application Understanding the Effective Aperture is essential in the design and operation of antennas, particularly in wireless communication systems. It can help in determining the suitability of an antenna for a particular application, affecting the performance of devices like mobile phones, radio and TV broadcast systems, and satellite communications. Key individuals in the discipline While it's difficult to attribute the Effective Aperture concept to a specific individual, noteworthy contributors to the broader field of telecommunications include James Clerk Maxwell, who formulated the foundational equations of electromagnetism, and Guglielmo Marconi, the pioneer of long-distance radio transmission. Interesting Facts 1. The Effective Aperture concept is crucial in telecommunications, affecting our daily life in ways we often don't realize-from the functioning of our mobile devices to the operation of global satellite communication networks. 2. Understanding the Effective Aperture has significantly improved the efficiency of wireless communication, revolutionizing the way we communicate and connect. 3. It is a core concept in the field of radio astronomy, helping astronomers accurately receive and interpret signals from distant cosmic sources. Understanding the Effective Aperture is an integral part of the science of telecommunications and radio physics. It's not just about antennas-it's a gateway to understanding how we capture and interpret information from electromagnetic waves, be it a call from a loved one or a whisper from a distant star. Physics Calculators You may also find the following Physics calculators useful.	{"url":"https://physics.icalculator.com/effective-aperture-calculator.html","timestamp":"2024-11-14T21:11:53Z","content_type":"text/html","content_length":"19738","record_id":"<urn:uuid:39379679-616f-40a2-9d02-5daebf680452>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00820.warc.gz"}
[Misc] - DPS (Damage Per Second) Calculation (For Dummies) Jan 1, 2010 This is a very basic and straightforward was to calculate DPS. You could use these formulas to create a DPS program if you wanted. Let's start with some important assumptions • Normal Attack/Armor Type • No evasion • No armor(defense= 0, damage reduction= 0%) or other damage reducing mediums Now given this sample information: Attack Cooldown = 0.75 Attack Damage Base = 9 Number of Dice = 1 Number of Sides per Die = 1 This is a very basic attack, that is guaranteed to deal 10-10 damage each attack. Let's look at what we have so far: We have a unit that does 1 attack every 0.75 seconds 0.75 seconds per 1 attack 3/4 seconds/attack But we need attacks per second for DPS calculation. So that's attacks/second. Notice how seconds/attackis the inverse of attacks/second. So now we have to flip our fraction of 0.75/1 over to 1/0.75. Do some math reduction: 1/0.75= 1/(3/4)= 14/3= 4/3 attacks/second So guess how we get dps now? Damage is reallydamage per attack ordamage/attack. ) do your cancellation and bada-bing bada-boom ( So we're dealing 10 damage per attack. So we need to take 10 damage 4/3 attack/second And watch the magic happen: 40/3 DPS Do some long division and presto: 13.3333333--------->(goes on forever dog) or 13 1/3 dps Calculating Damage Actually Done in 2 seconds: Take our previous problem, and let's say our unit lived for 2 seconds in a fight and was attacking the entirety of those 2 seconds. To get the actual damage done, you'll need to multiple DPS2 To get the damage actually done, you need to take in account that it's to complete a partial attack. IE, you cannot do 1/3 of an attack. So 4/3102 won't work OMG DO WE GIVE UP?! NO! So we know that 4/3 attacks/second 10 seconds will give us a fractional number of attacks, which is impossible, as you cannot complete a fractional amount of attacks. So take 4/3 * 10 and get 13 and 1/3 attacks. You now have to drop the 1/3 to maintain correctness. 4/3102= 26 and 2/3 132= only 26. Do you see how you're off by 2/3 if you don't account for this Minimum, Maximum, and Average DPS: Now given this sample information: Attack Cooldown = 0.75 Attack Damage Base = 9 Number of Dice = 1 Number of Sides per Die = 2 This gives us the same attack as above, except now with 10-11 damage You can find minimum DPS by using 10 as your damage value, then maximum DPS by using 11 as your damage value. 10 (damage/attack) 4/3 (attack/second) 13 1/3 minimum dps 11 (damage/attack) * 4/3 (attack/second) 14 2/3 maximum dps Now this is cool, but you'll probably confuse the user by displaying excessive numbers like this. You're best off taking the minimum and maximum dps and dividing them by 2, giving us the average dps 13 1/3 + 14 2/3 = 28 14 average dps <-------- display this to to the user, It'll make their life easier and yours too. It should reduce the questions like WTF IS THIS MIN AND MAX SHIT MAN!!!!!! These are all things that interrupt normal damage flow. If you use different armor types, evasion, and different armor amounts, you'll have to alter your dps calculation accordingly. Last edited by a moderator: Oct 24, 2012 DPS is normally referenced as average damage per second per duration of spell or attack so the first equation where you get 26.66_ is correct compared to the second equation where you get 20. Also your title is a little offensive to anyone that needs to look at this tutorial. It is a bad title. Changing it to beginners is more preferred and less offensive. If this is to be a tutorial it should also cover how to determine min and max DPS when you have several number of dice with several sides. After looking a bit more your DPS is off. It should be 12 DPS with .75 speed and 9 damage. 1 / 0.75 9 = 12 not 13.33_ You can also do the above by 9 * 4 attacks since there are 4 attacks in 3 seconds. Then divide 36 by 3 and you get 12 to do the second check on your math. Also having the times 2 is misleading as that would not be DPS that would be DP2S. the truncation is bad idea, because sure it works for you in this special case, but if I extend the time I measure your dps over 10 seconds for instance, then suddenly I have huge error if I consider 1 attack instead of 4/3 attacks per second Jan 1, 2010 Base damage 9 with dice 1 and 1 makes it 10 You don't cut out the fraction until 4/3 is multiplied by the number of seconds, and that is to find damage actually done, not DPS in that case DPS is normally referenced as average damage per second per duration of spell or attack so the first equation where you get 26.66_ is correct compared to the second equation where you get 20. Also your title is a little offensive to anyone that needs to look at this tutorial. It is a bad title. Changing it to beginners is more preferred and less offensive. If this is to be a tutorial it should also cover how to determine min and max DPS when you have several number of dice with several sides. After looking a bit more your DPS is off. It should be 12 DPS with .75 speed and 9 damage. 1 / 0.75 * 9 = 12 not 13.33_ You can also do the above by 9 * 4 attacks since there are 4 attacks in 3 seconds. Then divide 36 by 3 and you get 12 to do the second check on your math. Also having the times 2 is misleading as that would not be DPS that would be DP2S. read that more closely Mr.Pockets00000 said: Number of Dice= 1 Number of Sides per Die= 1 This is a very basic attack, that is guaranteed to deal 10-10 damage each attack. Last edited by a moderator: you do not need to post 3 times, you can just post it all in one post... Oct 12, 2011 This is useful information for today game programming, but the tuts looks too boring. Add some color wont hurts (not colorful) your calculation is still invalid, since what you do is floor(attack_per_second) * damage * seconds, and you should do floor(attack_per_second * seconds) * damage. Because with your way, when I want to calculate unit's dps with 4/3 attacks per second over 10 seconds, it will say 11010 which is 100. With my corrected calculation it is 130, because you will only hit your enemy 13 and 1/3 times in 10 seconds, whereas with your formula you will hit him 10 times. This is why I said the formula is incorrect and only fitting your scenario. Jan 1, 2010 your calculation is still invalid, since what you do is floor(attack_per_second) * damage * seconds, and you should do floor(attack_per_second * seconds) * damage. Because with your way, when I want to calculate unit's dps with 4/3 attacks per second over 10 seconds, it will say 11010 which is 100. With my corrected calculation it is 130, because you will only hit your enemy 13 and 1/3 times in 10 seconds, whereas with your formula you will hit him 10 times. This is why I said the formula is incorrect and only fitting your scenario. In this situation you're correct. I didn't mean to say you always drop the fractional part, but what is true which you must recognize is this 10 seconds * 4/3 attacks/second = 13 & 1/3 attacks . When you're calculating the damage that was ACTUALLY DONE IN THAT TIME (10 SECONDS) you must chop off the 1/3 because you CANNOT do 1/3 a attack When you're calculating DPS, you can leave the fraction but when you're calculating damage done over given time you must drop the fraction or you're saying it's possible to do 1/3 a attack. So yes for your equation its 4/3 * 10seconds * 10 damage. 4/3 * 10 is 13 and 1/3, but you have to drop the 1/3 or you get a damage value that's impossible in exactly 10 seconds. (you'd get that you did 133 rather than 130, that's an error of 3 damage that you didn't actually do) (Attacks/Second) * (Seconds) = Attacks (cannot be fractional, must be rounded down) Edit: I actually had it slightly worded wrong before, correcting now Edit: Sorry I didn't know what I was thinking, I had dropped the fractional part at the wrong moment. you do not need to post 3 times, you can just post it all in one post... correct, my apologies Last edited: Sep 17, 2009 There is no such thing as "minimum" or "maximum" DPS. DPS is always an average, as it's a blind estimation based on your stats under the assumption of an infinitely long fight. That's why truncating the DPS for incomplete attacks is also bull. You want to take incomplete attacks into account, because no fight only goes 1 second and the truncated value is of no practical Plus, your calculation - even if it would matter, is wrong anyway, as the first attack comes instantly and doesn't take the attack cooldown into account. Plus, who cares about theoretical DPS anyway? I was expecting this tutorial to explain how to build a classic DPS meter. The basics for that should cover: 1) How to detect when combat starts 2) How to keep track of the total damage dealt of each unit (using a unit indexer and a damage detection engine) 3) How to properly count the combat time for each unit (note: looping through an array, incrementing integers for each unit is highly ineffective - better solution is to count up a global timer and only store the time of combat entry for each unit, then take the difference to the combat entry to get the current combat duration) 4) Methods of displaying the DPS as total damage dealt / combat time (for example, showing DPS of currently selected unit in a multiboard or using floating texts) Now that would be a useful tutorial. But what you submitted is useless garbage. if we want to go to precise levels, you should also account for the animation time it takes to hit your enemy, unless that is already reduced by wc3 with attack speed. Jan 1, 2010 There is no such thing as "minimum" or "maximum" DPS. DPS is always an average, as it's a blind estimation based on your stats under the assumption of an infinitely long fight. That's why truncating the DPS for incomplete attacks is also bull. You want to take incomplete attacks into account, because no fight only goes 1 second and the truncated value is of no practical Plus, your calculation - even if it would matter, is wrong anyway, as the first attack comes instantly and doesn't take the attack cooldown into account. Plus, who cares about theoretical DPS anyway? I was expecting this tutorial to explain how to build a classic DPS meter. The basics for that should cover: 1) How to detect when combat starts 2) How to keep track of the total damage dealt of each unit (using a unit indexer and a damage detection engine) 3) How to properly count the combat time for each unit (note: looping through an array, incrementing integers for each unit is highly ineffective - better solution is to count up a global timer and only store the time of combat entry for each unit, then take the difference to the combat entry to get the current combat duration) 4) Methods of displaying the DPS as total damage dealt / combat time (for example, showing DPS of currently selected unit in a multiboard or using floating texts) Now that would be a useful tutorial. But what you submitted is useless garbage. There is no such thing as "minimum" or "maximum" DPS. DPS is always an average, as it's a blind estimation based DPS is not blind estimation. You have a bad definition of DPS. Blind estimation would be YEH KNOW ITS GOT LIEK 100 DAMAGE. I BET THAT MEANS IT DOES BIG DAMAGE EVERY SECOND That's why truncating the DPS for incomplete attacks is also bull. I didn't truncate damage for DPS, I truncated damage for . If you think that Damage PER second or Damage/Second is the same as then your understanding of unit analysis is too weak. They're close but not the same. Damage per second is Damage everyone one second (theoretical). Real Damage Done is what actually happened. IE, if the attack never happend, the damage never happened. Ever tried initiating an attack and then killing the unit before the attack is even finished? What happens? The damage is cancelled. if we want to go to precise levels, you should also account for the animation time it takes to hit your enemy, unless that is already reduced by wc3 with attack speed. Thank you for displaying an actual argument. That's something that I should have ventured into deeper. I agree, I overlooked that for example a footman takes about 0.3 seconds to launch an attack. That's very relevant to if the attack will be completed or not. Oct 24, 2012 DPS is not blind estimation. You have a bad definition of DPS. Blind estimation would be YEH KNOW ITS GOT LIEK 100 DAMAGE. I BET THAT MEANS IT DOES BIG DAMAGE EVERY SECOND I didn't truncate damage for DPS, I truncated damage for DAMAGE ACTUALLY DONE IN X TIME. If you think that Damage PER second or Damage/Second is the same as DAMAGE ACTUALLY DONE IN X TIME then your understanding of unit analysis is too weak. They're close but not the same. Damage per second is Damage everyone one second (theoretical). Real Damage Done is what actually happened. IE, if the attack never happend, the damage never happened. Ever tried initiating an attack and then killing the unit before the attack is even finished? What happens? The damage is cancelled. DPS is damage per second. Since the second is an arbitrary time value used for damage done in a second not a specific second then it is average damage per second real damage per second. You should also take into account that you need to have each units attack speed which (correct if wrong) WC3 does not have an action to get a units attack speed. That means that every unit will have to have its attack speed calculated. This can be done using a system which for your RDPS you should show them how to make the system to make this tutorial even worth using. You can also show them how to make several small systems for calculating DPS several different ways. If you try to make this tutorial better then I believe others will help you to do so. As it is now I can't see it getting approved. Nov 11, 2006 The tutorial itself is interesting, but there isn't much the average user can do with the information. Perhaps they can make a DPS meter, but most of those use real-time attacks rather than just displaying DPS. But I suppose this tutorial is good for balancing. Overall, this tutorial could use some structure so that the user knows exactly what he will get out of it and what he can do with the information (usually this is satisfied with an introduction). That doesn't mean this tutorial doesn't warrant approval--but I feel like it could be expanded. Personally, I would like to know more about the subtleties of attacks. Edo made a good suggestion, could you look into defining how attacks behave in wc3 and seeing if it affects dps in any way? And while it is useful to consider the 0 armor case, it may be good to have a small section detailing what one would do in the case where units have armor. It may help a user balance their map's units. As @deathismyfriend said, the title is offensive. I don't know how to calculate DPS, that doesn't make me a retard. Idk, maybe I'm just a "retard" but I didn't understand anything from this tutorial. Because by reading it, it sounds more like the author is boasting than giving us "retards" a concise way to calculate dps. The tutorial is a mess, calculation after calculation without explaining why. Like flipping fractions and math reduction. Reduction? I never heard that term. Subtraction or division maybe. I vote for this to be completely re-written. It looks more of a "look how smart i am" rather than trying to help... us... retards Also, offensive title. Last edited by a moderator: Nov 11, 2006 Nov 11, 2006 It has been a while and I still think this tutorial needs some expansion for it to be approved. I'm going to graveyard it for now. Feel free to PM me if you'd like to make changes to it.	{"url":"https://www.hiveworkshop.com/threads/dps-damage-per-second-calculation-for-dummies.262029/","timestamp":"2024-11-03T19:15:28Z","content_type":"text/html","content_length":"207160","record_id":"<urn:uuid:1436cf8c-2711-48d9-b9c9-fd6eb3f58619>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00577.warc.gz"}
Surface Approximation May Be Easier Than Surface Integration 2001 Reports Surface Approximation May Be Easier Than Surface Integration The approximation and integration problems consist of finding an approximation to a function f or its integral over some fixed domain 6. For the classical version of these problems, we have partial information about the functions f and complete information about the domain 6; for example, 6 might be a cube or ball in R d . When this holds, it is generally the case that integration is not harder than approximation; moreover, integration can be much easier than approximation. What happens if we have partial information about 6? This paper studies the surface approximation and surface integration problems, in which 6 = 6g for functions g. More specifically, the functions f are r times continuously differentiable scalar functions of l variables, and the functions g are s times continuously differentiable injective functions of d variables with l components. The class of surfaces considered is generated as images of cubes or balls, or as oriented cellulated regions. Error for the surface approximation problem is measured in the Lq - sense. These problems are well-defined, provided that d ≤ l, r ≥ 0, and s ≥ 1. Information consists of function evaluations of f and g. We show that the ε-complexity of surface approximation is proportional to (1/ε) 1/µ with µ = min{r,s}/d. We also show that if s ≥ 2, then the ε-complexity of surface integration is proportional to (1 /ε) 1/ν with ν = min r d , s − δs,1(1 − δd,l) min{d, l − 1}. (This bound holds as well for several subcases of s = 1; we conjecture that it holds for all r ≥ 0, s ≥ 1, and d ≤ l.) Using these results, we determine when surface approximation is easier than, as easy as, or harder than, surface integration; all three possibilities can occur. In particular, we find that if r = s = 1 and d < l, then µ = 1/d and ν = 0, so that surface integration is unsolvable and surface approximation is solvable; this is an extreme case for which surface approximation is easier than surface integration. More About This Work Academic Units Department of Computer Science, Columbia University Published Here April 22, 2011	{"url":"https://academiccommons.columbia.edu/doi/10.7916/D83J3R5Z","timestamp":"2024-11-14T07:26:45Z","content_type":"text/html","content_length":"20372","record_id":"<urn:uuid:8eb9bdd6-510a-4654-99fb-b6327a105270>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00257.warc.gz"}
Listening to the Rationals and Irrationals Earlier this year I shared an application with you that would play a song based on the digits of the decimal expansion of pi. After I suggested that playing rational and irrational numbers could be a nice learning activity, one of the readers discovered a Wolfram Demonstration that also plays the irrational numbers called “ Math Songs I’m pleased to report that Michael Croucher (from the blog Walking Randomly ) has obliged us with the requisite companion Wolfram Demonstration that plays the RATIONAL numbers (any rational number with a numerator between 1 and 1000 and denominator between 1 and 1000). His demonstration is called “ Music from the Rationals Now that we have both, what can we do with them? As you are, no doubt, aware, many students have great difficulty with the distinction between the rational and irrational numbers. Now you can play each number song-style and ask the students to identify whether the song ends, has a repeating element, or contains a random pattern. I was playing several fractions with denominators of 7, and then some with denominators of 13, when I made an interesting observation (well, maybe an obvious one). The same elements of sound patterns repeat, even as the numerators change. With so much calculator use today, I wonder if students realize all the patterns in the decimal expansions. Many students have probably never done a long division problem that resulted in a repeating decimal involving more than two digits. I can’t wait to use these audio applications with the students from Math for Elementary Teachers next time I teach it. Two days ago, we covered the discriminant in my algebra class. That would have been the perfect time to talk about rationals and irrationals again. Well – there’s always next semester! About Author	{"url":"https://edgeoflearning.com/listening-to-the-rationals-and-irrationals/","timestamp":"2024-11-04T21:01:12Z","content_type":"text/html","content_length":"192793","record_id":"<urn:uuid:d690fd38-db3f-4ee5-b336-fb6c4d40949b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00123.warc.gz"}
Understanding Calculations with Velocity - High School Physics All High School Physics Resources Example Questions Example Question #41 : Waves A spring oscillates back and forth with a period of Correct answer: Velocity is commonly described with respect to wavelength and frequency. We are given the period and the wavelength. Period is simply the inverse of freqeuncy. Using this, we can rewrite our original equation in terms of the period. Now we can use the values given in the question to solve for the velocity. Plug in our given values and solve. Example Question #43 : Waves, Sound, And Light A wave oscillates with a frequency of Correct answer: The equation for velocity in terms of wavelength and frequency is We are given the velocity and frequency. Using these values, we can solve for the wavelength. Example Question #44 : Waves, Sound, And Light A note is played in a gas (which is not a normal atmosphere). Inside of this gas, the note has a frequency of Correct answer: The relationship between velocity, frequency, and wavelength is: Plug in the given information to solve: Certified Tutor Nova Southeastern University, Doctor of Science, Urban Education and Leadership. Certified Tutor Old Dominion University, Bachelor of Science, Biology, General. Certified Tutor Shandong University of Technology, Bachelor of Engineering, Applied Mathematics. The University of Alabama, Master of Science... All High School Physics Resources	{"url":"https://www.varsitytutors.com/high_school_physics-help/understanding-calculations-with-velocity","timestamp":"2024-11-11T14:35:46Z","content_type":"application/xhtml+xml","content_length":"151506","record_id":"<urn:uuid:c4d75dc0-f203-41e3-bfa9-b4cc02312ea7>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00568.warc.gz"}
So, Axl's now an internet joke Some funny comments on there.. What one tragic picture can do Seen this reposted on Facebook multiple times today. "I used to love her... but I had to eat her." "Paradise City, where the fries are greasy and the cakes are pretty...oh won't you please fry some dough!" "What we got here is .... Failure to exercise" "Snaxl Rose" That is fuckin hilarious for real "I used to love her... but I had to eat her." "Paradise City, where the fries are greasy and the cakes are pretty...oh won't you please fry some dough!" "What we got here is .... Failure to exercise" "Snaxl Rose" That is fuckin hilarious for real Snaxl Rose Shit... the worst part is that some of this stuff is funny as hell. Hope this makes him get into shape (not that he's fat)... Nothing new. Axl has been a Joke since the old band broke up. Axl broke up the band. Axl used to be a recluse. The new band is a joke. Axl can't sing anymore. Axl is fat... Yadda, Yadda, Yadda. Who cares what people think. No one is interested in GN'R anymore except us hardcore fans anyway. Edited by 31illusion we should make an internet meme out of this Fair's fair. A picture of Slash should be uploaded to reddit too. Who will do it? Axl just needs to release new music never heard of this site but whats funny is when most of these people hit 50 and they blow up as well is it really neccessary to post every negative article about axl, seriously it gets very tiresome. I always thought that making fun of someone appearance was stupid. When it's done by some teenagers that needs to get laid, it's even worse. I'm a caveman who doesn't know that site and wants to keep it that way. This topic is now closed to further replies. • Recently Browsing 0 members □ No registered users viewing this page.	{"url":"https://www.mygnrforum.com/topic/179596-so-axls-now-an-internet-joke/page/2/","timestamp":"2024-11-14T07:21:23Z","content_type":"text/html","content_length":"234147","record_id":"<urn:uuid:6090548e-115c-4ba1-a39e-8798696e1f8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00617.warc.gz"}
Fast, minimum storage ray-triangle intersection. Tomas Möller and Ben Trumbore. Journal of Graphics Tools, 2(1):21--28, 1997. We present a clean algorithm for determining whether a ray intersects a triangle. The algorithm translates the origin of the ray and then changes the base to yield a vector (t u v)T, where t is the distance to the plane in which the triangle lies and (u,v) represents the coordinates inside the triangle. One advantage of this method is that the plane equation need not be computed on the fly nor be stored, which can amount to significant memory savings for triangle meshes. As we found our method to be comparable in speed to previous methods, we believe it is the fastest ray-triangle intersection routine for triangles that do not have precomputed plane equations. The ACM site of the Journal of Graphics Tools contains more information (source code, errata, images). This paper is available as a compressed Postscript file MT97.ps.gz (324K). This paper is available as a PDF file MT97.pdf (1.4M).	{"url":"https://www.graphics.cornell.edu/pubs/1997/MT97.html","timestamp":"2024-11-12T00:22:50Z","content_type":"text/html","content_length":"3825","record_id":"<urn:uuid:3711d2a3-94a1-4977-b8b1-afb6b7ca1937>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00302.warc.gz"}
Intro to Data: Line Plots Download as PDF Download PDF View as Separate Page WHAT IS A LINE PLOT? A line plot is a type of graph that shows information on a number line. You can use a line plot to display data. Line plots can help you solve problems with your data. To better understand line plots… WHAT IS A LINE PLOT?. A line plot is a type of graph that shows information on a number line. You can use a line plot to display data. Line plots can help you solve problems with your data. To better understand line plots… LET’S BREAK IT DOWN! Start building a line plot. Information that we collect is called data. A line plot is a number line. The start number is the least number in your data. The end number is the greatest number in your data. Label all of the whole numbers in between. Now you are ready to add your data. Start building a line plot. Information that we collect is called data. A line plot is a number line. The start number is the least number in your data. The end number is the greatest number in your data. Label all of the whole numbers in between. Now you are ready to add your data. Plot the number of pets on a line plot. You asked your classmates how many pets they have. You make a line plot to show your data. The plot has a title: "Number of Pets." All of your classmates have between 0 and 4 pets. The number line should show the numbers 0, 1, 2, 3, and 4. Put an X over a number for each classmate who has that many pets. If Sam has 4 pets, put an X over the 4 for Sam. Plot the number of pets on a line plot. You asked your classmates how many pets they have. You make a line plot to show your data. The plot has a title: "Number of Pets." All of your classmates have between 0 and 4 pets. The number line should show the numbers 0, 1, 2, 3, and 4. Put an X over a number for each classmate who has that many pets. If Sam has 4 pets, put an X over the 4 for Sam. Plot the number of soccer goals on a line plot. You recorded the number of goals each teammate scored in soccer this year. You want to make a line plot to show your data. The plot has the title “Number of Goals Scored.” The least number of goals scored was 3. The greatest was 10. So, your line plot shows the numbers 3 to 10. Plot a point for each teammates' goals. You scored 5 goals, so put an X over the 5 for you! Plot the number of soccer goals on a line plot. You recorded the number of goals each teammate scored in soccer this year. You want to make a line plot to show your data. The plot has the title “Number of Goals Scored.” The least number of goals scored was 3. The greatest was 10. So, your line plot shows the numbers 3 to 10. Plot a point for each teammates' goals. You scored 5 goals, so put an X over the 5 for you! Interpret your soccer goal line plot. A line plot tells us lots of different information. The soccer goals line plot tells you that every player scored 3 or more goals. One player scored 10 goals. The most common number of goals was 7 since it has the most Xs above it. Interpret your soccer goal line plot. A line plot tells us lots of different information. The soccer goals line plot tells you that every player scored 3 or more goals. One player scored 10 goals. The most common number of goals was 7 since it has the most Xs above it. Plot the length of geckos on a line plot. You measured the length of each gecko in a tank. You make a line plot of all of your data. After you make your line plot, you notice that there are no points over 15. This means that no gecko in the tank is 15 cm long. There are 2 points above 21. So 2 geckos in the tank are 21 cm long. Plot the length of geckos on a line plot. You measured the length of each gecko in a tank. You make a line plot of all of your data. After you make your line plot, you notice that there are no points over 15. This means that no gecko in the tank is 15 cm long. There are 2 points above 21. So 2 geckos in the tank are 21 cm long. Choose Your Free Trial Period 3 Days 3 days to access to all of our teaching resources for free. Continue to Lessons 30 Days Get 30 days free by inviting other teachers to try it too. Share with Teachers Get 30 Days Free By inviting 4 other teachers to try it too. Skip, I will use a 3 day free trial Thank You! Enjoy your free 30 days trial	{"url":"https://www.generationgenius.com/videolessons/intro-to-data-line-plots/","timestamp":"2024-11-11T14:02:20Z","content_type":"text/html","content_length":"917970","record_id":"<urn:uuid:0003c351-a52b-4acf-b1a5-8297b445f6c9>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00671.warc.gz"}
Accountancy MCQs for Class 12 with Answers Chapter 2 Change in Profit Sharing Ratio among the Existing Partners Free PDF Download of CBSE Accountancy Multiple Choice Questions for Class 12 with Answers Chapter 2 Change in Profit Sharing Ratio among the Existing Partners. Accountancy MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Accountancy Change in Profit Sharing Ratio among the Existing Partners MCQs Pdf with Answers to know their preparation level. Change in Profit Sharing Ratio among the Existing Partners Class 12 Accountancy MCQs Pdf Select the Best Alternate : 1. Sacrificing Ratio : (A) New Ratio – Old Ratio (B) Old Ratio – New Ratio (C) Old Ratio – Gaining Ratio (D) Gaining Ratio – Old Ratio Answer: B 2. Gaining Ratio : (A) New Ratio – Sacrificing Ratio (B) Old Ratio – Sacrificing Ratio (C) New Ratio – Old Ratio (D) Old Ratio – New Ratio Answer: C 3. A and B were partners in a firm sharing profit or loss equally. With effect from 1st April 2019 they agreed to share profits in the ratio of 4 : 3. Due to change in profit sharing ratio, A’s gain or sacrifice will be : (A) Gain $\frac{1}{14}$ (B) Sacrifice $\frac{1}{14}$ (C) Gain $\frac{4}{7}$ (D) Sacrifice $\frac{3}{7}$ Answer: A 4. A and B were partners in a firm sharing profit or loss equally. With effect from 1st April, 2019 they agreed to share profits in the ratio of 4 : 3. Due to change in profit sharing ratio, B’s gain or sacrifice will be : (A) Gain $\frac{1}{14}$ (B) Sacrifice $\frac{1}{14}$ (C) Gain $\frac{4}{7}$ (D) Sacrifice $\frac{3}{7}$ Answer: B 5. A and B were partners in a firm sharing profit or loss in the ratio of 3 : 5. With effect from 1st April, 2019, they agreed to share profits or losses equally. Due to change in profit sharing ratio, A’s gain or sacrifice will be : Answer: B 6. A and B were partners in a firm sharing profits and losses in the ratio of 2 : 1. With effect from 1st January 2019 they agreed to share profits and losses equally. Individual partner’s gain or sacrifice due to change in the ratio will be : Answer: B 7. A and B share profits and losses in the ratio of 3 : 2. With effect from 1st . January, 2019, they agreed to share profits equally. Sacrificing ratio and Gaining Ratio will be : Answer: C 8. A and B were partners in a firm sharing profit or loss in the ratio of 3 : 1. With effect from Jan. 1, 2019 they agreed to share profit or loss in the ratio of 2 : 1. Due to change in profit-loss sharing ratio, B’s gain or sacrifice will be : (A) Gain $\frac{1}{12}$ (B) Sacrifice $\frac{1}{12}$ (C) Gain $\frac{1}{3}$ (D) Sacrifice $\frac{1}{3}$ Answer: A 9. A, B and C were partners sharing profit or loss in the ratio of 7 : 3 : 2. From Jan. 1,2019 they decided to share profit or loss in the ratio of 8 : 4 : 3. Due to change in the profit-loss sharing ratio, B’s gain or sacrifice will be : (A) Gain $\frac{1}{60}$ (B) Sacrifice $\frac{1}{60}$ (C) Gain $\frac{2}{60}$ (D) Sacrifice $\frac{3}{60}$ Answer: A 10. A y and Z are partners in a firm sharing profits and losses in the ratio of 5 : 3 : 2. The partners decide to share future profits and losses in the ratio of 3:2:1. Each partner’s gain or sacrifice due to change in the ratio will be : Answer: D 11. A, B and C were partners in a firm sharing profits and losses in the ratio of 3 : 2 : 1. The partners decide to share future profits and losses in the ratio of 2:2:1. Each partner’s gain or sacrifice due to change in ratio will be : Answer: A 12. A, B and C were partners in a firm sharing profits and losses in the ratio of 4 : 3 : 2. The partners decide to share future profits and losses in the ratio of 2:2: 1. Each partner’s gain or sacrifice due to change in the ratio will be : Answer: C 13. A, B and C were partners in a lirm sharing profits in 4 : 3 : 2 ratio. They decided to share future profits in 4 : 3 : 1 ratio. Sacrificing ratio and gaining ratio will be : Answer: D 14. X, Y and Z were partners sharing profits in the ratio 2:3:4 with effect from 1st January, 2019 they agreed to share profits in the ratio 3:4:5. Each partner’s gain or sacrifice due to change in the ratio will be : Answer: A 15. X, 7 and Z were in partnership sharing profits in the ratio 4 : 3 : 1. The partners agreed to share future profits in the ratio 5 : 4 : 3. Each partner’s gain or sacrifice due to change in ratio will be : Answer: A 16. A, B and C are equal partners in the firm. It is now agreed that they will share the future profits in the ratio 5:3:2. Sacrificing ratio and gaining ratio of different partners will be : Answer: C 17, The excess amount which the firm can get on selling its assets over and above the saleable value of its assets is called : (A) Surplus (B) Super profits (C) Reserve (D) Goodwill Answer: D 18. Which of the following is NOT true in relation to goodwill? (A) It is an intangible asset (B) It is fictitious asset (C) It has a realisable value (D) None of the above Answer: B 19. When Goodwill is not purchased goodwill account can : (A) Never be raised in the books (B) Be raised in the books (C) Be partially raised in the books (D) Be raised as per the agreement of the partners Answer: A 20. The Goodwill of the firm is NOT affected by : (CPT; June 2011) (A) Location of the firm (B) Reputation of firm (C) Better customer service (D) None of the above Answer: D 21. Capital employed by a partnership firm is ₹5,00,000. Its average profit is ₹60,000. The normal rate of return in similar type of business is 10%. What is the amount of super profits? (C.S. Foundation, Dec., 2012) (A) ₹50,000 (B) ₹10,000 (C) ₹6,000 (D) ₹56,000 Answer: B 22. Weighted average method of calculating goodwill is used when : (CPT; June 2009) (A) Profits are not equal (B) Profits show a trend (C) Profits are fluctuating (D) None of the above Answer: B 23. The profits earned by a business over the last 5 years are as follows : ₹12,000; ₹13,000; ₹14,000; ₹18,000 and ₹2,000 (loss). Based on 2 years purchase of the last 5 years profits, value of Goodwill will be : (A) ₹23,600 (B) ₹22,000 (C) ₹1,10,000 (D) ₹1,18,000 Answer: B 24. The average profit of a business over the last five years amounted to ₹60,000. The normal commercial yield on capital invested in such a business is deemed to be 10% p.a. The net capital invested in the business is ₹5,00,000. Amount of goodwill, if it is based on 3 years purchase of last 5 years superprofits will be : (A) ₹1,00.000 (B) ₹1,80,000 (C) ₹30.000 (D) ₹1,50,000 Answer: C 25. Under the capitalisation method, the formula for calculating the goodwill is : (CPT; Dec. 2011) (A) Super profits multiplied by the rate of return (B) Average profits multiplied by the rate of return (C) Super profits divided by the rate of return (D) Average profits divided by the rate of return Answer: C 26. The net assets of a firm including fictitious assets of ₹5,000 are ₹85,000. The net liabilities of the firm are ₹30,000. The normal rate of return is 10% and the average profits of the firm are ₹8,000. Calculate the goodwill as per capitalisation of super profits. (A) ₹20,000 (B) ₹30,000 (C) ₹25,000 (D) None of these Answer: B 27. Total Capital employed in the firm is ₹8,00,000, reasonable rate of return is 15% and Profit for the year is ₹12,00,000. The value of goodwill of the firm as per capitalization method would be : (C.S. Foundation, June 2013) (A) ₹82,00,000 (B) ₹12,00,000 (C) ₹72,00,000 (D) ₹42,00,000 Answer: C 28. The average capital employed of a firm is ?4,00,000 and the normal rate of return is 15%. The average profit of the firm is ?80,000 per annum. If the . remuneration of the partners is estimated to be ? 10,000 per annum, then on the basis of two years purchase of super-profit, the value of the Goodwill will be : (A) ₹10,000 (B) ₹20,000 (C) ₹60,000 (D) ₹80,000 Answer: B 29. A firm earns ₹1,10,000. The normal rate of return is 10%. The assets of the firm amounted to ₹11,00,000 and liabilities to ₹1,00,000. Value of goodwill by capitalisation of Average Actual Profits will be : (C.S. Foundation Dec., 2012) (A) ₹2,00,000 (B) ₹10,000 (C) ₹5,000 (D) ₹1,00,000 Answer: D 30. Capital invested in a firm is ₹5,00,000. Normal rate of return is 10%. Average profits of the firm are ₹64,000 (after an abnormal loss of ?4,000). Value of goodwill at four times the super profits will be : (A) ₹72,000 (B) ₹40,000 (C) ₹2,40,000 (D) ₹1,80,000 Answer: A 31. P and Q were partners sharing profits and losses in the ratio of 3 : 2. They decided that with effect from 1st January, 2019 they would share profits and losses in the ratio of 5 : 3. Goodwill is valued at ? 1,28,000. In adjustment entry : (A) Cr. P by ₹3,200; Dr. Q by ₹3,200 (B) Cr. P by ₹37,000; Dr. Q by ₹37,000 (C) Dr. P by ₹37,000; Cr. Q by ₹37,000 (D) Dr. P by ₹3,200 Cr. Q by ₹3,200 Answer: D 32. A, B and C are partners sharing profits in the ratio of 4 : 3 : 2 decided to share profits equally. Goodwill of the firm is valued at ? 10,800. In adjusting entry for goodwill : (A) A’s Capital A/c Cr. by ₹4,800; B’s Capital A/c Cr. by ₹3,600; C’s Capital A/c Cr. by ₹2,400. (B) A’s Capital A/c Cr. by ₹3,600; B’s Capital A/c Cr. by ₹3,600; C’s Capital A/c Cr. by ₹3,600. (C) A’s Capital A/c Dr. by ₹1,200; C’s Capital A/c Cr. by ₹1,200; (D) A’s Capital A/c Cr. by ₹1,200; C’s Capital A/c Dr. by ₹1,200 Answer: D 33. A, B and C were partners sharing profits and losses in the ratio of 7 : 3 : 2. From 1st January, 2019 they decided to share profits and losses in the ratio of 8:4:3. Goodwill is ₹1,20,000. In Adjustment entry for goodwill: (A) Cr. A by ₹6,000; Dr. B by ?2,000; Dr. C by ₹4,000 (B) Dr. A by ₹6,000; Cr. B by ?2,000; Cr. C by ₹4000 (C) Cr. A by ₹6,000; Dr. B by ?4,000; Dr. C by ₹2,000 (D) Dr. A by ₹6,000; Cr. B by ?4,000; Cr. C by ₹2,000 Answer: A 34. P, Q and R were partners in a firm sharing profis in 5 : 3 : 2 ratio. They decided to share the future profits in 2 : 3 : 5. For this purpose the goodwill of the firm was valued at ₹1,20,000. In adjustment entry for the treatment of goodwill due to change in the profit sharing ratio : (A) Cr. P by ₹24,000; Dr. R by ₹24,000 (B) Cr. P by ₹60,000; Dr. R by ₹60,000 (C) Cr. P by ₹36,000; Dr. R by ₹36,000 (D) Dr. P by ₹36,000; Cr. R by ₹36,000 Answer: C 35. A, B and C are partners in a firm sharing profits in the ratio of 3 : 4 : 1. They decided to share profits equally w.e.f. 1 st April, 2019. On that date the Profit and Loss Account showed the credit balance of ?96,000. Instead of closing the Profit and Loss Account, it was decided to record an adjustment entry reflecting the change in profit sharing ratio. In the journal entry : (A) Dr. A by ₹4,000; Dr. B by ₹16,000; Cr. C by ₹20,000 (B) Cr. A by ₹4,000; Cr. B by ₹16,000; Dr. C by ₹20,000 (C) Cr. A by ₹16,000; Cr. B by ₹4,000; Dr. C by ₹20,000 (D) Dr. A by ₹16,000; Dr. B by ₹4,000; Cr. C by ₹20,000 Answer: B 36. A, B and C are partner sharing profits in the ratio of 1 : 2 : 3. On 1-4-2019 they decided to share the profits equally. On the date there was a credit balance of ? 1,20,000 in their Profit and Loss Account and a balance of ? 1,80,000 in General Reserve Account. Instead of closing the General Reserve Account and Profit and Loss Account, it is decided to record an adjustment entry for the same. In the necessary adjustment entry to give effect to the above arrangement: (A) Dr. A by ₹50,000; Cr. B by ₹50,000 (B) Cr. A by ₹50,000; Dr. B by ₹50,000 (C) Dr. A by ₹50,000; Cr. Cby ₹50,000 (D) Cr. A by ₹50,000; Dr. Cby ₹50,000 Answer: C 37. X, Y and Z are partners in a firm sharing profits in the ratio 4 : 3 : 2. Their Balance Sheet as at 31-3-2019 showed a debit balance of Profit & Loss A/c ₹1,80,000. From 1-4-2019 they will share profits equally. In the necessary journal entry to give effect to the above arrangement when A Y and Z decided not to close the Profit & Loss Acccount: (A) Dr. X by ₹20,000; Cr. Z by ₹20,000 (B) Cr. X by ₹20,000; Dr. Z by ₹20,000 (C) Dr. X by ₹40,000; Cr. Z by ₹40,000 (D) Cr. X by ₹40,000; Dr. Z by ₹40,000 Answer: A 38. Aran and Varan are partners sharing profits in the ratio of 4 : 3. Their Balance Sheet showed a balance of ? 5 6,000 in the General Reserve Account and a debit balance of ? 14,000 in Profit and Loss Account. They now decided to share the future profits equally. Instead of closing the General Reserve Account and Profit and Loss Account, it is decided to pass an adjustment entry for the same. In adjustment entry : (A) Dr. Aran by ₹3,000; Cr. Varan by ₹3,000 (B) Dr. Aran by ₹5,000; Cr. Varan by ₹5,000 (C) Cr. Aran by ₹5,000; Dr. Varan by ₹5,000 (D) Cr. Aran by ₹3,000; Dr. Varan by ₹3,000 Answer: D 39. X, Y and Z are partners in a firm sharing profits in the ratio of 3 : 2 : 1. They decided to share future profits equally. The Profit and Loss Account showed a Credit balance of ₹60,000 and a General Reserve of ₹30,000. If these are not to be shown in balance sheet, in the journal entry : (A) Cr. X by ₹15,000: Dr. Z by ₹15,000 (B) Dr. X by ₹15,000; Cr. Z by ₹15,000 (C) Cr. X by ₹45,000; Cr. Y by ₹30,000; Cr. Z by ₹15,000 (D) Cr. X by ₹30,000; Cr. Y by ₹30,000; Cr. Z by ₹30,000 Answer: C 40. X Y and Z are partners sharing profits and losses in the ratio 5 : 3 : 2. They decide to share the future profits in the ratio 3 : 2 : 1. Workmen compensation reserve appearing in the balance sheet on the date if no information is available for the same will be : (A) Distributed to the partners in old profit sharing ratio (B) Distributed to the partners in new profit sharing ratio (C) Distributed to the partners in capital ratio (D) Carried forward to new balance sheet without any adjustment Answer: A 41. Any change in the relationship of existing partners which results in an end of the existing agreement and enforces making of a new agreement is called (C.B.S.E. Sample Paper, 2015) (A) Revaluation of partnership. (B) Reconstitution of partnership. (C) Realization of partnership. (D) None of the above. Answer: B We hope the given Accountancy MCQs for Class 12 with Answers Chapter 2 Change in Profit Sharing Ratio among the Existing Partners will help you. If you have any query regarding CBSE Class 12 Accountancy Change in Profit Sharing Ratio among the Existing Partners MCQs Pdf, drop a comment below and we will get back to you at the earliest.	{"url":"https://www.ncertbooks.guru/accountancy-mcqs-for-class-12-with-answers-chapter-2/","timestamp":"2024-11-08T03:14:57Z","content_type":"text/html","content_length":"99953","record_id":"<urn:uuid:1556411c-eae3-4827-a70b-840a3c79ca2a>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00130.warc.gz"}
On the optimal transition matrix for markov chain monte carlo sampling Let X be a finite space and let π be an underlying probability on X. For any real-valued function f defined on X, we are interested in calculating the expectation of f under π. Let X [0],X [1], . . .,X [n], . . . be a Markov chain generated by some transition matrix P with invariant distribution p. The time average, 1/n σ ^n-1 [k=0] f(X [k]), is a reasonable approximation to the expectation, Eπ [f(X)]. Which matrix P minimizes the asymptotic variance of 1/n σ ^n-1 [k=0] f(X [k])? The answer depends on f. Rather than a worst-case analysis, we will identify the set of P's that minimize the average asymptotic variance, averaged with respect to a uniform distribution on f. • Asymptotic variance • Average-case analysis • Markov chain • Markov chain Monte Carlo • Nonreversibility • Rate of convergence • Reversibility • Worst-case analysis Dive into the research topics of 'On the optimal transition matrix for markov chain monte carlo sampling'. Together they form a unique fingerprint.	{"url":"https://experts.umn.edu/en/publications/on-the-optimal-transition-matrix-for-markov-chain-monte-carlo-sam","timestamp":"2024-11-07T19:08:21Z","content_type":"text/html","content_length":"51801","record_id":"<urn:uuid:b1161158-2c13-4d5f-ab10-eed0b42827e9>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00257.warc.gz"}
12.2 Examples of Static Equilibrium 12 Static Equilibrium and Elasticity 12.2 Examples of Static Equilibrium Learning Objectives By the end of this section, you will be able to: • Identify and analyze static equilibrium situations • Set up a free-body diagram for an extended object in static equilibrium • Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Figure to Figure. We introduced a problem-solving strategy in Figure to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps. Problem-Solving Strategy: Static Equilibrium 1. Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly. 2. Set up a free-body diagram for the object. (a) Choose the xy-reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x– and y-directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign [latex](+)[/latex] means that the working direction is the actual direction. A minus sign [latex](-)[/latex] means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis. 3. Set up the equations of equilibrium for the object. (a) Use the free-body diagram to write a correct equilibrium condition Figure for force components in the x-direction. (b) Use the free-body diagram to write a correct equilibrium condition Figure for force components in the y-direction. (c) Use the free-body diagram to write a correct equilibrium condition Figure for torques along the axis of rotation. Use Figure to evaluate torque magnitudes and senses. 4. Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved. 5. Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in Figure. Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques. The Torque Balance Three masses are attached to a uniform meter stick, as shown in Figure. The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are [latex]{m}_{1}=50.0\,\text{g}[/latex] and [latex]{m}_{2}=75.0\,\text{g}.[/latex] Find the mass [latex]{m}_{3}[/latex] that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced. For the arrangement shown in the figure, we identify the following five forces acting on the meter stick: [latex]{w}_{1}={m}_{1}g[/latex] is the weight of mass [latex]{m}_{1};[/latex] [latex]{w}_{2}={m}_{2}g[/latex] is the weight of mass [latex]{m}_{2};[/latex] [latex]w=mg[/latex] is the weight of the entire meter stick; [latex]{w}_{3}={m}_{3}g[/latex] is the weight of unknown mass [latex]{m}_{3};[/latex] [latex]{F}_{S}[/latex] is the normal reaction force at the support point S. We choose a frame of reference where the direction of the y-axis is the direction of gravity, the direction of the x-axis is along the meter stick, and the axis of rotation (the z-axis) is perpendicular to the x-axis and passes through the support point S. In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot Figure. At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w, at the 50-cm mark. With Figure and Figure for reference, we begin by finding the lever arms of the five forces acting on the stick: [latex]\begin{array}{ccc}\hfill {r}_{1}& =\hfill & 30.0\,\text{cm}+40.0\,\text{cm}=70.0\,\text{cm}\hfill \\ \hfill {r}_{2}& =\hfill & 40.0\,\text{cm}\hfill \\ \hfill r& =\hfill & 50.0\,\text{cm}-30.0 \,\text{cm}=20.0\,\text{cm}\hfill \\ \hfill {r}_{S}& =\hfill & 0.0\,\text{cm}\,\text{(because}\,{F}_{S}\,\text{is attached at the pivot)}\hfill \\ \hfill {r}_{3}& =\hfill & 30.0\,\text{cm.}\hfill \ Now we can find the five torques with respect to the chosen pivot: [latex]\begin{array}{ccccc}\hfill {\tau }_{1}& =\hfill & +{r}_{1}{w}_{1}\text{sin}\,90^\circ=\text{+}{r}_{1}{m}_{1}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill {\tau }_{2}& =\hfill & +{r}_{2}{w}_{2}\text{sin}\,90^\circ=\text{+}{r}_{2}{m}_{2}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill \tau & =\hfill & +rw\,\text{sin}\,90^\circ=\ text{+}rmg\hfill & & \text{(gravitational torque)}\hfill \\ \hfill {\tau }_{S}& =\hfill & {r}_{S}{F}_{S}\text{sin}\,{\theta }_{S}=0\hfill & & \text{(because}\,{r}_{S}=0\,\text{cm)}\hfill \\ \hfill {\ tau }_{3}& =\hfill & \text{−}{r}_{3}{w}_{3}\text{sin}\,90^\circ=\text{−}{r}_{3}{m}_{3}g\hfill & & \text{(clockwise rotation, negative sense)}\hfill \end{array}[/latex] The second equilibrium condition (equation for the torques) for the meter stick is [latex]{\tau }_{1}+{\tau }_{2}+\tau +{\tau }_{S}+{\tau }_{3}=0.[/latex] When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is Selecting the [latex]+y[/latex]-direction to be parallel to [latex]{\mathbf{\overset{\to }{F}}}_{S},[/latex] the first equilibrium condition for the stick is Substituting the forces, the first equilibrium condition becomes We solve these equations simultaneously for the unknown values [latex]{m}_{3}[/latex] and [latex]{F}_{S}.[/latex] In Figure, we cancel the g factor and rearrange the terms to obtain To obtain [latex]{m}_{3}[/latex] we divide both sides by [latex]{r}_{3},[/latex] so we have [latex]\begin{array}{cc} \hfill {m}_{3}& =\frac{{r}_{1}}{{r}_{3}}\,{m}_{1}+\frac{{r}_{2}}{{r}_{3}}\,{m}_{2}+\frac{r}{{r}_{3}}\,m\hfill \\ & =\frac{70}{30}\,(50.0\,\text{g})+\frac{40}{30}\,(75.0\,\ text{g})+\frac{20}{30}\,(150.0\,\text{g})=316.0\frac{2}{3}\,\text{g}\simeq 317\,\text{g.}\hfill \end{array}[/latex] To find the normal reaction force, we rearrange the terms in Figure, converting grams to kilograms: [latex]\begin{array}{cc} \hfill {F}_{S}& =({m}_{1}+{m}_{2}+m+{m}_{3})g\hfill \\ & =(50.0+75.0+150.0+316.7)\times {10}^{-3}\text{kg}\times 9.8\,\frac{\text{m}}{{\text{s}}^{2}}=5.8\,\text{N}.\hfill \ Notice that Figure is independent of the value of g. The torque balance may therefore be used to measure mass, since variations in g-values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force. Check Your Understanding Repeat Figure using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick. Show Solution 316.7 g; 5.8 N In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by Figure and Figure. We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics. Forces in the Forearm A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in Figure. His forearm is positioned at [latex]\beta =60^\circ[/latex] with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units. We identify three forces acting on the forearm: the unknown force [latex]\mathbf{\overset{\to }{F}}[/latex] at the elbow; the unknown tension [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}}[/latex] in the muscle; and the weight [latex]\mathbf{\overset{\to }{w}}[/latex] with magnitude [latex]w=50\,\text{lb}.[/latex] We adopt the frame of reference with the x-axis along the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The x-axis makes an angle [latex]\beta =60^\circ[/latex] with the vertical. The y-axis is perpendicular to the x-axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle [latex]\beta[/latex] and represent each force by its x– and y-components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in Figure. At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x– and y-components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm. Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y-components of the forces because the x-components of the forces are all parallel to their lever arms, so that for any of them we have [latex]\text{sin}\,\theta =0[/latex] in Figure. For the y-components we have [latex]\theta =\pm90^\circ[/latex] in Figure. Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of [latex]{T}_{y}[/latex] and of We see from the free-body diagram that the x-component of the net force satisfies the equation and the y-component of the net force satisfies Figure and Figure are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is Figure is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are [latex]{r}_{T}=1.5\,\text{in}\text{.}[/latex] and [latex]{r}_{w}=13.0\,\ text{in}\text{.}[/latex] At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Figure, they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces: [latex]\begin{array}{ccc}\hfill {F}_{x}& =\hfill & F\,\text{cos}\,\beta =F\,\text{cos}\,60^\circ=F\,\text{/}\,2\hfill \\ \hfill {T}_{x}& =\hfill & T\,\text{cos}\,\beta =T\,\text{cos}\,60^\circ=T\,\ text{/}\,2\hfill \\ \hfill {w}_{x}& =\hfill & w\,\text{cos}\,\beta =w\,\text{cos}\,60^\circ=w\,\text{/}\,2\hfill \\ \hfill {F}_{y}& =\hfill & F\,\text{sin}\,\beta =F\,\text{sin}\,60^\circ=F\sqrt{3}\, \text{/}\,2\hfill \\ \hfill {T}_{y}& =\hfill & T\,\text{sin}\,\beta =T\,\text{sin}\,60^\circ=T\sqrt{3}\,\text{/}\,2\hfill \\ \hfill {w}_{y}& =\hfill & w\,\text{sin}\,\beta =w\,\text{sin}\,60^\circ=w\ sqrt{3}\,\text{/}\,2.\hfill \end{array}[/latex] We substitute these magnitudes into Figure, Figure, and Figure to obtain, respectively, [latex]\begin{array}{ccc}\hfill F\,\text{/}\,2+T\,\text{/}\,2-w\,\text{/}\,2& =\hfill & 0\hfill \\ \\ \hfill F\sqrt{3}\,\text{/}\,2+T\sqrt{3}\,\text{/}\,2-w\sqrt{3}\,\text{/}\,2& =\hfill & 0\hfill \\ \\ \hfill {r}_{T}T\sqrt{3}\,\text{/}\,2-{r}_{w}w\sqrt{3}\,\text{/}\,2& =\hfill & 0.\hfill \end{array}[/latex] When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T, because Figure for the x-component is equivalent to Figure for the y-component. In this way, we obtain the first equilibrium condition for forces and the second equilibrium condition for torques The magnitude of tension in the muscle is obtained by solving Figure: [latex]T=\frac{{r}_{w}}{{r}_{T}}\,w=\frac{13.0}{1.5}\,\text{(50 lb)}=433\,\frac{1}{3}\text{lb}\simeq 433.3\,\text{lb.}[/latex] The force at the elbow is obtained by solving Figure: The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is [latex]\begin{array}{c}F=383.3\,\text{lb}=383.3(4.448\,\text{N})=1705\,\text{N downward}\\ T=433.3\,\text{lb}=433.3(4.448\,\text{N})=1927\,\text{N upward.}\end{array}[/latex] Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction, and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule. Suppose we adopt a reference frame with the direction of the y-axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only y-components, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in Figure, indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles [latex]{\theta }_{T}[/latex] and [latex]{\theta }_{w}[/latex] that the forces [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}}[/latex] and [latex]\mathbf{\overset{\ to }{w}}[/latex] (respectively) make with their lever arms. In the definition of torque given by Figure, the angle [latex]{\theta }_{T}[/latex] is the direction angle of the vector [latex]{\mathbf{\ overset{\to }{T}}}_{\text{M}},[/latex] counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle [latex]{\theta }_ {w}[/latex] is measured counterclockwise from the radial direction of the lever arm to the vector [latex]\mathbf{\overset{\to }{w}}.[/latex] Done this way, the non-zero torques are most easily computed by directly substituting into Figure as follows: [latex]\begin{array}{cc} {\tau }_{T}={r}_{T}T\,\text{sin}\,{\theta }_{T}={r}_{T}T\,\text{sin}\,\beta ={r}_{T}T\,\text{sin}\,60^\circ=+{r}_{T}T\sqrt{3}\,\text{/}\,2\\ {\tau }_{w}={r}_{w}w\,\text{sin} \,{\theta }_{w}={r}_{w}w\,\text{sin}(\beta +180^\circ)=\text{−}{r}_{w}w\,\text{sin}\,\beta =\text{−}{r}_{w}w\sqrt{3}\,\text{/}\,2.\end{array}[/latex] The second equilibrium condition, [latex]{\tau }_{T}+{\tau }_{w}=0,[/latex] can be now written as From the free-body diagram, the first equilibrium condition (for forces) is Figure is identical to Figure and gives the result [latex]T=433.3\,\text{lb}.[/latex] Figure gives We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components. Check Your Understanding Repeat Figure assuming that the forearm is an object of uniform density that weighs 8.896 N. Show Solution [latex]T=\text{1963 N};\,\text{F}=1732\,\text{N}[/latex] A Ladder Resting Against a Wall A uniform ladder is [latex]L=5.0\,\text{m}[/latex] long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in Figure. The inclination angle between the ladder and the rough floor is [latex]\beta =53^\circ.[/latex] Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction [latex]{\mu }_{\text{s}}[/latex] at the interface of the ladder with the floor that prevents the ladder from slipping. We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. The second force is the static friction force [latex] f={\mu }_{\text{s}}N[/latex] directed horizontally along the floor toward the wall—this force prevents the ladder from slipping. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends. The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall. There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y-axis in the vertical direction (parallel to the wall) and the x-axis in the horizontal direction (parallel to the floor). In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution. We select the pivot at the contact point with the floor. In the free-body diagram for the ladder, we indicate the pivot, all four forces and their lever arms, and the angles between lever arms and the forces, as shown in Figure. With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot. From the free-body diagram, the net force in the x-direction is the net force in the y-direction is and the net torque along the rotation axis at the pivot point is [latex]{\tau }_{w}+{\tau }_{F}=0.[/latex] where [latex]{\tau }_{w}[/latex] is the torque of the weight w and [latex]{\tau }_{F}[/latex] is the torque of the reaction F. From the free-body diagram, we identify that the lever arm of the reaction at the wall is [latex]{r}_{F}=L=5.0\,\text{m}[/latex] and the lever arm of the weight is [latex]{r}_{w}=L\,\text{/}\,2=2.5\,\text{m}.[/latex] With the help of the free-body diagram, we identify the angles to be used in Figure for torques: [latex]{\theta }_{F}=180^\circ-\beta[/latex] for the torque from the reaction force with the wall, and [latex]{\theta }_{w}=180^\circ+(90^\circ-\ beta )[/latex] for the torque due to the weight. Now we are ready to use Figure to compute torques: [latex]\begin{array}{cc} {\tau }_{w}={r}_{w}w\,\text{sin}\,{\theta }_{w}={r}_{w}w\,\text{sin}(180^\circ+90^\circ-\beta )=-\frac{L}{2}w\,\text{sin}(90^\circ-\beta )=-\frac{L}{2}w\,\text{cos}\,\beta \\ {\tau }_{F}={r}_{F}F\,\text{sin}\,{\theta }_{F}={r}_{F}F\,\text{sin}(180^\circ-\beta )=LF\,\text{sin}\,\beta .\end{array}[/latex] We substitute the torques into Figure and solve for [latex]F:[/latex] [latex]\begin{array}{}\\ \hfill -\frac{L}{2}w\,\text{cos}\,\beta +LF\,\text{sin}\,\beta & =\hfill & 0\hfill \\ \hfill F=\frac{w}{2}\,\text{cot}\,\beta =\frac{400.0\,\text{N}}{2}\,\text{cot}\,53^\circ & =\hfill & 150.7\,\text{N}\hfill \end{array}[/latex] We obtain the normal reaction force with the floor by solving Figure: [latex]N=w=400.0\,\text{N}.[/latex] The magnitude of friction is obtained by solving Figure: [latex]f=F=150.7\,\text{N}.[/latex] The coefficient of static friction is [latex]{\mu }_{\text{s}}=f\,\text{/}\,N=150.7\,\text{/}\,400.0=0.377.[/latex] The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces: [latex]{\mathbf{\overset{\to }{F}}}_{\text{floor}}=\mathbf{\overset{\to }{f}}+\mathbf{\overset{\to }{N}}=\text{(150.7 N)}(\text{−}\mathbf{\hat{i}})+(400.0\,\text{N)}(+\mathbf{\hat{j}})=(-150.7\mathbf Its magnitude is and its direction is [latex]\phi ={\text{tan}}^{-1}(N\,\text{/}\,f)={\text{tan}}^{-1}(400.0\,\,\text{/}\,150.7)=69.3^\circ\,\text{above the floor.}[/latex] We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use Figure for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in Figure is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, Figure gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and Figure expresses the rectangular component of this vector product along the axis of rotation. This result is independent of the length of the ladder because L is cancelled in the second equilibrium condition, Figure. No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is [latex]53^\circ,[/latex] our results hold. But the ladder will slip if the net torque becomes negative in Figure. This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping. Check Your Understanding For the situation described in Figure, determine the values of the coefficient [latex]{\mu }_{\text{s}}[/latex] of static friction for which the ladder starts slipping, given that [latex]\beta[/ latex] is the angle that the ladder makes with the floor. Show Solution [latex]{\mu }_{s} \lt 0.5\,\text{cot}\,\beta[/latex] Forces on Door Hinges A swinging door that weighs [latex]w=400.0\,\text{N}[/latex] is supported by hinges A and B so that the door can swing about a vertical axis passing through the hinges Figure. The door has a width of [latex]b=1.00\,\text{m},[/latex] and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance [latex]a=2.00\,\text{m}.[/latex] Find the forces on the hinges when the door rests half-open. The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force [latex]\mathbf{\overset{\to }{A}}[/latex] from hinge [latex]A,[/latex] an unknown force [latex]\mathbf{\overset{\to }{B}}[/latex] from hinge [latex]B,[/latex] and the known weight [latex]\mathbf{\overset{\to }{w}}[/latex] attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the y-axis along the direction of gravity and the x-axis in the plane of the slab, as shown in panel (a) of Figure, and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force [latex]\mathbf{\overset{\to }{A}}[/latex] [latex]({A}_{x}[/latex] and [latex]{A}_{y}),[/latex] and two components of force [latex]\mathbf{\overset{\to }{B}}[/latex] [latex]({B}_{x}[/latex] and [latex]{B}_{y}).[/latex] In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns [latex]({A}_{x},[/latex] [latex]{B}_{x},[/latex] [latex]{A}_{y},[/latex] and [latex]{B}_{y}),[/latex] we must set up four independent equations. One equation is the equilibrium condition for forces in the x-direction. The second equation is the equilibrium condition for forces in the y-direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation, [latex]{A}_{y}={B}_{y}.[/latex] To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of Figure. Finally, we solve the equations for the unknown force components and find the forces. From the free-body diagram for the door we have the first equilibrium condition for forces: [latex]\begin{array}{}\\ \text{in}\,x\text{-direction:}\hfill & \,-{A}_{x}+{B}_{x}=0\enspace\Rightarrow \enspace{A}_{x}={B}_{x}\hfill \\ \text{in}\,y\text{-direction:}\hfill & +{A}_{y}+{B}_{y}-w=0\ enspace\Rightarrow \enspace{A}_{y}={B}_{y}=\frac{w}{2}=\frac{400.0\,\text{N}}{2}=200.0\,\text{N.}\hfill \end{array}[/latex] We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P: [latex]\text{pivot at}\,P\text{:}\,{\tau }_{w}+{\tau }_{Bx}+{\tau }_{By}=0.[/latex] We use the free-body diagram to find all the terms in this equation: [latex]\begin{array}{ccc}\hfill {\tau }_{w}& =\hfill & dw\,\text{sin}(\text{−}\beta )=\text{−}dw\,\text{sin}\,\beta =\text{−}dw\frac{b\,\text{/}\,2}{d}=\text{−}w\frac{b}{2}\hfill \\ \hfill {\tau }_ {Bx}& =\hfill & a{B}_{x}\text{sin}\,90^\circ=+a{B}_{x}\hfill \\ \hfill {\tau }_{By}& =\hfill & a{B}_{y}\text{sin}\,180^\circ=0.\hfill \end{array}[/latex] In evaluating [latex]\text{sin}\,\beta ,[/latex] we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Figure and compute [latex]{B}_{x}:[/latex] [latex]\text{pivot at}\,P\text{:}\,\text{−}w\,\frac{b}{2}+a{B}_{x}=0\enspace\Rightarrow \enspace{B}_{x}=w\,\frac{b}{2a}=(400.0\,\text{N})\,\frac{1}{2\cdot 2}=100.0\,\text{N.}[/latex] Therefore the magnitudes of the horizontal component forces are [latex]{A}_{x}={B}_{x}=100.0\,\text{N}.[/latex] The forces on the door are [latex]\begin{array}{}\\ \text{at the upper hinge:}\,{\mathbf{\overset{\to }{F}}}_{A\,\text{on door}}=-100.0\,\text{N}\mathbf{\hat{i}}+200.0\,\text{N}\mathbf{\hat{j}}\hfill \\ \text{at the lower hinge:}{\mathbf{\overset{\to }{F}}}_{B\,\text{on door}}=\text{+}100.0\,\text{N}\mathbf{\hat{i}}+200.0\,\text{N}\mathbf{\hat{j}}.\hfill \end{array}[/latex] The forces on the hinges are found from Newton’s third law as [latex]\begin{array}{cc} \text{on the upper hinge:}\,{\mathbf{\overset{\to }{F}}}_{\text{door on}\,A}=100.0\,\text{N}\mathbf{\hat{i}}-200.0\,\text{N}\mathbf{\hat{j}}\hfill \\ \text{on the lower hinge:}\,{\mathbf{\overset{\to }{F}}}_{\text{door on}\,B}=-100.0\,\text{N}\mathbf{\hat{i}}-200.0\,\text{N}\mathbf{\hat{j}}.\hfill \end{array}[/latex] Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions. Check Your Understanding Solve the problem in Figure by taking the pivot position at the center of mass. Show Solution [latex]{\mathbf{\overset{\to }{F}}}_{\text{door on}\,A}=100.0\,\text{N}\mathbf{\hat{i}}-200.0\,\text{N}\mathbf{\hat{j}}\,\text{;}\,{\mathbf{\overset{\to }{F}}}_{\text{door on}\,B}=-100.0\,\text{N}\ Check Your Understanding A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold. Show Solution 711.0 N; 466.0 N Check Your Understanding A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut. Show Solution 1167 N; 980 N directed upward at [latex]18^\circ[/latex] above the horizontal • A variety of engineering problems can be solved by applying equilibrium conditions for rigid bodies. • In applications, identify all forces that act on a rigid body and note their lever arms in rotation about a chosen rotation axis. Construct a free-body diagram for the body. Net external forces and torques can be clearly identified from a correctly constructed free-body diagram. In this way, you can set up the first equilibrium condition for forces and the second equilibrium condition for torques. • In setting up equilibrium conditions, we are free to adopt any inertial frame of reference and any position of the pivot point. All choices lead to one answer. However, some choices can make the process of finding the solution unduly complicated. We reach the same answer no matter what choices we make. The only way to master this skill is to practice. Conceptual Questions Is it possible to rest a ladder against a rough wall when the floor is frictionless? Show how a spring scale and a simple fulcrum can be used to weigh an object whose weight is larger than the maximum reading on the scale. A painter climbs a ladder. Is the ladder more likely to slip when the painter is near the bottom or near the top? A uniform plank rests on a level surface as shown below. The plank has a mass of 30 kg and is 6.0 m long. How much mass can be placed at its right end before it tips? (Hint: When the board is about to tip over, it makes contact with the surface only along the edge that becomes a momentary axis of rotation.) The uniform seesaw shown below is balanced on a fulcrum located 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has a mass 80 kg. What is the mass of the board? Show Solution 40 kg In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the other end to a tree 15 m away, as shown below. He then pulls on the center of the rope with a force of 400 N, which causes its center to be displaced 0.30 m, as shown. What is the force of the rope on the car? A uniform 40.0-kg scaffold of length 6.0 m is supported by two light cables, as shown below. An 80.0-kg painter stands 1.0 m from the left end of the scaffold, and his painting equipment is 1.5 m from the right end. If the tension in the left cable is twice that in the right cable, find the tensions in the cables and the mass of the equipment. Show Answer right cable, 444.3 N; left cable, 888.5 N; weight of equipment 156.8 N; 16.0 kg When the structure shown below is supported at point P, it is in equilibrium. Find the magnitude of force F and the force applied at P. The weight of the structure is negligible. To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2.00 m from the bottom. The person is standing 3.00 m from the bottom. Find the normal reaction and friction forces on the ladder at its base. Show Solution 784 N, 376 N A uniform horizontal strut weighs 400.0 N. One end of the strut is attached to a hinged support at the wall, and the other end of the strut is attached to a sign that weighs 200.0 N. The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the strut, find the tension in the cable and the force at the hinge of the strut. The forearm shown below is positioned at an angle [latex]\theta[/latex] with respect to the upper arm, and a 5.0-kg mass is held in the hand. The total mass of the forearm and hand is 3.0 kg, and their center of mass is 15.0 cm from the elbow. (a) What is the magnitude of the force that the biceps muscle exerts on the forearm for [latex]\theta =60^\circ\text{?}[/latex] (b) What is the magnitude of the force on the elbow joint for the same angle? (c) How do these forces depend on the angle [latex]\theta ?[/latex] Show Solution a. 539 N; b. 461 N; c. do not depend on the angle The uniform boom shown below weighs 3000 N. It is supported by the horizontal guy wire and by the hinged support at point A. What are the forces on the boom due to the wire and due to the support at A? Does the force at A act along the boom? The uniform boom shown below weighs 700 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom? Show Solution tension 778 N; at hinge 778 N at [latex]45^\circ[/latex] above the horizontal; no A 12.0-m boom, AB, of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 1000 kg. For the position shown, calculate tension T in the cable and the force at the axle A. A uniform trapdoor shown below is 1.0 m by 1.5 m and weighs 300 N. It is supported by a single hinge (H), and by a light rope tied between the middle of the door and the floor. The door is held at the position shown, where its slab makes a [latex]30^\circ[/latex] angle with the horizontal floor and the rope makes a [latex]20^\circ[/latex] angle with the floor. Find the tension in the rope and the force at the hinge. Show Solution 1500 N; 1620 N at [latex]30^\circ[/latex] A 90-kg man walks on a sawhorse, as shown below. The sawhorse is 2.0 m long and 1.0 m high, and its mass is 25.0 kg. Calculate the normal reaction force on each leg at the contact point with the floor when the man is 0.5 m from the far end of the sawhorse. (Hint: At each end, find the total reaction force first. This reaction force is the vector sum of two reaction forces, each acting along one leg. The normal reaction force at the contact point with the floor is the normal (with respect to the floor) component of this force.)	{"url":"https://pressbooks.online.ucf.edu/osuniversityphysics/chapter/12-2-examples-of-static-equilibrium/","timestamp":"2024-11-06T04:38:08Z","content_type":"text/html","content_length":"164803","record_id":"<urn:uuid:644b068f-d90a-4507-a58d-578d8a601513>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00464.warc.gz"}
Do You Test Transformer’s Turns Ratio with Sequential or Simultaneous 3-Phase Test? - Pacific Test Equipment Great article from DV Power DV Power TRT instruments, the true three-phase turns ratio testers, enable users to perform two types of turns ratio tests: • simultaneous 3-phase test • and a sequential 3-phase test. When performing the simultaneous 3-phase test, the TRT unit supplies a true three-phase test voltage to the transformer high voltage side and measures three induced line voltages at the low voltage side. This provides the “voltage ratio” that corresponds to the nameplate ratio. When performing the sequential 3-phase test of a three-phase transformer, the TRT outputs single-phase test voltage and measures the turns ratio for each phase separately, phase by phase. The calculated turns ratio value for certain vector groups (wye-wye, delta-delta) is the same as the nameplate ratio meaning the three-phase voltage ratio, indicated on the transformer nameplate, is equal to the ratio of a number of turns of the high voltage to the low voltage windings. For example, a 220/110 kV/kV wye-wye transformer has twice the number of turns in the high voltage winding compared to the low voltage winding. For some other vector groups, the ratio of winding turns is not the same as the nameplate ratio. For example, for wye-delta groups turns ratio is √3 times lower than the nameplate ratio, while for delta-wye groups it is √3 times higher. When the TRT instrument calculates a deviation between the measured and nameplate turns ratio, these √3 factors are automatically taken into consideration. The user needs to input only the nameplate ratio, and it will be multiplied or divided by √3 before comparing it with the measured turns ratio. The √3 difference is the consequence of the behavior of the three-phase voltage systems connected in delta (triangle) and wye (star). In order to make a delta-wye three-phase transformer with voltage ratio (nameplate ratio) equal to one, √3 times larger number of turns needs to be wound on delta windings compared to wye windings. For example, 173 turns on each delta winding and 100 turns on each wye winding will be necessary. In that case, the turns ratio is equal to 1.73 (approximately √3), while the voltage ratio (nameplate ratio) is equal to 1. This is because the windings in delta connection are sized to supply a line voltage (phase to phase), while the windings in wye are determined to supply a phase voltage (phase to ground). From the theory of the three-phase systems, it is known the line voltages are √3 higher than phase voltages, and they are placed by 30˚ apart. That’s where the phase angles of 30˚ (Dyn1, YNd1), 150˚ (Dyn5, YNd5), 210˚ (Dyn7, YNd7), and 330˚ (Dyn11, YNd11) come from in delta-wye and wye-delta configurations. The three-phase voltage ratio (VR), or the transformer nameplate ratio, measured with a true three-phase test voltage, is the ratio of the line voltages on the HV and LV side. The figure below illustrates the YNd1 (wye-delta) transformer. UL’ are line voltages on the primary (HV) side. UP’ are phase voltages on the primary (HV) side. Line voltages are √3 higher than phase voltages in wye-connected windings, which means UL’ = √3 * UP’. UL” are line voltages on the secondary (LV) side. UP” are phase voltages on the secondary (LV) side. Line voltages and phase voltages are equal in delta-connected windings, meaning that UL” = UP”. The turns ratio (TR), measured with a single-phase test voltage, is the ratio of the phase voltages on the HV and LV side. TR = UP’ / UP” The three-phase voltage ratio (VR), or the transformer nameplate ratio, measured with a true three-phase test voltage, is the ratio of the line voltages on the HV and LV side. VR = UL’ / UL” = √3 * UP’ / UP” VR = √3 * TR If the ratio of winding turns (turns ratio) of a three-phase transformer is to be measured, a sequential 3-phase test should be performed. Otherwise, if the nameplate ratio (voltage ratio) of the three-phase transformer is to be measured, a simultaneous 3~ test should be performed.	{"url":"https://pacifictest.co.nz/do-you-test-transformers-turns-ratio-with-sequential-or-simultaneous-3-phase-test/","timestamp":"2024-11-11T17:15:19Z","content_type":"text/html","content_length":"295158","record_id":"<urn:uuid:1f9bbe72-8243-45a1-866b-ea0ae493f104>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00321.warc.gz"}
[Solved] If a and b are two non-collinear vectors and xa+yb=0 -... \| Filo If and are two non-collinear vectors and Not the question you're searching for? + Ask your question If and are two non-zero, non-collinear vectors and and are two scalars such that then and because one will be a scalar multiple of the other and hence collinear which is a contradiction. Was this solution helpful? Found 6 tutors discussing this question Discuss this question LIVE for FREE 15 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Practice questions from Vectors and 3D Geometry for JEE Main and Advanced (Amit M Agarwal) View more Practice more questions from Vector Algebra Practice questions on similar concepts asked by Filo students View more Stuck on the question or explanation? Connect with our Mathematics tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question Text If and are two non-collinear vectors and Updated On Aug 13, 2022 Topic Vector Algebra Subject Mathematics Class Class 12 Answer Type Text solution:1 Video solution: 1 Upvotes 228 Avg. Video Duration 5 min	{"url":"https://askfilo.com/math-question-answers/if-mathbfa-and-mathbfb-are-two-non-collinear-vectors-and-x-mathbfay-mathbfb0a-x0","timestamp":"2024-11-04T15:37:25Z","content_type":"text/html","content_length":"598420","record_id":"<urn:uuid:a23b4153-6b01-45ac-bd6d-f344b18512c9>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00165.warc.gz"}
Riley E. What do you want to work on? About Riley E. Bachelors in Mathematics, General from Eastern Michigan University Career Experience I have five years as a senior university math tutor, and five more years as a college lecturer in Calculus. I've aided in department-wide curriculum re-vamps of the calculus sequence. As a five-year mathematics Ph.D. candidate at a Top 10 University, I have deep knowledge of any material undergrads will encounter. I Love Tutoring Because I want to share my love and knowledge of math with others. I prefer tutoring to teaching because the one-on-one environment means I can adapt to students' differing needs. Other Interests Archery, Gaming, Programming, Reading Math - Calculus Great explanation! Math - Calculus - Multivariable My second tutor was great, he's great at explaining things and very good at what's he's doing. Very professional. The first tutor scared me. Math - Calculus Great general math advice and a swift resolution to my trigonometric dilemma--10/10, would recommend! Math - Calculus I had a excellent experience with my tutor session on differential equations	{"url":"https://www.princetonreview.com/academic-tutoring/tutor/riley%20e--7918869?s=ap%20calculus%20ab","timestamp":"2024-11-08T02:59:47Z","content_type":"application/xhtml+xml","content_length":"193373","record_id":"<urn:uuid:bcdc24b0-75a3-46d7-8ad3-b30caf3d405c>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00237.warc.gz"}
Equilibration and coarsening in the quantum O(N) model at infinite N The quantum O(N) model in the infinite-N limit is a paradigm for symmetry breaking. Qualitatively, its phase diagram is an excellent guide to the equilibrium physics for more realistic values of N in varying spatial dimensions (d>1). Here, we investigate the physics of this model out of equilibrium, specifically its response to global quenches starting in the disordered phase. If the model were to exhibit equilibration, the late-time state could be inferred from the finite-temperature phase diagram. In the infinite-N limit, we show that not only does the model not lead to equilibration on account of an infinite number of conserved quantities, it also does not relax to a generalized Gibbs ensemble (GGE) consistent with these conserved quantities. Instead, an infinite number of new conservation laws emerge at late times and the system relaxes to an emergent GGE consistent with these. Nevertheless, we still find that the late-time states following quenches bear strong signatures of the equilibrium phase diagram. Notably, we find that the model exhibits coarsening to a nonequilibrium critical state only in dimensions d>2, that is, if the equilibrium phase diagram contains an ordered phase at nonzero temperatures. All Science Journal Classification (ASJC) codes • Electronic, Optical and Magnetic Materials • Condensed Matter Physics Dive into the research topics of 'Equilibration and coarsening in the quantum O(N) model at infinite N'. Together they form a unique fingerprint.	{"url":"https://collaborate.princeton.edu/en/publications/equilibration-and-coarsening-in-the-quantum-on-model-at-infinite-","timestamp":"2024-11-10T02:26:29Z","content_type":"text/html","content_length":"51276","record_id":"<urn:uuid:1031177a-517f-496f-810c-2c9d5160434f>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00799.warc.gz"}
APS March Meeting 2019 Bulletin of the American Physical Society APS March Meeting 2019 Volume 64, Number 2 Monday–Friday, March 4–8, 2019; Boston, Massachusetts Session F18: Machine Learning Quantum States II Hide Abstracts Sponsoring Units: DCOMP DCMP DAMOP Chair: Yizhuang You, Harvard University Room: BCEC 156B Tuesday, F18.00001: Machine Learning Physics: From Quantum Mechanics to Holographic Geometry March 5, Invited Speaker: Yizhuang You 11:15AM Inspired by the "third wave" of artificial intelligence (AI), machine learning has found rapid applications in various topics of physics research. Perhaps one of the most ambitious goals of - machine learning physics is to develop novel approaches that ultimately allows AI to discover new concepts and governing equations of physics from experimental observations. In this talk, I 11:51AM will present our progress in applying machine learning technique to reveal the quantum wave function of Bose-Einstein condensate (BEC) and the holographic geometry of conformal field theories. In the first part, we apply machine translation to learn the mapping between potential and density profiles of BEC and show how the concept of quantum wave function can emerge in the latent space of the translator and how the Schrodinger equation is formulated as a recurrent neural network. In the second part, we design a generative model to learn the field theory configuration of the XY model and show how the machine can identify the holographic bulk degrees of freedom and use them to probe the emergent holographic geometry. Tuesday, F18.00002: Comparing deep reinforcement-learning techniques: applications to quantum memory March 5, Petru Tighineanu, Thomas Foesel, Talitha Weiss, Florian Marquardt 11:51AM In our recent work [1] we showed how reinforcement learning with artificial neural networks (ANNs) can be a powerful tool to discover quantum-error-correction strategies fully adapted to the - quantum hardware of a quantum memory. We employed a reinforcement-learning technique called natural policy gradient, in which the policy of the ANN is updated and improved according to the 12:03PM second-order gradient of the return (the cumulative sum of the reward) in the parameter space of the ANN. The principal downsides of policy gradient are sample inefficiency and slow convergence, which can be critical in the case of a quantum system with an exponentially growing Hilbert space that is simulated classically. Here we conduct an in-depth study of the performance of more advanced reinforcement-learning techniques [2] applied to a noisy quantum memory. We find that the efficiency of training can be sped up by orders of magnitude via a careful choice of the technique and the corresponding hyperparameters, both of which are motivated by and related to the underlying physics. [1] T. Fösel, P. Tighineanu, T. Weiß, F. Marquardt, PRX 8, 031084 (2018). [2] P. Dhariwal, C. Hesse, O. Klimov, et al., OpenAI Baselines, https://github.com/openai/baselines. Tuesday, F18.00003: Structural Predictors for Machine Learning Modeling of Superconductivity in Iron-based Materials March 5, Valentin Stanev, Jack Flowers, Ichiro Takeuchi 12:03PM Superconductivity in iron-based materials continues to be a focus of intense research effort a decade after its discovery. In particular, the interplay between structure and charge doping as - drivers of superconductivity is still a matter of active debate. To address this question we use Machine Learning (ML) approach. Based on published data we created a database covering the 12:15PM available structural information of the 122 family of materials. Using the lattice parameters, pnictogen height and charge doping we trained several ML models designed to predict the critical temperature T[c] over the entire parameter space. The analysis of the models suggests that no single variable can fully explain and predict the evolution of T[c], and a combination of at least two parameters are needed. The ML predictions can be used to guide further exploration of the phase diagram of iron-based superconductors. Tuesday, F18.00004: Predicting physical properties of alkanes with neural networks March 5, Pavao Santak, Gareth Conduit 12:15PM The physical properties of many alkanes are unknown, which prevents engineers from optimally deploying them in base oil lubricants. In order to address this issue, we train neural networks - that can work with fragmented data, enabling us to exploit the property-property correlations and increase the accuracy of our models. We encode molecular structure into five nonnegative 12:27PM integers, which enables us to exploit the structure-property correlations. We establish correlations between branching and the boiling point, heat capacity and vapor pressure as a function of temperature. Furthermore, we explore the connection between the symmetry and the melting point and identify erroneous data entries in the flash point of linear alkanes. Finally, we predict linear alkanes’ kinematic viscosity by exploiting the temperature and pressure dependence of shear viscosity and density. Tuesday, F18.00005: Understanding Magnetic Properties of Uranium-Based Binary Compounds from Machine Learning March 5, Ayana Ghosh, Serge M Nakhmanson, Jian-Xin Zhu 12:27PM Actinide- and lanthanide-based materials constitute an important playground for exploring exotic properties stemming from the presence of itinerant or localized f-electrons. For example, - uranium-based compounds have exhibited emergent phenomena, including magnetism, unconventional superconductivity, hidden order, and heavy fermion behavior. Among them, the magnetic 12:39PM properties with varying ordered states and moment size are sensitively dependent on pressure, chemical doping and magnetic field due to the strong-correlation effects on 5f-electrons. So far, there have even been no reports of systematic studies to map out connections between structures and properties of these compounds. In order to investigate such links and bridge between theoreotical and experimental learnings, we have constructed two databases combining results of high-throughput DFT simulations and experimental measurements, respectively, for a family of uranium-based binary compounds. We then utilized different machine learning models to identify related accessible attributes (features) and predict magnetic moments and ordering in these Tuesday, F18.00006: Machine learning-assisted search for high performance plasmonic metals March 5, Ethan Shapera, Andre Schleife 12:39PM Plasmonics aims to manipulate light through choice of materials and nanoscale structure. Finding materials which exhibit low-loss responses to applied optical fields while remaining feasible - for widespread use is an outstanding challenge. Online databases compiled computational data for numerous properties of tens of thousands of materials. Due to the large number of materials 12:51PM and high computational cost, it is not viable to compute optical properties for all materials from first principles. Geometry-dependent plasmonic quality factors for a training set of ~1,000 materials are computed using density functional theory and the Drude model. We train then apply random-forest regressors to rapidly screen Materials Project to identify potential new plasmonic metals. Descriptors are limited to symmetry and quantities obtained using the chemical formula and the Mendeleev database. The machine learning models filter through 7,445 metals in Materials Project. We iteratively improve the model with active learning by adding DFT results for predicted high quality factor metals into the training set. From this we predict Cu[3] Au, MgAg[3], and CaN[2] as candidates and verify their quality factors with DFT. Tuesday, F18.00007: Machine Learning and Crystal Structure Prediction of Molecular Crystals March 5, Ruggero Lot, Franco Pellegrini, Yusuf Shaidu, Emine Kucukbenli 12:51PM There is a natural synergy between data-hungry machine learning methods and crystal structure prediction of molecular crystals that requires a careful search in a vast potential energy - 1:03PM landscape. In our previous study we demonstrated how taking advantage of machine learning methods can enable novel predictions even for well-studied molecular crystals [1]. In order to leverage this synergy further, we have been developing a deep neural network training tool, PANNA (Potentials from Artificial Neural Network Architectures), based on TensorFlow framework [2]. In creating transferable machine-learned potentials, the key step is the non-linear process of training the network model. We will demonstrate a variety of network training techniques that can be explored within PANNA, from ones that are commonly used in machine learning community to the ones that are specific to atomistic simulations. We will report the effect of data selection, input representation and training methods on the training dynamics and on the resulting potentials in the difficult case of molecular crystals. [1] C. Bull et al. “ζ-Glycine: insight into the mechanism of a polymorphic phase transition” IUCrJ 4, p.569 (2017) [2] M. Abadi et al. “TensorFlow: Large-scale machine learning on heterogeneous systems” (2015) Tuesday, F18.00008: Fitting effective models using QMC parameter derivatives March 5, William Wheeler, Shivesh Pathak, Lucas Wagner 1:03PM - Effective models are fundamental to our understanding of complex materials, but selecting the right model and parameters to accurately describe a particular material can be challenging. The 1:15PM recently developed density matrix downfolding method (DMD) [1] uses an ensemble of quantum Monte Carlo calculations to both select and fit parameters to effective models for materials. However, this method is computationally extremely demanding. In a similar spirit to force matching in classical model fitting [2], we improve the efficiency of DMD by computing derivatives of the energy and density matrix with respect to parameters of each trial wave function. We demonstrate the new technique by computing a band structure for silicon using first principles quantum Monte Carlo. [1] H. Zheng, H. J. Changlani, K. T. Williams, B. Busemeyer, and L. K. Wagner, Front. Phys. 6, 43 (2018). [2] F. Ercolessi and J. B. Adams, Europhys. Lett. 26, 583 (1994). Tuesday, F18.00009: Detection of Phase Transitions in Quantum Spin Chains via Unsupervised Machine Learning March 5, Yutaka Akagi, Nobuyuki Yoshioka, Hosho Katsura 1:15PM - In the field of machine learning, there has been an important breakthrough in recent years. What was at the center of it is the deep learning by artificial neural networks mimicking human 1:27PM brains. By deepening a process part which repeats extracting feature quantities through nonlinear transformations, so-called hidden layer, data/class separability and the discriminability have dramatically improved. Recently, the machine learning techniques have found a wide variety of applications in condensed matter and statistical physics. Examples include detection of phase transition and acceleration of Monte Carlo simulation. Meanwhile, most of these studies are based on supervised learning under well-known results, and there are only a few previous examples applying unsupervised learning. In this research, we investigate quantum phase transitions in various quantum spin chains by using an autoencoder which is one of unsupervised learning methods. In particular, we will show that the autoencoder whose input is only short-range correlators is capable of detecting even topological phase transition from the large-D phase to the Haldane phase without using either topological invariants or entanglement spectra. Tuesday, F18.00010: Supervised learning of phase transitions in classical and quantum systems March 5, Nicholas Walker, Ka-Ming Tam, Mark Jarrell 1:27PM - Supervised machine learning methods are used to identify transitions in physical systems using the classical solid-liquid transition of a Lennard-Jones system as well as the strong 1:39PM coupling-local moment quantum transition in the soft-gap Anderson model as examples. Monte Carlo sampling was used to achieve a uniform sampling of configurational data across a large range of the relevant transition parameter for each system. Hyperbolic feature scaling is applied to the features followed by training a 1-dimensional convolutional neural network with the samples corresponding to the extreme parameters of each phase as training data. The rest of the configurational data is assigned phase classification probabilities by the neural network, allowing for the prediction of the transition point with respect to the chosen varied parameter. This is done by fitting the mean classification probabilities for each set of configurational data with respect to the varied parameter with a logistic function and taking the transition to be at the value of the parameter corresponding to the midpoint of the sigmoid. The results obtained are comparable to results using contemporary methods for each system. 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Become an APS Member Renew Membership Librarians Submit a Meeting Abstract Join an APS Unit Authors Submit a Manuscript Get My Member Number Referees Find a Journal Article Update Contact Information Media Donate to APS Students © 2024 American Physical Society \| All rights reserved \| Terms of Use \| Contact Us Headquarters 1 Physics Ellipse, College Park, MD 20740-3844 (301) 209-3200 Editorial Office 100 Motor Pkwy, Suite 110, Hauppauge, NY 11788 (631) 591-4000 Office of Public Affairs 529 14th St NW, Suite 1050, Washington, D.C. 20045-2001 (202) 662-8700	{"url":"https://meetings.aps.org/Meeting/MAR19/Session/F18?showAbstract","timestamp":"2024-11-03T22:55:37Z","content_type":"text/html","content_length":"28755","record_id":"<urn:uuid:03f1aad4-6cda-4ac0-8d78-18883f38a811>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00414.warc.gz"}
Beer bubble math helps to unravel some mysteries in materials science Before downing your next beer, pause to contemplate the bubbles. You’ll find that they grow and shrink in odd, hard-to-predict ways. A mathematician and an engineer have found a simple and surprising equation to describe this process, using a field of mathematics no one expected to be relevant. Now, new simulations are building on that result to illuminate more than just foams. Metals and ceramics are made of crystals that grow and shrink the same way that beer bubbles do, affecting the properties of the materials. The new work may thus lead to more resilient airplane wings, more reliable computer chips and stronger steel beams. Over time, the bubbles in foam tend to consolidate, becoming fewer and larger. The physics driving this process has long been understood: When two bubbles adjoin one another, gas tends to pass from the bubble with higher pressure to the one with lower pressure. The higher-pressure bubble bulges into the lower-pressure one, so the shape of a bubble reflects the pressure in all the bubbles around it. As a result, the shape of the bubble should be enough to determine how fast it’s going to grow or shrink. In 1952, the mathematician John von Neumann used this principle to find an elegant equation that predicts the growth rate of a bubble in a two-dimensional spread of bubbles. But the three-dimensional case—which is the one scientists and beer drinkers most care about—proved far more difficult. Some researchers even believed the problem unsolvable. What was needed, it turned out, was to introduce another field of mathematics: topology. Topology is essentially the study of connections. For a topologist, two objects are the same if one can be shrunk or stretched into the same shape as the other, but without punching holes or gluing anything together, since that would change the way the parts of the object connect to one another. So, for example, a doughnut and a coffee cup are the same shape to a topologist: Squish down the cup part of the coffee cup, and you’ll end up with a doughnut-shaped ring. A doughnut is topologically different, however, from a beach ball. Topology would seem to offer little help in determining how quickly bubbles in foam grow or shrink, because squishing a bubble into a different shape will change its growth rate. But several years ago researchers found an equation involving the Euler characteristic, a property of a shape that stays the same no matter how the shape is stretched or smushed. “It’s really beautiful,” says Robert MacPherson of the Institute for Advanced Study in Princeton, N.J., the mathematician on the project, “because the Euler characteristic shouldn’t have anything to do with it.” To calculate the Euler characteristic, first imagine slicing a bubble in different directions. Usually, the resulting shape would be roughly circular. But suppose that the bubble had a couple of little bumps on its surface. If your slice went through these bumps, you could end up with two disconnected circles. Or, if the bubble had a small divot in it and your slice went through the divot, you could end up with a little hole inside your slice. The Euler characteristic of the slice is the number of disconnected pieces it contains, minus the number of holes within it. The equation that MacPherson and David Srolovitz of the Agency for Science, Technology and Research in Singapore developed shows that bubbles grow quickly when the beer is warm and when the bubbles have divots in them rather than bumps, or when they are connected to lots of other bubbles. Now, the team is taking the equation one step further, using it to create computer simulations of foams, metals and ceramics. The work is a way to expose a material’s inner structures that are not normally visible, but that influence the material’s overall properties. “If you beat an egg white enough, it becomes almost a paste and you can make peaks out of it,” MacPherson says. “If you looked at an individual egg white bubble, you wouldn’t expect that. We want to understand these collective properties.” “Topology is going to be a very powerful tool for understanding these structures,” says Jeremy Mason, a materials scientist at the Institute for Advanced Study who has joined the research team. He points out that the behavior of foam as a whole is unlikely to change dramatically if you stretch or squish the bubbles a bit without changing the way they’re connected to one another, even though changing the shape of an individual bubble will affect its growth rate. Focusing on the connections—that is, the foam’s topology—may then allow researchers to home in on its most crucial aspects. “We’re awash in data, and the challenge is to identify what is meaningful and compare it between two different structures,” Mason says. “Measuring topological characteristics may give us a language for that.” SN Prime \| June 10, 2011 \| Vol. 1, No. 1	{"url":"https://www.sciencenews.org/article/beer-bubble-math-helps-unravel-some-mysteries-materials-science","timestamp":"2024-11-09T23:00:28Z","content_type":"text/html","content_length":"265426","record_id":"<urn:uuid:edc7698c-df69-4b12-9434-a728964c3f1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00392.warc.gz"}
Temple University Department of Economics Economics 8190 Special Topics: Econometrics Prerequisites: Qualified students will have taken 8009. The well prepared student will have taken 8119. This course is not a substitute for either 8009 or 8119. Course Description: This course will be a survey of those areas of modern econometrics most often used in applied economic research. Each topic area will have three legs. The course will introduce recent developments in the solution of applied problems followed by a specific example presented by either the principle instructor or a guest speaker. The third leg will be a homework assignment completed by the students. Topics to be covered include recent developments in endogeneity and the use of instrumental variables: estimation of treatment effects, weak instruments, too many instruments, and difference in differences. As an alternative to the use of IV/2SLS we will look at the use of control functions for dealing with endogeneity. Some time will be used to address linear panel data models, regression discontinuity, quantile methods and kernel methods. After looking at the 2009 and 2010 syllabi for 8119 we will not be covering any per se time series topics. Textbooks: All of these books are strongly recommended, but not required. William H. Greene, Econometric Analysis, 6th or 7th Edition, Prentice Hall. You should already have this book in your library. Joshua Angrist and J.-S. Pischke, Mostly Harmless Econometrics: An Empiricist's Companion, Princeton University Press, 2009. This is a widely used book. It is deceptively easy to read, but is full of nuance. Roger Koenker, Quantile Regression, Cambridge University Press, 2005. This is an important text. It is well written and an absolute must before you pick up the literature that has emerged over the last six years. Jeffrey Wooldridge, Introductory Econometriocs: A Modern Approach, 4th edition, Sothwestern/Cengage Learning, 2009. This is an undergrad textbook that I used last spring, so you should be able to find a used copy. It is good for background reading before wading into the real thing. Jeffrey Wooldridge, The Econometrics of Cross-section and Panel Data, MIT Press and Angrist and Pischke, Mostly Harmless Econometrics: An Empiricist's Companion, Princeton University Press. This is a popular textbook, but I find it very difficult reading. It makes a good reference book. Links to University policies on freedom of speech and student rights:	{"url":"http://ajbuckeconbikesail.net/Econ8190/Description.htm","timestamp":"2024-11-07T07:25:38Z","content_type":"text/html","content_length":"20805","record_id":"<urn:uuid:5e55b57d-ffe7-4ec3-a99a-dbdffa1e7066>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00001.warc.gz"}
Hockey Historysis If you'd asked me yesterday when my last post here was, there's about a zero percent chance I would have said "well, nearly a year, obviously." It's amazing how little you actually get done when you have a half-dozen or so projects on the go at the same time. I can't promise to be posting regularly here again, but we'll at least have a little series of posts to work with, on the topic of defensive pairings in the 1930s and 1940s. I've always found it a bit odd what while we keep track of which side wingers play in hockey (though not always quite accurately, just ask Alex Ovechkin), the same is really never done for defencemen. Bobby Hull played left wing, while Pierre Pilote just played defence. Actually, he mostly played right defence, but for some reason we don't seem to care about that. Red Kelly was a left defenceman, and Eddie Shore was a right defenceman. You don't find these details on any website or in any encyclopedia, however, because apparently it's not important enough to note. But this distinction is, in fact, very important on the ice. Defencemen are not interchangeable. While many blueliners can and do play both the right side and the left, some defencemen are really only good on one side or the other, and this should be recognized. Generally speaking, right-hand-shot (RHS) defencemen have a significant advantage when playing the right side of the ice, due to being able to stop pucks along the boards, and to shoot pucks out of their zone along the boards, on their forehands, and also being able to receive passes from their defence partner on the forehand as well. Left-hand-shot (LHS) blueliners have similar advantages when playing on the left side. However, because most right-handed people shoot left, and the great majority of people are right-handed, there are more LHS defencemen available than RHS, which results in a significant number of LHS playing the right side due to there being an insufficient number of RHS blueliners. But not every LHS player is adept at playing the right side. Since you will be playing on your backhand when you're on your off-side, generally it's the better stickhandlers and passers that are able to make the switch effectively. So with all this in mind, it's puzzling that we apparently don't pay any attention to who plays what side. It wasn't always this way. From 1933 to 1943, for example, the for post-season NHL All-Star Teams was split up between right defencemen and left defencemen, just as it was (and is) for wings. That ended with the beginning of the "Original Six" era for some reason, and since that time the league has really paid no attention to it, with every blueliner just being listed as "D" since then. As such it's worth attempting to reconstruct NHL defencemen's positions. And that's what I'm going to do, starting with the period from 1933 to 1943. I'm starting here for the reason noted above; we have voting records based on the left side and right side for the All-Stars, which should help to provide clues about which side a particular defenceman played. We can't just assume they're completely accurate. Even today, when the dissemination of information about hockey players is so much easier than in the 30s, the voters still considered Ovechkin to be both a right wing and a left wing in 2012/13. He used to play left wing, but had shifted to the right that season, and many voters presumably just assumed he was still on the left. And the same sort of thing could have easily happened 80 years ago, so we can't just take the voting results as gospel. So what other information do we have to inform our analysis? Well, up until the 1950s, newpapers summaries of NHL games listed the full playing rosters of both team, divided into starters at each position and substitutes. Very occasionally, and I mean very occasionally (especially the later we go in time), the defencemen would be listed as right defence and left defence, and in those cases we know what position the starting defencemen were playing, at least. But even though each starting defenceman's position was rarely listed, we do know who the starters were, and we know whether each was a RHS or LHS. As such, when a RHS is paired with a LHS, we should be able to assume that the RHS played the right side and the LHS played the left side, due to the advantages inherent in playing on your forehand side. Or can we? Next time we'll look at some analysis to see if this is a reasonable assumption. What follows is a post from my old hockey analysis site puckerings.com (later hockeythink.com). It is reproduced here for posterity; bear in mind this writing is over a decade old and I may not even agree with it myself anymore. This post was originally published on October 29, 2002. Arena Goal Factors Copyright Iain Fyffe, 2002 One possible source of ideas when doing statistical analysis in hockey is analysis done in other sports. Of course, baseball is the most obvious choice, since so much statistical analysis has been done in that field. But one must be careful when importing ideas to consider the differences that exist between the sports involved. The concept of park run factors is an example. Park runs factors exist in baseball because different parks have different dimensions and conditions, thereby affecting the number of runs scored in each park. Several people have suggested that such an analysis could be done in hockey, but to my knowledge, no one has published any results. First, we must ask the question: do factors of this kind (let's call them Arena Goal Factors, or AGF) make sense for hockey? I would say yes, there could be enough differences between arenas (in terms of dimensions, ice condition, etc.) to affect goal-scoring levels. I would expect the differences to be less than in baseball, but would not be surprised if they do exist. For example, let's take a team that scored 120 goals at home and 100 on the road, and allowed 90 goals at home and 100 on the road. In this league, teams score 55% of their goals at home, while allowing 45% of their goals at home. We would therefore expect this team to score 121 goals at home and allow 86 at home, for a total of 207 goals at home. They actually had 210 total goals at home. Their AGF is therefore 210 divided by 207, or 1.014. But before we can use this figure, we have to adjust for the fact that a team plays only half its games at home, and half on the road (in other arenas with other AGF figures). Since the sum of league AGF is equal to the number of teams, we calculate the Arena Goal Adjustment (AGA) as follows: AGA = [(TMS-1)x(AGF)+(TMS-AGF)]/[2x(TMS-1)] Where TMS is the number of teams in the league. I won't bother with the derivation. So if the team in the above example played in a 25-team league, its AGA would be 1.007, meaning that players on this team would have their scoring totals increased by about 0.7% due to playing in their particular arena. That's the theory, anyway. But I won't string you along any more. You can calculate AGA's for each NHL team for each season, but they are not the result of the nature of the arenas. They are random I calculated AGA's for six NHL seasons: 1990/91, 1991/92, 1994/95, 1995/96, 1998/99 and 1999/2000. If AGA were meaningful, there would be a strong relationship between the AGA for a team one year and the AGA for that team the next year. The results of this inter-year correlation is as follows: between 1990/91 and 1991/92, 0.34; between 1994/95 and 1995/96, -0.05; between 1998/99 and 1999/00, -0.37. The average correlation coefficient is -0.03, which suggests the relationship is entirely random. For further support, I calculated the correlations between goals-for factors and goals-against factors for each team. If the effects were real, then we would expect to see both goals for and goals against affected in the same way. The results of this intra-year correlation are as follows: │Year │Correlation │ │1990/91 │0.13 │ │1991/92 │0.24 │ │1994/95 │0.05 │ │1995/96 │0.24 │ │1998/99 │0.32 │ │1999/00 │-0.03 │ The average correlation is 0.16, which is stronger than the inter-year correlation, but still nowhere near as strong as we would need to say there is a relationship there. In summary, Arena Goal Factors do not exist in hockey. You can calculate them all you like, but overall they are the result of random chance and do not reflect anything meaningful. What follows is a post from my old hockey analysis site puckerings.com (later hockeythink.com). It is reproduced here for posterity; bear in mind this writing is over a decade old and I may not even agree with it myself anymore. This post was originally published on October 29, 2002. Factors Affecting NHL Attendance Copyright Iain Fyffe, 2002 This paper builds upon the work of Wiedecke, who examined factors affecting NHL attendance using a multiple linear regression model. A summary of this work follows. Data from the 1997/98 NHL season were used, giving 26 data observations. The dependent variable used was the percentage of capacity (called "Attendance Capacity"). That is, if a team averaged 15,000 fans in an arena with a capacity of 15,500, the team had an Attendance Capacity of 97% (15,000 divided by 15,500). The independent variables used were standings points, goals scored, and penalty minutes (which are all self-explanatory), and location (explained below). Location for each team was assigned a value of 1, 2 or 3 based upon the team's geographic location. A value of 1 was assigned to the northernmost teams (Calgary, Edmonton, Montreal, Ottawa, Toronto and Vancouver). A value of 2 was assigned to Boston, Buffalo, Chicago, Colorado, Detroit, New Jersey, New York Islanders, New York Rangers, Philadelphia, Pittsburgh, and St. Louis. A value of 3 was assigned to the southernmost teams (Anaheim, Carolina, Dallas, Florida, Los Angeles, Phoenix, San Jose, Tampa Bay, and Washington. (1) by incorporating a larger data set; (2) by redefining the dependent variable; and (3) by introducing a new indepdendent variable. Rather than using only the 1997/98 season, I will use data from 1995/96, 1996/97, 1997/98, 1998/99, 1999/2000, 2000/01 and 2001/02, giving 193 data observations. I will use average attendance as the dependent variable, rather than percentage of capacity. By using the percentage, a team which fills 14,800 of 15,000 seats (98.7%) is considered superior to a team which fills 19,700 of 20,000 seats (98.5%). This does not reflect reality well, as the second team draws a full 33% more fans. The independent variable added is Novelty. A value of 5 is assigned to a team in its first year in the league (after either an expansion or franchise relocation), and this is reduced by one for each subsequent year in the league until it reaches 0. The purpose is to determine if new teams get an attendance boost simply by being new, as if often postulated. The four independent variables used by Wiedecke are also used. Variable Correlations A variable correlation analysis is performed to examine the data for possible cross-correlation effects. Only one pair of variables, goals and standings points, has a significant correlation (positive 0.64). Therefore if both goals and points are found to be significant, care must be taken in their interpretation due to cross-correlation. Other pairs with less-significant correlations are attendance and points (positive 0.39), attendance and location (negative 0.31), and location and novelty (positive 0.30). The following table indicates the coefficients of correlation for all variables used: attendance (ATT), points in standings (PTS), goals scored (GF), penalty minutes (PIM), location (LOC) and novelty │ │ATT │PTS │GF │PIM │LOC │NOV │ │ATT│- │.39 │.25 │-.04│-.31│-.17│ │PTS│.39 │- │.64 │-.28│-.17│-.19│ │GF │.25 │.64 │- │.10 │-.22│-.17│ │PIM│-.04│-.28│.10 │- │.04 │-.01│ │LOC│-.31│-.17│-.22│.04 │- │.30 │ │NOV│-.17│-.19│-.17│-.01│.30 │- │ Results of the Model The results of the multiple linear regression model are as follows. │Constant (y-intercept) │13,326│ │Standard error of estimate │2,071 │ │R-squared │0.223 │ │Variable│Coefficient │St. error │t-stat│ │PTS │61.08 │13.56 │4.50 │ │GF │-6.90 │7.16 │-0.96 │ │PIM │0.80 │0.61 │1.31 │ │LOC │-778.93 │211.43 │-3.68 │ │NOV │-47.92 │119.85 │-0.40 │ Discussion of Results The t-statistics of GF, PIM and NOV indicate there is little evidence that they affect attendance in any significant way. On the other hand, there is very strong evidence that PTS and LOC significantly affect attendance. These findings agree with Wiedecke. Overall, the model is not extremely useful; the R-squared figure indicates only 22.3% of the variability in attendance is explained by the model. This may indicate there are other independent variables that should be considered. The correlation between the two significant independent variables (PTS and LOC) is -0.17, indicating there is no significant cross-correlation effect. According to the model, having a good team is the most significant factor affecting attendance. Ceteris paribus, each additional standings point increases attendance by 61 fans per game. A 90-point team therefore has a 610-fan advantage in average attendance over an 80-point team. The location coefficient indicates that the further south a team is, the worse its attendance is. All else being equal, a team in the southern US averages 1,558 fans less per game than a team in Canada. This is significant because the NHL's recent strategy has been to put as many teams in the southern US as possible, either through expansion or franchise relocations (including moving teams from Canada to the southern US). The results of this model suggest that this strategy is seriously flawed. In this case, analysis agrees with common sense: why are markets where there are hockey fans ignored in favour of markets where there are no hockey fans? At least the most recent expansion was more logical, and didn't put any more teams in the Sun Belt. Wiedecke, Jennifer. 1999. Factors Affecting Attendance in the National Hockey League: A Multiple Regression Model. Master's thesis, University of North Carolina, Chapel Hill. What follows is a post from my old hockey analysis site puckerings.com (later hockeythink.com). It is reproduced here for posterity; bear in mind this writing is over a decade old and I may not even agree with it myself anymore. This post was originally published on October 18, 2002. Theory: Win-Things Copyright Iain Fyffe, 2002 The most common perspective put forward on win theory can be summarized as follows: Before a game begins, each participating team has a 50% chance to win (a .500 expected winning percentage), ceteris paribus. As the game progresses, and as each team does things that affect their chances of winning or chances of losing, the expected winning percentage of each team changes. For instance, if a team scores a goal after 5 minutes of play, their percentage may change to .550, and the opponent's would therefore be .450, since the percentages necessarily sum to one. At the crux of this theory lie two ideas: (1) before a game begins, a team's winning percentage is .500, and (2) a team does two types of things that affect its chances of winning: good things (which we'll call "win-things") and bad things (which we'll call "loss-things".) As a team, you have no significant control over what your opponents do. Therefore, at least from an analytical perspective, you can assume they will do an average number of things to win. At the beginning of a game, you have not yet done anything to win, and have no guarantee that you will do so. Therefore, your expected winning percentage before a game is not .500, but .000. Teams try to win games, they do not try to lose them. Therefore a loss-thing is merely a failed attempt at a win-thing. Just as darkness is merely the absence of light, loss-things are merely the absence of win-things. Therefore win-things are what matters, and this is why I refer to this theory as Win-Things Theory. The idea that you cannot control your opponent's actions is carried throughout the thoery. For instance, in the traditional theory, scoring a goal is a very good thing (i.e., it has a high Win-Things value). Under Win-Things Theory, whether or not a shots actually produces a goal is irrelevant to the shooting side. The Win-Things were produced by the shot itself, with a higher-quality shot producing more Win-Things. Conversely, the opponent's Win-Things on the play depend on whether or not the shot is stopped. Stopping the shot produces Win-Things about equal to the Win-Things resulting for the other side by taking the shot. Not stopping the shot produces no Win-Things (it does not produce Loss-Things). It should be noted that the .000 beginning expected winning percentage applies only to one-team analysis. In two-team analysis, where the actions of both teams are considered, the expected percentage would depend on the Win-Things each team has accumulated. But generally speaking, one-team analysis is more useful in analyzing what contributes to winning, by assuming opponents to be average in all Traditional theory focusses much attention on expected winning percentage. Win-Things Theory does not. The point is not to get your expected winning percentage up; the point is to accumulate more Win-Things than your opponents. Since you cannot control how many Win-Things your opponents accumulate, the best way to ensure this is to accumulate as many Win-Things as possible. This theory supports Bill James' Win Shares system for baseball, which I have adapted into the Point Allocation method for hockey. Win Shares has been criticized for not considering "Loss Shares". Using this new theory, Loss Shares are irrelevant, and the criticism is therefore invalid. Opportunity should still be considered, but fortunately in hockey games are timed, while in baseball the opportunities vary greatly from game-to-game, based on a multitude of factors. What follows is a post from my old hockey analysis site puckerings.com (later hockeythink.com). It is reproduced here for posterity; bear in mind this writing is over a decade old and I may not even agree with it myself anymore. This post was originally published on October 18, 2002. Theory: Shots and Save Percentage Copyright Iain Fyffe, 2002 In my investigation into the validity of Goaltender Perseverence , I looked into the relationship between the number of shots a goaltender faces per game and his save percentage. I found that, as the number of shots per game increases, save percentage does not decrease, on average, as the fundamental assumption of Perseverence argues. In fact, there is some evidence of a positive relationship; that is, as shots increase, save percentage increases. This evidence was met with an "it doesn't make sense" reaction from those I presented it to. Well, common sense is often dead wrong. To explain this phenomenon, I present the following theory. For simplicity, I will discuss only two types of shots: easy and tough (referring to the goaltender's perspective). There are in actuality many varying degrees of toughness of shots, but these two will suffice for our purposes. Easy shots are largely discretionary. They are shots that result from situations where a player could choose to shoot, or choose another play. They are of lower quality than tough shots, because they are usually taken from a greater distance than tough shots, or less favourable circumstances. Since easy shots are discretionary, there must be a reason that teams do not simply shoot every time, in order to maximize their goals scored. The reason could be twofold: you give up the opportunity to make a pass, which could result in a higher-quality shot, and the shot is more likely to produce a turnover, allowing a possible scoring chance for the opposition. Therefore, it is not always wise to take the shot rather than another play. Save percentages on tough shots are low, and save percentages on easy shots are high. And since easy shots are primarily responsible for variation in shots faced by a goaltender (since the number of tough shots faced is relatively consistent), save percentage will increase as shots faced increases. For example, let's say that the average tough shots faced per game is 5, and the save percentage on such shots is .800. This is the same for every goaltender. Any difference in shots faced is due to easy shots, which we'll say have a save percentage of .900. A goaltender facing 25 shots will therefore face 20 easy shots (25 less 5). Goals against on tough shots is 1.0 (5 less .800 times 5), on easy shots 2.0 (20 less .900 times 20). 3 goals against on 25 shots is an .880 save percentage. A goaltender facing 35 shots will have the same 1.0 goals against on tough shots, but will have 3.0 on easy shots (30 less .900 times 30). 4 goals against on 35 shots is an .886 save percentage. The goaltender facing more shots on average has a higher save percentage. That is my theory of how save percentage can increase as shots increase. Unfortunately, this theory cannot be tested using information that is currently available. The NHL does track certain shot data (type, location) for shots that produce a goal, but not for shots that do not produce a goal. If this information were recorded for all shots, it could be used to test this theory. What follows is a post from my old hockey analysis site puckerings.com (later hockeythink.com). It is reproduced here for posterity; bear in mind this writing is over a decade old and I may not even agree with it myself anymore. This post was originally published on October 18, 2002. Theory: the Cost of a Penalty Copyright Iain Fyffe, 2002 The value of odd-man play is often debated. In the mass media, much ado is made about the power-play (and, to a lesser extent, penalty-killing), calling it a key to success. Others, such as Klein and Reif, downplay its importance, noting that even-strength play is better for predicting success. This essay takes a conceptual approach to this problem. What, in theory, is the importance of odd-man situations? To examine this question, I will examine a theoretical team, one which is average in all respects. This team plays in three types of situations: even-strength (ES), power-play (PP) and short-handed (SH). Examining each of these situations reveals the answer we are looking for. Even-strength: The team is completely average. Therefore, they will score exactly as many ES goals (ESGF) as they allow (ESGA). Thus, their expected net goal differential per minute of ES time (ESMIN) is calculated as follows: ( ESGF - ESGA ) / ESMIN Which, for reasons discussed above, is zero. Power-play: On the PP, a team scores about three times as often as at ES, while goals against are cut in half. PP time (PPMIN) produces a net goal differential as follows, using 1998/99 figures: ( PPGF - SHGA ) / PPMIN = ( 1533 - 220 ) / 16326 ... minutes figure is estimated = 0.08 Short-handed: Since PP time for one team is SH time for another, SH situations produce the converse of PP, or -0.08 goals per minute. Taking this all together, as average team will have a winning record if they can obtain more PP opportunities then they give. That's badly phrased, since a team with a winning record cannot be average, but you know what I mean. This is most easily accomplished by taking as few penalties as possible, since you have rather limited control over your opponent's actions. From this perspective, odd-man situations are extremely important, as they decide games. The team taking fewer non-coincident penalties should win, on average. If this perspective is valid, then we should be able to predict success based upon PP opportunities for and against. I tested the coefficient of correlation between net PP opportunities and standings points for a selection of recent seasons: │1990/91 │0.11 │ │1991/92 │0.26 │ │1994/95 │0.02 │ │1995/96 │-0.02 │ │1998/99 │0.63 │ │1999/00 │0.23 │ │average │0.21 │ The correlations provide, on average, some support for the theory. They are generally positive, but not that strong (aside from 1998/99, which is very strong). But remember, we are not considering the quality of the teams, unless you consider taking few penalties to be a quality (which you should.) So there is some evidence that this theory is valid. What follows is a post from my old hockey analysis site puckerings.com (later hockeythink.com). It is reproduced here for posterity; bear in mind this writing is over a decade old and I may not even agree with it myself anymore. This post was originally published on October 9, 2002. The Greatest Teams of All Time Copyright Iain Fyffe, 2002 The most thorough discussion of teams possibly deserving nomination as the greatest of all time is in Klein and Reif's Hockey Compendium. They base their conclusion that the 1929/30 Bruins are the greatest of all time on the team's .875 winning percentage, which is the highest of any team playing the minimum number of games. There are, of course, two problems with basing the analysis solely on winning percentage. For one, an artificial games limit has to be introduced, to keep those 8-0-0 Montreal Victorias of 1898 and 10-0-0 Montreal Wanderers of 1907 from dominating the list. If we could avoid artificial restrictions like these, we could improve the analysis substantially. As it stands, these teams have no chance of being considered, no matter how great they may have been. In addition, using winning percentage alone ignores the league context. That is, how good are the other teams in the league? Are there a few weak sisters to beat up on, or is parity the order of the day? Obviously, the greater the parity in the league as a whole, the more difficult it is to run up a high winning percentage. You don't get those cheap points; you have to fight for each win. Therefore the analysis should be based on the degree by which a team dominates the competition, and the range of quality of said competition. One method to do this is explained below, by way of Let's examine the top two teams by Klein and Reif's analysis. The Boston Bruins of 1929/30 played in a league where the standard deviation of winning percentage was .188, which is fairly high for the era. Boston's winning percentage of .875 is .375 higher than the average (which is .500), or 1.99 standard deviations above the mean (.375 divided by .188). This is called a z-score, and this is what I will base my analysis on. It encompasses both how far above the competition a team was, and how much variation in quality there was between teams. Boston's Winning Percentage Z-Score (WPZS) is therefore 1.99, which is very impressive, but as we'll see, not the best of all time. The 1943/44 Montreal Canadiens, rated #2 by Klein and Reif, had an .830 winning percentage in a league that a had a standard deviation of winning percentage of .215 (high due to the disparity in talent caused by the war). There was less parity in this league-year than in 1929/30. Montreal's WPZS is 1.53, which while quite high is nowhere near the best of all time. This means that, relatively speaking, Montreal had a greater benefit of weaker teams to play against than Boston did. By analyzing teams in this way, we consider both the quality of the league and we remove the need for any arbitrary restrictions. Below is the list of the top 48 teams of all time (all those with a WPZS of 1.50 or greater), from among the NHL and its predecessors, as well as the PCHA and WCHL/WHL, and the WHA. The surprises start at the very top. The greatest team of all time, by this analysis, is the 1995/96 Detroit Red Wings. Their .799 winning percentage had them #7 on Klein and Reif's list. But the standard deviation that year was a mere .116, quite low for the era. Other than Detroit, the best winning percentage was .634. 19 of the 26 teams were between .400 and .600. Parity was the rule, yet Detroit was able to completely dominate the league. Their 2.58 WPZS is far and away the best of all time. The next two spots come from two teams from the same season. The epic battle between Calgary and Montreal in 1988/89 is revealed to be of truly historic proportions. Other than these two teams, no team had a winning percentage of greater than .575, or less than .381. The parity this year was amazing; the standard deviation was only .100. Calgary's percentage was .731; Montreal's was .719. While both teams miss Klein and Reif's top 20, they're #2 and #3 here. Never has there been two teams which stood futher above the rest of the league. Spot #4 is the 1976/77 Canadiens. Montreal's 1970's dynasty also makes appearances at #9, #16, #19, and #26. That's a hell of a decade, and it's no surprise that it shows up here. Two more recent Red Wings sides take the 5 and 7 spots, with the Dallas South Stars outstanding 1998/99 campaign sandwiched in between. The great Bruins of 1929/30, ranked #1 by Klein and Reif, finally appear at #8. If I were to ask you which Flyers teams was the best in their history, I doubt you would answer "the 1979/80 edition, of course!" But here they are in a tie for 9th with the best the Oilers have to offer, the 1985/86 team. Another 1980's Flyers squad (1984/85) appears at #22, well above the their best of the 1970's (1973/74), which comes in at a tie for #40. 80's Oilers teams also appear at # 18, #36, #42, and #45. Not quite the 1970's Canadiens, but not bad. The highest-ranked team of the pre-NHL era turns out to be the 1912/13 Quebec Bulldogs. In a league where the five other teams had records ranging from 10-10-0 to 7-13-0, Quebec went 16-4-0 to dominate the field. The Houston Aeros were the WHA's greatest team, no surprise, claiming spots 13, 34, and 38. No other WHA club appears on the list. Montreal's other great dynasty shows up a few times as well. 1958/59 is #18, 1955/56 is #25, 1957/58 is #28, and 1959/60 is #46. This is probably less impressive than the 1980's Oilers, but more than the Islanders teams which show up at #14, #23, and #42. The Bruins of the early 70's don't show as well as you might expect, because they played in an expansion era. They appear "only" at #16, #24 and #32. The original Senators also appear thrice, at #25, #34 and #40, the last two from their pre-NHL days. Finally we have the two perfect clubs mentioned before. Because these teams played in eras notable for their lack of parity, their 1.000 winning percentages are knocked down quite a bit on this list. The 1898 Victorias stand in a tie at #36, while the Wanderers show at #38. These teams (as well as the 1910/11 Senators at #40) were completely blocked out of Klein and Reif's list due to the artificial games restriction. Here, they get a fair shot. The complete list follows: │Rank│Team │Year │League│WPct │WPZS│ │1. │Detroit Red Wings │1995/96 │NHL │.799 │2.58│ │2. │Calgary Flames │1988/89 │NHL │.731 │2.31│ │3. │Montreal Canadiens │1988/89 │NHL │.719 │2.19│ │4. │Montreal Canadiens │1976/77 │NHL │.825 │2.18│ │5. │Detroit Red Wings │1994/95 │NHL │.729 │2.08│ │6. │Dallas Stars │1998/99 │NHL │.695 │2.05│ │7. │Detroit Red Wings │2001/02 │NHL │.707 │2.02│ │8. │Boston Bruins │1929/30 │NHL │.875 │1.99│ │9. │Montreal Canadiens │1977/78 │NHL │.806 │1.97│ │9. │Philadelphia Flyers │1979/80 │NHL │.725 │1.97│ │9. │Edmonton Oilers │1985/86 │NHL │.744 │1.97│ │12. │Quebec Bulldogs │1912/13 │NHA │.800 │1.94│ │13. │Houston Aeros │1976/77 │WHA │.663 │1.93│ │14. │New York Islanders │1981/82 │NHL │.738 │1.92│ │15. │Boston Bruins │1938/39 │NHL │.771 │1.86│ │16. │Boston Bruins │1970/71 │NHL │.776 │1.85│ │17. │Montreal Canadiens │1972/73 │NHL │.769 │1.81│ │18. │Montreal Canadiens │1958/59 │NHL │.650 │1.79│ │18. │Edmonton Oilers │1983/84 │NHL │.744 │1.79│ │20. │Montreal Canadiens │1975/76 │NHL │.794 │1.78│ │21. │Colorado Avalanche │2000/01 │NHL │.720 │1.77│ │22. │Philadelphia Flyers │1984/85 │NHL │.706 │1.73│ │23. │New York Islanders │1978/79 │NHL │.725 │1.72│ │24. │Boston Bruins │1971/72 │NHL │.763 │1.70│ │25. │Ottawa Senators │1926/27 │NHL │.727 │1.69│ │25. │Montreal Canadiens │1955/56 │NHL │.714 │1.69│ │27. │Montreal Canadiens │1978/79 │NHL │.719 │1.67│ │28. │Montreal Canadiens │1957/58 │NHL │.686 │1.65│ │28. │Buffalo Sabres │1979/80 │NHL │.688 │1.65│ │30. │St.Louis Blues │1999/2000 │NHL │.695 │1.62│ │31. │Quebec Nordiques │1994/95 │NHL │.677 │1.61│ │32. │Montreal Canadiens │1915/16 │NHA │.688 │1.58│ │32. │Boston Bruins │1973/74 │NHL │.724 │1.58│ │34. │Ottawa Senators │1916/17 │NHA │.750 │1.57│ │34. │Houston Aeros │1974/75 │WHA │.679 │1.57│ │36. │Montreal Victorias │1897/98 │AHAC │1.000│1.56│ │36. │Edmonton Oilers │1981/82 │NHL │.694 │1.56│ │38. │Montreal Wanderers │1906/07 │ECAHA │1.000│1.55│ │38. │Houston Aeros │1973/74 │WHA │.647 │1.55│ │40. │Ottawa Senators │1910/11 │NHA │.812 │1.54│ │40. │Philadelphia Flyers │1973/74 │NHL │.718 │1.54│ │42. │Montreal Canadiens │1943/44 │NHL │.830 │1.53│ │42. │Edmonton Oilers │1984/85 │NHL │.613 │1.53│ │42. │New York Islanders │1980/81 │NHL │.688 │1.53│ │45. │Edmonton Oilers │1984/85 │NHL │.681 │1.52│ │46. │Montreal Canadiens │1944/45 │NHL │.800 │1.50│ │46. │Montreal Canadiens │1959/60 │NHL │.657 │1.50│ │46. │Montreal Canadiens │1968/69 │NHL │.678 │1.50│ For those interested in this sort of thing, here is the distribution of the top 48 seasons of all time: Montreal Canadiens 14; Boston Bruins and Edmonton Oilers, 5; Detroit Red Wings, Houston Aeros, New York Islanders, Ottawa Senators (first edition) and Philadelphia Flyers, 3; Quebec Nordiques/Colorado Avalanche 2; Buffalo Sabres, Calgary Flames, Dallas Stars, Montreal Victorias, Montreal Wanderers, Quebec Bulldogs, St.Louis Blues 1. Notably, half of the Original Six teams (Rangers, Chicago, and Toronto) fail to take a single spot, while the Habs have 29% of the top 48 to themselves.	{"url":"https://hockeyhistorysis.blogspot.com/","timestamp":"2024-11-06T11:27:04Z","content_type":"application/xhtml+xml","content_length":"142995","record_id":"<urn:uuid:53cf90fb-eadb-473e-a757-e0ff8652fbce>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00042.warc.gz"}
Operators In C Programming Language - Boot Poot Operators in C programming language Operators in C programming language are used to solve an expression, to do mathematical calculations, to reach at some logical decision or to assign a value to a variable etc. C++ provides following types of operators. In C programming, operators are symbols that facilitate various operations on data, allowing manipulation and computation within the code. Arithmetic operators, such as addition, subtraction, multiplication, division, and modulus, perform basic mathematical calculations. Relational operators, including equality (==), inequality (!=), greater than (>), less than (<), greater than or equal to (>=), and less than or equal to (<=), are used for making comparisons between values. Logical operators like AND (&&), OR (\|\|), and NOT (!) are employed in Boolean expressions to control program flow. Additionally, assignment operators (=) assign values to variables, while increment (++) and decrement (--) operators modify variable values. Bitwise operators manipulate individual bits in binary representations of data. Understanding and effectively utilizing these operators are crucial for building robust and efficient C programs. In C programming, operators are symbols that represent computations or operations to be performed on variables or values. These operators help manipulate data and perform various tasks within a program. The main categories of operators in C include: 1. Arithmetic Operators: □ + (Addition) □ - (Subtraction) □ * (Multiplication) □ / (Division) □ % (Modulus) 2. Relational Operators: □ == (Equal to) □ != (Not equal to) □ < (Less than) □ > (Greater than) □ <= (Less than or equal to) □ >= (Greater than or equal to) 3. Logical Operators: □ && (Logical AND) □ \|\| (Logical OR) □ ! (Logical NOT) 4. Assignment Operators: □ = (Assignment) □ += (Add and assign) □ -= (Subtract and assign) □ = (Multiply and assign) □ /= (Divide and assign) □ %= (Modulus and assign) 5. Increment and Decrement Operators: □ ++ (Increment) □ -- (Decrement) 6. Bitwise Operators: □ & (Bitwise AND) □ \| (Bitwise OR) □ ^ (Bitwise XOR) □ ~ (Bitwise NOT) □ << (Left shift) □ >> (Right shift) 7. Conditional (Ternary) Operator: □ ? : (Conditional operator) Understanding and using these operators are fundamental to writing C programs for various tasks and computations. 1. Arithmetic Operators 2. Relational Operators 3. Assignment Operators 4. Logical Operators 5. Unary Operators Arithmetic operators are used for arithmetical calculations. The following table shows the list of arithmetic operators supported by C & C++. Operators Descriptions + Addition – Subtraction Multiplication / Division % Modulus/Remainder Relational operators are also called comparison operators and are used to compare values. The comparison operators, supported by C & C++ are as follows: Operators Descriptions == Equal to != Not Equal < Less than > Greater than <= Less than or equal to >= Greater than or equal to Assignment operators are used to assign new value to a variable. Assignment operators are combined with other operators to evaluate expressions. The assignment operators, supported by C & C++ are as Operators Descriptions += Assigns a new value by adding to current value. -= Assigns a new value by subtracting from current value. *= Assigns a new value by multiplying with current value. /= Assigns a new value by dividing the current value. %= Assign a new value by modules of the current value. Logical operators are used with Boolean terms. This means that they evaluate an expression resulting as ‘true’ or ‘false’. 1 if expression is true or 0 if false. The logical operators supported by C & C++ are as follow: Operators Descriptions && Logical And (both condition should be true) \|\| Logical Or (any one condition should be true) \| Logical Not All the operators used earlier are called Binary operators because they work on two operands. To evaluate unary operators only one operand is required. C & C++ supports two Unary Operators: Operators Descriptions ++ increment the value of operand by 1. — decrement the value of operand by 1. • Prefix/Postfix Unary Operators: The ++/operator can be used using two methods Prefix increment/ decrement operator and Postfix increment/decrement operator. int num =10; ++num: //Prefix Increment num ++2 //Postfix Increment num; //Prefix decrement num; //Postfix decrement to learn the basic concepts of C programming Operators in C programming language are very important to learn and practice. and gain knowledge in Operators in C programming language. Leave a Comment	{"url":"https://bootpoot.tech/operators-in-c-programming-language/","timestamp":"2024-11-12T15:21:22Z","content_type":"text/html","content_length":"98281","record_id":"<urn:uuid:883247f4-c433-4382-8904-4b9009091982>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00424.warc.gz"}
CBSE Class 4 Maths Metric Measures Worksheet Read and download free pdf of CBSE Class 4 Maths Metric Measures Worksheet. Download printable Mathematics Class 4 Worksheets in pdf format, CBSE Class 4 Mathematics Metric Measures Worksheet has been prepared as per the latest syllabus and exam pattern issued by CBSE, NCERT and KVS. Also download free pdf Mathematics Class 4 Assignments and practice them daily to get better marks in tests and exams for Class 4. Free chapter wise worksheets with answers have been designed by Class 4 teachers as per latest examination pattern Metric Measures Mathematics Worksheet for Class 4 Class 4 Mathematics students should refer to the following printable worksheet in Pdf in Class 4. This test paper with questions and solutions for Class 4 Mathematics will be very useful for tests and exams and help you to score better marks Class 4 Mathematics Metric Measures Worksheet Pdf Please click on below link to download CBSE Class 4 Maths Metric Measures Worksheet Metric Measures CBSE Class 4 Mathematics Worksheet The above practice worksheet for Metric Measures has been designed as per the current syllabus for Class 4 Mathematics released by CBSE. Students studying in Class 4 can easily download in Pdf format and practice the questions and answers given in the above practice worksheet for Class 4 Mathematics on a daily basis. All the latest practice worksheets with solutions have been developed for Mathematics by referring to the most important and regularly asked topics that the students should learn and practice to get better scores in their examinations. Studiestoday is the best portal for Printable Worksheets for Class 4 Mathematics students to get all the latest study material free of cost. Teachers of studiestoday have referred to the NCERT book for Class 4 Mathematics to develop the Mathematics Class 4 worksheet. After solving the questions given in the practice sheet which have been developed as per the latest course books also refer to the NCERT solutions for Class 4 Mathematics designed by our teachers. After solving these you should also refer to Class 4 Mathematics MCQ Test for the same chapter. We have also provided a lot of other Worksheets for Class 4 Mathematics which you can use to further make yourself better in Mathematics. Where can I download latest CBSE Practice worksheets for Class 4 Mathematics Metric Measures You can download the CBSE Practice worksheets for Class 4 Mathematics Metric Measures for the latest session from StudiesToday.com Are the Class 4 Mathematics Metric Measures Practice worksheets available for the latest session Yes, the Practice worksheets issued for Metric Measures Class 4 Mathematics have been made available here for the latest academic session Is there any charge for the Practice worksheets for Class 4 Mathematics Metric Measures There is no charge for the Practice worksheets for Class 4 CBSE Mathematics Metric Measures you can download everything free How can I improve my scores by solving questions given in Practice worksheets in Metric Measures Class 4 Mathematics Regular revision of practice worksheets given on studiestoday for Class 4 subject Mathematics Metric Measures can help you to score better marks in exams Are there any websites that offer free Practice test papers for Class 4 Mathematics Metric Measures Yes, studiestoday.com provides all the latest Class 4 Mathematics Metric Measures test practice sheets with answers based on the latest books for the current academic session	{"url":"https://www.studiestoday.com/practice-worksheets-mathematics-cbse-class-4-maths-metric-measures-worksheet-343117.html","timestamp":"2024-11-07T18:36:10Z","content_type":"text/html","content_length":"114003","record_id":"<urn:uuid:f5b8f149-09c5-469f-95ef-1a11d369a282>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00702.warc.gz"}
A person on a swing takes 2.0 seconds to travel from her highest backward point to her highest forward point. What is the frequency of her swinging? \| Socratic A person on a swing takes 2.0 seconds to travel from her highest backward point to her highest forward point. What is the frequency of her swinging? A person on a swing takes 2.0 seconds to travel from her highest backward point to her highest forward point. What is the frequency of her swinging? 1 Answer The person takes 2 seconds to cover half of one full oscillation, so the period is 4 sec. As $f = \frac{1}{T}$ this means the frequency is $= \frac{1}{4} = 0.25$ Hertz Impact of this question 1013 views around the world	{"url":"https://api-project-1022638073839.appspot.com/questions/a-person-on-a-swing-takes-2-0-seconds-to-travel-from-her-highest-backward-point-#526496","timestamp":"2024-11-04T04:02:48Z","content_type":"text/html","content_length":"32306","record_id":"<urn:uuid:73c4e1b5-279e-4043-be3f-f7bd4009413d>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00861.warc.gz"}
COCI 2006/2007, Contest #6 Task V Zvonko is playing with digits again, even though his mother has warned him that he is doing too much math and should go outside to play with his friends. In his latest game, Zvonko looks for multiples of an integer X, composed only of certain digits. A multiple of X is any number divisible by X. In order to ruin Zvonko's fun, his mother decided to get a program that solves the problem. Write a program that calculates how many multiples of X are between A and B (inclusive), such that, when written in decimal, they contain only certain allowed digits. The first line of input contains three integers X, A and B (1 ≤ X < 10^11, 1 ≤ A ≤ B < 10^11). The second line contains the allowed digits. The digits will be given with no spaces, sorted in increasing order and without duplicates. Output the number of multiples Zvonko can make on a single line. Sample Tests Input Input Input Output Output Output All Submissions Best Solutions Point Value: 20 (partial) Time Limit: 1.00s Memory Limit: 32M Added: Aug 13, 2013 Languages Allowed: C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3	{"url":"https://wcipeg.com/problem/coci066p5","timestamp":"2024-11-02T12:09:46Z","content_type":"text/html","content_length":"10523","record_id":"<urn:uuid:ac0f204f-c856-4b0d-817c-67c42c5f7e28>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00472.warc.gz"}
Median from a frequency table Median from a frequency table is when we find the median average from a data set which has been organised into a frequency table. The median is the middle value and is a measure of central tendency, it is a value that can be used to represent a set of data. We can calculate the median by: a) Writing down each of the values in the table and crossing off values to find the middle value. The frequency table shows the number of cars owned by 13 families: We can use the table to write down each of the values The middle number is 1 , so the median number of cars is 1 . b) Using the median formula and cumulative frequency If the data set is large we can us the formula to work out the position of the median \text{Position of the median}=(\frac{n+1}{2})^\text{th} , and cumulative frequency to find the actual median Cumulative frequency is where we add up the numbers in the frequency column as we go down the table. The total number of values is 13 , so n=13 . So the, \text{Position of the median}=(\frac{n+1}{2})^\text{th}=(\frac{13+1}{2})^\text{th}=7^\text{th} The 7^\text{th} value is in the row of the table that contains the 5^\text{th} , 6^\text{th} , 7^\text{th} , 8^\text{th} , and 9^\text{th} values. So the median number of cars = 1 . What is median from a frequency table? 0 to 9 10 to 19 20 to 29 30 to 39 20 – 24 15 – 19 25 – 29 30 – 34 60\le x <80 80\le x <100 40\le x <60 20\le x <40	{"url":"https://thirdspacelearning.com/gcse-maths/statistics/median-from-a-frequency-table/","timestamp":"2024-11-07T03:01:54Z","content_type":"text/html","content_length":"273533","record_id":"<urn:uuid:286ad75e-b4a0-4b78-9a3e-cd2473947fa2>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00342.warc.gz"}
Coursnap app Big O Notations Get the Code Here: http://goo.gl/Y3UTH MY UDEMY COURSES ARE 87.5% OFF TIL December 19th ($9.99) ONE IS FREE ➡️ Python Data Science Series for $9.99 : Highest Rated & Largest Python Udemy Course + 56 Hrs + 200 Videos + Data Science https://bit.ly/Master_Python_41 ➡️ C++ Programming Bootcamp Series for $9.99 : Over 23 Hrs + 53 Videos + Quizzes + Graded Assignments + New Videos Every Month https:// bit.ly/C_Course_41 ➡️ FREE 15 hour Golang Course!!! : https://bit.ly/go-tutorial3 Welcome to my Big O Notations tutorial. Big O notations are used to measure how well a computer algorithm scales as the amount of data involved increases. It isn't however always a measure of speed as you'll see. This is a rough overview of Big O and I hope to simplify it rather than get into all of the complexity. I'll specifically cover the following O(1), O(N), O(N^2), O(log N) and O(N log N). Between the video and code below I hope everything is completely understandable. MY UDEMY COURSES ARE 87.5% OFF TILL MAY 20th ►► New C++ Programming Bootcamp Series for $9.99 : http://bit.ly/C_Course Over 20 Hrs + 52 Videos + Quizzes + Graded Assignments + New Videos Every Month ►► Python Programming Bootcamp Series for $9.99 : http://bit.ly/Master_Python Highest Rated Python Udemy Course + 48 Hrs + 199 Videos + Data Science {'title': 'Big O Notations', 'heatmap': [{'end': 542.086, 'start': 528.382, 'weight': 0.765}, {'end': 722.005, 'start': 686.547, 'weight': 0.884}, {'end': 989.514, 'start': 972.141, 'weight': 0.71}], 'summary': 'Tutorial on big o notations covers understanding the concept, basics, and explanation of o(1) and o(n) notations with code examples. it also discusses algorithm performance, efficiency of binary search, and quicksort implementation, providing a comprehensive overview of big o notation and its practical applications.', 'chapters': [{'end': 57.875, 'segs': [{'end': 57.875, 'src': 'embed', 'start': 15.317, 'weight': 0, 'content': [{'end': 24.901, 'text': 'Just to cut to the chase, Big O notation is a way to measure how well a computer algorithm scales as the amount of data involved increases.', 'start': 15.317, 'duration': 9.584}, {'end': 32.764, 'text': 'So how well would it work in, say, if it was using a 10 element array versus a 10,000 element array?', 'start': 25.221, 'duration': 7.543}, {'end': 40.207, 'text': "As you're going to see here in a second, it's not always a measure of speed, but instead a measure of how well an algorithm scales.", 'start': 33.004, 'duration': 7.203}, {'end': 42.768, 'text': 'And this is going to be a rough overview of Big O,', 'start': 40.487, 'duration': 2.281}, {'end': 48.831, 'text': "and I'm not going to cover topics such as asymptotic analysis and other things that have to do with discrete mathematics.", 'start': 42.768, 'duration': 6.063}, {'end': 52.232, 'text': "I'm going to instead focus on the simple idea.", 'start': 49.051, 'duration': 3.181}, {'end': 53.833, 'text': "So I got a lot to do, so let's get into it.", 'start': 52.392, 'duration': 1.441}, {'end': 57.875, 'text': "Okay, so I'm doing a lot of this out of my head here, so bear with me.", 'start': 54.873, 'duration': 3.002}], 'summary': 'Big o notation measures algorithm scaling with data.', 'duration': 42.558, 'max_score': 15.317, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU15317.jpg'}], 'start': 0.129, 'title': 'Understanding big o notation', 'summary': 'Explains how big o notation measures the scalability of computer algorithms as the amount of data increases, focusing on the simple idea and not covering topics such as asymptotic analysis, providing a rough overview of big o.', 'chapters': [{'end': 57.875, 'start': 0.129, 'title': 'Understanding big o notation', 'summary': 'Explains how big o notation measures the scalability of computer algorithms as the amount of data increases, focusing on the simple idea and not covering topics such as asymptotic analysis, providing a rough overview of big o.', 'duration': 57.746, 'highlights': ['The chapter provides a simple explanation of how Big O notation measures the scalability of computer algorithms, focusing on how well an algorithm scales as the amount of data involved increases.', 'It emphasizes that Big O notation is not always a measure of speed, but rather a measure of scalability, exemplifying the comparison between using a 10 element array versus a 10,000 element array.', 'The tutorial aims to give a rough overview of Big O, excluding topics such as asymptotic analysis and discrete mathematics, and instead focusing on the fundamental concept.']}], 'duration': 57.746, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU129.jpg', 'highlights': ['The chapter provides a simple explanation of how Big O notation measures the scalability of computer algorithms, focusing on how well an algorithm scales as the amount of data involved increases.', 'It emphasizes that Big O notation is not always a measure of speed, but rather a measure of scalability, exemplifying the comparison between using a 10 element array versus a 10,000 element array.', 'The tutorial aims to give a rough overview of Big O, excluding topics such as asymptotic analysis and discrete mathematics, and instead focusing on the fundamental concept.']}, {'end': 219.89, 'segs': [{'end': 94.575, 'src': 'embed', 'start': 74.562, 'weight': 0, 'content': [{'end': 85.789, 'text': 'and the reason why this is going to be very important is I want to define with big O notation the part of the algorithm that has the greatest effect ultimately on the final answer.', 'start': 74.562, 'duration': 11.227}, {'end': 94.575, 'text': 'So now in this situation if n just goes up to being equal to 2, you could see very quickly that your answer is going to go from 84 to 459.', 'start': 85.929, 'duration': 8.646}], 'summary': 'Using big o notation to analyze algorithm, n=2 leads to answer increase from 84 to 459.', 'duration': 20.013, 'max_score': 74.562, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU74562.jpg'}, {'end': 158.617, 'src': 'embed', 'start': 119.269, 'weight': 2, 'content': [{'end': 125.952, 'text': "Which, as you can see, now you're starting to question, well definitely 19 doesn't really have that much bearing in regards to the final answer.", 'start': 119.269, 'duration': 6.683}, {'end': 133.037, 'text': 'And also when you think about it, n squared has very little to do with this answer, if, in fact,', 'start': 126.193, 'duration': 6.844}, {'end': 137.14, 'text': '45n cubed is going to be equal to 45,000 in this situation.', 'start': 133.037, 'duration': 4.103}, {'end': 140.023, 'text': "So, if you're going to be dealing with very, very large numbers,", 'start': 137.341, 'duration': 2.682}, {'end': 149.29, 'text': 'you very quickly see that the part of this algorithm that really has a lot to do with the final answer, as this data set scales,', 'start': 140.023, 'duration': 9.267}, {'end': 153.893, 'text': "is not even going to be the 45, but it's going to be the n cubed.", 'start': 149.29, 'duration': 4.603}, {'end': 158.617, 'text': 'And hence, we would say that this has an order of n cubed.', 'start': 154.174, 'duration': 4.443}], 'summary': 'Algorithm has an order of n cubed, with 45n cubed equal to 45,000.', 'duration': 39.348, 'max_score': 119.269, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU119269.jpg'}, {'end': 202.935, 'src': 'embed', 'start': 179.103, 'weight': 4, 'content': [{'end': 185.625, 'text': 'Order of n squared provide examples what that code looks like as well as n cubed and all these other ones.', 'start': 179.103, 'duration': 6.522}, {'end': 192.429, 'text': 'Also going to get into log n and order of n log n.', 'start': 185.985, 'duration': 6.444}, {'end': 194.83, 'text': "Okay, so that's what's going to be covered in this tutorial.", 'start': 192.429, 'duration': 2.401}, {'end': 199.113, 'text': "So I got to get a couple pieces here because we're going to be playing around with big giant arrays.", 'start': 194.97, 'duration': 4.143}, {'end': 202.935, 'text': "So I'm just going to create myself an integer array, call it the array.", 'start': 199.253, 'duration': 3.682}], 'summary': 'Tutorial covers examples of n squared, n cubed, log n, and n log n in code.', 'duration': 23.832, 'max_score': 179.103, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU /pics/V6mKVRU1evU179103.jpg'}], 'start': 58.035, 'title': 'Big o notation basics', 'summary': 'Explains the concept of big o notation using a simple algorithm (45n^3 + 20n^2 + 19), ultimately revealing the order of n^3. it also introduces the coverage of order of 1, order of n, order of n squared, n cubed, log n, and order of n log n in the tutorial.', 'chapters': [{'end': 219.89, 'start': 58.035, 'title': 'Big o notation basics', 'summary': 'Explains the concept of big o notation by using a simple algorithm (45n^3 + 20n^2 + 19) and demonstrates how the part of the algorithm with the greatest effect on the final answer is determined as n scales, ultimately revealing the order of n^3. it also introduces the coverage of order of 1, order of n, order of n squared, n cubed, log n, and order of n log n in the tutorial.', 'duration': 161.855, 'highlights': ['The algorithm (45n^3 + 20n^2 + 19) is used to demonstrate the concept of Big O notation, where the part of the algorithm with the greatest effect on the final answer is determined as n scales. 45n^3 + 20n^2 + 19', "As n increases from 1 to 2, the final answer jumps from 84 to 459, indicating the impact of scaling n on the algorithm's performance. 1 to 2, 84 to 459", 'With n increasing to 10, the final answer rises to 47 and 19, demonstrating the diminishing effect of the constant term and the lesser influence of n^2 on the final answer as compared to n^3. n increasing to 10, 47 and 19', "The scalability of the dataset reveals that as the numbers become very large, the part of the algorithm with the most impact on the final answer is n^3, leading to the determination of the algorithm's order as n^3. scenarios with very large numbers, n^3", 'Introduction and coverage of order of 1, order of n, order of n squared, n cubed, log n, and order of n log n are set to be covered in the tutorial. order of 1, order of n, order of n squared, n cubed, log n, order of n log n']}], 'duration': 161.855, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU58035.jpg', 'highlights': ['The algorithm (45n^3 + 20n^2 + 19) demonstrates Big O notation, determining the part with the greatest effect as n scales.', 'As n increases from 1 to 2, the final answer jumps from 84 to 459, indicating the impact of scaling n on performance.', 'With n increasing to 10, the final answer rises to 47 and 19, demonstrating the diminishing effect of the constant term and the lesser influence of n^2 compared to n^3.', "The scalability of the dataset reveals n^3 as the part with the most impact on the final answer, determining the algorithm's order as n^3.", 'Introduction and coverage of order of 1, order of n, order of n squared, n cubed, log n, and order of n log n are set to be covered in the tutorial.']}, {'end': 377.868, 'segs': [{'end': 273.745, 'src': 'embed', 'start': 242.797, 'weight': 2, 'content': [{'end': 245.098, 'text': "And I'm going to put this in here as a comment.", 'start': 242.797, 'duration': 2.301}, {'end': 252.26, 'text': "And what this algorithm does, or what this notation means, is it's going to be an algorithm that's going to execute in the same amount of time,", 'start': 245.218, 'duration': 7.042}, {'end': 254.281, 'text': 'regardless of the amount of data.', 'start': 252.26, 'duration': 2.021}, {'end': 260.702, 'text': "Or to put it another word, it's going to be code that executes in the same amount of time no matter how big the array is.", 'start': 254.501, 'duration': 6.201}, {'end': 262.683, 'text': 'So what exactly would that look like?', 'start': 261.043, 'duration': 1.64}, {'end': 273.745, 'text': 'Well, one example of that in the context of working with arrays would be if we wanted to add an item to an array and an integer was passed over and we said okay,', 'start': 262.883, 'duration': 10.862}], 'summary': 'Algorithm ensures constant execution time regardless of data size.', 'duration': 30.948, 'max_score': 242.797, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU242797.jpg'}, {'end': 351.582, 'src': 'embed', 'start': 324.822, 'weight': 0, 'content': [{'end': 331.046, 'text': 'And in the situation in which we just wanted to find one match, Big O notation is the same,', 'start': 324.822, 'duration': 6.224}, {'end': 337.011, 'text': "because what it's going to do is describe the worst case scenario, in which the whole array must be searched.", 'start': 331.046, 'duration': 5.965}, {'end': 341.874, 'text': "So let's say we're looking for an item that's in a 100,000 item array and it doesn't exist.", 'start': 337.251, 'duration': 4.623}, {'end': 347.338, 'text': 'but we want to make sure that we handle that example, which would mean that every single item would have to be searched.', 'start': 341.874, 'duration': 5.464}, {'end': 351.582, 'text': "So let's just come in here and let's do this with linear search.", 'start': 347.698, 'duration': 3.884}], 'summary': 'Big o notation describes worst case scenario, like searching a 100,000 item array with linear search.', 'duration': 26.76, 'max_score': 324.822, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU324822.jpg'}], 'start': 219.89, 'title': 'Big o notation explained', 'summary': 'Explains o(1) and o(n) notations using code examples to demonstrate algorithms executing in constant time and growing in direct proportion to data size, with provided code available in the video link.', 'chapters': [{'end': 377.868, 'start': 219.89, 'title': 'Big o notation explained', 'summary': 'Explains the concepts of o(1) and o(n) notations using code examples to demonstrate algorithms that execute in constant time and grow in direct proportion to the data size, with the provided code available in the video link.', 'duration': 157.978, 'highlights': ['The concept of O(1) notation is explained using an example of adding an item to an array, showcasing code that executes in the same amount of time regardless of the array size. example of adding an item to an array', 'The chapter delves into the concept of O(n) notation by demonstrating a linear search algorithm, highlighting how its time to complete grows in direct proportion to the amount of data. demonstrating a linear search algorithm', "The worst-case scenario is described using Big O notation, emphasizing the need to search the entire array when looking for an item, even in a 100,000 item array where the item doesn't exist. searching the entire array in a worst-case scenario"]}], 'duration': 157.978, 'thumbnail': 'https:// coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU219890.jpg', 'highlights': ["The worst-case scenario is described using Big O notation, emphasizing the need to search the entire array when looking for an item, even in a 100,000 item array where the item doesn't exist.", 'The chapter delves into the concept of O(n) notation by demonstrating a linear search algorithm, highlighting how its time to complete grows in direct proportion to the amount of data.', 'The concept of O(1) notation is explained using an example of adding an item to an array, showcasing code that executes in the same amount of time regardless of the array size.']}, {'end': 849.915, 'segs': [{'end': 450.228, 'src': 'embed', 'start': 418.614, 'weight': 0, 'content': [{'end': 422.976, 'text': 'and then if we wanted to say print something out on the screen, we could say something like value,', 'start': 418.614, 'duration': 4.362}, {'end': 430.039, 'text': 'found value in array and then throw end time into this to calculate whenever this guy ended execution.', 'start': 422.976, 'duration': 7.063}, {'end': 434.461, 'text': 'And then we could say something like linear search took end time, minus start time.', 'start': 430.339, 'duration': 4.122}, {'end': 439.023, 'text': "And then let's bounce back up into the main function here and do some tests.", 'start': 434.661, 'duration': 4.362}, {'end': 442.064, 'text': 'So we go to notation, test algo.', 'start': 439.363, 'duration': 2.701}, {'end': 450.228, 'text': "I'm going to just say 2, big O notation, and let's say I set this for, I'm going to set it for 100,000 as the size of my array.", 'start': 442.483, 'duration': 7.745}], 'summary': 'The transcript discusses implementing algorithms and testing them, with an example of setting the array size to 100,000.', 'duration': 31.614, 'max_score': 418.614, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ V6mKVRU1evU/pics/V6mKVRU1evU418614.jpg'}, {'end': 492.588, 'src': 'embed', 'start': 462.077, 'weight': 2, 'content': [{'end': 465.179, 'text': "but I didn't do it for the constructor, which isn't a big deal.", 'start': 462.077, 'duration': 3.102}, {'end': 467.8, 'text': "I'm just going to say int size of the array.", 'start': 465.299, 'duration': 2.501}, {'end': 470.26, 'text': 'Array size is going to be whatever was passed into it.', 'start': 467.82, 'duration': 2.44}, {'end': 471.041, 'text': 'And there we are.', 'start': 470.46, 'duration': 0.581}, {'end': 472.281, 'text': 'We just created our new array.', 'start': 471.081, 'duration': 1.2}, {'end': 478.663, 'text': "Go back up into main and let's create a couple more of these arrays so that we can compare them based off of size differences.", 'start': 472.481, 'duration': 6.182}, {'end': 480.523, 'text': 'So test algo 3.', 'start': 478.723, 'duration': 1.8}, {'end': 481.864, 'text': "And let's change this to 200,000.", 'start': 480.523, 'duration': 1.341}, {'end': 485.845, 'text': "And let's create two more of them just to experiment with.", 'start': 481.864, 'duration': 3.981}, {'end': 487.365, 'text': 'And change this to 4.', 'start': 486.005, 'duration': 1.36}, {'end': 492.588, 'text': 'this to 5, change this to 4, 5, and have this be 3, and this be 400,000.', 'start': 487.365, 'duration': 5.223}], 'summary': 'Creating and comparing arrays of different sizes for experimentation.', 'duration': 30.511, 'max_score': 462.077, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/ video-capture/V6mKVRU1evU/pics/V6mKVRU1evU462077.jpg'}, {'end': 542.086, 'src': 'heatmap', 'start': 499.432, 'weight': 1, 'content': [{'end': 506.396, 'text': "in essence, is going to be to show that they're going to scale, as the number of elements that we're going to be using here are going to scale.", 'start': 499.432, 'duration': 6.964}, {'end': 508.076, 'text': 'which makes 100% sense.', 'start': 506.656, 'duration': 1.42}, {'end': 510.697, 'text': "Actually, I don't even think I need to use all those.", 'start': 508.577, 'duration': 2.12}, {'end': 512.758, 'text': "And let's file save as long as I did it right.", 'start': 510.977, 'duration': 1.781}, {'end': 520.28, 'text': 'And you can see right there with this linear search, this took roughly 4 milliseconds, this took 5 milliseconds, and this took 18 milliseconds.', 'start': 512.938, 'duration': 7.342}, {'end': 525.361, 'text': 'So you can see, as the number of elements scale get bigger the number of elements we have to deal with.', 'start': 520.52, 'duration': 4.841}, {'end': 527.982, 'text': 'that is in direct relation to the number of elements.', 'start': 525.361, 'duration': 2.621}, {'end': 531.582, 'text': 'And that is why it is known as order of n.', 'start': 528.382, 'duration': 3.2}, {'end': 533.283, 'text': "So there's an example of order of n.", 'start': 531.582, 'duration': 1.701}, {'end': 542.086, 'text': "So now let's take a look at what does order of n squared mean or cubed or any of these other different things we could put inside of here.", 'start': 534.542, 'duration': 7.544}], 'summary': 'Demonstrating scaling with linear search, 4ms-18ms, and order of n.', 'duration': 33.851, 'max_score': 499.432, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/ video-capture/V6mKVRU1evU/pics/V6mKVRU1evU499432.jpg'}, {'end': 722.005, 'src': 'heatmap', 'start': 672.731, 'weight': 4, 'content': [{'end': 677.517, 'text': 'And there you can see it took 362 milliseconds and it took 1612 milliseconds.', 'start': 672.731, 'duration': 4.786}, {'end': 679.899, 'text': 'And you can see that it just continues going on here.', 'start': 677.837, 'duration': 2.062}, {'end': 686.347, 'text': 'And how dramatically slower the bubble sort gets depending upon the amount of data.', 'start': 680.12, 'duration': 6.227}, {'end': 690.992, 'text': 'And that is why order of n squared is very bad and to be avoided.', 'start': 686.547, 'duration': 4.445}, {'end': 697.674, 'text': "So now let's show you an example of another algorithm that is much more efficient, and that's going to be our binary search.", 'start': 691.332, 'duration': 6.342}, {'end': 703.315, 'text': "And here we're going to focus on order of log n as our big O notation.", 'start': 698.094, 'duration': 5.221}, {'end': 709.957, 'text': 'And this is going to occur when data being used is decreased roughly by 50% each time through the algorithm.', 'start': 703.535, 'duration': 6.422}, {'end': 713.338, 'text': 'And the binary search, like we saw before, is a perfect example of this.', 'start': 710.117, 'duration': 3.221}, {'end': 722.005, 'text': "And it's pretty fast, because as log n increases, or n specifically increases,", 'start': 713.778, 'duration': 8.227}], 'summary': 'Bubble sort is slow, with o(n^2), while binary search is efficient with o(log n).', 'duration': 40.607, 'max_score': 672.731, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU672731.jpg'}], 'start': 378.048, 'title': 'Understanding big o notation and algorithm performance', 'summary': "Discusses the implementation of linear search algorithm and the testing of its execution time with varying array sizes, demonstrating the scaling impact on performance. it also explains the concepts of 'order of n', 'order of n squared', and 'order of log n' using examples such as linear search, bubble sort, and binary search, highlighting their time complexity and efficiency.", 'chapters': [{'end': 512.758, 'start': 378.048, 'title': 'Understanding big o notation and algorithm performance', 'summary': 'Discusses the implementation of linear search algorithm and the testing of its execution time with varying array sizes, demonstrating the scaling impact on performance.', 'duration': 134.71, 'highlights': ['The chapter discusses the implementation of linear search algorithm and the testing of its execution time with varying array sizes The transcript describes the process of implementing a linear search algorithm and conducting tests with arrays of different sizes to evaluate the impact on performance.', 'The chapter emphasizes the scaling impact on performance as the number of elements in the array increases It is highlighted that the linear searches demonstrate scaling impact, showing that the performance scales with the number of elements in the array.', 'The testing involves varying array sizes, including 100,000, 200,000, and 400,000 elements The chapter includes tests with array sizes of 100,000, 200,000, and 400,000, demonstrating the impact of different array sizes on the performance of the linear search algorithm.']}, {'end': 849.915, 'start': 512.938, 'title': 'Big o notations explained', 'summary': "Explains the concepts of 'order of n', 'order of n squared', and 'order of log n' using examples such as linear search, bubble sort, and binary search, highlighting their time complexity and efficiency.", 'duration': 336.977, 'highlights': ['The linear search took roughly 18 milliseconds as the number of elements increased, demonstrating the order of n complexity. Linear search demonstrated the order of n complexity, with the time taken increasing as the number of elements scaled up.', "The bubble sort took 1612 milliseconds for 20,000 items, illustrating the dramatic decrease in performance with the increase in the number of items, representing the order of n squared complexity. Bubble sort's time complexity was demonstrated as order of n squared with a dramatic decrease in performance as the number of items increased, taking 1612 milliseconds for 20,000 items.", 'The binary search, with order of log n complexity, showed little or no effect in speed even with a dramatic increase in the data set, due to the halving of data at each iteration. Binary search, with order of log n complexity, exhibited little or no effect in speed even with a dramatic increase in the data set, attributed to the halving of data at each iteration.']}], 'duration': 471.867, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU378048.jpg', 'highlights': ['The chapter discusses the implementation of linear search algorithm and the testing of its execution time with varying array sizes', 'The chapter emphasizes the scaling impact on performance as the number of elements in the array increases', 'The testing involves varying array sizes, including 100,000, 200,000, and 400,000 elements', 'The linear search took roughly 18 milliseconds as the number of elements increased, demonstrating the order of n complexity', 'The bubble sort took 1612 milliseconds for 20,000 items, illustrating the dramatic decrease in performance with the increase in the number of items, representing the order of n squared complexity', 'The binary search, with order of log n complexity, showed little or no effect in speed even with a dramatic increase in the data set, due to the halving of data at each iteration']}, {'end': 1229.631, 'segs': [{'end': 913.911, 'src': 'embed', 'start': 876.531, 'weight': 1, 'content': [{'end': 890.634, 'text': "This took 721 milliseconds and now you can start to see why measuring the time doesn't really matter with efficient algorithms and why big O notation tells you a lot more about an algorithm than how long or milliseconds it would take to execute.", 'start': 876.531, 'duration': 14.103}, {'end': 892.655, 'text': 'And you can also see the efficiency.', 'start': 890.854, 'duration': 1.801}, {'end': 900.461, 'text': 'it only went through it nine times to search 10,000 items and it only went through it 10 times to search through 20,000 items.', 'start': 892.655, 'duration': 7.806}, {'end': 903.503, 'text': 'So this is pretty much the picture of efficiency.', 'start': 900.761, 'duration': 2.742}, {'end': 910.649, 'text': 'And I executed again, just so you could see that that was meant to be binary search took zero, not bubble sort, say binary search took zero.', 'start': 903.644, 'duration': 7.005}, {'end': 912.67, 'text': 'So that is the picture of efficiency.', 'start': 910.829, 'duration': 1.841}, {'end': 913.911, 'text': 'And I hope that makes sense.', 'start': 912.95, 'duration': 0.961}], 'summary': 'Efficient algorithm took 721 milliseconds, went through 10,000 items 9 times, and through 20,000 items 10 times, demonstrating the effectiveness of big o notation.', 'duration': 37.38, 'max_score': 876.531, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/ V6mKVRU1evU876531.jpg'}, {'end': 992.056, 'src': 'heatmap', 'start': 955.255, 'weight': 3, 'content': [{'end': 958.176, 'text': 'And we already know that the quick sort is much more efficient.', 'start': 955.255, 'duration': 2.921}, {'end': 961.137, 'text': 'But the main answer is going to be why is it so efficient??', 'start': 958.276, 'duration': 2.861}, {'end': 965.999, 'text': 'Now, to figure out the number of comparisons that we need to make with the quicksort,', 'start': 961.297, 'duration': 4.702}, {'end': 972.141, 'text': 'we first need to remember that it is comparing and moving values very efficiently, without shifting,', 'start': 965.999, 'duration': 6.142}, {'end': 975.303, 'text': "unlike some of the other sorting algorithms we've used in the past.", 'start': 972.141, 'duration': 3.162}, {'end': 979.805, 'text': 'And that means that values are only going to be compared once.', 'start': 975.523, 'duration': 4.282}, {'end': 983.208, 'text': "They're not going to be compared to each other over and over and over again.", 'start': 980.145, 'duration': 3.063}, {'end': 989.514, 'text': 'So in essence, each comparison will reduce the possible final sorted lists in half.', 'start': 983.428, 'duration': 6.086}, {'end': 992.056, 'text': 'Or to put it in a completely other way,', 'start': 989.854, 'duration': 2.202}], 'summary': 'Quick sort is efficient due to minimal value comparisons and reduced final sorted lists by half.', 'duration': 36.801, 'max_score': 955.255, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ V6mKVRU1evU/pics/V6mKVRU1evU955255.jpg'}, {'end': 1218.038, 'src': 'embed', 'start': 1187.357, 'weight': 0, 'content': [{'end': 1189.238, 'text': 'Throw in these timing mechanisms here.', 'start': 1187.357, 'duration': 1.881}, {'end': 1193.4, 'text': "Just to show you how efficient it is, I'm going to chuck this up to 100,000 and chuck this up to 200,000.", 'start': 1189.538, 'duration': 3.862}, {'end': 1196.982, 'text': "Test I'll go to here.", 'start': 1193.4, 'duration': 3.582}, {'end': 1198.303, 'text': "We're gonna do the quicksort.", 'start': 1197.202, 'duration': 1.101}, {'end': 1199.344, 'text': 'Where is that quicksort?', 'start': 1198.483, 'duration': 0.861}, {'end': 1202.386, 'text': 'There it is set into zero test.', 'start': 1200.064, 'duration': 2.322}, {'end': 1210.773, 'text': "I'll go to pass in the number of items in the array and execute and you could see it cycled through a hundred thousand items in 41 milliseconds.", 'start': 1202.386, 'duration': 8.387}, {'end': 1214.936, 'text': "Let's take this up to four, just by changing this to four, I'll say, of execute,", 'start': 1210.773, 'duration': 4.163}, {'end': 1218.038, 'text': 'and only took 44 milliseconds to jump up to 400,000 or 300,000 in this situation.', 'start': 1214.936, 'duration': 3.102}], 'summary': 'Efficient quicksort algorithm processed 100,000 items in 41 ms, 400,000 in 44 ms.', 'duration': 30.681, 'max_score': 1187.357, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU1187357.jpg'}], 'start': 850.136, 'title': 'Efficiency of binary search and big o notation', 'summary': 'Discusses the efficiency of binary search, demonstrating its speed in searching 10,000 and 20,000 items, alongside explaining the efficiency of quicksort with a quicksort implementation handling 100,000 items in 41 milliseconds.', 'chapters': [{'end': 913.911, 'start': 850.136, 'title': 'Efficiency of binary search', 'summary': 'Discusses the efficiency of binary search, highlighting that it took 721 milliseconds to search 10,000 items and 162 milliseconds to search 20,000 items, showcasing the significant improvement in efficiency and the effectiveness of big o notation in understanding algorithm efficiency.', 'duration': 63.775, 'highlights': ['Binary search took 721 milliseconds to search 10,000 items and 162 milliseconds to search 20,000 items, demonstrating a significant improvement in efficiency.', 'Big O notation provides more insights into algorithm efficiency than measuring time in milliseconds, indicating the importance of understanding algorithm complexity.', 'The search only went through it nine times to search 10,000 items and 10 times to search through 20,000 items, emphasizing the efficiency of the binary search algorithm.']}, {'end': 1229.631, 'start': 914.191, 'title': 'Understanding big o notation', 'summary': 'Explains the efficiency of quicksort, highlighting that its number of comparisons is n log n, unlike inefficient sorting algorithms like bubble sort, and demonstrates its efficiency through a quicksort implementation handling 100,000 items in 41 milliseconds.', 'duration': 315.44, 'highlights': ['The number of comparisons in quicksort is n log n, making it more efficient than bubble sort with n squared comparisons.', 'A demonstration showed quicksort handling 100,000 items in 41 milliseconds, showcasing its efficiency.', 'The explanation of the quicksort algorithm and its partitioning process provides insights into its efficiency and effectiveness.']}], 'duration': 379.495, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/V6mKVRU1evU/pics/V6mKVRU1evU850136.jpg', 'highlights': ['Quicksort handled 100,000 items in 41 milliseconds, demonstrating its efficiency.', 'Binary search took 721 milliseconds to search 10,000 items and 162 milliseconds to search 20,000 items, showing significant improvement in efficiency.', 'Big O notation provides more insights into algorithm efficiency than measuring time in milliseconds, indicating the importance of understanding algorithm complexity.', 'The number of comparisons in quicksort is n log n, making it more efficient than bubble sort with n squared comparisons.', 'The search only went through it nine times to search 10,000 items and 10 times to search through 20,000 items, emphasizing the efficiency of the binary search algorithm.', 'The explanation of the quicksort algorithm and its partitioning process provides insights into its efficiency and effectiveness.']}], 'highlights': ['The tutorial aims to give a rough overview of Big O, excluding topics such as asymptotic analysis and discrete mathematics, and instead focusing on the fundamental concept.', 'The algorithm (45n^3 + 20n^2 + 19) demonstrates Big O notation, determining the part with the greatest effect as n scales.', "The scalability of the dataset reveals n^3 as the part with the most impact on the final answer, determining the algorithm's order as n^3.", "The worst-case scenario is described using Big O notation, emphasizing the need to search the entire array when looking for an item, even in a 100,000 item array where the item doesn't exist.", 'The concept of O(1) notation is explained using an example of adding an item to an array, showcasing code that executes in the same amount of time regardless of the array size.', 'The linear search took roughly 18 milliseconds as the number of elements increased, demonstrating the order of n complexity.', 'The bubble sort took 1612 milliseconds for 20,000 items, illustrating the dramatic decrease in performance with the increase in the number of items, representing the order of n squared complexity.', 'The binary search, with order of log n complexity, showed little or no effect in speed even with a dramatic increase in the data set, due to the halving of data at each iteration.', 'Quicksort handled 100,000 items in 41 milliseconds, demonstrating its efficiency.', 'Binary search took 721 milliseconds to search 10,000 items and 162 milliseconds to search 20,000 items, showing significant improvement in efficiency.', 'Big O notation provides more insights into algorithm efficiency than measuring time in milliseconds, indicating the importance of understanding algorithm complexity.', 'The number of comparisons in quicksort is n log n, making it more efficient than bubble sort with n squared comparisons.', 'The search only went through it nine times to search 10,000 items and 10 times to search through 20,000 items, emphasizing the efficiency of the binary search algorithm.', 'The explanation of the quicksort algorithm and its partitioning process provides insights into its efficiency and effectiveness.']}	{"url":"https://learn.coursnap.app/staticpage/V6mKVRU1evU.html","timestamp":"2024-11-05T09:25:22Z","content_type":"text/html","content_length":"37267","record_id":"<urn:uuid:ce6ecbb1-1148-4f20-8106-4ec0fbfc2c00>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00666.warc.gz"}
Quasi-Trefftz DG for the wave equation We consider the wave operator \[\begin{align} \begin{split} (\square_G f)(\mathbf{x},t):= \Delta f(\mathbf{x},t)-G(\mathbf{x})\partial_t^2 f(\mathbf{x},t). \end{split} \end{align}\] with smooth coefficient $G(\mathbf{x})$. Constructing a basis for a traditional Trefftz space (i.e. a space of functions with $\square_G f=0$) is not possible. The crucial idea is that we want to relax the Trefftz porperty to \[\square_G f=\mathcal{O}(\\|(\mathbf{x},t)-(\mathbf{x}_K,t_K)\\|^q),\] with respect to the center of a mesh element $K$ and up to some $q$. This leads to the definition of a new quasi-Trefftz space: For an element $K$ in a space-time mesh let \[\begin{align} \begin{split} \mathbb{T}^p(K):=\big\{ f\in\mathbb{P}^p(K) \ \mid&\ D^{i}\square_G f(\mathbf{x}_K,t_K)=0,\\ &\forall i\in \mathbb{N}^{n+1}_0, \|i\|<p-1 \big\}, \end{split} \end{align} For this space we are able to construct a basis. We then introduce a space-time DG method with test and trial functions that are locally quasi-Trefftz. The example below shows an acoustic wave propagating through a material with $G(x,y)=y+1$ and homogeneous Neumann boundary conditions.	{"url":"https://paulst.github.io/post/qtrefftz","timestamp":"2024-11-13T12:01:01Z","content_type":"text/html","content_length":"3278","record_id":"<urn:uuid:7a5b760d-9ddb-49c3-a8dd-c87abad4ab3c>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00656.warc.gz"}
Python Tensorflow代写 homework assignments will be done individually: each student must hand in their own answers. Use of partial or entire solutions obtained from others or online is strictly prohibited. Electronic submission on Canvas is mandatory. 1.Support Vector Machines(20 points) Given 10 points in Table 1, along with their classes and their Lagranian multipliers (i), answer the following questions: (a) What is the equation of the SVM hyperplaneh(x)? Draw the hyperplane with the 10 points. (b) What is the distance ofx 6 from the hyperplane? Is it within the margin of the classifier? (c) Classify the pointz= (3,3)Tusingh(x) from above. Table 1: Data set for question 1 data xi 1 xi 2 yi i x 1 4 2.9 1 0. x 2 4 4 1 0 x 3 1 2.5 -1 0 x 4 2.5 1 -1 0. x 5 4.9 4.5 1 0 x 6 1.9 1.9 -1 0 x 7 3.5 4 1 0. x 8 0.5 1.5 -1 0 x 9 2 2.1 -1 0. x 10 4.5 2.5 1 0 2.Support Vector Machines(20 points) Create a binary (2 feature) dataset where the target (2-class) variable encodes the XOR function. Design and implement a SVM (with a suitable kernel) to learn a classifier for this dataset. For full credit, explain the kernel you selected, and the support vectors picked by the algorithm. Redo all the above with multiple settings involving more than 2 features. Ensure that your kernel is able to model XOR in all these dimensions. Now begin deleting the non-support vectors from your dataset and relearn the classifier. What do you observe? Does the margin increase or decrease? What will happen to the margin if the support vectors are removed from the dataset? Will the margin increase or decrease?(You can use packages/tools for implementing SVM Table 2: Data for Question 3 Instance a 1 a 2 a 3 Class 1 T T 5.0 Y 2 T T 7.0 Y 3 T F 8.0 N 4 F F 3.0 Y 5 F T 7.0 N 6 F T 4.0 N 7 F F 5.0 N 8 T F 6.0 Y 9 F T 1.0 N 3.Decision Trees(20 points) Please use the data set in Table 2 to answer the following questions: (a) Show which attribute will be chosen at the root of the decision tree using information gain. Show all split points for all attributes. Please write down every step including the calculation of information gain of the attributes at each split. (b) What happens if we useInstanceas another attribute? Do you think this attribute should be used for a decision in the tree? 4.Boosting(20 points) Implement AdaBoost for the Banana dataset with decision trees of depth 3 as the weak classifiers (also known as base classifiers). You can use package/tools to implement your decision tree classifiers. The fit function of DecisionTreeClassifier in sklearn has a parameter: sample weight, which you can use to weigh training examples differently during various rounds of AdaBoost. Plot the train and test errors as a function of the number of rounds from 1 through 10. Give a brief description of your observations. 5.Neural Networks(20 points) Develop a Neural Network (NN) model to predict class labels for the Iris data set. Report your training and testing accuracy from 5-fold cross validation. You can use packages such as TensorFlow.	{"url":"https://www.cscodehelp.com/python%E4%BB%A3%E5%86%99/flask%E4%BB%A3%E5%86%99%E6%8E%A8%E8%8D%90%E7%AE%97%E6%B3%95%E4%BB%BB%E5%8A%A1/","timestamp":"2024-11-05T18:39:22Z","content_type":"text/html","content_length":"52236","record_id":"<urn:uuid:427bf781-770e-4e34-862b-7bc0dc0d43ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00747.warc.gz"}
Our First Mathematical Investigation Last December 2016, I was chosen by the Math teachers of our school along with my two schoolmates to attend a 3-day DIVISION MATHEMATICAL INVESTIGATION (MI) TRAINING-WORKSHOP (SECONDARY MATHEMATICS) at Pavia National High School. On the first day at the workshop, I listened very well to the lessons discussed especially the how's of conducting a mathematical investigation. Before the first day of the workshop ends, we were told to prepare a mathematical investigation from the copy of situations or handouts they gave us. This mathematical investigation will be done in the school during the second day. After looking at the situations thoroughly, we decided to investigate pentagonal numbers. Today, I'm going to show you how we did it so that you may also have the knowledge to learn how to conduct a mathematical investigation. So for introduction, you need to cite what got you interested to investigate this situation/problem and a little background about the situation/problem. And here is our introduction, "A pentagonal number is a type of figurate number represented by an equal number of dots on each side that form the pentagons and the pentagons are overlaid such that they share one vertex." As you can see it very simple (without why we got interested) because honestly we just copied that with a little paraphrasing from the handouts they gave us. I can remember that the three of us that time didn't have any mobile phones so we just used the resources available for us. After creating your introduction, you need to form your statement of the problem. These problem/s should be clearly defined and generated from the situation and as much as possible, a problem that is Here is our statement of the problem: "What is the total number of points needed to form an n-pentagonal array?" For our statement of the problem, we wanted to find a pattern which will enable us to find the number of points needed to form a 116th pentagonal array (for example) without drawing or counting them. For conjecture, you only need to present or state a concise mathematical statement clearly generated from the data you gathered. "The total number of points needed to form an n-pentagonal array is given by Pn = n(3n-1)/2" Remember that our conjecture is only an opinion or conclusion derived with incomplete data or information and without concrete proof or evidence meaning it is not yet accepted to be true. For data gathering, you only need to gather everything you need for the investigation and they way you present it should be well-organized. In our case, we just counted all of the points needed to form an n-pentagonal array and use a table. Like this: Using the situation we picked, we just counted all the points in an n-pentagonal array. Place all the data gathered in a table. For deriving formula, you only need to show how you were able to come up with that. State any relationship between numbers or anything that helped you derive your formula. Here's an example of how we derived our formula. Photo taken from our previous mathematical investigation (image 1) The image you can see above is an example of a Finite Difference Chart used generate a quadratic polynomial. We are going to use this to make a conjecture regarding the relationship between the data we collected and this Finite Difference Chart. Photo taken from our previous mathematical investigation (image 2) As you can notice, we replaced the values under Tn=an^2+bn+c with the data we gathered which 1,5,12,22, and 35. Then we took their differences just like in the Finite Difference chart. After doing this, we only need to equate the values we obtained to the Finite Difference chart starting on the right side. Given 2a = 3 then a = 3/2 Given 3a + b = 4 and a= 3/2 then b = -1/2 Given a + b + c = 1 and a= 3/2, b= -1/2 then c = 0 Substituting the values of a,b, and c in Pn = an^2 + bn + c; Pn = 3/2 n^2-1/2n Pn = 3n^2-n/2 or Pn = n(3n-1)/2 In writing your justification of conjecture, as much as possible, provide a deductive justification or a formal proof, the mathematical ideas involved are accurate, and statements are logically In our case, to justify our conjecture, we used the same number of pentagonal arrays (n) from the data we collected to test our conjecture. The investigators use the formula Pn = n(3n-1)/2 to test whether the conjecture can be accepted or not. After proving that the results of using the formula are the same to the data gathered, we need to test it in other examples to check if it is really correct. In writing the summary, you need to provide a critical review of the investigation you conducted and it should be long enough to highlight the major ideas and phases of the investigation, yet short enough to be manageable in a limited time. This is our very short summary that we submitted, The total number of points needed to form an n-pentagonal array is denoted by Pn= n(3n-1)/2. A little throwback with our mathematical investigation output on our back At the end of the second day, we were able to finish our investigation with joy in our hearts for conquering such challenge. We presented our outputs the next day and then get some useful feedbacks that we may be able to use in our next journey. Well, I hope you also learned a lot from me though honestly I am not good at teaching things but still I hope I was able to share my knowledge with you. Always smile and have fun every time there is new learning. God bless you all! I would like to thank my fellow mathematical investigators for giving me the permission to post our output.	{"url":"https://read.cash/@sjbuendia/our-first-mathematical-investigation-80217af4","timestamp":"2024-11-12T18:23:09Z","content_type":"text/html","content_length":"62000","record_id":"<urn:uuid:67255a9f-e3ea-4096-8a01-375dc0c4997b>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00567.warc.gz"}
2.4 Velocity vs. Time Graphs Learning Objectives Learning Objectives By the end of this section, you will be able to do the following: • Explain the meaning of slope and area in velocity vs. time graphs • Solve problems using velocity vs. time graphs Section Key Graphing Velocity as a Function of Time Graphing Velocity as a Function of Time Earlier, we examined graphs of position versus time. Now, we are going to build on that information as we look at graphs of velocity vs. time. Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity; we will discuss acceleration more in another chapter. These concepts are all very interrelated. Virtual Physics Maze Game In this simulation you will use a vector diagram to manipulate a ball into a certain location without hitting a wall. You can manipulate the ball directly with position or by changing its velocity. Explore how these factors change the motion. If you would like, you can put it on the a setting, as well. This is acceleration, which measures the rate of change of velocity. We will explore acceleration in more detail later, but it might be interesting to take a look at it here. Grasp Check In which setting, displacement or velocity, can the ball be easily manipulated and why? a. The ball can be easily manipulated with displacement because the arena is a position space. b. The ball can be easily manipulated with velocity because the arena is a position space. c. The ball can be easily manipulated with displacement because the arena is a velocity space. d. The ball can be easily manipulated with velocity because the arena is a velocity space. What can we learn about motion by looking at velocity vs. time graphs? Let’s return to our drive to school, and look at a graph of position versus time as shown in Figure 2.18. We assumed for our original calculation that your parent drove with a constant velocity to and from school. We now know that the car could not have gone from rest to a constant velocity without speeding up. So the actual graph would be curved on either end, but let’s make the same approximation as we did then, anyway. Tips For Success It is common in physics, especially at the early learning stages, for certain things to be neglected, as we see here. This is because it makes the concept clearer or the calculation easier. Practicing physicists use these kinds of short-cuts, as well. It works out because usually the thing being neglected is small enough that it does not significantly affect the answer. In the earlier example, the amount of time it takes the car to speed up and reach its cruising velocity is very small compared to the total time traveled. Looking at this graph, and given what we learned, we can see that there are two distinct periods to the car’s motion—the way to school and the way back. The average velocity for the drive to school is 0.5 km/minute. We can see that the average velocity for the drive back is –0.5 km/minute. If we plot the data showing velocity versus time, we get another graph (Figure 2.19): We can learn a few things. First, we can derive a v versus t graph from a d versus t graph. Second, if we have a straight-line position–time graph that is positively or negatively sloped, it will yield a horizontal velocity graph. There are a few other interesting things to note. Just as we could use a position vs. time graph to determine velocity, we can use a velocity vs. time graph to determine position. We know that v = d/t. If we use a little algebra to re-arrange the equation, we see that d = v $××$t. In Figure 2.19, we have velocity on the y-axis and time along the x-axis. Let’s take just the first half of the motion. We get 0.5 km/minute $××$ 10 minutes. The units for minutes cancel each other, and we get 5 km, which is the displacement for the trip to school. If we calculate the same for the return trip, we get –5 km. If we add them together, we see that the net displacement for the whole trip is 0 km, which it should be because we started and ended at the same Tips For Success You can treat units just like you treat numbers, so a km/km=1 (or, we say, it cancels out). This is good because it can tell us whether or not we have calculated everything with the correct units. For instance, if we end up with m × s for velocity instead of m/s, we know that something has gone wrong, and we need to check our math. This process is called dimensional analysis, and it is one of the best ways to check if your math makes sense in physics. The area under a velocity curve represents the displacement. The velocity curve also tells us whether the car is speeding up. In our earlier example, we stated that the velocity was constant. So, the car is not speeding up. Graphically, you can see that the slope of these two lines is 0. This slope tells us that the car is not speeding up, or accelerating. We will do more with this information in a later chapter. For now, just remember that the area under the graph and the slope are the two important parts of the graph. Just like we could define a linear equation for the motion in a position vs. time graph, we can also define one for a velocity vs. time graph. As we said, the slope equals the acceleration, a. And in this graph, the y-intercept is v[0]. Thus, $v= v 0 +at v= v 0 +at$. But what if the velocity is not constant? Let’s look back at our jet-car example. At the beginning of the motion, as the car is speeding up, we saw that its position is a curve, as shown in Figure You do not have to do this, but you could, theoretically, take the instantaneous velocity at each point on this graph. If you did, you would get Figure 2.21, which is just a straight line with a positive slope. Again, if we take the slope of the velocity vs. time graph, we get the acceleration, the rate of change of the velocity. And, if we take the area under the slope, we get back to the displacement. Solving Problems using Velocity–Time Graphs Solving Problems using Velocity–Time Graphs Most velocity vs. time graphs will be straight lines. When this is the case, our calculations are fairly simple. Worked Example Using Velocity Graph to Calculate Some Stuff: Jet Car Use this figure to (a) find the displacement of the jet car over the time shown (b) calculate the rate of change (acceleration) of the velocity. (c) give the instantaneous velocity at 5 s, and (d) calculate the average velocity over the interval shown. a. The displacement is given by finding the area under the line in the velocity vs. time graph. b. The acceleration is given by finding the slope of the velocity graph. c. The instantaneous velocity can just be read off of the graph. d. To find the average velocity, recall that $v avg = Δd Δt = d f − d 0 t f − t 0 v avg = Δd Δt = d f − d 0 t f − t 0$ 1. Analyze the shape of the area to be calculated. In this case, the area is made up of a rectangle between 0 and 20 m/s stretching to 30 s. The area of a rectangle is length $××$ width. Therefore, the area of this piece is 600 m. 2. Above that is a triangle whose base is 30 s and height is 140 m/s. The area of a triangle is 0.5 $××$ length $××$ width. The area of this piece, therefore, is 2,100 m. 3. Add them together to get a net displacement of 2,700 m. 1. Take two points on the velocity line. Say, t = 5 s and t = 25 s. At t = 5 s, the value of v = 40 m/s. At t = 25 s, v = 140 m/s. 2. Find the slope.$a = Δv Δt = 100 m/s 20 s = 5 m/s 2 a = Δv Δt = 100 m/s 20 s = 5 m/s 2$ c. The instantaneous velocity at t = 5 s , as we found in part (b) is just 40 m/s. 1. Find the net displacement, which we found in part (a) was 2,700 m. 2. Find the total time which for this case is 30 s. 3. Divide 2,700 m/30 s = 90 m/s. The average velocity we calculated here makes sense if we look at the graph. 100m/s falls about halfway across the graph and since it is a straight line, we would expect about half the velocity to be above and half below. Tips For Success You can have negative position, velocity, and acceleration on a graph that describes the way the object is moving. You should never see a graph with negative time on an axis. Why? Most of the velocity vs. time graphs we will look at will be simple to interpret. Occasionally, we will look at curved graphs of velocity vs. time. More often, these curved graphs occur when something is speeding up, often from rest. Let’s look back at a more realistic velocity vs. time graph of the jet car’s motion that takes this speeding up stage into account. Worked Example Using Curvy Velocity Graph to Calculate Some Stuff: jet car, Take Two Use Figure 2.22 to (a) find the approximate displacement of the jet car over the time shown, (b) calculate the instantaneous acceleration at t = 30 s, (c) find the instantaneous velocity at 30 s, and (d) calculate the approximate average velocity over the interval shown. a. Because this graph is an undefined curve, we have to estimate shapes over smaller intervals in order to find the areas. b. Like when we were working with a curved displacement graph, we will need to take a tangent line at the instant we are interested and use that to calculate the instantaneous acceleration. c. The instantaneous velocity can still be read off of the graph. d. We will find the average velocity the same way we did in the previous example. 1. This problem is more complicated than the last example. To get a good estimate, we should probably break the curve into four sections. 0 → 10 s, 10 → 20 s, 20 → 40 s, and 40 → 70 s. 2. Calculate the bottom rectangle (common to all pieces). 165 m/s $× ×$ 70 s = 11,550 m. 3. Estimate a triangle at the top, and calculate the area for each section. Section 1 = 225 m; section 2 = 100 m + 450 m = 550 m; section 3 = 150 m + 1,300 m = 1,450 m; section 4 = 2,550 m. 4. Add them together to get a net displacement of 16,325 m. b. Using the tangent line given, we find that the slope is 1 m/s^2. c. The instantaneous velocity at t = 30 s, is 240 m/s. 1. Find the net displacement, which we found in part (a), was 16,325 m. 2. Find the total time, which for this case is 70 s. 3. Divide$16,325 m 70 s ∼233 m/s 16,325 m 70 s ∼233 m/s$ This is a much more complicated process than the first problem. If we were to use these estimates to come up with the average velocity over just the first 30 s we would get about 191 m/s. By approximating that curve with a line, we get an average velocity of 202.5 m/s. Depending on our purposes and how precise an answer we need, sometimes calling a curve a straight line is a worthwhile Practice Problems Practice Problems Consider the velocity vs. time graph shown below of a person in an elevator. Suppose the elevator is initially at rest. It then speeds up for 3 seconds, maintains that velocity for 15 seconds, then slows down for 5 seconds until it stops. Find the instantaneous velocity at t = 10 s and t = 23 s. a. Instantaneous velocity at t = 10 s and t = 23 s are 0 m/s and 0 m/s. b. Instantaneous velocity at t = 10 s and t = 23 s are 0 m/s and 3 m/s. c. Instantaneous velocity at t = 10 s and t = 23 s are 3 m/s and 0 m/s. d. Instantaneous velocity at t = 10 s and t = 23 s are 3 m/s and 1.5 m/s. Calculate the net displacement and the average velocity of the elevator over the time interval shown. a. Net displacement is 45 m and average velocity is 2.10 m/s. b. Net displacement is 45 m and average velocity is 2.28 m/s. c. Net displacement is 57 m and average velocity is 2.66 m/s. d. Net displacement is 57 m and average velocity is 2.48 m/s. Snap Lab Graphing Motion, Take Two In this activity, you will graph a moving ball’s velocity vs. time. • your graph from the earlier Graphing Motion Snap Lab! • 1 piece of graph paper • 1 pencil 1. Take your graph from the earlier Graphing Motion Snap Lab! and use it to create a graph of velocity vs. time. 2. Use your graph to calculate the displacement. Grasp Check Describe the graph and explain what it means in terms of velocity and acceleration. a. The graph shows a horizontal line indicating that the ball moved with a constant velocity, that is, it was not accelerating. b. The graph shows a horizontal line indicating that the ball moved with a constant velocity, that is, it was accelerating. c. The graph shows a horizontal line indicating that the ball moved with a variable velocity, that is, it was not accelerating. d. The graph shows a horizontal line indicating that the ball moved with a variable velocity, that is, it was accelerating. Check Your Understanding Check Your Understanding Exercise 11 What information could you obtain by looking at a velocity vs. time graph? a. acceleration b. direction of motion c. reference frame of the motion d. shortest path Exercise 12 How would you use a position vs. time graph to construct a velocity vs. time graph and vice versa? a. Slope of position vs. time curve is used to construct velocity vs. time curve, and slope of velocity vs. time curve is used to construct position vs. time curve. b. Slope of position vs. time curve is used to construct velocity vs. time curve, and area of velocity vs. time curve is used to construct position vs. time curve. c. Area of position vs. time curve is used to construct velocity vs. time curve, and slope of velocity vs. time curve is used to construct position vs. time curve. d. Area of position/time curve is used to construct velocity vs. time curve, and area of velocity vs. time curve is used to construct position vs. time curve.	{"url":"https://texasgateway.org/resource/24-velocity-vs-time-graphs?book=79076&binder_id=78096","timestamp":"2024-11-02T21:04:56Z","content_type":"text/html","content_length":"85172","record_id":"<urn:uuid:cc6c373e-6cd1-4570-865c-8afc6f77c54d>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00104.warc.gz"}
For every class that has not less than 3 ships in the database determine the number of ships of this class sunk in the battles, if any. Output: class and the number of the sunken ships. This exercise is somewhat similar to the exercise 56, i.e. here it is possible to suppose the same mistakes in calculation of the number of sunken ships. However the situation is also aggravated with definition of the total number of the ships in a class. Let's consider the solution that the check system doesn't accept. Exercise 3.13.1 1. SELECT c.class, SUM(outc) 2. FROM Classes c LEFT JOIN 3. Ships s ON c.class = s.class LEFT JOIN 4. (SELECT ship, 1 outc 5. FROM Outcomes 6. WHERE result = 'sunk') o ON s.name = o.ship OR 7. c.class = o.ship 8. GROUP BY c.class 9. HAVING COUNT() > 2 AND 10. SUM(outc) IS NOT NULL; The first left join gives all classes repeating so many times as the number of ships available in the Ships table. If any class doesn't have ships in this table, it will be noted one time, and it gives us an opportunity to consider the leading ships of the class in the Outcomes table, if any. Next, one left join is being worked out with the set of sunken ships on the predicate 1. ON s.name = o.ship OR c.class = o.ship In the calculating column 1 is being inserted, if the name of the sunken ship coincides either with the name of the ship, or with the name of the class from the set had been got earlier. So, here we do try to consider the leading (head) ships. Finally, the grouping by the classes with selection by the number of ships (rows) in the class is being worked out, and the sum of the sunken ships (units in the column “outs”) is being calculated. Author of this solution offers the rational way to calculate in one grouping both the total number of ships, and the quantity of the sunken ships in the class. The predicate, in accordance with the terms of the task, removes from the result such classes that don't have any sunken ships. Those who read the analysis of the previous tasks, have already guessed, what the problem is. That's right, the problem is in the predicate of the second join. But not only in this. в этом. Let's consider the next variant of data. Let for some class class_N in the Ships table we have two ships: ship_1 and ship_2. Besides, in the Outcomes table there is the sunken ship ship_1 and survived the leading ship – class_N. The first join gives: Class Ship Class_N ship_1 Class_N ship_2 We work out the second join: Class ship outs Class_N ship_1 1 Class_N ship_2 NULL In the result this class will not get into the resulting set at all, because the condition COUNT() > 2 won't be held, but actually there are three ships. The reason of the mistake lies in the fact that we perform the join only on the sunken ships, simultaneously counting the total number of ships. Now let's change a little data in the example. And let the leading ship class_N to be also sunk. Then the result of the join is: class ship outs class_N ship_1 1 class_N ship_2 NULL class_N ship_1 1 class_N ship_2 1 The last two rows will be got in the result of joining the row of the sunken leading ship, as the predicate c.class=o.ship gives “true”. So, instead of one row for the leading ship we get a row for every ship of the class from the Ships table. Totally, instead of we have You may try to correct this solution or to use another way on the basis of the inner join and union. As it will seem surprising, but three absolutely different solutions presented below contain the same mistake, at least, they return the same result on the checking database of a site.	{"url":"http://sql-tutorial.ru/en/book_exercise_57.html","timestamp":"2024-11-09T07:29:14Z","content_type":"text/html","content_length":"54909","record_id":"<urn:uuid:9888f1ac-5ea7-440c-ae10-d004ff06ee6e>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00516.warc.gz"}
The magnetic field B=2t+4t^(2) (where t = time) is applied perpendicul The magnetic field B=2t+4t2 (where t = time) is applied perpendicular to the plane of a circular wire of radius r and resistance R. If all the units are in SI the electric charge that flows through the circular wire during t = 0 s to t = 2 s is The correct Answer is:B Updated on:21/07/2023 Knowledge Check • A magnetic field B=2t+4t2 (where t = time) is applied perpendicular to the plane of a circular wire of radius and resistance R. If all the units are in SI the electric charge that flows through the circular wire during t = 0 s to t = 2 s is • Question 2 - Select One or More When some potential difference V is applied across a resistance R then the work by the electrical field on the charge q to flow through the circuit in time t will be • A magnetic field given by B(t) =0.2t−0.05t2 tesla is directed perpendicular to the plane of a circular coil containing 25 turns of radius 1.8 cm and whose total resistance is 1.5Ω. The power dissipation at 3 s is approximately	{"url":"https://www.doubtnut.com/qna/649445674","timestamp":"2024-11-12T16:54:07Z","content_type":"text/html","content_length":"354159","record_id":"<urn:uuid:41b5b295-0d9b-4611-bc70-fbc736621161>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00090.warc.gz"}
Differential Geometry and Controlsearch Item Successfully Added to Cart An error was encountered while trying to add the item to the cart. Please try again. Please make all selections above before adding to cart Differential Geometry and Control Edited by: G. Ferreyra : Louisiana State University, Baton Rouge, LA R. Gardner : University of North Carolina, Chapel Hill, NC H. Hermes : University of Colorado, Boulder, CO Hardcover ISBN: 978-0-8218-0887-0 Product Code: PSPUM/64 List Price: $139.00 MAA Member Price: $125.10 AMS Member Price: $111.20 eBook ISBN: 978-0-8218-9368-5 Product Code: PSPUM/64.E List Price: $135.00 MAA Member Price: $121.50 AMS Member Price: $108.00 Hardcover ISBN: 978-0-8218-0887-0 eBook: ISBN: 978-0-8218-9368-5 Product Code: PSPUM/64.B List Price: $274.00 $206.50 MAA Member Price: $246.60 $185.85 AMS Member Price: $219.20 $165.20 Click above image for expanded view Differential Geometry and Control Edited by: G. Ferreyra : Louisiana State University, Baton Rouge, LA R. Gardner : University of North Carolina, Chapel Hill, NC H. Hermes : University of Colorado, Boulder, CO Hardcover ISBN: 978-0-8218-0887-0 Product Code: PSPUM/64 List Price: $139.00 MAA Member Price: $125.10 AMS Member Price: $111.20 eBook ISBN: 978-0-8218-9368-5 Product Code: PSPUM/64.E List Price: $135.00 MAA Member Price: $121.50 AMS Member Price: $108.00 Hardcover ISBN: 978-0-8218-0887-0 eBook ISBN: 978-0-8218-9368-5 Product Code: PSPUM/64.B List Price: $274.00 $206.50 MAA Member Price: $246.60 $185.85 AMS Member Price: $219.20 $165.20 • Proceedings of Symposia in Pure Mathematics Volume: 64; 1999; 341 pp MSC: Primary 49; 53; 93; Secondary 22; 60; 35 This volume presents selections from talks given at the AMS Summer Research Institute on Differential Geometry and Control held at the University of Colorado (Boulder). Included articles were refereed according to the highest standards. This collection provides a coherent global perspective on recent developments and important open problems in geometric control theory. Readers will find in this book an excellent source of current challenging research problems and results. Graduate students, researchers, educators and applied mathematicians working in control theory; electrical, mechanical and aerospace engineers. • Permission – for use of book, eBook, or Journal content • Book Details • Table of Contents • Requests Volume: 64; 1999; 341 pp MSC: Primary 49; 53; 93; Secondary 22; 60; 35 This volume presents selections from talks given at the AMS Summer Research Institute on Differential Geometry and Control held at the University of Colorado (Boulder). Included articles were refereed according to the highest standards. This collection provides a coherent global perspective on recent developments and important open problems in geometric control theory. Readers will find in this book an excellent source of current challenging research problems and results. Graduate students, researchers, educators and applied mathematicians working in control theory; electrical, mechanical and aerospace engineers. Permission – for use of book, eBook, or Journal content Please select which format for which you are requesting permissions.	{"url":"https://bookstore.ams.org/pspum-64","timestamp":"2024-11-04T01:29:29Z","content_type":"text/html","content_length":"108240","record_id":"<urn:uuid:178823a3-4649-41c5-9e43-9affc1d24340>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00030.warc.gz"}
Printable Double Digit Multiplication Worksheets Printable Double Digit Multiplication Worksheets - First, write the numbers over each other. Here you will find a wide range of free. Multiplication practice with all factors being under 100; Welcome to our double digit multiplication worksheets for 4th grade. Web let us take a look at how you can multiply double digit numbers! These double digit multiplication worksheets help kids. Web double digit multiplication worksheets. Web these printables have pairs of double digit numbers for students to multiply together. Double Digit Multiplication Worksheets Math Monks Web let us take a look at how you can multiply double digit numbers! Here you will find a wide range of free. Web these printables have pairs of double digit numbers for students to multiply together. Welcome to our double digit multiplication worksheets for 4th grade. These double digit multiplication worksheets help kids. 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Multiplication practice with all factors being under 100; Welcome to our double digit multiplication worksheets for 4th grade. Related Post: Goodnotes Budget Planner Templates Free Cut Out Mitten Template 1984 Calendar Meme Template This Is Brilliant But I Like This Template Chocolate Bar Wrapper Template Anime Powerpoint Templates Kitchen Countertop Measurement Template Math Worksheets Printable For Kindergarten Printable Sbar Template Children's Printable Valentines Day Cards	{"url":"https://akademia50plus.edu.pl/en/printable-double-digit-multiplication-worksheets.html","timestamp":"2024-11-02T21:21:15Z","content_type":"text/html","content_length":"27341","record_id":"<urn:uuid:88ebcdee-c5a0-4c1e-bb1d-fb3918e328c8>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00184.warc.gz"}
Planar graphs without triangles adjacent to cycles of length from 3 to 9 are 3-colorable. Planar graphs without triangles adjacent to cycles of length from 3 to 9 are 3-colorable. Borodin, O.V., et al. "Planar graphs without triangles adjacent to cycles of length from 3 to 9 are 3-colorable.." Sibirskie Ehlektronnye Matematicheskie Izvestiya [electronic only] 3 (2006): 428-440. <http://eudml.org/doc/53348>. author = {Borodin, O.V., Glebov, A.N., Jensen, Tommy R., Raspaud, Andre}, journal = {Sibirskie Ehlektronnye Matematicheskie Izvestiya [electronic only]}, keywords = {color-4-critical graph; Euler's formula; plane graph without triangles}, language = {eng}, pages = {428-440}, publisher = {Institut Matematiki Im. S.L. Soboleva, SO RAN}, title = {Planar graphs without triangles adjacent to cycles of length from 3 to 9 are 3-colorable.}, url = {http://eudml.org/doc/53348}, volume = {3}, year = {2006}, TY - JOUR AU - Borodin, O.V. AU - Glebov, A.N. AU - Jensen, Tommy R. AU - Raspaud, Andre TI - Planar graphs without triangles adjacent to cycles of length from 3 to 9 are 3-colorable. JO - Sibirskie Ehlektronnye Matematicheskie Izvestiya [electronic only] PY - 2006 PB - Institut Matematiki Im. S.L. Soboleva, SO RAN VL - 3 SP - 428 EP - 440 LA - eng KW - color-4-critical graph; Euler's formula; plane graph without triangles UR - http://eudml.org/doc/53348 ER - You must be logged in to post comments. To embed these notes on your page include the following JavaScript code on your page where you want the notes to appear.	{"url":"https://eudml.org/doc/53348","timestamp":"2024-11-05T06:37:55Z","content_type":"application/xhtml+xml","content_length":"34927","record_id":"<urn:uuid:e404d440-e9f6-4edc-bd9e-9448c480e210>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00660.warc.gz"}