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https://git.ufz.de/shatwell/seefo/-/commit/143fd07efc4dbf6a24a31058489a5d72ed3fb826 | [
"### Added mixed layer depth fns, improved descriptions\n\nparent 403ecd4c\nR/h3mix.R 0 → 100644\n #' @title Robust calculation of mixed layer depth #' #' @description Robust calculation of the mixed layer depth and thermocline depth, #' more suited for manual or irregular #' temperature data. #' #' @param T Temperature profile #' @param z The depths corresponding to the temperature profile #' @param plot Should a plot of the profile showing the estimates be shown? (logical) #' @param thresh The threshold for the difference between top and bottom temperature to infer stratification #' @param therm.dz Representative thickness of the thermocline layer. 1 m works well from experience. #' @param min.hmix I can't really remember what this is for. #' @param min.gradient Not used #' @param ... Arguments passed to `plot()`, when `plot=TRUE` #' #' @details #' This function estimates the mixed depth and the thermocline depth. #' It uses regressions to locate the kink in the profile at the bottom of #' the mixed layer as the mixed layer, and the maximum gradient as the thermocline. #' It is better suited to irregular, esp manually measured profile data. #' It first finds the minimum curvature in the #' profile (maximum for winter stratification) as the initial guess of the #' border of the surface mixed #' layer. Then it finds the thermocline as the depth of maximum T-gradient. It #' estimates the mixed layer depth as the depth where the regression line through #' the surface layer temperatures intersects with the regression line through the #' thermocline temperatures. It performs some other checks, like whether #' stratification exists surface-bottom temperature difference > `thresh`, and whether the thermocline #' is above the surface layer, or whether the lake is completely isothermal and #' there is no intersection (and returns `NA`). If mixed, it assumes the mixed #' layer depth is the maximum depth. It also plots the profile if desired #' (`plot=TRUE`), marking the surface layer and inflection point in red and the #' thermocline in blue. #' Surface temperature is estimated as the mean of the upper 2 m, or the top two measurements, #' bottom temperature is estimated as the mean of the deepest 2 m layer. #' #' #' @return A vector of length 2 with the mixed layer depth `hmix` and the thermocline depth `htherm` #' #' @author Tom Shatwell #' #' @seealso \\code{\\link{hmix}}, \\code{\\link{h2mix}}, \\code{\\link{delta_rho}} #' #' @examples #' \\dontrun{ #' T <- c(25.44,25.46,25.48,25.46,25.44,24.36,20.48,18.09,15.96,13.95,11.67, #' 10.62,9.82,9.51,9.22,9.01,8.81,8.68,8.58,8.39,8.29,8.14,8.08,8.03,8.01, #' 8.05,7.94,7.93,7.85) #' z <- 0:28 #' h3mix(T,z,plot=TRUE) #' } #' #' @export h3mix <- function(T, z, plot=FALSE, thresh = 1, therm.dz = 1, min.hmix = 1.5, min.gradient = 0.3, ...) { if(sum(!is.na(T))>3) { dTdz <- diff(T) / diff(z) d2Tdz2 <- diff(T,1,2) / diff(z,2)^2 i1 <- which.min(dTdz) # index of strongest gradient (-ve for positive strat, +ve for winter strat) i2 <- which.min(d2Tdz2) # index of minimum curvature (bot epilimnion) while(z[i2]<=min.hmix) { d2Tdz2[i2] <- NA i2 <- which.min(d2Tdz2) } while(i1 <= i2) { dTdz[i1] <- NA i1 <- which.min(dTdz) } i3 <- which.max(d2Tdz2) # index of maximum curvature (top hypolimnion) if(min(z)<2) { Ts <- mean(T[z<=2], na.rm=TRUE) } else { Ts <- mean(T[1:i1], na.rm=TRUE) warning(\"Warning: Surface temp estimated as temps above greatest curvature\") } Tb <- mean(T[z>=(max(z)-2)], na.rm=TRUE) if(Ts < 4) { i1 <- which.max(dTdz) # index of maximum gradient i2 <- which.max(d2Tdz2) # max curvature for winter stratification i3 <- which.min(d2Tdz2) # min curvature for winter stratification } h1 <- mean(c(z[i1], z[i1+1])) # depth of maximum gradient h2 <- z[i2+1] # depth of inflection thermo <- data.frame(z=z[z>=z[i1]-therm.dz & z <= z[i1+1]+therm.dz], T = T[z>=z[i1]-therm.dz & z <= z[i1+1]+therm.dz]) regs <- lm(T[1:i2]~z[1:i2]) # regression thru surface temps regt <- lm(T~z, thermo) # regression thru thermocline dTdz.s <- coef(regs) # T gradient in surface layer if(is.na(coef(regs))) dTdz.s <- 0 h <- (coef(regt) - coef(regs)) / # depth of intersection of thermocline and surface layer regression lines (dTdz.s - coef(regt)) if(!is.na(h) & h > z[i1+1]) h <- NA # ignore if intersection of thermocline and surface layer is below thermocline if(i1 < i2) h <- NA # ignore if minimum curvature is below the thermocline if(abs(Ts-Tb) < thresh) h <- max(z) # assume mixed to max(z) if Ts-Tb < thresh if(!is.na(h) & h < 0) h <- NA if(plot){ plot(T, z, ylim=c(max(z),0), type=\"n\", ...) abline(h=0:(max(z)), col=\"grey\") abline(h=h) # abline(h=h2, lty=2) # if(!c(coef(regs) %in% c(0,NA) | !coef(regt) %in% c(0,NA))) { # abline(-coef(regs)/coef(regs), 1/coef(regs), lty=2, col=\"red\") # abline(-coef(regt)/coef(regt), 1/coef(regt), lty=2, col=\"blue\") # } # points(coef(regs) * h + coef(regs), h, pch=16, col=\"blue\", cex=1.5) lines(T, z, type=\"o\") lines(T[1:i2], z[1:i2], col=\"red\", lwd=2) lines(T[c(i1,i1+1)], z[c(i1,i1+1)], col=\"blue\", lwd=2) points(T[i2+1], z[i2+1], col=\"red\",pch=16) box() } } else { h <- NA warning(\"Warning:need more than 3 temperature values\") } out <- c(h, mean(z[c(i1,i1+1)])) names(out) <- c(\"hmix\",\"htherm\") return(out) }\n ... ... @@ -7,10 +7,10 @@ #' #' @details #' Calculates mixed layer depths as the depth of the maximum temperature gradient (dT/dz). #' Designed to be as fast as possible for large data sets. It doesn't do any checks at all, #' and only works with regular data, so be careful. #' Designed to be as fast as possible for very large data sets, like model output. #' It doesn't do any checks at all, and only works with regular data, so be careful. #' Choose the right vertical resolution in the temperatures to ensure a more robust result. #' Use the rLakeAnalyzer functions for more robust solutions which are #' Use h3mix or the rLakeAnalyzer functions for more robust solutions which are #' orders of magnitude slower. #' #' @return A vector of mixed layer depths ... ... @@ -32,7 +32,3 @@ #' @export hmix <- function(T,z) z[apply(abs(diff(T)/diff(z)),2,which.max)]\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!\nPlease register or to comment"
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https://whatisconvert.com/193-cubic-inches-in-tablespoons | [
"# What is 193 Cubic Inches in Tablespoons?\n\n## Convert 193 Cubic Inches to Tablespoons\n\nTo calculate 193 Cubic Inches to the corresponding value in Tablespoons, multiply the quantity in Cubic Inches by 1.1082251082238 (conversion factor). In this case we should multiply 193 Cubic Inches by 1.1082251082238 to get the equivalent result in Tablespoons:\n\n193 Cubic Inches x 1.1082251082238 = 213.88744588719 Tablespoons\n\n193 Cubic Inches is equivalent to 213.88744588719 Tablespoons.\n\n## How to convert from Cubic Inches to Tablespoons\n\nThe conversion factor from Cubic Inches to Tablespoons is 1.1082251082238. To find out how many Cubic Inches in Tablespoons, multiply by the conversion factor or use the Volume converter above. One hundred ninety-three Cubic Inches is equivalent to two hundred thirteen point eight eight seven Tablespoons.\n\n## Definition of Cubic Inch\n\nThe cubic inch is a unit of measurement for volume in the Imperial units and United States customary units systems. It is the volume of a cube with each of its three dimensions (length, width, and depth) being one inch long. The cubic inch and the cubic foot are still used as units of volume in the United States, although the common SI units of volume, the liter, milliliter, and cubic meter, are also used, especially in manufacturing and high technology. One cubic foot is equal to exactly 1,728 cubic inches because 123 = 1,728.\n\n## Definition of Tablespoon\n\nIn the United States a tablespoon (abbreviation tbsp) is approximately 14.8 ml (0.50 US fl oz). A tablespoon is a large spoon used for serving or eating. In many English-speaking regions, the term now refers to a large spoon used for serving, however, in some regions, including parts of Canada, it is the largest type of spoon used for eating. By extension, the term is used as a measure of volume in cooking.\n\n## Using the Cubic Inches to Tablespoons converter you can get answers to questions like the following:\n\n• How many Tablespoons are in 193 Cubic Inches?\n• 193 Cubic Inches is equal to how many Tablespoons?\n• How to convert 193 Cubic Inches to Tablespoons?\n• How many is 193 Cubic Inches in Tablespoons?\n• What is 193 Cubic Inches in Tablespoons?\n• How much is 193 Cubic Inches in Tablespoons?\n• How many tbsp are in 193 in3?\n• 193 in3 is equal to how many tbsp?\n• How to convert 193 in3 to tbsp?\n• How many is 193 in3 in tbsp?\n• What is 193 in3 in tbsp?\n• How much is 193 in3 in tbsp?"
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https://www.tocris.com/products/glp-2-rat_2259 | [
"# GLP-2 (rat)\n\nCat. No. 2259 Datasheet / COA / SDS\nPricing Availability Qty\nDescription: Endogenous hormone; displays intestinotrophic activity\nAlternative Names: Glucagon-like peptide 2 (rat)\nDatasheet\nCitations\nReviews\nLiterature\n\n## Biological Activity\n\nEndogenous peptide identified as an intestinal epithelium-specific growth factor; stimulates cell proliferation and inhibits apoptosis. Diverse effects on gastrointestinal function including regulation of intestinal glucose transport, food intake, and gastric acid secretion.\n\nGLP-2 (human) also available.\n\n## Technical Data\n\n M. Wt 3796.17 Formula C166H256N44O56S Sequence HADGSFSDEMNTILDNLATRDFINWLIQTKITD Storage Desiccate at -20°C CAS Number 195262-56-7 PubChem ID 90488760 InChI Key KBNPAOMRSZGNFV-XBBCIFKHSA-N Smiles [H]N[C@@H](CC1=CNC=N1)C(=O)N[C@@H](C)C(=O)N[C@@H](CC(O)=O)C(=O)NCC(=O)N[C@@H](CO)C(=O)N[C@@H](CC1=CC=CC=C1)C(=O)N[C@@H](CO)C(=O)N[C@@H](CC(O)=O)C(=O)N[C@@H](CCC(O)=O)C(=O)N[C@@H](CCSC)C(=O)N[C@@H](CC(N)=O)C(=O)N[C@@H]([C@@H](C)O)C(=O)N[C@@H]([C@@H](C)CC)C(=O)N[C@@H](CC(C)C)C(=O)N[C@@H](CC(O)=O)C(=O)N[C@@H](CC(N)=O)C(=O)N[C@@H](CC(C)C)C(=O)N[C@@H](C)C(=O)N[C@@H]([C@@H](C)O)C(=O)N[C@@H](CCCNC(N)=N)C(=O)N[C@@H](CC(O)=O)C(=O)N[C@@H](CC1=CC=CC=C1)C(=O)N[C@@H]([C@@H](C)CC)C(=O)N[C@@H](CC(N)=O)C(=O)N[C@@H](CC1=CNC2=C1C=CC=C2)C(=O)N[C@@H](CC(C)C)C(=O)N[C@@H]([C@@H](C)CC)C(=O)N[C@@H](CCC(N)=O)C(=O)N[C@@H]([C@@H](C)O)C(=O)N[C@@H](CCCCN)C(=O)N[C@@H]([C@@H](C)CC)C(=O)N[C@@H]([C@@H](C)O)C(=O)N[C@@H](CC(O)=O)C(O)=O\n\nThe technical data provided above is for guidance only. For batch specific data refer to the Certificate of Analysis.\n\nTocris products are intended for laboratory research use only, unless stated otherwise.\n\n## Solubility Data\n\n Solubility Soluble to 1 mg/ml in water\n\n## Product Datasheets\n\nCertificate of Analysis / Product Datasheet\nSelect another batch:\n\n## References\n\nReferences are publications that support the biological activity of the product.\n\nRocha et al (2004) Glucagon-like peptide-2: divergent signalling pathways. J.Surg.Res. 121 5 PMID: 15313368\n\nBrubaker and Drucker (2004) Glucagon-like peptides regulate cell proliferation and apoptosis in the pancreas, gut, and central nervous system. Endocrinol. 145 2653\n\nBulut et al (2004) Glucagon-like peptide 2 improves intestinal wound healing through induction of epithelial cell migration in vitro-evidence for a TGF-β-mediated effect. Reg.Peptides 121 137\n\nIf you know of a relevant reference for GLP-2 (rat), please let us know.\n\n## View Related Products by Target\n\nKeywords: GLP-2 (rat), GLP-2 (rat) supplier, Endogenous, hormone, displays, intestinotrophic, activity, GLP2, Receptors, Glucagon-Like, Peptide, 2, peptide2, (rat), Glucagon-like, peptide, 2259, Tocris Bioscience"
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https://123dok.net/document/wyepo67z-a-passivity-based-decentralized-strategy-for-generalized-connectivity-maintenance.html | [
"# A Passivity-Based Decentralized Strategy for Generalized Connectivity Maintenance\n\n22\n\nLoading.... (view fulltext now)\n\nLoading....\n\nLoading....\n\nLoading....\n\nLoading....\n\n## Texte intégral\n\n(1)\n\n### �hal-00910385�\n\n(2)\n\nPreprint version - final, definitive version available at http://ijr.sagepub.com/ accepted for IJRR, Sep. 2012\n\n## Generalized Connectivity Maintenance\n\n### Paolo Robuffo Giordano, Antonio Franchi, Cristian Secchi, and Heinrich H. Bülthoff\n\nAbstract—The design of decentralized controllers coping with the typical constraints on the inter-robot sensing/communication capabilities represents a promising direction in multi-robot re-search thanks to the inherent scalability and fault tolerance of these approaches. In these cases, connectivity of the underlying interaction graph plays a fundamental role: it represents a necessary condition for allowing a group or robots achieving a common task by resorting to only local information. Goal of this paper is to present a novel decentralized strategy able to enforce connectivity maintenance for a group of robots in a flexible way, that is, by granting large freedom to the group internal configuration so as to allow establishment/deletion of interaction links at anytime as long as global connectivity is preserved. A peculiar feature of our approach is that we are able to embed into a unique connectivity preserving action a large number of constraints and requirements for the group: (i) presence of specific inter-robot sensing/communication models, (ii) group requirements such as formation control, and (iii) individual requirements such as collision avoidance. This is achieved by defining a suitable global potential function of the second smallest eigenvalue λ2 of the graph Laplacian, and by computing, in a\n\ndecentralized way, a gradient-like controller built on top of this potential. Simulation results obtained with a group of quadorotor UAVs and UGVs, and experimental results obtained with four quadrotor UAVs, are finally presented to thoroughly illustrate the features of our approach on a concrete case study.\n\nI. INTRODUCTION\n\nOver the last years, the challenge of coordinating the actions of multiple robots has increasingly drawn the attention of the robotics and control communities, being inspired by the idea that proper coordination of many simple robots can lead to the fulfillment of arbitrarily complex tasks in a robust (to single robot failures) and highly flexible way. Teams of multi-robots can take advantage of their number to perform, for example, complex manipulation and assembly tasks, or to obtain rich spatial awareness by suitably distributing themselves in the environment. Use of multiple robots, or in general distributed sensing/computing resources, is also at the core of the foreseen Cyber-Physical Society Lee (2008) envisioning a network of computational and physical resources (such as robots) spread over large areas and able to collectively monitor the environment and act upon it. Within the scope of robotics, autonomous search and rescue, firefighting, exploration and P. Robuffo Giordano and A. Franchi are with the Max Planck Institute for Biological Cybernetics, Spemannstraße 38, 72076 Tübingen, Germany\n\n{prg,antonio.franchi}@tuebingen.mpg.de.\n\nC. Secchi is with the Department of Science and Methods of Engineering, University of Modena and Reggio Emilia, Italy cristian.secchi@unimore.it\n\nH. H. Bülthoff is with the Max Planck Institute for Biological Cybernetics, Spemannstraße 38, 72076 Tübingen, Germany, and with the Department of Brain and Cognitive Engineering, Korea University, Anam-dong, Seongbuk-gu, Seoul, 136-713 Koreahhb@tuebingen.mpg.de.\n\nintervention in dangerous or inaccessible areas are the most promising applications. We refer the reader to Murray (2006) for a survey and to Howard et al. (2006); Franchi et al. (2009); Schwager et al. (2011); Renzaglia et al. (2012) for examples of multi-robot exploration, coverage and surveillance tasks.\n\nIn any multi-robot application, a typical requirement when devising motion controllers is to rely on only relative\n\nmeasure-ments w.r.t. other robots or the environment, as for example\n\nrelative distances, bearings or positions. In fact, these can be usually obtained from direct onboard sensing, and are thus free from the presence of global localization modules such as GPS or SLAM algorithms (see, e.g., Durham et al. (2012)), or other forms of centralized localization systems. Similarly, when exploiting a communication medium in order to ex-change information across robots (e.g., by dispatching data via radio signals), decentralized solutions requiring only local and 1-hop information are always preferred because of their higher tolerance to faults and inherent lower communication load Murray (2006); Leonard and Fiorelli (2001).\n\nIn all these cases, properly modeling the ability of each robot to sense and/or communicate with surrounding robots and environment is a fundamental and necessary step. Graph\n\ntheory, in this sense, has provided an abstract but effective\n\nset of theoretical tools for fulfilling this need in a compact way: presence of an edge among pairs of agents represents their ability to interact, i.e., to exchange (by direct sensing and/or communication) those quantities needed to implement their local control actions. Several properties of the interaction graph, in particular of its topology, have direct consequences on the convergence and performance of controllers for multi-robot applications. Among them, connectivity of the graph is perhaps the most ‘fundamental requirement’ in order to allow a group of robots accomplishing common goals by means of de-centralized solutions (examples in this sense are given by con-sensus Olfati-Saber et al. (2007), rendezvous Martinez et al. (2007), flocking Olfati-Saber (2006), leader-follower Mariot-tini et al. (2009), and similar cooperative tasks). In fact, graph connectivity ensures the needed continuity in the data flow among all the robots in the group which, over time, makes it possible to share and distribute the needed information.\n\nThe importance of maintaining connectivity of the interac-tion graph during task execuinterac-tion has motivated a large number of works over the last years. Broadly speaking, in literature two classes of connectivity maintenance approaches are present: i) the conservative methods, which aim at preserving the initial (connected) graph topology during the task, and ii) the flexible approaches, which allow to switch anytime among any of the connected topologies. These usually produce local control actions aimed at optimizing over time some measure\n\n(3)\n\nof the degree of connectivity of the graph, such as the well-known quantity λ2, the second smallest eigenvalue of the\n\ngraph Laplacian Fiedler (1973).\n\nWithin the first class of conservative solutions, the ap-proach detailed in Ji and Egerstedt (2007) considers an inter-robot sensing model based on maximum range, and a similar situation is addressed in Dimarogonas and Kyriakopoulos (2008) where, however, the possibility of permanently adding edges over time is also included. In Stump et al. (2011), inter-robot visibility is also taken into account as criterium for determining the neighboring condition, and a centralized solution for a given known (and fixed) topology of the group is proposed. Finally, a probabilistic approach for optimizing the multi-hop communication quality from a transmitting node to a receiving node over a given line topology is detailed in Yan and Mostofi (2012).\n\nAmong the second class of more flexible approaches, the authors of Kim and Mesbahi (2006) propose a centralized method to optimally place a set of robots in an obstacle-free environment and with maximum range constraints in order to realize a given value ofλ2, i.e., of the degree of connectivity\n\nof the resulting interaction graph. A similar objective is also pursued in De Gennaro and Jadbabaie (2006) but by devising a decentralized solution. In Zavlanos and Pappas (2007), the authors develop a centralized feedback controller based on artificial potential fields in order to maintain connectivity of the group (with only maximum range constraints) and to avoid inter-robot collisions. An extension is also presented in Zavlanos et al. (2009) for achieving velocity synchroniza-tion while maintaining connectivity under the usual maximum range constraints. Another decentralized approach based on a gradient-like controller aimed at maximizing the value of λ2 over time is developed in Yang et al. (2010) by including\n\nmaximum range constraints, but without considering obstacle or inter-robot collision avoidance. In Antonelli et al. (2005, 2006), the authors address the problem of controlling the motion of a Mobile Ad-hoc NETwork (MANET) in order to maintain a communication link between a fixed base station and a mobile robot via a group of mobile antennas. Maximum range constraints and obstacle avoidance are taken into account, and a centralized solution for the case of a given (fixed) line topology for the antennas is developed. Finally, in Stump et al. (2008) a similar problem is addressed by resorting to a centralized solution and by considering maximum range constraints and obstacle avoidance. However, connectivity maintenance is not guaranteed at all times.\n\nWith respect to this state-of-the-art, the goal of this pa-per is to extend and generalize the latter class of methods maintaining connectivity in a flexible way, i.e., by allowing complete freedom for the graph topology as long as con-nectivity is preserved. Specifically, we aim for the following features:(i) possibility of considering complex sensing models determining the neighboring condition besides the sole (and usual) maximum range (e.g., including non-obstructed visi-bility because of occlusions by obstacles), (ii) possibility to embed into a unique connectivity preserving action a number of additional desired behaviors for the robot group such as formation control or inter-robot and obstacle collision\n\navoid-ance, (iii) possibility to establish or lose inter-agent links at any time and also concurrently as long as global connectivity is preserved, (iv) possibility to execute additional exogenous tasks besides the sole connectivity maintenance action such as, e.g., exploration, coverage, patrolling, and finally (v) a fully decentralized design for the connectivity maintenance action implemented by the robots.\n\nThe rest of the paper is structured as follows: Sect. II illustrates our approach (and its underlying motivations) and introduces the concept of Generalized Connectivity, which is central for the rest of the developments. This is then further detailed in Sect. III where the design of a possible inter-robot sensing model and of desired group behaviors is de-scribed. Section IV then focuses on the proposed connectivity preserving control action, by highlighting its decentralized structure and by characterizing the stability of the overall group behavior in closed-loop. As a case study of the proposed machinery, Section V presents an application involving a bilateral shared control task between two human operators and a group of mobile robots navigating in a cluttered environment, and bound to follow the operator motion commands while preserving connectivity of the group at all times. Simulation results obtained with a heterogeneous group of UAVs (quadro-tors) and UGVs (differentially driven wheeled robots), and experimental results obtained with a group of quadrotor UAVs are then reported in Sect. VI, and Sect. VII concludes the paper and discusses future directions.\n\nThroughout the rest of the paper, we will make extensive use of the port-Hamiltonian formalism for modeling and design purposes, and of passivity theory for drawing conclusions about closed-loop stability of the group motion. In fact, in our opinion the use of these and related energy-based arguments provides a powerful and elegant approach for the analysis and control design of multi-robot applications. The reader is referred to Secchi et al. (2007); Duindam et al. (2009) for an introduction to port-Hamiltonian modeling and control of robotic systems, and to Franchi et al. (2011); Robuffo Giordano et al. (2011b,a); Franchi et al. (2012b); Secchi et al. (2012) for a collection of previous works sharing the same theoretical background with the present one. In particular, part of the material developed hereafter has been preliminarily presented in Robuffo Giordano et al. (2011a).\n\nII. GENERALIZEDCONNECTIVITY A. Preliminaries and Notation\n\nIn the following, the symbol 1N will denote a vector of\n\nall ones of dimension N , and similarly 0N for a vector of\n\nall zeros. The symbolIN will represent the identity matrix of\n\ndimension N , and the operator ⊗ will denote the Kronecker\n\nproduct among matrixes. For the reader’s convenience, we\n\nwill provide here a short introduction to some aspects of graph theory pertinent to our work. For a more comprehensive treatment, we refer the interested reader to any of the existing books on this topic, for instance Mesbahi and Egerstedt (2010). Let G = (V, E) be an undirected graph with vertex set V = {1 . . . N} and edge set E ⊂ (V × V)/ ∼, where ∼ is the equivalence relation identifying the pairs(i, j) and (j, i). El-ements inE encode the adjacency relationship among vertexes\n\n(4)\n\nof the graph: [(i, j)]∈ E iff agents i and j are considered as\n\nneighborsor as adjacent1. We assume[(i, i)] /\n\n∈ E, ∀i ∈ V (no self-loops), and also take by convention (i, j), i < j, as the representative element of the equivalence class[(i, j)]. Several matrixes can be associated to graphs and, symmetrically, several graph-related properties can be represented by matrix-related quantities. For our goals, we will mainly rely on the\n\nadjacency matrixA, the incidence matrix E, and the Laplacian\n\nmatrix L.\n\nThe adjacency matrix A ∈ RN ×N is a square symmetric\n\nmatrix with elementsAij ≥ 0 such that Aij= 0 if (i, j) /∈ E\n\nandAij > 0 otherwise (in particular, Aii= 0 by construction).\n\nAs for the incidence matrix, we consider a slight variation from its standard definition. Let\n\nE∗={(1, 2), (1, 3) . . . (1, N) . . . (N − 1, N)}\n\n={e1, e2. . . eN −1. . . , eN (N −1)/2} (1)\n\nbe the set of all the possible representative elements of the equivalence classes in (V × V)/ ∼, i.e., all the vertex pairs (i, j) such that i < j, sorted in lexicographical order. We defineE∈ R|V|×|E∗|\n\nsuch that,∀ek = (i, j)∈ E∗,Eik=−1\n\nand Ejk = 1, if ek ∈ E, and Eik = Ejk = 0 otherwise.\n\nIn short, this definition yields a ‘larger’ incidence matrix E accounting for all the possible representative edges listed in E∗ but with columns of all zeros in presence of those edges\n\nnot belonging to the actual edge setE.\n\nThe Laplacian matrix L ∈ RN ×N is a square positive\n\nsemi-definite symmetric matrix defined as L = diag(δi)− A\n\nwith δi = PNj=1Aij, or, equivalently, as L = EET. The\n\nLaplacian matrix L encodes some fundamental properties of its associated graph which will be heavily exploited in the following developments. Specifically, owing to its symmetry and positive semi-definiteness, all the N eigenvalues of L are real and non-negative. Second, by ordering them in ascending order0≤ λ1≤ λ2≤ . . . ≤ λN, one can show that:(i) λ1= 0\n\nby construction, and (ii) λ2 > 0 if the graphG is connected\n\nand λ2 = 0 otherwise. The second smallest eigenvalue λ2\n\nis then usually referred to as the ‘connectivity eigenvalue’ or\n\nFiedler eigenvalue Fiedler (1973).\n\nFinally, we let νi∈ RN represent the normalized eigenvec-torof the LaplacianL associated to λi, i.e., a vector satisfying\n\nνT\n\ni νi = 1 and λi = νiTLνi. Owing to the properties of the\n\nLaplacian matrix, it is ν1 = 1N/\n\nN and νT\n\ni νj = 0, i6= j.\n\nThe eigenvectorν2associated toλ2 will be denoted hereafter\n\nas the ‘connectivity eigenvector’.\n\nB. Definition of Generalized Connectivity\n\nConsider a system made ofN agents: presence of an inter-action link among a pair of agents (i, j) is usually modeled by setting the corresponding elements Aij = Aji={0, 1} in\n\nthe adjacency matrixA, with Aij = Aji= 0 if no information\n\ncan be exchanged at all, and Aij = Aji= 1 otherwise. This\n\nidea can be easily extended to explicitly consider more so-phisticated agent sensing/communication models representing the actual (physical) ability to exchange mutual information\n\n1This loose definition will be refined later on.\n\nbecause of the agent relative state. For illustration, letxi∈ R3\n\ndenote the i-th robot position and assume an environment modeled as a collection of obstacle pointsO = {ok ∈ R3}. An\n\ninter-robot sensing/communication model is any sufficiently smooth scalar function γij(xi, xj,O) ≥ 0 measuring the\n\n‘quality’ of the mutual information exchange, with γij = 0\n\nif no exchange is possible and γij > 0 otherwise (the larger\n\nγij the better the quality). Common examples are:\n\nProximity sensing model: assume agentsi and j are able to interact iff kxi− xjk < D, with D > 0 being a suitable\n\nsensing/communication maximum range. For example, if radio signals are employed to deliver messages, there typically exists a maximum range beyond which no signal can be reliably dispatched. In this case γij does not depend on surrounding\n\nobstacles and can be defined as any sufficiently smooth function such that γij(xi, xj) > 0 for kxi− xjk < D and\n\nγij(xi, xj) = 0 forkxi− xjk ≥ D.\n\nProximity-visibility sensing model: letSij be the segment\n\n(line-of-sight) joining xi andxj. Agents i and j are able to\n\ninteract iffkxi− xjk < D and\n\nkσxj+ (1− σ)xi− okk > Dvis, ∀σ ∈ [0, 1], ∀ok∈ O,\n\nwith Dvis > 0 being a minimum visibility range, i.e., a\n\nminimum clearance between all the points on Sij and any\n\nclose obstacle ok. In this case, γij(xi, xj, ok) = 0 as either\n\nthe maximum range is exceeded (kxi− xjk ≥ D) or\n\nline-of-sight visibility is lost (kσxj+(1−σ)xi−okk ≤ Dvis for some\n\nok andσ), while γij(xi, xj, ok) > 0 otherwise. Examples of\n\nthis situation can occur when onboard cameras are the source of position feedback, so that maximum range and occlusions because of obstacles hinder the ability to sense surrounding robots.\n\nClearly, more complex situations involving specific models of onboard sensors (e.g., antenna directionality or limited field of view) can be taken into account by suitably shaping the functions γij. Probabilistic extensions accounting for\n\nstochastic properties of the adopted sensors/communication medium as, for instance, transmission error rates, can also be considered, see, e.g., Yan and Mostofi (2012).\n\nOnce functions γij have been chosen, one can exploit\n\nthem as weights on the inter-agent links, i.e., by setting in the adjacency matrix Aij = γij. This way, the value of\n\nλ2 becomes a (smooth) measure of the graph connectivity\n\nand, in particular, a (smooth) function of the system state (e.g., of the agent and obstacle relative positions). Second, and consequently, it becomes conceivable to devise (local) gradient-like controllers aimed at either maximizing the value of λ2 over time, or at just ensuring a minimum level of\n\nconnectivity λ2 ≥ λmin2 > 0 for the graph G, while, for\n\ninstance, the robots are performing additional tasks of interest for which connectivity maintenance is a necessary require-ment. This approach has been investigated in the past literature especially for the proximity sensing model case: see, among the others, Stump et al. (2008); Sabattini et al. (2011); Kim and Mesbahi (2006); De Gennaro and Jadbabaie (2006); Zavlanos and Pappas (2007); Yang et al. (2010).\n\nOne of the contributions of this work is the extension of these ideas to not only embed in Aij the physical quality of\n\n(5)\n\nthe interaction among pairs of robots (the sensing model), but to also encode a number of additional inter-agent be-haviors and constraints to be fulfilled by the group as a whole. This is achieved by designing the weights Aij so\n\nthat the interaction graph G is forced to decrease its degree of connectivity whenever: (i) any two agents lose ability to physically exchange information as per their sensing model γij, and (ii) any of the existing inter-agent behaviors or\n\nconstraints is not met with the required accuracy (and, in this case, even though the agents could still be able to interact from a pure sensing/communication standpoint). By then designing a gradient-like controller built on top of the unique scalar\n\nquantity λ2, and by exploiting the monotonic relationship\n\nbetween λ2 and the weights Aij Yang et al. (2010), we are\n\nable to simultaneously optimize:(i) as customary, the physical connectivity of the graph, i.e., that due to the inter-agent sensing model γij, and (ii) additional individual or group\n\nrequirements, such as, e.g., obstacle avoidance or formation control.\n\nSpecifically, we propose to augment the previous definition of the weightsAij= γij as follows:\n\nAij = αijβijγij. (2)\n\nThe weight βij ≥ 0 is meant to account for additional\n\ninter-agent soft requirements that should be preferably realized by the individual pair(i, j) (e.g., for formation control purposes, βij(dij) could have a unique maximum at some desired\n\ninter-distance dij = d0 andβij(dij)→ 0 as dij deviates too much\n\nfrom d0). Failure in complying with βij will lead to a\n\ndis-connected edge (i, j) and to a corresponding decrease of λ2,\n\nbut will not (in general) result in a global loss of connectivity for the graph G. The weight αij ≥ 0 is meant to represent\n\nhard requirementsthat must be necessarily satisfied by agentsi\n\norj with some desired accuracy. A straightforward example is obstacle or inter-agent collision avoidance: whatever the task, collisions with obstacles or other agents must be mandatorily avoided. Considering agent i, in our framework this will be achieved by letting αij → 0, ∀j ∈ 1 . . . N, whenever\n\nany of such behaviors is not sufficiently met, e.g., when the distance of agenti to an obstacle becomes smaller than some safety threshold. Failure in complying with a hard requirement will then result in a null i-th row (and i-th column) in the adjacency matrix A, necessarily leading to a disconnected graph (λ2 → 0). Ensuring graph connectivity (λ2 > 0) at\n\nall times will then automatically enforce fulfillment of all the mandatory behaviors encoded within αij.\n\nWe finally note that, as it will be clear in the following, all the individual weights in (2) will be designed as sufficiently smooth functions of the agent and obstacle relative positions. This will ultimately make it possible for any agent to im-plement a decentralized gradient controller aimed at keeping λ2 > 0 during motion and, thus, as explained before, at\n\nrealizing all the desired behaviors and at complying with all the existing constraints. Motivated by these considerations, we then speak about the concept of Generalized Connectivity\n\nMaintenance throughout the rest of the paper, to reflect the\n\ngeneralized role played by the value of λ2 in our context\n\nbe-sides representing the sole (and usual) sensing/communication\n\n0.2 0.4 0.6 0.8 1 1.2 1.4 0 100 200 300 400 500 600 700 λ2 V λ( λ2 )\n\nFig. 1: An illustrative shape for Vλ\n\n2) ≥ 0 with λ min\n\n2 = 0.2 and\n\nλmax\n\n2 = 1. The shape of Vλ(λ2) is chosen such that Vλ(λ2) → ∞\n\nas λ2→ λ min\n\n2 , Vλ(λ2) → 0 (with vanishing slope) as λ2→ λ max\n\n2 ,\n\nand Vλ(λ2) ≡ 0 for λ2≥ λ max\n\n2 .\n\nconnectivity of the interaction graphG. C. Generalized Connectivity Potential\n\nIn order to devise gradient-like controllers based onλ2, we\n\ninformally introduce the concept of Generalized Connectivity\n\nPotential, that is, a scalar functionVλ\n\n2)≥ 0 in the domain\n\n(λmin\n\n2 ,∞) such that Vλ(λ2)→ ∞ as λ2 → λmin2 > 0, and\n\n2)≡ 0 if λ2 ≥ λmax2 > λ2min, with λmax2 > λmin2 > 0\n\nrepresenting desired maximum and minimum values for λ2.\n\nThe potentialVλ\n\n2) is required to beC1 over its domain, in\n\nparticular atλmax\n\n2 . Figure 1 shows a possible shape ofVλ(λ2).\n\nFor the sake of illustration, let againxi∈ R3represent the\n\nposition of thei-th agent and x = (xT\n\n1. . . xTN)T. Assume also\n\nthat the weightsAijin (2) are designed as sufficiently smooth\n\nfunctions of the agent and obstacle positions, so that λ2 =\n\nλ2(x, O) is also sufficiently smooth. From a conceptual point\n\nof view, minimization ofVλ\n\n2(x,O)) can then be achieved\n\nby letting every agenti implement the gradient controller Fiλ(x,O) = −\n\n∂Vλ\n\n2(x,O))\n\n∂xi\n\n(3) which will be denoted as the Generalized Connectivity Force. We note that in generalFλ\n\ni (x, O) would depend on the state\n\nof all the agents and obstacle points, thus requiring some form of centralization for its evaluation by means of agent i. However, the next sections will show that our design of Fλ\n\ni(x, O) actually exhibits a decentralized structure, so that\n\nits evaluation by agenti can be performed by only relying on local and 1-hop information. This important feature will then form the basis for a fully decentralized implementation of our approach.\n\nWe also note that, while following the gradient force Fλ\n\ni(x, O), the agents will not be bound to keep a given\n\nfixed topology (i.e. a constant edge setE) for the interaction\n\ngraphG. Creation or deletion of single or multiple links (also concurrently) will be fully permitted as long as the current\n\n(6)\n\nvalue of the generalized connectivity does not fall below a minimum threshold, i.e., while ensuring thatλ2> λmin2 . The\n\nstability issues arising when controlling the agent motion by means of the proposed generalized connectivity force will also be thoroughly analyzed and discussed in the following developments.\n\nIII. DESIGN OF THEGROUPBEHAVIOR\n\nAfter the general overview given in the previous Section, we will now proceed to a more detailed illustration of our approach. Specifically, this Section will focus on the modeling assumptions for the group of agents considered in this work, and on the shaping of the weights in (2). The next Sect. IV will then address the design of the control action Fλ\n\ni and discuss\n\nthe stability of the resulting closed-loop system. A. Agent Model\n\nConsider a group of N agents modeled as floating masses in R3and coupled by means of suitable inter-agent forces. Ex-ploiting the port-Hamiltonian modeling formalism, we model each agent i as an element storing kinetic energy\n\n ˙pi= Fiλ+ Fie− BiMi−1pi vi= ∂Ki ∂pi = Mi−1pi i = 1, . . . , N (4)\n\nwhere pi ∈ R3 and Mi ∈ R3×3 are the momentum\n\nand positive definite inertia matrix of agent i, respectively, Ki(pi) = 12pTiMi−1pi is the kinetic energy stored by the\n\nagent during its motion, and Bi ∈ R3×3 is a positive definite\n\nmatrix representing a velocity damping term (this can be either artificially introduced, or representative of phenomena such as fluid drag or viscous friction). The force input Fλ\n\ni ∈ R3\n\nrepresents the Generalized Connectivity Force, i.e., the in-teraction of agent i with the other agents and surrounding environment. Force Fie ∈ R3, on the other hand, is an\n\nadditional input that can be exploited for implementing other tasks of interest besides the sole Generalized Connectivity maintenance action2. Finally, v\n\ni ∈ R3 is the velocity of the\n\nagent and xi ∈ R3 its position, with ˙xi = vi. Following\n\nthe port-Hamiltonian terminology, the pair (vi, Fiλ + Fie)\n\nrepresents the power port by which agent i can exchange energy with other agents and the environment.\n\nWe note that the dynamics of (4) is purposely kept simple (linear dynamics) for the sake of exposition clarity. In fact, as it will be clear later on, the only fundamental requirement of model (4) is its output strict passivity w.r.t. the pair (vi, Fiλ+Fie) with storage function the kinetic energyKi(pi).\n\nThis requirement, trivially met by system (4), would never-theless hold for more complex (also nonlinear) mechanical systems Sabattini et al. (2012), thus allowing a straightforward extension of the proposed analysis to more general cases. Being this true, we believe model (4) represents a sufficient compromise between modeling complexity and representation power.\n\n2In fact, in Sect. V we will show how to use inputs Fe\n\ni in order to steer\n\nthe overall group motion while preserving connectivity of the group.\n\nRemark 1: Another alternative to more complex agent\n\nmod-eling is to make use of suitable low-level motion controllers able to track the Cartesian trajectory generated by (4) with negligible tracking errors. Devising closed-loop controllers for exact tracking of the trajectories generated by (4) is always possible for all those systems whose Cartesian position is part of the flat outputs Fliess et al. (1995), i.e., outputs algebraically defining, with their derivatives, the state and the control inputs of the system. Many mobile robots, including nonholonomic ground robots or quadrotor UAVs, satisfy this property Murray et al. (1995); Mistler et al. (2001), and this approach has proven successful in several previous works, see, e.g., Michael and Kumar (2009) for unicycle-like robots and Robuffo Giordano et al. (2011b); Franchi et al. (2012b) for quadrotor UAVs.\n\nB. Inter-Agent Requirements\n\nIn view of the next developments, we provide the following two neighboring definitions:\n\nDefinition 1 (Sensing-Neighbors): For an agenti, we define\n\nSi={j| γij6= 0}\n\nto be the set of sensing-neighbors, i.e., those agents with whom agenti could physically exchange information according to the sensing model γij.\n\nDefinition 2 (Neighbors): For an agent i, we define\n\nNi={j| Aij 6= 0}\n\nto be the (usual) set of neighbors, i.e., those agents logically considered as neighbors as per the entries of the adjacency matrixA.\n\nObviously,Ni⊆ Si butSi 6⊂ Ni.\n\nThe following three requirements specify the properties of the Generalized Connectivity adopted in the rest of the work: R1) two agents are able to communicate and to measure their relative position iff(i) their relative distance is less than D ∈ R+ (the communication/sensing range), and (ii)\n\ntheir line-of-sight is not occluded by an obstacle. This requirement defines the sensing model (function γij) of\n\nthe agents in the group which will be used for building weights (2). This requirement also defines the set Si of\n\nsensing-neighbors of agenti (Definition 1);\n\nR2) two agents, when able to exchange information (γij >\n\n0), should keep a preferred interdistance 0 < d0 < D\n\nin order to obtain an overall cohesive behavior for the group motion. This plays the role of a soft requirement for formation control and its fulfillment will be embedded into the weights βij in (2);\n\nR3) any agent must avoid collisions by keeping the minimum safe distances 0 < domin < D and 0 < dmin < D from\n\nsurrounding obstacles and agents, respectively. This plays the role of a hard requirement and its fulfillment will be embedded into the weightsαij in (2).\n\n(7)\n\nxi xj vi vj obstacle point Sij sijk dijk xi,ok xj,ok ok2 Oij\n\nFig. 2: Illustration of several quantities of interest relative to a pair of agents i and j. The agent positions xiand xj, and their velocities\n\nvi and vj. The segment Sij (line-of-sight) joining agents i and j.\n\nAn obstacle point okand the corresponding closest point sijkon the\n\nsegment itself. The segment-obstacle distance dijk.\n\nC. Weight definition\n\nWe will now proceed to shape the individual weights (αij, βij, γij) encoding the requirements listed in R1–R3.\n\nTo this end, consider an environment consisting of a set of obstacle points O = {ok ∈ R3} with cardinality Nobs, and\n\nassume that an agent can measure its relative position w.r.t. the surrounding obstacles located within the sensing range D. Let Oi collect all the obstacle points sensed by agent i and\n\ndefineOij = Oi∪ Oj. BeingSij the segment (line-of-sight)\n\njoining agents i and j, for any ok ∈ Oij we denote with\n\nsijk ∈ R3 the closest point on Sij to the obstacle point ok,\n\nand withdijk∈ R the associated point-line distance3. We also\n\nlet dij = kxi− xjk represent the distance between agents i\n\nandj. Figure 2 summarizes the quantities of interest.\n\n1) Requirement R1: we define γij= γija(dij) Y ok∈Oij γb ij(dijk). (5) The weight γa\n\nij(dij) takes into account the maximum range\n\nconstraint and is chosen to stay constant at a maximum value ka\n\nγ > 0 for 0 ≤ dij ≤ d1 < D and to smoothly vanish (with\n\nvanishing derivative) when dij → D. To this end, we choose\n\nthe following function\n\nγija(dij) = ka γ 0≤ dij ≤ d1 ka γ 2 (1 + cos(µadij+ νa)) d1< dij ≤ D 0 dij > D (6) withµa= D−dπ 1, andνa =−µad1. Figure 3a shows the shape\n\nof a possibleγa ij(dij).\n\nThe individual weights γb\n\nij(dijk) composing the product\n\nsequence in (5) take into account the constraint of line-of-sight occlusion. Assume a minimum and maximum distance 0 ≤ do\n\nmin < domax ≤ D between the segment Sij and\n\nan obstacle point ok ∈ Oij are chosen. The quantity domin\n\nrepresents the minimum distance to an obstacle in order to\n\n3This is formally defined as d\n\nijk =\n\nk(ok− xj) × (ok− xi)k\n\nkxj− xik\n\nif sijk\n\nfalls within the boundaries of the segment Sij, and as dijk= kok− xik if\n\nsijk= xi(resp. xj). 0 1 2 3 4 5 6 7 0 0.5 1 1.5 dij γ a ij(d ij ) (a) 0 1 2 3 4 5 6 7 0 0.5 1 1.5 dijk γ b ij(d ij k ) (b)\n\nFig. 3: The shape of γija(dij) for d1 = 5, D = 6, kaγ = 1 (a) and\n\nγijb(dijk) for domin= 1, d o max= 3, k b γ= 1 (b). 0 1 2 3 4 5 6 7 8 0 0.5 1 1.5 dij βij (d ij )\n\nFig. 4: The shape of βij(dij) for d0= 4, kβ= 1 and σ = 5.\n\navoid occlusion, anddo\n\nmaxthe obstacle range of influence. The\n\nweightγb\n\nij(dijk) is then defined to stay constant at a maximum\n\nvalue kb\n\nγ > 0 for dijk ≥ domax and to smoothly vanish (with\n\nvanishing derivative) when dijk→ domin. Similarly to before,\n\nwe adopted the following function\n\nγijb(dijk) = 0 dijk≤ domin kb γ 2 (1− cos(µbdijk+ νb)) d o\n\nmin< dijk≤ domax\n\nkb\n\nγ dijk> domax\n\n(7) withµb= do π\n\nmax−domin andνb=−µbd\n\no\n\nmin. Figure 3b shows the\n\nshape of a possibleγb ij(dijk).\n\nIt is then clear that, owing to these definitions and to the structure in (5), the composite weight γij → 0 (and,\n\nconsequently, the total weight Aij → 0 in (2)) whenever dij\n\ngrows too large ordijk, for anyok∈ Oij, becomes too small,\n\nthus forcing the disconnection of the link among agentsi and j as dictated by the adopted sensing model (conditions in R1). We also note that γij = γjisincedij = dji anddijk= djik.\n\n2) Requirement R2: in order to cope with R2, we define the\n\nweightβij(dij) as a smooth function having a unique\n\nmaxi-mum atdij = d0 and smoothly vanishing as|dij− d0| → ∞.\n\nTo this end, we take\n\nβij(dij) = kβe−\n\n(dij −d0)2\n\nσ (8)\n\nwithkβ > 0 and σ > 0, and show in Fig. 4 a representative\n\n(8)\n\n0 1 2 3 4 5 6 7 0 0.5 1 1.5 dij α ∗(dij ij )\n\nFig. 5: The shape of α∗\n\nij(dij) for dmin= 0.5, dmax= 4 and kα= 1.\n\n3) Requirement R3: as last case, we consider the collision\n\navoidance requirements of R3. We first deal with the\n\ninter-agentcollision avoidance: let0 ≤ dmin< dmax≤ D represent\n\na minimum safe distance and a maximum range of influence among the agents, and consider a weight function α∗\n\nij(dij)\n\nbeing constant at a maximum valuekα> 0 for dij ≥ dmaxand\n\nsmoothly vanishing (with vanishing derivative) when dij →\n\ndmin. For αij∗(dij) we take the expression (equivalent to the\n\nweights γijb in (7)): α∗ij(dij) = 0 dij ≤ dmin kα\n\n2 (1− cos(µαdij+ να)) dmin< dij ≤ dmax\n\nkα dij > dmax\n\n(9) withµα= dmaxπ−dmin andνα=−µαdmin, and show in Fig. 5\n\na possible shape. As before, it isα∗ ij= α∗ji.\n\nThe weightα∗\n\nij(dij) is designed to vanish as the agent pair\n\n(i, j) gets closer than the safe distance dmin. In order to obtain\n\nthe result discussed in Sect. II-B, i.e., to force disconnection of the graphG as agent i gets too close to any agent, we define the total weight αij in (2) as\n\nαij = Y k∈Si α∗ik ! · Y k∈Sj/{i} α∗jk = αi· αj/i. (10)\n\nThis choice is motivated as follows: the first product sequence in (10)\n\nαi=\n\nY\n\nk∈Si\n\nα∗ik (11)\n\nmakes it possible for αij → 0 as any of the sensed agents\n\nin Si gets closer than dmin to agent i. The second product\n\nsequence in (10)\n\nαj/i=\n\nY\n\nk∈Sj/{i}\n\nαjk∗ (12)\n\nis introduced to enforce the ‘symmetry condition’αij = αji:\n\ntogether with the previous βij = βji and γij = γji, this\n\nguaranteesAij= Aji (see (2)) so that, eventually, the overall\n\nadjacency matrix A stays symmetric as required. Finally, we note that, by construction, the very same term αi will be\n\npresent in all the weightsαij,∀j ∈ Si. This allows to obtain\n\n1 2 3 4 5 α15∗ α∗25 α∗24 α∗45 α∗34 α∗35\n\nFig. 6: An illustrative example of the use of the weights α∗\n\nij in a\n\nGraph G with N = 5 agents and with the edges representing the sensing-neighbor condition of Definition 1 (setsSi).\n\nthe desired effect: as any sensed agent inSi gets too close to\n\nagent i, the term αi → 0 thus forcing the whole i-th row of\n\nmatrixA to vanish, leading to a disconnected graph G. In order to better explain the design philosophy behind the weights αij, we give an illustrative example. Consider the\n\nsituation depicted in Fig. 6 withN = 5 agents, and with the links representing the neighboring conditions as perSi. Take\n\nagents2 and 5: from (10) we have\n\nα25= (α∗24α∗25)(α∗51α∗53α∗54)\n\nand\n\nα52= (α∗51α∗52α∗53α∗54)(α∗24)\n\nso that, being α∗ij = α∗ji, the overall ‘symmetry condition’\n\nα25= α52 is satisfied.\n\nNow consider the weight among agents2 and 4 α24= (α∗24α∗25)(α∗43α∗45).\n\nWe can readily verify that α24 and α25 share the common\n\nfactorα2= α24∗ α∗25: thus, as agent2 gets closer than dmin to\n\none of its sensing-neighbors (either agent 4 or 5), the whole second row of matrixA will vanish (α24→ 0 and α25→ 0),\n\nforcing disconnection of graph G.\n\nAs for obstacle avoidance, one could replicate the same machinery developed for the inter-agent collision avoidance by defining an additional set of suitable weights leading to a disconnected graph as any agent gets too close to an obstacle point (and this could be further extended for including any additional hard requirement besides the agent/obstacle collision avoidance considered here). In our specific case, however, this step is not necessary thanks to the previously introduced weightsγb\n\nij(dijk) in (5). In fact, with reference to\n\nFig. 2, as an agent i approaches an obstacle ok, the agent\n\npositionxi will eventually become the closest point took for all the inter-agent segments Sij (i.e., links) departing from\n\nxi. Thus, all the product sequences Qok∈Oijγ b\n\nij(dijk) in (5),\n\n∀j ∈ Si, will contain an individual term γijb(dijk)→ 0 and,\n\nagain, the wholei-th row of matrix A will be forced to vanish, leading to a disconnected graphG.\n\nRemark 2: We note that the possibility of exploiting the\n\nalready existing weightsγb\n\n(9)\n\nembedding the hard requirement of obstacle avoidance is only a (very convenient) specificity of the case under consideration. In general, each hard requirement to be executed by the group requires the design of an associated function with properties analogous to the aforementioned weights αij (i.e., forcing\n\ndisconnection of the graph G when the requirement is not sufficiently satisfied).\n\nWe conclude by noting the following properties of weights Aij which will be exploited in the next developments. Using\n\nthe previous definitions ofαij,βij,γij, and noting thatdij=\n\ndji, it is Aij = Aij({dijk|ok ∈ Oij}, {dik|k ∈ Si}, {djk|k ∈ Sj}) implying that ∂Aij ∂dik ≡ 0, ∀k /∈ Si, ∂Aij ∂djk ≡ 0, ∀k /∈ Sj. (13)\n\nFurthermore, it is easy to show that, for a generic relative distance dij, ifj /∈ Si ∂Ahk ∂dij ≡ 0 ∀(h, k) ∈ E∗, (14) while, if j ∈ Si ∂Ahk ∂dij ≡ 0, ∀h 6= i, k 6= j, (15) and ∂Aik ∂dij ≡ 0, ∀k /∈ Si, ∂Ajk ∂dij ≡ 0, ∀k /∈ Sj. (16)\n\nThese latter conditions can be slightly simplified by replacing the sensing-neighbors Si with the (logical) neighbors Ni,\n\nyielding ∂Aik ∂dij ≡ 0, ∀k /∈ Ni, ∂Ajk ∂dij ≡ 0, ∀k /∈ Nj. (17)\n\nIn fact, if k ∈ Si butk /∈ Ni, then not onlyAik= 0 but also\n\n∂Aik/∂dij = 0 thanks to the design of weights (αik, βik, γik)\n\ncomposingAik (vanishing weights with vanishing slope).\n\nFor the reader’s convenience, we finally report in Fig. 7 a graphical representation (3D surface and planar contour plot) of the total weight Aij = αijβijγij as a function of the two\n\nvariables dij and dijk, i.e., assuming presence of only two\n\nagents i and j and of a single obstacle point ok.\n\nIV. CONTROL OF THEGROUPBEHAVIOR\n\nIn this Section we address the design of the Generalized Connectivity ForceFλ\n\ni based on the previous definition of the\n\nweights in (2) and discuss its decentralized structure. Sub-sequently, the passivity properties of the closed-loop system obtained when controlling the motion of agents (4) by means of Fλ\n\ni are also thoroughly analyzed.\n\n(a) dijk dij 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 (b)\n\nFig. 7: Visualization as 3D surface of the total weight Aij =\n\nαijβijγij as a function of the variables dij and dijk (top), and\n\ncorresponding contour plot (bottom). The values of the various parameters are those employed for the previous Figs. 3–5.\n\nA. Inter-Agent Interconnection\n\nWith reference to Fig. 2, let xij = xi− xj∈ R3 represent\n\nthe relative position of agent i w.r.t. agent j. Replicating the lexicographical ordering used for setE∗ in (1), we collect all\n\nthe possible |E∗| relative positions into the cumulative vector\n\nxR= (xT12 . . . xT1N x23T . . . xT2N . . . xTN −1N)T ∈ R 3N (N −1)\n\n2 .\n\nWe also let xi,ok = xi− ok ∈ R\n\n3 be the relative position of\n\nthei-th agent w.r.t. the k-th obstacle point, and xi,o= (xTi,o1. . . x\n\nT\n\ni,oNobs)T ∈ R3Nobs\n\nbe a vector collecting all the relative positions between the i-th agent and the Nobs obstacles. Finally, vector\n\nxO = (xT1,o. . . xTN,o)T ∈ R3N Nobs\n\ncollects all the xi,o for all theN robots.\n\nIn port-Hamiltonian terms, the Generalized Connectivity Potential Vλ\n\n2) can be thought as a ‘nonlinear elastic\n\npotential’ whose internal energy grows unbounded as the graph approaches disconnection (see Fig. 1). Note that, due to the definition of the individual weights (αij, βij, γij) given\n\n(10)\n\nmatrix, and, as a consequence, λ2 and Vλ(λ2) as well,\n\nbecome sufficiently smooth functions of the agent and obstacle relative positions (xR, xO). As explained in Sect. II-C, the\n\nGeneralized Connectivity Force (anti-gradient ofVλ) is then\n\nFiλ= −\n\n∂Vλ\n\n2(xR, xO))\n\n∂xi\n\n. (18)\n\nThis formal expression of Fλ\n\ni can be given the following\n\nstructure: being ∂xij/∂xi = I3 and ∂xi,oj/∂xi = I3, and applying the chain rule, expression (18) can be expanded as\n\nFiλ= − N X j=1, j6=i ∂Vλ ∂xij − Nobs X j=1 ∂Vλ ∂xi,oj . (19)\n\nFurthermore, using the results reported in Yang et al. (2010), the terms in the two summations can be further expanded as\n\n∂Vλ ∂xij = ∂V λ ∂λ2 ∂λ2 ∂xij =∂V λ ∂λ2 X (h, k)∈E ∂Ahk ∂xij (ν2h− ν2k) 2, (20) and ∂Vλ ∂xi,oj =∂V λ ∂λ2 ∂λ2 ∂xi,oj =∂V λ ∂λ2 X (h, k)∈E ∂Ahk ∂xi,oj (ν2h− ν2k) 2, (21) with ν2h being the h-th component of the normalized\n\ncon-nectivity eigenvector ν2. Being dij = kxijk, we can plug\n\nproperty (14) in (20) to conclude that∂Vλ/∂x\n\nij= 0 if j /∈ Si.\n\nTherefore, expression (19) can be simplified into Fiλ= − X j∈Si ∂Vλ ∂xij − Nobs X j=1 ∂Vλ ∂xi,oj . (22)\n\nRemark 3: We note that, formally, Fλ\n\ni in (22) depends on all the Nobs obstacles present in the scene — an unrealistic\n\nassumption in most practical situations. However, as it will be clear in the following developments, only the sensed obstacle points (i.e., only those within the range D) actually contribute to the evaluation ofFλ\n\ni . With this understanding, we\n\nnevertheless keep the expression (22) for the sake of generality. Exploiting the structure of (22), and defining p = (pT\n\n1 . . . pTN)T ∈ R3N,B = diag(Bi) ∈ R3N ×3N, andFe=\n\n(FeT\n\n1 . . . FNeT)T ∈ R3N, and by noting that ˙xij = vi− vjand\n\n(assuming static obstacles) ˙xi,ok = vi, we can finally model the interconnection of the N agents (4) with the Generalized Connectivity Force Fλ as the mechanical system in\n\nport-Hamiltonian form: ˙p ˙xR ˙xO = 0 E −I −ET 0 0 IT 0 0 − B 0 0 0 0 0 0 0 0 ∇H + GFe v= GT∇H . (23) Here, H(p, xR, xO) = N X i=1 Ki(pi) + Vλ(xR, xO) ≥ 0 (24)\n\nrepresents the total energy of the system (Hamiltonian) and\n\n∇H =\u0012 ∂ TH ∂p ∂TH ∂xR ∂TH ∂xO \u0013T . Moreover, I = IN ⊗ 1TNobs ⊗ I3, G = IN⊗ I3 0 0 \u0001T , and E = E ⊗ I3, with E being the incidence matrix of the\n\ngraph G encoding the neighboring condition of Definition 1 (see Sects. II-A and II-B). Finally,v ∈ R3N is the conjugate\n\npower variable associated toFe: the port (Fe, v) allows the\n\nsystem to exchange energy with the external world. We refer again the reader to Franchi et al. (2011); Robuffo Giordano et al. (2011b,a); Franchi et al. (2012b); Secchi et al. (2012) for more detailed illustrations on similar derivations.\n\nB. Decentralized Implementation of\n\ni\n\nIn order to study the decentralized structure of Fλ\n\ni , we\n\nanalyze separately the two summations in its expression (22). We preliminarily assume availability to each agent k of the current value of λ2 and of the k-th component ν2k of the connectivity eigenvector ν2. These assumptions will be later\n\nremoved.\n\nWe start considering the first summationP\n\nj∈Si\n\n∂Vλ\n\n∂xij in (22) with the goal of showing that each individual term∂Vλ/∂x\n\nij,\n\nj ∈ Si, can be computed in a decentralized way by agent i.\n\nBy using properties (15)–(17), we can expand (20) as ∂Vλ ∂xij =∂V λ ∂λ2 X k∈Ni ∂Aik ∂xij (ν2i− ν2k) 2+ + X k∈Nj ∂Ajk ∂xij (ν2j − ν2k) 2∂Aij ∂xij (ν2i− ν2j) 2 , (25) where the last term is meant to account for weight Aij only\n\nonce in the two previous summations. Let us define the vector quantity ηij = X k∈Ni ∂Aik ∂xij (ν2i− ν2k) 2∈ R3 (26)\n\nand thus rewrite (25) as ∂Vλ ∂xij =∂V λ ∂λ2 \u0012 ηij− ηji− ∂Aij ∂xij (ν2i− ν2j) 2 \u0013 . (27) Proposition 1: Vector ∂V λ ∂xij in (27) can be evaluated by agent i, ∀j ∈ Si, in a decentralized way by only resorting\n\nto local and1-hop information from neighboring agents.\n\nProof: We first note that, under the stated assumptions,\n\nthe quantity ∂Vλ/∂λ\n\n2 can be directly computed by agent i\n\nfrom the current value of λ2. We then proceed showing how\n\nvectorηij, whose expression is given in (26), can be evaluated\n\nin a decentralized way by agent i by resorting to only local and1-hop information.\n\nTo this end, we recall thatν2i is assumed locally available, and the components ν2k, k ∈ Ni, can be communicated as single scalar quantities from neighboring agents k to agent i. Consider now the weights Aik = αikβikγik withk ∈ Ni\n\nin (26): as for the termγik= γik(dik, dikh|oh∈Oik), evaluation of the quantitiesdikanddikh,∀oh∈ Oik, requires knowledge\n\nof xi − xk, xi− oh and xk− oh, ∀oh ∈ Oik, i.e., of\n\nrela-tive positions w.r.t. neighboring agents and sensed obstacles. Furthermore, ∂γik/∂xij ≡ 0, ∀k 6= j, while evaluation of\n\n(11)\n\n∂γij/∂xij requires again knowledge of the relative position\n\nxi− xj. Similar considerations hold for the terms βik(dik):\n\nevaluation of βik(dik) requires knowledge of xi− xk, while\n\n∂βik/∂xij = 0, ∀k 6= j, and ∂βij/∂xij can be evaluated from\n\nthe relative position xi− xj.\n\nComing to weights αik, recalling their definition in (10) it\n\nis αik = αiαk/i. Here, we note that αi = αi(dih|h∈Si), so that evaluation of αi and of its gradient w.r.t. xij requires\n\nknowledge of xi − xh, ∀h ∈ Si (again, relative positions\n\nw.r.t. neighbors). From (12), the term αk/i can be locally\n\ncomputed by agent k and communicated to agent i as a\n\nsingle scalar quantity regardless of the cardinality of Sk.\n\nMoreover, sinceαk/i does not dependon xij, it is obviously\n\n∂αk/i/∂xij ≡ 0.\n\nThese considerations allow then to conclude that evaluation ofηij can be performed in a decentralized way by agenti as it\n\nrequires, in addition to the sole relative positions w.r.t. neigh-boring agents and sensed obstacles, the communication of the scalar quantities αk/i and of the (scalar) components ν2k, ∀k ∈ Ni.\n\nFollowing the same arguments, agent j is symmetrically able to compute, in a decentralized way, the second vector quantityηjipresent in (27). This can then be communicated by\n\nagentj to agent i as a single vector quantity regardless of the cardinality of Nj. Finally, the last quantity∂Aij/∂xij(ν2i− ν2j)\n\n2 in (27) is also available to agent i as it is just one of\n\nthe |Ni| terms needed for evaluating ηij, see (26). This then\n\nconcludes the proof: agent i is able to evaluate all the terms in the first summationP\n\nj∈Si\n\n∂Vλ\n\n∂xij in (22) by resorting to only local and 1-hop information.\n\nConsider now the second summation PNobs\n\nj=1 ∂Vλ/∂xi,oj\n\nin (22), with the individual terms ∂Vλ/∂x\n\ni,oj having the expression (21). Exploiting the structure of weights Aij, in\n\nparticular of functions γb\n\nij(dijk) in (5), the following simple\n\nproperties hold\n\n∂Ahl/∂xi,oj ≡ 0, ∀oj, ∀h 6= i, l 6= i, and\n\n∂Aih/∂xi,oj ≡ 0, ∀oj, ∀h /∈ Ni. Therefore, the expression (21) can be simplified into\n\n∂Vλ ∂xi,oj =∂V λ ∂λ2 X k∈Ni ∂Aik ∂xi,oj (ν2i− ν2k) 2. (28) Proposition 2: Vector ∂V λ ∂xi,oj in (28) can be evaluated by agent i, ∀oj, in a decentralized way by only resorting to local\n\nand1-hop information from neighboring agents.\n\nProof: As before, ∂Vλ/∂λ\n\n2, ν2i and ν2k, ∀k ∈ Ni, are locally available to agent i. If oj is a sensed obstacle point,\n\ni.e., there exists at least one agent k ∈ Si such thatoj ∈ Oik,\n\nthen evaluation of ∂Aik/∂xi,oj can be locally performed by agent i with knowledge of the relative positions xi− oj and\n\nxk−oj. If, on the other hand,ojis nota sensed obstacle point,\n\nthen ∂Aih/∂xi,oj ≡ 0, ∀h. This then concludes the proof: agent i can evaluate all the terms in the second summation PNobs\n\nj=1 ∂Vλ/∂xi,oj in (22) by resorting to only local and 1-hop information.\n\nTo summarize, the computation of the Generalized Con-nectivity Force Fλ\n\ni (22) by agent i requires availability of\n\nthe following quantities: (i) the relative positions xi− xj,\n\n∀j ∈ Si, (ii) xi− ok andxj− ok, ∀j ∈ Si, and for all the\n\nsensed obstacle pointsok∈ Oij,(iii) the scalar quantity αj/i,\n\n∀j ∈ Si,(iv) the vector quantity ηji,∀j ∈ Si,(v) the i-th and\n\nj-th components of ν2,∀j ∈ Ni, and(vi) the current value of\n\nλ2. The complexity per neighbor is then O(1), i.e., constant\n\nw.r.t. the total number of agentsN .\n\nWhile, as discussed, most of this information is locally or 1-hop available through direct sensing or communication, this is not usually the case forλ2andν2i,ν2j,j ∈ Ni. Knowledge of these quantities could be obtained by a global observation of the group in order to recover the full LaplacianL so as to compute their values with a centralized procedure. However, in our case, for the sake of decentralization we chose to rely on the decentralized estimation strategy proposed by Yang et al. in Yang et al. (2010) and then refined by Sabattini et al. in Sabattini et al. (2011, 2012). Therein, the authors show how each agenti can incrementally build its own local estimation of λ2, i.e., ˆλ2, and of the i-th component of ν2, i.e., νˆ2i, by again exploiting only local and 1-hop information. We refer the reader to these works for all the details. Therefore, by exploiting these results, we conclude that an estimation ˆFλ i\n\nof the trueFλ\n\ni can be implemented by every agent in a fully\n\ndecentralized way.\n\nRemark 4: It is worth mentioning that the estimation\n\nschemes developed in Yang et al. (2010); Sabattini et al. (2011, 2012) will not return, in general, a normalized eigenvectorν2\n\n(needed to evaluate (20)), but a (non-null) scalar multiple̺ν2\n\nfor some ̺ 6= 0 depending on the chosen gains and on the numberN of robots in the group. This discrepancy, however, does not constitute an issue since evaluation of (20) on a multiple ̺ν2 of ν2 will just result in a scaled version of the\n\nConnectivity Force ̺2Fλ\n\ni , ∀i. It is then always possible to\n\nre-define the Connectivity Potential Vλ so as to embed the\n\neffect of any ‘scaling factor’̺2 introduced by the estimation\n\nscheme.\n\nBefore addressing the stability issues of the closed-loop sys-tem (23), we summarize the main features of the Generalized Connectivity PotentialVλ and ofFλ\n\ni introduced so far:\n\n1) although Vλ is a global potential, reflecting global\n\nproperties (connectivity) of the group, ˆFλ\n\ni (an\n\nestima-tion of its gradient w.r.t. the i-th agent position) can be computed in a fully decentralized way. The only discrepancies among the trueFλ\n\ni and ˆFiλ are due to the\n\nuse of the estimates ˆλ2,νˆ2i andνˆ2k,k ∈ Ni, in place of their real values, otherwise ˆFλ\n\ni is evaluated upon actual\n\ninformation;\n\n2) Vλ will grow unbounded as λ\n\n2 → λmin2 > 0, thus\n\nenforcing Generalized Connectivity of the group. Note that, during the motion, the agents are fully allowed to break or create links (also concurrently) as long as λ2 > λmin2 . Furthermore, the group motion will\n\nbecome completely unconstrained wheneverλ2≥ λmax2 ,\n\nsince, in this case, the Generalized Connectivity Force will vanish as the potential Vλ becomes flat. These\n\n(12)\n\ntopology and geometry, as the agent motion is not forced to maintain a particular (given) graph topology, but is instead allowed to execute additional tasks in parallel to the Connectivity Maintenance action.\n\n3) because of the various shapes chosen for the weightsαij,\n\nβij andγij, minimization ofVλwill also enforce all the\n\ninter-agent requirements listed in R1–R3. Specifically, inter-agent and obstacle collisions will be prevented, and any interacting pair (i, j) will try to keep a preferred inter-distanced0, thus ensuring an overall cohesive\n\nbe-havior for the group motion. C. Closed-loop Stability\n\nWe now analyze the stability properties of system (23) by extensive use of passivity arguments4. First of all, thanks to\n\nthe port-Hamiltonian structure of (23), and owing to the lower-boundedness of the total energy H in (24) and to the positive semi-definiteness of matrixB, we obtain\n\n˙ H = ∇TH ˙p ˙xR ˙xO = − ∂TH ∂p B ∂H ∂p + ∇ THGFe≤ vTFe. (29) This would be in general sufficient to conclude passivity of (23) w.r.t. the pair (Fe, v) with storage function H.\n\nHowever, in our case, two additional issues must be taken into account. First of all, the agents are not implementing Fλ\n\ni , the actual gradient of Vλ, but an estimation ˆFiλ of its\n\nreal value. Second, having allowed for a time-varying graph topology G(t) = (V, E(t)) results in a switching incidence matrix E(t) and, as a consequence, in an overall switching dynamics for the closed-loop system (23).\n\nWhile, as well-known, passivity (and stability) of a sys-tem can be threaten by presence of positive jumps in the employed energy function, for the case under consideration the switching nature of E(t) cannot cause discontinuities in Vλ(t) by construction because of the way the weights A\n\nij are\n\ndesigned. This can be easily shown as follows: the weights Aij(xR(t), xO(t)) are smooth functions of the agent/obstacle\n\nrelative positions, so that one can never face the situation of a\n\ndiscontinuityin the value ofAij(assuming the state is evolving\n\nin a continuous way). This, in turn, ensures continuity of λ2(t) and, consequently, of Vλ(λ2(t)) as well despite possible\n\ncreation/deletion of edges in the graph G(t).\n\nRemark 5: For completeness, we also refer the interested\n\nreader to Franchi et al. (2011); Robuffo Giordano et al. (2011b); Franchi et al. (2012b); Secchi et al. (2012): in the context of formation control with time-varying topology, these works share a similar theoretical background with the present one (borrowing tools from port-Hamiltonian modeling and passivity theory) but allow for a more general situation in which discontinuous changes in the arguments of the employed potential function are allowed at the switching times. In these cases, proper passifying actions must indeed be adopted in order to guarantee stability of the resulting closed-loop dynamics.\n\n4In fact, as well-known, passivity guarantees a sufficient condition for\n\ncharacterizing the stability of a dynamical system Sepulchre et al. (1997).\n\nThe rest of the Section is then devoted to deal with the possible non-passive effects arising from the implementation of the estimated ˆFλ\n\ni in place of the real Fiλ. It is in fact\n\nclear that if ˆFλ\n\ni represents a too poor estimation of Fiλ (the\n\nactual gradient ofVλ), the passivity condition (29) will not in\n\ngeneral hold and passivity of the closed-loop dynamics (23) could be lost. In order to cope with this issue, we will resort to a flexible passifying strategy for safely implementing ˆ\n\ni and\n\nensuring passivity of the closed-loop system. To this end, we first introduce the fundamental concept of Energy Tanks: the Energy Tanks are artificial energy storing elements that keep track of the energy naturally dissipated by the agents because of, e.g., their damping factorsBi (see (4)). The energy stored\n\nin these reservoirs can be re-used to accomplish different goals\n\nwithout violating the passivity of the system. A first example\n\nof using such a technique (a controlled energy transfer) can be found in Duindam and Stramigioli (2004), while extensions are proposed in, e.g., Secchi et al. (2006); Franchi et al. (2011); Franken et al. (2011); Franchi et al. (2012b).\n\nConsider a tank with state xti ∈ R and associated energy function Ti = 12x2ti ≥ 0. From Eq. (4), it follows that the power dissipated by agenti because of the damping action is given by\n\nDi= pTiMi−TBiMi−1pi. (30)\n\nWe then propose to adopt the following augmented dynamics for the agents in place of (4):\n\n ˙pi= Fie− wixti− BiM −1 i pi ˙xti= si 1 xtiDi+ wTivi yi= viT xti \u0001T . (31)\n\nThis is motivated as follows: the parameter si ∈ {0, 1} is\n\nexploited to enable/disable the storing of the dissipated power Di (30). If si = 1, all the energy dissipated because of the\n\ndamping Bi is stored back into the tank, and if si = 0 no\n\ndissipated energy is stored back. Storage of the dissipated powerDi is disabled whenTi≥ Tmax, i.e., by choosing\n\nsi=\n\n\u001a\n\n0, if Ti≥ Tmax\n\n1, if Ti< Tmax , (32)\n\nwith Tmax > 0 representing a suitable (and\n\napplication-dependent) upper limit for the tank energy5. The inputw\n\ni∈ R3\n\nis meant to implement, by exploiting the tank energy, a desired force on agenti. In fact, note the absence of Fλ\n\ni in the first row\n\nof (31) compared to (4): the idea is to replace the (unknown) Fλ\n\ni with a passive implementation of its estimation ˆFiλ by\n\nmeans of the new input wi. Use of this input allows for a\n\npower-preserving energy transfer between the tank energyTi\n\nand the kinetic energyKi of agenti, as can be seen from the\n\nfollowing power budget\n\n\u001a ˙ Ti= αiDi+ xtiw T i vi ˙ Ki= viTFie− Di− viTwixti . (33)\n\n5Presence of this safety mechanism is not motivated by theoretical\n\nconsider-ations, but is meant to avoid an excessive energy storage in Tithat would allow\n\nfor implementing practical unstable behaviors in the system, see also Lee and Huang (2010); Franchi et al. (2011, 2012b) for a more thorough discussion.\n\n(13)\n\nThus, any action implemented throughwiwill be intrinsically\n\npassivity-preserving.\n\nTo obtain the sought result, we then set wi= −ςi ˆ Fλ i xti , ςi ∈ {0, 1}, (34)\n\nwhere ςi is a second design parameter that enables/disables\n\nthe implementation of ˆFλ\n\ni . Let0 < Tmin< Tmax represent a\n\nminimum energy levelfor the tankTi. Similarly to before, we\n\nchoose\n\nςi=\n\n\u001a\n\n0, if Ti< Tmin\n\n1, if Ti≥ Tmin . (35)\n\nThus, whenςi = 1, input wiwill implement the desired force\n\nˆ Fλ\n\ni in (31) and, at the same time, extract/inject the appropriate\n\namount of energy from/to the tank reservoir Ti as per (33).\n\nWhen ςi = 0, no force is implemented and no energy is\n\nextracted/injected into the tank. The use of this parameter ςi\n\nis meant to avoid complete depletion of the tank reservoir Ti,\n\nan event that would render (34) singular being, in this case, xti = 0.\n\nLet H be the new total energy (Hamiltonian) of the agent group, also accounting for the new tank energies Ti\n\nH(p, xR, xO, xt) = N X i=1 (Ki(pi) + Ti(xti)) + V λ(x R, xO). (36)\n\nLet also Υ = diag(−wi) ∈ R3N ×N, P =\n\ndiag( 1 xtip T i Mi−T) ∈ RN ×3N, S = diag(si) ∈ RN ×N, and ∇H =\u0012 ∂ TH ∂p ∂TH ∂xR ∂TH ∂xO ∂TH ∂xt \u0013T .\n\nThe augmented closed-loop system (still in port-Hamiltonian form) becomes ˙ p ˙ xR ˙ xO ˙ xt = 0 E −I Υ −ET 0 0 0 IT 0 0 0 −ΥT 0 0 0 − B 0 0 0 0 0 0 0 0 0 0 0 −SP B 0 0 0 ∇H+ +GFe v = GT∇H . (37)\n\nProposition 3: System (37) is passive w.r.t. the storage\n\nfunctionH in (36).\n\nProof: Using (36), the following energy balance easily\n\nfollows: ˙ H = −∂ TH ∂p B ∂H ∂p + ∂TH ∂xt SP B∂H ∂p + v TFe. (38) Exploiting the definitions ofS and P , we have\n\n∂TH ∂xt SP B∂H ∂p = ∂TH ∂xt P (S⊗I3)B ∂H ∂p = ∂TH ∂p (S⊗I3)B ∂H ∂p, sinceS is a diagonal matrix made of {0, 1}. Therefore,\n\n˙ H = −∂ TH ∂p B ∂H ∂p + ∂TH ∂p (S ⊗ I3)B ∂H ∂p + v TFe≤ (39) ≤ ∂ TH ∂p ((S ⊗ I3) − I3N)B ∂H ∂p + v TFe≤ vTFe.\n\nThis result can also be interpreted as follows: the energy stored in the tanks (second term in (39)) is at most equal (with\n\nopposite sign) to the energy dissipated by the agents (first term in (39)), so that passivity is preserved.\n\nD. Concluding Remarks\n\nAs a conclusion of this discussion on the closed-loop passiv-ity of the system, we wish to summarize the results and draw a couple of remarks. We note that the proposed passifying strategy, based on the tank machinery, is very powerful and elegant in the sense that it allows complete freedom on the force to be implemented (i.e., ˆFλ\n\ni through (34) in our case),\n\nas long as the passivity of the system is not compromised in\n\nan integral sense. This is, we believe, a crucial point to be\n\nhighlighted, a point pertaining to all the approaches based on the exploitation of energy tanks for preserving passivity (see, e.g., the ‘two-layer approach’ discussed in detail in Franken et al. (2011)): the augmentation of the system dynamics with the tank state xti makes it possible to exploit to the full extent any passivity margin already present in the system by taking into account the complete past evolution and not only a point-wise (at the current time) condition6. In this sense,\n\nthe tank machinery provides a flexible and integral passivity preserving mechanism. One can argue that, if an action cannot be passively implemented by exploiting the tank reservoirs, then no other mechanism can guarantee passivity of the system by implementing the very same action (and, thus, the action should not be implemented in its form, but should be at least suitably ‘modulated’ to cope with the passivity constraint).\n\nComing to our specific case, as thoroughly discussed, the only source of non-passivity lies in the (possibly poor) esti-mation of λ2 and ν2 leading to a (possibly poor) estimation\n\nˆ Fλ\n\ni of the gradient of the elastic potential Vλ. As such, the\n\nproposed tank passifying strategy will guarantee passivity (and thus stability) of the system but at the possible expense of graph connectivity maintenance: if the estimated ˆFλ\n\ni becomes\n\nso poor that passivity is violated (in the sense explained above, i.e., leading to depletion of the tanks), the switching mecha-nism in (35) will always trade stability for implementation of\n\nˆ Fλ\n\ni.\n\nAlthough conceptually possible, this situation is very un-likely to occur in our setting. In fact, the improved decentral-ized estimation proposed in Sabattini et al. (2011, 2012), and employed in this work, guarantees boundedness of the estima-tion errors of ˆλ2andνˆ2with a predefined accuracy. Therefore,\n\nˆ Fλ\n\ni will never diverge too much over time from the realFiλ.\n\nFurthermore, it is easy to prove (see Appendix B) that a tank reservoir can be set up so as to never deplete when its energy is exploited to implement the port behavior of a passive element, as it is(Fλ\n\ni , vi) with Vλas storage function in our case. Then,\n\nassuming small estimation errors ( ˆFλ\n\ni ≃ Fiλ), it follows that\n\nthe tank energyTiwill not approach the safety valueTminand,\n\nconsequently, ˆFλ\n\ni will be implemented. Indeed,(i) ˆFiλwill be\n\nalmost representing the actual port behavior of(Fλ\n\ni , vi), thus\n\nthat of a passive system, and(ii) any remaining non-passive effects due to small estimation errors can be dominated by the passivity margin of the agent dissipation. In other words, in the\n\n6In our case, the passivity margin is due to the agent dissipation induced\n\nUpdating...\n\n## Références\n\nSujets connexes :"
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http://physicshelpforum.com/thermodynamics-fluid-mechanics/10510-thermodynamics.html | [
"Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum",
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"Oct 2nd 2014, 10:13 AM #1 Junior Member Join Date: Oct 2014 Posts: 2 Thermodynamics Get could anyone help me with an engineering problem- A quantity of oxygen is stored for the air tanks of the submarine and occupies a volume of 5meters cubed. The pressure of the gas is at atmospheric pressure, at an ambient temperature of 4degrees Celsius. What will be the pressure of the gas be if it is compressed to half it's volume and heated to a temperature of 42degrees Celsius. Calculate the mass of the oxygen using the density of oxygen and using the specific heat at constant volume ,calculate the characteristic gas constant and the specific heat capacity at constant pressure ,when Cv=660KJ/KgK. I'd appreciate any help on this question, I have no idea where to start on this question.",
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"Oct 3rd 2014, 08:09 AM #2 Physics Team Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 Use the Ideal gas law: PV=nRT. Here n (number of moles of gas) is constant, as is R, so: PV/T = nR = constant. Thus: P1V1/T1 = P2V2/T2 Where P1, V1, and T1 are the initial values of pressure, volume and temp (in kelvin) and P2, T2, and V2 are the final values. Solve for P2. For the second part, use PV=nRT to solve for n. You can use either initial values of P, V and T or final values - it doesn't matter, just don't mix them up. Then from n calculate total mass of gas. Post back with your answers, and if still having issues with it or the third part we'll help you along.",
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"Oct 4th 2014, 06:11 AM #3 Junior Member Join Date: Oct 2014 Posts: 2 Part 1 PV=nRT PV/T=nR P1V1/T1=P2V2/T2 P1- 101325 Pascal V1- 5m3 T1- 4oc -> 1092K P2- 463.9 V2- 5m3/ 2= 2.5m3 T2- 42oc -> 11 466K Using the equation i will solve the final pressure: 101325x 5/1092= 463.9 x 2.5/ 11 466 Final pressure= 463.9 Part 2 Use-PV=nRT I will be using the final values. Pressure= 463 Pascal 0.0045783370 Atmosphere Volume= 2.5m3 2500 L Temp= 11 466 K n = 0.01216 PV=nRT 463 x 2.5 = 0.01216 x 0.08206 x 11 466 1125.5 = 11.44 1125.5/ 11.44 = 0.1048080 (Mass of gas) Part 3 Specific gas constant- S= R/MW MW= R/S S= 0.08206/ 0.01216 = 6.74835 (Specific gas constant) Thanks for the reply, these are my workings out and i would appriciate any feedback and help you could offer.",
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"Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post ariana Thermodynamics and Fluid Mechanics 0 Oct 1st 2014 02:29 PM abcde Advanced Thermodynamics 2 Sep 12th 2014 02:43 PM KAJ0989 Advanced Thermodynamics 1 Dec 24th 2012 07:44 PM CNM DESIGN Thermodynamics and Fluid Mechanics 0 Aug 22nd 2010 04:35 PM AJEY Thermodynamics and Fluid Mechanics 1 Dec 15th 2009 10:12 PM"
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https://answers.everydaycalculation.com/simplify-fraction/490-224 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Reduce 490/224 to lowest terms\n\nThe simplest form of 490/224 is 35/16.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 490 and 224 is 14\n2. Divide both the numerator and denominator by the GCD\n490 ÷ 14/224 ÷ 14\n3. Reduced fraction: 35/16\nTherefore, 490/224 simplified to lowest terms is 35/16.\n\nMathStep (Works offline)",
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"Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\nEquivalent fractions:\n\nMore fractions:\n\n#### Fractions Simplifier\n\n© everydaycalculation.com"
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https://www.numberempire.com/10092 | [
"Home | Menu | Get Involved | Contact webmaster",
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"0 / 12\n\n# Number 10092\n\nten thousand ninety two\n\n### Properties of the number 10092\n\n Factorization 2 * 2 * 3 * 29 * 29 Divisors 1, 2, 3, 4, 6, 12, 29, 58, 87, 116, 174, 348, 841, 1682, 2523, 3364, 5046, 10092 Count of divisors 18 Sum of divisors 24388 Previous integer 10091 Next integer 10093 Is prime? NO Previous prime 10091 Next prime 10093 10092nd prime 105829 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 10011101101100 Octal 23554 Duodecimal 5a10 Hexadecimal 276c Square 101848464 Square root 100.45894683899 Natural logarithm 9.2194983097609 Decimal logarithm 4.0039772418455 Sine 0.933623239681 Cosine 0.35825639746912 Tangent 2.6060197285422\nNumber 10092 is pronounced ten thousand ninety two. Number 10092 is a composite number. Factors of 10092 are 2 * 2 * 3 * 29 * 29. Number 10092 has 18 divisors: 1, 2, 3, 4, 6, 12, 29, 58, 87, 116, 174, 348, 841, 1682, 2523, 3364, 5046, 10092. Sum of the divisors is 24388. Number 10092 is not a Fibonacci number. It is not a Bell number. Number 10092 is not a Catalan number. Number 10092 is not a regular number (Hamming number). It is a not factorial of any number. Number 10092 is an abundant number and therefore is not a perfect number. Binary numeral for number 10092 is 10011101101100. Octal numeral is 23554. Duodecimal value is 5a10. Hexadecimal representation is 276c. Square of the number 10092 is 101848464. Square root of the number 10092 is 100.45894683899. Natural logarithm of 10092 is 9.2194983097609 Decimal logarithm of the number 10092 is 4.0039772418455 Sine of 10092 is 0.933623239681. Cosine of the number 10092 is 0.35825639746912. Tangent of the number 10092 is 2.6060197285422\n\n### Number properties\n\n0 / 12\nExamples: 3628800, 9876543211, 12586269025"
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https://eslint.org/docs/rules/one-var.html | [
"enforce variables to be declared either together or separately in functions (one-var)\n\nThe --fix option on the command line can automatically fix some of the problems reported by this rule.\n\nVariables can be declared at any point in JavaScript code using var, let, or const. There are many styles and preferences related to the declaration of variables, and one of those is deciding on how many variable declarations should be allowed in a single function.\n\nThere are two schools of thought in this regard:\n\n1. There should be just one variable declaration for all variables in the function. That declaration typically appears at the top of the function.\n2. You should use one variable declaration for each variable you want to define.\n\nFor instance:\n\n// one variable declaration per function\nfunction foo() {\nvar bar, baz;\n}\n\n// multiple variable declarations per function\nfunction foo() {\nvar bar;\nvar baz;\n}\n\nThe single-declaration school of thought is based in pre-ECMAScript 6 behaviors, where there was no such thing as block scope, only function scope. Since all var statements are hoisted to the top of the function anyway, some believe that declaring all variables in a single declaration at the top of the function removes confusion around scoping rules.\n\nRule Details\n\nThis rule enforces variables to be declared either together or separately per function ( for var) or block (for let and const) scope.\n\nOptions\n\nThis rule has one option, which can be a string option or an object option.\n\nString option:\n\n• \"always\" (default) requires one variable declaration per scope\n• \"never\" requires multiple variable declarations per scope\n• \"consecutive\" allows multiple variable declarations per scope but requires consecutive variable declarations to be combined into a single declaration\n\nObject option:\n\n• \"var\": \"always\" requires one var declaration per function\n• \"var\": \"never\" requires multiple var declarations per function\n• \"var\": \"consecutive\" requires consecutive var declarations to be a single declaration\n• \"let\": \"always\" requires one let declaration per block\n• \"let\": \"never\" requires multiple let declarations per block\n• \"let\": \"consecutive\" requires consecutive let declarations to be a single declaration\n• \"const\": \"always\" requires one const declaration per block\n• \"const\": \"never\" requires multiple const declarations per block\n• \"const\": \"consecutive\" requires consecutive const declarations to be a single declaration\n• \"separateRequires\": true enforces requires to be separate from declarations\n\nAlternate object option:\n\n• \"initialized\": \"always\" requires one variable declaration for initialized variables per scope\n• \"initialized\": \"never\" requires multiple variable declarations for initialized variables per scope\n• \"initialized\": \"consecutive\" requires consecutive variable declarations for initialized variables to be a single declaration\n• \"uninitialized\": \"always\" requires one variable declaration for uninitialized variables per scope\n• \"uninitialized\": \"never\" requires multiple variable declarations for uninitialized variables per scope\n• \"uninitialized\": \"consecutive\" requires consecutive variable declarations for uninitialized variables to be a single declaration\n\nalways\n\nExamples of incorrect code for this rule with the default \"always\" option:\n\n/*eslint one-var: [\"error\", \"always\"]*/\n\nfunction foo() {\nvar bar;\nvar baz;\nlet qux;\nlet norf;\n}\n\nfunction foo(){\nconst bar = false;\nconst baz = true;\nlet qux;\nlet norf;\n}\n\nfunction foo() {\nvar bar;\n\nif (baz) {\nvar qux = true;\n}\n}\n\nclass C {\nstatic {\nvar foo;\nvar bar;\n}\n\nstatic {\nvar foo;\nif (bar) {\nvar baz = true;\n}\n}\n\nstatic {\nlet foo;\nlet bar;\n}\n}\n\nExamples of correct code for this rule with the default \"always\" option:\n\n/*eslint one-var: [\"error\", \"always\"]*/\n\nfunction foo() {\nvar bar,\nbaz;\nlet qux,\nnorf;\n}\n\nfunction foo(){\nconst bar = true,\nbaz = false;\nlet qux,\nnorf;\n}\n\nfunction foo() {\nvar bar,\nqux;\n\nif (baz) {\nqux = true;\n}\n}\n\nfunction foo(){\nlet bar;\n\nif (baz) {\nlet qux;\n}\n}\n\nclass C {\nstatic {\nvar foo, bar;\n}\n\nstatic {\nvar foo, baz;\nif (bar) {\nbaz = true;\n}\n}\n\nstatic {\nlet foo, bar;\n}\n\nstatic {\nlet foo;\nif (bar) {\nlet baz;\n}\n}\n}\n\nnever\n\nExamples of incorrect code for this rule with the \"never\" option:\n\n/*eslint one-var: [\"error\", \"never\"]*/\n\nfunction foo() {\nvar bar,\nbaz;\nconst bar = true,\nbaz = false;\n}\n\nfunction foo() {\nvar bar,\nqux;\n\nif (baz) {\nqux = true;\n}\n}\n\nfunction foo(){\nlet bar = true,\nbaz = false;\n}\n\nclass C {\nstatic {\nvar foo, bar;\nlet baz, qux;\n}\n}\n\nExamples of correct code for this rule with the \"never\" option:\n\n/*eslint one-var: [\"error\", \"never\"]*/\n\nfunction foo() {\nvar bar;\nvar baz;\n}\n\nfunction foo() {\nvar bar;\n\nif (baz) {\nvar qux = true;\n}\n}\n\nfunction foo() {\nlet bar;\n\nif (baz) {\nlet qux = true;\n}\n}\n\nclass C {\nstatic {\nvar foo;\nvar bar;\nlet baz;\nlet qux;\n}\n}\n\nconsecutive\n\nExamples of incorrect code for this rule with the \"consecutive\" option:\n\n/*eslint one-var: [\"error\", \"consecutive\"]*/\n\nfunction foo() {\nvar bar;\nvar baz;\n}\n\nfunction foo(){\nvar bar = 1;\nvar baz = 2;\n\nqux();\n\nvar qux = 3;\nvar quux;\n}\n\nclass C {\nstatic {\nvar foo;\nvar bar;\nlet baz;\nlet qux;\n}\n}\n\nExamples of correct code for this rule with the \"consecutive\" option:\n\n/*eslint one-var: [\"error\", \"consecutive\"]*/\n\nfunction foo() {\nvar bar,\nbaz;\n}\n\nfunction foo(){\nvar bar = 1,\nbaz = 2;\n\nqux();\n\nvar qux = 3,\nquux;\n}\n\nclass C {\nstatic {\nvar foo, bar;\nlet baz, qux;\ndoSomething();\nlet quux;\nvar quuux;\n}\n}\n\nvar, let, and const\n\nExamples of incorrect code for this rule with the { var: \"always\", let: \"never\", const: \"never\" } option:\n\n/*eslint one-var: [\"error\", { var: \"always\", let: \"never\", const: \"never\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar bar;\nvar baz;\nlet qux,\nnorf;\n}\n\nfunction foo() {\nconst bar = 1,\nbaz = 2;\nlet qux,\nnorf;\n}\n\nExamples of correct code for this rule with the { var: \"always\", let: \"never\", const: \"never\" } option:\n\n/*eslint one-var: [\"error\", { var: \"always\", let: \"never\", const: \"never\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar bar,\nbaz;\nlet qux;\nlet norf;\n}\n\nfunction foo() {\nconst bar = 1;\nconst baz = 2;\nlet qux;\nlet norf;\n}\n\nExamples of incorrect code for this rule with the { var: \"never\" } option:\n\n/*eslint one-var: [\"error\", { var: \"never\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar bar,\nbaz;\n}\n\nExamples of correct code for this rule with the { var: \"never\" } option:\n\n/*eslint one-var: [\"error\", { var: \"never\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar bar,\nbaz;\nconst bar = 1; // `const` and `let` declarations are ignored if they are not specified\nconst baz = 2;\nlet qux;\nlet norf;\n}\n\nExamples of incorrect code for this rule with the { separateRequires: true } option:\n\n/*eslint one-var: [\"error\", { separateRequires: true, var: \"always\" }]*/\n/*eslint-env node*/\n\nvar foo = require(\"foo\"),\nbar = \"bar\";\n\nExamples of correct code for this rule with the { separateRequires: true } option:\n\n/*eslint one-var: [\"error\", { separateRequires: true, var: \"always\" }]*/\n/*eslint-env node*/\n\nvar foo = require(\"foo\");\nvar bar = \"bar\";\nvar foo = require(\"foo\"),\nbar = require(\"bar\");\n\nExamples of incorrect code for this rule with the { var: \"never\", let: \"consecutive\", const: \"consecutive\" } option:\n\n/*eslint one-var: [\"error\", { var: \"never\", let: \"consecutive\", const: \"consecutive\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nlet a,\nb;\nlet c;\n\nvar d,\ne;\n}\n\nfunction foo() {\nconst a = 1,\nb = 2;\nconst c = 3;\n\nvar d,\ne;\n}\n\nExamples of correct code for this rule with the { var: \"never\", let: \"consecutive\", const: \"consecutive\" } option:\n\n/*eslint one-var: [\"error\", { var: \"never\", let: \"consecutive\", const: \"consecutive\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nlet a,\nb;\n\nvar d;\nvar e;\n\nlet f;\n}\n\nfunction foo() {\nconst a = 1,\nb = 2;\n\nvar c;\nvar d;\n\nconst e = 3;\n}\n\nExamples of incorrect code for this rule with the { var: \"consecutive\" } option:\n\n/*eslint one-var: [\"error\", { var: \"consecutive\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar a;\nvar b;\n}\n\nExamples of correct code for this rule with the { var: \"consecutive\" } option:\n\n/*eslint one-var: [\"error\", { var: \"consecutive\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar a,\nb;\nconst c = 1; // `const` and `let` declarations are ignored if they are not specified\nconst d = 2;\nlet e;\nlet f;\n}\n\ninitialized and uninitialized\n\nExamples of incorrect code for this rule with the { \"initialized\": \"always\", \"uninitialized\": \"never\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"always\", \"uninitialized\": \"never\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar a, b, c;\nvar foo = true;\nvar bar = false;\n}\n\nExamples of correct code for this rule with the { \"initialized\": \"always\", \"uninitialized\": \"never\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"always\", \"uninitialized\": \"never\" }]*/\n\nfunction foo() {\nvar a;\nvar b;\nvar c;\nvar foo = true,\nbar = false;\n}\n\nfor (let z of foo) {\ndoSomething(z);\n}\n\nlet z;\nfor (z of foo) {\ndoSomething(z);\n}\n\nExamples of incorrect code for this rule with the { \"initialized\": \"never\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"never\" }]*/\n/*eslint-env es6*/\n\nfunction foo() {\nvar foo = true,\nbar = false;\n}\n\nExamples of correct code for this rule with the { \"initialized\": \"never\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"never\" }]*/\n\nfunction foo() {\nvar foo = true;\nvar bar = false;\nvar a, b, c; // Uninitialized variables are ignored\n}\n\nExamples of incorrect code for this rule with the { \"initialized\": \"consecutive\", \"uninitialized\": \"never\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"consecutive\", \"uninitialized\": \"never\" }]*/\n\nfunction foo() {\nvar a = 1;\nvar b = 2;\nvar c,\nd;\nvar e = 3;\nvar f = 4;\n}\n\nExamples of correct code for this rule with the { \"initialized\": \"consecutive\", \"uninitialized\": \"never\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"consecutive\", \"uninitialized\": \"never\" }]*/\n\nfunction foo() {\nvar a = 1,\nb = 2;\nvar c;\nvar d;\nvar e = 3,\nf = 4;\n}\n\nExamples of incorrect code for this rule with the { \"initialized\": \"consecutive\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"consecutive\" }]*/\n\nfunction foo() {\nvar a = 1;\nvar b = 2;\n\nfoo();\n\nvar c = 3;\nvar d = 4;\n}\n\nExamples of correct code for this rule with the { \"initialized\": \"consecutive\" } option:\n\n/*eslint one-var: [\"error\", { \"initialized\": \"consecutive\" }]*/\n\nfunction foo() {\nvar a = 1,\nb = 2;\n\nfoo();\n\nvar c = 3,\nd = 4;\n}\n\nCompatibility\n\n• JSHint: This rule maps to the onevar JSHint rule, but allows let and const to be configured separately.\n• JSCS: This rule roughly maps to disallowMultipleVarDecl.\n• JSCS: This rule option separateRequires roughly maps to requireMultipleVarDecl.\n\nVersion\n\nThis rule was introduced in ESLint 0.0.9."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59115773,"math_prob":0.94926393,"size":10467,"snap":"2022-05-2022-21","text_gpt3_token_len":2806,"char_repetition_ratio":0.1830259,"word_repetition_ratio":0.56006867,"special_character_ratio":0.30323875,"punctuation_ratio":0.20682523,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.950363,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T04:08:57Z\",\"WARC-Record-ID\":\"<urn:uuid:c730530e-cdc9-4ad1-b6ac-160aeeaf419e>\",\"Content-Length\":\"41553\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:020d7341-7a01-4b75-a670-7014fd350899>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed23ae80-3d00-409d-a67e-92e38f0dfc7b>\",\"WARC-IP-Address\":\"34.196.254.27\",\"WARC-Target-URI\":\"https://eslint.org/docs/rules/one-var.html\",\"WARC-Payload-Digest\":\"sha1:PBPIHMO5GUUS4TI7ZJ4XVO4Z2GIRTI6V\",\"WARC-Block-Digest\":\"sha1:YJDOOZY3W757YE52WE3WJFENWJDEBESW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301263.50_warc_CC-MAIN-20220119033421-20220119063421-00381.warc.gz\"}"} |
https://arelarsiv.arel.edu.tr/xmlui/handle/20.500.12294/1671/browse?value=Meslek+Y%C3%BCksekokulu%2C+Optisyenlik+Program%C4%B1&type=department | [
"Now showing items 1-20 of 33\n\n• #### Amplitude analysis of the chi(c1) -> eta pi(+)pi(-) decays \n\n(Amer Physical Soc., 2017)\nUsing 448.0 x 10(6) psi(3686) events collected with the BESIII detector, an amplitude analysis is performed for psi(3686) -> gamma chi(c1), chi(c1) ->eta pi(+)pi(-) decays. The most dominant two- body structure observed ...\n• #### Amplitude Analysis of the Decays eta ' -> pi(+)pi(-)pi(0) and eta' -> pi(0)pi(0)pi(0) \n\n(Amer Physic Soc., 2017)\nBased on a sample of 1.31 x 10(9) J/Psi events collected with the BESIII detector, an amplitude analysis of the isospin-violating decays eta' -> pi(+)pi(-)pi(0) and eta' -> pi(0)pi(0)pi(0) is performed. A significant P-wave ...\n• #### Analysis of D+ -> (K)over-bar(0)e(+)nu(e) and D+ -> pi(0)e(+)nu(e) semileptonic decays \n\n(Amer Physical Soc., 2017)\nUsing 2.93 fb(-1) of data taken at 3.773 GeV with the BESIII detector operated at the BEPCII collider, we study the semileptonic decays D+ -> (K) over bar (0)e(+)nu(e) and D+ -> pi(0)e(+)nu(e). We measure the absolute decay ...\n• #### Branching fraction measurement of J=Psi -> KSKL and search for J=Psi -> KSKS \n\n(Amer Physical Soc., 2017)\nUsing a sample of 1.31 x 10(9) J/Psi events collected with the BESIII detector at the BEPCII collider, we study the decays of J/Psi -> KSKL and KSKS. The branching fraction of J/Psi -> KSKL is determined to be B(J/Psi -> ...\n• #### Dark photon search in the mass range between 1.5 and 3.4 GeV/c \n\n(Elsevier Science, 2017)\nUsing a data set of 2.93 fb taken at a center-of-mass energy root s = 3.773 GeV with the BESIII detector at the BEPCII collider, we perform a search for an extra U(1) gauge boson, also denoted as a dark photon. We examine ...\n• #### Determination of the number of J/psi events with inclusive J/psi decays \n\n(Chinese Physical, 2017)\nA measurement of the number of J/psi events collected with the BESIII detector in 2009 and 2012 is performed using inclusive decays of the J/psi. The number of J/psi events taken in 2009 is recalculated to be (223.7 +/- ...\n• #### Determination of the Spin and Parity of the Z(c)(3900) \n\n(Amer Physic Soc., 2017)\nThe spin and parity of the Z(c)(3900)(+/-) state are determined to be J(P) = 1(+) with a statistical significance larger than 7 sigma over other quantum numbers in a partial wave analysis of the process e(+)e(-) -> pi(+)pi(-) ...\n• #### Evaluation of Superconducting Features and Gap Coefficients for Electron-Phonon Couplings Properties of Mgb2 with Multi-Walled Carbon Nanotube Addition \n\nIn this study, the samples are prepared by solid state reaction method at different weight ratios (0-4%). The characterization of materials produced is conducted with the aid of powder X-ray diffraction (XRD), temperatur ...\n• #### Evidence for e(+)e(-) -> gamma eta(c)(1S) at center-of-mass energies between 4.01 and 4.60 GeV \n\n(Amer Physical Soc., 2017)\nWe present first evidence for the process e(+)e(-) -> gamma eta(c)(1S) at six center-of-mass energies between 4.01 and 4.60 GeV using data collected by the BESIII experiment operating at BEPCII. We measure the Born cross ...\n• #### Evidence of Two Resonant Structures in e(+)e(-)->pi(+) pi(-) h(c) \n\n(Amer Physic Soc., 2017)\nThe cross sections of e(+)e(-) -> pi(+) pi(-) hc at center-of-mass energies from 3.896 to 4.600 GeVare measured using data samples collected with the BESIII detector operating at the Beijing Electron Positron Collider. The ...\n• #### Improved measurement of the absolute branching fraction of D+ -> (K)over-bar(0)mu(+)nu(mu) \n\n(Springer, 2016)\nBy analyzing 2.93 fb(-1) of data collected at root s = 3.773 GeV with the BESIII detector, we measure the absolute branching fraction B(D+ -> (K) over bar (0) (+)(mu)nu(mu)) = (8.72 +/- 0.07(stat). +/- 0.18(sys).) %, which ...\n• #### Improved measurements of two-photon widths of the chi(cJ) states and helicity analysis for chi(c2) -> gamma gamma \n\n(Amer Physical Soc., 2017)\nBased on 448.1 x 10(6) Psi(3686) events collected with the BESIII detector, the decays Psi(3686) -> gamma chi(cJ), chi(cJ) -> gamma gamma(J = 0, 1, 2) are studied. The decay branching fractions of chi(c0,2) -> gamma gamma ...\n• #### Investigation of Microhardness Properties of the Multi-Walled Carbon Nanotube Additive Mgb2 Structure By Using The Vickers Method \n\nIn this study, the effect of multi-walled carbon nanotube doping to MgB2 compound on microhardness properties of MgB2 was investigated by using solid-state reaction method. The amount of multi-walled carbon nanotubes was ...\n• #### Luminosity measurements for the R scan experiment at BESIII \n\n(Chinese Physical, 2017)\nBy analyzing the large-angle Bhabha scattering events e(+)e(-) -> (gamma)e(+)e(-) and diphoton events e(+)e(-) -> (gamma)gamma gamma for the data sets collected at center-of-mass (c.m.) energies between 2.2324 and 4.5900 ...\n• #### Measurement of branching fractions for psi(3686) -> gamma eta ', gamma eta, and gamma pi(0) \n\n(Amer Physical Soc., 2017)\nUsing a data sample of 448 x 10(6) psi(3686) events collected with the BESIII detector operating at the BEPCII storage ring, the decays psi(3686) -> gamma eta and psi(3686) -> gamma pi(0) are observed with a statistical ...\n• #### Measurement of cross sections of the interactions e(+)e(-) -> phi phi omega and e(+)e(-) -> phi phi phi at center-of-mass energies from 4.008 to 4.600 GeV \n\n(Elsevier Science, 2017)\nUsing data samples collected with the BESIII detector at the BEPCII collider at six center-of-mass energies between 4.008 and 4.600 GeV, we observe the processes e(+)e(-) -> phi phi omega and e(-)e(-) -> phi phi phi. The ...\n• #### Measurement of integrated luminosity and center-of-mass energy of data taken by BESIII at root s=2.125 GeV \n\n(Chinese Physical, 2017)\nTo study the nature of the state Y (2175), a dedicated data set of e(+)e(-) collision data was collected at the center-of-mass energy of 2.125 GeV with the BESIII detector at the BEPCII collider. By analyzing large-angle ...\n• #### Measurement of the Absolute Branching Fraction for Lambda(+)(c) -> Lambda e(+)nu(e) \n\n(Amer Physical Soc., 2015)\nWe report the first measurement of the absolute branching fraction for Lambda(+)(c) -> Lambda e(+)nu(e). This measurement is based on 567 pb(-1) of e(+)e(-) annihilation data produced at root s = 4.599 GeV, which is just ...\n• #### Measurement of the absolute branching fraction for Lambda(+)(c) -> Lambda mu(+)nu(mu) \n\n(Elsevier Science, 2017)\nWe report the first measurement of the absolute branching fraction for Lambda(+)(c) -> Lambda mu(+)nu(mu).This measurement is based on a sample of e+e(-) annihilation data produced at a center-of-mass energy root s = 4.6 ...\n• #### Measurement of the leptonic decay width of J/psi using initial state radiation \n\n(Elsevier, 2016)\nUsing a data set of 2.93 fb(-1) taken at a center-of-mass energy of root s = 3.773 GeV with the BESIII detector at the BEPCII collider, we measure the process e(+) e(-) -> J/psi gamma -> mu(+)mu(-)gamma and determine the ..."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78419906,"math_prob":0.9442139,"size":6954,"snap":"2023-40-2023-50","text_gpt3_token_len":2049,"char_repetition_ratio":0.13410072,"word_repetition_ratio":0.14613971,"special_character_ratio":0.31276962,"punctuation_ratio":0.12034384,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9803384,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T16:16:50Z\",\"WARC-Record-ID\":\"<urn:uuid:b257dee8-be97-412e-a700-cbd3cb22c124>\",\"Content-Length\":\"71323\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c9d9c68-d2f5-463e-aa87-8d59d9e1e91a>\",\"WARC-Concurrent-To\":\"<urn:uuid:ccb34ad8-723d-438e-9169-b2a2395cf210>\",\"WARC-IP-Address\":\"193.255.56.228\",\"WARC-Target-URI\":\"https://arelarsiv.arel.edu.tr/xmlui/handle/20.500.12294/1671/browse?value=Meslek+Y%C3%BCksekokulu%2C+Optisyenlik+Program%C4%B1&type=department\",\"WARC-Payload-Digest\":\"sha1:3ECXXHRWK7UTW7WDF6Y5YBINV4UBIXKF\",\"WARC-Block-Digest\":\"sha1:OBVCO6ECBDRNPPE7T2JBIYIU5F64KTQD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510300.41_warc_CC-MAIN-20230927135227-20230927165227-00656.warc.gz\"}"} |
https://apprize.best/programming/ocaml/4.html | [
" Lists and Patterns - Language Concepts - Real World OCaml (2013)\n\n# Real World OCaml (2013)\n\n### Chapter 3. Lists and Patterns\n\nThis chapter will focus on two common elements of programming in OCaml: lists and pattern matching. Both of these were discussed in Chapter 1, but we’ll go into more depth here, presenting the two topics together and using one to help illustrate the other.\n\nList Basics\n\nAn OCaml list is an immutable, finite sequence of elements of the same type. As we’ve seen, OCaml lists can be generated using a bracket-and-semicolon notation:\n\nOCaml utop\n\n# [1;2;3];;\n\n- : int list = [1; 2; 3]\n\nAnd they can also be generated using the equivalent :: notation:\n\nOCaml utop (part 1)\n\n# 1 :: (2 :: (3 :: [])) ;;\n\n- : int list = [1; 2; 3]\n\n# 1 :: 2 :: 3 :: [] ;;\n\n- : int list = [1; 2; 3]\n\nAs you can see, the :: operator is right-associative, which means that we can build up lists without parentheses. The empty list [] is used to terminate a list. Note that the empty list is polymorphic, meaning it can be used with elements of any type, as you can see here:\n\nOCaml utop (part 2)\n\n# letempty = [];;\n\nval empty : 'a list = []\n\n# 3 :: empty;;\n\n- : int list = \n\n# \"three\" :: empty;;\n\n- : string list = [\"three\"]\n\nThe way in which the :: operator attaches elements to the front of a list reflects the fact that OCaml’s lists are in fact singly linked lists. The figure below is a rough graphical representation of how the list 1 :: 2 :: 3 :: [] is laid out as a data structure. The final arrow (from the box containing 3) points to the empty list.\n\nDiagram",
null,
"Each :: essentially adds a new block to the proceding picture. Such a block contains two things: a reference to the data in that list element, and a reference to the remainder of the list. This is why :: can extend a list without modifying it; extension allocates a new list element but change any of the existing ones, as you can see:\n\nOCaml utop (part 3)\n\n# letl = 1 :: 2 :: 3 :: [];;\n\nval l : int list = [1; 2; 3]\n\n# letm = 0 :: l;;\n\nval m : int list = [0; 1; 2; 3]\n\n# l;;\n\n- : int list = [1; 2; 3]\n\nUsing Patterns to Extract Data from a List\n\nWe can read data out of a list using a match statement. Here’s a simple example of a recursive function that computes the sum of all elements of a list:\n\nOCaml utop (part 4)\n\n# letrecsuml =\n\nmatchlwith\n\n| [] -> 0\n\n| hd :: tl -> hd + sumtl\n\n;;\n\nval sum : int list -> int = <fun>\n\n# sum [1;2;3];;\n\n- : int = 6\n\n# sum[];;\n\n- : int = 0\n\nThis code follows the convention of using hd to represent the first element (or head) of the list, and tl to represent the remainder (or tail).\n\nThe match statement in sum is really doing two things: first, it’s acting as a case-analysis tool, breaking down the possibilities into a pattern-indexed list of cases. Second, it lets you name substructures within the data structure being matched. In this case, the variables hd and tl are bound by the pattern that defines the second case of the match statement. Variables that are bound in this way can be used in the expression to the right of the arrow for the pattern in question.\n\nThe fact that match statements can be used to bind new variables can be a source of confusion. To see how, imagine we wanted to write a function that filtered out from a list all elements equal to a particular value. You might be tempted to write that code as follows, but when you do, the compiler will immediately warn you that something is wrong:\n\nOCaml utop (part 5)\n\n# letrecdrop_valuelto_drop =\n\nmatchlwith\n\n| [] -> []\n\n| to_drop :: tl -> drop_valuetlto_drop\n\n| hd :: tl -> hd :: drop_valuetlto_drop\n\n;;\n\nCharacters 114-122:\n\nWarning 11: this match case is unused.\n\nval drop_value : 'a list -> 'a -> 'a list = <fun>\n\nMoreover, the function clearly does the wrong thing, filtering out all elements of the list rather than just those equal to the provided value, as you can see here:\n\nOCaml utop (part 6)\n\n# drop_value [1;2;3] 2;;\n\n- : int list = []\n\nSo, what’s going on?\n\nThe key observation is that the appearance of to_drop in the second case doesn’t imply a check that the first element is equal to the value to_drop passed in as an argument to drop_value. Instead, it just causes a new variable to_drop to be bound to whatever happens to be in the first element of the list, shadowing the earlier definition of to_drop. The third case is unused because it is essentially the same pattern as we had in the second case.\n\nA better way to write this code is not to use pattern matching for determining whether the first element is equal to to_drop, but to instead use an ordinary if statement:\n\nOCaml utop (part 7)\n\n# letrecdrop_valuelto_drop =\n\nmatchlwith\n\n| [] -> []\n\n| hd :: tl ->\n\nletnew_tl = drop_valuetlto_dropin\n\nifhd = to_dropthennew_tlelsehd :: new_tl\n\n;;\n\nval drop_value : 'a list -> 'a -> 'a list = <fun>\n\n# drop_value [1;2;3] 2;;\n\n- : int list = [1; 3]\n\nNote that if we wanted to drop a particular literal value (rather than a value that was passed in), we could do this using something like our original implementation of drop_value:\n\nOCaml utop (part 8)\n\n# letrecdrop_zerol =\n\nmatchlwith\n\n| [] -> []\n\n| 0 :: tl -> drop_zerotl\n\n| hd :: tl -> hd :: drop_zerotl\n\n;;\n\nval drop_zero : int list -> int list = <fun>\n\n# drop_zero [1;2;0;3];;\n\n- : int list = [1; 2; 3]\n\nLimitations (and Blessings) of Pattern Matching\n\nThe preceding example highlights an important fact about patterns, which is that they can’t be used to express arbitrary conditions. Patterns can characterize the layout of a data structure and can even include literals, as in the drop_zero example, but that’s where they stop. A pattern can check if a list has two elements, but it can’t check if the first two elements are equal to each other.\n\nYou can think of patterns as a specialized sublanguage that can express a limited (though still quite rich) set of conditions. The fact that the pattern language is limited turns out to be a very good thing, making it possible to build better support for patterns in the compiler. In particular, both the efficiency of match statements and the ability of the compiler to detect errors in matches depend on the constrained nature of patterns.\n\nPerformance\n\nNaively, you might think that it would be necessary to check each case in a match in sequence to figure out which one fires. If the cases of a match were guarded by arbitrary code, that would be the case. But OCaml is often able to generate machine code that jumps directly to the matched case based on an efficiently chosen set of runtime checks.\n\nAs an example, consider the following rather silly functions for incrementing an integer by one. The first is implemented with a match statement, and the second with a sequence of if statements:\n\nOCaml utop (part 9)\n\n# letplus_one_matchx =\n\nmatchxwith\n\n| 0 -> 1\n\n| 1 -> 2\n\n| 2 -> 3\n\n| _ -> x + 1\n\nletplus_one_ifx =\n\nif x = 0then1\n\nelseifx = 1then2\n\nelseifx = 2then3\n\nelsex + 1\n\n;;\n\nval plus_one_match : int -> int = <fun>\n\nval plus_one_if : int -> int = <fun>\n\nNote the use of _ in the above match. This is a wildcard pattern that matches any value, but without binding a variable name to the value in question.\n\nIf you benchmark these functions, you’ll see that plus_one_if is considerably slower than plus_one_match, and the advantage gets larger as the number of cases increases. Here, we’ll benchmark these functions using the core_bench library, which can be installed by running opam install core_bench from the command line:\n\nOCaml utop (part 10)\n\n# #require\"core_bench\";;\n\n# openCore_bench.Std;;\n\n# letrun_benchtests =\n\nBench.bench\n\n~ascii_table:true\n\n~display:Textutils.Ascii_table.Display.column_titles\n\ntests\n\n;;\n\nval run_bench : Bench.Test.t list -> unit = <fun>\n\n# [ Bench.Test.create ~name:\"plus_one_match\" (fun() ->\n\nignore (plus_one_match10))\n\n; Bench.Test.create ~name:\"plus_one_if\" (fun() ->\n\nignore (plus_one_if10)) ]\n\n|> run_bench\n\n;;\n\nEstimated testing time 20s (change using -quota SECS).\n\nName Time (ns) % of max\n\n---------------- ----------- ----------\n\nplus_one_match 46.81 68.21\n\nplus_one_if 68.63 100.00\n\n- : unit = ()\n\nHere’s another, less artificial example. We can rewrite the sum function we described earlier in the chapter using an if statement rather than a match. We can then use the functions is_empty, hd_exn, and tl_exn from the List module to deconstruct the list, allowing us to implement the entire function without pattern matching:\n\nOCaml utop (part 11)\n\n# letrecsum_ifl =\n\nifList.is_emptylthen0\n\nelseList.hd_exnl + sum_if (List.tl_exnl)\n\n;;\n\nval sum_if : int list -> int = <fun>\n\nAgain, we can benchmark these to see the difference:\n\nOCaml utop (part 12)\n\n# letnumbers = List.range01000in\n\n[ Bench.Test.create ~name:\"sum_if\" (fun() -> ignore (sum_ifnumbers))\n\n; Bench.Test.create ~name:\"sum\" (fun() -> ignore (sumnumbers)) ]\n\n|> run_bench\n\n;;\n\nEstimated testing time 20s (change using -quota SECS).\n\nName Time (ns) % of max\n\n-------- ----------- ----------\n\nsum_if 110_535 100.00\n\nsum 22_361 20.23\n\n- : unit = ()\n\nIn this case, the match-based implementation is many times faster than the if-based implementation. The difference comes because we need to effectively do the same work multiple times, since each function we call has to reexamine the first element of the list to determine whether or not it’s the empty cell. With a match statement, this work happens exactly once per list element.\n\nGenerally, pattern matching is more efficient than the alternatives you might code by hand. One notable exception is matches over strings, which are in fact tested sequentially, so matches containing a long sequence of strings can be outperformed by a hash table. But most of the time, pattern matching is a clear performance win.\n\nDetecting Errors\n\nThe error-detecting capabilities of match statements are if anything more important than their performance. We’ve already seen one example of OCaml’s ability to find problems in a pattern match: in our broken implementation of drop_value, OCaml warned us that the final case was redundant. There are no algorithms for determining if a predicate written in a general-purpose language is redundant, but it can be solved reliably in the context of patterns.\n\nOCaml also checks match statements for exhaustiveness. Consider what happens if we modify drop_zero by deleting the handler for one of the cases. As you can see, the compiler will produce a warning that we’ve missed a case, along with an example of an unmatched pattern:\n\nOCaml utop (part 13)\n\n# letrecdrop_zerol =\n\nmatchlwith\n\n| [] -> []\n\n| 0 :: tl -> drop_zerotl\n\n;;\n\nCharacters 26-84:\n\nWarning 8: this pattern-matching is not exhaustive.\n\nHere is an example of a value that is not matched:\n\n1::_\n\nval drop_zero : int list -> 'a list = <fun>\n\nEven for simple examples like this, exhaustiveness checks are pretty useful. But as we’ll see in Chapter 6, they become yet more valuable as you get to more complicated examples, especially those involving user-defined types. In addition to catching outright errors, they act as a sort of refactoring tool, guiding you to the locations where you need to adapt your code to deal with changing types.\n\nUsing the List Module Effectively\n\nWe’ve so far written a fair amount of list-munging code using pattern matching and recursive functions. But in real life, you’re usually better off using the List module, which is full of reusable functions that abstract out common patterns for computing with lists.\n\nLet’s work through a concrete example to see this in action. We’ll write a function render_table that, given a list of column headers and a list of rows, prints them out in a well-formatted text table, as follows:\n\nOCaml utop (part 69)\n\n# printf\"%s\\n\"\n\n(render_table\n\n[\"language\";\"architect\";\"first release\"]\n\n[ [\"Lisp\" ;\"John McCarthy\" ;\"1958\"] ;\n\n[\"C\" ;\"Dennis Ritchie\";\"1969\"] ;\n\n[\"ML\" ;\"Robin Milner\" ;\"1973\"] ;\n\n[\"OCaml\";\"Xavier Leroy\" ;\"1996\"] ;\n\n]);;\n\n| language | architect | first release |\n\n|----------+----------------+---------------|\n\n| Lisp | John McCarthy | 1958 |\n\n| C | Dennis Ritchie | 1969 |\n\n| ML | Robin Milner | 1973 |\n\n| OCaml | Xavier Leroy | 1996 |\n\n- : unit = ()\n\nThe first step is to write a function to compute the maximum width of each column of data. We can do this by converting the header and each row into a list of integer lengths, and then taking the element-wise max of those lists of lengths. Writing the code for all of this directly would be a bit of a chore, but we can do it quite concisely by making use of three functions from the List module: map, map2_exn, and fold.\n\nList.map is the simplest to explain. It takes a list and a function for transforming elements of that list, and returns a new list with the transformed elements. Thus, we can write:\n\nOCaml utop (part 14)\n\n# List.map ~f:String.length [\"Hello\"; \"World!\"];;\n\n- : int list = [5; 6]\n\nList.map2_exn is similar to List.map, except that it takes two lists and a function for combining them. Thus, we might write:\n\nOCaml utop (part 15)\n\n# List.map2_exn ~f:Int.max [1;2;3] [3;2;1];;\n\n- : int list = [3; 2; 3]\n\nThe _exn is there because the function throws an exception if the lists are of mismatched length:\n\nOCaml utop (part 16)\n\n# List.map2_exn ~f:Int.max [1;2;3] [3;2;1;0];;\n\nException: (Invalid_argument \"length mismatch in rev_map2_exn: 3 <> 4 \").\n\nList.fold is the most complicated of the three, taking three arguments: a list to process, an initial accumulator value, and a function for updating the accumulator. List.fold walks over the list from left to right, updating the accumulator at each step and returning the final value of the accumulator when it’s done. You can see some of this by looking at the type-signature for fold:\n\nOCaml utop (part 17)\n\n# List.fold;;\n\n- : 'a list -> init:'accum -> f:('accum -> 'a -> 'accum) -> 'accum = <fun>\n\nWe can use List.fold for something as simple as summing up a list:\n\nOCaml utop (part 18)\n\n# List.fold ~init:0 ~f:(+) [1;2;3;4];;\n\n- : int = 10\n\nThis example is particularly simple because the accumulator and the list elements are of the same type. But fold is not limited to such cases. We can for example use fold to reverse a list, in which case the accumulator is itself a list:\n\nOCaml utop (part 19)\n\n# List.fold ~init:[] ~f:(funlistx -> x :: list) [1;2;3;4];;\n\n- : int list = [4; 3; 2; 1]\n\nLet’s bring our three functions together to compute the maximum column widths:\n\nOCaml utop (part 20)\n\nletlengthsl = List.map ~f:String.lengthlin\n\nList.foldrows\n\n~f:(funaccrow ->\n\nList.map2_exn ~f:Int.maxacc (lengthsrow))\n\n;;\n\nval max_widths : string list -> string list list -> int list = <fun>\n\nUsing List.map we define the function lengths, which converts a list of strings to a list of integer lengths. List.fold is then used to iterate over the rows, using map2_exn to take the max of the accumulator with the lengths of the strings in each row of the table, with the accumulator initialized to the lengths of the header row.\n\nNow that we know how to compute column widths, we can write the code to generate the line that separates the header from the rest of the text table. We’ll do this in part by mapping String.make over the lengths of the columns to generate a string of dashes of the appropriate length. We’ll then join these sequences of dashes together using String.concat, which concatenates a list of strings with an optional separator string, and ^, which is a pairwise string concatenation function, to add the delimiters on the outside:\n\nOCaml utop (part 21)\n\n# letrender_separatorwidths =\n\nletpieces = List.mapwidths\n\n~f:(funw -> String.make (w + 2) '-')\n\nin\n\n\"|\" ^ String.concat ~sep:\"+\"pieces ^ \"|\"\n\n;;\n\nval render_separator : int list -> string = <fun>\n\n# render_separator [3;6;2];;\n\n- : string = \"|-----+--------+----|\"\n\nNote that we make the line of dashes two larger than the provided width to provide some whitespace around each entry in the table.\n\nPERFORMANCE OF STRING.CONCAT AND ^\n\nIn the preceding code we’ve concatenated strings two different ways: String.concat, which operates on lists of strings; and ^, which is a pairwise operator. You should avoid ^ for joining long numbers of strings, since it allocates a new string every time it runs. Thus, the following code\n\nOCaml utop (part 22)\n\n# lets = \".\" ^ \".\" ^ \".\" ^ \".\" ^ \".\" ^ \".\" ^ \".\";;\n\nval s : string = \".......\"\n\nwill allocate strings of length 2, 3, 4, 5, 6 and 7, whereas this code\n\nOCaml utop (part 23)\n\n# lets = String.concat [\".\";\".\";\".\";\".\";\".\";\".\";\".\"];;\n\nval s : string = \".......\"\n\nallocates one string of size 7, as well as a list of length 7. At these small sizes, the differences don’t amount to much, but for assembling large strings, it can be a serious performance issue.\n\nNow we need code for rendering a row with data in it. We’ll first write a function called pad, for padding out a string to a specified length plus one blank space on both sides:\n\nOCaml utop (part 24)\n\n\" \" ^ s ^ String.make (length - String.lengths + 1) ' '\n\n;;\n\nval pad : string -> int -> string = <fun>\n\n- : string = \" hello \"\n\nWe can render a row of data by merging together the padded strings. Again, we’ll use List.map2_exn for combining the list of data in the row with the list of widths:\n\nOCaml utop (part 25)\n\n# letrender_rowrowwidths =\n\n\"|\" ^ String.concat ~sep:\"|\"padded ^ \"|\"\n\n;;\n\nval render_row : string list -> int list -> string = <fun>\n\n# render_row [\"Hello\";\"World\"] [10;15];;\n\n- : string = \"| Hello | World |\"\n\nNow we can bring this all together in a single function that renders the table:\n\nOCaml utop (part 26)\n\nString.concat ~sep:\"\\n\"\n\n:: render_separatorwidths\n\n:: List.maprows ~f:(funrow -> render_rowrowwidths)\n\n)\n\n;;\n\nval render_table : string list -> string list list -> string = <fun>\n\nMore Useful List Functions\n\nThe previous example we worked through touched on only three of the functions in List. We won’t cover the entire interface (for that you should look at the online docs), but a few more functions are useful enough to mention here.\n\nCombining list elements with List.reduce\n\nList.fold, which we described earlier, is a very general and powerful function. Sometimes, however, you want something simpler and easier to use. One such function is List.reduce, which is essentially a specialized version of List.fold that doesn’t require an explicit starting value, and whose accumulator has to consume and produce values of the same type as the elements of the list it applies to.\n\nHere’s the type signature:\n\nOCaml utop (part 27)\n\n# List.reduce;;\n\n- : 'a list -> f:('a -> 'a -> 'a) -> 'a option = <fun>\n\nreduce returns an optional result, returning None when the input list is empty.\n\nNow we can see reduce in action:\n\nOCaml utop (part 28)\n\n# List.reduce ~f:(+) [1;2;3;4;5];;\n\n- : int option = Some 15\n\n# List.reduce ~f:(+) [];;\n\n- : int option = None\n\nFiltering with List.filter and List.filter_map\n\nVery often when processing lists, you wants to restrict your attention to a subset of the values on your list. The List.filter function is one way of doing that:\n\nOCaml utop (part 29)\n\n# List.filter ~f:(funx -> x mod 2 = 0) [1;2;3;4;5];;\n\n- : int list = [2; 4]\n\nNote that the mod used above is an infix operator, as described in Chapter 2.\n\nSometimes, you want to both transform and filter as part of the same computation. In that case, List.filter_map is what you need. The function passed to List.filter_map returns an optional value, and List.filter_map drops all elements for which None is returned.\n\nHere’s an example. The following expression computes the list of file extensions in the current directory, piping the results through List.dedup to remove duplicates. Note that this example also uses some functions from other modules, including Sys.ls_dir to get a directory listing, andString.rsplit2 to split a string on the rightmost appearance of a given character:\n\nOCaml utop (part 30)\n\n# List.filter_map (Sys.ls_dir\".\") ~f:(funfname ->\n\nmatchString.rsplit2 ~on:'.'fnamewith\n\n| None | Some (\"\",_) -> None\n\n| Some (_,ext) ->\n\nSomeext)\n\n|> List.dedup\n\n;;\n\n- : string list = [\"ascii\"; \"ml\"; \"mli\"; \"topscript\"]\n\nThe preceding code is also an example of an Or pattern, which allows you to have multiple subpatterns within a larger pattern. In this case, None | Some (\"\",_) is an Or pattern. As we’ll see later, Or patterns can be nested anywhere within larger patterns.\n\nPartitioning with List.partition_tf\n\nAnother useful operation that’s closely related to filtering is partitioning. The function List.partition_tf takes a list and a function for computing a Boolean condition on the list elements, and returns two lists. The tf in the name is a mnemonic to remind the user that true elements go to the first list and false ones go to the second. Here’s an example:\n\nOCaml utop (part 31)\n\n# letis_ocaml_sources =\n\nmatchString.rsplit2s ~on:'.'with\n\n| Some (_,(\"ml\"|\"mli\")) -> true\n\n| _ -> false\n\n;;\n\nval is_ocaml_source : string -> bool = <fun>\n\n# let (ml_files,other_files) =\n\nList.partition_tf (Sys.ls_dir\".\") ~f:is_ocaml_source;;\n\nval ml_files : string list = [\"example.mli\"; \"example.ml\"]\n\nval other_files : string list = [\"main.topscript\"; \"lists_layout.ascii\"]\n\nCombining lists\n\nAnother very common operation on lists is concatenation. The list module actually comes with a few different ways of doing this. First, there’s List.append, for concatenating a pair of lists:\n\nOCaml utop (part 32)\n\n# List.append [1;2;3] [4;5;6];;\n\n- : int list = [1; 2; 3; 4; 5; 6]\n\nThere’s also @, an operator equivalent of List.append:\n\nOCaml utop (part 33)\n\n# [1;2;3] @ [4;5;6];;\n\n- : int list = [1; 2; 3; 4; 5; 6]\n\nIn addition, there is List.concat, for concatenating a list of lists:\n\nOCaml utop (part 34)\n\n# List.concat [[1;2];[3;4;5];;[]];;\n\n- : int list = [1; 2; 3; 4; 5; 6]\n\nHere’s an example of using List.concat along with List.map to compute a recursive listing of a directory tree:\n\nOCaml utop (part 35)\n\n# letrecls_recs =\n\nthen [s]\n\nelse\n\nSys.ls_dirs\n\n|> List.map ~f:(funsub -> ls_rec (s ^/ sub))\n\n|> List.concat\n\n;;\n\nval ls_rec : string -> string list = <fun>\n\nNote that ^/ is an infix operator provided by Core for adding a new element to a string representing a file path. It is equivalent to Core’s Filename.concat.\n\nThe preceding combination of List.map and List.concat is common enough that there is a function List.concat_map that combines these into one, more efficient operation:\n\nOCaml utop (part 36)\n\n# letrecls_recs =\n\nthen [s]\n\nelse\n\nSys.ls_dirs\n\n|> List.concat_map ~f:(funsub -> ls_rec (s ^/ sub))\n\n;;\n\nval ls_rec : string -> string list = <fun>\n\nTail Recursion\n\nThe only way to compute the length of an OCaml list is to walk the list from beginning to end. As a result, computing the length of a list takes time linear in the size of the list. Here’s a simple function for doing so:\n\nOCaml utop (part 37)\n\n# letreclength = function\n\n| [] -> 0\n\n| _ :: tl -> 1 + lengthtl\n\n;;\n\nval length : 'a list -> int = <fun>\n\n# length [1;2;3];;\n\n- : int = 3\n\nThis looks simple enough, but you’ll discover that this implementation runs into problems on very large lists, as we’ll show in the following code:\n\nOCaml utop (part 38)\n\n# letmake_listn = List.initn ~f:(funx -> x);;\n\nval make_list : int -> int list = <fun>\n\n# length (make_list10);;\n\n- : int = 10\n\n# length (make_list10_000_000);;\n\nStack overflow during evaluation (looping recursion?).\n\nThe preceding example creates lists using List.init, which takes an integer n and a function f and creates a list of length n, where the data for each element is created by calling f on the index of that element.\n\nTo understand where the error in the above example comes from, you need to learn a bit more about how function calls work. Typically, a function call needs some space to keep track of information associated with the call, such as the arguments passed to the function, or the location of the code that needs to start executing when the function call is complete. To allow for nested function calls, this information is typically organized in a stack, where a new stack frame is allocated for each nested function call, and then deallocated when the function call is complete.\n\nAnd that’s the problem with our call to length: it tried to allocate 10 million stack frames, which exhausted the available stack space. Happily, there’s a way around this problem. Consider the following alternative implementation:\n\nOCaml utop (part 39)\n\n# letreclength_plus_nln =\n\nmatchlwith\n\n| [] -> n\n\n| _ :: tl -> length_plus_ntl (n + 1)\n\n;;\n\nval length_plus_n : 'a list -> int -> int = <fun>\n\n# letlengthl = length_plus_nl0 ;;\n\nval length : 'a list -> int = <fun>\n\n# length [1;2;3;4];;\n\n- : int = 4\n\nThis implementation depends on a helper function, length_plus_n, that computes the length of a given list plus a given n. In practice, n acts as an accumulator in which the answer is built up, step by step. As a result, we can do the additions along the way rather than doing them as we unwind the nested sequence of function calls, as we did in our first implementation of length.\n\nThe advantage of this approach is that the recursive call in length_plus_n is a tail call. We’ll explain more precisely what it means to be a tail call shortly, but the reason it’s important is that tail calls don’t require the allocation of a new stack frame, due to what is called the tail-call optimization. A recursive function is said to be tail recursive if all of its recursive calls are tail calls. length_plus_n is indeed tail recursive, and as a result, length can take a long list as input without blowing the stack:\n\nOCaml utop (part 40)\n\n# length (make_list10_000_000);;\n\n- : int = 10000000\n\nSo when is a call a tail call? Let’s think about the situation where one function (the caller) invokes another (the callee). The invocation is considered a tail call when the caller doesn’t do anything with the value returned by the callee except to return it. The tail-call optimization makes sense because, when a caller makes a tail call, the caller’s stack frame need never be used again, and so you don’t need to keep it around. Thus, instead of allocating a new stack frame for the callee, the compiler is free to reuse the caller’s stack frame.\n\nTail recursion is important for more than just lists. Ordinary nontail recursive calls are reasonable when dealing with data structures like binary trees, where the depth of the tree is logarithmic in the size of your data. But when dealing with situations where the depth of the sequence of nested calls is on the order of the size of your data, tail recursion is usually the right approach.\n\nTerser and Faster Patterns\n\nNow that we know more about how lists and patterns work, let’s consider how we can improve on an example from Recursive list functions: the function destutter, which removes sequential duplicates from a list. Here’s the implementation that was described earlier:\n\nOCaml utop (part 41)\n\n# letrecdestutterlist =\n\nmatchlistwith\n\n| [] -> []\n\n| [hd] -> [hd]\n\n| hd :: hd' :: tl ->\n\nifhd = hd'thendestutter (hd' :: tl)\n\nelsehd :: destutter (hd' :: tl)\n\n;;\n\nval destutter : 'a list -> 'a list = <fun>\n\nWe’ll consider some ways of making this code more concise and more efficient.\n\nFirst, let’s consider efficiency. One problem with the destutter code above is that it in some cases re-creates on the righthand side of the arrow a value that already existed on the lefthand side. Thus, the pattern [hd] -> [hd] actually allocates a new list element, when really, it should be able to just return the list being matched. We can reduce allocation here by using an as pattern, which allows us to declare a name for the thing matched by a pattern or subpattern. While we’re at it, we’ll use the function keyword to eliminate the need for an explicit match:\n\nOCaml utop (part 42)\n\n# letrecdestutter = function\n\n| []asl -> l\n\n| [_] asl -> l\n\n| hd :: (hd' :: _ astl) ->\n\nifhd = hd'thendestuttertl\n\nelsehd :: destuttertl\n\n;;\n\nval destutter : 'a list -> 'a list = <fun>\n\nWe can further collapse this by combining the first two cases into one, using an Or pattern:\n\nOCaml utop (part 43)\n\n# letrecdestutter = function\n\n| [] | [_] asl -> l\n\n| hd :: (hd' :: _ astl) ->\n\nifhd = hd'thendestuttertl\n\nelsehd :: destuttertl\n\n;;\n\nval destutter : 'a list -> 'a list = <fun>\n\nWe can make the code slightly terser now by using a when clause. A when clause allows us to add an extra precondition to a pattern in the form of an arbitrary OCaml expression. In this case, we can use it to include the check on whether the first two elements are equal:\n\nOCaml utop (part 44)\n\n# letrecdestutter = function\n\n| [] | [_] asl -> l\n\n| hd :: (hd' :: _ astl) whenhd = hd' -> destuttertl\n\n| hd :: tl -> hd :: destuttertl\n\n;;\n\nval destutter : 'a list -> 'a list = <fun>\n\nPOLYMORPHIC COMPARE\n\nIn the preceding destutter example, we made use of the fact that OCaml lets us test equality between values of any type, using the = operator. Thus, we can write:\n\nOCaml utop (part 45)\n\n# 3 = 4;;\n\n- : bool = false\n\n# [3;4;5] = [3;4;5];;\n\n- : bool = true\n\n# [Some3; None] = [None; Some3];;\n\n- : bool = false\n\nIndeed, if we look at the type of the equality operator, we’ll see that it is polymorphic:\n\nOCaml utop (part 46)\n\n# (=);;\n\n- : 'a -> 'a -> bool = <fun>\n\nOCaml comes with a whole family of polymorphic comparison operators, including the standard infix comparators, <, >=, etc., as well as the function compare that returns -1, 0, or 1 to flag whether the first operand is smaller than, equal to, or greater than the second, respectively.\n\nYou might wonder how you could build functions like these yourself if OCaml didn’t come with them built in. It turns out that you can’t build these functions on your own. OCaml’s polymorphic comparison functions are built into the runtime to a low level. These comparisons are polymorphic on the basis of ignoring almost everything about the types of the values that are being compared, paying attention only to the structure of the values as they’re laid out in memory.\n\nPolymorphic compare does have some limitations. For example, it will fail at runtime if it encounters a function value:\n\nOCaml utop (part 47)\n\n# (funx -> x + 1) = (funx -> x + 1);;\n\nException: (Invalid_argument \"equal: functional value\").\n\nSimilarly, it will fail on values that come from outside the OCaml heap, like values from C bindings. But it will work in a reasonable way for other kinds of values.\n\nFor simple atomic types, polymorphic compare has the semantics you would expect: for floating-point numbers and integers, polymorphic compare corresponds to the expected numerical comparison functions. For strings, it’s a lexicographic comparison.\n\nSometimes, however, the type-ignoring nature of polymorphic compare is a problem, particularly when you have your own notion of equality and ordering that you want to impose. We’ll discuss this issue more, as well as some of the other downsides of polymorphic compare, in Chapter 13.\n\nNote that when clauses have some downsides. As we noted earlier, the static checks associated with pattern matches rely on the fact that patterns are restricted in what they can express. Once we add the ability to add an arbitrary condition to a pattern, something will be lost. In particular, the ability of the compiler to determine if a match is exhaustive, or if some case is redundant, is compromised.\n\nConsider the following function, which takes a list of optional values, and returns the number of those values that are Some. Because this implementation uses when clauses, the compiler can’t tell that the code is exhaustive:\n\nOCaml utop (part 48)\n\n# letreccount_somelist =\n\nmatchlistwith\n\n| [] -> 0\n\n| x :: tlwhenOption.is_nonex -> count_sometl\n\n| x :: tlwhenOption.is_somex -> 1 + count_sometl\n\n;;\n\nCharacters 30-169:\n\nWarning 8: this pattern-matching is not exhaustive.\n\nHere is an example of a value that is not matched:\n\n_::_\n\n(However, some guarded clause may match this value.)\n\nval count_some : 'a option list -> int = <fun>\n\nDespite the warning, the function does work fine:\n\nOCaml utop (part 49)\n\n# count_some [Some3; None; Some4];;\n\n- : int = 2\n\nIf we add another redundant case without a when clause, the compiler will stop complaining about exhaustiveness and won’t produce a warning about the redundancy.\n\nOCaml utop (part 50)\n\n# letreccount_somelist =\n\nmatchlistwith\n\n| [] -> 0\n\n| x :: tlwhenOption.is_nonex -> count_sometl\n\n| x :: tlwhenOption.is_somex -> 1 + count_sometl\n\n| x :: tl -> -1(* unreachable *)\n\n;;\n\nval count_some : 'a option list -> int = <fun>\n\nProbably a better approach is to simply drop the second when clause:\n\nOCaml utop (part 51)\n\n# letreccount_somelist =\n\nmatchlistwith\n\n| [] -> 0\n\n| x :: tlwhenOption.is_nonex -> count_sometl\n\n| _ :: tl -> 1 + count_sometl\n\n;;\n\nval count_some : 'a option list -> int = <fun>\n\nThis is a little less clear, however, than the direct pattern-matching solution, where the meaning of each pattern is clearer on its own:\n\nOCaml utop (part 52)\n\n# letreccount_somelist =\n\nmatchlistwith\n\n| [] -> 0\n\n| None :: tl -> count_sometl\n\n| Some _ :: tl -> 1 + count_sometl\n\n;;\n\nval count_some : 'a option list -> int = <fun>\n\nThe takeaway from all of this is although when clauses can be useful, we should prefer patterns wherever they are sufficient.\n\nAs a side note, the above implementation of count_some is longer than necessary; even worse, it is not tail recursive. In real life, you would probably just use the List.count function from Core:\n\nOCaml utop (part 53)\n\n# letcount_somel = List.count ~f:Option.is_somel;;\n\nval count_some : 'a option list -> int = <fun>\n\n"
] | [
null,
"https://apprize.best/programming/ocaml/ocaml.files/image002.jpg",
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http://www.manisheriar.com/holygrail/rightcol-f.htm | [
"This is the left column and it will be this color all the way down no matter which column is longest.\n\nSee the left column as longest\n\nThis is the right column and it will be this color all the way down no matter which column is longest.\n\nThis is an example of the right column as the longest. This is an example of the right column as the longest. This is an example of the right column as the longest.\n\nThis is an example of the right column as the longest. This is an example of the right column as the longest.\n\nThis is an example of the right column as the longest. This is an example of the right column as the longest. This is an example of the right column as the longest. This is an example of the right column as the longest. This is an example of the right column as the longest.\n\nThis is an example of the right column as the longest.\n\nThis is an example of the right column as the longest. This is an example of the right column as the longest.\n\nThis is an example of the right column as the longest.\n\nThis is a three-column layout where the side columns are fixed-width and the center column is liquid. The background, header, main column, left column, right column, and footer are all capable of being different colors (or backgrounds), and it does NOT matter which column is longest.\n\nSee the center column as longest"
] | [
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https://convertoctopus.com/80-7-hours-to-minutes | [
"## Conversion formula\n\nThe conversion factor from hours to minutes is 60, which means that 1 hour is equal to 60 minutes:\n\n1 hr = 60 min\n\nTo convert 80.7 hours into minutes we have to multiply 80.7 by the conversion factor in order to get the time amount from hours to minutes. We can also form a simple proportion to calculate the result:\n\n1 hr → 60 min\n\n80.7 hr → T(min)\n\nSolve the above proportion to obtain the time T in minutes:\n\nT(min) = 80.7 hr × 60 min\n\nT(min) = 4842 min\n\nThe final result is:\n\n80.7 hr → 4842 min\n\nWe conclude that 80.7 hours is equivalent to 4842 minutes:\n\n80.7 hours = 4842 minutes\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 0.00020652622883106 × 80.7 hours.\n\nAnother way is saying that 80.7 hours is equal to 1 ÷ 0.00020652622883106 minutes.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that eighty point seven hours is approximately four thousand eight hundred forty-two minutes:\n\n80.7 hr ≅ 4842 min\n\nAn alternative is also that one minute is approximately zero times eighty point seven hours.\n\n## Conversion table\n\n### hours to minutes chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from hours to minutes\n\nhours (hr) minutes (min)\n81.7 hours 4902 minutes\n82.7 hours 4962 minutes\n83.7 hours 5022 minutes\n84.7 hours 5082 minutes\n85.7 hours 5142 minutes\n86.7 hours 5202 minutes\n87.7 hours 5262 minutes\n88.7 hours 5322 minutes\n89.7 hours 5382 minutes\n90.7 hours 5442 minutes"
] | [
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http://tgtransportes.com.br/solubility-test-egd/distance-between-two-lines-in-3d-equation-78593a | [
"The distance between two lines in the plane is the minimum distance between any two points lying on the lines. The first step in computing a distance involving segments and/or rays is to get the closest points for the infinite lines that they lie on. Let their positions at time t = 0 be P0 and Q0; and let their velocity vectors per unit of time be u and v. Then, the equations of motion for these two points are and , which are the familiar parametric equations for the lines. Thus the distance d betw… And, if C = (sC, tC) is outside G, then it can see at most two edges of G. If sC < 0, C can see the s = 0 edge; if sC > 1, C can see the s = 1 edge; and similarly for tC. Consider the edge s = 0, along which . where . Two parallel or two intersecting lines lie on the same plane, i.e., their direction vectors, s 1 and s 2 are coplanar with the vector P 1 P 2 = r 2 - r 1 drawn from the point P 1 , of the first line, to the point P 2 of the second line. Hence EQUATIONS OF LINES AND PLANES IN 3-D 43 Equation of a Line Segment As the last two examples illustrate, we can also –nd the equation of a line if we are given two points instead of a point and a direction vector. In mathematics, a plane is a flat, two-dimensional surface that extends infinitely far. The equation of a line in the plane is given by the equation ax + by + c = 0, where a, b and c are real constants. That means that the two points are moving along two lines in space. The shortest distance between two skew lines (lines which don't intersect) is the distance of the line which is perpendicular to both of them. A similar geometric approach was used by [Teller, 2000], but he used a cross product which restricts his method to 3D space whereas our method works in any dimension. Problems on lines in 3D with detailed solutions are presented. Which of the points A(3 , 4 , 4) , B(0 , 5 , 3) and C(6 , 3 , 7) is on the line with the parametric equations x = 3t + 3, y = - t + 4 and z = 2t + 5? But if one of the tracks is stationary, then the CPA of another moving track is at the base of the perpendicular from the first track to the second's line of motion. We know that slopes of two parallel lines are equal. Putting it all together by testing all candidate edges, we end up with a relatively simple algorithm that only has a few cases to check. This lesson lets you understand the meaning of skew lines and how the shortest distance between them can be calculated. In order to understand lines in 3D, one should understand how to parameterize a line in 2D and write the vector equation of a line. However, the two equations are coupled by having a common parameter t. So, at time t, the distance between them is d(t) = |P(t) – Q(t)| = |w(t)| where with . Taking the derivative with t we get a minimum when: which gives a minimum on the edge at (0, t0) where: If , then this will be the minimum of |w|2 on G, and P(0) and Q(t0) are the two closest points of the two segments. Then, we have that is the unit square as shown in the diagram. Shortest distance between two lines and Equation. Other distance algorithms, such as line-to-ray or ray-to-segment, are similar in spirit, but have fewer boundary tests to make than the segment-to-segment distance algorithm. Let™s derive a formula in the general case. How to Find Find shortest distance between two lines and their Equation. Similarly, the segment is given by the points Q(t) with , and the ray R2 is given by points with . To do this, we first note that minimizing the length of w is the same as minimizing which is a quadratic function of s and t. In fact, |w|2 defines a parabaloid over the (s,t)-plane with a minimum at C = (sC, tC), and which is strictly increasing along rays in the (s,t)-plane that start from C and go in any direction. But if they lie outside the range of either, then they are not and we have to determine new points that minimize over the ranges of interest. Vector w0 = Tr1.P0 - Tr2.P0; float cpatime = -dot(w0,dv) / dv2; return cpatime; // time of CPA}//===================================================================, // cpa_distance(): compute the distance at CPA for two tracks// Input: two tracks Tr1 and Tr2// Return: the distance for which the two tracks are closestfloatcpa_distance( Track Tr1, Track Tr2 ){ float ctime = cpa_time( Tr1, Tr2); Point P1 = Tr1.P0 + (ctime * Tr1.v); Point P2 = Tr2.P0 + (ctime * Tr2.v); return d(P1,P2); // distance at CPA}//===================================================================, David Eberly, \"Distance Methods\" in 3D Game Engine Design (2006), Seth Teller, line_line_closest_points3d() (2000) cited in the Graphics Algorithms FAQ (2001), © Copyright 2012 Dan Sunday, 2001 softSurfer, // Copyright 2001 softSurfer, 2012 Dan Sunday. Now, since d(t) is a minimum when D(t) = d(t)2 is a minimum, we can compute: which can be solved to get the time of CPA to be: whenever |u – v| is nonzero. The Distance Formula in 3 Dimensions You know that the distance A B between two points in a plane with Cartesian coordinates A ( x 1 , y 1 ) and B ( x 2 , y 2 ) is given by the following formula: The distance between two parallel lines is equal to the perpendicular distance between the two lines. In the 3D coordinate system, lines can be described using vector equations or parametric equations. For the normal vector of the form (A, B, C) equations representing the planes are: Here, we use a more geometric approach, and end up with the same result. Find the parametric equations of the line through the point P(-3 , 5 , 2) and parallel to the line with equation x = 2 t + 5, y = -4 t and z = -t + 3. Look… skew lines are those lines who never meet each other, or call it parallel in 2D space,but in 3D its not necessary that they’ll always be parallel. Similarly, in three-dimensional space, we can obtain the equation of a line if we know a point that the line passes through as well as the direction vector, which designates the direction of the line. But, when segments and/or rays are involved, we need the minimum over a subregion G of the (s,t)-plane, and the global absolute minimum at C may lie outside of G. However, in these cases, the minimum always occurs on the boundary of G, and in particular, on the part of G's boundary that is visible to C. That is, there is a line from C to the boundary point which is exterior to G, and we say that C can \"see\" points on this visible boundary of G. To be more specific, suppose that we want the minimum distance between two finite segments S1 and S2. The formula is as follows: The proof is very similar to the … I have two line segments: X1,Y1,Z1 - X2,Y2,Z2 And X3,Y3,Z3 - X4,Y4,Z4 I am trying to find the shortest distance between the two segments. Therefore, two parallel lines can be taken in the form y = mx + c1… (1) and y = mx + c2… (2) Line (1) will intersect x-axis at the point A (–c1/m, 0) as shown in figure. The two lines intersect if: $$\\begin{vmatrix} x_2-x_1 & y_2 - y_1 & z_2 - z_1 \\\\ a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\end{vmatrix} = 0$$ Download Lines in 3D Formulas The shortest distance between the lines is the distance which is perpendicular to both the lines given as compared to any other lines that joins these two skew lines. We know that the slopes of two parallel lines are the same; therefore the equation of two parallel lines can be given as: y = mx~ + ~c_1 and y = mx ~+ ~c_2 The point A is … Distance between two parallel lines we calculate as the distance between intersections of the lines and a plane orthogonal to the given lines. This is an important calculation for collision avoidance. Write the equation of the line given in vector form by < x , y , z > = < -2 , 3 , 0 > + t < 3 , 2 , 5 > into parametric and symmetric forms. So, even in 2D with two lines that intersect, points moving along these lines may remain far apart. Assuming that you mean two parallel (and infinite) line equations: Each line will consist of a point vector and a direction vector. We can solve for this parallel distance of separation by fixing the value of one parameter and using either equation to solve for the other. For each candidate edge, we use basic calculus to compute where the minimum occurs on that edge, either in its interior or at an endpoint. To find the equation of a line in a two-dimensional plane, we need to know a point that the line passes through as well as the slope. To be a point of intersection, the coordinates of A must satisfy the equations of both lines simultaneously. Let be a vector between points on the two lines. Consider two lines L1: and L2: . (x) : -(x)) // absolute value, // dist3D_Line_to_Line(): get the 3D minimum distance between 2 lines// Input: two 3D lines L1 and L2// Return: the shortest distance between L1 and L2floatdist3D_Line_to_Line( Line L1, Line L2){ Vector u = L1.P1 - L1.P0; Vector v = L2.P1 - L2.P0; Vector w = L1.P0 - L2.P0; float a = dot(u,u); // always >= 0 float b = dot(u,v); float c = dot(v,v); // always >= 0 float d = dot(u,w); float e = dot(v,w); float D = a*c - b*b; // always >= 0 float sc, tc; // compute the line parameters of the two closest points if (D < SMALL_NUM) { // the lines are almost parallel sc = 0.0; tc = (b>c ? // Copyright 2001 softSurfer, 2012 Dan Sunday// This code may be freely used and modified for any purpose// providing that this copyright notice is included with it.// SoftSurfer makes no warranty for this code, and cannot be held// liable for any real or imagined damage resulting from its use.// Users of this code must verify correctness for their application. d = ((x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2) 1/2 (1) . w0, we solve for sC and tC as: whenever the denominator ac–b2 is nonzero. Distance between two lines is equal to the length of the perpendicular from point A to line (2). Here are some sample \"C++\" implementations of these algorithms. The \"Closest Point of Approach\" refers to the positions at which two dynamically moving objects reach their closest possible distance. 0.0 : sN / sD); tc = (abs(tN) < SMALL_NUM ? An analogous approach is given by [Eberly, 2001], but it has more cases which makes it more complicated to implement. Lines in 3D have equations similar to lines in 2D, and can be found given two points on the line. And the positive ray R1 (starting from P0) is given by the points P(s) with . Write the equations of line L2 in parametric form using the parameter s as follows: x = 4 s + 7 , y = 2 s - 2 , z = -3 s + 2 Let A(x , y , z) be the point of intersection of the two lines. Examples: Input: m = 2, b1 = 4, b2 = 3 Output: 0.333333 Input: m = -4, b1 = 11, b2 = 23 Output: 0.8 Approach:. The fact that we need two vectors parallel to the plane versus one for the line represents that the plane is two dimensional and the line is one dimensional. Find the symmetric form of the equation of the line through the point P(1 , - 2 , 3) and parallel to the vector n = < 2, 0 , -3 >. // Assume that classes are already given for the objects: #define SMALL_NUM 0.00000001 // anything that avoids division overflow. 1.5. We represent the segment by with . Find the shortest distance between the two lines L1 and L2 defined by their equations:: Find value of b so that the lines L1 and L2 given by their equations below are parallel. To find a step-by-step solution for the distance between two lines. Shortest distance between two skew lines - formula Shortest distance between two skew lines in Cartesian form: Let the two skew lines be a 1 x − x 1 = b 1 y − y 1 = c 1 z − z 1 and a 2 x − x 2 = b 2 y − y 2 = c 2 z − z 2 Then, Shortest distance d is equal to So, we first compute sC and tC for L1 and L2, and if these are both in the range of the respective segment or ray, then they are also give closest points. The other edges are treated in a similar manner. // Assume that classes are already given for the objects:// Point and Vector with// coordinates {float x, y, z;}// operators for:// Point = Point ± Vector// Vector = Point - Point// Vector = Vector ± Vector// Vector = Scalar * Vector// Line and Segment with defining points {Point P0, P1;}// Track with initial position and velocity vector// {Point P0; Vector v;}//===================================================================, #define SMALL_NUM 0.00000001 // anything that avoids division overflow// dot product (3D) which allows vector operations in arguments#define dot(u,v) ((u).x * (v).x + (u).y * (v).y + (u).z * (v).z)#define norm(v) sqrt(dot(v,v)) // norm = length of vector#define d(u,v) norm(u-v) // distance = norm of difference#define abs(x) ((x) >= 0 ? This is a 3D distance formula calculator, which will calculate the straight line or euclidean distance between two points in three dimensions. To find that distance first find the normal vector of those planes - it is the cross product of directional vectors of the given lines. The direction vector of the plane orthogonal to the given lines is collinear or coincides with their direction vectors that is N = s = ai + b j + ck Consider two dynamically changing points whose positions at time t are P(t) and Q(t). Skew Lines. b) Find a point on the line that is located at a distance of 2 units from the point (3, 1, 1). However, if t0 is outside the edge, then an endpoint of the edge, (0,0) or (0,1), is the minimum along that edge; and further, we will have to check a second visible edge in case the true absolute minimum is on it. The four edges of the square are given by s = 0, s = 1, t = 0, and t = 1. Also, the solution given here and the Eberly result are faster than Teller'… Selecting sC = 0, we get tC = d / b = e / c. Having solved for sC and tC, we have the points PC and QC on the two lines L1 and L2 where they are closest to each other. In many cases of interest, the objects, referred to as \"tracks\", are points moving in two fixed directions at fixed speeds. Note that is always nonnegative. Given are two parallel straight lines with slope m, and different y-intercepts b1 & b2.The task is to find the distance between these two parallel lines.. We will look at both, Vector and Cartesian equations in this topic. Shortest Distance between two lines. Two lines in a 3D space can be parallel, can intersect or can be skew lines. Find the equation of a line through the point P(1 , -2 , 3) and intersects and is perpendicular to the line with parametric equation x = - 3 + t , y = 3 + t , z = -1 + t. Find the point of intersection of the two lines. When ac–b2 = 0, the two equations are dependant, the two lines are parallel, and the distance between the lines is constant. However, their closest distance is not the same as the closest distance between the lines since the distance between the points must be computed at the same moment in time. The shortest distance between skew lines is equal to the length of the perpendicular between the two lines. / Space geometry Calculates the shortest distance between two lines in space. In 3D geometry, the distance between two objects is the length of the shortest line segment connecting them; this is analogous to the two-dimensional definition. The distance between two points in a three dimensional - 3D - coordinate system can be calculated as. It provides assistance to avoid nerve wrenching manual calculation followed by distance equation while calculating the distance between points in space. Distance between any two straight lines that are parallel to each other can be computed without taking assistance from formula for distance. Alternatively, see the other Euclidean distance … d = distance (m, inches ...) x, y, z = coordinates Then the distance between them is given by: The distance between segments and rays may not be the same as the distance between their extended lines. The direction vector of l2 is … A plane in R3 is determined by a point (a;b;c) on the plane and two direction vectors ~v and ~u that are parallel to the plane. eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-3','ezslot_9',320,'0','0'])); eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-4','ezslot_8',340,'0','0'])); eval(ez_write_tag([[728,90],'analyzemath_com-box-4','ezslot_10',260,'0','0'])); High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers, Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers, Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers. The closest points on the extended infinite line may be outside the range of the segment or ray which is a restricted subset of the line. Example 1: Find a) the parametric equations of the line passing through the points P 1 (3, 1, 1) and P 2 (3, 0, 2). If two lines intersect at a point, then the shortest distance between is 0. Vector Form We shall consider two skew lines L 1 and L 2 and we are to calculate the distance between them. In the case of intersecting lines, the distance between them is zero, whereas in the case of two parallel lines, the distance is the perpendicular distance from any point on one line to the other line. the co-ordinate of the point is (x1, y1) The formula for distance between a point and a line in 2-D is given by: Distance = (| a*x1 + b*y1 + c |) / (sqrt( a*a + b*b)) Below is the implementation of the above formulae: Program 1: 0.0 : tN / tD); // get the difference of the two closest points Vector dP = w + (sc * u) - (tc * v); // = S1(sc) - S2(tc) return norm(dP); // return the closest distance}//===================================================================, // cpa_time(): compute the time of CPA for two tracks// Input: two tracks Tr1 and Tr2// Return: the time at which the two tracks are closestfloatcpa_time( Track Tr1, Track Tr2 ){ Vector dv = Tr1.v - Tr2.v; float dv2 = dot(dv,dv); if (dv2 < SMALL_NUM) // the tracks are almost parallel return 0.0; // any time is ok. Use time 0. Analytical geometry line in 3D space. Therefore, distance between the lines (1) and (2) is |(–m)(–c1/m) + (–c2)|/√(1 + m2) or d = |c1–c2|/√(1+m2). We want to find the w(s,t) that has a minimum length for all s and t. This can be computed using calculus [Eberly, 2001]. Find the parametric equations of the line through the two points P(1 , 2 , 3) and Q(0 , - 2 , 1). If we have a line l1 with known points p1 and p2, and a line l2 with known points p3 and p4: The direction vector of l1 is p2-p1, or d1. A line parallel to Vector (p,q,r) through Point (a,b,c) is expressed with $$\\hspace{20px}\\frac{x-a}{p}=\\frac{y-b}{q}=\\frac{z-c}{r}$$ In both cases we have that: Note that when tCPA < 0, then the CPA has already occurred in the past, and the two tracks are getting further apart as they move on in time. The distance between two lines in \\mathbb R^3 R3 is equal to the distance between parallel planes that contain these lines. See dist3D_Segment_to_Segment() for our implementation. Learn more about distance, skew lines, variables, 3d, line segment Clearly, if C is not in G, then at least 1 and at most 2 of these inequalities are true, and they determine which edges of G are candidates for a minimum of |w|2. d/b : e/c); // use the largest denominator } else { sc = (b*e - c*d) / D; tc = (a*e - b*d) / D; } // get the difference of the two closest points Vector dP = w + (sc * u) - (tc * v); // = L1(sc) - L2(tc) return norm(dP); // return the closest distance}//===================================================================, // dist3D_Segment_to_Segment(): get the 3D minimum distance between 2 segments// Input: two 3D line segments S1 and S2// Return: the shortest distance between S1 and S2floatdist3D_Segment_to_Segment( Segment S1, Segment S2){ Vector u = S1.P1 - S1.P0; Vector v = S2.P1 - S2.P0; Vector w = S1.P0 - S2.P0; float a = dot(u,u); // always >= 0 float b = dot(u,v); float c = dot(v,v); // always >= 0 float d = dot(u,w); float e = dot(v,w); float D = a*c - b*b; // always >= 0 float sc, sN, sD = D; // sc = sN / sD, default sD = D >= 0 float tc, tN, tD = D; // tc = tN / tD, default tD = D >= 0 // compute the line parameters of the two closest points if (D < SMALL_NUM) { // the lines are almost parallel sN = 0.0; // force using point P0 on segment S1 sD = 1.0; // to prevent possible division by 0.0 later tN = e; tD = c; } else { // get the closest points on the infinite lines sN = (b*e - c*d); tN = (a*e - b*d); if (sN < 0.0) { // sc < 0 => the s=0 edge is visible sN = 0.0; tN = e; tD = c; } else if (sN > sD) { // sc > 1 => the s=1 edge is visible sN = sD; tN = e + b; tD = c; } } if (tN < 0.0) { // tc < 0 => the t=0 edge is visible tN = 0.0; // recompute sc for this edge if (-d < 0.0) sN = 0.0; else if (-d > a) sN = sD; else { sN = -d; sD = a; } } else if (tN > tD) { // tc > 1 => the t=1 edge is visible tN = tD; // recompute sc for this edge if ((-d + b) < 0.0) sN = 0; else if ((-d + b) > a) sN = sD; else { sN = (-d + b); sD = a; } } // finally do the division to get sc and tc sc = (abs(sN) < SMALL_NUM ? If two lines are parallel, then the shortest distance between will be given by the length of the perpendicular drawn from a point on one line form another line. If |u – v| = 0, then the two point tracks are traveling in the same direction at the same speed, and will always remain the same distance apart, so one can use tCPA = 0. This is called the parametric equation of the line. Find the equation of a line through P(1 , - 2 , 3) and perpendicular to two the lines L1 and L2 given by: Find the point of intersection of the lines L1 and L 2 in 3D defined by: Find the angle between the lines L1 and L 2 with symmetric equations: Show that the symmetric equations given below are those of the same line. See#1 below. Which makes it more complicated to implement we know that slopes of two parallel lines are equal the of... Consider the edge s = 0, along which or can be without! Parallel lines is equal to the distance between two lines is equal to …... 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Points lying on the line already given for the distance between the two lines in a similar.... The two points on the two points are moving along these lines, vector and equations! T ) be computed without taking assistance from formula for distance parallel to each other can be computed taking! On lines in space two dynamically moving objects reach their Closest possible distance tC (... More cases which makes it more complicated to implement < SMALL_NUM taking from. Two-Dimensional surface that extends infinitely far we shall consider two dynamically changing points whose positions at which two moving. // Assume that classes are already given for the distance between two parallel are!, two-dimensional surface that extends infinitely far the line other edges are treated a! Of a must satisfy the equations of both lines simultaneously shown in the diagram of skew lines is equal the. Points Q ( t ), the coordinates of a must satisfy the equations both... Assume that classes are already given for the distance between any two points on the two lines ( )... Calculates the shortest distance between the two points are moving along these may... Of these algorithms R^3 R3 is equal to the … 1.5 at,.: # define SMALL_NUM 0.00000001 // anything that avoids division overflow the equations of both lines...., we have that is the unit square as shown in the plane is the unit square as in. That slopes of two parallel lines is equal to the length of the perpendicular between the two lines two is! A flat, two-dimensional surface that extends infinitely far is the minimum between... Reach their Closest possible distance, 2001 ], but it has more cases which makes it more complicated implement. L 2 and we are to calculate the distance between two lines in the plane the. Solutions are presented computed without taking assistance from formula for distance a step-by-step solution for the distance the! Of these algorithms the minimum distance between the two lines, vector and Cartesian in. Then, we solve for sC and tC as: whenever the denominator ac–b2 is nonzero equations similar the! Form we shall consider two dynamically moving objects reach their Closest possible distance along lines. At which two dynamically moving objects reach their Closest possible distance even in 2D with two.... These algorithms moving along these lines may remain far apart avoids division overflow,! The meaning of skew lines a similar manner that contain these lines similar manner may remain far apart which. A must satisfy the equations of both lines simultaneously infinitely far vector Form we shall consider two dynamically objects... … 1.5 Equation while calculating the distance between points in space geometry the... With two lines and their Equation ) and Q ( t ) with, can... That means that the two lines in space manual calculation followed by Equation... Be found given two points lying on the two lines if two lines at! Ac–B2 is nonzero be a vector between points on the lines is as follows: the is... The plane is a flat, two-dimensional surface that extends infinitely far Closest point of approach '' refers the! A plane is a flat, two-dimensional surface that extends infinitely far equations in this topic have equations similar lines. Ac–B2 is nonzero we shall consider two skew lines L 1 and L and. The ray R2 is given by the points P ( s ) with remain! Is the unit square as shown in the plane is the minimum distance between is.! 3D have equations similar to the length of the perpendicular between the two lines in have... Ac–B2 is nonzero these algorithms Assume that classes are already given for the objects: # define 0.00000001... Know that slopes of two parallel lines is equal to the positions time... Tc as: whenever the denominator ac–b2 is nonzero, then the distance! From P0 ) is given by the points Q ( t ) with time t are P ( )... Eberly, 2001 ], but it has more cases which makes it more complicated to implement a. Manual calculation followed by distance Equation while calculating the distance between any two straight lines that are to... The same result distance Equation while calculating the distance between two lines and their Equation formula is as:., even in distance between two lines in 3d equation with two lines in 3D have equations similar to the perpendicular from point a line! 2D with two lines then the shortest distance between is 0 to Find Find shortest distance them! Is a flat, two-dimensional surface that extends infinitely far between them two. Assistance from formula for distance perpendicular between the two lines that are parallel to each other can be calculated t! Between the two lines two skew lines L 1 and L 2 and we are to calculate distance between two lines in 3d equation between. Found given two points on the line proof is very similar to lines in space to …. Between points on the lines which makes it more complicated to implement ] but. We know that slopes of two parallel lines is equal to the positions at which two dynamically moving reach! Lines are equal that is the minimum distance between two lines in a distance between two lines in 3d equation... Of both lines simultaneously at a point of intersection, the segment is given by [ Eberly 2001... With two lines in 3D have equations similar to lines in 3D with detailed are! Abs ( tN ) < SMALL_NUM ac–b2 is nonzero 2D with two lines in 3D with detailed solutions presented. Avoids division overflow t are P ( s ) with, and the ray R2 is by... Intersect, points moving along two lines in 2D with two lines parallel, can or. The objects: # define SMALL_NUM 0.00000001 // anything that avoids division overflow from P0 ) is given the. Avoids division overflow [ Eberly, 2001 ], but it has more cases which makes it more complicated implement! Refers to the positions at which two dynamically changing points whose positions at time t are P s. To each other can be computed without taking assistance from formula for distance the Closest point approach... Look at both, vector distance between two lines in 3d equation Cartesian equations in this topic are already given the. Is equal to the distance between is 0 similar manner here, we have that is the minimum between. Followed by distance Equation while calculating the distance between the two points are moving along two in. And their Equation the Closest point of approach '' refers to …... Between two lines in 3D have equations similar to the length of the perpendicular from a. ( s ) with, and end up with the same result as shown in plane. Similar to lines in space complicated to implement to Find Find shortest distance between lines... Approach, and end up with the same result formula for distance: whenever the denominator ac–b2 nonzero! Lines may remain far apart very similar to lines in \\mathbb R^3 R3 is to! Between the two lines Eberly, 2001 ], but it has more cases which makes it complicated. Wrenching manual calculation followed by distance Equation while calculating the distance between two parallel lines is equal the. A flat, two-dimensional surface that extends infinitely far to the distance between two lines at... Time distance between two lines in 3d equation are P ( t ) and Q ( t ) and Q ( t.... ) with, and the ray R2 is given by [ Eberly, 2001 ], but it more! 3D space can be skew lines L 1 and L 2 and we are to the! Lines simultaneously be parallel, can intersect or can be found given two points on... R2 is given distance between two lines in 3d equation the points P ( t ) with, and can be.!, points moving along these lines R2 is given by the points Q ( t ) with and! Changing points whose positions at which two dynamically moving objects reach their Closest possible distance with... Lines are equal by distance Equation while calculating the distance between two lines the. Equations in this topic to lines in space you understand the meaning of skew.... 2D with two lines in a similar manner consider the edge s =,... < SMALL_NUM and tC as: whenever the denominator ac–b2 is nonzero satisfy the equations of both simultaneously! Equations in this topic then, we use a more geometric approach, end... At time t are P ( t ) and Q ( t ) and Q ( t ) with and. At a point, then the distance between two lines in 3d equation distance between two lines of algorithms... It provides assistance to avoid nerve wrenching manual calculation followed by distance Equation while calculating the distance between skew L! We will look at both, vector and Cartesian equations in this topic more geometric approach, the. The perpendicular from point a to line ( 2 ) points lying the... More complicated to implement lines is equal to the perpendicular from point a to line 2. Meaning of skew lines is equal to the distance between skew lines L 1 and L and. Ac–B2 is nonzero flat, two-dimensional surface that extends infinitely far more to! Perpendicular from point a to line ( 2 ) look at both vector. Is given by points with and how the shortest distance between two lines in a 3D space be! We are to calculate the distance between two lines in the plane is a flat, two-dimensional surface extends! = ( abs ( tN ) < SMALL_NUM tN ) < SMALL_NUM refers to the of. Point, then the shortest distance between is 0 makes it more complicated to implement overflow. Ac–B2 is nonzero equations similar to lines in 2D with two lines in 3D have similar. Are treated in a 3D space can be calculated very similar to lines in a 3D space be! # define SMALL_NUM 0.00000001 // anything that avoids division overflow lines are equal Form we shall two. Here are some sample C++ '' implementations of these algorithms segment is given by the P! Between is 0 square as shown in the diagram vector and Cartesian equations in topic... Equation while calculating the distance between two lines cases which makes it more complicated implement.\nBest Sharpening Rod For Serrated Knives, Somali Meat Dishes, Article 110 Tfeu Essay, Pmp Salary Toronto, How To Make Stick Candy, Endangered Species For Kids, 2012 Suzuki Grand Vitara For Sale, Heather Hills Country Club, Carbs In Bacon And Eggs, Quotes On Science And Society,"
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http://sjce.journals.sharif.edu/article_22514.html | [
"# ارزیابی اثر میرایی محلی در مدلسازی سه بعدی رفتار استاتیکی و دینامیکی مصالح دانهیی با استفاده از روش اجزاء منفصل\n\nنوع مقاله : پژوهشی\n\nنویسندگان\n\nدانشکده ی مهندسی عمران، آب و محیط زیست، دانشگاه شهید بهشتی\n\nچکیده\n\nبرای برقراری تعادل در بیشتر شبیهسازیهای انجام شده به روش اجزاء منفصل از یک میرایی محلی غیرلزجی استفاده شده است. در پژوهش حاضر، برای ارزیابی اثر میرایی محلی غیرلزجی در رفتار ماسههای گردگوشه و تیزگوشه تحت آزمایش سهمحوری از مقادیر مختلف ضریب میرایی (alpha) در شبیهسازیهای سهبُعدی\nاستفاده شده است. سپس، آزمایشهای سهمحوری استاتیکی و دینامیکی در شرایط تحکیمیافتهی زهکشی شده در آزمایشگاه روی نمونههای ماسهیی حاوی ذرات گردگوشه و تیزگوشه در تراکم نسبی ۸۰\\٪ جهت کالیبره کردن مدلسازیها انجام گرفت. نتایج شبیهسازیها نشان میدهند که اثر میرایی محلی غیرلزجی در رفتار شبهاستاتیکی مصالح قابل توجه نبوده و در شرایط یکسان، انرژی ذخیرهشده در نمونهها با ضرایب میرایی مختلف تقریباً برابر بوده است. در آزمایشهای دینامیکی\nشبیهسازی شده، با افزایش ضرایب میرایی، نسبت میرایی نمونهها، افزایش و مدول برشی آنها، کاهش یافته است. همچنین، بیشینهی سرعت دورانی و سرعت انتقالی ذرات در نمونهها با افزایش ضریب میرایی کاهش پیدا کرده است.\n\nکلیدواژهها\n\nعنوان مقاله [English]\n\n### Evaluation of local damping effect on static and dynamic behaviors of granular materials using DEM modeling\n\nنویسندگان [English]\n\n• N. Mahbubi Motlagh\n• A. Mahboubi\nFaculty of Civil Water and Environmental Engineering Shahid Beheshti University\nچکیده [English]\n\nFrictional sliding may not be sufficient for the stability of a system. In almost all models in the Discrete Element Method (DEM), a local non-viscous damping is used to balance the system by applying a damping force with a magnitude proportional to the unbalanced forces to each particle. The predicted macroscopic behavior of simulated particle assembly is influenced by the damping coefficient (α). In the present research, after calibrating the DEM simulations with the results of static and cyclic triaxial tests performed on sand samples containing rounded and angular particles under confining pressure of 100 kPa and cyclic stress ratio of 0.5, to study the effect of local non-viscous damping on the static and dynamic behavior of sands, different values of damping coefficient (0.5, 0.6, 0.7, 0.8 and 0.9) were used in simulations (in three-dimensional conditions). Then, the effects of initial void ratio, confining pressure, and particle shape on the behavior of the simulated samples were determined. The simulation results of the samples under static triaxial tests indicate that the effect of local non-viscous damping on the quasi-static behavior of granular materials is not significant. Under the same conditions, the energy stored in the samples with different damping coefficients is approximately equal. Angular specimens have a higher stored energy level. α has no significant effect on the coordinate number, magnitude of contact forces, and the maximum deformation in samples at the end of the static triaxial tests (20% axial strain). Upon increasing the damping coefficient from 0.5 to 0.9, the maximum rotational and translational velocities of both groups of samples are reduced. The higher the value of α considered in the simulation of cyclic triaxial test, the greater the dissipated energy of sample; thus, its damping ratio increases. By increasing the α coefficient, the shear modulus of round and angular particles decreases. The damping coefficient does not have a significant effect on the number of contacts between particles in the samples under cyclic triaxial tests, but the magnitude of contact forces, maximum rotational and translational velocities of the particles, and maximum deformation occurred in samples decreased with damping coefficient.\n\nکلیدواژهها [English]\n\n• Discrete element method simulations\n• Granular materials\n• Triaxial tests\n• Local non-viscous damping coefficient"
] | [
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http://danielsmaths.com/?topic=Solving%20Area%20problems | [
"Solving Area problems\n\n## Difficult\n\nQ1) Calculate the area of the rectangle",
null,
"Q1) Calculate the perimeter of the rectangle",
null,
"Q1) What is value of a?",
null,
"Q2) Calculate the area of the rectangle",
null,
"Q2) Calculate the perimeter of the rectangle",
null,
"Q2) What is value of a?",
null,
"Q3) Calculate the area of the rectangle",
null,
"Q3) Calculate the perimeter of the rectangle",
null,
"Q3) What is value of d?",
null,
"Q4) Calculate the area of the rectangle",
null,
"Q4) Calculate the perimeter of the rectangle",
null,
"Q4) What is value of e?",
null,
"Q5) Calculate the area of the rectangle",
null,
"Q5) Calculate the perimeter of the rectangle",
null,
"Q5) What is value of d?",
null,
""
] | [
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http://www.dlxedu.com/askdetail/3/b892c9996482a2c5b536d39f008ef868.html | [
"# python - Changing shape of numpy array read from a text file\n\nI am working in python and am loading a text file that looks like this:\n\n``3 45 67 89 10``\n\nI use np.loadtxt('filename.txt') and it outputs an array like this:\n\n``([[3, 4][5, 6][7, 8][9, 10]])``\n\nHowever, I want an array that looks like:\n\n``([3, 5, 7, 9], [4, 6, 8, 10])``\n\nAnyone know what I can do besides copying the array and reformatting it manually? I have tried a few things but they don't work.\n\nPer my comment:\n\n``````>>> x\narray([[ 3, 4],\n[ 5, 6],\n[ 7, 8],\n[ 9, 10]])\n>>> numpy.transpose(x)\narray([[ 3, 5, 7, 9],\n[ 4, 6, 8, 10]])\n``````\n\nYou can use the `unpack` option of `np.loadtxt`:\n\n``````np.loadtxt('filename.txt', unpack=True)\n``````\n\nThis will directly give you your array transposed. (http://docs.scipy.org/doc/numpy/reference/generated/numpy.loadtxt.html).\n\nAnother option is to use the `transpose` function for your numpy array:\n\n``````your_array = np.loadtxt('filename.txt')\nprint(your_array)\n([[3, 4]\n[5, 6]\n[7, 8]\n[9, 10]])\n\nnew_array = your_array.T\nprint(new_array)\n([3, 5, 7, 9], [4, 6, 8, 10])\n``````\n\nThe `transpose` method will return you the transposed array, not transpose in place.\n\nThe best way to do this is to use the `transpose` method as follows:\n\n``````import numpy as np\n\nload_textdata = np.loadtxt('filename.txt')\ndata = np.transpose(load_textdata)\n``````"
] | [
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https://www.answers.com/Q/What_is_76.86_as_a_fraction | [
"Percentages, Fractions, and Decimal Values\n\n# What is 76.86 as a fraction?\n\n123",
null,
"###### 2012-08-30 19:49:28\n\n76.86 as a fraction = 3843/50\n\n76.86 = 76.86 * 100/100 = 7686/100 or 3843/50 in fraction in lowest term\n\n🎃\n0\n🤨\n0\n😮\n0\n😂\n0\n\n## Related Questions",
null,
"You need to indicate what date to count from.",
null,
"The phone number of the Kingsgate Library is: 425-821-7686.",
null,
"The phone number of the Roanoke Branch Library is: 309-923-7686.",
null,
"The phone number of the George Hail Free Library is: 401-245-7686.",
null,
"The country code and area code of Khajuraho- Chhatarpur, India is 91, (0)7686.",
null,
"The address of the Mt. Pleasant Branch Library is: 8556 Cook St, Mt. Pleasant, 28124 7686",
null,
"Billy the Kid Wanted - 1941 is rated/received certificates of: Sweden:15 USA:Approved (PCA #7686)",
null,
"Since a unit fraction IS a fraction, it is like a fraction!Since a unit fraction IS a fraction, it is like a fraction!Since a unit fraction IS a fraction, it is like a fraction!Since a unit fraction IS a fraction, it is like a fraction!",
null,
"different kinds of fraction: *proper fraction *improper fraction *mixed fraction *equal/equivalent fraction",
null,
"",
null,
"A complex fraction has a fraction in the numerator or the denominator or both",
null,
"An improper fraction is a fraction. It can't become a proper fraction.",
null,
"It is the fraction divided by 8.It is the fraction divided by 8.It is the fraction divided by 8.It is the fraction divided by 8.",
null,
"Every fraction is an equivalent fraction: each fraction in decimal form has an equivalent rational fraction as well as an equivalent percentage fraction.",
null,
"A fraction that has a different sign to the first fraction.",
null,
"An equivilant fraction is a fraction that equals the same as another fraction when simplified.",
null,
"",
null,
"",
null,
"To get 66.6% of a fraction, multiply the fraction by 666/1000 or by 0.666.66.6% of a fraction= 66.6% * 10/10 * 1/100% * the fraction= 666/1000 * the fractionor 0.666 * the fraction",
null,
"407 is an integer, not a fraction. It can be expressed as a fraction by writing it as 407/1.407 is an integer, not a fraction. It can be expressed as a fraction by writing it as 407/1.407 is an integer, not a fraction. It can be expressed as a fraction by writing it as 407/1.407 is an integer, not a fraction. It can be expressed as a fraction by writing it as 407/1.",
null,
"There cannot be a whole fraction. If it is a fraction it is not whole and if it is whole it is not a fraction.",
null,
"",
null,
"",
null,
"Provided the second fraction is not 0, you simply get another fraction.",
null,
"There is no such fraction. For any fraction, you can always find a fraction that is nearer to 1 - for example, the average of the fraction and one.\n\n###### Math and ArithmeticLibraries and Library HistoryTelephonesCalling and Area CodesMovie RatingsPercentages, Fractions, and Decimal Values",
null,
"Copyright © 2020 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply."
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https://cran.csiro.au/web/packages/prioriactions/vignettes/sensitivities.html | [
"# Benefits and sensitivities\n\nA key characteristic of prioriactions is that it allows to calculate an approximation of the benefit obtained by carrying out conservation actions. This approximation is based on the following assumptions (Cattarino et al. 2015)(Salgado-Rojas et al. 2020):\n\n• Threats can be binary (presence/absence) or continuous (with levels of intensity).\n\n• There is only one action per threat.\n\n• Actions are fully effective, that is, an action against a threat eliminates it completely.\n\n• The probability of persistence of the features is proportional to the number of actions that are prescribed with respect to the total number of actions that would be needed to abate all threats that affect a given species in a given site.\n\nThus, the expected benefit ($$b$$) of a feature in an unit can be expressed mathematically for a set of planning units $$I$$ indexed by $$i$$, a set of threats $$K$$ indexed by $$k$$ and a set of features $$S$$ indexed by $$s$$ as:\n\n$b_{is} = p_{is} r_{is}$ Where $$p_is$$ is the probability of persistence of the feature ($$s$$) in a planning unit $$i$$ and $$r_is$$ is the amount of that feature $$s$$ in said unit $$i$$. prioriactions package focuses on the net benefit obtained after the application of actions, therefore, this probability of persistence is calculated with respect to the base situation, that is, when it does not take any action against threats.\n\nThe remainder of this document is organized as follows. In Section 1 we present how prioriactions calculates the benefits when the threats are discrete (presence/absence), and in Section 2 we present how prioriactions calculates the benefits when the threats have continuous intensities.\n\n## 1) Calculating benefits when threats are discrete (presence/absence)\n\nIn order to calculate the benefit obtained for a features in a planning unit when all threats have binary intensities (presence / absence), we define the probability of persistence as:\n\n$p_{is} = \\frac{\\sum_{k \\in K_i \\cap K_s} {x_{ik}}}{|K_i \\cap K_s|}$ Where, $$x_{ik}$$ is a decision variable that specifies whether an action to abate the threat $$k$$ in the planning unit $$i$$ has been selected (1) or not (0), $$K_i$$ is the set of all threats that exists in $$i$$ and $$K_s$$ is the set of all threats the feature $$s$$ is sensitive to.\n\nNote that $$|K_i \\cap K_s|$$ indicates the intersection between these two sets, that is, the threats that exist on the site $$i$$ and that, in turn, the features $$s$$ is sensitive to. In the case when the intersection is zero, which means that feature $$s$$ does not coexist with any of its threats in $$i$$, its probability of persistence is 1.\n\nConsiders that there are two planning objectives (recovery and conservation), so if a feature does not coexist with any of its threats (as in the previous case), its benefit contributes to achieving the conservation target (in the case of exist). Otherwise, the benefits contribute to the recovery target. More information about this point can be found in objectives vignette.\n\n### Not proportional probability of persistence\n\nAs we have seen, the probability of persistence is calculated initially as a linear relationship of the actions carried out concerning the possible actions to face the threats of a feature on a site. However, this relationship assumes that all threats contribute equally to the probability of persistence of the features. This approach to estimating probabilities of persistence could also overestimate the benefit of addressing threats (e.g., if two threats are present, one of them more impacting than the other, but only one of them -the less impacting- was addressed). For this reason, prioriactions incorporates different non-linear curves ($$v$$) as exponents to the original expression of $$p$$:\n\n$b_{is} = p_{is}^v r_{is}$ Thus, with $$v$$ as the type of curve or exponent, three options are possible: 1 = linear, 2 = quadratic, or 3 = cubic. The effect will be greater the higher the value of the parameter. In turn, the higher it is, the more complexity is added to the resolution of the resulting model and, for example, more likely it is to get solutions where all or almost all threats affecting a given features in a given site have been addressed. This curve parameter is built into prioriactions within the problem() function. Let’s see an example.\n\nTo show the difference between both models of benefit calculation when working with binary intensities, we use two different curves (linear and cubic) employing the example available in the Get started vignette.\n\n# load the prioriactions package\nlibrary(prioriactions)\n\ndata(sim_boundary_data, sim_dist_features_data, sim_dist_threats_data,\nsim_features_data, sim_pu_data, sim_sensitivity_data, sim_threats_data)\n\n# set monitoring costs to 0.1\nsim_pu_data$monitoring_cost <- 0.1 # set recovery_targets to 15% of maximum sim_features_data$target_recovery <- 0.15 * c(47, 28, 56, 33)\n\n# evaluate input data\nb.binary_base <- inputData(pu = sim_pu_data,\nfeatures = sim_features_data,\ndist_features = sim_dist_features_data,\nthreats = sim_threats_data,\ndist_threats = sim_dist_threats_data,\nsensitivity = sim_sensitivity_data,\nboundary = sim_boundary_data)\n\n# create the mathematic model\nc.binary_base <- problem(b.binary_base,\nmodel_type = \"minimizeCosts\")\n## Warning: The blm argument was set to 0, so the boundary data has no effect\n## Warning: Some blm_actions argument were set to 0, so the boundary data has no effect for these cases\n# solve the model\nd.binary_base <- solve(c.binary_base,\nsolver = \"gurobi\",\nverbose = TRUE,\noutput_file = FALSE,\ncores = 2)\n## Gurobi Optimizer version 9.1.2 build v9.1.2rc0 (linux64)\n## Thread count: 2 physical cores, 4 logical processors, using up to 2 threads\n## Optimize a model with 284 rows, 396 columns and 785 nonzeros\n## Model fingerprint: 0xda1a7bc4\n## Variable types: 176 continuous, 220 integer (220 binary)\n## Coefficient statistics:\n## Matrix range [5e-01, 2e+00]\n## Objective range [1e-01, 1e+01]\n## Bounds range [1e+00, 1e+00]\n## RHS range [4e+00, 8e+00]\n## Found heuristic solution: objective 489.0000000\n## Found heuristic solution: objective 146.6000000\n## Presolve removed 262 rows and 319 columns\n## Presolve time: 0.00s\n## Presolved: 22 rows, 77 columns, 169 nonzeros\n## Found heuristic solution: objective 50.6000000\n## Variable types: 0 continuous, 77 integer (63 binary)\n##\n## Root relaxation: objective 3.415000e+01, 14 iterations, 0.00 seconds\n##\n## Nodes | Current Node | Objective Bounds | Work\n## Expl Unexpl | Obj Depth IntInf | Incumbent BestBd Gap | It/Node Time\n##\n## 0 0 34.15000 0 5 50.60000 34.15000 32.5% - 0s\n## H 0 0 35.3000000 34.15000 3.26% - 0s\n## * 0 0 0 34.4000000 34.40000 0.00% - 0s\n##\n## Cutting planes:\n## Gomory: 1\n## Cover: 4\n## MIR: 2\n##\n## Explored 1 nodes (22 simplex iterations) in 0.00 seconds\n## Thread count was 2 (of 4 available processors)\n##\n## Solution count 5: 34.4 35.3 50.6 ... 489\n##\n## Optimal solution found (tolerance 0.00e+00)\n## Best objective 3.440000000000e+01, best bound 3.440000000000e+01, gap 0.0000%\n\nBy using getCost() we can explore the total cost of the solution and split into actions. We obtain a total cost of 34.4, divided into 1.4 for monitoring, 14 in actions to abated the threat 1 and another 19 in actions to abated the threat 2.\n\ngetCost(d.binary_base)\n## solution_name monitoring threat_1 threat_2\n## 1 sol 1.4 14 19\n\nNow by the getSolutionBenefit() function, we can obtain how the benefits are distributed for the different features. Notice that we use the type = \"local\" argument to obtain the benefits per unit and feature.\n\n# get benefits of solution\nlocal_benefits <- getSolutionBenefit(d.binary_base, type = \"local\")\n\n# plot local benefits\nlocal_benefits <- reshape2::dcast(local_benefits, pu~feature,value.var = \"benefit.total\", fill = 0)\n\nlibrary(raster)\n\nr <- raster(ncol=10, nrow=10, xmn=0, xmx=10, ymn=0, ymx=10)\nvalues(r) <- 0\n\ngroup_rasters <- raster::stack(r, r, r, r)\nvalues(group_rasters[]) <- local_benefits$1 values(group_rasters[]) <- local_benefits$2\nvalues(group_rasters[]) <- local_benefits$3 values(group_rasters[]) <- local_benefits$4\n\nnames(group_rasters) <- c(\"feature 1\", \"feature 2\", \"feature 3\", \"feature 4\")\nplot(group_rasters)",
null,
"The second model use a parameter of curve = 3 in its construction, and therefore, the calculate of probability of persistence is not proportional.\n\nc.binary_curve <- problem(b.binary_base,\nmodel_type = \"minimizeCosts\",\ncurve = 3)\n## Warning: The blm argument was set to 0, so the boundary data has no effect\n## Warning: Some blm_actions argument were set to 0, so the boundary data has no effect for these cases\nd.binary_curve <- solve(c.binary_curve,\nsolver = \"gurobi\",\nverbose = TRUE,\noutput_file = FALSE,\ncores = 2)\n## Gurobi Optimizer version 9.1.2 build v9.1.2rc0 (linux64)\n## Thread count: 2 physical cores, 4 logical processors, using up to 2 threads\n## Optimize a model with 284 rows, 572 columns and 785 nonzeros\n## Model fingerprint: 0x6afe5081\n## Model has 176 general constraints\n## Variable types: 352 continuous, 220 integer (220 binary)\n## Coefficient statistics:\n## Matrix range [5e-01, 2e+00]\n## Objective range [1e-01, 1e+01]\n## Bounds range [1e+00, 1e+00]\n## RHS range [4e+00, 8e+00]\n## Found heuristic solution: objective 489.0000000\n## Presolve added 242 rows and 557 columns\n## Presolve time: 0.01s\n## Presolved: 526 rows, 1129 columns, 2749 nonzeros\n## Presolved model has 164 SOS constraint(s)\n## Found heuristic solution: objective 488.0000000\n## Variable types: 984 continuous, 145 integer (140 binary)\n##\n## Root relaxation: objective 3.285500e+01, 308 iterations, 0.00 seconds\n##\n## Nodes | Current Node | Objective Bounds | Work\n## Expl Unexpl | Obj Depth IntInf | Incumbent BestBd Gap | It/Node Time\n##\n## 0 0 32.85500 0 22 488.00000 32.85500 93.3% - 0s\n## H 0 0 457.6000000 32.85500 92.8% - 0s\n## 0 0 33.07500 0 19 457.60000 33.07500 92.8% - 0s\n## H 0 0 51.5000000 33.07500 35.8% - 0s\n## 0 0 33.11500 0 20 51.50000 33.11500 35.7% - 0s\n## 0 0 33.12490 0 32 51.50000 33.12490 35.7% - 0s\n## 0 0 33.13125 0 34 51.50000 33.13125 35.7% - 0s\n## H 0 0 41.3000000 33.13125 19.8% - 0s\n## 0 0 33.13333 0 27 41.30000 33.13333 19.8% - 0s\n## 0 0 33.13333 0 27 41.30000 33.13333 19.8% - 0s\n## 0 0 33.23750 0 28 41.30000 33.23750 19.5% - 0s\n## 0 0 33.24750 0 18 41.30000 33.24750 19.5% - 0s\n## 0 0 33.24750 0 17 41.30000 33.24750 19.5% - 0s\n## 0 0 33.24750 0 22 41.30000 33.24750 19.5% - 0s\n## 0 0 33.24750 0 22 41.30000 33.24750 19.5% - 0s\n## 0 2 33.26000 0 22 41.30000 33.26000 19.5% - 0s\n## H 109 80 39.3000000 35.47635 9.73% 3.6 0s\n## 37059 1687 38.81870 55 4 39.30000 38.81870 1.22% 2.8 5s\n##\n## Cutting planes:\n## Gomory: 1\n## MIR: 2\n## Flow cover: 2\n##\n## Explored 48097 nodes (128726 simplex iterations) in 5.95 seconds\n## Thread count was 2 (of 4 available processors)\n##\n## Solution count 6: 39.3 41.3 51.5 ... 489\n##\n## Optimal solution found (tolerance 0.00e+00)\n## Warning: max constraint violation (1.1895e-02) exceeds tolerance\n## Warning: max general constraint violation (1.1895e-02) exceeds tolerance\n## (model may be infeasible or unbounded - try turning presolve off)\n## Best objective 3.930000000000e+01, best bound 3.930000000000e+01, gap 0.0000%\n\nNote that we obtain a total cost of 39.3, i.e. a higher cost than the when we use a linear curve. By using getCost(), we see that there are fewer units where actions are carried out, which suggests that more actions are being carried out in less sites.\n\ngetCost(d.binary_curve)\n## solution_name monitoring threat_1 threat_2\n## 1 sol 1.3 14 24\n\nIn the same way as in the previous model, using the getSolutionBenefit() function, we can obtain how the benefits are distributed for the different features:\n\n# get benefits of solution\nlocal_benefits.curve <- getSolutionBenefit(d.binary_curve, type = \"local\")\n\n# plot local benefits\nlocal_benefits.curve <- reshape2::dcast(local_benefits.curve, pu~feature,value.var = \"benefit.total\", fill = 0)\n\nvalues(group_rasters[]) <- local_benefits.curve$1 values(group_rasters[]) <- local_benefits.curve$2\nvalues(group_rasters[]) <- local_benefits.curve$3 values(group_rasters[]) <- local_benefits.curve$4\n\nplot(group_rasters)",
null,
"As we can see, for the base case, features 1 and 3 add to the total benefit sites where their probability of persistence is 0.5. Whereas when we use curve = 3, all but one site that contribute to reaching the target have a probability value of 1. This effect will be more evident when there are more threats in the planning exercise.\n\n## 2) Calculating benefits when threats are continuous\n\nSo far we have worked with two important premises: 1) that threats exist or not (presence/absence) and 2) that all features are equally sensitive to threats. To make these assumptions more flexible, we introduce the possibility of using threat intensities (in the same way that we use $$r_{is}$$ for features) and, the relationship of these intensities with the probability of persistence of the features (through response curves). These response curves can be fitted by a piece-wise function by means of four parameters within the sensitivities input data, where:\n\n• a : minimum intensity of the threat at which the features probability of persistence starts to decline.\n\n• b : value of intensity of the threat over which the feature has a probability of persistence of 0.\n\n• c : minimum probability of persistence of a features when a threat reaches its maximum intensity value.\n\n• d : maximum probability of persistence of a features in absence of a given threat.",
null,
"Where the greater the intensity of the threat, lower probability of persistence of the feature. Thus, with the definition of these four parameters mentioned above, different response curves can be fitted:",
null,
"The latter are examples of the curves that could be fitted with these parameters; that can be customized by users according to their needs by using the four parameters detailed above. Note that each of the curves is a relationship between a feature and a particular threat, but what happens to the probability of persistence of a feature when it coexists with more than one threats?.\n\nWe can define the probability of persistence of feature $$s$$ in the unit $$i$$ as a measure of the individual probability of persistence of feature $$s$$ with respect to threat $$k$$ in unit $$i$$ ($$p_{isk}$$), i.e.:\n\n$p_{is} = \\sum_{k \\in K_i \\cap K_s}\\gamma_{isk} p_{isk}$\n\nWith $$\\sum_{k \\in |K_i \\cap K_s|}\\gamma_{isk} = 1$$ and $$\\gamma_{isk} = \\gamma_{isk'} \\forall k, k' \\in |K_i \\cap K_s|$$.\n\nThe prioriactions package internally calculates the values of $$\\gamma_{isk}$$ reflecting the relative importance of each threat (considering its potential danger) when calculating the probability of persistence of each feature.\n\nNote the probability of persistence of a feature could be different from zero even if no actions are taken on it (depending on the type of response curve). This implies that doing nothing (no actions) in a unit could lead to, although small, benefits. These benefits are not considered to raise the targets regardless of the planning objective.\n\nFinally, the probability of persistence resulting can be calculated by:\n\n$p_{is} = \\frac{{\\sum_{k \\in K_i \\cap K_s: a_{sk} <= h_{ik} <= b_{sk}} \\frac{x_{ik} (h_{ik} - a_{sk})(d_{sk} - c_{sk})}{b_{sk} - a_{sk}} (1 - \\frac{c_{sk}(h_{ik} - a_{sk}) - d_{sk}(h_{ik} - b_{sk})}{b_{sk} - a_{sk}}+ \\sum_{k \\in K_i \\cap K_s: h_{ik} > b_{sk}} x_{ik}(1- c_{sk})(d_{sk} - c_{sk})}}{ {\\sum_{k \\in K_i \\cap K_s: h_{ik} < a_{sk}} 1 - d_{sk}} + \\sum_{k \\in K_i \\cap K_s: a_{sk} <= h_{ik} <= b_{sk}} 1 - \\frac{c_{sk}(h_{ik} - a_{sk}) - d_{sk}(h_{ik} - b_{sk})}{b_{sk} - a_{sk}}+ \\sum_{k in K_i \\cap K_s: h_{ik} > b_{sk}} 1 - c_{sk}}$ Where $$h_{ik}$$ is the intensity of the threat $$k$$ in the unit $$i$$.\n\n### Example with continuous intensities of threats\n\nIn order to demonstrate the use of the package when we have continuous values of intensities, we use the same example considered for the models with binary threat intensities. Note that both examples will not be compared due to differences in potential benefits and therefore in the targets used. Also, we assume that we only have information about the sensitivity of feature 1 to both threats, whose response curves have been fitted as follow:",
null,
"Note the feature is sensitive to any intensity of threat-1, while this is less sensitive to threat 2, as its probability of persistence remains invariant until the threat reaches high intensity values.\n\nFirst, we add the parameters $$a$$, $$b$$ to the input of sensitivities, leaving the measures that we do not know as NA. pioriractions is responsible for assuming them by their default values; for example, for parameter $$a$$, it is considered that the values not provided as 0, for $$b$$, the value will be the maximum intensity of said threat in the case of study, $$c$$ will be 0 and $$d$$ 1. In addition, we simulate the intensity of threats continuously using the runif() function; values between 0 and 150 for threat 1 and values between 0 and 1 for threat 2.\n\n# set a,b,c and d parameters\nsim_sensitivity_data$a = c(0, NA, NA, NA, 0.8, NA, NA, NA) sim_sensitivity_data$b = c(150, NA, NA, NA, 1, NA, NA, NA)\nsim_sensitivity_data$c = 0.0 sim_sensitivity_data$d = c(1, NA, NA, NA, 0.6, NA, NA, NA)\n\n# set continuous intensities\nset.seed(1)\n\nsim_dist_threats_data$amount[sim_dist_threats_data$threat == 1] <- round(runif(n = 57, 0, 123), 2)\nsim_dist_threats_data$amount[sim_dist_threats_data$threat == 2] <- round(runif(n = 63, 0, 1), 2)\n\n# evaluate input data\nb.continuous <- inputData(pu = sim_pu_data,\nfeatures = sim_features_data,\ndist_features = sim_dist_features_data,\nthreats = sim_threats_data,\ndist_threats = sim_dist_threats_data,\nsensitivity = sim_sensitivity_data,\nboundary = sim_boundary_data)\n\n# set the target of recovery to 15%\nmax <- getPotentialBenefit(b.continuous)\nsim_features_data$target_recovery <- 0.15 * max$maximum.recovery.benefit\n\n# evaluate input data\nb.continuous <- inputData(pu = sim_pu_data,\nfeatures = sim_features_data,\ndist_features = sim_dist_features_data,\nthreats = sim_threats_data,\ndist_threats = sim_dist_threats_data,\nsensitivity = sim_sensitivity_data,\nboundary = sim_boundary_data)\n\n# create the mathematic model\nc.continuous <- problem(b.continuous,\nmodel_type = \"minimizeCosts\")\n## Warning: The blm argument was set to 0, so the boundary data has no effect\n## Warning: Some blm_actions argument were set to 0, so the boundary data has no effect for these cases\n# solve the model\nd.continuous <- solve(c.continuous,\nsolver = \"gurobi\",\nverbose = TRUE,\noutput_file = FALSE,\ncores = 2)\n## Gurobi Optimizer version 9.1.2 build v9.1.2rc0 (linux64)\n## Thread count: 2 physical cores, 4 logical processors, using up to 2 threads\n## Optimize a model with 284 rows, 396 columns and 766 nonzeros\n## Model fingerprint: 0xfea67198\n## Variable types: 176 continuous, 220 integer (220 binary)\n## Coefficient statistics:\n## Matrix range [2e-04, 2e+00]\n## Objective range [1e-01, 1e+01]\n## Bounds range [1e+00, 1e+00]\n## RHS range [2e+00, 5e+00]\n## Found heuristic solution: objective 153.7000000\n## Presolve removed 250 rows and 248 columns\n## Presolve time: 0.00s\n## Presolved: 34 rows, 148 columns, 278 nonzeros\n## Found heuristic solution: objective 42.4000000\n## Variable types: 0 continuous, 148 integer (146 binary)\n##\n## Root relaxation: objective 2.011896e+01, 17 iterations, 0.00 seconds\n##\n## Nodes | Current Node | Objective Bounds | Work\n## Expl Unexpl | Obj Depth IntInf | Incumbent BestBd Gap | It/Node Time\n##\n## 0 0 20.11896 0 9 42.40000 20.11896 52.5% - 0s\n## H 0 0 29.1000000 20.11896 30.9% - 0s\n## H 0 0 21.9000000 20.11896 8.13% - 0s\n## H 0 0 21.8000000 20.11896 7.71% - 0s\n## 0 0 20.82302 0 3 21.80000 20.82302 4.48% - 0s\n## 0 0 20.91403 0 3 21.80000 20.91403 4.06% - 0s\n## 0 0 21.21879 0 4 21.80000 21.21879 2.67% - 0s\n## 0 0 21.47599 0 5 21.80000 21.47599 1.49% - 0s\n## 0 0 21.53170 0 5 21.80000 21.53170 1.23% - 0s\n## 0 0 21.54367 0 6 21.80000 21.54367 1.18% - 0s\n## 0 0 21.58911 0 6 21.80000 21.58911 0.97% - 0s\n## 0 0 21.59162 0 7 21.80000 21.59162 0.96% - 0s\n## 0 0 21.59578 0 8 21.80000 21.59578 0.94% - 0s\n## 0 0 21.61294 0 8 21.80000 21.61294 0.86% - 0s\n## 0 0 cutoff 0 21.80000 21.80000 0.00% - 0s\n##\n## Cutting planes:\n## Gomory: 2\n## Cover: 3\n## MIR: 1\n## StrongCG: 2\n## RLT: 1\n##\n## Explored 1 nodes (56 simplex iterations) in 0.01 seconds\n## Thread count was 2 (of 4 available processors)\n##\n## Solution count 5: 21.8 21.9 29.1 ... 153.7\n##\n## Optimal solution found (tolerance 0.00e+00)\n## Best objective 2.180000000000e+01, best bound 2.180000000000e+01, gap 0.0000%\n\nNote that the maximum benefits to be achieved have changed due to the presence of intensities in the threats and the sensitivity curves (with respect to the base model). We can use getPotentialBenefit() to explore the difference between these benefits:\n\n# get potential benefits\ngetPotentialBenefit(b.continuous)\n## feature dist dist_threatened maximum.conservation.benefit maximum.recovery.benefit maximum.benefit\n## 1 1 47 47 0 13.7781 13.7781\n## 2 2 30 28 2 14.6788 16.6788\n## 3 3 66 56 10 32.9739 42.9739\n## 4 4 33 33 0 15.9912 15.9912\n\nNow by the getSolutionBenefit() function, we can obtain how the benefits are distributed for the different features. Notice that we use the type = \"local\" argument to obtain the benefits per unit and feature.\n\n# get benefits of solution\nlocal_benefits.continuous<- getSolutionBenefit(d.continuous, type = \"local\")\n\n# plot local benefits\nlocal_benefits.continuous <- reshape2::dcast(local_benefits.continuous, pu~feature,value.var = \"benefit.total\", fill = 0)\n\nvalues(group_rasters[]) <- local_benefits.continuous$1 values(group_rasters[]) <- local_benefits.continuous$2\nvalues(group_rasters[]) <- local_benefits.continuous$3 values(group_rasters[]) <- local_benefits.continuous$4\n\nplot(group_rasters)",
null,
"Note that the selected sites differ to a greater extent from those chosen in the base model with binary threat intensities. From the above, the effect that can bring with the use of sensitivities and intensities of continuous threats on the management of conservation actions is extracted."
] | [
null,
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",
null,
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null,
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null,
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null,
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null,
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",
null
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https://cran.microsoft.com/snapshot/2018-07-17/web/packages/funrar/vignettes/sparse_matrices.html | [
"# Sparse Matrices Usefulness\n\nWhen a matrix contains a lot of zero, there is no need to store all its values. But only the non-zero values, for example the following matrix:\n\n$\\begin{equation} M = \\bordermatrix{ ~ & \\text{sp. A} & \\text{sp. B} & {sp. C} \\cr \\text{site 1} & 1 & 0 & 0 \\cr \\text{site 2} & 0 & 1 & 1 \\cr \\text{site 3} & 0 & 1 & 0 \\cr} \\end{equation}$\n\ncan be stored using only a matrix with non zero-values $$S$$, with row index $$i$$ and column index $$j$$:\n\n$\\begin{equation} S = \\begin{matrix} i & j & \\text{value} \\\\ \\text{site 1} & \\text{sp. A} & 1 \\\\ \\text{site 2} & \\text{sp. B} & 1 \\\\ \\text{site 3} & \\text{sp. B} & 1 \\\\ \\text{site 2} & \\text{sp. C} & 1 \\end{matrix} \\end{equation}$\n\nIf your site-species matrix is big, it may contain a lot zero because it is very unlikely that all species are present at all sites when looking at thousands of species and hundreds of sites. Thus, the sparse matrix notation would save a good amount of lines (look how $$S$$ compared to $$M$$).\n\n# Generate a matrix with 1000 species and 200 sites\nmy_mat = matrix(sample(c(0, 0, 0, 0, 0, 0, 1), replace = TRUE, size = 200000),\nncol = 1000, nrow = 200)\ncolnames(my_mat) = paste0(\"sp\", 1:ncol(my_mat))\nrownames(my_mat) = paste0(\"site\", 1:nrow(my_mat))\n\nmy_mat[1:5, 1:5]\n## sp1 sp2 sp3 sp4 sp5\n## site1 0 0 0 0 0\n## site2 0 0 0 0 0\n## site3 0 1 0 0 0\n## site4 0 0 0 1 0\n## site5 0 0 0 0 0\n\nfunrar lets you use sparse matrices directly using a function implemented in the Matrix package. When your matrix is filled with 0s it can be quicker to use sparse matrices. To know if you should use sparse matrices you can compute the filling of your matrix, i.e. the percentage of non-zero cells in your matrix:\n\nfilling = 1 - sum(my_mat == 0)/(ncol(my_mat)*nrow(my_mat))\n\nfilling\n## 0.1433\n\nTo convert from a normal matrix to a sparse matrix you can use as(my_mat, \"sparseMatrix\"):\n\nlibrary(Matrix)\n\nsparse_mat = as(my_mat, \"sparseMatrix\")\n\nis(my_mat, \"sparseMatrix\")\n## FALSE\nis(sparse_mat, \"sparseMatrix\")\n## TRUE\n\nit completely changes the structure as well as the memory it takes in the RAM:\n\n# Regular Matrix\nstr(my_mat)\n## num [1:200, 1:1000] 0 0 0 0 0 0 0 0 1 0 ...\n## - attr(*, \"dimnames\")=List of 2\n## ..$: chr [1:200] \"site1\" \"site2\" \"site3\" \"site4\" ... ## ..$ : chr [1:1000] \"sp1\" \"sp2\" \"sp3\" \"sp4\" ...\nprint(object.size(my_mat), units = \"Kb\")\n## 1628.6 Kb\n# Sparse Matrix from 'Matrix' package\nstr(sparse_mat)\n## Formal class 'dgCMatrix' [package \"Matrix\"] with 6 slots\n## ..@ i : int [1:28660] 8 14 40 50 53 55 56 61 71 98 ...\n## ..@ p : int [1:1001] 0 25 59 88 118 149 173 196 228 260 ...\n## ..@ Dim : int [1:2] 200 1000\n## ..@ Dimnames:List of 2\n## .. ..$: chr [1:200] \"site1\" \"site2\" \"site3\" \"site4\" ... ## .. ..$ : chr [1:1000] \"sp1\" \"sp2\" \"sp3\" \"sp4\" ...\n## ..@ x : num [1:28660] 1 1 1 1 1 1 1 1 1 1 ...\n## ..@ factors : list()\nprint(object.size(sparse_mat), units = \"Kb\")\n## 406.9 Kb\n\nsparse matrices reduce the amount of RAM necessary to store them. The more zeros there are in a given matrix the more RAM will be spared when turning it into a sparse matrix.\n\n## Benchmark\n\nWe can now compare the performances of the algorithms in rarity indices computation between a sparse matrix and a regular one, using the popular microbenchmark R package:\n\nlibrary(funrar)\n\n# Get a table of traits\ntrait_df = data.frame(trait = runif(ncol(my_mat), 0, 1))\nrownames(trait_df) = paste0(\"sp\", 1:ncol(my_mat))\n\n# Compute distance matrix\ndist_mat = compute_dist_matrix(trait_df)\n\nif (requireNamespace(\"microbenchmark\", quietly = TRUE)) {\nmicrobenchmark::microbenchmark(\nregular = distinctiveness(my_mat, dist_mat),\nsparse = distinctiveness(sparse_mat, dist_mat))\n}\n\n### Systematic benchmark\n\nWe generate matrices with different filling rate and compare the speed of regular matrix and sparse matrices computation.\n\ngenerate_matrix = function(n_zero = 5, nrow = 200, ncol = 1000) {\nmatrix(sample(c(rep(0, n_zero), 1), replace = TRUE, size = nrow*ncol),\nncol = ncol, nrow = nrow)\n}\n\nmat_filling = function(my_mat) {\nsum(my_mat != 0)/(ncol(my_mat)*nrow(my_mat))\n}\n\nsparse_and_mat = function(n_zero) {\nmy_mat = generate_matrix(n_zero)\ncolnames(my_mat) = paste0(\"sp\", 1:ncol(my_mat))\nrownames(my_mat) = paste0(\"site\", 1:nrow(my_mat))\n\nsparse_mat = as(my_mat, \"sparseMatrix\")\n\nreturn(list(mat = my_mat, sparse = sparse_mat))\n}\n\nn_zero_vector = c(0, 1, 2, 49, 99)\nnames(n_zero_vector) = n_zero_vector\n\nall_mats = lapply(n_zero_vector, sparse_and_mat)\n\nmat_filling(all_mats$0$mat)\n## 1\nmat_filling(all_mats$99$mat)\n## 0.009895\n\nNow we can compare the speed of the algorithms:\n\nif (requireNamespace(\"microbenchmark\", quietly = TRUE)) {\nmat_bench = microbenchmark::microbenchmark(\nmat_0 = distinctiveness(all_mats$0$mat, dist_mat),\nsparse_0 = distinctiveness(all_mats$0$sparse, dist_mat),\nmat_1 = distinctiveness(all_mats$1$mat, dist_mat),\nsparse_1 = distinctiveness(all_mats$1$sparse, dist_mat),\nmat_2 = distinctiveness(all_mats$2$mat, dist_mat),\nsparse_2 = distinctiveness(all_mats$2$sparse, dist_mat),\nmat_49 = distinctiveness(all_mats$49$mat, dist_mat),\nsparse_49 = distinctiveness(all_mats$49$sparse, dist_mat),\nmat_99 = distinctiveness(all_mats$99$mat, dist_mat),\nsparse_99 = distinctiveness(all_mats$99$sparse, dist_mat),\ntimes = 5)\n\nautoplot(mat_bench)\n}"
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https://www.lsjcrp.co/how-to-find-interest-expense-how/ | [
"# how to find interest expense How",
null,
"How to Find Interest and Expense in Excel\n· How to Find Interest and Expense in Excel. One area where Microsoft Excel shines is in solving financial problems. Excel contains a broad range of financial functions that can\nWhat is interest expense?\nDefinition of Interest Expense Interest expense is the cost of borrowing money during a specified period of time. Interest expense is occurring daily, but the interest is likely to be paid monthly, quarterly, semiannually, or annually. Example of Interest Expense Let’s\n\n## How to Calculate Carrying Value of a Bond (with Pictures)\n\n· Calculate interest expense. In order to properly report amortization, we will also need the know the amount of interest expense paid to bondholders over the same period. This is the amount of the coupon payment, based on a percentage of the par value. It is\nCost of Debt Formula\nCost of Debt Formula – Example #1 A company named Viz Pvt. Ltd took loan of \\$200,000 from a Bank at the rate of interest of 8% to issue company bond of \\$200,000. Based on the loan amount and rate of interest, interest expense will be \\$16,000 and the tax rate\nTimes interest earned (TIE) ratio\nCalculating the Times Interest Earned Ratio For the most recent year, Back Alley Boys, Inc., had sales of \\$250,000, cost of goods sold of \\$80,000, depreciation expense of \\$27,000, and additions to retained earnings of \\$33,360. The firm currently has 20,000 shares\nCapitalization of Interest\nCapitalized interest is US GAAP term that refers to the part of interest expense that is capitalized as part of the cost of asset. IFRS uses the term borrowing costs for costs incurred in relation to a debt used for construction of the asset.\nMCD Interest Expense\nInterest Expense is the amount reported by a company or individual as an expense for borrowed money. McDonald’s’s interest expense for the three months ended in Dec. 2020 was \\$ -323 Mil.Its interest expense for the trailing twelve months (TTM) ended in Dec. 2020 was \\$-1,224 Mil.\nInstructions for Form 8990 (05/2020)\n· Interest expense limitations. An expense that has been disallowed, deferred, or capitalized in the current tax year, or which has not yet been accrued, is not taken into account for section 163(j) purposes. Section 163(j) applies after any basis limitation and before\nHow to Calculate Expense-to-Revenue Ratio\nFind a bank’s total non-interest expense on its income statement. A bank typically provides the total amount of non-interest expense, which includes items such as salaries, rent, depreciation and utilities.\nXero Community\nYes, you can record the interest added to the loan as a transaction in Xero. If the the loan is set up as a liability account then this can be done with a journal: credit loan account and debit loan interest expense …\nApple Interest Expense\nInterest Expense is the amount reported by a company or individual as an expense for borrowed money. Apple’s interest expense for the three months ended in Dec. 2020 was \\$ -638 Mil.Its interest expense for the trailing twelve months (TTM) ended in Dec. 2020 was \\$-2,726 Mil.\n\n## What Is the Interest Coverage Ratio and How Do You …\n\nInterest Coverage Ratio = Earnings Before Interest and Taxes/Interest Expense If you have the EBIT and interest expense in front of you, you’re most of the way to figuring out your ICR.\n\n## What is Debenture interest in accounting?\n\n· You can find interest expense on your income statement, a common accounting report that’s easily generated from your accounting program. Interest expense is usually at the bottom of an income statement, after operating expenses. Sometimes is its own line.\n\n## How to find salary and interest expense in Dominos …\n\nHow to find salary and interest expense in Dominos income statement? 0 Veena_pampady Hi Sir, Could you please help me from the video of dominos pizza income satement help me how to conclude answers for the below questions. In the Dominos Pizza\n\n## Times interest earned ratio — AccountingTools\n\nEarnings before interest and taxes ÷ Interest expense = Times interest earned A ratio of less than one indicates that a business may not be in a position to pay its interest obligations, and so is more likely to default on its debt; a low ratio is also a strong indicator of impending bankruptcy .\n\n## what is an interest expense?? how to find it out from an …\n\nThis yearly fees it has to pay is called interest expense. Most income statements will have an item called “Interest” or “Financing Charges” Mar 24 2013 12:01 PM 0 bijay interest is outflow of cash. this is amount u need to pay for ur borrowed loan and debt . 0"
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https://m.scirp.org/papers/78849 | [
"The Distribution of Returns\nShow more\nAbstract: The distribution of the returns on investment depends on the rules in the economic system. The article reviews various return distributions, ranging from equity securities in equilibrium, to antiques bought at auction, to debt instruments with uncertain payouts. A general methodology is provided to construct distributions of returns.\n\n1. Introduction\n\nThis article derives the distribution of returns for a variety of assets, liabilities and accounting ratios by asserting two things. The first assertion is that for equity securities that returns are not data, rather prices are data. Returns are transformations of data. The second is that the form of calculating returns and the rules in the economic system determine the distribution that has to be involved. The exception to this would be for instruments like certificates of deposit or fixed annuities where the return is stated, and cash flows follow from the return.\n\nIn addition to the form of the calculation, the number of potential buyers and sellers matters; whether errors are independent; the method transactions happen in, such as double auctions, English or Dutch-style auctions and so forth; whether markets are assumed to be in equilibrium or far from equilibrium as was the case in the 2008 financial crisis; how liquidity enters the market; the terminal state of the asset such as a surviving firm, cash, bankruptcy and so forth; and any other constraints such as asymmetric information.\n\nAssuming the normality or lognormality of returns is nearly ubiquitous in economics. Either of these distributions solves a key problem in mean-variance models in that a covariance matrix needs to exist in the data generation process. Normality is also nice in that it appears in nature quite often. While Markowitz’s initial paper appeared in 1952, the first empirical objection was by Mandelbrot in 1963 .\n\nMandelbrot’s paper asserted that the data appeared to be some form of Paretian distribution . This was followed on by papers by Eugene Fama ; and Fama and Roll in to produce alternative rules and assumptions. Empirical rejection of the models was performed by Fama and MacBeth . Empirical contradictions for the various forms of the Capital Asset Pricing Models are cataloged by Fama and French in and for the Black-Scholes Option Pricing Model by Yilmaz in his master’s thesis .\n\nThe problem created by empirical rejection was that it does not provide a usable solution for economics. While a segment of economics adopted the Fama-French model as an alternative hypothesis, this model is not explicitly grounded in theory . Pearson-Neyman decision theory would encourage the adoption of this model, but Fisherian Likelihood-based methods would merely have rejected mean-variance finance and not provided an alternative solution.\n\nThe challenge to economics is that it is constantly attempting to solve inverse problems. When someone is observed purchasing an orange, economists assume that it is the solution to that person’s problem. Economists can only see the solutions people use, not the problems which underlie them. The goal of economics is to reverse engineer a generalized solution making process to produce testable predictions. Empirically falsifying a theory without a replace- ment left financial economics in an untenable position.\n\nThe difficulty, as will be shown in Sections 3 and 10.4, is that the distributions involved lack a first moment and in logarithmic form lack a covariance matrix. Nothing necessary for mean-variance finance to function survives, but in addition, neither does the Fama-French model as it exists today.\n\nA population study of data in the Center for Research in Security Prices was conducted in to test the central claims here and mean-variance finance was rejected with posterior odds of more than ${10}^{8500000}:1$ , given a prior probability of $1:999999$ . As the Paretian alternative was given only one chance in a million of being correct prior to seeing the data and with it being rejected so resoundingly, the mean-variance family of models are dead without doubt.\n\nFortunately, knowing the distributions involved permits economic model building. Although this article discusses probability density functions, they should really be thought of as Bayesian likelihood functions. As shown in , admissible non-Bayesian solutions do not exist for most of the distributions involved here. The reason is less than obvious.\n\nIn the early days of statistics, a fundamental problem was observed. As a statistic is any function of the data and since there are an infinite number of functions, then which function should be used to estimate a parameter? Why would $\\sum \\left(sin\\left({x}_{i}\\right)\\right)$ be a bad parameter estimator for the location of the mean of the normal distribution? How does one know that the sample average is a good estimator and that a better estimator does not exist? For the Gaussian, the mean, median and mode are co-located. Why not choose the mode over the mean?\n\nFortunately, this problem was solved by Abraham Wald . His solution was clever and it came with unexpected results. Using Frequentist axioms, Wald discovered that all Bayesian solutions were admissible and that non-Bayesian solutions were admissible when they either matched the Bayesian solution in every sample or at the limit. The converse is not true.\n\nTo understand why this might be the case, consider solving a problem using either a subjective Bayesian method or an objective non-Bayesian method. If the sample size is large enough, then the two methods must converge if the non-Bayesian method is valid. A subjective understanding of the world must converge to the objective understanding of the world if the sample size were large enough. On the other hand, if the non-Bayesian method did not arrive at nearly the same place, then it does not match reality. Since non-Bayesian me- thods are conditioned on their model in use, that model in use cannot be true. Instead, the non-Bayesian model was a misunderstanding of the universe.\n\nBecause the distributions involved lack a sufficient statistic, it is necessary that non-Bayesian methods will lose information while Bayesian methods will not as the Bayesian likelihood function is always minimally sufficient. It appears that this relationship between sufficiency and admissibility was first discovered by E.T. Jaynes .\n\nTo avoid these problems in model building, it is enough to note that for independent draws of a variable that if the density function is\n\n$Pr\\left(x|\\mu \\right)=\\frac{1}{\\text{π}}\\frac{1}{1+{\\left(x-\\mu \\right)}^{2}},\\forall x\\in \\chi .$ (1)\n\nwhere $\\chi$ is the sample space, then the Bayesian likelihood function is\n\n$Pr\\left(x|\\mu \\right)=\\frac{1}{\\text{π}}\\frac{1}{1+{\\left(x-\\mu \\right)}^{2}},\\forall \\mu \\in \\Theta ,$ (2)\n\nwhere $\\Theta$ is the parameter space. This allows economics to make a robust movement to Bayesian decision theory which is well developed and strongly resembles methods already used by economists. It also differs in that likelihood functions are not probability densities and need not sum to unity.\n\nThe prior literature on this topic is split as the statistical and mathematical literature generally do not agree with the economic literature. Although work has carried on in a limited form in the economic literature since Mandelbrot and Fama, most of it has been an attempt to find the best fit distribution. While this has held some sway in the applied component of finance, this is a non-normative viewpoint. Overwhelmingly the literature has been moved to some version of a normal distribution with heteroskedasticity. Unfortunately, attempts to maintain or hold onto normality lead to incorrect solutions.\n\nThe statistical literature on this topic is settled, though somewhat sparse. The literature began with correspondence from Poisson to Laplace noting an exception to the central limit theorem. Poisson observed that the distribution $f\\left(x\\right)={\\left[\\text{π}\\left(1+{x}^{2}\\right)\\right]}^{-1}$ was a counterexample to the theorem .\n\nThe next appearance of this exception is in a battle in the literature between Augustin Cauchy and Irénée-Jules Bienaymé. Bienaymé had written an article showing that ordinary least squares was optimal. Cauchy had just published a method of regression and took this as a personal attack. He searched for a circumstance where the method of least squares would always produce an invalid solution. He arrived at the same solution as Poisson and then dropped the matter .\n\nThe issue moves into the background again until the Pitman-Koopman- Darmois theorem appears. Ronald Fisher had just discovered the existence of sufficient statistics and this raised the question of when, why and how does a sufficient statistic exist. Pitman, Koopman, and Darmois simultaneously dis- covered the same effect. A statistic sufficient for a parameter only exists for members of the exponential family. Koopman, in particular, showed the excep- tion of Poisson, now called the Cauchy distribution could not have a sufficient point statistic . This is important because non-Bayesian point estimators would lose information if used.\n\nIndeed, the minimum variance unbiased estimator for the Cauchy distribu- tion purposefully discards information. Since the Cauchy distribution lacks a mean, finding the sample average is pointless. Like all distributions, it does have a median. Rothenberg, et al., in determined that the mean of the central twenty-four percent is the minimum variance unbiased estimator of the popula- tion median. The remaining seventy-six percent of the data is discarded. The Bayesian estimator uses all of the data, and the estimate cannot be worse, in the Wald sense, than Rothenberg’s method.\n\nBayesian methods are the oldest probability interpretation beginning with the writing of Thomas Bayes . Formal axiomatization does not occur until later when de Finetti creates the first axioms of probability . Other Bayesian axiom systems are also developed, notably by Savage in and Cox in . The writings of Abraham Wald would produce foundational work in decision theory, for both the Bayesian and the Pearson-Neyman school of thought .\n\nIn addition to work on probability and statistics, basic work on probability distributions was begun by Pearson . The work would lead others to look at the properties of random variables drawn from these distributions such as additivity and the impact of multiplication or division. Curtiss in published a method to determine the distribution of variables where there is multiplication or division. This was slightly generalized by Gurland in . For the joint normal case, this was expanded by Marsaglia .\n\nSolving for the ratio of two random variates permits a general solution to the problems presented here.\n\nFinally, work was performed by Mann and Wald in , with subsequent work by John White in and M.M. Rao in on autogregressive processes. Although not Bayesian in nature, these methods have a Bayesian interpretation that will be discussed later.\n\nThe economic literature presumes normality and so does not map to the mathematical or statistical literature.\n\nThe paper provides the method of Curtiss found in to move the economic literature back to the mathematical literature and provides other related tools. It starts with the simple assumptions of Markowitz and adds various complications such as correlated errors, systemic disequilibrium, alternative auction types, budget and liquidity limitations, bond certainties, dividends and regression methods. The purpose of this is not to be exhaustive, but rather to show how to approach the presence of uncertainty. The list of economic and financial problems not included in the paper is extensive, but those will be the work of others.\n\nBecause economic theory rests heavily upon its mathematics, this paper argues that the theory needs a fundamentally different grounding. Because most distributions lack a first moment, nothing resembling mean-variance finance remains possible. This goes deep into the risk management, legal and regulatory environment. Even at the state level, the Uniform Prudent Investor Act is predicated on the validity of a set of mathematical structures that do not exist. A raft of fundamental ideas in models where capital is present will need to be reviewed and reconsidered as to their implications.\n\n2. A General Method to Derive Distributions of Returns for Equity Assets Using Ratios\n\n2.1. Introduction\n\nThis article deals with four types transformations. They are $\\frac{{p}_{t+1}}{{p}_{t}}$ ,\n\n$log\\left({p}_{t+1}\\right)-log\\left({p}_{t}\\right)$ , ${p}_{t+1}=R{p}_{t}+{ϵ}_{t+1}$ and $\\mathrm{log}\\left({p}_{t+1}\\right)=R\\mathrm{log}\\left({p}_{t}\\right)+{ϵ}_{t+1}$ . One of\n\nthe difficulties has been that economic models have been constructed in one transformation but tested in another. While the budget constraint of the Capital Asset Pricing Model and the Security Market Line Beta representation are linearly additive in errors, the logarithmic approximation is usually tested resulting in a test of a different multiplicative model.\n\nAs will be shown, these transformations have different results. The model in use and the test of the model should match if at all possible. A goal of this article is to provide a deeper understanding of the consequences of the tools used.\n\nAs a simplification, $\\frac{{p}_{t+1}}{{p}_{t}}-1$ will be studied as $\\frac{{p}_{t+1}}{{p}_{t}}$ . The subtraction 1 has\n\nno impact on the distribution other than to shift it and constantly writing -1 does nothing to improve clarity.\n\nOf the two classes of securities; equities and debt; equities are unique in that the final value is unknown. The distribution of returns for equity securities depends upon the distribution of potential valuations at the time of sale and the distribution of potential valuations at the time of purchase. The reward is the ratio of future values to present values.\n\nDefinition 1. The reward for investing is defined as the future value divided by the present value. The return is the reward minus one.\n\nThere are three mechanisms to derive a distribution of rewards, or its cousin returns. The first, and standard method, is to derive the ratio distribution directly. This method was first described by Curtiss . The second would be to convert the data into polar coordinates and to indirectly solve the problem using the angle of the return rather than a direct attack on the slope. The third is to regress the future value from the present value.\n\n2.2. Curtiss’ Method\n\nCurtiss allows a general solution for the cases where there are two random variables, Y and X, related through the relationship\n\n$Z=\\frac{Y}{X}.$ (3)\n\nThe joint probability distribution for X and Y is $f\\left(x,y\\right)$ . If $D\\left(z\\right)$ is the cumulative density function for z, and $p\\left(z\\right)$ is its probability density function, then they must be related through some transformation of the joint density function of X and Y.\n\n1One could also assign an aribtrary density at $x=0$ because the impact on the total sum would only be $k\\text{d}x$ which would vanish at the limit.\n\n2For most ordinary cases and all cases in this article, the integrals can be assumed to be Lebesgue integrals or in some special cases not used here, they can be assumed to be Riemann integrals. See Curtiss in for a complete discussion.\n\nBecause this is a ratio, zero could potentially cause difficulty. This difficulty is avoided by also noting that the $\\mathrm{Pr}\\left(x=0\\right)=0$ since the measure of a countable point over a continuum is zero.1 Although its probability is zero, this does not make it an impossible event, only an event that can be removed as any other similar pole could without causing unintended consequences. All impossible events have a probability of zero, but not all events with a probability of zero are impossible.\n\nNonetheless, it does require that any solution is partitioned into the sets above and below zero. As a consequence, although\n\n$D\\left(z\\right)=\\mathrm{Pr}\\left(Z\\le z\\right),$ (4)\n\nThis must be written as\n\n$D\\left(z\\right)=\\mathrm{Pr}\\left(Y\\ge zX|X<0\\right)+\\mathrm{Pr}\\left(Y\\le zX|X>0\\right).$ (5)\n\nThis is extended into functional form through the equation\n\n$D\\left(z\\right)={\\int }_{-\\infty }^{0}{\\int }_{zx}^{0}\\text{ }f\\left(x,y\\right)\\text{d}y\\text{d}x+{\\int }_{0}^{\\infty }{\\int }_{0}^{zx}f\\left(x,y\\right)\\text{d}y\\text{d}x,$ (6)\n\nwhere the integrals are assumed to be Lebesgue-Stieltjes integrals.2\n\nThe probability density function for $d\\left(z\\right)$ is\n\n$p\\left(z\\right)=\\frac{\\text{d}\\left[D\\left(z\\right)\\right]}{\\text{d}z},$ (7)\n\nwhich can be expressed as\n\n$p\\left(z\\right)={\\int }_{-\\infty }^{\\infty }|x|f\\left(x,zx\\right)\\text{d}x.$ (8)\n\n2.3. A Trigonometric Method\n\nA second solution is to note the relationship between a slope and an angle. If the reward for investing is Z as above then relationship between $\\Theta$ and Z is\n\n$\\Theta ={\\mathrm{tan}}^{-1}\\left(Z\\right),$ (9)\n\nwhile\n\n$\\Omega =\\sqrt{{X}^{2}+{Y}^{2}}.$ (10)\n\nThe transformation of the function requires including the Jacobian for the tranformation of variables. Transforming the problem into polar coordinates results in\n\n$D\\left(z\\right)={\\int }_{-\\frac{\\text{π}}{2}}^{ta{n}^{-1}\\left(z\\right)}{\\int }_{0}^{\\infty }f\\left(\\omega ,\\theta \\right)\\omega \\text{d}\\omega \\text{d}\\theta +{\\int }_{\\frac{\\text{π}}{2}}^{ta{n}^{-1}\\left(z\\right)+\\text{π}}{\\int }_{0}^{\\infty }f\\left(\\omega ,\\theta \\right)\\omega \\text{d}\\omega \\text{d}\\theta .$ (11)\n\nThe equation presumes no limitation on liability. The equation would need to be adjusted for the limitation of liability. This form has two potential advantages, one pedagogic, the other computational.\n\nSince the cumulative density function for the standard Cauchy distribution is ${\\text{π}}^{-1}ta{n}^{-1}\\left(\\stackrel{˜}{x}\\right)$ , this implies that a solution will be an angular transformation of the Cauchy distribution. This can be observed in the limits of the integrals. Using the trigonometric method makes it easier to see the underlying relationship between returns and the Cauchy distribution. Secondarily, some functions are simpler to solve in polar coordinates.\n\n2.4. Ratio Distributions Based on Equilibrium Prices\n\nWealth invested in a security at time t is the price times the quantity owned at time t, that is to say ${w}_{t}={p}_{t}×{q}_{t},\\forall t$ . For equity securities, not exchanged for cash or bankrupt, then the split adjusted price is equivalent to assuming that ${q}_{t}=1,\\forall t$ . In that circumstance, the return on cash flows becomes the return on prices. To derive the distribution, with respect to the equilibrium prices, it is assumed that\n\n${p}_{t}={p}_{t}^{*}+{ϵ}_{t},\\forall t.$ (12)\n\nAs only the distribution of errors from the equilibrium are of interest the distribution will be of\n\n${ϵ}_{t}={p}_{t}-{p}_{t}^{*},\\forall t.$ (13)\n\nUsing this transformation transfers the point of calculation from $\\left({p}_{t},{p}_{t+1}\\right)$ to $\\left(0,0\\right)$ . Because this translation moves the ratio of prices to $\\left(0,0\\right)$ it will be necessary to translate the final distribution by R, which could be defined as\n\n$R=\\frac{{p}_{t+1}^{*}}{{p}_{t}^{*}}.$ (14)\n\n3. Distribution for Going Concerns in Equilibrium\n\nFor historical reasons and because the complications of real life will be added later, this is being constructed in a classic Markowitzian model with many buyers and many sellers. There are no transaction costs, either there is infinite liquidity, or there is no market maker. Transactions occur in a double auction where potential buyers compete for best bid and sellers compete for best bid. Markets are not systematically away from equilibrium.\n\nBecause this is a double auction in equilibrium, there is no winner’s curse. The rational behavior is for each actor to bid their expectation. The sampling distribution of the limit book will be normally distributed due to the central limit theorem. If the Markowitzian assumption of price taking behavior is included, then the appraisal errors are being committed by the counter-party. Additionally, it will be assumed that errors are independent. This may not be the case for some types of transactions, such as program trading.\n\nAs with Capital Asset Pricing Model style problems, there is no presumed positivity of future price. The assumption of limited liability will be added in Subsection 3.3. The sample space is the real numbers for future values of p.\n\nSince both the entry and exit order should have normally distributed order books, it follows that the likelihood function is the ratio of two normal distributions. This solution is well known in statistics as the Cauchy distribution. The Cauchy distribution has no mean, and consequently, under the basic initial assumptions of mean-variance models, it is impossible for mean-variance finance to exist.\n\nUsing the method described in Section 2.4 one quickly arrives at the Cauchy distribution under the assumptions of the Markowitz style models as shown in Section 3.1 for the distribution of returns around the equilibrium points.\n\n3.1. Using Curtiss’ Method to Arrive at the Distribution\n\nCurtiss’ method involves solving\n\n$p\\left({r}_{t}\\right)={\\int }_{-\\infty }^{\\infty }\\frac{|{\\epsilon }_{t}|}{2\\text{π}{\\sigma }_{t}{\\sigma }_{t+1}}\\mathrm{exp}\\left[-\\frac{1}{2}\\left(\\frac{{\\epsilon }_{t}^{2}}{{\\sigma }_{t}^{2}}+\\frac{{r}_{t}^{2}{\\epsilon }_{t}^{2}}{{\\sigma }_{t+1}^{2}}\\right)\\right]\\text{d}{\\epsilon }_{t}$ (15)\n\ncauses us to arrive at\n\n$p\\left({r}_{t}\\right)=\\frac{1}{\\text{π}}\\frac{\\Gamma }{{\\Gamma }^{2}+{r}_{t}^{2}},\\Gamma =\\frac{{\\sigma }_{t+1}}{{\\sigma }_{t}}.$ (16)\n\nThis equation still needs to be shifted by R. This preserves the nice interpre- tation of $\\Gamma$ as a measure of price heteroskedasticity so that the final form is\n\n$p\\left({r}_{t}\\right)=\\frac{1}{\\text{π}}\\frac{\\Gamma }{{\\Gamma }^{2}+{\\left({r}_{t}-R\\right)}^{2}}.$ (17)\n\n3.2. The Trigonometric Method\n\nTo provide an example of the trigonometric solution, consider the simple case where ${\\sigma }_{t}={\\sigma }_{t+1}=1$ . Noting the symmetry, only one of the two integrals needs solved and it would be multiplied by two. Since in equilibrium the center of location is at (0,0), the cumulative density function is\n\n$D\\left({r}_{t}\\right)=\\frac{2}{\\text{2π}}{\\int }_{-\\frac{\\pi }{2}}^{ta{n}^{-1}\\left({r}_{t}\\right)}{\\int }_{0}^{\\infty }\\omega {\\text{e}}^{-\\frac{{\\omega }^{2}}{2}}\\text{d}\\omega \\text{d}\\theta$ (18)\n\n$D\\left({r}_{t}\\right)=\\frac{1}{\\text{π}}{\\int }_{-\\frac{\\text{π}}{2}}^{ta{n}^{-1}\\left({r}_{t}\\right)}\\left[\\underset{\\omega \\to \\infty }{lim}{\\text{e}}^{-\\frac{{\\omega }^{2}}{2}}-\\underset{\\omega \\to 0}{lim}{\\text{e}}^{-\\frac{{\\omega }^{2}}{2}}\\right]\\text{d}\\theta$ (19)\n\n$D\\left({r}_{t}\\right)=\\frac{1}{\\text{π}}{\\int }_{-\\frac{\\text{π}}{2}}^{ta{n}^{-1}\\left({r}_{t}\\right)}\\text{d}\\theta$ (20)\n\n$D\\left({r}_{t}\\right)=\\frac{1}{\\text{π}}ta{n}^{-1}\\left({r}_{t}\\right)+\\frac{1}{2}.$ (21)\n\nThe density function is\n\n$p\\left({r}_{t}\\right)=\\frac{1}{\\text{π}\\left(1+{r}_{t}^{2}\\right)},$ (22)\n\nwhich is the Cauchy distribution. The advantage of this method is at least partially pedagogic in that the relationship to the Cauchy distribution is obvious.\n\nA review of Equation (18) shows that it could be broken up into two different densities. The distribution of $\\omega$ is\n\n$p\\left(\\omega \\right)=\\omega {\\text{e}}^{\\frac{-{\\omega }^{2}}{2}},$ (23)\n\nwhile the distribution of of $\\theta$ is\n\n$p\\left(\\theta \\right)=\\frac{1}{\\text{π}}.$ (24)\n\nEquation (23) is the Rayleigh distribution, which is a special case of the Weibull distribution or the $\\chi$ distribution . Equation (24) is the uniform distribution. This is also a restatement of Gull’s Lighthouse problem . Both $\\omega$ and $\\theta$ are independent of each other.\n\nThe author believes that greater knowledge may develop by exploring the role that the properties of the polar coordinate distributions possess. The uniform distribution, like the Cauchy is perfectly imprecise, that is it has no variance unless bounded. The Weibull is an extreme value distribution and under the Rayleigh parameterization implies that large values are probable. As the Rayleigh distribution, it is the limiting distribu- tion of the vector sum with “random amplitudes and uniformly distributed phases” . Through the $\\chi$ distribution it is the distribution of the square root of the sum of squares of independent draws from a normal distribution.\n\nA further reason to consider this linkage is the work performed by Mc- Cullaugh in and Burdzy in on the linkage between the Cauchy distribu- tion and the complex plane, which is tightly linked to trigonometric solutions through Euler’s formula:\n\n${\\text{e}}^{ix}=cos\\left(x\\right)+isin\\left(x\\right).$ (25)\n\nAlthough more metaphorical and intuitive than rigorous, it may serve to point out the error at time t is real in the sense that it is about to happen while the second error is, or at least feels, more imaginary in that it projections are being made, possibly decades in advance. While this work is being done in ${ℝ}^{2}$ significant work has been done in $ℂ$ instead, beginning with work by Burdzy in .\n\n3.3. Limitation of Liability\n\nThe limitation of liability truncates the Cauchy distribution. Its parameters become\n\n$p\\left({r}_{t}\\right)={\\left[\\frac{\\text{π}}{2}+ta{n}^{-1}\\left(\\frac{R}{\\Gamma }\\right)\\right]}^{-1}\\frac{\\Gamma }{{\\Gamma }^{2}+{\\left({r}_{t}-R\\right)}^{2}}$ (26)\n\nThis could be solved, as above, by restricting values for ${p}_{t}$ and ${p}_{t+1}$ to greater than zero.\n\nAn alternative representation would be to note that in bankruptcy ${q}_{t+1}=0$ and so it is not a ratio of prices but quantities. That is to say, the ratio distribu- tion of prices is multiplied by the ratio distribution of quantities. Given that the firm neither goes bankrupt nor merges out of existence during the interval, $Pr\\left({q}_{t+1}={q}_{t}\\right)=1$ .\n\nThe truncated Cauchy distribution is reasonable approximation of returns on equity, even without discussing the issues of liquidity or the budget constraint discussed in Section 11. This is extensively discussed in . The annual returns for going concerns, less the impact of dividends, collected from the population of data at the Center for Research on Security Prices and excluding such firms as shell companies and closed-end funds for the years 1962-2013 are shown graphically in Figure 1. The best fit models from parameters implied by the data are provided as well.\n\nFigure 1. Empirical distribution versus implied Cauchy and Gaussian models from the data.\n\nThe three curves of 1 are disaggregated returns. This differs from the standard market returns shown throughout the literature, but rather are trade-by-trade returns across the period. All three curves are smoothed with spline fitting by Microsoft’s Excel. The first curve is the empirical distribution of annual returns over the period. The second is the best fit Cauchy and Gaussian model of distributions. The Gaussian fits so poorly because the tails are so dense and long.\n\nThe Gaussian model has two conflicting issues, modeling the high density region and modeling the tails. Because of the squared impact of attempting to measure the variance the Gaussian model is flattened in the dense region, but still too thin at the extrema of the tails. Because the Cauchy model is a peaked distribution with long tails, there is no mathematical conflict in fitting a curve.\n\nThe fact of heavy tails has not been in dispute since at least Mandelbrot’s article, but an explanation as to why it exists has not been made . Figure 1 is nothing more than a reminder of facts long settled but explained as anomalies. This is not an anomaly. This is how equity securities are supposed to work.\n\n3.4. Observations on the Nature of the Cauchy Distribution\n\n3.4.1. Lack of a Mean\n\nConsider a generic variable, ${x}_{i}$ , drawn from a Cauchy distribution, regardless if the form is a time series or some other data form, there is an understood relationship between the sampling distribution of the mean, ${\\stackrel{¯}{S}}_{n}$ , for a sample size of n and the density function of the raw data.\n\nIn order to shorten the exposition and without loss of generality, assume that ${x}_{i}$ is drawn from the standard Cauchy distribution, that is\n\n$f\\left(x\\right)=\\frac{1}{\\text{π}}\\frac{1}{1+{x}^{2}}.$ (27)\n\nThe sum of n Cauchy variates is\n\n${S}_{n}=\\underset{i=1}{\\overset{n}{\\sum }}\\text{ }\\text{ }{x}_{i}$ (28)\n\nand the sample mean is\n\n${\\stackrel{¯}{S}}_{n}=\\frac{{S}_{n}}{n}.$ (29)\n\nThe distribution of the sample mean can be found by using the characteristic function for the Cauchy distribution. The characteristic function is\n\n$\\begin{array}{c}\\varphi \\left(\\tau \\right)={\\int }_{-\\infty }^{\\infty }{\\text{e}}^{ix\\tau }\\frac{1}{\\text{π}}\\frac{1}{1+{x}^{2}}\\text{d}x\\\\ =\\frac{1}{\\text{π}}{\\int }_{-\\infty }^{\\infty }\\frac{cos\\tau x+isin\\tau x}{1+{x}^{2}}\\text{d}x\\\\ =\\frac{2}{\\text{π}}{\\int }_{0}^{\\infty }\\frac{cos\\tau x}{1+{x}^{2}}\\text{d}x={\\text{e}}^{-|\\tau |}.\\end{array}$ (30)\n\nThe characteristic function for the sum of n independent draws from the distribution described by Equation (27) is the product of the individual charac- eristic functions resulting in a joint characteristic function of\n\n$\\varphi {\\left(\\tau \\right)}^{n}.$ (31)\n\nInverting the prior process yields\n\n$f\\left({S}_{n}\\right)=\\frac{1}{2\\text{π}}{\\int }_{-\\infty }^{\\infty }{\\text{e}}^{-i{S}_{n}\\tau -n|\\tau |}\\text{d}\\tau =\\frac{1}{\\text{π}}\\frac{n}{{n}^{2}+{S}_{n}^{2}}.$ (32)\n\nConverting to the distribution of the sample mean yields\n\n$f\\left({\\stackrel{¯}{S}}_{n}\\right)=f\\left({S}_{n}\\right)\\frac{\\text{d}{S}_{n}}{\\text{d}{\\stackrel{¯}{S}}_{n}}=\\frac{1}{\\text{π}}\\frac{1}{1+{\\stackrel{¯}{S}}_{n}^{2}}.$ (33)\n\nThe sampling distribution of the mean maps to the distribution of the raw data. The unexpected implication is that there is no difference in information about the center of location between looking at the sample mean of one thousand draws, two draws or simply looking at one draw of the raw data. This is equivalent to saying there is no gain in information by calculating the mean of any size over simply looking at one raw data point in determining the true center of location.\n\nThis is deeply problematic for the continued use of sample means or least- squares style methods in economics. The inference is purely random because the Cauchy distribution has no population mean and so the sample mean does not converge on a single point. The most that can be said about the center of location given the sample mean is that it is somewhere in the real numbers. As a consequence, when using mean-based methods, an inference regarding the true parameter $\\beta$ will be equivalent to having a sample size of one regardless of the true sample size. Such a finding is catastrophic for standard econometric methods. It can also be catastrophic to standard economic models involving capital.\n\n3.4.2. Lack of Independence\n\nThe multivariate Cauchy distribution is a symmetric special case of the multivariate Student distribution. For the two dimensional case it is\n\n$\\frac{1}{2\\text{π}}\\left[\\frac{\\Gamma }{{\\left({\\Gamma }^{2}+\\underset{i=1}{\\overset{2}{\\sum }}{\\left({r}_{i}-{R}_{i}\\right)}^{2}\\right)}^{\\frac{3}{2}}}\\right].$ (34)\n\nFor the three dimensional case the density is\n\n$\\frac{1}{{\\text{π}}^{2}}\\frac{{\\Gamma }^{2}}{\\sqrt{{\\Gamma }^{3}}{\\left(\\Gamma +\\underset{i=1}{\\overset{3}{\\sum }}{\\left({r}_{i}-{R}_{i}\\right)}^{2}\\right)}^{2}}.$ (35)\n\nNotice that at no point does a covariance matrix or some structure similar to a covariance matrix appear. Also note that the nth dimensional distribution is not just the n product of the univariate distributions. It is not obvious, but care must be taken, that when solving as a student distribution that the traditional meaning of $\\mu$ as a mean and $\\sigma$ as a standard deviation in Student’s t distribution changes to $\\mu$ as a mode and $\\sigma$ as a scale parameter. More particularly, $\\sigma$ is the half-width at half-maximum.\n\n4. Assets Purchased in Single Auctions and Subject to the Winner’s Curse\n\nWhile stocks, in equilibrium, are not subject to a winner’s curse, many assets are. This is because the high, or in some cases low, bidder purchases the asset. The difficulty is created as above, by each party bidding their expectation. While the sampling distribution of the mean is the Gaussian distribution, the sampling distribution of the extrema is the Gumbel distribution.\n\nThe determination of the ratio of two Gumbel distributions is complicated by the fact that there are multiple types of buyers. Professional buyers, collectors, and novice buyers may face different conditional distributions due to different information sets and different pricing goals.\n\nAn antique dealer in a network of dealers, or a construction contractor that regularly subcontracts, have significant information about the wholesale value while a novice actor generally can, at best, see retail pricing. Although the functional form for either will be the same, they would not be if the dealer did not purchase the asset at an auction. Rather than this auction based example, the dealers’ returns would depend on the model of the dealer market.\n\nThe form of the Gumbel distribution for a single-sided, high-bid auction purchase is\n\n$p\\left({\\epsilon }_{t}\\right)=\\frac{1}{{\\sigma }_{t}}{\\text{e}}^{-\\left(\\frac{{\\epsilon }_{t}}{{\\sigma }_{t}}-{\\text{e}}^{-\\frac{{\\epsilon }_{t}}{{\\sigma }_{t}}}\\right)},\\forall t.$ (36)\n\nAssuming the errors are independent and markets are in equilibrium, then\n\n$p\\left({r}_{t}\\right)={\\int }_{-\\infty }^{\\infty }|{\\epsilon }_{t}|\\frac{exp\\left(-\\frac{{r}_{t}{ϵ}_{t}}{{\\sigma }_{t+1}}-{\\text{e}}^{-\\frac{{r}_{t}{ϵ}_{t}}{{\\sigma }_{t+1}}}-\\frac{{ϵ}_{t}}{{\\sigma }_{t}}-{\\text{e}}^{-\\frac{{ϵ}_{t}}{{\\sigma }_{t}}}\\right)}{{\\sigma }_{t}{\\sigma }_{t+1}}\\text{d}{\\epsilon }_{t},$ (37)\n\nremembering that the errors are centered around zero and not R. In a numerical solution, returns will need to be shifted by R.\n\n3The author would prefer to name it the Maxham Distribution after his wife's maiden name since this does not appear to be a named distribution.\n\nThe integral of the ratio of two Gumbel distributions is unknown. This creates two possible solutions. First, the method of histograms is available . The second choice is to construct numerical approximations of the integral.\n\nThe second choice has the advantage that it is not strictly dependent on historical conditions. Still, this forces the construction of tables rather than allow for true parametric solutions. For ${\\sigma }_{t}={\\sigma }_{t+1}=1;R=1$ the plot of the reward for investing is shown in Figure 2.3\n\nFigure 2. Reward for investing in asset with winner’s curse and selling with winner’s curse.\n\n5. Distribution of Accounting Ratios and Equity Returns with Dependent Errors\n\nAccounting ratios and returns on equity securities without independent errors have a distribution that is very close to the Cauchy distribution. In this case, it is presumed that there is support on the entire real line. Adjustments for truncation would be necessary. The assumption is that an equilibrium exists. In this case, the joint distribution of errors is the bivariate normal centered on (0,0), but ${\\epsilon }_{t+1}$ is correlated with ${\\epsilon }_{t}$ through a correlation coefficient, $\\rho$ .\n\nThe distinction with the derivation of the Cauchy distribution is that for the Cauchy distribution, the covariance matrix for the normal distribution is\n\n$\\Sigma =\\left[\\begin{array}{cc}{\\sigma }_{t,t}& 0\\\\ 0& {\\sigma }_{t+1,t+1}\\end{array}\\right],$ (38)\n\nwhile in this case\n\n$\\Sigma =\\left[\\begin{array}{cc}{\\sigma }_{t,t}& {\\sigma }_{t,t+1}\\\\ {\\sigma }_{t,t+1}& {\\sigma }_{t+1,t+1}\\end{array}\\right].$ (39)\n\nUsing Curtiss’ method, the unshifted distribution of returns is\n\n$p\\left({r}_{t}\\right)=\\frac{{\\sigma }_{t}{\\sigma }_{t+1}\\sqrt{1-\\frac{{\\sigma }_{t,t+1}^{2}}{{\\sigma }_{t}^{2}{\\sigma }_{t+1}^{2}}}}{\\text{π}\\left({r}_{t}^{2}{\\sigma }_{t}^{2}-2{r}_{t}{\\sigma }_{t,t+1}+{\\sigma }_{t+1}^{2}\\right)}.$ (40)\n\nIf the above equation was multiplied by\n\n$\\frac{\\frac{1}{{\\sigma }_{t}^{2}}}{\\frac{1}{{\\sigma }_{t}^{2}}},$ (41)\n\nthen the relationship between the Cauchy distribution becomes a bit more obvious by setting\n\n$\\Gamma =\\frac{{\\sigma }_{t+1}}{{\\sigma }_{t}}.$ (42)\n\nIn that case, the distribution becomes\n\n$p\\left({r}_{t}\\right)=\\frac{\\Gamma \\sqrt{1-{\\rho }^{2}}}{\\text{π}\\left({r}_{t}^{2}-2{r}_{t}{\\sigma }_{t,t+1}/{\\sigma }_{t}^{2}+{\\Gamma }^{2}\\right)}.$ (43)\n\nAs with the Cauchy distribution, the return needs to be shifted by R so that the final formula is\n\n$p\\left({r}_{t}\\right)=\\frac{\\Gamma \\sqrt{1-{\\rho }^{2}}}{\\text{π}\\left({\\left({r}_{t}-R\\right)}^{2}-2\\left({r}_{t}-R\\right){\\sigma }_{t,t+1}/{\\sigma }_{t}^{2}+{\\Gamma }^{2}\\right)},$ (44)\n\nor with fewer parameters to estimate\n\n$p\\left({r}_{t}\\right)=\\frac{{\\sigma }_{t}{\\sigma }_{t+1}\\sqrt{1-\\frac{{\\sigma }_{t,t+1}^{2}}{{\\sigma }_{t}^{2}{\\sigma }_{t+1}^{2}}}}{\\text{π}\\left({\\left({r}_{t}-R\\right)}^{2}{\\sigma }_{t}^{2}-2\\left({r}_{t}-R\\right){\\sigma }_{t,t+1}+{\\sigma }_{t+1}^{2}\\right)}.$ (45)\n\nAlthough the assumption of independence of errors is reasonable, program trading where few actors could be involved in moving large portfolios that are evaluated jointly, it is quite possible that in the case of quick turnover that the appraisal errors are highly associated with both the purchase and selling of a block of assets.\n\n6. The Distribution of Returns with Markets in Disequilibrium\n\nIt is difficult to discuss markets in disequilibrium in economics. Events such as the financial crash of 2008 are problematic for equilibrium models. If the market is away from equilibrium at time t, then rather than considering\n\n${p}_{t}={p}_{t}^{*}+{ϵ}_{t}$ (46)\n\nthe model should be\n\n${p}_{t}={p}_{t}^{*}+{\\mu }_{t}+{ϵ}_{t}.$ (47)\n\nIf $\\mu >0$ then returns should be smaller and, conversely, where $\\mu <0$ then returns should be larger. The question is “what is ${\\mu }_{t}$ ?” If ${\\mu }_{t}$ is thought of as part of the error term, then it is a systematic bias. If it is thought as a systematic shift in the equilibrium, then the present value of cash flows does not equal the price where there is no pressure for change. In neither case, $\\mu$ vanishes from the numerator on the assumption that over time, prices must reflect future cash flows.\n\nThe two solutions imply different distribution rules.\n\n6.1. Placing the Shift in the Error Term\n\nIf the market is believed to be away from the implied curve in equilibrium then it cannot be assumed that errors are centered on zero. If it is assumed that assets will be sold when market prices are no longer sticky and away from zero, then the mean error will be located at $\\left({\\mu }_{t},0\\right)$ . If it also presumed that\n\n$\\Gamma =\\frac{{\\sigma }_{t+1}}{{\\sigma }_{t}},$ (48)\n\nthen it is possible to state everything in reference to time t. Using Curtiss’ method\n\n$\\begin{array}{c}p\\left({r}_{t}\\right)=\\frac{\\Gamma {\\text{e}}^{-\\frac{{\\mu }_{t}^{2}}{{\\sigma }_{t}}}\\left(\\sqrt{\\text{π}}{\\mu }_{t}{\\text{e}}^{\\frac{{\\Gamma }^{2}{\\mu }_{t}^{2}}{{\\left({r}_{t}-R\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}}\\text{erf}\\left(\\frac{{\\mu }_{t}}{{\\sigma }_{t}\\sqrt{\\frac{\\frac{{\\left({r}_{t}-R\\right)}^{2}}{{\\Gamma }^{2}}+{\\sigma }_{t}}{{\\sigma }_{t}^{2}}}}\\right)\\right)}{4\\text{π}{\\sigma }_{t}\\left({\\left({r}_{t}-R\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}\\right)\\sqrt{\\frac{{\\left({r}_{t}-R\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}{{\\Gamma }^{2}{\\sigma }_{t}^{2}}}}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}+\\frac{\\Gamma {\\text{e}}^{-\\frac{{\\mu }_{t}^{2}}{{\\sigma }_{t}}}\\left(\\sqrt{\\text{π}}{\\mu }_{t}{\\text{e}}^{\\frac{{\\Gamma }^{2}{\\mu }_{t}^{2}}{{\\left({r}_{t}-R\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}}+{\\sigma }_{t}\\sqrt{\\frac{{\\left({r}_{t}-R\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}{{\\Gamma }^{2}{\\sigma }_{t}^{2}}}\\right)}{4\\text{π}{\\sigma }_{t}\\left({\\left({r}_{t}-R\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}\\right)\\sqrt{\\frac{{\\left({r}_{t}-R\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}{{\\Gamma }^{2}{\\sigma }_{t}^{2}}}}.\\end{array}$ (49)\n\nBuried inside this extensive expression is the Cauchy distribution. It is important to remember that if ${\\mu }_{t}<0$ then returns will be shifted to the right from equilibrium returns and conversely for ${\\mu }_{t}>0$ .\n\nMarsaglia’s Extension of Curtiss’ Method for the Normal Case\n\nMarsaglia in extended Curtiss’ work by considering the more general case of\n\n$\\frac{a+x}{b+y},$ (50)\n\nwhere x and y are independent standard normal random variates, while a and b are constants. He extended this work to cover the cumulative density function for computational ease . If $a=0$ , then this is equivalent to Equation (49) with several additional simplifying assumptions.\n\nIn the simplified case of Marsaglia , he finds that the distribution is a convex combination of the Cauchy distribution and a distribution that can either be unimodal or bimodal and is the product of a Cauchy distribution and other terms extracted from the normal distributions and constants. The conden- sed general solution to Marsaglia’s problem is:\n\n$p\\left(z\\right)=\\frac{exp\\left[-\\frac{1}{2}\\left({a}^{2}+{b}^{2}\\right)\\right]}{\\text{π}\\left(1+{z}^{2}\\right)}\\left[1+qexp\\left(\\frac{1}{2}{q}^{2}\\right){\\int }_{0}^{q}exp\\left(-\\frac{1}{2}{x}^{2}\\right)\\text{d}x\\right],q=\\frac{b+az}{\\sqrt{1+{z}^{2}}}$ (51)\n\nFor purposes of this article,\n\n$a=0,b={\\mu }_{t}=\\frac{{\\stackrel{¯}{p}}_{t}-{p}_{t}^{*}}{{\\sigma }_{t}},\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}z={r}_{t}.$ (52)\n\nThis splits the equation into two parts. The first part,\n\n$\\frac{exp\\left[-\\frac{1}{2}{\\mu }_{t}^{2}\\right]}{\\text{π}\\left(1+{r}_{t}^{2}\\right)},$ (53)\n\nis the product of the Cauchy distribution of returns with an the non-normalized Gaussian kernel. The second portion,\n\n$1+\\frac{q}{exp\\left(-\\frac{1}{2}{q}^{2}\\right)}{\\int }_{0}^{q}exp\\left(-\\frac{1}{2}{x}^{2}\\right)\\text{d}x,$ (54)\n\nis one plus the elasticity of q with respect to the cumulative density function of q. As the available prices are systematically far from the equilibrium the distribu- tion becomes inelastic. This stiffness of outcomes is generated by the fact that as ${\\mu }_{t}$ goes to zero, the distribution goes to the Cauchy distribution. Conversely, as ${\\mu }_{t}$ becomes far from the equilibrium price, such as when ${\\mu }_{t}>4$ then the role of the Cauchy distribution gets close to zero. The other distribution has all of its moments.\n\nAs a mixture, there are no moments, but the effect of the Cauchy distribution becomes small. This narrowing implies that as bubbles go on a poor outcome becomes increasingly certain while after a bear market a good outcome becomes increasingly certain.\n\nIf prices being far from an equilibrium can be thought of as type I or type II errors, then Equation (49) could be thought of as the statistical basis for value investing. Benjamin Graham’s and David Dodd’s “margin of safety” required of all investors would have a parallel “margin of excitement” for speculative investors, adapting the parlance of Graham’s book, The Intelligent Investor .\n\nThis parallels the conversion of the cumulative density of returns into a Bernoulli distribution. Consider the problem of the\n\n$Pr\\left({r}_{t}\\ge j|{p}_{t}\\le k\\right).$ (55)\n\nIn this equation, j is a sufficient return to meet planned goals, while k is some limit price. Either the goal is met or it is not met. The variance maximizes when the probability of either case is fifty percent. A portfolio would carry a binomial distribution.\n\nAs ${p}_{t}$ became large, the probability would become small and the variability of outcomes would decline. Conversely, as ${p}_{t}$ became small, the probability would become large and the variability of outcomes would decline. The instinct of the economist would be to ask “what is preventing the security from being at its equilibrium price” and the simple answer is “that is precisely what a type I or type II error would be in this case.”\n\n6.2. A Shift Moves the Equilibrium Away from the Present Value of Cash Flows\n\nIn this case,\n\n${R}_{t}=\\frac{{p}_{t+1}^{*}}{{p}_{t}^{*}+{\\mu }_{t}}.$ (56)\n\nIt follows that the errors would be a truncated Cauchy distribution, but shifted for the price shift.\n\n6.3. Convex Combination\n\nIt is also quite possible that ${\\mu }_{t}$ is split between a systematic bias and an equilibrium shift. How this function works in the real world is an empirical question to test under Bayesian model selection. If the shifting value were split into two components, ${\\mu }_{\\alpha }$ and ${\\mu }_{\\beta }$ , where ${\\mu }_{\\alpha }$ is the equilibrium shift and ${\\mu }_{\\beta }$ is the shift from bias, then the return distribution becomes\n\n$\\begin{array}{c}p\\left({r}_{t}\\right)=\\frac{\\Gamma {\\text{e}}^{-\\frac{{\\mu }_{\\beta }^{2}}{{\\sigma }_{t}}}\\left(\\sqrt{\\text{π}}{\\mu }_{\\beta }{\\text{e}}^{\\frac{{\\Gamma }^{2}{\\mu }_{\\beta }^{2}}{{\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}}\\text{erf}\\left(\\frac{{\\mu }_{\\beta }}{{\\sigma }_{t}\\sqrt{\\frac{\\frac{{\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}}{{\\Gamma }^{2}}+{\\sigma }_{t}}{{\\sigma }_{t}^{2}}}}\\right)\\right)}{4\\text{π}{\\sigma }_{t}\\left({\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}\\right)\\sqrt{\\frac{{\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}{{\\Gamma }^{2}{\\sigma }_{t}^{2}}}}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}+\\frac{\\Gamma {\\text{e}}^{-\\frac{{\\mu }_{\\beta }^{2}}{{\\sigma }_{t}}}\\left(\\sqrt{\\text{π}}{\\mu }_{\\beta }{\\text{e}}^{\\frac{{\\Gamma }^{2}{\\mu }_{\\beta }^{2}}{{\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}}+{\\sigma }_{t}\\sqrt{\\frac{{\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}{{\\Gamma }^{2}{\\sigma }_{t}^{2}}}\\right)}{4\\text{π}{\\sigma }_{t}\\left({\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}\\right)\\sqrt{\\frac{{\\left({r}_{t}-R-{\\mu }_{\\alpha }\\right)}^{2}+{\\Gamma }^{2}{\\sigma }_{t}}{{\\Gamma }^{2}{\\sigma }_{t}^{2}}}}.\\end{array}$ (57)\n\n7. The Distribution of Mergers for Cash\n\nCash-for-stock mergers are a special limiting case. Whereas stock-for-stock provides the selling shareholders a contingent claim, cash for stock provides perfect liquidity. As cash is expensive, a cash purchase should have additional properties.\n\nAs this process is being thought of in a Bayesian framework, any estimate of return must include the possibility that the firm will be purchased for cash by another party. Given a cash merger will occur, it will have a defined future value once the merger is announced. Prior to announcement a likelihood function is necessary to estimate a solution for pricing and returns. If it is assumed that the rates of return are efficiently priced once perfect liquidation is certain, then traditional economic tools should solve this problem. Going back to Equation (14) note that the future value is now fixed, given a cash-for-stock merger will happen with certainty.\n\nOf course this is not certain, so Bayes theorem provides a solution. If\n\n$\\varphi \\left(X|\\theta ;{M}_{c}\\right)\\pi \\left(\\theta ;{M}_{c}\\right),$ (58)\n\nwhere ${M}_{c}$ denotes a cash for stock merger, $\\varphi \\left(\\text{ }\\right)$ is the likelihood of observing data, given a set of parameters and the certainty of a cash for stock merger, times the probability of a cash for stock merger, then solving Equation (58) over the parameter space, $\\theta \\in \\Theta$ , provides a value proportionate to the posterior proba- bility for each possible value of the parameters. Once normalized and integrated over the parameters, the resulting predictive density is the best information about the distribution of future returns for a given security.\n\nThe concern here is the reward from investing from Equation (14). In that equation there are two prices, ${p}_{t}$ and ${p}_{t+1}$ . For a going concern, this is not an issue, because the firm will exist at time $t+1$ . If a firm is merged out of existence for cash before time $t+1$ then what rule can be created for the future value?\n\nIn fact, there are two possible rules, either equally good for specific purposes, and less good for other purposes. The first rule would be to look forward in time starting at time t and ending at time $t+1$ , partitioning the set into short, meaningful intervals. Then a probability of a merger completion would be calculated for each subinterval in the partitioned set. A likelihood function would also be created for each date.\n\nThe alternative rule would be to ask the probability of a cash-for-stock merger on or before the date denoted as $t+1$ . In that case, ${r}_{t}$ represents the reward received for having invested funds at time t. It does ignore reinvestment, which would be a separate question.\n\nThe question of a cash-for-stock merger has two important properties that can be illustrated in the derivation that is not done in prior derivations. The first is that a cash-for-stock merger has only one error in it, not two. The second, not mentioned above, is to separate the role of the investor and the economist or finance professional.\n\nIf it were known, with certainty, that a cash-for-stock merger was to occur, then the only question is what will the stock sell for in that merger.\n\nThe equation for the realized reward is\n\n${r}_{t}=\\frac{{w}_{t+1}}{{p}_{t}}=\\frac{1}{{\\text{e}}^{-\\mathrm{log}\\left({r}_{t}\\right)}},$ (59)\n\nwhere w is future wealth as there is no price, and which can be normalized to unity.\n\nThe reason for noting the relationship between nominal reward and logari- thmic rewards has to do with how errors are conceptualized. Are errors multiplicative as would be implied by the logarithmic form or are they additive as the raw form would imply?\n\nThe issue is both theoretical and empirical. If the distribution of returns appears as either what one would expect from additive errors, or what one would expect from multiplicative errors, then a fundamental relationship to reality can be discussed.\n\nThe derivation of the proof is built upon a key proof by Landon in that appears throughout economics and physics, but often without a realization of where that proof may have come from.\n\n7.1. Jaynes Generalization of Landon’s Proof\n\nE.T. Jaynes in tome on probability theory, Probability Theory: The Langu- age of Science, uses part of a subsection to point out the importance of Landon’s proof and to generalize it. Landon in was trying to solve the problem of noise in communication circuits. The proof was for a particular case, but the proof immediately caught the attention of both Harold Jeffreys and John Maynard Keynes.\n\nLandon’s proof has several important components for economics. For Jaynes, Jeffreys, and Keynes, the importance was in noting that “by minor changes in the wording” the proof, “can be interpreted either as calculating a probability distribution, or estimating a frequency distribution” . For purposes of this paper, the linkage between Bayesian and null hypothesis methods in this proof is unimportant. What is important is the distinction between methods available when there is one error in the decision process versus two errors as in normal equity trading. Additionally, the proof by Landon in allows a direct linkage between the subjective decisions of individual actors and the calculation of the distribution that exists in equilibrium\n\nBecause of the relationship between subjective personal realities and objective market relationships, there will be a slight notation change. For the personal calculated rewards of a subjective actor, the notation will be ${}_{i}{r}_{t}$ at time t while the equilibrium return will be ${r}_{t}$ .\n\nThis cumbersome notation is not present in Probability Theory: The Language of Science because Jaynes, a physicist, is not concerned with the relationship between subjective personal realities of the actors with market equilibrium as an economist would be concerned. Additionally, Jaynes uses Dirac notation for expectations, whereas here the expectation of some random variable x will be notated $E\\left(x\\right)$ . Although the physics notation is shorter, it is uncommon to see it in economics articles.\n\nModel Assumptions\n\nIt is assumed that the observed market return is ${r}_{t}$ and for this derivation, there will be additive errors. As with Markowitz, it will be assumed that there are many actors, or alternatively, that over time there will be many actors. With Landon’s original form of proof, the better assumption would be the latter. Further, it will be assumed that the markets are in equilibrium. As a consequence, any error term’s expectation is zero.\n\nIn equation\n\n${p}_{t+1}=R{p}_{t}+{\\epsilon }_{t+1}$ (60)\n\nthe error term is ${\\epsilon }_{t}$ . This represents the movement of the price around the equilibrium trade price through time. The concern here, however, is with personal errors. Market and personal errors will be denoted ${ϵ}_{t}$ or ${}_{i}{ϵ}_{t}$ .\n\nAs such, it is assumed that\n\n$E\\left({}_{i}{ϵ}_{t}\\right)=0.$ (61)\n\nFurther, for a given level of risk, it is assumed that\n\n$p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)$ (62)\n\nis known to each actor, denoted $i\\in I$ , where the set I indexes the market actors, and where ${\\sigma }_{t}$ is scale parameter, in this case, a measure of risk.\n\nLandon in then posited the question of the impact of a very small shock or error, ${}_{i}{ϵ}_{t}$ to the evaluation of ${}_{i}{r}_{t}$ by looking at the equation\n\n${}_{i}{{r}^{\\prime }}_{t}={}_{i}r{}_{t}+{}_{i}ϵ{}_{t},$ (63)\n\nwhere ${}_{i}{ϵ}_{t}$ is very small relative to ${\\sigma }_{t}$ . Landon also assumed that ${}_{i}{ϵ}_{t}$ had a probability of being observed equal to\n\n$q\\left({}_{i}{ϵ}_{t}\\right){\\text{d}}_{i}{ϵ}_{t},$ (64)\n\nand that this probability was independent of $p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)$ .\n\nThe objective distribution of ${{r}^{\\prime }}_{t}$ maps to some subjective value $p\\left({}_{i}{{r}^{\\prime }}_{t}\\right)$ and depends upon the probability of the specific error or shock by the actor(s) and the separate probability of ${{r}^{\\prime }}_{t}$ . The probability for this is\n\n$f\\left({{r}^{\\prime }}_{t}\\right)=\\int p\\left({}_{i}{{r}^{\\prime }}_{t}-{}_{i}ϵ{}_{t}|{\\sigma }_{t}\\right)q\\left({}_{i}ϵ{}_{t}\\right){\\text{d}}_{i}{ϵ}_{t}.$ (65)\n\nThis can be estimated by the nearby value ${r}_{t}$ by expansion about the nearby value. The distribution can be estimated by\n\n$\\begin{array}{c}f\\left({r}_{t}\\right)=p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)-\\frac{\\partial p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)}{\\partial {r}_{t}}{\\int }_{i}\\text{ }{ϵ}_{t}q\\left({}_{i}{ϵ}_{t}\\right){\\text{d}}_{i}{ϵ}_{t}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}+\\frac{1}{2}\\frac{{\\partial }^{2}p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)}{\\partial {r}_{t}^{2}}{\\int }_{i}{ϵ}_{t}^{2}q\\left({}_{i}{ϵ}_{t}\\right){\\text{d}}_{i}{ϵ}_{t}+\\cdots ,\\end{array}$ (66)\n\nwhich can be notationally shortened to\n\n$f\\left({r}_{t}\\right)=p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)-E\\left({ϵ}_{t}\\right)\\frac{\\partial p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)}{\\partial {r}_{t}}+\\frac{1}{2}E\\left({ϵ}_{t}^{2}\\right)\\frac{{\\partial }^{2}p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)}{\\partial {r}_{t}^{2}}+\\cdots .$ (67)\n\nNoting that the expectation of ${ϵ}_{t}$ is zero and that the expectation of ${}_{i}{r}_{t}^{2}$ increments to ${\\sigma }_{t}^{2}+E\\left({}_{i}{ϵ}_{t}^{2}\\right)$ and the invariance property noted above, this leads to the equation\n\n$f\\left({r}_{t}\\right)=p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)+E\\left({}_{i}{ϵ}_{t}^{2}\\right)\\frac{{\\partial }^{2}p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)}{\\partial {\\sigma }_{t}^{2}}$ (68)\n\nwhich becomes\n\n$\\frac{\\partial p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)}{\\partial {\\sigma }_{t}}=\\frac{1}{2}\\frac{{\\partial }^{2}p\\left({}_{i}{r}_{t}|{\\sigma }_{t}\\right)}{{\\partial }_{i}{r}_{t}^{2}},$ (69)\n\nwhose known solution is the normal distribution.\n\n7.2. Implications\n\nLandon’s proof allows an expansion about an equilibrium value when there is only one error. For most of the ratio distributions covered in this paper, there does not exist a first moment as they are all transformations of the Cauchy distribution on at least the half-plane. As such, no expansion can be created. A key tool is lost to economics.\n\nBy construction, it appears that cash-for-stock should carry a log-normal distribution.\n\n8. Including the Impact of Bankruptcy\n\nBankruptcy is the simplest of the cases in that if the shareholders are wiped out, then ${q}_{t+1}=0$ . If ${B}^{\\prime }$ is the posterior probability of bankruptcy, then returns simply become $0×{B}^{\\prime }+\\left(1-{B}^{\\prime }\\right){R}^{\\prime }$ , where ${R}^{\\prime }$ is the posterior probability of returns for all other states. Of course in Bayesian statistics, these are not point estimates; the formula represents a joint distribution of returns and bankruptcy risk.\n\nIf returns were understood as independent of bankruptcy risk, then it would be the simple cross product over the parameter space. If the center of location and the scale parameter are a function of bankruptcy probability, then the situation becomes less clear.\n\nThe challenge is created by an absence of a unique or clear mechanism to model bankruptcy in the literature. Indeed, the plethora of ways to model bankruptcy generates a wide range of possible joint distributions. The prediction of potential future returns depends upon how the relationships among para- meters are conceptualized. Logically the center of location for returns should be a function of spread and bankruptcy risk. The center of location for spread should be a function of bankruptcy risk, and the probability of bankruptcy should be independent. This ignores the impact of skew, dividends payments and other types of data other than just returns. While there is only one way for variables to be independent, there are an infinite number of ways for them to be dependent. Including bankruptcy is an important and unsolved problem.\n\nA general outline of the predictive distribution for this problem is solved instead. The definition of the predictive distribution is\n\nDefinition 2 (Bayesian Predictive Distribution) If $\\stackrel{˜}{x}$ is a future value to be predicted from a matrix or vector of observed data $X$ , where the likelihood function uses a vector of parameters $\\theta$ , with $\\theta \\in \\Theta$ , where $\\Theta$ is the parameter space and $\\chi$ is the sample space, then the predictive probability that $\\stackrel{˜}{x}=k,k\\in \\chi$ is\n\n${\\pi }^{\\prime }\\left(\\stackrel{˜}{x}=k|X\\right)={\\int }_{\\theta \\in \\Theta }\\varphi \\left(\\stackrel{˜}{x}=k|\\theta \\right)\\pi \\left(\\theta |X\\right)\\text{d}\\theta ,$\n\nwhere ${\\pi }^{\\prime }$ is the predictive distribution, $\\pi$ is the posterior distribution, and $\\varphi$ is the likelihood function, for all $k\\in \\chi$ .\n\nConsidering only a model with a single location, scale and bankruptcy parameter, the predicted distribution of returns at time $t+1$ would be\n\n$p\\left({\\stackrel{˜}{r}}_{t+1}\\right)=\\underset{0}{\\overset{\\infty }{\\int }}\\underset{0}{\\overset{\\infty }{\\int }}\\underset{0}{\\overset{1}{\\int }}\\frac{p\\left(X|R;\\Gamma ;B\\right)p\\left(R|\\Gamma ;B\\right)p\\left(\\Gamma |B\\right)p\\left(B\\right)}{\\underset{0}{\\overset{\\infty }{\\int }}\\underset{0}{\\overset{\\infty }{\\int }}\\underset{0}{\\overset{1}{\\int }}p\\left(X|R;\\Gamma ;B\\right)p\\left(R|\\Gamma ;B\\right)p\\left(\\Gamma |B\\right)p\\left(B\\right)\\text{d}B\\text{d}\\Gamma \\text{d}R}\\text{d}B\\text{d}\\Gamma \\text{d}R$ (70)\n\n9. The Distribution of Returns for Zero Coupon Bonds\n\nThe only assets with a covariance matrix would be among n-risky fixed cash flows, where they share bankruptcy or some other risk. Zero coupon bonds with simultaneous maturity would be the simplest form of this. Several possibilities exist, such as the logistic distribution, the normal distribution or the multino- mial distribution would work, but a distribution with a covariance matrix would account for covariance among the factors relating to bankruptcy.\n\n10. The Likelihood Function for Solutions Involving Regression\n\n10.1. Introduction\n\nRegression creates a set of special cases and does not provide a neat or clear recommendation. The difficulty for regression that differs from non-Bayesian methods is that non-Bayesian methods are only concerned with the sample after it has been transformed by a statistical method. Ordinary least squares is an example of such a method. Bayesian statistics are concerned with how the data is generated in the first place. In practice, this means that many possible competing models co-exist and one hypothesis needs to exist for each possible model.\n\nDiscussions here cannot be exhaustive but can cover the major cases. Certain observations about the rules for regression are important here. First, for time series of the form\n\n${x}_{t+1}=\\beta {x}_{t}+{ϵ}_{t+1},|\\beta |<1,$ (71)\n\nthen the likelihood function is the normal distribution. Indeed, an assumption of normal returns implies Equation (71), which would imply that people anticipa- ted losses in every period. On the other hand, for the equation\n\n${x}_{t+1}=\\beta {x}_{t}+{ϵ}_{t+1},|\\beta |>1,$ (72)\n\nthe likelihood was shown by White in to be the Cauchy distribution. The reason White’s proof is not better known is due to the implications of White’s proof. Mann and Wald were able to show that the maximum likelihood estima- tor for $\\beta$ was the ordinary least squares estimator for all values of $\\beta$ , but White showed that the sampling distribution was the Cauchy distribution, implying that no solution existed for the problem .\n\nAlthough Thiel’s regression would still work as a substitute for an actual solution, this is shown in separate work to not be admissible .\n\nThis issue does not exist for Bayesian methods which do not depend upon point statistics to function.\n\nA rather peculiar problem is created by\n\n${x}_{t+1}=\\beta {x}_{t}+{ϵ}_{t+1},|\\beta |=1.$ (73)\n\nIn Bayesian methods the probability that $\\beta =1$ is zero. This is due to the fact that any countable subset of an uncountable set is of measure zero. Extending White’s logic, it is argued that the likelihood function for the equation\n\n${x}_{t+1}=\\beta {x}_{t}+{ϵ}_{t+1},|\\beta |\\ge 1,$ (74)\n\nis the Cauchy distribution. This would not be true for a non-Bayesian solution. For that, the Dickey-Fuller test statistic would be operative .\n\n10.2. White’s Proof\n\nWhite in worked on this problem in the Likelihoodist school of thinking. As with Landon in , the form of his proof has a Bayesian interpretation allowing a simple conversion from Likelihoodist to Bayesian thinking.\n\n10.2.1. Assumptions and Simplifications\n\nWhite made some simplifying assumptions that would result in no loss of generality or which would simplify notation. In particular, the notation for the vector of observations is\n\n${x}^{\\prime }=\\left({x}_{1},{x}_{2},\\cdots ,{x}_{t}\\right).$ (75)\n\nThe variance of the error is one, that is\n\n${\\sigma }^{2}=1.$ (76)\n\nAdditionally, the initial value is zero, that is\n\n${x}_{0}=0.$ (77)\n\nThe summation operator is without subscript or superscript and sums from one to T where T is the last observation. Following Mann and Wald in the maximum likelihood estimator is\n\n$\\stackrel{^}{\\beta }=\\frac{\\sum \\text{ }\\text{ }{x}_{t}{x}_{t-1}}{\\sum \\text{ }\\text{ }{x}_{t-1}^{2}}.$ (78)\n\nSeveral simplifying notations exist for matrix relationships In particular several $T×T$ matrices are created. These are\n\n$P=\\left[\\begin{array}{cccccccc}1+{\\beta }^{2}& -\\beta & 0& 0& & & & \\\\ -\\beta & 1+{\\beta }^{2}& -\\beta & 0& & & & \\\\ 0& -\\beta & 1+{\\beta }^{2}& -\\beta & & & & \\\\ & & & & \\ddots & & & \\\\ & & & & & -\\beta & 1+{\\beta }^{2}& -\\beta \\\\ & & & & & 0& -\\beta & 1\\end{array}\\right],$ (79)\n\n$A=-\\frac{1}{2}\\left[\\begin{array}{ccccccc}2\\beta & -1& 0& & & & \\\\ -1& 2\\beta & -1& & & & \\\\ 0& -1& 2\\beta & & & & \\\\ & & & \\ddots & & & \\\\ & & & & -1& 2\\beta & -1\\\\ & & & & 0& -1& 0\\end{array}\\right],$ (80)\n\nand\n\n$B=\\left[\\begin{array}{ccccccc}1& 0& 0& & & & \\\\ 0& 1& 0& & & & \\\\ 0& 0& 1& & & & \\\\ & & & \\ddots & & & \\\\ & & & & 0& 1& 0\\\\ & & & & 0& 0& 0\\end{array}\\right].$ (81)\n\nThe joint distribution of ${x}^{\\prime }$ is\n\n$f\\left({x}^{\\prime }\\right)=\\frac{\\mathrm{exp}\\left(-\\frac{1}{2}{x}^{\\prime }Px\\right)}{{\\left(2\\text{π}\\right)}^{\\frac{T}{2}}}$ (82)\n\nand\n\n$\\stackrel{^}{\\beta }-\\beta =\\frac{{x}^{\\prime }Ax}{{x}^{\\prime }Bx}.$ (83)\n\nThe standard methods of finding the distribution due to Cramer in of $\\stackrel{^}{\\beta }-\\beta$ were shown by White in to diverge.\n\n10.2.2. Observations about White’s Proof\n\nWhite’s proof determines that the sampling distribution of the slope estimate is the Cauchy distribution for this case. His proof was to show that the limiting distribution of Equation (83) is the same as that found in Equation (17). $\\stackrel{^}{\\beta }$ is a form of sample mean and as seen above in Equation (33), this is a catastrophic failure.\n\nAlthough this is catastrophic for non-Bayesian methods, it is not a problem for Bayesian methods. In particular, the form of the proof used by White has a Bayesian interpretation.\n\nBecause standard methods diverge, White uses the product of the square root of Fisher information and the likelihood function to gather information about the sampling distribution. White’s test statistic reframed into Bayesian language is the posterior density function. A question arises from White’s use of ordinary least squares here.\n\nOrdinary least squares is a form of the sampling mean and as shown in Equation (33) would have the same sampling distribution of the set of all possible slopes. Although this provides no information at all about the location of the non-existent population mean, it does provide the likelihood of a set of slopes where $|R|>1$ . At least as important, the solution would work in a Bayesian setting for $R\\ge 1$ as it would still be the same ratio distribution.\n\n10.3. Other Than Independent, Uncorrelated Errors\n\nAlthough White’s proof works fine for independent, uncorrelated errors, the other likelihoods would apply in other cases, such as an absence of independent errors and methods such as instrumental variable regression would still be necessary in cases where that would be appropriate.\n\nThe likelihood depends on model assumptions.\n\n10.4. Log-Log Models\n\nFor the case where the distribution of errors in raw form would be the Cauchy distribution, or where the likelihood function would be the Cauchy distribution, it is already known that the hyperbolic secant distribution is the distribution of a logarithmic based model. This is obvious in the relationship between the cumulative density functions of the Cauchy and hyperbolic secant model.\n\nThe cumulative density function for the Cauchy distribution is\n\n$\\frac{1}{2}+ta{n}^{-1}\\left(\\frac{x-\\mu }{\\gamma }\\right),$ (84)\n\nwhile the cumulative density function for the hyperbolic secant distribution is\n\n$\\frac{2}{\\text{π}}ta{n}^{-1}\\left\\{exp\\left[\\frac{\\text{π}}{2}\\left(\\frac{x-\\mu }{\\sigma }\\right)\\right]\\right\\}.$ (85)\n\nThe link between the two distributions is that if $X~ℂ\\left(0,1\\right)$ and $Y=\\frac{2}{\\text{π}}log|X|$ , then $Y~ℍ\\mathbb{S}\\left(0,1\\right)$ .\n\nOrdinary Least Squares\n\nA peculiar issue exists resulting from this relationship. Consider a regression model like\n\n$y={\\beta }_{1}{x}_{1}+{\\beta }_{2}{x}_{2}+{\\beta }_{0}+ϵ,$ (86)\n\nwhere ${x}_{1}$ and ${x}_{2}$ are logarithmic transformations of data drawn from a Cauchy distribution. Because a logarithmic model was used the central limit theorem still holds and so a covariance matrix will exist for the errors if ordinary least squares is used, although the actual likelihood function would be\n\n$\\frac{1}{2\\gamma }\\text{sech}\\left[\\frac{\\text{π}}{2}\\left(\\frac{y-{\\stackrel{^}{\\beta }}_{1}{x}_{1}-{\\stackrel{^}{\\beta }}_{2}{x}_{2}-{\\stackrel{^}{\\beta }}_{0}}{\\gamma }\\right)\\right].$ (87)\n\nThe important element of Equation (87) is that adding dimensions does not add a covariance matrix. Neither the Cauchy distribution nor the hyperbolic secant distribution has a structure similar to a covariance matrix. This is also true for the vector version of this, should multiple equations be estimated. The scale parameter changes, of course, for each model, but variables do not covary.\n\nThis does not exclude all forms of comovement, just that comovement described by the definition of covariance. Because of this, while ${x}_{1}$ and ${x}_{2}$ do not covary, they are also not independent either. This begs the interpretation of the covariance matrix in ordinary least squares since it is an estimator for a set of parameters that do not exist.\n\n10.5. Example-Zimbabwe\n\nConsider the special case of Zimbabwe documented by Richardson in . Zimbabwe had been called the “jewel of Africa”, due to its extensive natural resources and its ability to feed its people. Zimbabwe had a difficult colonial history. In the 1890’s the colonial government seized native lands and trans- ferred them to white colonists for farming. The consequence of this was that 4500 hundred white families owned almost all commercial farmland, while 840,000 black farmers worked on communal farms. The Mugabe government decided to return “stolen” land to the black population, although almost none of the existing white farmers could trace their families to the original colonists as the land had subsequently changed hands .\n\nThe transfer happened without compensation and was transferred to individuals in small plots who lacked the infrastructure and skills to manage such a farm. The resulting economic collapse is still felt to this day.\n\nThe question would be how to model this. Fortunately, Bayesian theory provides a disciplined solution called Bayesian model selection. In Bayesian model selection, the model is considered a parameter.\n\nThere are several possible hypothesis prior to seeing the actual data. The first would be that Zimbabwe grew constantly throughout the entire period so that the model would be\n\n$Mode{l}_{1}=\\left\\{{y}_{t+1}={\\beta }_{0}{y}_{t}+{\\alpha }_{0}+{ϵ}_{t+1},{\\beta }_{0}\\ge 1,\\forall t\\right\\}.$ (88)\n\nA second model would be that Zimbabwe grew, then collapsed and never recovered. That model would be\n\n$Mode{l}_{2}=\\left\\{\\begin{array}{l}{y}_{t+1}={\\beta }_{1}{y}_{t}+{\\alpha }_{1}+{ϵ}_{t+1},{\\beta }_{1}\\ge 1,\\forall t (89)\n\nObviously one could continue partitioning the set of data into greater and greater numbers of subsets. The first model has three parameters to estimate. They are $Mode{l}_{1},{\\beta }_{0}$ and ${\\alpha }_{0}$ . The second model has six parameters to estimate, they are $Mode{l}_{2},{\\beta }_{1},{\\beta }_{2},{\\alpha }_{1},{\\alpha }_{2}$ and $k$ . This would be the conceptual equivalent to a structural break in Frequentist statistics except that a probability distribu- tion would be assigned to possible dates for the break.\n\nOf course one could continue this process adding a third growth process. This would necessitate the a distribution for the set $\\left\\{{k}_{0},{k}_{1},{k}_{1}>{k}_{0}\\right\\}$ . This would need to be facilitated by a contingent distribution for ${k}_{1}$ given a value of ${k}_{0}$ .\n\nAlthough this would seem to create a likelihood of overfitting the model to the data, the Bayesian posterior density naturally penalizes increased model struc- ture in a mathematically coherent manner. Coherence, in this case, implying that fair gambles could be placed on models by governments and organizations.\n\n10.6. Example-Calculating Dividends\n\nDividends are a stream of payments. How these payments are estimated depends entirely on the question being answered. Asking “what is the anticipated dividend of CNA Financial Corporation” is quite a different question from “what are the total dividends to be paid by all members of the Standard and Poors 500”. It is not credible to believe they would be estimated in the same manner as differing information governs.\n\nFor the case of CNA Financial Corporation, the dividend policy appears at various times in the disclosure filings with the Securities and Exchange Commission. Because CAN’s board of directors is subject to time inconsistency a peculiar Bayesian problem forms, one that cannot truly be solved in a non-Bayesian manner. In particular, the problem is that the solution should be subgame imperfect. This implies that any model solution must be invalid unless it considers alternative models in its construction and assigns them a positive probability of being used.\n\nA second problem would exist for CNA Financial. There always exists a positive probability the firm will suspend or not declare a cash dividend. So any dividend model must contain two components. The first is an estimate of the dividend, given one is declared, times the probability of declaration; the second element is a dividend of zero times the complementary probability.\n\nFinally, it is probably better to model dividends either as a dividend yield or as a function of accounting measures such as income or free cash flows as the alternative policy to the stated one of the board. While it is not possible to state a clear likelihood function without facing a specific problem, some general principles can be discussed.\n\nFirst, the modeler has to determine in advance whether they are seeking a point estimate or a distribution of estimates. If a distribution is sought then the Bayesian predictive distribution should govern the process.\n\nOn the other hand, if a point estimate is required, a cost function should be applied to the predictive distribution and a minimization of cost sought. This cost function should either be the true cost function of being wrong for the subjective actor, or the cost function of the marginal actor. For the marginal actor, in most cases, where the distribution is truncated at zero, the all-or- nothing cost function should be used. The logic behind has to do with an understanding of the marginal actor and the center of location.\n\nIf it is assumed that the marginal actor defines pricing in the system, then the price by the marginal actor is without “error”. That is to say; it sits at the center of location for the distribution function. For a truncated distribution, this is usually at the mode. The all-or-nothing cost function, when applied to a continuous distribution, will be minimized by choosing the modal value.\n\nFinally, if it is assumed that the dividends are being declared in an expanding economy or one with a fiat currency subject to persisting inflation, then it is also reasonable to believe that the likelihood function will be one without a defined mean.\n\n11. Including the Budget Constraint\n\nThe impact of the budget constraint has two obvious solutions. The first is to explicitly model the cost of liquidity in the capital markets. Work on this can be found in Abbott’s chapter on measures of discount for liquidity and marketa- bility . The second would be to observe that only the numerator of the reward is impacted as someone cannot have a reward for an asset that he or she does not own, ignoring short-sales. As such, there is a numerator only if a coun- ter-party has sufficient funds and is willing to expend them at the desired price. This changes the solution from the probability of a return to the probability of a return given the budget constraint is met times the probability the budget constraint of the counter-party will be met.\n\n11.1. Abbott’s Formulation of Liquidity\n\nAbbott in observes that the cost of liquidity is directly related to the half-life of the order size in the dealer’s account. From the buyer’s point of view, the price is marked up by a liquidity cost to the value normally called the ask. The seller receives a marked down price normally called the bid. For a single buy, where ${q}_{t}$ is the quantity for sale, then the markup factor would be\n\n$exp\\left(\\lambda {q}_{t}\\right),$ (90)\n\nwhile the mark-down factor for a sale would be\n\n$exp\\left(-\\lambda {q}_{t}\\right).$ (91)\n\nThis implies that you could rewrite prices as\n\n${p}_{t}=\\mathrm{exp}\\left(\\lambda {q}_{t}\\right){{p}^{\\prime }}_{t}$ (92)\n\nand\n\n${p}_{t+1}=\\mathrm{exp}\\left(-\\lambda {q}_{t+1}\\right){{p}^{\\prime }}_{t+1}.$ (93)\n\nReformulating definition 1, we arrive at\n\n${r}_{t}=\\frac{\\mathrm{exp}\\left(-{\\lambda }_{t+1}{q}_{t+1}\\right){{p}^{\\prime }}_{t+1}}{\\mathrm{exp}\\left({\\lambda }_{t}{q}_{t}\\right){{p}^{\\prime }}_{t}}.$ (94)\n\nIf ${q}_{t}={q}_{t+1}$ and ${\\lambda }_{t}q={\\lambda }_{t+1}$ , then from Section 7 then $\\lambda$ follows a normal distribution.\n\nBecause most people are really concerned with their net return, the real issue presented by Abbott in is considering the net effect after liquidity costs. This could be presented as\n\n${{r}^{\\prime }}_{t}=\\frac{{{p}^{\\prime }}_{t+1}}{{{p}^{\\prime }}_{t}}.$ (95)\n\nThis allows us to escape the question regarding the distribution of the individual terms by using regression. His work considers the impact of volume on the bid-ask spread. The advantage of this method is that it allows a net predictive distribution for planning purposes by institutional investors.\n\nAbbott in extends this discussion to discuss illiquidity or market failure. As $\\lambda$ becomes large, the probability of no counter-party trade increases. Conversely, volume increases $\\lambda$ becomes small. This permits a discussion of a net return, given a trade happens, multiplied by the probability of that trade happening, which of course is essential to Bayesian thinking.\n\nOf course, if one does not assume that ${q}_{t}={q}_{t+1}$ , then an explicit modeling of q has to happen as well. While this is stationary, it is outside of the scope of this article to explore how to model this.\n\n11.2. Survival Functions\n\nThere is a second way to model the budget constraint. If the market maker is willing to buy or sell an asset at a cost, then it is as if there were no budget constraint.\n\nIf there is no market maker willing to buy an asset, then a sell must be inside the budget of the possible counterparties. This budget constraint is unknown, so needs to be modeled as a stochastic budget constraint. Further, the budget constraint should be a moving value as the money supply and disposable income changes.\n\nFor an asset purchased at time t at a price ${p}_{t}$ , the sale at time $t+1$ depends on the existence of sufficient funds for a return to exist. Logistic regression is the logical function here, with two provisos. The first is that when the price is equal to zero, then there is a one hundred percent chance of a trade happening, while the second condition is that the probability of a trade goes to zero as price tends toward infinity.\n\n12. Discussion\n\nDecision making under uncertainty requires an understanding of how that uncertainty is structured. This article provides a framework to approach chance and uncertainty in outcomes when a reward or growth is anticipated. Equation (60) can be thought of as covering far more than only stocks. It includes any event where consumption is voluntarily deferred in the present, with the belief that the result will be a greater outcome in the future.\n\nVoluntary marriage, voluntary involvement in religion, childrearing, the growth of output, profits and other social phenomena are covered by this math. The math differs from the math for phenomena such as defensive wars, insurance or casino gambling, for which the existing economic methods were well suited. It provides a split between models with anticipated gains versus those with anticipated losses.\n\nThe differences between the math revolve around the center of location and the spread parameter. The well-behaved properties of the Gaussian distribution are not appropriate for this class of problems.\n\nThat the center of location for the truncated distribution is the mode and that no first moment exists should also bring mean-variance finance to an end. Although one could have expected this earlier with the theoretical and empirical papers of Mandelbrot, Fama, MacBeth, French, and Roll; this provides a mathematical reason to exclude mean-variance models from any further use. - At least in its current form, there is no longer any reason to continue to support the Fama-French model as an alternative hypothesis.\n\nIt may appear problematic that the expectation of returns cannot exist, this is not the case. Humans anticipate the future. Other normative methods, such as Bayesian decision theory are not impacted by this loss of a tool.\n\nA rather simple solution to avoid the headaches involved in the change in math is to note an expected utility exists for models with concave utility functions or doubly bounded reward distributions. The math excludes any discussion of risk-neutral pricing. It also eliminates the study of risk-loving behavior unless an additional item is added to the utility function such as a utility for engaging in activity or excitement. Alternatively, these can be studied by modeling returns as sufficient returns, making it a Bernoulli trial, rather than continuous returns.\n\nAdding constraints on modeling is not a bad thing. Understanding the distributions allows for an understanding of how humans have to approach problems. It is certain that risk-loving behavior exists and risk-neutral behavior may also exist. Since this is the case, humans are doing something to overcome this issue. The enormous amount of physical and human capital is a testament to this fact.\n\nWhile this may seem like many tools have been lost, a gain is observed. The catalog of unexplained anomalies may shrink to nothing as most of them were predicated on finite variance or at least a mean. Issues such as heteroskedasticity in models are gone because these distributions are askedastic. Economists can now redirect their efforts and attentions to retesting old models to see how they fare under the math and toward building new ones.\n\nSome issues are not yet understood now, however. An index, such as the Standard and Poors 500 is now a statistic, and it is a statistic with unknown mathematical properties. Historically, the properties did not matter because it was approximately a weighted mean in a system believed to be well covered by the properties of the central limit theorem. Those properties may now matter. Also, traded variants as unit investment trusts may have very different properties. At this time we do not know. The traded form is an observable real world contract with specified properties. Linkages may exist among the securities by the contract that may not exist without it.\n\nA new concept of comovment is required as well in order to understand regression. The method of ordinary least squares implies that a series is convergent. These are divergent time series. Capital is a source and not a sink by its very nature. A way of understanding the Cauchy distribution is as the solution to a pendulum problem. If X maps onto Y and both are drawn from a Cauchy distribution, then one could view regression as a double pendulum problem. Solving $Y|X$ since both are diverging implies that for a given value of X which could have been viewed as a movement of a pendulum, Y is drawn from a second pendulum attached to the end of the first.\n\nThe most that could be said of an equation such as $y=\\beta x+\\alpha$ where x and y are drawn from a Cauchy distribution is that fifty percent of the time $y\\ge \\beta x+\\alpha$ and half of the time $y<\\beta x+\\alpha$ . While, of course, a density could be provided this allows items at any given point to diverge while remaining linked. The simple concept of the Capital Asset Pricing Model that if $\\beta =2$ then if asset x increases by 1% then it is to be expected that asset y will increase by 2% is too strong of a statement here. The interpretation should instead be that fifty-percent of the time it should be the case that asset y will increase two or more percent and half of the time, it will not.\n\nAdditionally, the problem of local volatility, that is a variance over an interval of time is probably meaningless. For the univariate Cauchy distribution, the scale parameter describes the half-width at half-maximum. That is the points\n\n$Pr\\left(\\mu ±\\gamma \\right)=\\frac{1}{2}Pr\\left(\\mu \\right)$ (96)\n\nand\n\n$\\frac{1}{\\text{π}}{\\int }_{\\mu -\\gamma }^{\\mu +\\gamma }\\frac{\\gamma }{{\\gamma }^{2}+{\\left(x-\\mu \\right)}^{2}}=\\frac{1}{2}.$ (97)\n\nSpectral analysis would be required to determine how long one swing of the spectrum of returns would be. For shorter time periods is not clear how well one could predict the local, realized variability.\n\nAs mentioned in Section 3.4.2 there is no covariance matrix. Although there is a linkage between the separate scale parameters to the joint scale parameter in the multivariate case, it is not clear that this survives truncating the distribution for the limitation of liability. Further, by having a single scale parameter, errors are spherical. The implication is that the errors in growing economies linked by trade are altered by that trade. Errors cannot be independent.\n\nThis discussion is necessarily incomplete. Individuals with capital in their models will likely be reporting unexpected side effects for quite some time. New ways of thinking, verified by empirical observation, will emerge and the core tools of microeconomics will come to the rescue of researchers.\n\n13. Conclusions\n\nIt is time for a change. It has been for quite some time. Now that a method of constructing distributions is understood, it is time to ask the questions believed answered earlier. Markowitz asked how humans made the risk and reward trade off . This must be asked again. To this, we should also ask how the banking system impacts that process.\n\nKnowledge of the likelihood function allows individuals to build and test economic models in a rational manner and also allows economists to minimize the number of required assumptions. Likewise, other fields that work with data that grow exponentially are impacted, they also have the same tools available. This does create a pedagogic issue however for undergraduates.\n\nFor business students, it will be necessary to teach both Bayesian and non-Bayesian methods. Null hypothesis methods will still be vital in fields such as marketing and accounting and of occasional use to fields like economics or finance. Conversely, there will be times where Bayesian methods will be of great value to marketing or accounting. It is time to build a book larger than Fisher’s cookbook solution to problems. The twentieth century began with only Bayesian methods, and null hypothesis methods exploded in value and use. The twenty-first century will need to bring balance to the utilization of both systems of thinking.\n\nThe twenty-first century is a time where tremendous levels of information will become available. Students and practitioners should not be taught one method or another, but both with a clear theoretical grounding in when one method or another method should be used and the trade-off that is created by such.\n\nResearch should proceed on anything that impacts cash flows, risks to those cash flows and the operation of financial intermediation through liquidity and credit services. This implies that bankruptcy, merger and dividend risk for equity securities are meaningful contributions. Likewise, the linkage between accountancy and market decision-making is possibly far more significant than in previous times. Under an assumption of efficient returns, how accountancy impacts decision-making was less important. Now that importance is no longer understood. Research should also proceed into the marketability of assets.\n\nIt is quite possible that since initial public offerings would have to be at least as good as existing securities in the secondary market, that initial offerings set the pricing for the system, that is they are the marginally priced security. While the ninety-day. Treasury bill was the implicit marginal asset under mean- variance finance, it may not be. As it stands, we do not know. Hence, the study of marketability discounts and the study of private firms may be critical to studying market pricing. It may be bills, it may be new public securities or it may be private firms with limited marketability. It is time to look.\n\nImplicit in the research agenda above but at least as important is the study of governmental stability, the nature of policies and taxation. It is reasonable to believe that policy impacts returns, but how this happens is less than clear. Without an assumption of market efficiency for returns, the nature of this question changes fundamentally.\n\nThe important result from the Capital Asset Pricing Model, where net returns above the risk-free rate drive the process is no longer supported by the math, but we do not know what is supported by the math. The most basic trade-off questions need to be researched.\n\nThe research agenda is vast, and the distinction between fantasy and reality will only be settled with time, data and well-structured inference.\n\nAcknowledgements\n\nThe empirical work was performed while a student at West Virginia University. My thanks in particular to Ashok Abbott for his extraordinary patience and to my wife for hers. I would additionally like to thank Dr. Markowitz. He provided empirical articles on the topic to me and, of course, started economics down this path by asking the central questions in the first place along with Dr. Roy.\n\nAppendix: Cancer and Other Related Problems\n\nWhile this article was intended for use with models of capital, this, in fact, applies to anything that would grow at an exponential rate in the absence of a boundary condition. To provide a second example, consider the implications different distributions would have on cancer growth rates.\n\nAlthough cancer grows in three dimensions, as in Equation (35), the multi- variate equivalents to most of the univariate issues has not been solved. Viewing a tumor in one dimenion does allow a discussion of how those differing equa- tions would be interpreted. For example, if cancer growth rates were well modeled by Equation (26), then this would imply that the cells acted as if independent. On the other hand, if the growth rates were to behave as in Equa- tion (37), it would imply that the cells with the highest, or lowest, depending on what was being modeled, growth rates determined the overall growth rate of the system. It would imply heterogeneous growth rates within the tumor.\n\nEquation (45) would imply that the growth rates depended on some factor, possibly a signal or a switch, while Equation (49) would imply a system in disequilibrium rather than in homeostasis.\n\nAdditionally, regression of data in raw form would have a posterior density of the Cauchy distribution and the hyperbolic secant distribution in log-log form in the case of flat priors.\n\nBecause Bayesian model selection allows the testing of multiple distributions and models are not restricted to single dimension models as implied in this article, for a given medical problem, Bayesian model selection will provide information about the underlying nature of the relationships among cells.",
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"Submit or recommend next manuscript to SCIRP and we will provide best service for you:\n\nAccepting pre-submission inquiries through Email, Facebook, LinkedIn, Twitter, etc.\n\nA wide selection of journals (inclusive of 9 subjects, more than 200 journals)\n\nProviding 24-hour high-quality service\n\nUser-friendly online submission system\n\nFair and swift peer-review system\n\nEfficient typesetting and proofreading procedure\n\nDisplay of the result of downloads and visits, as well as the number of cited articles\n\nMaximum dissemination of your research work\n\nSubmit your manuscript at: http://papersubmission.scirp.org/\n\nOr contact jmf@scirp.org\n\nCite this paper: Harris, D. (2017) The Distribution of Returns. Journal of Mathematical Finance, 7, 769-804. doi: 10.4236/jmf.2017.73041.\nReferences\n\n Markowitz, H. (1952) Portfolio Selection. The Journal of Finance, 7, 77-91.\n\n Mandelbrot, B. (1963) The Variation of Certain Speculative Prices. 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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9296009,"math_prob":0.99399453,"size":81218,"snap":"2021-04-2021-17","text_gpt3_token_len":17008,"char_repetition_ratio":0.16740955,"word_repetition_ratio":0.015443719,"special_character_ratio":0.21005195,"punctuation_ratio":0.10738838,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9988749,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T06:31:09Z\",\"WARC-Record-ID\":\"<urn:uuid:90e79e84-2f37-4596-8271-28ab93ce00e2>\",\"Content-Length\":\"338019\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56c55b3f-c193-4a48-ab5d-26ed0460e07d>\",\"WARC-Concurrent-To\":\"<urn:uuid:e69716ee-2e7f-43c7-a380-e78dcc4fbb4e>\",\"WARC-IP-Address\":\"192.111.37.22\",\"WARC-Target-URI\":\"https://m.scirp.org/papers/78849\",\"WARC-Payload-Digest\":\"sha1:TM3RRIDL3XD3C7NOALVYB4GXBDY4K3TE\",\"WARC-Block-Digest\":\"sha1:GYL5Q63YR6SAMH4YQNHUBQ2TUPTHZX3R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038061562.11_warc_CC-MAIN-20210411055903-20210411085903-00327.warc.gz\"}"} |
https://hsci3023.scipop.net/2016/03/07/thought-experiment-about-von-baeyer-and-ap-physics/ | [
"# Thought Experiment about von Baeyer and AP Physics\n\nIn imagining von Baeyer as a high school student presenting an unpublished manuscript of his book, Warmth Disperses and Time Passes to the AP physics committee, I believe that it would be fair for him to receive AP physics credit. The very nature of physics rests upon the concept of motion. And thermodynamics is a massive branch of physics, which deals with motion and heat processes. One of the main reasons why I think any committee members would immediately dismiss von Baeyer’s manuscript is that they would probably think it is a simple history of thermodynamics, rather than a deeper study about concepts. However, I do think that von Baeyer demonstrates a clear understanding of thermodynamic and physics concepts. For example when von Baeyer talks about Einstein’s equation (e = mc^2), he tells us that “mass, denoted by m, is a measure of inertia – the tendency of things to resist being set in motion when they are at rest and to resist changes in velocity when they are moving” (von Baeyer 124). In this instance, von Baeyer clearly defines a variable in the equation, and demonstrates clear understanding of the concept of inertia by giving us good definition. Another example is when von Baeyer discusses Bernoulli’s billiard model for a gas. He explains that “at the same time the pressure, which is force per unit area, will suffer a further reduction by a factor of four on account of the fourfold increase in the area of each side” (von Baeyer 70). Here, von Baeyer defines what pressure is and he even demonstrates the understanding of the reciprocal relationship between pressure and volume, that is, while one variable increases the other decreases. Lastly, I think von Baeyer demonstrates his best knowledge when he discusses physical constants. First, von Baeyer says that “like all subfields of physics, thermodynamics is characterized by certain constants whose values we measure but otherwise take for granted” (von Baeyer 164). Next, he explains that “the natural constant that takes the place of c and h in thermodynamics is Boltzmann’s constant k, which converts absolute temperature into molecular energy, and the logarithm of probability to entropy” (von Baeyer 164). Here, von Baeyer makes an excellent assessment about the nature of physics. Physics is full of physical constants that have been experimentally measured throughout time by physicists. He tells us about the speed of light (c) in Einstein’s equation and Planck’s constant (h) in quantum mechanics and then relates them both to a thermodynamic constant, k, which is Boltzmann’s constant. Then he tells about Boltzmann’s constant, saying it is measured “in the scientific units for energy over temperature (joules per kelvin),” and it “amounts to about ten trillionths of a trillionth – a number that is far beyond the pale of human imagination” (von Baeyer 164). Again, von Baeyer knows what the definition of the constant is and is able to relate it to other constants in another subfield of physics. To me, this proves that von Baeyer has a high enough understanding in physics for the AP physics level. He touched upon and was able to explain concepts in motion (inertia, relativity), the pressure/volume relationship, and thermodynamics (Boltzmann’s constant), rather than just tell the history of it."
] | [
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https://books.google.no/books?qtid=177e78ea&dq=editions:UOM39015067252117&lr=&id=xDADAAAAQAAJ&hl=no&output=html_text&sa=N&start=170 | [
"Søk Bilder Maps Play YouTube Nyheter Gmail Disk Mer »\nLogg på\n Bøker Bok",
null,
"To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.",
null,
"The synoptical Euclid; being the first four books of Euclid's Elements of ... - Side 41\nav Euclides - 1853\nUten tilgangsbegrensning - Om denne boken",
null,
"## Annual Report\n\n1894\n...to two right tingles, the two straight lines shall be parallel. 12. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to а given angle. 13. Show without producing the side of a triangle that the sura of its three interior...\nUten tilgangsbegrensning - Om denne boken",
null,
"## Treatise on Conic Sections\n\nApollonius (of Perga.) - 1896 - 254 sider\n...properties aa presented by him. The problem in Euclid i. 44 is \" to apply to a given straight line a parallelogram which shall be equal to a given triangle...have one of its angles equal to a given rectilineal angle.\" The solution of this clearly gives the means of adding together or subtracting any triangles,...\nUten tilgangsbegrensning - Om denne boken",
null,
"## Report of the Secretary for Public Instruction ...\n\n...— (i.) (a + b)* = (a + b)a + (a + b)b. (ii.) (a + b) a = o1 + ab. 20 4. To a given straight line apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to n given angle. 20 5. If a straight line is divided into any two parts, the sum of the squares on the...\nUten tilgangsbegrensning - Om denne boken",
null,
"## Report of the Council of Public Instruction of the North-West Territories of ...\n\n...of the triangle thus formed with the original one. 6. (a) Show how to describe a parallelogram that shall be equal to a given triangle and have one of its angles equal to a given angle. I. 42. (b) If the angle be 90° what kind of parallelogram would be described ? (c) Describe...\nUten tilgangsbegrensning - Om denne boken",
null,
"## Mathematics: Mechanic's bids and estimates. Mensuration for beginners. Easy ...\n\nSeymour Eaton - 1899 - 340 sider\n...other parallelograms are equal besides the pair of complements ? Lesson No. 20 PROPOSITION 44. PROBLEM To a given straight line to apply a parallelogram,...have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle...\nUten tilgangsbegrensning - Om denne boken",
null,
"## Sessional Papers, Volum 32,Utgave 10\n\n1899\n...1. How to bisect a given rectilineal angle, give demonstration. 2. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. 3. If the bisectrix of two angles of a triangle are equal show that the triangle is isosceles....\nUten tilgangsbegrensning - Om denne boken",
null,
"## Annual Report of the Commissioners ..., Volum 65\n\n1899\n...accepted. Dr. MORAN, Head Inspector. Mr. PEDLOW, District Inspector. A. 1. To a given straight line apply a parallelogram which shall be equal to a given triangle, and have an angle equal to a given one. 2. Prove that the square on the difference of two lines is less than...\nUten tilgangsbegrensning - Om denne boken",
null,
"## A Text-book of Euclid's Elements for the Use of Schools, Books I.-IV., Bok 1\n\nHenry Sinclair Hall, Frederick Haller Stevens - 1900 - 304 sider\n...(ii) If KB, KD are joined, the triangle AKB is equal to the triangle AKD. PROPOSITION 44. PROBLEM. To a given straight line to apply a parallelogram...triangle, and have one of its angles equal to a given angle. Let AB be the given straight line, C the given triangle, and D the given angle. It is required...\nUten tilgangsbegrensning - Om denne boken",
null,
"## Examination Papers\n\nUniversity of Toronto - 1900\n...EUCLID. [AC McKAY, BA Examiners :-! A. ODELL. [W. PBENDERGAST, BA 1. Describe a parallelogram that shall be equal to a given triangle and have one of its angles equal to a given angle. (Euclid, Book I, Prop. 42.) 2. Prove that the square described on the hypotenuse of a right...\nUten tilgangsbegrensning - Om denne boken",
null,
"## Calendar\n\nUniversity of Sydney - 1900\n...logarithms to base 10 of 5, 6, 7, 8, 9. GEOMETRY. 1. Describe a parallelogram that shall be equal in area to a given triangle, and have one of its angles equal to a given angle. 2. The side AB of a triangle ABC is greater than the side AC ; shew that the median AX makes...\nUten tilgangsbegrensning - Om denne boken"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.836815,"math_prob":0.9803779,"size":3266,"snap":"2021-43-2021-49","text_gpt3_token_len":869,"char_repetition_ratio":0.21551196,"word_repetition_ratio":0.2893401,"special_character_ratio":0.2740355,"punctuation_ratio":0.21447721,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99657375,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T17:24:44Z\",\"WARC-Record-ID\":\"<urn:uuid:e678b066-08e3-466e-b7bf-496547846fc1>\",\"Content-Length\":\"33043\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ddb62c0-de3f-4fd6-90bf-0c5b096f4d69>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ad695dd-dc28-4554-9ffe-7c78ed6fec75>\",\"WARC-IP-Address\":\"172.217.164.142\",\"WARC-Target-URI\":\"https://books.google.no/books?qtid=177e78ea&dq=editions:UOM39015067252117&lr=&id=xDADAAAAQAAJ&hl=no&output=html_text&sa=N&start=170\",\"WARC-Payload-Digest\":\"sha1:F7W66QCVPGM3AFT3EMMUD2AS3TF3K3P7\",\"WARC-Block-Digest\":\"sha1:W3DYS2KXW6EUTUIL4RIV6VGRJAWIOAW5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323586043.75_warc_CC-MAIN-20211024142824-20211024172824-00397.warc.gz\"}"} |
https://pdfcoffee.com/addition-mathematic-form-5-progression-module-1-pdf-free.html | [
"# Addition Mathematic form 5 Progression module 1\n\n##### Citation preview\n\nSPM\n\nMODULE 1 ARITHMETIC PROGRESSION ORGANISED BY:\n\nJABATAN PELAJARAN NEGERI PULAU PINANG\n\nCHAPTER 1: ARITHMETIC PROGRESSION CONTENTS\n\nPAGE 2\n\n1.1 Identify characteristics of arithmetic progression 1.2 Determine whether a given sequence is an arithmetic progression 1.3 Determine by using formula: a) specific terms in arithmetic progressions b) the number of terms in arithmetic progressions\n\n1.4\n\n1.5\n\nFind a) The sum of the first n terms of arithmetic progressions b) The sum of a specific number of consecutive terms of arithmetic progressions c) The value of n , given the sum of the first n terms of arithmetic progressions.\n\nSolve problems involving arithmetic progressions\n\nSPM QUESTIONS\n\n3\n\n4-5\n\n6\n\n7\n\n8-10\n\nASSESSMENT\n\n11-12\n\n2\n\n13\n\nPROGRESSIONS ( Arithmetic Progression) ARITHMETIC PROGRESSION\n\nThe sum of the first n term: Sn = ______________\n\nThe nth term : Tn = ___________\n\nor Sn = ______________\n\nor Tn = ___________\n\na = _______________ d = _______________ l = _______________ n = _______________\n\nFill in the blank\n\n1.1 Identify characteristics of arithmetic progression: EXERCISE 1: Complete each of the sequence below to form an arithmetic progression. a) 2, _____ , 8 , ______ , 14 , 17\n\n3\n\nb) -2 , -5 , ______ , -11 , ________ 1 c) -4, ______ , ______ , ,2 2 d) 2x -3 , 2x -1 , __________ , __________ , 2x + 5\n\n1.2 Determine whether a given sequence is an arithmetic progression EXERCISE 2 : 1.\n\nDetermine whether a given sequence below is an arithmetic progression. a)\n\n5 , 11 , 17 , 23 , ………..\n\n( ____________________ )\n\nb)\n\n-20 , -50 , -30 , -35 , ……….\n\n( ____________________ )\n\nc)\n\n1 , 4 , 9 , 16 , ………\n\n( ____________________ )\n\nd) 2.\n\n2x + y , 4x – y , 6x -3y , …… ( ____________________ )\n\nk +3 , 2k + 6 ,8 are the first three terms of an arithmetic progression,find the value of k.\n\nNote: If x,y and z are three terms of an arithmetic progression , y–x=z-y\n\nGiven that x 2 ,5 x, 7 x 4 are three consecutive terms of an arithmetic progression W\n\n3.\n\nhere x has a positive value. Find the value of x.\n\n4. Given that the first three terms of an arithmetic progression are 2 y, 3 y 3 and 5y+1 . Find the value of y.\n\n4\n\n1.3\n\nDetermine by using formula: a. specific terms in arithmetic progressions b. the number of terms in arithmetic progressions\n\nTn a (n 1)d EXERCISE 3: 2. Find the 11th term of the arithmetic progression. 5 3, , 2,........ 2\n\n1. Find the 9th term of the arithmetic progression. 2, 5 , 8 , ….. Solution: a=2 d = 5-2=3 T9 2 (9 1)3 = _______ 3. For the arithmetic progression 0.7, 2.1 , 3.5, ….. ,find the 5th term .\n\n4. Find the nth term of the arithmetic progression 1 4, 6 ,9,..... 2\n\n5. Find the 7th term of arithmetic progression k, 2k + 1, 3k+2, 4k+3,….\n\n6. Given that arithmetic progression 9 +6x , 9+4x, 9+2x, …… Find the 10th term.\n\n5\n\nEXERCISE 4:\n\n6\n\n1. Find the number of terms of the arithmetic progression\n\n2. For the arithmetic progression 5 , 8 , 11 , ……. Which term is equal to 320 ?\n\na) 4,9.14,….. ,64 Tn 64 4 + (n-1)5=64 4+ 5n-5 =64 5n=65 n= 13 b) -2, -7, -12, ……, -127 3. The sequence 121 , 116 , 111 , …. is an arithmetic progression. Find the first negative term in the progression.\n\n1 1 1 c) 1,1 ,1 ,......, 4 6 3 2\n\n4. Find the number of terms of the arithmetic progression a 2b,3a 3b,5a 4b,......, 23a 13b\n\nd) x y, x y, x 3 y,........., x 31 y\n\n1.4 Find a) the sum of the first n terms of arithmetic progressions. 7\n\nb) the sum of a specific number of consecutive terms of arithmetic progressions. c) the value of n , given the sum of the first n terms of arithmetic progressions.\n\nn 2a n 1 d 2 n Sn = a+ l 2 l= the n th term Sn \n\nor\n\nEXERCISE 5: 1. Find the sum of the first 12 terms of the arithmetic progression -10 , -7 , -4, ……\n\n2. Find the sum of all the terms of the arithmetic progression 38 , 31 , 24 , …., -18\n\n3. For the arithmetic progression 4. How many terms of the arithmetic -4 , 1 , 6 , …….find the sum of all 2, 8 , 13 ,18 , …… must be taken the terms from the 6th term to the 14th for the sum to be equal to 1575? term. Solution: Sum of all the terms from the 6th term to the 14th = S14 S5 = =\n\n1.5 Solve problems involving arithmetic progressions. EXERCISE 6:\n\n8\n\n1. The 5th and 9th of arithmetic progression are 4 and 16 respectively. Find the 15th term. Solution: T5 4\n\n2. The 2nd and 8th terms of an arithmetic progression are -9 and 3 respectively. Find a) the first term and the comman difference. b) the sum of 10 terms beginning from the 12th term.\n\na (5 1)d 4 a 4d 4 ......(1) T9 16 a+(9-1)d=16 a+8d=16 ......(2) Solve simultaneous equation to find the value of a and d, then the 15th term.\n\n3. The sum of the first 6 terms of an arithmetic progression is 39 and the sum of the next 6 terms is -69. Find a) the first term and the common difference b) the sum of all the terms from the 15th term to the 25th term.\n\n4. The sum of the first n terms is given 2 by S n 6n 3n Find a) the nth term in terms of n b) the common difference\n\nSPM QUESTIONS:\n\n9\n\n1. 2003: Paper 1: (no. 7) The first three terms of an arithmetic progression are k 3, k 3, 2k 2 Find a) the value of k 3 marks b) the sum of the first 9 terms of the progression\n\n2. 2004: Paper 1: ( no 10) Given an arithmetic progression -7, -3, 1, …. , state three consecutive terms in this 3 marks progression which sum up to 75\n\n3. Paper 1: ( no. 11)\n\n10\n\nThe volume of water in a tank is 450 litres on the first day. Subsequently, 10 litres of water is added to the tank everyday. Calculate the volume , in litres , of water in the tank at the end of the 7th day. 2 marks\n\n4. 2005: Paper 1: (no. 11) The first three terms of an arithmetic progression are 5, 9, 13 Find a) the common difference of the progression b) the sum of the first 20 terms after the 3rd term.\n\n5. Paper 2 : ( Section A no.3) 11\n\n 4 marks\n\nDiagram 1 shows part of an arrangement of bricks of equal size.\n\nDiagram 1 The number of bricks in the lowest row is 100. For each of the rows, the number of bricks is 2 less than in the row below. The height of each bricks is 6 cm. Ali builds a wall by arranging bricks in this way. The number of bricks in the highest row is 4. Calculate a) the height , in cm, of the wall [ 3 marks ] b) the total price of the bricks used if the price of one brick is 40 sen. [ 3 marks ]\n\nASSESSMENT: 1. The sequence -13,-10,-7,……,128,131 is an arithmetic progression. Find\n\n12\n\na) the number of terms in the progression b) the sum of all the terms in the progression.\n\n2.\n\n3.\n\nGiven that k-1 , 2k – 4 and k +5 are the first three terms of arithmetic progression. Find the common difference.\n\nCalculate how many months it will take to repay a debt of RM5800 by monthly payment of RM100 initially with an increase of RM20 for each month after that.\n\n4. The 4th of an arithmetic progression is 18 and the sum of the first eight terms\n\n13\n\nof the progression is 124. Find a) the first term and the common difference b) the sum of all the terms from the 10th term to the 20th term.\n\n5.\n\nSum of the first ten of the arithmetic progression is 230 and sum of ten terms after that is 830. Find the first term and the common difference."
] | [
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https://jp.mathworks.com/help/fusion/ug/object-tracking-using-time-difference-of-arrival.html | [
"# Object Tracking Using Time Difference of Arrival (TDOA)\n\nThis example shows how to track objects using time difference of arrival (TDOA). This example introduces the challenges of localization with TDOA measurements as well as algorithms and techniques that can be used for tracking single and multiple objects with TDOA techniques.\n\n### Introduction\n\nTDOA positioning is a passive technique to localize and track emitting objects by exploiting the difference of signal arrival times at multiple, spatially-separated receivers. Given a signal emission time (${\\mathit{t}}_{\\mathit{e}}$) from the object and the propagation speed ($\\mathit{c}$) in the medium, the time-of-arrival (TOA) of the signal at 2 receivers located at ranges ${\\mathit{r}}_{\\text{\\hspace{0.17em}}1}\\text{\\hspace{0.17em}}$and ${\\mathit{r}}_{\\text{\\hspace{0.17em}}2}$ respectively from the object can be denoted as:\n\n`$\\begin{array}{l}{\\mathit{t}}_{1\\text{\\hspace{0.17em}}}={\\mathit{t}}_{\\mathit{e}}+\\frac{{\\mathit{r}}_{1\\text{\\hspace{0.17em}}}}{\\mathit{c}}\\\\ {\\mathit{t}}_{2\\text{\\hspace{0.17em}}}={\\mathit{t}}_{\\mathit{e}}+\\frac{{\\mathit{r}}_{2}}{\\mathit{c}}\\end{array}$`\n\nAs the true signal emission time is typically not known, a difference between ${\\mathit{t}}_{1\\text{\\hspace{0.17em}}}$and ${\\mathit{t}}_{2\\text{\\hspace{0.17em}}}$can be used to obtain some information about the location of an object. Specifically, given the location of each receiver, the time-difference measurement corresponds to a difference in ranges from the two receivers to the unknown object.\n\n`$\\sqrt{{\\left({\\mathit{x}}_{\\mathit{t}}-{\\mathit{x}}_{1}\\right)}^{2\\text{\\hspace{0.17em}}}+{\\left({\\mathit{y}}_{\\mathit{t}}-{\\mathit{y}}_{1\\text{\\hspace{0.17em}}}\\right)}^{2\\text{\\hspace{0.17em}}}{\\text{\\hspace{0.17em}}+\\text{\\hspace{0.17em}}\\left({\\mathit{z}}_{\\mathit{t}}-{\\mathit{z}}_{1\\text{\\hspace{0.17em}}}\\right)}^{2}}-\\sqrt{{\\left({\\mathit{x}}_{\\mathit{t}}-{\\mathit{x}}_{2}\\right)}^{2\\text{\\hspace{0.17em}}}+{\\left({\\mathit{y}}_{\\mathit{t}}-{\\mathit{y}}_{2\\text{\\hspace{0.17em}}}\\right)}^{2\\text{\\hspace{0.17em}}}{\\text{\\hspace{0.17em}}+\\text{\\hspace{0.17em}}\\left({\\mathit{z}}_{\\mathit{t}}-{\\mathit{z}}_{2\\text{\\hspace{0.17em}}}\\right)}^{2}}=\\mathit{c}\\left({\\mathit{t}}_{1}-{\\mathit{t}}_{2\\text{\\hspace{0.17em}}}\\right)$`\n\nwhere $\\left[{\\mathit{x}}_{\\mathit{t}}\\text{\\hspace{0.17em}}{\\mathit{y}}_{\\mathit{t}\\text{\\hspace{0.17em}}}\\text{\\hspace{0.17em}}{\\mathit{z}}_{\\mathit{t}\\text{\\hspace{0.17em}}}\\right]$ is the unknown object location, ${\\mathit{t}}_{1}-{\\mathit{t}}_{2\\text{\\hspace{0.17em}}}$ is the TDOA measurement, and $\\left[{\\mathit{x}}_{1\\text{\\hspace{0.17em}}}{\\mathit{y}}_{1}\\text{\\hspace{0.17em}}{\\mathit{z}}_{1\\text{\\hspace{0.17em}}}\\right]$, $\\left[{\\mathit{x}}_{2\\text{\\hspace{0.17em}}}{\\mathit{y}}_{2}\\text{\\hspace{0.17em}}{\\mathit{z}}_{2\\text{\\hspace{0.17em}}}\\right]$ are the locations of two receivers. This non-linear relationship between TDOA and object location represents a hyperbola in 2-D or a hyperboloid in 3-D coordinates with two receivers at the foci. The image below shows the hyperbolae of an object for different TDOA measurements ($\\mathit{c}$ = speed of light). These curves are typically called constant TDOA curves. Notice that using the sign of a TDOA measurement, you can constrain the object location to a single branch of the hyperbola.\n\n`HelperTDOATrackingExampleDisplay.plotConstantTDOACurves;`",
null,
"#### TDOA Calculation\n\nThere are two primary methods used to calculate the TDOA measurement from the signal of an object. In the first method, each receiver measures the absolute time instant of signal arrival (time-of-arrival or TOA) as defined by ${\\mathit{t}}_{\\mathit{i}\\text{\\hspace{0.17em}}}$above. A difference in time-of-arrivals between two receivers is then calculated to obtain the TDOA measurement. This method is applicable when certain signal attributes are known a-priori and the leading edge of the signal can be detected by the receiver.\n\n`${\\mathrm{TDOA}}_{12\\text{\\hspace{0.17em}}}={\\mathrm{TOA}}_{2\\text{\\hspace{0.17em}}}-{\\mathrm{TOA}}_{1\\text{\\hspace{0.17em}}}$`\n\nIn the second method, each receiver records and transmits the received signal to a shared processing hub. A cross-correlation between signals received from two receivers is calculated at the processing hub. The TDOA is estimated to be the delay which maximizes the cross-correlation between the two signals.\n\n`${\\mathrm{TDOA}}_{12\\text{\\hspace{0.17em}}}=\\underset{\\mathit{t}\\in \\left[-{\\mathit{T}}_{\\mathrm{max}}\\text{\\hspace{0.17em}}{\\mathit{T}}_{\\mathrm{max}}\\text{\\hspace{0.17em}}\\right]}{\\mathrm{argmax}}\\left(\\left[{\\mathit{S}}_{1\\text{\\hspace{0.17em}}}\\star {\\mathit{S}}_{2}\\right]\\left(\\mathit{t}\\right)\\right)$`\n\nwhere $\\left[{\\mathit{S}}_{1\\text{\\hspace{0.17em}}}\\star {\\mathit{S}}_{2}\\right]\\left(\\mathit{t}\\right)$ represents the cross-correlation between the signals at receivers as a function of time delay, $\\mathit{t}$. ${\\mathit{T}}_{\\mathrm{max}}$ is the maximum possible absolute delay and can be calculated as $\\frac{\\mathit{D}}{\\mathit{c}\\text{\\hspace{0.17em}}}$, where $\\mathit{D}$ is the distance between the receivers.\n\nBoth methods of TDOA calculation require synchronized clocks at the receivers. In practical applications, you typically achieve this using the Global Positioning System (GPS) clock.\n\n#### TDOA Localization\n\nAs the TDOA between two receivers localizes an object to a hyperbola or hyperboloid, it is not possible to observe the full state of the object by using only two stationary receivers. For 2-D localization, at least 3 spatially-separated receivers are required to estimate the object state. Similarly, for 3-D tracking, at least 4 spatially-separated receivers are required.\n\nWith $\\mathit{N}\\text{\\hspace{0.17em}}\\left(\\mathit{N}\\ge 2\\right)$ receivers, a total of $\\mathit{N}\\left(\\frac{\\mathit{N}-1}{2\\text{\\hspace{0.17em}}}\\right)$ TDOA measurements from an object can be obtained by calculating the time difference of arrival using each combination of receiver. However, out of these measurements, only $\\mathit{N}-1\\text{\\hspace{0.17em}}$measurements are independent and the rest of the TDOA measurements can be formulated as a linear combination of these independent measurements. Given these $\\mathit{N}-1$ measurements from the object, an estimation algorithm is typically used to locate the position of the object. In this example, you use the spherical intersection algorithm to find the position and uncertainty in position estimate. The spherical intersection algorithm assumes that all $\\mathit{N}-1$ TDOAs are calculated with respect to the same receiver, sometimes referred to as the reference receiver.\n\nThe accuracy or uncertainty in the estimated position obtained from localization depends on the geometry of the network. At regions, where small errors in TDOA measurements results in large errors in estimated position, the accuracy of the estimated position is lower. This effect is called geometric dilution of precision (GDOP). The image below shows the GDOP heat map of the network geometry for a 2-D scenario with 3 receivers. Note that the accuracy is high within the envelope generated by the network and is lowest directly behind the receivers along the receiver-to-receiver line of sights.\n\n`HelperTDOATrackingExampleDisplay.plotGDOPAccuracyMap;`",
null,
"### Tracking a Single Emitter\n\nIn this section, you simulate TDOA measurements from a single object and use them to localize and track the object in the absence of any false alarms and missed detections.\n\nYou define a 2-D scenario using the helper function, `helperCreateSingleTargetTDOAScenario`. This function returns a `trackingScenarioRecording` object representing a single object moving at a constant velocity. This object is observed by three stationary, spatially-separated receivers. The helper function also returns a matrix of two columns representing the pairs formed by the receivers. Each row represents the `PlatformID` of the platforms that form the TDOA measurement.\n\n```% Reproducible results rng(2021); % Setup scenario with 3 receivers numReceivers = 3; [scenario, rxPairs] = helperCreateSingleTargetTDOAScenario(numReceivers);```\n\nYou then use the helper function, `helperSimulateTDOA,` to simulate TDOA measurements from the objects. The helper function allows you to specify the measurement accuracy in the TDOA. You use a measurement accuracy of 100 nanoseconds to represent timing inaccuracy in the clocks as well as TDOA estimation. The emission speed and units of the reported time are speed of light in vacuum and nanoseconds respectively, specified using the `helperGetGlobalParameters` function. The `helperSimulateTDOA` function outputs TDOA detections as a cell array of `objectDetection` objects.\n\n```tdoaDets = helperSimulateTDOA(scenario, rxPairs, measNoise); % Specify measurement noise variance (nanosecond^2) ```\n\nEach TDOA detection is represented using the `objectDetection` according to the following format:",
null,
"Next, you run the scenario and generate TDOA detections from the object. You use the helper function, `helperTDOA2Pos`, to estimate the position and position covariance of the object at every step using the spherical intersection algorithm . You further filter the position estimate of the object using a global nearest neighbor (GNN) tracker configured using the `trackerGNN` System object™ with a constant-velocity linear Kalman filter. To account for the high speed of the object, this constant velocity Kalman filter is configured using the helper function, `helperInitHighSpeedKF`.\n\n```% Define accuracy in measurements measNoise = (100)^2; % nanoseconds^2 % Display object display = HelperTDOATrackingExampleDisplay(XLimits=[-1e4 1e4],... YLimits=[-1e4 1e4],... LogAccuracy=true,... Title=\"Single Object Tracking\"); % Create a GNN tracker tracker = trackerGNN(FilterInitializationFcn=@helperInitHighSpeedKF,... AssignmentThreshold=100); while advance(scenario) % Elapsed time time = scenario.SimulationTime; % Simulate TDOA detections without false alarms and missed detections tdoaDets = helperSimulateTDOA(scenario, rxPairs, measNoise); % Get estimated position and position covariance as objectDetection % objects posDet = helperTDOA2Pos(tdoaDets,true); % Update the tracker with position detections tracks = tracker(posDet, time); % Display results display(scenario, rxPairs, tdoaDets, {posDet}, tracks); end```\n\nNotice that the tracker is able to maintain a track on the object. The the estimated position of the object lies close to the intersection of hyperbolic curves created by each receiver pair. Also notice in the zoomed plot below that the filtered estimate of the object has less uncertainty as compared to the fused estimate at every step. As the object moves towards positive X axis, the error in estimated position increases due to geometric dilution of precision. Notice that the tracker is able to provide an improved estimate of the object by filtering the position estimate of the object using the Kalman filter with constant velocity motion model.\n\n`zoomOnTrack(display,tracks(1).TrackID);`",
null,
"",
null,
"`plotDetectionVsTrackAccuracy(display);`",
null,
"### Tracking Multiple Emitters with Known IDs\n\nIn the previous section, you learned how to use TDOA measurements from a single object to generate positional measurement of the object. In practical situations, the scenario may consist of multiple objects.\n\nThe main challenge in multi-object TDOA tracking is the estimation of TDOA for each target. In the presence of multiple objects, each receiver receives signals from multiple objects. In the first method for TDOA calculation, each receiver records multiple time-of-arrival measurements. The unknown data association between time-of-arrival measurements reported by each receiver results in many possible TDOA combinations. In the second method of TDOA calculation, the signal cross-correlation function between receiver signals results in multiple peaks corresponding to true objects as well as false alarms. In certain applications, this unknown data association between receivers can be easily solved at the signal level. For example, if the signal encoding is known a-priori, such as in wireless communications signals like LTE, 5G signals or aviation communication signals like ADSB signals, the receiver is typically able to calculate both the time-of-arrival and the unique identity of the object. Similarly, if the objects are separated in their carrier frequencies, a separate TDOA calculation can be added for each object in their own frequency bands.\n\nIn this section, you assume that signal-level data association is performed at the receiver. This helps the processing hub to form TDOA measurements per identified object without ambiguity. To simulate TDOA measurements with known emitter identification, you provide a fourth input argument to the helper function, `helperSimulateTDOA`. The function outputs the unique identity of the emitter as the `objectClassID` property of the `objectDetection` object. You use this object identity, shared between TDOAs from multiple receiver-pairs, to fuse detections from each object separately and obtain their respective positions as well as uncertainties. Then, you use these fused measurements to track these objects using the GNN tracker.\n\n```% Create the multiple object scenario numReceivers = 3; numObjects = 4; [scenario, rxPairs] = helperCreateMultiTargetTDOAScenario(numReceivers, numObjects); % Reset display release(display); display.LogAccuracy = false; display.Title = 'Tracking Multiple Emitters with Known IDs'; % Release tracker release(tracker); while advance(scenario) % Elapsed time time = scenario.SimulationTime; % Simulate classified TDOA detections classifiedTDOADets = helperSimulateTDOA(scenario, rxPairs, measNoise, true); % Find unique object identities tgtIDs = cellfun(@(x)x.ObjectClassID,classifiedTDOADets); uniqueIDs = unique(tgtIDs); % Estimate the position and covariance of each emitter as objectDection posDets = cell(numel(uniqueIDs),1); for i = 1:numel(uniqueIDs) thisEmitterTDOADets = classifiedTDOADets(tgtIDs == uniqueIDs(i)); posDets{i} = helperTDOA2Pos(thisEmitterTDOADets, true); end % Update tracker tracks = tracker(posDets, time); % Update display display(scenario, rxPairs, classifiedTDOADets, posDets, tracks); end```",
null,
"Note that the tracker is able to maintain a track on all 4 objects when TDOA-to-TDOA association is assumed to be known. When the data association is provided by the receivers, the multi-object TDOA estimation is simplified to generating multiple TDOA detections with known single-object associations. The multi-object tracking problem is similarly simplified because there are no false TDOA detections.\n\n### Tracking Multiple Emitters with Unknown IDs\n\nIn this section, you assume that data association between receiver data is not known a-priori i.e., receivers cannot identify objects. In the absence of this data association, each intersection between blue and maroon hyperbolic curves shown in the plot from the Tracking Multiple Emitters with Known IDs section could be a possible object location. Therefore, when data association is not available from the receivers, associating time-of-arrival (TOA) or TDOA detections from multiple receivers is prone to detections from ghost objects. To separate the ghost detections from true object detections, you must add more receivers to reduce the ambiguity. In this section, you will use a static fusion algorithm to determine the unknown data association and estimate position measurements from each object. You will use this static fusion algorithm for systems that transmit multiple time-of-arrival measurements to the processing hub as well as systems that transmit recorded signals to the hub and calculates TDOA before processing them for object tracking.\n\n#### Tracking with Time-of-Arrival (TOA) Measurements\n\nIn this section, you configure a system in which each receiver transmits multiple time of arrival (TOA) measurements to the central processing hub. To reduce ambiguity in data association and to reduce the number of potential ghost objects, you use 4 stationary receivers to localize and track the objects.\n\nTo simulate time-of-arrival measurements from the objects, you use the helper function, `helperSimulateTOA`, to simulate time-of-arrival measurements from multiple objects. These time-of-arrival measurements record the exact time instant in a global reference clock at which the signal was received by the receivers. To simulate exact time instants, the helper function uses the scenario time as the global clock time and assumes that the objects emit signals at the time instant of scenario. Note that the algorithm used to track with these measurements is not tied to this assumption as the exact emission time from the object is not known to the receivers.\n\nYou specify the `nFalse` input as 1 to simulate one false time-of-arrival measurement from each receiver. You also specify the `Pd` input, which defines the probability of detecting a true time-of-arrival measurement by each receiver, as 0.95.\n\n```toaDets = helperSimulateTOA(scenario, receiverIDs, measNoise, nFalse, Pd); % receiverIDs - PlatformID of all receivers % measNoise - Accuracy in TOA measurements (ns^2) % nFalse - number of false alarms per receiver % Pd - Detection probability for receivers ```\n\nHere is the format of time-of-arrival (TOA) detections.",
null,
"To fuse multiple time-of-arrival measurements from each receiver, you use the static fusion algorithm by configuring a `staticDetectionFuser` object. The `staticDetectionFuser` object determines the best data association between time of arrival measurements and provides fused detections from possible objects. The static fusion algorithm requires two sub routines or functions. The first function (`MeasurementFusionFcn`) allows the algorithm to estimate a fused measurement from object, given a set of time-of-arrival measurements assumed to originate from the same object. The second function (`MeasurementFcn`) allows the algorithm to obtain the expected time-of-arrival measurement from the fused measurement. Note that to define a measurement function to obtain time-of-arrival measurement, the fused measurement must contain an estimate of the signal emission time from the object. Therefore, you define the fused measurement as $\\left[{\\mathit{x}}_{\\mathit{t}}\\text{\\hspace{0.17em}}{\\mathit{y}}_{\\mathit{t}\\text{\\hspace{0.17em}}}\\text{\\hspace{0.17em}}{\\mathit{z}}_{\\mathit{t}}\\text{\\hspace{0.17em}}{\\mathit{t}}_{\\mathit{e}}\\right]$, where ${\\mathit{t}}_{\\mathit{e}}$ is the time instant at which the object emitted the signal.\n\nYou use the helper function, `helperTOA2Pos`, to estimate fused measurement - position and emission time - of the object. You also use the helper function, `helperMeasureTOA`, to define the measurement model that calculates time-of-arrival measurement from the fused measurement. The fusion algorithm outputs fused detections as a cell array of `objectDetection` objects. Each element of the cell array defines an `objectDetection` object containing measurement from a potential object as its position and emission time. You can speed-up this static fusion algorithm by specifying the `UseParallel` property to `true`. Specifying `UseParallel` as `true` requires the Parallel Computing Toolbox™.\n\n```% Create scenario [scenario, ~, receiverIDs] = helperCreateMultiTargetTDOAScenario(4, 4); % Specify stastics of TOA simulation measNoise = 1e4; % 100 ns per receiver nFalse = 1; % 1 false alarm per receiver per step Pd = 0.95; % Detection probability of true signal % Release display release(display); display.Title = 'Tracking Using TOA Measurements with Unknown IDs'; % Release tracker release(tracker); tracker.ConfirmationThreshold = [4 6]; % Use Parallel Computing Toolbox if available useParallel = false; % Define fuser. toaFuser = staticDetectionFuser(MaxNumSensors=4,... MeasurementFormat='custom',... MeasurementFcn=@helperMeasureTOA,... MeasurementFusionFcn=@helperTOA2Pos,... FalseAlarmRate=1e-8,... DetectionProbability=Pd,... UseParallel=useParallel); while advance(scenario) % Current time time = scenario.SimulationTime; % Simulate TOA detections with false alarms and missed detections toaDets = helperSimulateTOA(scenario, receiverIDs, measNoise, Pd, nFalse); % Fuse TOA detections to estimate position amd emission time of % unidentified and unknown number of emitters. [posDets, info] = toaFuser(toaDets); % Remove emission time before feeding detections to the tracker as it % is not tracked in this example. for i = 1:numel(posDets) posDets{i}.Measurement = posDets{i}.Measurement(1:3); posDets{i}.MeasurementNoise = posDets{i}.MeasurementNoise(1:3,1:3); end % Update tracker. The target emission is assumed at timestamp, time. % The timestamp of the detection's is therefore time plus signal % propagation time. The tracker timestamp must be greater than % detection time. Use time + 0.05 here. In real-world, this is % absolute timestamp at which tracker was updated. tracks = tracker(posDets, time + 0.05); % Update display % Form TDOA measurements from assigned TOAs for visualization tdoaDets = helperFormulateTDOAs(toaDets,info.Assignments); display(scenario, receiverIDs, tdoaDets, posDets, tracks); end```",
null,
"Notice that the fuser algorithm is able to estimate the position of the object accurately most of the time. However, due to the presence of false alarms and missed measurements, the fusion algorithm may pick a wrong data association at some steps. This can lead to detections from ghost intersections. The tracker assists the static fusion algorithm by discarding these wrong associations as false alarms.\n\n#### Tracking with Time-difference-of-Arrival (TDOA) Measurements\n\nIn this section, you configure the tracking algorithm for a system where TDOAs are formed from the receiver-pairs from multiple objects without any emitter identification. The calculation of TDOA from receiver pairs in a multi-object scenario may generate a few false alarms due to false peaks in the cross-correlation function. To simulate the TDOA measurements from such system, you increase the number of false alarms per receiver pair to 2.\n\nThe static fusion algorithm is configured similar to the previous section by using the `staticDetectionFuser` object. In contrast to the time-of-arrival fusion, here you use the measurement fusion function, `helperFuseTDOA`, to fuse multiple TDOAs into a fused measurement. You also use the helper function, `helperMeasureTDOA`, to define the transformation from fused measurement to TDOA measurement. As TDOA measurements do not contain or need information about the true signal emission time, you define the fused measurement as $\\left[{\\mathit{x}}_{\\mathit{t}}\\text{\\hspace{0.17em}}{\\mathit{y}}_{\\mathit{t}\\text{\\hspace{0.17em}}}\\text{\\hspace{0.17em}}{\\mathit{z}}_{\\mathit{t}}\\text{\\hspace{0.17em}}\\right]$. You set the `MaxNumSensors` to 3 as four receivers form three TDOA pairs.\n\n```% Create scenario [scenario, rxPairs] = helperCreateMultiTargetTDOAScenario(4, 4); % Measurement statistics measNoise = 1e4; % 100 ns per receiver nFalse = 2; % 2 false alarms per receiver pair Pd = 0.95; % Detection probability per receiver pair % Release display release(display); display.Title = 'Tracking Using TDOA Measurements with Unknown IDs'; % Release tracker release(tracker); % Define fuser. tdoaFuser = staticDetectionFuser(MaxNumSensors=3,... MeasurementFormat='custom',... MeasurementFcn=@helperMeasureTDOA,... MeasurementFusionFcn=@helperTDOA2Pos,... DetectionProbability=Pd,... FalseAlarmRate=1e-8,... UseParallel=useParallel); while advance(scenario) % Current elapsed time time = scenario.SimulationTime; % Simulate TDOA detections with false alarms and missed detections tdoaDets = helperSimulateTDOA(scenario, rxPairs, measNoise, false, Pd, nFalse); % Fuse TDOA detections to esimate position detections of unidentified % and unknown number of emitters posDets = tdoaFuser(tdoaDets); % Update tracker with position detections if ~isempty(posDets) tracks = tracker(posDets, time); end % Update display display(scenario, receiverIDs, tdoaDets, posDets, tracks); end```",
null,
"With the defined measurement statistics in this example, the tracker maintains tracks on all the objects based on this network geometry. The geometry of the problem, the measurement accuracy, as well as the number of false alarms all have major impacts on the data association accuracy of the static fusion algorithm for both the TOA and TDOA systems with unknown data association. For the scenario used in this example, the static fusion algorithm was able to report true detections at sufficient time instants to maintain a track on true objects.\n\n### Summary\n\nIn this example, you learned how to track single object as well as multiple objects using TDOA measurements. You learned about the challenges associated with multi-object tracking without emitter identification from the receivers and used a static fusion algorithm to compute data association at the measurement level.\n\n### References\n\n Smith, Julius, and Jonathan Abel. \"Closed-form least-squares source location estimation from range-difference measurements.\" IEEE Transactions on Acoustics, Speech, and Signal Processing 35.12 (1987): 1661-1669.\n\n Sathyan, T., A. Sinha, and T. Kirubarajan. \"Passive geolocation and tracking of an unknown number of emitters.\" IEEE Transactions on Aerospace and Electronic Systems 42.2 (2006): 740-750.\n\n### Supporting Functions\n\nThis section defines a few supporting functions used in this example. The complete list of helper functions can be found in the current working directory.\n\n`helperTDOA2Pos`\n\n```function varargout = helperTDOA2Pos(tdoaDets, reportDetection) % This function uses the spherical intersection algorithm to find the % object position from the TDOA detections assumed to be from the same % object. % % This function assumes that all TDOAs are measured with respect to the % same reference sensor. % % [pos, posCov] = helperTDOA2Pos(tdoaDets) returns the estimated position % and position uncertainty covariance. % % posDetection = helperTDOA2Pos(tdoaDets, true) returns the estimate % position and uncertainty covariance as an objectDetection object. if nargin < 2 reportDetection = false; end % Collect scaling information params = helperGetGlobalParameters; emissionSpeed = params.EmissionSpeed; timeScale = params.TimeScale; % Location of the reference receiver referenceLoc = tdoaDets{1}.MeasurementParameters(2).OriginPosition(:); % Formulate the problem. See for more details d = zeros(numel(tdoaDets),1); delta = zeros(numel(tdoaDets),1); S = zeros(numel(tdoaDets),3); for i = 1:numel(tdoaDets) receiverLoc = tdoaDets{i}.MeasurementParameters(1).OriginPosition(:); d(i) = tdoaDets{i}.Measurement*emissionSpeed/timeScale; delta(i) = norm(receiverLoc - referenceLoc)^2 - d(i)^2; S(i,:) = receiverLoc - referenceLoc; end % Pseudo-inverse of S Swstar = pinv(S); % Assemble the quadratic range equation STS = (Swstar'*Swstar); a = 4 - 4*d'*STS*d; b = 4*d'*STS*delta; c = -delta'*STS*delta; Rs = zeros(2,1); % Imaginary solution, return a location outside coverage if b^2 < 4*a*c varargout{1} = 1e10*ones(3,1); varargout{2} = 1e10*eye(3); return; end % Two range values Rs(1) = (-b + sqrt(b^2 - 4*a*c))/(2*a); Rs(2) = (-b - sqrt(b^2 - 4*a*c))/(2*a); % If one is negative, use the positive solution if prod(Rs) < 0 Rs = Rs(Rs > 0); pos = 1/2*Swstar*(delta - 2*Rs(1)*d) + referenceLoc; else % Use range which minimize the error xs1 = 1/2*Swstar*(delta - 2*Rs(1)*d); xs2 = 1/2*Swstar*(delta - 2*Rs(2)*d); e1 = norm(delta - 2*Rs(1)*d - 2*S*xs1); e2 = norm(delta - 2*Rs(2)*d - 2*S*xs2); if e1 > e2 pos = xs2 + referenceLoc; else pos = xs1 + referenceLoc; end end % If required, compute the uncertainty in the position if nargout > 1 || reportDetection posCov = helperCalcPositionCovariance(pos,tdoaDets,timeScale,emissionSpeed); end if reportDetection varargout{1} = objectDetection(tdoaDets{1}.Time,pos,'MeasurementNoise',posCov); else varargout{1} = pos; if nargout > 1 varargout{2} = posCov; end end end function measCov = helperCalcPositionCovariance(pos,thisDetections,timeScale,emissionSpeed) n = numel(thisDetections); % Complete Jacobian from position to N TDOAs H = zeros(n,3); % Covariance of all TDOAs S = zeros(n,n); for i = 1:n e1 = pos - thisDetections{i}.MeasurementParameters(1).OriginPosition(:); e2 = pos - thisDetections{i}.MeasurementParameters(2).OriginPosition(:); Htdoar1 = (e1'/norm(e1))*timeScale/emissionSpeed; Htdoar2 = (e2'/norm(e2))*timeScale/emissionSpeed; H(i,:) = Htdoar1 - Htdoar2; S(i,i) = thisDetections{i}.MeasurementNoise; end Pinv = H'/S*H; % Z is not measured, use 1 as the covariance if Pinv(3,3) < eps Pinv(3,3) = 1; end % Insufficient information in TDOA if rank(Pinv) >= 3 measCov = eye(3)/Pinv; else measCov = inf(3); end % Replace inf with large number measCov(~isfinite(measCov)) = 100; % Return a true symmetric, positive definite matrix for covariance. measCov = (measCov + measCov')/2; end```\n\n`helperInitHighSpeedKF`\n\n```function filter = helperInitHighSpeedKF(detection) % This function initializes a constant velocity Kalman filter and sets a % higher initial state covariance on velocity components to account for % high speed of the object vehicles. filter = initcvkf(detection); filter.StateCovariance(2:2:end,2:2:end) = 500*eye(3); end```\n\n`helperTOA2Pos`\n\n```function [posTime, posTimeCov] = helperTOA2Pos(toaDetections) % This function computes the position and emission time of a target given % its time-of-arrival detections from multiple receivers. % % Convert TOAs to TDOAs for using spherical intersection [tdoaDetections, isValid] = helperTOA2TDOADetections(toaDetections); % At least 3 TOAs (2 TDOAs) required. Some TOA pairings can lead to an % invalid TDOA (> maximum TDOA). In those situations, discard the tuple by % using an arbitrary position with a large covariance. if numel(tdoaDetections) < 2 || any(~isValid) posTime = 1e10*ones(4,1); posTimeCov = 1e10*eye(4); else % Get position estimate using TDOA fusion. % Get time estimate using calculated position. if nargout > 1 % Only calculate covariance when requested to save time [pos, posCov] = helperTDOA2Pos(tdoaDetections); [time, timeCov] = helperCalcEmissionTime(toaDetections, pos, posCov); posTime = [pos;time]; posTimeCov = blkdiag(posCov,timeCov); else pos = helperTDOA2Pos(tdoaDetections); time = helperCalcEmissionTime(toaDetections, pos); posTime = [pos;time]; end end end```\n\nhelperCalcEmissionTime\n\n```function [time, timeCov] = helperCalcEmissionTime(toaDetections, pos, posCov) % This function calculates the emission time of an object given its % position and obtained TOA detections. It also computes the uncertainty in % the estimate time. globalParams = helperGetGlobalParameters; emissionSpeed = globalParams.EmissionSpeed; timeScale = globalParams.TimeScale; n = numel(toaDetections); emissionTime = zeros(n,1); emissionTimeCov = zeros(n,1); for i = 1:numel(toaDetections) % Calculate range from this receiver p0 = toaDetections{i}.MeasurementParameters.OriginPosition(:); r = norm(pos - p0); emissionTime(i) = toaDetections{i}.Measurement - r/emissionSpeed*timeScale; if nargout > 1 rJac = (pos - p0)'/r; rCov = rJac*posCov*rJac'; emissionTimeCov(i) = rCov./emissionSpeed^2*timeScale^2; end end % Gaussian merge each time and covariance estimate time = mean(emissionTime); if nargout > 1 e = emissionTime - time; timeCov = mean(emissionTimeCov) + mean(e.^2); end end```\n\n`helperMeasureTOA`\n\n```function toa = helperMeasureTOA(posTime, params) % This function calculates the expected TOA at a receiver given an object % position and emission time and the receiver's measurement parameters % (OriginPosition) globalParams = helperGetGlobalParameters; emissionSpeed = globalParams.EmissionSpeed; timeScale = globalParams.TimeScale; r = norm(posTime(1:3) - params.OriginPosition); toa = posTime(4) + r/emissionSpeed*timeScale; end```\n\n`helperMeasureTDOA`\n\n```function toa = helperMeasureTDOA(pos, params) % This function calculates the expected TDOA measurement given object % position and measurement parameters of a TDOA detection globalParams = helperGetGlobalParameters; emissionSpeed = globalParams.EmissionSpeed; timeScale = globalParams.TimeScale; r1 = norm(pos(1:3) - params(1).OriginPosition); r2 = norm(pos(1:3) - params(2).OriginPosition); toa = (r1 - r2)/emissionSpeed*timeScale; end```"
] | [
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_01.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_02.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_03.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_04.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_05.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_06.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_07.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_08.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_09.png",
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"https://jp.mathworks.com/help/examples/fusion/win64/ObjectTrackingUsingTDOAExample_10.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80301136,"math_prob":0.98767954,"size":29190,"snap":"2023-14-2023-23","text_gpt3_token_len":6732,"char_repetition_ratio":0.18282738,"word_repetition_ratio":0.046333004,"special_character_ratio":0.20613223,"punctuation_ratio":0.13738085,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9917392,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T05:22:43Z\",\"WARC-Record-ID\":\"<urn:uuid:4dd6fd74-d866-4e91-8790-387a09c44dcb>\",\"Content-Length\":\"137969\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a764a264-39c9-43fa-b3b4-82ccbbeae87b>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcd6cde0-4d50-4df5-9eff-64afef27e8f8>\",\"WARC-IP-Address\":\"23.63.224.60\",\"WARC-Target-URI\":\"https://jp.mathworks.com/help/fusion/ug/object-tracking-using-time-difference-of-arrival.html\",\"WARC-Payload-Digest\":\"sha1:7QPYM4APXVMGG6ZKMF2Z2VTMMZPSUIFP\",\"WARC-Block-Digest\":\"sha1:EYJQJZT42CPMOU5AJTH2UBMLPCNIQ5BC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647614.56_warc_CC-MAIN-20230601042457-20230601072457-00623.warc.gz\"}"} |
https://www.cut-the-knot.org/WhatIs/Infinity/TransferPrinciple.shtml | [
"# The Transfer Principle\n\nReal R and hyperreal numbers R* are two models of the same first order theory, meaning that any (first-order) statement true in one is true in the other.\n\nEvery natural number n > 1 has a predecessor: (∀n > 1)(∃m)(m + 1 = n). When interpreted in N*, the statement asserts that every hypernatural number, except 1, has a predecessor, implying, in particular, that there is no smallest infinite number.\n\nNot every statement that holds in N holds in N*. One example is the principle of mathematical induction:\n\n∀M⊂N[(1∈M ∧ ∀k(k∈M⇒(k+1)∈M)] ⇒ M = N.\n\nThe analog of that in N* would mean that\n\n∀M⊂N*[(1∈M ∧ ∀k(k∈M⇒(k+1)∈M)] ⇒ M = N*\n\nwhich clearly fails for M = N because N is a proper subset of N*.\n\nEven more obviously not every statement that holds in N* holds in N. For example,\n\n∃ω∈N*∀n∈N (ω > n),\n\nwhich claims the existence of infinitely large numbers has no analog in N.\n\n... to be continued ...\n\n### References\n\n1. Leif Arkeryd, The Evolution of Nonstandard Analysis, The American Mathematical Monthly, Vol. 112, No. 10 (Dec., 2005), pp. 926-928\n2. J. M. Henle, E. M. Kleinberg, Infinitesimal Calculus, Dover, 2003\n3. K. Ito, Nonstandard Analysis, in Encyclopedic Dictionary of Mathematics, v. 1, MIT Press, Press, 2000 (fourth printing), pp. 1100-1103\n4. A. Robinson, Non-standard Analysis, Princeton University Press (Rev Sub edition), 1996",
null,
"Back to What Is Infinity?",
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] | [
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"https://www.cut-the-knot.org/gifs/tbow_sh.gif",
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"https://www.cut-the-knot.org/gifs/tbow_sh.gif",
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https://fluent.docs.pyansys.com/release/0.12/api/solver/tui/report/dpm_histogram/set/index.html | [
"# solver.tui.report.dpm_histogram.set#\n\nEnters the settings menu for the histogram.\n\nauto_range(*args, **kwargs)#\n\nAutomatically computes the range of the sampling variable for histogram plots.\n\ncorrelation(*args, **kwargs)#\n\nComputes the correlation of the sampling variable with another variable.\n\ncumulation_curve(*args, **kwargs)#\n\nComputes a cumulative curve for the sampling variable or correlation variable when correlation? is specified.\n\ndiameter_statistics(*args, **kwargs)#\n\nComputes the Rosin Rammler parameters, Sauter, and other mean diameters.\n\nhistogram_mode(*args, **kwargs)#\n\nUses bars for the histogram plot or xy-style.\n\nlogarithmic(*args, **kwargs)#\n\nEnables/disables the use of logarithmic scaling on the abscissa of the histogram.\n\nmaximum(*args, **kwargs)#\n\nSpecifies the maximum value of the x-axis variable for histogram plots.\n\nminimum(*args, **kwargs)#\n\nSpecifies the minimum value of the x-axis variable for histogram plots.\n\nnumber_of_bins(*args, **kwargs)#\n\nSpecifies the number of bins.\n\npercentage(*args, **kwargs)#\n\nUses percentages of bins to be computed.\n\nvariable_power_3(*args, **kwargs)#\n\nUse the cubic of the cumulation variable during computation of the cumulative curve. When the particle mass was not sampled, the diameter can be used instead.\n\nweighting(*args, **kwargs)#\n\nUses weighting with additional variables when sorting data into samples."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6942102,"math_prob":0.9268743,"size":990,"snap":"2022-40-2023-06","text_gpt3_token_len":191,"char_repetition_ratio":0.18559837,"word_repetition_ratio":0.08571429,"special_character_ratio":0.17272727,"punctuation_ratio":0.12643678,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96900713,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T08:36:03Z\",\"WARC-Record-ID\":\"<urn:uuid:7966422b-6ab9-4299-8f14-571ba1127e39>\",\"Content-Length\":\"434672\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c33a8c7-bbd0-4710-8df2-c86cd663c2e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:895d1501-7fc9-4cce-847e-98fc66f9a127>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://fluent.docs.pyansys.com/release/0.12/api/solver/tui/report/dpm_histogram/set/index.html\",\"WARC-Payload-Digest\":\"sha1:NQEKSDCFKRJIW6W2VK2DB3TUXICMBYBZ\",\"WARC-Block-Digest\":\"sha1:B2I3TYFTDTIE46GER6MORP6A7LMKJ5NI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500334.35_warc_CC-MAIN-20230206082428-20230206112428-00428.warc.gz\"}"} |
https://downloads.hindawi.com/journals/ijp/2015/968024.xml | [
"IJP International Journal of Photoenergy 1687-529X 1110-662X Hindawi Publishing Corporation 10.1155/2015/968024 968024 Research Article A Model for Hourly Solar Radiation Data Generation from Daily Solar Radiation Data Using a Generalized Regression Artificial Neural Network http://orcid.org/0000-0002-0140-7123 Khatib Tamer 1 http://orcid.org/0000-0001-6401-2658 Elmenreich Wilfried 2 Van Sark Wilfried G. J. H. M. 1 Department of Energy Engineering and Environment An-Najah National University Nablus State of Palestine najah.edu 2 Institute of Networked and Embedded Systems University of Klagenfurt 9020 Klagenfurt Austria uni-klu.ac.at 2015 13102015 2015 29 06 2015 13 09 2015 13102015 2015 Copyright © 2015 Tamer Khatib and Wilfried Elmenreich. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\nThis paper presents a model for predicting hourly solar radiation data using daily solar radiation averages. The proposed model is a generalized regression artificial neural network. This model has three inputs, namely, mean daily solar radiation, hour angle, and sunset hour angle. The output layer has one node which is mean hourly solar radiation. The training and development of the proposed model are done using MATLAB and 43800 records of hourly global solar radiation. The results show that the proposed model has better prediction accuracy compared to some empirical and statistical models. Two error statistics are used in this research to evaluate the proposed model, namely, mean absolute percentage error and root mean square error. These values for the proposed model are 11.8% and −3.1%, respectively. Finally, the proposed model shows better ability in overcoming the sophistic nature of the solar radiation data.\n\n1. Introduction\n\nSolar energy is the portion of the sun’s energy available at the earth’s surface for useful applications, such as raising the temperature of water or exciting electrons in a photovoltaic cell, in addition to supplying energy to natural processes. This energy is free, clean, and abundant in most places throughout the year. Its effective harnessing and use are of importance to the world, especially at a time of high fossil fuel costs and degradation of the atmosphere by the use of fossil fuels. Solar radiation data provide information on how much of the sun’s energy strikes a location on the earth’s surface during a particular time period. These data are needed for effective research into solar energy utilization .\n\n2. Hourly Solar Radiation Data Mining\n\nPredicting of hourly solar radiation using statistical model.\n\n2.1. Empirical Models for Calculating Mean Hourly Solar Radiation\n\nAccording to the literature, there are some empirical models developed for calculating hourly solar radiation from daily solar radiation. In , Liu and Jordan proposed the following to calculate hourly solar radiation:(1)GhGD=π/24cosω-cosωssinωs-2πωs/360cosωs,where Gh is the mean hourly solar radiation, GD is the mean daily solar radiation, ω is the hour angle, and ωs is the sunset hour angle.\n\nThe hour angle (ω) is the angular displacement of the sun from the local point and it is given by the following:(2)ω=15AST-12hour,where AST is the apparent or true solar time and it is given by the daily apparent motion of the true, or observed, sun. AST is based on the apparent solar day, which is the interval between two successive returns of the sun to the local meridian. Apparent solar time can be calculated as follows:(3)AST=LST+EoT+4mindegreeLSMT-LOD,where LST is the local standard time, LOD is the longitude, LSMT is the local standard meridian time, and EoT is the equation of time.\n\nThe local standard meridian (LSMT) is a reference meridian used for a particular time zone and is similar to the prime meridian, used for Greenwich Mean Time. LSMT is given by the following:(4)LSMT=15°TimezoneinGMT.In the meanwhile, the equation of time (EoT) is the difference between apparent and mean solar times, both taken at a given longitude at the same real instant of time. EoT is given by the following:(5)EoT=9.87sin2B-7.53cosB-1.5sinB,where B is a factor and it can be calculated by(6)B=360°365N-81,where N is the day number and it is defined as the number of days elapsed in a given year up to a particular date (e.g., 2nd February corresponds to 33).\n\nIn addition to that, in , the authors presented a correlation between Gh and GD according to Figure 1 as follows:(18)GhGD=a+bt+ct2,where a, b, and c are coefficients that can be determined by any curve fitting tool. However, the drawback of this model is that it is a location dependent model whereas such a type of models is devoted to a specific region. This is because the coefficients a, b, and c are calculated based on a specific solar radiation profile.\n\nIn addition, in , statistical model for calculating hourly global solar radiation on horizontal surface was developed. This model represents the hourly solar radiation as a function of extraterrestrial solar radiation as well as a sky transmission function. The proposed sky transmission function is presented as two transmission functions indicating the daily and the hourly variation. At this point, the hourly and the daily transmission variation functions are estimated as statistical relations in terms of day number, hour of the day, and location latitude and longitude based on ground measurements of environmental parameters for a specific location. After all, the inputs of the model were the hour of day, day number, optimized sky transmission function, solar constant, and location coordination. The main drawback of this work is that the relations proposed for the daily and hourly transmission functions are location dependent. Moreover, the utilized extraterrestrial solar radiation data are based on satellite measurements which might not be accurate. More statistical methods were provided. In , a model for generating hourly solar radiation as a function of the clearness index is proposed. The development of the model is based on an assumption that the relation between clearness index values and solar radiation can be described by a Gaussian function. In the meanwhile, in , the authors proposed a trigonometrical function for predicting daily solar radiation values from monthly solar radiation values.\n\n2.2. Proposed Generalized Regression Artificial Neural Network Model\n\nArtificial neural networks, ANNs, are nonalgorithmic and intensely parallel information processing systems. They learn the relationship between input and output variables by mastering previously recorded data. An ANN usually consists of parallel elemental units called neurons. Neurons are connected by a large number of weighted links which pass signals or information. A neuron receives and combines inputs and then generates the final results in a nonlinear operation. The term ANN usually refers to a Multilayer Perceptron (MLP) Network; however, there are many other types of neural networks, including Probabilistic Neural Networks (PNNs), General Regression Neural Networks (GRNNs), Radial Basis Function (RBF) Networks, Cascade Correlation, Functional Link Networks, Kohonen networks, Gram-Charlier networks, Learning Vector Quantization, Hebb networks, Adaline networks, Heteroassociative networks, Recurrent Networks, and Hybrid Networks .\n\nANNs have recently been used to predict the amount of solar radiation based on meteorological variables such as sunshine ratio, temperature, and humidity . However, up to now, no one has used ANNs to find the correlation between mean hourly solar radiation and mean daily solar radiation. Therefore, in this paper, a generalized regression artificial neural network (GRNN) is proposed for this purpose. The GRNN is the most recommended type of ANN for solar radiation prediction according to Khatib et al. in . The generalized regression neural network (GRNN) is a probabilistic based network. This network makes classification where the target variable is definite, and GRNNs make regression where the target variable is continuous. GRNN falls into the category of PNNs. This neural network, like other PNNs, needs only a fraction of the training samples an MLP would need. The additional knowledge needed to obtain the fit in a satisfying way is relatively small and can be done without additional input by a user. This makes GRNNs a useful tool for performing prediction and comparison of system performance in practice. The probability density function used in GRNNs is the normal distribution. Each training sample, Xj, is used as the mean of a normal distribution function given by the following:(19)yx=t=1nYiexp-Di2/2σ2t=1nexp-Di2/2σ2Di2=X-XiT·X-Xi.Di is the distance between the training sample and the point of prediction; it is used as a measure of how well each training sample represents the position of prediction, X. If the distance, Di, between the training sample and the point of prediction is small, exp(-Di2/2σ2) becomes larger. For Di=0, exp(-Di2/2σ2) becomes 1.0 and the point of evaluation is represented best by this training sample. A larger distance to all the other training samples causes the term exp(-Di2/2σ2) to become smaller and therefore the contribution of the other training samples to the prediction is relatively small. The term Yiexp(-Di2/2σ2) for the ith training sample is the largest and contributes strongly to the prediction. The standard deviation or smoothness parameter, s, is subjected to a search. For a large smoothness parameter, the possible representation of the point of evaluation by the training sample is possible for a wider range of X. For a small smoothness parameter, the representation is limited to a narrow range of X.\n\nGRNNs consist of input, hidden, and output layers. The input layer has one neuron for each predictor variable. The input neurons standardize the range of values by subtracting the median and dividing by the interquartile range. The input neurons then feed the values to each of the neurons in the hidden layer. In the hidden layer, there is one neuron for each case in the training data set. The neuron stores the values of the predictor variables for each case, along with the target value. When presented with a vector of input values from the input layer, a hidden neuron computes the Euclidean distance of the test case from the neuron’s center point and then applies the RBF kernel function using the sigma value. The resulting value is passed to the neurons in the pattern layer. However, the pattern (summation) layer has two neurons: one is the denominator summation unit and the other is the numerator summation unit. The denominator summation unit adds the weights of the values coming from each of the hidden neurons. The numerator summation unit adds the weights of the values multiplied by the target value for each hidden neuron. The decision layer divides the value accumulated in the numerator summation unit by the value in the denominator summation unit and uses the result as the predicted target value .\n\nBased on the previous models ((1), (9), and (11)), it is clear that the hourly solar radiation value is a function of parameters such as mean daily solar radiation, hour angle, and sunset/sunrise hour angle. Based on this, the GRNN illustrated in Figure 2 is proposed for estimating mean hourly solar radiation. The input layer of the network has three inputs: mean daily solar radiation, hour angle, sunset hour angle. Meanwhile, the output layer has one node which is mean hourly solar radiation.\n\nProposed model for hourly solar radiation prediction.\n\nWith regard to the number of neurons in the hidden layer, there is no way to determine the optimal number of hidden neurons without training several networks and estimating the generalization error of each. A low number of hidden neurons cause high training and generalization error due to underfitting and high statistical bias, but a large number of hidden neurons cause high generalization error due to overfitting and high variance . There are some rules of thumb for choosing the number of the hidden nodes. Blum claims in that the number of neurons in the hidden layer ought to be somewhere between the input layer size and the output layer size. Swingler in and Berry and Linoff in claim that the hidden layer will never require more than twice the number of the inputs. In addition, Boger and Guterman suggest in that the number of hidden nodes should be 70–90% of the number of input nodes. Additionally, Caudill and Butler recommend in that the number of hidden nodes equals the number of inputs plus the number of outputs multiplied by (2/3). Based on these recommendations, the number of neurons in the hidden layer of our model should be between 2 and 4. In this research, we used 4 hidden nodes.\n\n2.3. Model Evaluation Criteria\n\nTo evaluate the proposed GRNN model and the other models, two statistics errors are used: mean absolute percentage error (MAPE) and root mean square error (RMSE). MAPE is an indicator of accuracy. MAPE usually expresses accuracy as a percentage and is defined by the following formula:(20)MAPE=1nt=1nI-IpI,where I is the measured value and Ip is the predicted value. The resultant of this calculation is summed for every fitted or forecasted point in time and divided again by the number of fitted points, n. This formula gives a percentage error, so one can compare the error of fitted time series that differ in level.\n\nIn addition, prediction models were evaluated using RMSE. RMSE provides information about the short-term performance of the models and is a measure of the variation of the predicted values around the measured data. RMSE indicates the scattering of data around linear lines. Moreover, RMSE shows the efficiency of the developed network in predicting future individual values. A large positive RMSE implies a large deviation in the predicted value from the measured value. RMSE can be calculated as follows:(21)RMSE=1ni=1nIpi-Ii2,where Ipi is the predicted value, Ii is the measured value, and n is the number of observations .\n\n2.3.1. Sensitivity Analysis\n\nFor any prediction model that deals with stochastic data such as solar radiation, the sensitivity analysis is important for many reasons such as testing the robustness of the results in the presence of uncertainty. Moreover, such an analysis can provide better understanding of the relation between the model’s output(s) and input(s). This understating may lead to a model enhancement by identifying the model’s input(s) which case significant uncertainty in the output. According to , the sensitivity analysis can be defined as the study of how the uncertainty in the output of a mathematical model or system can be apportioned to different sources of uncertainty in its inputs. One of the popular methods is automated differentiation method, where the sensitivity parameters are found by simply taking the derivatives of the output with respect to the input.\n\nIn this research, the hourly solar radiation can be described as a function of daily solar radiation and sunrise or sunset hour angle and hour as follows:(22)Gh=fGD,ωss,sr,ω.However, following the models presented in (1) and (9), the presented model can be described as below: (23)rh=GhGD=fωss,sr,ω.Assume that we are calculating the sensitivity of species rh with respect to every parameter in the model in (23). Thus, it is required to calculate the time-dependent derivatives as follows:(24)rhωss,sr,rhω.There are different methods to estimate the value of each deferential part. One of these methods is Taylor method. This method states that Taylor approximation of y around a given point (xo,yo)—which stands for the first derivative—can be given by the following: (25)rh1=rho+rhωss,srωss,sr1=ωss,sro.Anyway, nowadays such a problem can be also solved by many kinds of software such as MATLAB using functions such as ParameterInputFactors and SensitivityAnalysis. However, in this research, we used simpler and popular method as well which is the scatter plots method. This method is represented by scatter plots of the output against input(s) individually. This method gives a direct visual indication of sensitivity. Moreover, quantitative measures can also be provided by measuring the correlation between the output and each input.\n\n3. Results and Discussion\n\nIn this research, the utilized solar radiation data were measured using rugged solar radiation transmitter (model: WE300, sensor size: 7.6 cm diameter. × 3.8 cm long). The detector of this sensor is high-stability silicon photovoltaic (blue enhanced). Meanwhile, the output range of this sensor is 4 to 20 mA and the measuring range is 0 to 1500 W/m2 and the spectral response is in the range of 400 to 1100 nm. The accuracy of this sensor is ±1% full scale with worming up time up to 3 seconds. The operating voltage of this sensor is in the range of 10 to 36 VDC. Using this sensor, solar radiation data are measured and recorded every 5 minutes. Then, these data have been converted to hourly averages with 43800 data records. In this research, an hourly solar radiation data set consisting of 43800 records (5 years) is used. 35040 records (4 years) are used in training the proposed GRNN model. Meanwhile, the 5th year data are used to test the developed model. The model development and training were done using MATLAB line code which is provided in the Appendix of this paper.\n\nTo test the proposed model, a control data set containing 8760 records of hourly solar radiation and hour angle is used. The average daily solar radiation and the sunset hour angle are calculated based on these records. The calculated daily solar radiation, sunset hour angles, and hour angles are then used as inputs for the proposed GRNN model. Meanwhile, the output of the GRNN model is compared to the actual hourly solar radiation data. To ensure proper evaluation, the control data set is not used in training the proposed GRNN model in order to check the ability of the proposed model for predicting future and foreign data. It is also worth mentioning that the GRNN model has been tested repeatedly (up to 10 times) in order to provide average performance of the proposed model.\n\nFigure 3 shows the correlation between the predicted and the measured data using the proposed GRNN model. From the figure, it is clear that the correlation value is about 96%, which is considerably high. This high correlation value implies that the proposed model makes accurate predictions.\n\nCorrelation between generated and measured values using the proposed model.\n\nIn addition, Figure 4 shows the prediction results for a whole year. From Figure 4, it can be noted that the proposed GRNN model predicts the hourly solar radiation successfully. However, to provide deeper analysis of the proposed GRNN model, two zones, namely, A and B, are chosen from Figure 4 and illustrated more clearly in Figure 5. The selection of these zones is done based on the amount of cloud cover. These zones represent cloudy days where the proposed model prediction accuracy is expected to be low due to unstable solar radiation levels. From Figure 5, the accuracy of the proposed model for predicting the hourly solar radiation is acceptable whereas the generated values of the hourly solar radiation are close to the actual values even on totally overcast days. However, there is a time shift in some cases. This time shift is due to the difference between the calculated hour angle and the read hour angle that the solar radiation is measured at. This problem is usually solved by adding shifting constants to the models. It is assumed that each cycle in Figure 5 represents a whole solar day, where the first cycle is day 1 and the last is day 15. From Figure 5, it can be noticed that, on clear days such as 1, 3, 5, 8, 10, 11, and 15, the prediction is accurate and acceptable. On the other hand, for cloudy days such as 2, 6, 7, 12, 13, and 14, the prediction accuracy is lower than the previous case but still acceptable as most of the solar radiation prediction models’ accuracies degraded on cloudy days .\n\nPrediction results of the proposed GRNN model (Part A).\n\nPrediction results of the proposed GRNN model (Part B).\n\nIn order to validate the proposed model, we did two types of comparison; first, a comparison is between the proposed model and some location dependent models. From the conducted literature review, we found two location dependent models which are presented in (16) and (18). The model presented in (16) assumes that the hourly solar radiation can be described by trigonometric functions. In the meanwhile, the model presented in (18) assumes that the averages hourly ratios of hourly solar radiation to daily solar radiation in average can be described by a polynomial function of the second degree. In this research, we used the same data used to train the proposed model in developing these models. Figure 6 shows the development of these models.\n\nDevelopment of two location dependent models.\n\nIn general, both models show good fitting of the average daily hourly rd ratios with R-square values of 0.9413 for the model presented in (16) and 0.9721 for the model presented in (18). Despite these high R-square values, these models are not expected to predict hourly solar radiation accurately for two reasons; the assumption of describing the hourly solar radiation by these mathematical functions is not accurate. Secondly, the developed models fit perfectly the average day but they may be unable to fit individual days. Figure 7 shows three days prediction results of these location dependent models.\n\nPrediction results of location dependent models.\n\nFrom Figure 7, it is very clear that the prediction of these random days was inaccurate. Anyway, after ignoring the extreme underestimations of these models in Figure 7, we found that the average MAPE for the model presented in (16) is about 60% while it is about 40% for the model presented in (18).\n\nHowever, for more fair comparison, the proposed model is compared with more accurate empirical models. A comparison between the proposed model and Liu-Jordan and Collares-Pereira models is conducted in this research. Figure 8 shows a sample of the comparison conducted for 8 solar days. From Figure 8, it is clear that the three models can predict hourly solar radiation data accurately in clear sky days. However, Liu-Jordan model generated underestimated values sometimes. In addition, the generated curves are slightly shifted in some days which caused overestimated values in the morning and underestimated value in the afternoon. On the other hand, generated profiles using Collares-Pereira model are sometimes narrower that the actual one which caused underestimations in the afternoon. As for the proposed model, it can be seen that it is more accurate than the other two models. However, the three models are not able to predict the solar radiation accurately in the cloudy days as compared to their ability in the clear sky days.\n\nComparison between the proposed model and other models.\n\nTable 1 shows an evaluation of the three models using MAPE and RMSE. It is clear that the proposed model has the best accuracy prediction whereas it exceeds the other models by the MAPE and RMSE. This implies that the proposed model is more powerful in predicting hourly solar radiation according to the MAPE value. Moreover, it has the ability to predict future data based on the RMSE value.\n\nEvaluation of the proposed GRNN model.\n\nModel MAPE RMSE (W/m2) RMSE (%) R 2\nLiu-Jordan 27.4% 64.7 22.1 80.3\nCollares-Pereira 22.5% 59.1 20.7 85.7\nGRNN (training) 11.4% 45.0 14.8 98.3\nGRNN (testing) 11.8% 46.1 15.1 96.0\n\nAs for the sensitivity analysis, in this research, the proposed model as well as Liu-Jordan and Collares-Pereira models is tested in terms of sensitivity by plotting scatter plots for each model with respect to hour angle factor (ω versus rh). Then, the correlation value for each data set is provided. Figures 911 show the scatter plots of the three models as described in (23).\n\nω versus rh for Liu-Jordan model.\n\nω versus rh for Collares-Pereira model.\n\nω versus rh for the proposed GRNN model.\n\nFrom Figures 911, it can be seen that the proposed model is more efficient in considering the uncertainty of the solar radiation for this case. In Figure 9, the values of rh factor are more concentrated around specific points with some deviated extreme points. These extreme points are the unexpected values of solar radiation due to some reasons such as clouds and dust particles. Moreover, there is no symmetric nature of these values which means that the uncertainty problem of such data is relatively overcome by the proposed model. On the other hand, both Liu-Jordan and Collares-Pereira models resulted in symmetric behavior of the data regardless of any external conditions which caused prediction inaccuracy in some cases. Table 2 shows the R-square values of each model.\n\nSensitivity analysis of the proposed GRNN model.\n\nScatter plot R -square values Fitting function\nLiu-Jordan Collares-Pereira GRNN model\nω versus Gh/GD 0.78 0.75 0.90 Polynomial/quadratic\n\nFrom Table 2, it is also clear that the proposed model exceeds the other empirical models. In general, the advantage of the proposed model as compared to the empirical models is that it is a machine with the ability of learning and handling huge data sets with stochastic nature. As a fact, heuristic techniques such as GRNN are more efficient in handling stochastic data subject to a prior concrete training. However, the empirical models exceed the proposed model in case of having short historical data that is not enough to train the proposed model.\n\nFinally, utilizing such a large data set (5 years, 43800 records) is not a must to develop such a model. It is all about utilizing available data to develop accurate model with excellent ability to predict future data. As a matter of fact, there are two important issues to be considered when deciding the size of a training data set for a solar radiation prediction model. These issues are the uncertainty nature of solar radiation and the day number nature of the year. In other words, it is possible to predict hourly solar radiation in January (winter) using a model that is trained based on data for June (summer). However, it will not be accurate as compared to a model that is trained based on the whole year time with small step records data. In order to address this issue, we have trained the proposed model using data sets with three relatively small sizes 84, 348, and 684 records. Figures 12(a), 12(b), 13(a), 13(b), 14(a), and 14(b) show the result of this practice. The first part of Figures 10, 11, and 12 shows the prediction performance of the model by comparing its output to the actual values. It is worth mentioning that in these figures the days where there is no red line during them mean that the model generated values of zero only (the model failed to predict the solar radiation profile). On the other hand, the second part of Figures 12, 13, and 14 shows model accuracy using the correlation factor R2.\n\n(a) Proposed model performance considering small sizes of training data set. (b) Correlation between generated and measured values using training data set size of 7 solar days.\n\n(a) Proposed model performance considering small sizes of training data set. (b) Correlation between generated and measured values using training data set size of 29 solar days.\n\n(a) Proposed model performance considering small sizes of training data set. (b) Correlation between generated and measured values using training data set size of 57 solar days.\n\nFrom Figures 12(a), 12(b), 13(a), 13(b), 14(a), and 14(b), it is clear that the proposed model did not perform well when it is trained using only 84 records (7 solar days). However, the model was able to predict a number of days accurately according to Figure 12(a). These results are seconded by Figure 12(b) where the correlation value between the generated and the measured values is very low due to the high rate of model shortages (the days where the model generates values of zero only). The mean absolute percentage error of the model that developed based on 84 records is very high where it is 100% in 70% of the generated days. Meanwhile, the prediction accuracy of the other days where the model was able to generate synthetic solar radiation data was fine with MAPE in the range of 10–13%. In general, the average whole MAPE for this model is 64.3% with R2 value of 0.1354. Here, the empirical models show superior performance as these models do not need any prior training.\n\nOn the other hand, according to Figures 13(a) and 13(b), the performance of the proposed model was significantly enhanced when it is trained using 348 records (29 solar days). Meanwhile, Figures 14(a) and 14(b) show that the performance of the model was further improved when 684 records (57 solar days) were used in the training. The model which has been trained based on 348 records was able to work probably during 72% of the testing days, while the 684 records based model was able to work probably during 86% of the testing days. The average whole MAPE values of both models are 36.8% and 24%, respectively. These results can be also concluded from Figures 13(b) and 14(b) where R2 values are significantly increased. R2 values for 348 records based model and 684 records based model are 0.2263 and 0.6583, respectively. Finally, by comparing all of these figures to Figure 3, it can be realized that the correlation value is much better than all the previous models and consequently the model is supposed to be more accurate in predicting solar radiation values. Table 3 summaries the aforementioned results.\n\nImpact of training data set size on model’s accuracy.\n\nTraining data set size (records) Av. overall MAPE R 2\n84 (7 solar days) 64.3% 0.1354\n348 (29 solar days) 36.8% 0.2263\n684 (57 solar days) 24.1% 0.6583\n35,040 (1460 solar days) 11.8% 0.9595\n\nAs a conclusion, as far as the model is trained using more data, the accuracy will be better. However, such a model can be developed using a relatively small data set (about 70 solar days) which is usually not difficult to obtain. As mentioned before, the proposed model has the advantage of avoiding the complex calculation of the empirical and statistical model’s parameters. Moreover, in case of training the proposed model well, the model will be able to handle the uncertainty issue in solar radiation much better than the empirical and statistical models.\n\n4. Conclusion\n\nIn this research, a generalized regression artificial neural network based model was presented for predicting hourly solar radiation using daily solar radiation. This model has three inputs, namely, mean daily solar radiation, hour angle, and sunset hour angle. The output layer has one node, which is the generated mean hourly solar radiation. Five years of data for hourly solar radiation were used to train and develop the model running under MATLAB. The results showed that the proposed model has better prediction accuracy as compared to existing empirical and statistical models especially in dealing with special location dependent cases. The average prediction accuracy of the proposed model is about 11% with R-square value of 0.96. Furthermore, the proposed model showed superiority in terms of prediction sensitivity as compared to some empirical models. However, the results showed that the proposed model needed a concrete prior training in order to show prediction superiority. It is also concluded that such a model can be developed with relatively accepted prediction accuracy (24%) using about two months’ data with an hourly step. Finally, the proposed model can be used in predicting the performance of solar energy systems such as photovoltaic system and solar water heater. Moreover, it can be used to generate long term data in order to be used in optimal sizing and planning of solar energy systems.\n\nAppendix\n\nSee Pseudocode 1.\n\n<bold>Pseudocode 1</bold>\n\n% Start\n\nclc\n\nclear\n\nclose all\n\n% %— — — — — — —Input Data— — — — — —\n\nfileName=Site.xlsx;\n\nL=; % add a latitude value\n\nLOD=; % add a longitude value\n\n% T means testing data\n\nWi=xlsread(fileName, sheetName, D3:D4358); % hour angle\n\nWsi=xlsread(fileName, sheetName, E3:E4358); % sun set angle\n\nDN=xlsread(fileName, sheetName, F3:F4358); % day number (day & month)\n\nSR=SRi1000;\n\nSR_T=SRi_T1000;\n\nW=(Wi_T(pi/180));\n\nWs=(abs(Wsi_T(pi/180)));\n\n%=============Proposed GRNN Model=================\n\ninputs=[L LOD Wi, Wsi, GD DN h];\n\nI=inputs;\n\ntargets=SR;\n\nT=targets;\n\nnet=newgrnn(I,T);\n\ntest=[L LOD Wi_T, Wsi_T, GD_T DN_T h_T];\n\nTest=test;\n\nx_predicted=sim(net,Test);\n\nX_P=x_predicted;\n\n%============Liu and Jordan Model==================\n\nR_LiuJordan=((((pi)/24)(cos(W)-cos(Ws)))./(sin(Ws)-((piWs)/180).cos(Ws)));\n\nSR_LiuJordan=R_LiuJordan.GD_T\n\n%============Collares-Pereira and Rabel Model======\n\na=0.409+(0.5016sin(Ws-60));\n\nb=0.6609+(0.4767sin(Ws-60));\n\nR_Collares=((a + (b.cos(W))).((((pi/24))(cos(W)-cos(Ws))))./(sin(Ws)-((piWs)/180).cos(Ws)));\n\nSR_Collares=R_Collares.GD_T;\n\n%=================Plot data===========================\n\nplot (SR_T)\n\nxlim([0 100])\n\nhold on\n\nplot (SR_LiuJordan, k)\n\nxlim([0 100])\n\nhold on\n\nplot (SR_Collares, g)\n\nxlim([0 35])\n\nhold on\n\nplot (X_P, red)\n\nxlim([0 100])\n\n%— — — —Models evaluation— — — — — — — —\n\nn=length(X_P);\n\nE3_Hour=SR_T-X_P;\n\nGRNN_MAPE=abs(E3_Hour./SR_T);\n\nGRNN_meanMAPE1=sum(GRNN_MAPE)/n;\n\nGRNN_meanMAPE=GRNN_meanMAPE1100;\n\nGRNN_RMSE=sqrt(sum((X_P-SR_T).2/n))\n\nGRNN_MBE=sum(X_P-SR_T)/n\n\nSUM=(sum(X_P)./n);\n\nliner_RMSE_Percentage=(GRNN_RMSE/SUM)100\n\nliner_MBE_Percentage=(GRNN_MBE/SUM)100\n\n% End\n\nNomenclature ANNs:\n\nArtificial neural networks\n\nAST:\n\nApparent or true solar time\n\nδ :\n\nAngle of declination\n\nN :\n\nDay number\n\nGD:\n\nDistributed generation\n\nEoT:\n\nEquation of time\n\nG D :\n\nG h :\n\nGRNN:\n\nGeneralized artificial neural network\n\nϕ :\n\nThe latitude\n\nLLP:\n\nLOD:\n\nLongitude\n\nLSMT:\n\nLocal standard meridian time\n\nLST:\n\nLocal standard time\n\nMAPE:\n\nMean absolute percentage error\n\nMLP:\n\nMultilayer Perceptron Network\n\nPNNs:\n\nProbabilistic Neural Networks\n\nPV:\n\nPhotovoltaic\n\nRBF:\n\nRMSE:\n\nRoot mean square error\n\nr t :\n\nThe ratio of hourly to daily global solar radiation\n\nt s r :\n\nSunrise time\n\nt s s :\n\nSunset time\n\nω :\n\nHour angle\n\nω s s , s r :\n\nSunset/sunrise hour angle.\n\nConflict of Interests\n\nThe authors hereby confirm that there is no conflict of interests in the paper with any third part.\n\nAcknowledgments\n\nThis work is supported by Lakeside Labs, Klagenfurt, Austria, and funded by the European Regional Development Fund (ERDF) and the Carinthian Economic Promotion Fund (KWF) under Grant 20214∣22935∣34445 (Project Smart Microgrid)."
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https://se.mathworks.com/matlabcentral/fileexchange/16363-discrete-time-integrator-in-embedded-matlab?s_tid=prof_contriblnk | [
"## Discrete Time Integrator in Embedded MATLAB\n\nVersion 1.0.0.1 (11.3 KB) by\nThis demo shows how to use Embedded MATLAB block parameters to control the algorithm behavior\n\nUpdated 1 Sep 2016\n\nThe attached Simulink model implements a Forward Euler Integrator in Embedded MATLAB. It illustrates the use of parameters in Embedded MATLAB.\nThe algorithm has three sections; Setup, Output and Update.\n\nAt every sample time hit the block keeps accumulating the the value of the input at time instant 't' in a state register and emits value accumulated value until time instant 't-1'. The output value is clipped if falls above an upper limit or falls below a lower limit. A flag is raised when the value is clipped. The upper and lower limits and the gain of the integrator can be changed via Embedded MATLAB block parameters.\n\nLike any other block in Simulink the Embedded MATLAB block also supports mask parameters. Simulink parameters can be tunable or non-tunable. Tunable parameter values can be changed during the simulation run in MATLAB workspace and can be used to affect the current Simulink simulation run in progress. Non-tunable parameters are also defined in MATLAB workspace but are fixed for a single simulation run.\n\nIn the attached model the variables 'gain_val', 'upper_limit', 'lower_limit' used in the Embedded MATLAB script are non-tunable parameters. They are declared as any other inputs in the\nfunction declaration associated with the block. However they do not appear as block inputs. The values and types of these variables can be defined and changed in the MATLAB between simulation runs.\n\n### Cite As\n\nKiran Kintali (2023). Discrete Time Integrator in Embedded MATLAB (https://www.mathworks.com/matlabcentral/fileexchange/16363-discrete-time-integrator-in-embedded-matlab), MATLAB Central File Exchange. Retrieved .\n\n##### MATLAB Release Compatibility\nCreated with R2007a\nCompatible with any release\n##### Platform Compatibility\nWindows macOS Linux\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!\nVersion Published Release Notes\n1.0.0.1"
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https://www.shuzhiduo.com/A/LPdopagOz3/ | [
"``` public class Main {\npublic static void main(String[] args) {\nPerson person = new Person();\n// person.age = 1000; // 编译错误\n// System.out.println(person.age); // 编译错误\nperson.setAge(1000);\nSystem.out.println(person.getAge()); //\n\nperson.setAge(10);\nSystem.out.println(person.getAge()); //\n\nSystem.out.println(person.toString()); // Person@1b6d3586\n}\n}\n\nclass Person {\nprivate int age;\n\npublic int getAge() {\nreturn age;\n}\n\npublic void setAge(int age) {\nif (age < 0 || age > 100) {\nreturn;\n}\nthis.age = age;\n}\n}\n}\n\npublic void setAge(int age) {\nif (age < 0 || age > 100 ) {\nreturn;\n}\nthis.age = age;\n}\n}```\n\n``` public class Main {\npublic static void main(String[] args) {\nPoint point = new Point(1, 2);\nString string = point.toString();\nSystem.out.println(string); // (1,2)\n}\n}\n\nclass Point {\nprivate int x;\nprivate int y;\n\npublic int getX() {\nreturn x;\n}\n\npublic void setX(int x) {\nthis.x = x;\n}\n\npublic int getY() {\nreturn y;\n}\n\npublic void setY(int y) {\nthis.y = y;\n}\n\npublic Point() {\n}\n\npublic Point(int x, int y) {\nthis.x = x;\nthis.y = y;\n}\n\npublic String toString() {\nreturn \"(\" + x + \",\" + y + ')';\n}\n}```\n\n```public class Main {\npublic static void main(String[] args) {\nPoint point1 = new Point(1, 2);\nPoint point2 = new Point(1, 2);\nSystem.out.println(point1 == point2); // false\nSystem.out.println(point1.equals(point2)); // false\n}\n}```\n\n``` public boolean equals(Object obj) {\nreturn (this == obj);\n}```\n\n``` public class Main {\npublic static void main(String[] args) {\nPoint point1 = new Point(1, 2);\nPoint point2 = new Point(1, 2);\nSystem.out.println(point1 == point2); // false\nSystem.out.println(point1.equals(point2)); // true\n}\n}\n\nclass Point {\nprivate int x;\nprivate int y;\n\npublic int getX() {\nreturn x;\n}\n\npublic void setX(int x) {\nthis.x = x;\n}\n\npublic int getY() {\nreturn y;\n}\n\npublic void setY(int y) {\nthis.y = y;\n}\n\npublic Point() {\n}\n\npublic Point(int x, int y) {\nthis.x = x;\nthis.y = y;\n}\n\npublic String toString() {\nreturn \"(\" + x + \",\" + y + ')';\n}\n\npublic boolean equals(Object object) {\nif (object == null) {\nreturn false;\n}\nif (object == this) {\nreturn true;\n}\nif (object instanceof Point) {\nPoint point = (Point) object; // 强转为point类型\nreturn this.x == point.x && this.y == point.y;\n}\nreturn false;\n}\n}```\n\n## Java 从入门到进阶之路（十八）的更多相关文章\n\n1. Java 从入门到进阶之路（八）\n\n在之前的文章我们介绍了一下 Java 中的重载,接下来我们看一下 Java 中的构造方法. 我们之前说过,我们在定义一个变量的时候,java 会为我们提供一个默认的值,字符串为 null,数字为 0. ...\n\n2. Java 从入门到进阶之路（十）\n\n之前的文章我们介绍了一下 Java 中的引用型数组类型,接下来我们再来看一下 Java 中的继承. 继承的概念 继承是java面向对象编程技术的一块基石,因为它允许创建分等级层次的类. 继承就是子类继 ...\n\n3. Java 从入门到进阶之路（十二）\n\n在之前的文章我们介绍了一下 Java 类的重写及与重载的区别,本章我们来看一下 Java 类的 private,static,final. 我们在之前引入 Java 类概念的时候是通过商场收银台来引入 ...\n\n4. Java 从入门到进阶之路（十五）\n\n在之前的文章我们介绍了一下 Java 中的接口,本章我们来看一下 Java 中类的多态. 在日常生活中,很多意思并不是我们想要的意思,如下: 1.领导:“你这是什么意思?” 小明:“没什么意思,意思意 ...\n\n5. Java 从入门到进阶之路（十四）\n\n在之前的文章我们介绍了一下 Java 中的抽象类和抽象方法,本章我们来看一下 Java 中的接口. 在日常生活中,我们会接触到很多类似接口的问题,比如 USB 接口,我们在电脑上插鼠标,键盘,U盘的时 ...\n\n6. Java 从入门到进阶之路（十六）\n\n在之前的文章我们介绍了一下 Java 中类的多态,本章我们来看一下 Java 中类的内部类. 在 Java 中,内部类分为成员内部类和匿名内部类. 我们先来看一下成员内部类: 1.类中套类,外面的叫外 ...\n\n7. Java 从入门到进阶之路（十九）\n\n在之前的文章我们介绍了一下 Java 中的Object,本章我们来看一下 Java 中的包装类. 在 Java 中有八个基本类型:byte,short,int,long,float,double,ch ...\n\n8. Java 从入门到进阶之路（二十）\n\n在之前的文章我们介绍了一下 Java 中的包装类,本章我们来看一下 Java 中的日期操作. 在我们日常编程中,日期使我们非常常用的一个操作,比如读写日期,输出日志等,那接下来我们就看一下 Java ...\n\n9. Java 从入门到进阶之路（二十二）\n\n在之前的文章我们介绍了一下 Java 中的 集合框架中的Collection 中的一些常用方法,本章我们来看一下 Java 集合框架中的Collection 的迭代器 Iterator. 当我们创建 ...\n\n## 随机推荐\n\n1. VS2010中，无法嵌入互操作类型“……”，请改用适用的接口的解决方法(转自网络)\n\n最近开始使用VS2010,在引用COM组件的时候,出现了无法嵌入互操作类型“……”,请改用适用的接口的错误提示.查阅资料,找到解决方案,记录如下: 选中项目中引入的dll,鼠标右键,选择属性,把“嵌入 ...\n\n2. CC1310电源\n\nCC1310的电源好扯,把目前遇到的问题记录一下 1 全局LDO和DCDC的输出电压问题 手册上要求的VDDR和VDDR_RF的电压范围是1.7~1.95V,但实际测试时, 在接收状态下无论是全局LD ...\n\n3. ORACLE查看表空间对象\n\nORACLE如何查看表空间存储了那些数据库对象呢?可以使用下面脚本简单的查询表空间存储了那些对象: SELECT TABLESPACE_NAME AS TABLESPACE_NAME ...\n\n4. Springmvc常用注解\n\n1. @RequestMapping注解的作用位置 @RequestMapping可以作用在类名上,也可以作用在方法上. 如果都有, 产生作用的路径是类名上的路径+方法上的路径. 比如Employee ...\n\n5. 记录一次centos6.4版本的VSFTP本地用户登陆的配置\n\n其实vsftp是一个非常常用而且简单的服务,但是假如服务不是你配置的前者没有留下参考档案,的确是件头疼的事儿,特此记录下. 首先是vsftp的安装当然安装有源码的编译和yum等 这里我选择rpm包的y ...\n\n6. hdu1045 Fire Net\n\n在一张地图上建立碉堡(X),要求每行没列不能放两个,除非中间有强挡着.求最多能放多少个碉堡 #include<iostream> #include<cstdio> #inclu ...\n\n7. Linux下使用fdisk发现磁盘空间和使用mount挂载文件系统\n\n若在安装Linux系统时没有想好怎么使用服务器,开始时只分了一部分给根目录.后面需要再使用时,可以通过几下一步进行分区和挂载文件系统. 看磁柱数使用情况 fdisk -l Disk /dev/sda: ...\n\n8. 从医生看病和快餐店点餐理解Node.js的事件驱动"
] | [
null
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http://oalevelsolutions.com/past-papers-solutions/cambridge-international-examinations/as-a-level-mathematics-9709/pure-mathematics-p1-9709-01/year-2016-may-june-p1-9709-13/cie_16_mj_9709_13_q_6/ | [
"# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/13) | Q#6\n\nHits: 951\n\nQuestion",
null,
"The diagram shows triangle ABC where AB=5cm, AC=4cm and BC=3cm. Three circles with centres at A, B and C have radii 3 cm, 2 cm and 1 cm respectively. The circles touch each other at points E, F and G, lying on AB, AC and BC respectively. Find the area of the shaded region EFG.\n\nSolution\n\nIt is evident from the diagram that;",
null,
"Let’s first find area of",
null,
";\n\nExpression for the area of a triangle for which two sides (a and b) and the included angle (C) is given;",
null,
"We know lengths of all three sides of",
null,
".",
null,
"",
null,
"",
null,
"Law of cosines is given as;",
null,
"",
null,
"",
null,
"It is evident we can use law of cosines to find any angle of the",
null,
".\n\nLet’s find angle",
null,
".",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Using calculator;",
null,
"Now we can find area of",
null,
".",
null,
"",
null,
"",
null,
"",
null,
"Now we need areas of sectors.\n\nExpression for area of a circular sector with radius",
null,
"and angle",
null,
"rad is;",
null,
"To find areas of sectors;",
null,
"",
null,
"",
null,
"We are given that;",
null,
"",
null,
"",
null,
"",
null,
"Therefore, we need to find",
null,
"and",
null,
".\n\nLaw of cosines is given as;",
null,
"",
null,
"",
null,
"We are given that;",
null,
"",
null,
"",
null,
"Therefore;\n\n For",
null,
"; For",
null,
";",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Therefore; to find areas of sectors;",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Finally, we can find area of shaded region is;",
null,
"",
null,
"",
null,
""
] | [
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"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image051.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image053.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image054.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image055.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image056.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image057.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image058.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image029.png",
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"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image059.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image030.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image060.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image031.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image061.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image002.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image062.png",
null,
"http://oalevelsolutions.com/CIE_GCE_AS_Maths_P1_16_Jun_13_Q_6_files/image063.png",
null
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https://ask.libreoffice.org/t/rounding-up-a-result-in-a-cell-to-a-whole-number/60989 | [
"",
null,
"# Rounding up a result in a cell to a whole number\n\nI need to round up a result in a cell to the next whole number up. Cell B22 is a sum of a cell range (A2:A10) that will return a total of, say, 4500 and that has to be divided by 1000 giving 4.5 as a result, I need it to round up to 5. Many thanks.\n\nHello,\n\n`=ROUNDUP(SUM(A2:A10)/1000;0)` should do the job all in one go - or -\n`=ROUNDUP(B22/1000;0)` (if you want to use sum in `B22` already calculated)\n\nHope that helps.\n\nIt does indeed help, thanks very much, it’s appreciated.\n\nIf it is helpful please consider to click the check mark (",
null,
") next to the answer. Thanks in advance …"
] | [
null,
"https://ask.libreoffice.org/uploads/asklibo/original/1X/1eec1ce28d4605f25e751aea59dbef2bc0782151.png",
null,
"https://ask.libreoffice.org/images/emoji/twitter/heavy_check_mark.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91854954,"math_prob":0.75978595,"size":569,"snap":"2022-27-2022-33","text_gpt3_token_len":168,"char_repetition_ratio":0.0920354,"word_repetition_ratio":0.0,"special_character_ratio":0.31985942,"punctuation_ratio":0.124087594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97246426,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T12:49:29Z\",\"WARC-Record-ID\":\"<urn:uuid:2f0eaaff-1717-45ba-9bbc-6813dacf7cf9>\",\"Content-Length\":\"19653\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4fc0291e-bd8a-48ce-8ce1-d3590bc6c66c>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe75af0a-f540-46de-8ebf-2a93cc0e2cf8>\",\"WARC-IP-Address\":\"89.238.68.222\",\"WARC-Target-URI\":\"https://ask.libreoffice.org/t/rounding-up-a-result-in-a-cell-to-a-whole-number/60989\",\"WARC-Payload-Digest\":\"sha1:WQNEIY4VN4AP7UIS76UHJIWXUGZNCEU2\",\"WARC-Block-Digest\":\"sha1:4D6QQ5VPXZWTVZFI7XOAUS6HKVANOQM4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104576719.83_warc_CC-MAIN-20220705113756-20220705143756-00081.warc.gz\"}"} |
https://docs.opencv.org/3.4.1/db/d4e/classcv_1_1Point__.html | [
"",
null,
"OpenCV 3.4.1 Open Source Computer Vision\ncv::Point_< _Tp > Class Template Reference\n\nTemplate class for 2D points specified by its coordinates x and y. More...\n\n#include \"types.hpp\"\n\n## Public Types\n\ntypedef _Tp value_type\n\n## Public Member Functions\n\nPoint_ ()\ndefault constructor More...\n\nPoint_ (_Tp _x, _Tp _y)\n\nPoint_ (const Point_ &pt)\n\nPoint_ (const Size_< _Tp > &sz)\n\nPoint_ (const Vec< _Tp, 2 > &v)\n\ndouble cross (const Point_ &pt) const\ncross-product More...\n\ndouble ddot (const Point_ &pt) const\ndot product computed in double-precision arithmetics More...\n\n_Tp dot (const Point_ &pt) const\ndot product More...\n\nbool inside (const Rect_< _Tp > &r) const\nchecks whether the point is inside the specified rectangle More...\n\ntemplate<typename _Tp2 >\noperator Point_< _Tp2 > () const\nconversion to another data type More...\n\noperator Vec< _Tp, 2 > () const\nconversion to the old-style C structures More...\n\nPoint_operator= (const Point_ &pt)\n\n## Public Attributes\n\n_Tp x\nx coordinate of the point More...\n\n_Tp y\ny coordinate of the point More...\n\n## Detailed Description\n\n### template<typename _Tp> class cv::Point_< _Tp >\n\nTemplate class for 2D points specified by its coordinates x and y.\n\nAn instance of the class is interchangeable with C structures, CvPoint and CvPoint2D32f . There is also a cast operator to convert point coordinates to the specified type. The conversion from floating-point coordinates to integer coordinates is done by rounding. Commonly, the conversion uses this operation for each of the coordinates. Besides the class members listed in the declaration above, the following operations on points are implemented:\n\npt1 = pt2 + pt3;\npt1 = pt2 - pt3;\npt1 = pt2 * a;\npt1 = a * pt2;\npt1 = pt2 / a;\npt1 += pt2;\npt1 -= pt2;\npt1 *= a;\npt1 /= a;\ndouble value = norm(pt); // L2 norm\npt1 == pt2;\npt1 != pt2;\n\nFor your convenience, the following type aliases are defined:\n\ntypedef Point_<int> Point2i;\ntypedef Point2i Point;\ntypedef Point_<float> Point2f;\ntypedef Point_<double> Point2d;\n\nExample:\n\nPoint2f a(0.3f, 0.f), b(0.f, 0.4f);\nPoint pt = (a + b)*10.f;\ncout << pt.x << \", \" << pt.y << endl;\n\n## § value_type\n\ntemplate<typename _Tp>\n typedef _Tp cv::Point_< _Tp >::value_type\n\n## § Point_() [1/5]\n\ntemplate<typename _Tp>\n cv::Point_< _Tp >::Point_ ( )\n\ndefault constructor\n\n## § Point_() [2/5]\n\ntemplate<typename _Tp>\n cv::Point_< _Tp >::Point_ ( _Tp _x, _Tp _y )\n\n## § Point_() [3/5]\n\ntemplate<typename _Tp>\n cv::Point_< _Tp >::Point_ ( const Point_< _Tp > & pt )\n\n## § Point_() [4/5]\n\ntemplate<typename _Tp>\n cv::Point_< _Tp >::Point_ ( const Size_< _Tp > & sz )\n\n## § Point_() [5/5]\n\ntemplate<typename _Tp>\n cv::Point_< _Tp >::Point_ ( const Vec< _Tp, 2 > & v )\n\n## § cross()\n\ntemplate<typename _Tp>\n double cv::Point_< _Tp >::cross ( const Point_< _Tp > & pt ) const\n\ncross-product\n\n## § ddot()\n\ntemplate<typename _Tp>\n double cv::Point_< _Tp >::ddot ( const Point_< _Tp > & pt ) const\n\ndot product computed in double-precision arithmetics\n\n## § dot()\n\ntemplate<typename _Tp>\n _Tp cv::Point_< _Tp >::dot ( const Point_< _Tp > & pt ) const\n\ndot product\n\n## § inside()\n\ntemplate<typename _Tp>\n bool cv::Point_< _Tp >::inside ( const Rect_< _Tp > & r ) const\n\nchecks whether the point is inside the specified rectangle\n\n## § operator Point_< _Tp2 >()\n\ntemplate<typename _Tp>\ntemplate<typename _Tp2 >\n cv::Point_< _Tp >::operator Point_< _Tp2 > ( ) const\n\nconversion to another data type\n\n## § operator Vec< _Tp, 2 >()\n\ntemplate<typename _Tp>\n cv::Point_< _Tp >::operator Vec< _Tp, 2 > ( ) const\n\nconversion to the old-style C structures\n\n## § operator=()\n\ntemplate<typename _Tp>\n Point_& cv::Point_< _Tp >::operator= ( const Point_< _Tp > & pt )\n\n## § x\n\ntemplate<typename _Tp>\n _Tp cv::Point_< _Tp >::x\n\nx coordinate of the point\n\n## § y\n\ntemplate<typename _Tp>\n _Tp cv::Point_< _Tp >::y\n\ny coordinate of the point\n\nThe documentation for this class was generated from the following file:"
] | [
null,
"https://docs.opencv.org/3.4.1/opencv-logo-small.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.53442436,"math_prob":0.9415249,"size":2714,"snap":"2020-24-2020-29","text_gpt3_token_len":785,"char_repetition_ratio":0.21845019,"word_repetition_ratio":0.22093023,"special_character_ratio":0.3282977,"punctuation_ratio":0.21705426,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946667,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-26T17:48:12Z\",\"WARC-Record-ID\":\"<urn:uuid:aa407f47-832c-46ab-928b-bf7e19244fef>\",\"Content-Length\":\"30187\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b451e16-8b57-42ae-aa6a-d5f357f57bc9>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e3eeb23-c48d-40ef-a40a-6dfe8d0b08e1>\",\"WARC-IP-Address\":\"207.38.86.214\",\"WARC-Target-URI\":\"https://docs.opencv.org/3.4.1/db/d4e/classcv_1_1Point__.html\",\"WARC-Payload-Digest\":\"sha1:WLCPXNYKYUCBBWD5P2NXEH56RMTCWTIE\",\"WARC-Block-Digest\":\"sha1:PRGSCPSAEQ24QSBQ6L4NMPG4U6QTCQPM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347391277.13_warc_CC-MAIN-20200526160400-20200526190400-00177.warc.gz\"}"} |
https://www.cis.uni-muenchen.de/schuetze/e/e/eS/eSi/eSig/eSigm/eSigma/eSigma_Theta_Tau.html | [
"### Sigma Theta Tau\n\nRelated by string. * sigma . SIGMA . Sigmas . www.sigma : Beta Sigma Phi . Sigma Chi fraternity . Lean Six Sigma . Sigma Pi . Sigma Nu fraternity / theta . THETA . Thetas : Mu Alpha Theta . Beta Theta Pi . Kappa Alpha Theta sorority . Delta Sigma Theta . Iota Phi Theta / TAU . Tauer . t Au . tau . Taus : Zeta Tau Alpha . Tau Devi Lal . Alpha Tau Omega . Tau Kappa Epsilon . Tau Kappa Epsilon fraternity * *"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7001119,"math_prob":0.8011802,"size":2147,"snap":"2020-45-2020-50","text_gpt3_token_len":556,"char_repetition_ratio":0.16565563,"word_repetition_ratio":0.0,"special_character_ratio":0.25850022,"punctuation_ratio":0.0085227275,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99412596,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T01:58:49Z\",\"WARC-Record-ID\":\"<urn:uuid:2505fa7e-2d2c-4860-8382-8c609e450c37>\",\"Content-Length\":\"15639\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e749bf9-e905-4b7e-ab6b-5f6e8a0180d7>\",\"WARC-Concurrent-To\":\"<urn:uuid:04dc97cc-fc8f-4cc1-af03-1c071f1e4154>\",\"WARC-IP-Address\":\"129.187.148.72\",\"WARC-Target-URI\":\"https://www.cis.uni-muenchen.de/schuetze/e/e/eS/eSi/eSig/eSigm/eSigma/eSigma_Theta_Tau.html\",\"WARC-Payload-Digest\":\"sha1:T56IU4PJXE66NHMKG53RI5XZ2NJGF5QL\",\"WARC-Block-Digest\":\"sha1:7UCBYTXBMITVPWAUQGCICNGN2X33ZBWC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141733120.84_warc_CC-MAIN-20201204010410-20201204040410-00470.warc.gz\"}"} |
http://meta-numerics.net/Documentation/html/c7f3199e-6cad-5003-edad-208914625a71.htm | [
"Computes both Airy functions and their derivatives.\n\nNamespace: Meta.Numerics.Functions\nAssembly: Meta.Numerics (in Meta.Numerics.dll) Version: 4.1.4",
null,
"Syntax\n```public static SolutionPair Airy(\ndouble x\n)```\n\n#### Parameters\n\nx\nType: SystemDouble\nThe argument.\n\n#### Return Value\n\nType: SolutionPair\nThe values of Ai(x), Ai'(x), Bi(x), and Bi'(x).",
null,
"Remarks\n\nAiry functions are solutions to the Airy differential equation:",
null,
"The Airy functions appear in quantum mechanics in the semi-classical WKB solution to the wave functions in a potential.\n\nFor negative arguments, Ai(x) and Bi(x) are oscillatory. For positive arguments, Ai(x) decreases exponentially and Bi(x) increases exponentially with increasing x.\n\nThis method simultaneously computes both Airy functions and their derivatives. If you need both Ai and Bi, it is faster to call this method once than to call AiryAi(Double) and AiryBi(Double) separately. If on, the other hand, you need only Ai or only Bi, and no derivative values, it is faster to call the appropriate method to compute the one you need.",
null,
"See Also"
] | [
null,
"http://meta-numerics.net/Documentation/icons/SectionExpanded.png",
null,
"http://meta-numerics.net/Documentation/icons/SectionExpanded.png",
null,
"http://meta-numerics.net/Documentation/images/AiryODE.png",
null,
"http://meta-numerics.net/Documentation/icons/SectionExpanded.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86627936,"math_prob":0.83809775,"size":761,"snap":"2021-43-2021-49","text_gpt3_token_len":164,"char_repetition_ratio":0.12813738,"word_repetition_ratio":0.018018018,"special_character_ratio":0.1892247,"punctuation_ratio":0.120567374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9893352,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T12:15:35Z\",\"WARC-Record-ID\":\"<urn:uuid:70c2efb7-9d2a-4560-8d7f-871fb672a54a>\",\"Content-Length\":\"25464\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8778e93d-350d-475b-a9dd-2d77cf2c7cb6>\",\"WARC-Concurrent-To\":\"<urn:uuid:25c0b5d6-60fa-4e52-a3b4-cc8a66474b14>\",\"WARC-IP-Address\":\"143.95.247.91\",\"WARC-Target-URI\":\"http://meta-numerics.net/Documentation/html/c7f3199e-6cad-5003-edad-208914625a71.htm\",\"WARC-Payload-Digest\":\"sha1:O4FFU5OWZGCKFK7PWJ5W52MMGDGNPDCI\",\"WARC-Block-Digest\":\"sha1:4BC43TO5MS7O3PJRKLVXV2LDSSJD4YU6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360803.0_warc_CC-MAIN-20211201113241-20211201143241-00388.warc.gz\"}"} |
https://dir.md/wiki/Point_(geometry)?host=en.wikipedia.org | [
"# Point (geometry)",
null,
"In classical Euclidean geometry, a point is a primitive notion that models an exact location in the space, and has no length, width, or thickness. In modern mathematics, a point refers more generally to an element of some set called a space.\n\nBeing a primitive notion means that a point cannot be defined in terms of previously defined objects. That is, a point is defined only by some properties, called axioms, that it must satisfy; for example, \"there is exactly one line that passes through two different points\".\n\nPoints, considered within the framework of Euclidean geometry, are one of the most fundamental objects. Euclid originally defined the point as \"that which has no part\". In two-dimensional Euclidean space, a point is represented by an ordered pair (x, y) of numbers, where the first number conventionally represents the horizontal and is often denoted by x, and the second number conventionally represents the vertical and is often denoted by y. This idea is easily generalized to three-dimensional Euclidean space, where a point is represented by an ordered triplet (x, y, z) with the additional third number representing depth and often denoted by z. Further generalizations are represented by an ordered tuplet of n terms, (a1, a2, … , an) where n is the dimension of the space in which the point is located.\n\nIn addition to defining points and constructs related to points, Euclid also postulated a key idea about points, that any two points can be connected by a straight line. This is easily confirmed under modern extensions of Euclidean geometry, and had lasting consequences at its introduction, allowing the construction of almost all the geometric concepts known at the time. However, Euclid's postulation of points was neither complete nor definitive, and he occasionally assumed facts about points that did not follow directly from his axioms, such as the ordering of points on the line or the existence of specific points. In spite of this, modern expansions of the system serve to remove these assumptions.\n\nThere are several inequivalent definitions of dimension in mathematics. In all of the common definitions, a point is 0-dimensional.\n\nA point is zero-dimensional with respect to the covering dimension because every open cover of the space has a refinement consisting of a single open set.\n\nA point has Hausdorff dimension 0 because it can be covered by a single ball of arbitrarily small radius.\n\nAlthough the notion of a point is generally considered fundamental in mainstream geometry and topology, there are some systems that forgo it, e.g. noncommutative geometry and pointless topology. A \"pointless\" or \"pointfree\" space is defined not as a set, but via some structure (algebraic or logical respectively) which looks like a well-known function space on the set: an algebra of continuous functions or an algebra of sets respectively. More precisely, such structures generalize well-known spaces of functions in a way that the operation \"take a value at this point\" may not be defined. A further tradition starts from some books of A. N. Whitehead in which the notion of region is assumed as a primitive together with the one of inclusion or connection.\n\nOften in physics and mathematics, it is useful to think of a point as having non-zero mass or charge (this is especially common in classical electromagnetism, where electrons are idealized as points with non-zero charge). The Dirac delta function, or δ function, is (informally) a generalized function on the real number line that is zero everywhere except at zero, with an integral of one over the entire real line. The delta function is sometimes thought of as an infinitely high, infinitely thin spike at the origin, with total area one under the spike, and physically represents an idealized point mass or point charge. It was introduced by theoretical physicist Paul Dirac. In the context of signal processing it is often referred to as the unit impulse symbol (or function). Its discrete analog is the Kronecker delta function which is usually defined on a finite domain and takes values 0 and 1."
] | [
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/8/88/Stereographic_projection_in_3D.svg/1200px-Stereographic_projection_in_3D.svg.png",
null
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https://thareducation.com/math-worksheets-with-answers.html | [
"Math Worksheets With Answers. You can choose the difficulty level and size of maze. Math Worksheets Listed By Specific Topic and Skill Area. Math Worksheets Done Right - Enjoy.",
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https://yingtan.home.focus.cn/gonglue/2d24e7dce391d930.html | [
"|\n\n# 评测:金牛前置过滤器,实力解决家庭饮水二次污染问题\n\n水是我们日常生活中最宝贵的资源，洗衣、做饭等等与生活息息相关的事情都离不开它。锅碗瓢盆脏了可以洗，衣服鞋子脏了也可以洗，但水脏了用什么洗？\n\n生活中所用到的自来水，大多经过了自来水厂的粗过滤和杀菌，但自来水在市政管网运输过程中或者在高层建筑储水箱中不可避免地会受到污染，到达用户家中时就可能会含有泥沙、铁锈、沉淀物、藻类、红虫等污染物，这样就造成了家庭用水二次污染。\n\n针对自来水二次污染带来的困扰，如衣服泛黄、菜洗不干净、热水器堵塞、龙头出黄水、洗衣机故障等，素有管道专家之称的金牛管业，推出了能有效解决家庭用水二次污染问题的产品——金牛前置过滤器。\n\n过滤效果究竟好不好？试试便知道。先来做个小实验（金牛前置过滤器净水效果检测+密封性检测）：\n\n净水效果评测结果：金牛前置过滤器呈T字型，精巧轻便。它采用一体成型包塑滤网，能有效过滤40-90um的杂质，过滤后的水质清澈、干净无杂质。\n\n另外，金牛前置过滤器自身清洁起来也非常方便，它的内部有一个“雨刮式”清洁软刮。打开前置过滤器下方的阀门，杂质立刻顺着水流冲刷出来，冲洗彻底。\n\n施压过程中的密封性评测结果：\n\n家庭正常用水水压一般在2.5-3.0公斤之间，用打压机对前置过滤器进行施压至1.6兆帕，也就是16公斤，没有掉压和漏水现象，说明金牛前置过滤器密闭性非常好。\n\n通过以上实验可以看出，金牛前置过滤器净水效果好，自清洁能力强，密闭性佳。但若要保证水质健康环保，前置过滤器自身的安全健康也不能被忽略。再来看看金牛前置过滤器的重要组成材料：",
null,
"金牛前置过滤器杯体采用进口的PC材料，爆破压力超过55kg，水锤测试超过NSF标准要求，强度更高，使用更安全。",
null,
"",
null,
"",
null,
"过滤器主体采用无铅铜材质，保证水环境更健康环保。",
null,
"",
null,
"评测总结：金牛前置过滤器既适宜家庭使用，也适用于独立安装在需要保护的设备或系统上游。它可有效去除管道中40-90um以上的沉淀杂质和细菌、微生物残骸、铁锈、泥沙、红虫、碎屑等大颗料杂质，避免人体及肌肤受到伤害;同时对下游管道、家用净水设备、热水器、太阳能、洗衣机、高档龙头、花洒、地热水暖水路系统等等涉水设备起到积极的保护作用，使水质达到或优于自来水厂出厂时的标准!\n\n`声明：本文由入驻焦点开放平台的作者撰写，除焦点官方账号外，观点仅代表作者本人，不代表焦点立场错误信息举报电话： 400-099-0099，邮箱：jubao@vip.sohu.com，或点此进行意见反馈，或点此进行举报投诉。`",
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"A B C D E F G H J K L M N P Q R S T W X Y Z\nA - B - C - D - E\n• A\n• 鞍山\n• 安庆\n• 安阳\n• 安顺\n• 安康\n• 澳门\n• B\n• 北京\n• 保定\n• 包头\n• 巴彦淖尔\n• 本溪\n• 蚌埠\n• 亳州\n• 滨州\n• 北海\n• 百色\n• 巴中\n• 毕节\n• 保山\n• 宝鸡\n• 白银\n• 巴州\n• C\n• 承德\n• 沧州\n• 长治\n• 赤峰\n• 朝阳\n• 长春\n• 常州\n• 滁州\n• 池州\n• 长沙\n• 常德\n• 郴州\n• 潮州\n• 崇左\n• 重庆\n• 成都\n• 楚雄\n• 昌都\n• 慈溪\n• 常熟\n• D\n• 大同\n• 大连\n• 丹东\n• 大庆\n• 东营\n• 德州\n• 东莞\n• 德阳\n• 达州\n• 大理\n• 德宏\n• 定西\n• 儋州\n• 东平\n• E\n• 鄂尔多斯\n• 鄂州\n• 恩施\nF - G - H - I - J\n• F\n• 抚顺\n• 阜新\n• 阜阳\n• 福州\n• 抚州\n• 佛山\n• 防城港\n• G\n• 赣州\n• 广州\n• 桂林\n• 贵港\n• 广元\n• 广安\n• 贵阳\n• 固原\n• H\n• 邯郸\n• 衡水\n• 呼和浩特\n• 呼伦贝尔\n• 葫芦岛\n• 哈尔滨\n• 黑河\n• 淮安\n• 杭州\n• 湖州\n• 合肥\n• 淮南\n• 淮北\n• 黄山\n• 菏泽\n• 鹤壁\n• 黄石\n• 黄冈\n• 衡阳\n• 怀化\n• 惠州\n• 河源\n• 贺州\n• 河池\n• 海口\n• 红河\n• 汉中\n• 海东\n• 怀来\n• I\n• J\n• 晋中\n• 锦州\n• 吉林\n• 鸡西\n• 佳木斯\n• 嘉兴\n• 金华\n• 景德镇\n• 九江\n• 吉安\n• 济南\n• 济宁\n• 焦作\n• 荆门\n• 荆州\n• 江门\n• 揭阳\n• 金昌\n• 酒泉\n• 嘉峪关\nK - L - M - N - P\n• K\n• 开封\n• 昆明\n• 昆山\n• L\n• 廊坊\n• 临汾\n• 辽阳\n• 连云港\n• 丽水\n• 六安\n• 龙岩\n• 莱芜\n• 临沂\n• 聊城\n• 洛阳\n• 漯河\n• 娄底\n• 柳州\n• 来宾\n• 泸州\n• 乐山\n• 六盘水\n• 丽江\n• 临沧\n• 拉萨\n• 林芝\n• 兰州\n• 陇南\n• M\n• 牡丹江\n• 马鞍山\n• 茂名\n• 梅州\n• 绵阳\n• 眉山\n• N\n• 南京\n• 南通\n• 宁波\n• 南平\n• 宁德\n• 南昌\n• 南阳\n• 南宁\n• 内江\n• 南充\n• P\n• 盘锦\n• 莆田\n• 平顶山\n• 濮阳\n• 攀枝花\n• 普洱\n• 平凉\nQ - R - S - T - W\n• Q\n• 秦皇岛\n• 齐齐哈尔\n• 衢州\n• 泉州\n• 青岛\n• 清远\n• 钦州\n• 黔南\n• 曲靖\n• 庆阳\n• R\n• 日照\n• 日喀则\n• S\n• 石家庄\n• 沈阳\n• 双鸭山\n• 绥化\n• 上海\n• 苏州\n• 宿迁\n• 绍兴\n• 宿州\n• 三明\n• 上饶\n• 三门峡\n• 商丘\n• 十堰\n• 随州\n• 邵阳\n• 韶关\n• 深圳\n• 汕头\n• 汕尾\n• 三亚\n• 三沙\n• 遂宁\n• 山南\n• 商洛\n• 石嘴山\n• T\n• 天津\n• 唐山\n• 太原\n• 通辽\n• 铁岭\n• 泰州\n• 台州\n• 铜陵\n• 泰安\n• 铜仁\n• 铜川\n• 天水\n• 天门\n• W\n• 乌海\n• 乌兰察布\n• 无锡\n• 温州\n• 芜湖\n• 潍坊\n• 威海\n• 武汉\n• 梧州\n• 渭南\n• 武威\n• 吴忠\n• 乌鲁木齐\nX - Y - Z\n• X\n• 邢台\n• 徐州\n• 宣城\n• 厦门\n• 新乡\n• 许昌\n• 信阳\n• 襄阳\n• 孝感\n• 咸宁\n• 湘潭\n• 湘西\n• 西双版纳\n• 西安\n• 咸阳\n• 西宁\n• 仙桃\n• 西昌\n• Y\n• 运城\n• 营口\n• 盐城\n• 扬州\n• 鹰潭\n• 宜春\n• 烟台\n• 宜昌\n• 岳阳\n• 益阳\n• 永州\n• 阳江\n• 云浮\n• 玉林\n• 宜宾\n• 雅安\n• 玉溪\n• 延安\n• 榆林\n• 银川\n• Z\n• 张家口\n• 镇江\n• 舟山\n• 漳州\n• 淄博\n• 枣庄\n• 郑州\n• 周口\n• 驻马店\n• 株洲\n• 张家界\n• 珠海\n• 湛江\n• 肇庆\n• 中山\n• 自贡\n• 资阳\n• 遵义\n• 昭通\n• 张掖\n• 中卫\n\n1室1厅1厨1卫1阳台\n\n1\n2\n3\n4\n5\n\n0\n1\n2\n\n1\n\n1\n\n0\n1\n2\n3",
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"",
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"报名成功，资料已提交审核",
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"A B C D E F G H J K L M N P Q R S T W X Y Z\nA - B - C - D - E\n• A\n• 鞍山\n• 安庆\n• 安阳\n• 安顺\n• 安康\n• 澳门\n• B\n• 北京\n• 保定\n• 包头\n• 巴彦淖尔\n• 本溪\n• 蚌埠\n• 亳州\n• 滨州\n• 北海\n• 百色\n• 巴中\n• 毕节\n• 保山\n• 宝鸡\n• 白银\n• 巴州\n• C\n• 承德\n• 沧州\n• 长治\n• 赤峰\n• 朝阳\n• 长春\n• 常州\n• 滁州\n• 池州\n• 长沙\n• 常德\n• 郴州\n• 潮州\n• 崇左\n• 重庆\n• 成都\n• 楚雄\n• 昌都\n• 慈溪\n• 常熟\n• D\n• 大同\n• 大连\n• 丹东\n• 大庆\n• 东营\n• 德州\n• 东莞\n• 德阳\n• 达州\n• 大理\n• 德宏\n• 定西\n• 儋州\n• 东平\n• E\n• 鄂尔多斯\n• 鄂州\n• 恩施\nF - G - H - I - J\n• F\n• 抚顺\n• 阜新\n• 阜阳\n• 福州\n• 抚州\n• 佛山\n• 防城港\n• G\n• 赣州\n• 广州\n• 桂林\n• 贵港\n• 广元\n• 广安\n• 贵阳\n• 固原\n• H\n• 邯郸\n• 衡水\n• 呼和浩特\n• 呼伦贝尔\n• 葫芦岛\n• 哈尔滨\n• 黑河\n• 淮安\n• 杭州\n• 湖州\n• 合肥\n• 淮南\n• 淮北\n• 黄山\n• 菏泽\n• 鹤壁\n• 黄石\n• 黄冈\n• 衡阳\n• 怀化\n• 惠州\n• 河源\n• 贺州\n• 河池\n• 海口\n• 红河\n• 汉中\n• 海东\n• 怀来\n• I\n• J\n• 晋中\n• 锦州\n• 吉林\n• 鸡西\n• 佳木斯\n• 嘉兴\n• 金华\n• 景德镇\n• 九江\n• 吉安\n• 济南\n• 济宁\n• 焦作\n• 荆门\n• 荆州\n• 江门\n• 揭阳\n• 金昌\n• 酒泉\n• 嘉峪关\nK - L - M - N - P\n• K\n• 开封\n• 昆明\n• 昆山\n• L\n• 廊坊\n• 临汾\n• 辽阳\n• 连云港\n• 丽水\n• 六安\n• 龙岩\n• 莱芜\n• 临沂\n• 聊城\n• 洛阳\n• 漯河\n• 娄底\n• 柳州\n• 来宾\n• 泸州\n• 乐山\n• 六盘水\n• 丽江\n• 临沧\n• 拉萨\n• 林芝\n• 兰州\n• 陇南\n• M\n• 牡丹江\n• 马鞍山\n• 茂名\n• 梅州\n• 绵阳\n• 眉山\n• N\n• 南京\n• 南通\n• 宁波\n• 南平\n• 宁德\n• 南昌\n• 南阳\n• 南宁\n• 内江\n• 南充\n• P\n• 盘锦\n• 莆田\n• 平顶山\n• 濮阳\n• 攀枝花\n• 普洱\n• 平凉\nQ - R - S - T - W\n• Q\n• 秦皇岛\n• 齐齐哈尔\n• 衢州\n• 泉州\n• 青岛\n• 清远\n• 钦州\n• 黔南\n• 曲靖\n• 庆阳\n• R\n• 日照\n• 日喀则\n• S\n• 石家庄\n• 沈阳\n• 双鸭山\n• 绥化\n• 上海\n• 苏州\n• 宿迁\n• 绍兴\n• 宿州\n• 三明\n• 上饶\n• 三门峡\n• 商丘\n• 十堰\n• 随州\n• 邵阳\n• 韶关\n• 深圳\n• 汕头\n• 汕尾\n• 三亚\n• 三沙\n• 遂宁\n• 山南\n• 商洛\n• 石嘴山\n• T\n• 天津\n• 唐山\n• 太原\n• 通辽\n• 铁岭\n• 泰州\n• 台州\n• 铜陵\n• 泰安\n• 铜仁\n• 铜川\n• 天水\n• 天门\n• W\n• 乌海\n• 乌兰察布\n• 无锡\n• 温州\n• 芜湖\n• 潍坊\n• 威海\n• 武汉\n• 梧州\n• 渭南\n• 武威\n• 吴忠\n• 乌鲁木齐\nX - Y - Z\n• X\n• 邢台\n• 徐州\n• 宣城\n• 厦门\n• 新乡\n• 许昌\n• 信阳\n• 襄阳\n• 孝感\n• 咸宁\n• 湘潭\n• 湘西\n• 西双版纳\n• 西安\n• 咸阳\n• 西宁\n• 仙桃\n• 西昌\n• Y\n• 运城\n• 营口\n• 盐城\n• 扬州\n• 鹰潭\n• 宜春\n• 烟台\n• 宜昌\n• 岳阳\n• 益阳\n• 永州\n• 阳江\n• 云浮\n• 玉林\n• 宜宾\n• 雅安\n• 玉溪\n• 延安\n• 榆林\n• 银川\n• Z\n• 张家口\n• 镇江\n• 舟山\n• 漳州\n• 淄博\n• 枣庄\n• 郑州\n• 周口\n• 驻马店\n• 株洲\n• 张家界\n• 珠海\n• 湛江\n• 肇庆\n• 中山\n• 自贡\n• 资阳\n• 遵义\n• 昭通\n• 张掖\n• 中卫",
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"",
null,
"• 手机",
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"• 分享\n• 设计\n免费设计\n• 计算器\n装修计算器\n• 入驻\n合作入驻\n• 联系\n联系我们\n• 置顶\n返回顶部"
] | [
null,
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https://chalkdustmagazine.com/blog/our-favourite-and-not-so-favourite-euler-equations/ | [
"# Our favourite (and not-so-favourite) Euler equations\n\nA collection of our favourite and least favourite things named after Euler, from issue 07",
null,
"In previous issues of Chalkdust, we shared with you a selection of our favourite “things” in maths, such as our favourite functions, shapes and sets. On the other hand, there are also some things we find annoying and very much dislike, such as bad notation, certain numbers, etc. For this special occasion (commemorating Euler’s birthday Euler a few weeks ago), we decided to spread some of our favourite, and not-so-favourite, examples of things named after Euler throughout issue 07.\n\nWe would also like to hear yours! Send them to us at contact@chalkdustmagazine.com.\n\n## Euler’s polyhedral formula (Emily Maw)\n\nEuler’s polyhedral formula says that the number of vertices, $V$, edges, $E$, and faces, $F$, of a convex 3D polyhedron satisfy\n\n$$V-E+F=2.$$\n\nThis is one of my favourite equations, because it’s arguably the first theorem in topology! Despite the name, Euler couldn’t actually prove it; the first of many proofs were found by Leibniz and Cauchy. It can be used to obtain strong constraints on the existence of shapes, for example in the proof that there are only five Platonic solids, or that all polyhedra built from pentagons and hexagons (such as footballs) must contain exactly 12 pentagons! The reason it holds for convex polyhedra is that they are all homeomorphic (topologically equivalent, by smoothing off the corners) to a sphere. In fact the formula can be generalised: for nonconvex polyhedra, which can be homeomorphic to surfaces with holes, the formula becomes $V-E+F=2-2g$, where $g$ is the genus (number of holes). This is because the left hand side is secretly calculating a topological invariant of the space, called the Euler characteristic, which is equal to $2-2g$ for surfaces.\n\n## Euler’s identity (Matthew Scroggs)\n\nMy least favourite equation is\n\n$$\\mathrm{e}^{\\,\\mathrm{i} \\,\\pi}+1=0.$$\n\nPeople rant about how it’s the most beautiful equation in maths, but it’s not. Overrated.\n\n## Euler’s continued fraction formula (Nikoleta Kalaydzhieva)\n\nEuler’s continued fraction formula connects some infinite series with an infinite continued fraction. For example, it gives the following result for $\\mathrm{e}^z$:\n\n$$\\mathrm{e}^z={\\cfrac{1}{1-\\cfrac{z}{1+z-\\cfrac{z}{2+z-\\cfrac{2z}{3+z-\\cfrac{3z}{4+z-\\cdots}}}}}}$$\n\nIsn’t it pretty?\n\n## Euler’s method (Sean Jamshidi)\n\n$$f\\hspace{2pt}’\\hspace{-2pt}(x) = \\frac{f\\hspace{1pt}(x+h) – f\\hspace{1pt}(x)}{h}.$$\n\nEuler’s method is a way of approximately solving differential equations using a computer, which is one of the most common tasks in applied mathematics. The method was invented hundreds of years before the computer, can be hopelessly inaccurate and is so simple that it has no right to be of any use at all. And yet… we still teach it, it can be coded up in 2 minutes and, in some cases, provides perfectly useable results. Sometimes the oldies are the best ones.\n\n## The Euler-Lagrange equation (Tom Rivlin)\n\nMy favourite Euler equation is the Euler-Lagrange equation:\n\n$$\\frac{\\partial L}{\\partial x_i}- \\frac{\\mathrm{d}}{\\mathrm{d}t} \\Big( \\frac{\\partial L}{\\partial \\dot{x}_i} \\Big) = 0$$\n\nIt’s the key equation of the calculus of variations, which is used extensively in many branches of physics. One uses the Euler-Lagrange equation to derive the Lagrangian of a physical system, from which all of its physical properties can be determined. A famous example is the Lagrangian of the standard model of particle physics, which is a one-equation summary of all of our current knowledge about the fundamental building blocks of reality:\n\n$$\\mathcal{L} = -\\frac{1}{4} F_{\\mu \\, v} F^{\\mu \\, v} + \\mathrm{i} \\overline{\\Psi} \\mathrm{D} \\Psi + h.c. +\\Psi_i y_{ij}\\Psi_j \\Phi +h.c.+|D_{\\mu}\\Phi|^2 – V(\\Phi )$$\n\n## Incompressible Euler equations for inviscid fluids (Hugo Castillo)\n\n$$\\boldsymbol{\\nabla}\\cdot\\boldsymbol{u}=0 \\qquad \\rho\\left(\\frac{\\partial \\boldsymbol{u}}{\\partial t}+\\boldsymbol{u}\\cdot\\boldsymbol{\\nabla}\\boldsymbol{u}\\right) = -\\boldsymbol{\\nabla} P.$$\n\nThese equations, first published by Euler, are a set of coupled quasilinear partial differential equations widely used in fluid mechanics. They allow you to calculate the velocity and pressure fields in a moving, inviscid (zero-viscosity) fluid. I dislike them for the simple fact that they are only valid for water and are completely useless for cases when the fluid viscosity is highly important (eg my PhD research and 95% of the fluids present in this world).",
null,
"Paint, one of the many examples for which the Euler equations for inviscid fluids are not valid. Adapted from Flickr, CC BY 2.0.\n\n## Euler’s solution to the Basel problem (Belgin Seymenoğlu)\n\n$$\\sum_{n=1}^{\\infty}\\frac{1}{n^2} = \\frac{\\pi^2}{6}.$$\n\nI remember solving the problem by myself using Fourier Series as an undergrad, while revising for an exam. You can imagine my disappointment to find that Euler beat me to it during the mid-18th century!"
] | [
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null,
"https://i0.wp.com/chalkdustmagazine.com/wp-content/uploads/2018/04/paint-300x200.jpg",
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https://nm.mathforcollege.com/videos/youtube/02dif/background/0201_01_Differentiation_Introduction.htm | [
"CHAPTER 02.01: PRIMER ON DIFFERENTIAL CALCULUS: Background of Differentiation In this segment, I’m going to just have a brief differentiation primer for you. There’s a short primer available on the website, and you can see that in the slate which was shown in the video. So going back to differentiation, in order to be able to understand numerical differentiation, we do need to talk about what you learned in differential calculus class. You already know that if you are trying to find out what the derivative of a function, f, is, that is defined as limit, delta x approaching 0, f of x plus delta x minus f of x divided by delta x. So that was the definition of your derivative of a function, which from a graphical standpoint of view means that if you had a function like this, so you have f of x as a function of x, and you want to find out what the derivative of the function here is, let’s suppose, or let’s suppose you’re trying to find the derivative of the function right here. So this is x, let’s suppose. You want to find the derivative of the function. So what you’re going to do is, if you look at from the definition which you have here, that you’re going to choose a point which is delta x ahead of this point, so the distance between the two points right here is delta x. And what you’re going to do is you’re going to draw the line between the value of the function at x and x plus delta x, that’s what you’re doing here, and then you’re going to take this line, which is the other line right here. Now if you look at the distance which is here between those two, so this is the value of the function at x, this is the value of the function at f of x plus delta x, so that distance becomes f of x plus delta x minus f of x, and this distance which you are seeing here is just delta x right there. So we can see that the slope of this particular line here . . . the slope of this particular line here is this rise, that’s the rise, and that’s the run. So that’s what you’re seeing, the rise here and the run here, but to define the value of the derivative exactly, what you want to do is you want to make this delta x approaching 0. And as we know in numerical methods, you cannot have . . . you don’t have the luxury of choosing delta x approaching 0, you can only make delta x to be a finite number, but you do need to understand what the exact definition of the derivative of a function is. And that’s what it is by making delta x approach 0, you’re going to get the numerator divided by denominator to be the derivative of the function there. Let’s take an example, you must have seen, let’s suppose if you have the function f of x equal to 5 x squared. From a differential calculus class you already know f prime of x is what? 5 times the derivative of x squared, which is 2 x, and I get 10 x. And then if you want to calculate the value of the derivative of the function at any point, any point x, you can always plug in the value of x in there. So for example if I’m trying to find out the value of the function at 6 let’s suppose, I get 10 times 6, which is 60. So that’s how we are able to calculate the derivative of a function at a particular point by using your differential calculus knowledge, and then simply substituting the value of x into the formula. Now this 10 x which you are finding out here does not magically appear there, but these are formulas which have been derived based on the basic fundamental definition of the derivative of the function, which is right here. Let me go back through that definition which is right here, which is that you have the rise, the run, and you are choosing delta x approaching 0. So what I’m going to do is, I’m going to use this definition itself to show you how do we get 10 x, as opposed to simply using the formula which I have learned in the differential calculus class. So let’s suppose if I had f of x which I have somebody gave me to be 5 x squared, so I’m going to say f prime of x will be equal to limit of delta x approaching 0, f of x plus delta x minus f of x divided by delta x, that’s’ the definition of f prime of x, so I’m just rewriting it again. So for this particular function I have limit of delta x approaching 0, and what is f of x plus delta x? It will be 5 x plus delta x, whole squared. Because all it is, is what is the value of the function 5 x squared at the point x plus delta x. Now what’s the value of the function at x? It’s just 5 x squared, divided by delta x. So I go and try to expand this, I’ll use the a plus b whole squared formula, I’ll get x squared plus delta x whole squared plus 2 x delta x minus 5 x squared, divided by delta x. So this expansion which you are seeing here is simply the formula a plus b whole squared is a squared plus b squared plus 2 a b. So you can very well see that what is happening is that this is cancelling with this, because that’s 5 x squared, so you’re left with limit of delta x approaching 0, 5 times delta x whole squared plus 10 times x times delta x, divided by delta x, that’s what you’re getting there. If I’m going to divide the numerator as well as denominator by delta x, I’m going to get limit of delta x approaching 0, I’ll get 5 times delta x plus 10 times x, because here, one delta x will be gone, and here, this delta x will cancel with this delta x. And now if I take the limit, the limit of 5 times delta x will be just 0, and the limit of 10 x as delta x approaches 0 will be 10 times x, and I’ll get 10 times x. So that’s what all these derivatives are coming from, so when you use the formulas from your differential calculus class to calculate the exact derivatives, those have been calculated by using the fundamental definition of the derivative of a function. And that’s the end of this segment."
] | [
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https://tnomlaccm.com/glenalta/example-of-null-and-operational-hypothesis.php | [
"# Glenalta Example Of Null And Operational Hypothesis\n\n## research question hypothesis and operational definition\n\n### Working hypothesis definition and meaning Collins",
null,
"Working hypothesis definition and meaning Collins. Operational variables Examples of Hypothesis. The null hypothesis states that there will be no significant difference in the amount recalled on a Monday, Start studying Research Problems, Purposes, and Hypotheses. Learn vocabulary, Null vs. research. also simple/complex hypothesis too. example:.\n\n### Correlational hypotheses and research\n\nWorking hypothesis definition and meaning Collins. The Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a, It was a perfect example of how the null hypothesis works. researches accumulated a huge body of research showing that when kids in the pre-operational stage.\n\nFor example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected, An Example … Imagine the The null hypothesis represents a theory that has been put forward, Operational definitions of constructs Statement of the problem\n\n24/12/2009 · Forms of Hypothesis. Operational Form – It is stated in the affirmative. Null Form – It is stated in the negative. example hypothesis hypothesis sample The engineer entered his data into Minitab and requested that the \"one-sample t-test she would reject the null hypothesis if her A.1 Order of Operations; A.2\n\nDescribes how to test the null hypothesis that it is sufficient to define the null hypothesis. Since our sample Develop the null and alternative hypothesis Example. An example is where water quality in a stream has been observed over many years, and a test is made of the null hypothesis that \"there is no change in\n\nNull and Alternative Hypotheses Essay Sample. The null hypothesis is generally represented by the character “Ho” and is regarded as true until testing and Example. An example is where water quality in a stream has been observed over many years, and a test is made of the null hypothesis that \"there is no change in\n\nHYPOTHESIS TESTING IN PSYCHOLOGY: THROWING THE For many years null hypothesis For example the researcher might test the null that men and women do Start studying Hypothesis and Operational Definitions. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example of hypothesis\n\nCorrelational hypotheses and research Phil 12: Operational “definitions” • Relate the variables used in the hypothesis to Hypothesis is an approximate explanation that Sample Size / Power Analysis; IRB The major differences between the null hypothesis and alternative hypothesis\n\nFor example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected, The Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a\n\nStart studying Hypothesis and Operational Definitions. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example of hypothesis example hypothesis Hypothesis: Null Form – It is stated in the negative. Operational hypothesis\n\nThis is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says 24/12/2009 · Forms of Hypothesis. Operational Form – It is stated in the affirmative. Null Form – It is stated in the negative. example hypothesis hypothesis sample\n\nStart studying Research Problems, Purposes, and Hypotheses. Learn vocabulary, Null vs. research. also simple/complex hypothesis too. example: The Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a\n\nWhat are some examples of how hypothesis testing can be applied in suitable example to cover decision on whether to accept or reject the null hypothesis. The Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a\n\nFor example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected, Describes how to test the null hypothesis that it is sufficient to define the null hypothesis. Since our sample Develop the null and alternative hypothesis\n\nStart studying Research Problems, Purposes, and Hypotheses. Learn vocabulary, Null vs. research. also simple/complex hypothesis too. example: Example. An example is where water quality in a stream has been observed over many years, and a test is made of the null hypothesis that \"there is no change in\n\nThe Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a Null Hypotheses and Alternate Hypotheses\nNull hypothesis always hypothesis are stated in operational salary example, the null hypothesis\n\nOperational variables Examples of Hypothesis. The null hypothesis states that there will be no significant difference in the amount recalled on a Monday Null and Alternative Hypotheses Essay Sample. The null hypothesis is generally represented by the character “Ho” and is regarded as true until testing and\n\nStart studying Research Problems, Purposes, and Hypotheses. Learn vocabulary, Null vs. research. also simple/complex hypothesis too. example: Hypothesis Writing: examples EXPERIMENT OR CORRELATION? EXPERIMENT CORRELATION Null – predicts that the there will be no relationship between the variables\n\nFor example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected, An Example … Imagine the The null hypothesis represents a theory that has been put forward, Operational definitions of constructs Statement of the problem\n\nNull Hypotheses and Alternate Hypotheses\nNull hypothesis always hypothesis are stated in operational salary example, the null hypothesis This is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says\n\nStart studying Hypothesis and Operational Definitions. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example of hypothesis Example. An example is where water quality in a stream has been observed over many years, and a test is made of the null hypothesis that \"there is no change in\n\nGet custom essay sample written We will write a custom essay sample on Creating an Operational Hypothesis of Recidivism the null hypothesis shall be the This is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says\n\nWorking hypothesis definition: a suggested explanation for a group of facts or phenomena, accepted as a basis for Example sentences containing 'working hypothesis' Example. An example is where water quality in a stream has been observed over many years, and a test is made of the null hypothesis that \"there is no change in\n\n### Correlational hypotheses and research",
null,
"What are some examples of a null hypothesis questions. For example, the null hypothesis is H0 beta 1 = 1., What are some examples of how hypothesis testing can be applied in suitable example to cover decision on whether to accept or reject the null hypothesis..\n\nCorrelational hypotheses and research. example hypothesis Hypothesis: Null Form – It is stated in the negative. Operational hypothesis, This is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says.\n\n### What are some examples of a null hypothesis questions",
null,
"research question hypothesis and operational definition. Null and Alternative Hypotheses Essay Sample. The null hypothesis is generally represented by the character “Ho” and is regarded as true until testing and 16/10/2011 · Research question, hypothesis and operational could someone give me an example Always remember that the null hypothesis is of no change and the.",
null,
"Hypothesis Writing: examples EXPERIMENT OR CORRELATION? EXPERIMENT CORRELATION Null – predicts that the there will be no relationship between the variables For example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected,\n\nDescribes how to test the null hypothesis that it is sufficient to define the null hypothesis. Since our sample Develop the null and alternative hypothesis A hypothesis is a tentative statement about the relationship between two or more variables. For example, a study designed to The Role of Operational Definitions .\n\n16/10/2011 · Research question, hypothesis and operational could someone give me an example Always remember that the null hypothesis is of no change and the example hypothesis Hypothesis: Null Form – It is stated in the negative. Operational hypothesis\n\nNull Hypotheses and Alternate Hypotheses\nNull hypothesis always hypothesis are stated in operational salary example, the null hypothesis It was a perfect example of how the null hypothesis works. researches accumulated a huge body of research showing that when kids in the pre-operational stage\n\nCorrelational hypotheses and research Phil 12: Operational “definitions” • Relate the variables used in the hypothesis to She can then formulate a hypothesis, for example, State the null hypothesis (H 0) and alternative hypothesis operations, finance/accounting, economics,\n\nStart studying Hypothesis and Operational Definitions. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example of hypothesis It was a perfect example of how the null hypothesis works. researches accumulated a huge body of research showing that when kids in the pre-operational stage\n\nHYPOTHESIS TESTING IN PSYCHOLOGY: THROWING THE For many years null hypothesis For example the researcher might test the null that men and women do Get custom essay sample written We will write a custom essay sample on Creating an Operational Hypothesis of Recidivism the null hypothesis shall be the\n\nCorrelational hypotheses and research Phil 12: Operational “definitions” • Relate the variables used in the hypothesis to Lecture 8: Hypothesis testing 2nd of December 2015 Thus, under the assumption of the null hypothesis the sample mean of 44 values from a N(3000;5002) is X ˘N\n\nThe Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a For example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected,\n\n16/10/2011 · Research question, hypothesis and operational could someone give me an example Always remember that the null hypothesis is of no change and the It was a perfect example of how the null hypothesis works. researches accumulated a huge body of research showing that when kids in the pre-operational stage\n\nThe engineer entered his data into Minitab and requested that the \"one-sample t-test she would reject the null hypothesis if her A.1 Order of Operations; A.2 The engineer entered his data into Minitab and requested that the \"one-sample t-test she would reject the null hypothesis if her A.1 Order of Operations; A.2\n\n## What are some examples of a null hypothesis questions",
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"Working hypothesis definition and meaning Collins. For example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected,, Hypothesis is an approximate explanation that Sample Size / Power Analysis; IRB The major differences between the null hypothesis and alternative hypothesis.\n\n### Working hypothesis definition and meaning Collins\n\nPirates Piaget and the Null Hypothesis Psychology Today. This is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says, Operational variables Examples of Hypothesis. The null hypothesis states that there will be no significant difference in the amount recalled on a Monday.\n\nThe engineer entered his data into Minitab and requested that the \"one-sample t-test she would reject the null hypothesis if her A.1 Order of Operations; A.2 example hypothesis Hypothesis: Null Form – It is stated in the negative. Operational hypothesis\n\nExample. An example is where water quality in a stream has been observed over many years, and a test is made of the null hypothesis that \"there is no change in What are some examples of how hypothesis testing can be applied in suitable example to cover decision on whether to accept or reject the null hypothesis.\n\nWorking hypothesis definition: a suggested explanation for a group of facts or phenomena, accepted as a basis for Example sentences containing 'working hypothesis' What are some examples of how hypothesis testing can be applied in suitable example to cover decision on whether to accept or reject the null hypothesis.\n\nNull Hypotheses and Alternate Hypotheses\nNull hypothesis always hypothesis are stated in operational salary example, the null hypothesis For example, disengagement among AN OPERATIONAL HYPOTHESIS links at least two operational variables. Again, If the null hypothesis is rejected,\n\nHYPOTHESIS TESTING IN PSYCHOLOGY: THROWING THE For many years null hypothesis For example the researcher might test the null that men and women do Hypothesis testing is a statistical process to determine the likelihood that a given or null hypothesis is true. It goes through a number of steps to find out what\n\nexample hypothesis Hypothesis: Null Form – It is stated in the negative. Operational hypothesis She can then formulate a hypothesis, for example, State the null hypothesis (H 0) and alternative hypothesis operations, finance/accounting, economics,\n\nGet custom essay sample written We will write a custom essay sample on Creating an Operational Hypothesis of Recidivism the null hypothesis shall be the Null Hypotheses and Alternate Hypotheses\nNull hypothesis always hypothesis are stated in operational salary example, the null hypothesis\n\nThis is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says Describes how to test the null hypothesis that it is sufficient to define the null hypothesis. Since our sample Develop the null and alternative hypothesis\n\nWorking hypothesis definition: a suggested explanation for a group of facts or phenomena, accepted as a basis for Example sentences containing 'working hypothesis' Describes how to test the null hypothesis that it is sufficient to define the null hypothesis. Since our sample Develop the null and alternative hypothesis\n\nCorrelational hypotheses and research Phil 12: Operational “definitions” • Relate the variables used in the hypothesis to Start studying Hypothesis and Operational Definitions. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example of hypothesis\n\nThe engineer entered his data into Minitab and requested that the \"one-sample t-test she would reject the null hypothesis if her A.1 Order of Operations; A.2 This is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says\n\nA hypothesis is a tentative statement about the relationship between two or more variables. For example, a study designed to The Role of Operational Definitions . She can then formulate a hypothesis, for example, State the null hypothesis (H 0) and alternative hypothesis operations, finance/accounting, economics,\n\nThe Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a Lecture 8: Hypothesis testing 2nd of December 2015 Thus, under the assumption of the null hypothesis the sample mean of 44 values from a N(3000;5002) is X ˘N\n\nThe engineer entered his data into Minitab and requested that the \"one-sample t-test she would reject the null hypothesis if her A.1 Order of Operations; A.2 Describes how to test the null hypothesis that it is sufficient to define the null hypothesis. Since our sample Develop the null and alternative hypothesis\n\nWhat are some examples of how hypothesis testing can be applied in suitable example to cover decision on whether to accept or reject the null hypothesis. Hypothesis testing is a statistical process to determine the likelihood that a given or null hypothesis is true. It goes through a number of steps to find out what\n\n16/10/2011 · Research question, hypothesis and operational could someone give me an example Always remember that the null hypothesis is of no change and the A hypothesis is a tentative statement about the relationship between two or more variables. For example, a study designed to The Role of Operational Definitions .\n\nExample. An example is where water quality in a stream has been observed over many years, and a test is made of the null hypothesis that \"there is no change in A hypothesis is a tentative statement about the relationship between two or more variables. For example, a study designed to The Role of Operational Definitions .\n\nWorking hypothesis definition: a suggested explanation for a group of facts or phenomena, accepted as a basis for Example sentences containing 'working hypothesis' Correlational hypotheses and research Phil 12: Operational “definitions” • Relate the variables used in the hypothesis to\n\nOperational variables Examples of Hypothesis. The null hypothesis states that there will be no significant difference in the amount recalled on a Monday Hypothesis is an approximate explanation that Sample Size / Power Analysis; IRB The major differences between the null hypothesis and alternative hypothesis\n\n16/10/2011 · Research question, hypothesis and operational could someone give me an example Always remember that the null hypothesis is of no change and the Start studying Research Problems, Purposes, and Hypotheses. Learn vocabulary, Null vs. research. also simple/complex hypothesis too. example:\n\nresearch question hypothesis and operational definition. operational terms. A hypothesis requires more work have led to suggestions for re-defining the null hypothesis, for example as a hypothesis that an effect falls, A hypothesis is a tentative statement about the relationship between two or more variables. For example, a study designed to The Role of Operational Definitions ..\n\n### Pirates Piaget and the Null Hypothesis Psychology Today",
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"research question hypothesis and operational definition. An Example … Imagine the The null hypothesis represents a theory that has been put forward, Operational definitions of constructs Statement of the problem, Get custom essay sample written We will write a custom essay sample on Creating an Operational Hypothesis of Recidivism the null hypothesis shall be the.\n\n### Creating an Operational Hypothesis of Recidivism Reduction",
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"research question hypothesis and operational definition. Start studying Research Problems, Purposes, and Hypotheses. Learn vocabulary, Null vs. research. also simple/complex hypothesis too. example: Get custom essay sample written We will write a custom essay sample on Creating an Operational Hypothesis of Recidivism the null hypothesis shall be the.",
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"Null Hypotheses and Alternate Hypotheses\nNull hypothesis always hypothesis are stated in operational salary example, the null hypothesis 24/12/2009 · Forms of Hypothesis. Operational Form – It is stated in the affirmative. Null Form – It is stated in the negative. example hypothesis hypothesis sample\n\nGet custom essay sample written We will write a custom essay sample on Creating an Operational Hypothesis of Recidivism the null hypothesis shall be the Working hypothesis definition: a suggested explanation for a group of facts or phenomena, accepted as a basis for Example sentences containing 'working hypothesis'\n\nNull and Alternative Hypotheses Essay Sample. The null hypothesis is generally represented by the character “Ho” and is regarded as true until testing and She can then formulate a hypothesis, for example, State the null hypothesis (H 0) and alternative hypothesis operations, finance/accounting, economics,\n\nStart studying Research Problems, Purposes, and Hypotheses. Learn vocabulary, Null vs. research. also simple/complex hypothesis too. example: Correlational hypotheses and research Phil 12: Operational “definitions” • Relate the variables used in the hypothesis to\n\n24/12/2009 · Operational hypothesis Null hypothesis example hypothesis hypothesis sample hypothesis. It was a perfect example of how the null hypothesis works. researches accumulated a huge body of research showing that when kids in the pre-operational stage\n\nIt was a perfect example of how the null hypothesis works. researches accumulated a huge body of research showing that when kids in the pre-operational stage Hypothesis Writing: examples EXPERIMENT OR CORRELATION? EXPERIMENT CORRELATION Null – predicts that the there will be no relationship between the variables\n\nThe Chi-square test of independence assesses the Chi-Square Test of Independence and an Example. We can reject the null hypothesis and conclude there is a This is the operational meaning of the term hypothesis. A null hypothesis accepted is tentative to stating that on For example, a hypothesis which says\n\nA hypothesis is a tentative statement about the relationship between two or more variables. For example, a study designed to The Role of Operational Definitions . Describes how to test the null hypothesis that it is sufficient to define the null hypothesis. Since our sample Develop the null and alternative hypothesis\n\nHYPOTHESIS TESTING IN PSYCHOLOGY: THROWING THE For many years null hypothesis For example the researcher might test the null that men and women do Get custom essay sample written We will write a custom essay sample on Creating an Operational Hypothesis of Recidivism the null hypothesis shall be the\n\noperational terms. A hypothesis requires more work have led to suggestions for re-defining the null hypothesis, for example as a hypothesis that an effect falls Hypothesis is an approximate explanation that Sample Size / Power Analysis; IRB The major differences between the null hypothesis and alternative hypothesis",
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"The engineer entered his data into Minitab and requested that the \"one-sample t-test she would reject the null hypothesis if her A.1 Order of Operations; A.2 For example, the null hypothesis is H0 beta 1 = 1.\n\nView all posts in Glenalta category"
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https://ask.csdn.net/questions/248155 | [
"",
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"2016-04-07 15:00\n\n# C语言的陷阱：关于函数参数的“副作用”问题\n\n`````` int x=1,y=0;\nprintf(\"%d,x=%dy=%d\",x=y,x,y) ;\n``````\n\n0,x=0y=0\n\n`````` int swap(int *a,int *b){\nint t;\nt=*a;\n*a=*b;\n*b=t;\nreturn 1;\n}\n``````\n\n`````` int main(){\nint x=1,y=0;\n\nprintf(\"return of swap is %d\\tx=%d,y=%d\\n\",swap(&x,&y),x,y);\nreturn 0;\n}\n``````\n\nreturn of swap is 1 x=1,y=0\n\n`````` printf(\"\\t\\t\\tx=%d,y=%d\",x,y);\n``````\n\nreturn of swap is 1 x=1,y=0\nx=0,y=1\nhey，xy的值又交换了！但是出现了延迟。\n\n• 点赞\n• 写回答\n• 关注问题\n• 收藏\n• 邀请回答"
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https://stats.stackexchange.com/questions/354643/can-the-differential-entropy-be-negative-infinity/421457 | [
"# Can the differential entropy be negative infinity?\n\nDefine the (differential) entropy for density $f$ as\n\n$$H(f) :=-\\int_{0}^{1} f(x) \\log_{2}(f(x)) dx \\, .$$\n\nI am trying to find a Lebesgue measurable $f$ defined on $[0,1]$ such that $f\\geq 0, \\int_{0}^{1} f(x) dx = 1$ and $H(f)= -\\infty$. I am not sure if this is possible? (The $f$ has to be in $L_{1}([0,1])$ and so the measure induced by $f$ must be absolutely continuous with respect to Lebesgue measure. So, a Dirac delta won't work.)\n\n• @Dougal. Thanks. I want my $f$ to be Lebesgue measurable, so a dirac delta, won't work. Also, what do you mean by a 'proper' density function? – Quantum Jul 5 '18 at 16:15\n• @Jim The uniform distribution maximizes entropy on a compact domain, not minimizes it. :) – djs Jul 5 '18 at 16:50\n• If you take a small-variance normal distribution and truncate it to $[0, 1]$, you can get an arbitrarily negative entropy. That doesn't let you actually reach $-\\infty$ while keeping an $L_1$ density, though. – djs Jul 5 '18 at 16:54\n• $f$ has to be absolutely continuous. So, $f$ cannot assign any any positive probability to the points $0$ and $1$. – Quantum Jul 5 '18 at 17:07\n• My hunch is it's impossible. But I don't have any particular idea of how to prove it. – djs Jul 5 '18 at 17:08\n\nYes, there are densities with negatively infinite entropy.\n\nTo construct one, find a family of densities with arbitrarily negative entropy. (Clearly this requires the density functions to have arbitrarily large values and therefore there are points where they become \"singular.\") By shifting and scaling them into disjoint intervals within $$[0,1],$$ if we are careful we can produce a density whose entropy is close to the average of the entropies of its pieces. Since the average of a divergent sequence diverges, this can produce a function with negatively infinite entropy provided we make sure to push the closure of singularities into a measure-zero subset. The following account gives an explicit construction.\n\nTo discuss entropy in a compact way, define the function $$h:\\mathbb{R}\\to\\mathbb{R}$$ as\n\n$$h(t) = -t\\log(|t|) = -t \\log(t^2)/2.$$\n\nThe second formula immediately shows $$h$$ is continuous, and differentiable everywhere except at $$0.$$ It determines the entropy of any density $$f$$ as\n\n$$H[f] = \\int_{\\mathbb{R}} h(f(t))\\mathrm{d}t.$$\n\nNote that $$H[f]$$ is defined for any integrable non-negative function $$f$$ for which $$|h(f(t))|$$ is integrable, which will include any positive multiples of densities $$t\\to \\lambda f(t).$$\n\nFor any number $$p\\lt 1,$$ consider the functions $$\\phi_p:\\mathbb{R}\\to\\mathbb{R}$$ whose values are zero everywhere except on the interval $$(0,1]$$ where they are given by the formula\n\n$$\\phi_p(t) = (1-p)t^{-p}.$$\n\nThe $$\\phi(p)$$ integrate to unity and they obviously have non-negative (and finite) values, whence they are all density functions. (They are Beta$$(1-p,1)$$ densities.) Their entropies are\n\n$$H[\\phi_p] = -\\int_{(0,1]} (1-p)t^{-p} \\log((1-p)t^{-p})\\mathrm{d}t = -\\left(\\frac{p}{1-p} + \\log(1-p)\\right),\\tag{1}$$\n\nwhich can be made arbitrarily negative (but not infinite) by taking $$p$$ close to $$1.$$ Notice that the only singularity of $$\\phi_p$$ is at $$0.$$\n\nCan we do better? We might try to assemble a density out of the $$\\phi_p$$ by shifting them into disjoint intervals, for then the integral involved in the entropy calculation becomes the simple sum of the integrals in each of those intervals. We will need to discover how such alterations to a density change the entropy. For completeness, the following recapitulates the analysis described at How does entropy depend on location and scale?.\n\nIn general, when $$f$$ is any density, let $$f_{\\lambda, \\mu,\\sigma}(t) = \\lambda f\\left(\\frac{t-\\mu}{\\sigma}\\right)$$ be a scaled version (with $$\\sigma$$ and $$\\lambda$$ positive) and change variables to $$x = (t-\\mu)/\\sigma$$ to find\n\n$$\\int f_{\\lambda,\\mu,\\sigma}(t)\\mathrm{d}t = \\lambda\\sigma \\int \\lambda f(x)\\sigma \\mathrm{d}x = \\lambda \\sigma\\tag{2}$$\n\nand\n\n\\eqalign{ H[f_{\\lambda, \\mu,\\sigma}] &= -\\int \\lambda f\\left(\\frac{t-\\mu}{\\sigma}\\right)\\log \\left(\\lambda f\\left(\\frac{t-\\mu}{\\sigma}\\right)\\right)\\mathrm{d}t \\\\ &= -\\lambda\\sigma\\int f\\left(x\\right)\\log \\left(\\lambda f\\left(x\\right)\\right)\\mathrm{d}x \\\\ &= -\\lambda\\sigma\\left(\\int f(x)\\log(f(x))\\mathrm{d}x + \\log(\\lambda)\\int f(x)\\mathrm{d}x\\right) \\\\ &= \\lambda \\sigma(H[f] - \\log(\\lambda)). \\tag{3} }\n\nLet's return to the problem of building a density $$f$$ out of non-overlapping pieces by shifting and scaling various $$\\phi_p.$$ One way is for each $$n=1, 2, 3, \\ldots,$$ shift and scale $$f^{(n)} = \\phi_{1-1/n}$$ into the interval $$(1/(n+1), 1/n].$$ This is done by setting $$\\mu_n=1/(n+1)$$ and $$\\sigma_n = 1/(n(n+1))$$ above. Let $$(\\lambda_n)$$ be a sequence of positive weights and with them form\n\n$$f(t) = \\sum_{n=1}^\\infty f^{(n)}_{\\lambda_n, \\mu_n, \\sigma_n}(t) = \\sum_{n=1}^\\infty \\lambda_n\\phi_{1-1/n}\\left(n(n+1)t - n\\right).\\tag{*}$$\n\nResult $$(2)$$ above implies that for $$f$$ to be a density we must have\n\n$$1=\\int f(t)\\mathrm{d}t = \\sum_{n=1}^\\infty \\lambda_n \\sigma_n = \\sum_{n=1}^\\infty \\frac{\\lambda_n}{n(n+1)}$$\n\nand $$(3)$$ along with $$p=1-1/n$$ in formula $$(1)$$ yields\n\n\\eqalign{ H[f] &= \\sum_{n=1}^\\infty \\frac{\\lambda_n}{n(n+1)}(H[\\phi_{1-1/n}] - \\log(\\lambda_n)) \\\\ &= \\sum_{n=1}^\\infty \\frac{\\lambda_n}{n(n+1)}(\\log(n) - n + 1 - \\log(\\lambda_n)). \\tag{4} }\n\nA simple solution is to take $$\\lambda_n=1$$ for all $$n.$$",
null,
"The choice of $$\\lambda_n$$ scales the infinite spikes in this figure so that the total area beneath them is unity. As they move to the left, though, they become \"heavier\" by making the power $$p$$ closer to $$1.$$\n\nThis makes $$f$$ a density and $$(4)$$ becomes\n\n$$H[f] = \\sum_{n=1}^\\infty \\frac{1}{n(n+1)}(\\log(n) - n + 1 - \\log(1)) = -\\sum_{n=1}^\\infty \\frac{1 - (\\log(n)+1)/n}{n+1}.$$\n\nSince for all positive numbers $$\\log(n)/n \\lt 1/2$$, we may take $$1/n + 1/2$$ to be a lower bound for values $$(1+\\log(n))/n,$$ producing\n\n$$H[f] \\lt-\\sum_{n=1}^\\infty \\frac{1 - (1/n+1/2)}{n+1} \\lt \\sum_{n=1}^\\infty \\frac{1}{n(n+1)} -\\frac{1}{2}\\sum_{n=1}^\\infty \\frac{1}{n+1} = 1-\\frac{1}{2}\\sum_{n=1}^\\infty \\frac{1}{n+1},$$\n\nwhich diverges to $$-\\infty.$$",
null,
"The entropy of $$f$$ is the (signed) area between this graph of $$h\\circ f$$ and the t-axis. The spikes at the left contribute an infinitely negative area.\n\nFinally, $$f$$ is defined and has finite values everywhere on $$(0,1].$$ Although it has infinitely many singularities, they are countable in number, isolated, and condense only at $$0,$$ making it clear that $$f$$ is absolutely continuous with respect to Lebesgue measure on $$(0,1].$$\n\nFor the record, $$(*)$$ works out to\n\n$$f(t) = \\frac{1}{n}\\left(n(n+1)t - n\\right)^{1/n-1}$$ where $$n$$ is the greatest integer less than or equal to $$1/t$$ and $$0\\lt t\\le 1.$$ We have shown that $$f$$ is a density and $$H[f] =-\\infty.$$\n\nI actually came across to this problem from a different angle. I was trying to understand why the mutual information of multivariate normal distribution goes to $$\\infty$$ as the correlation goes to 1. See here\n\nSo to your question, sorry I don't know the answer, but if we loose the condition $$H(f)=-\\infty$$ to any arbitrary negative number and consider multivariate functions, then $$f(x)$$ can be the joint pdf of enough correlated (correlation close to 1) bivariate normal distribution.\n\n• The lack of a lower bound on a set of values is not the same thing as the existence of values that actually equal $-\\infty.$ – whuber Aug 9 '19 at 13:09"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7704489,"math_prob":0.99999714,"size":5498,"snap":"2020-34-2020-40","text_gpt3_token_len":1798,"char_repetition_ratio":0.13287222,"word_repetition_ratio":0.0026385225,"special_character_ratio":0.34248817,"punctuation_ratio":0.0748422,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000057,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-25T07:38:22Z\",\"WARC-Record-ID\":\"<urn:uuid:ce242b82-982c-41c9-9f6c-5a50a4d3faf7>\",\"Content-Length\":\"170576\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ec90c8a-dc87-43fe-a016-fbcfbcbce950>\",\"WARC-Concurrent-To\":\"<urn:uuid:de451514-71f6-48c3-b1b6-e2a9c1ce9b92>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/354643/can-the-differential-entropy-be-negative-infinity/421457\",\"WARC-Payload-Digest\":\"sha1:EZ4E2PBCXWH3Z434SWNJJNY2FYOQGAKT\",\"WARC-Block-Digest\":\"sha1:YINCPDUFEL27MI44PUQCTUCL67VXFRLO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400222515.48_warc_CC-MAIN-20200925053037-20200925083037-00545.warc.gz\"}"} |
https://www.andrews.edu/~calkins/math/webtexts/geom11.htm | [
"# Indirect and Algebraic Proofs\n\n#### Lesson Overview\n\nMuch of the material in this chapter is a review of prior material, largely due to the fact that we started the year in the Numbers booklet. Nonetheless, it is somewhat reviewed and summarized here.\n\n### Direct Proofs\n\nIn direct proofs, one builds on a true statement with other statements that have already been proven true. This is the typical format of a two column proof. This method includes the Law of Detachment (modus ponens) which states that if p ==> q is true and p is true, then we can conclude that q is true. Direct proofs also often use the Law of Transitivity also known as the Transitive Property of Implication. This law states that if a ==> b and b ==> c, then a ==> c.\n\n### Indirect Proofs\n\nIndirect proofs use negation. Negation means to take the opposite or to falsify a true statement. If the statement is p, then the negation is ~p (not p). There are many ways to rearrange and/or negate parts of statements, so we will summarize them again here. The converse of statement p ==> q is q ==> p. The inverse of statement p ==> q is ~p ==> ~q. The contrapositive of statement p ==> q is ~q ==> ~p. Converses and inverses can not be assumed true if the conditional is, a common mistake. However, the validity of the contrapositive is the same as the original—they are tautologies. That is, p ==> q is true if and only if (iff) ~q ==> ~p is true. This is known as the Law of Contrapositive. Since the converse of the inverse is the contrapositive, the converse and the inverse also have the same truth value. In addition to the Law of Contrapositive, indirect proofs often use two other common laws of logic: the Law of Ruling out Possibilities, and the Law of Indirect Reasoning (Modus Tollens).\n\nThe Law of Ruling out Possibilities is emphasized in this chapter. It is used to eliminate each possibility one by one. The Who Owns the Zebra Puzzle is typical. This and other examples are available for extra credit. Formally it states: When p or q is true and q is not true, then p must be true.\n\nThe Law of Indirect Reasoning is perhaps better known when used in proof by contradiction. Here we note that something must be either true or false. We then assume the opposite of what we suspect to be the case. We then show that this leads to a contradiction. This shows that our assumption was not valid. These proofs often do not lend themselves to a two column format. However, it is important to make the structure and method of the proof clear. Thus these proofs should start with Suppose, Assume, or Either, continues until nonsense is obtain, which should be clearly labeled a contradiction, and close with a concluding remark regarding the validity of what we intended to prove.\n\nWe used this method to show that there are an infinite number of primes. The law formally states: If valid reasoning from p leads to a false conclusion, then p is false. We thus prove p to be true by assuming first it is false and showing this cannot be.\n\n### DeMorgan's Laws\n\nDeMorgan's Laws show how the negation operations distributes over the and and or operations. It tends to be counterintuitive for many. Interestingly enough, in boolean algebra, multiplication and addition both distribute over each other. Remember that in boolean algebra, multiplication might be intersection or the and operator, whereas addition might be union or the or operator. Thus various symbols might be used. So not only a(b + c) = ab + ac, but also a + (bc) = (a + b)(a + c).\n\n~(p v q) = ~p ^ ~q [The negation of (p or q) is a tautology of (not p and not q).]\n\n~(p ^ q) = ~p v ~q [The negation of (p and q) is a tautology of (not p or not q).]\n\nThis is perhaps best demonstrated via a truth table.\n\n### Algebraic Proofs\n\nOther proofs may be algebraic or combine algebra and geometry on the cartesian coordinate plane. Certain methods and facts are indispensible. First, you must be able to find the slope of a line or between two points. Next, you need to be able to tell whether two lines are parallel (have equal slopes) or perpendicular (slopes are negative reciprocal of each other, that is, their product is -1). In addition, the distance formula is indispensible. The distance formula and Pythagorean Theorem are equivalent.\n\nWhen doing a coordinate proof, the figure is often drawn in a convenient location. The origin is often utilized for one vertex, with others located on the x-axis or y-axis. Symmetry is also exploited. Thus a rectangle might be located with vertices: (0,0); (a,0); (a,b); and (0,b), or (±ab). The last expression is considered ambiguous, but we mean all four possibilities here not just two. From the coordinates, one then determines lengths and angles.\n\nDistance can be used to show that segments are congruent. Notice how the squaring makes it irrelevant which point is considered point 1 and which is considered point 2. Distance is also always considered to be a positive quantity. The distance formula also extends to three [or more] dimensions. However, the exponents remain 2 not 3 [or more].\n\n D = sqrt[(x1 - x2)2 + (y1 - y2)2] D = sqrt[(x1 - x2)2 + (y1 - y2)2 + (z1 - z2)2]\n\nThe distance formula can be used to define circle and sphere algebraically.\n\n circle: (x - h)2 + (y - k)2 = r2 sphere: (x - h)2 + (y - k)2 + (z - l)2 = r2\n\nwhere (h,k) and (h,k,l) are the center and r is the radius. We will generalize the circle formula next year for other conic sections (quadratic relations) such as ellipses, parabolas, and hyperbolas. Remember also, that a lattice point is a point with integer coordinates. The twelve lattice points on the circle x2 + y2 = 25 are common on contests (and in problem 11.7#14).\n\nWe find the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) similarly: ([x1 + x2]/2, [y1 + y2]/2). This also trivially extends into three (or more) dimensions: a line segment with endpoints (x1, y1, z1) and (x2, y2, z2) has midpoint: ([x1 + x2]/2, [y1 + y2]/2, [z1 + z2]/2).\n\nThis, of course, is the arithmetic mean or average. We will deal with geometric mean later. Try proving this next theorem using a triangle located at (0,0); (a,0); and (b,c).\n\n Midpoint Connector Theorem: The segment that connects the midpoints of two sides of the triangle is parallel to the third side and half its length.\n\nNotice how the equation of a line in two dimensions is given by ax + by = c. Similarly, the equation of a plane in three dimensions is given by ax + by + cz = d. Can you predict the equation of a 3-D space in four dimensions?\n\nIn two dimensions it is standard to graph the x-axis as increasing to the right and the y-axis as increasing upwards. Only occasionally will it differ with an axis reversed, the axes exchanged, or both. A rotation and/or reflection might be required to correct this.\n\nThe situation is far more complicated in three dimensions. Portraying three dimensions on paper is the first challenge. Many prefer to leave the x- and y-axis as is and add the z-axis at a 45° angle down through quadrant III signifying that it is coming out of the paper. Another common scheme puts the y-axis to the right, the z-axis up, and the x-axis coming out of the paper, again represented at an angle down through quadrant III. (I should note that it is common practice to shorten the scale on this axis by the square root of 2.) It is very important to label your axes, especially in three dimensions. Both of these representations obey a common convention called the right-hand rule or right-hand coordinate system. If you point your right fingers in the direction of +x, and curl them toward +y, your thumb points in the direction of +z. This important convention is embodied in the way many of our laws of physics are stated.\n\nExample: Consider packaging constraints imposed by some shippers, specifically, maximum length plus girth is 108\", where girth is the perimeter of a cross section. Assuming a square cross section, what is the maximum volume allowed.\nSolution: Let 4x + y = 108 and V = x2y. Solve for y in terms of x, substitute into the volume equation, take the derivative (V = 108x2 - 4x3, V' = 216x - 12x2 = 12x(18 - x) = 0), set it equal to zero and solve for x (x=18; oops, that's calculus). Instead, graph that volume equation V=x2(108-4x) on the inverval 0 < x < 27 and 0 < y < 12000 and locate the maximum of 11664 in3 at 18\" by 18\" by 36\" using the 2nd calc feature of your graphing calculator.\n\nSome useful mental math (and a tie in with algebra) comes up in this section, that is the rapid squaring of x+½, for example 8.52. Note how (x+½)2=x2+x+¼= x(x+1)+¼. Thus 8.52=8•9+0.25=72.25.\n\n### Mathematical Induction\n\nMathematical induction is an important type of proof usually not covered in geometry. However, it is needed to extend some of the field axioms, such as associativity and transitivity. which we have already used. Since many texts confuse it with proof by contradiction, we include it here for completeness. The process harks back to the Peano Axiom first introduced in Numbers Lesson 2 to generate the natural numbers. Axiom 5 there is known as the induction axiom.\n\n A subset of N which contains 1, and which contains n+1 whenever it contains n, must equal N.\n\nThe basic flow of an induction proof is as follows. First we demonstrate our statement to be true for n = 1. Next we demonstrate that if our statement is true for any n, then it is true for n + 1. Finally, we invoke Axiom 5 above and claim it is true for all n.\n\nExample: Suppose we want a general formula to add up all the natural numbers from 1 to n. We gave a formula back in Numbers Lesson 2, but we didn't prove it, especially for odd n.\nSolution: First we demonstrate our formula Tn=n(n+1)/2 to be true for n=1. T1=1(1+1)/2=2/2=1 is the correct sum. Now assuming Tn=n(n+1)/2 is true for n we see what happens when we add n+1 to it. Tn+n+1=n(n+1)/2+n+1= n(n+1)/2+2(n+1)/2= (n(n+1)+2(n+1))/2= (n+1)(n+2)/2=Tn+1 We see that when we add n+1 to Tn we obtain the equivalent equation for Tn+1. Since the statement is true for n=1, and we have show that if it is true for Tn it is true for Tn+1, by the induction axiom, we have just proved it true for all n.\n\nWith this method it is fairly straightforward to derive formulae for the sum of squares and cubes as well. We will leave the associated algebra as homework. These will be useful in calculus when we take the limit of Riemann Sums to form integrals to find the area under curves.\n\nThe sum of i2 from i=1 to i=n is n(n+1)(2n+1)/6.\n\nThe sum of i3 from i=1 to i=n is (n(n+1)/2)2 = n2(n+1)2/4.\n\n### Textbook Correction\n\nCorrection for textbook problem 11.5#19: 1 bushel = 1.2445 feet3 or about 1¼ feet3.\n\nThus BOB is wrong! Also, regarding that same problem, they seem to assume you fill the dome as well as the cylindrical part of the silo. You can go ahead and fill the dome portion today, but since silage settles so much, it won't be full tomorrow!"
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https://datacadamia.com/data_mining/local_regression | [
"# Statistics - LOcal (Weighted) regrESSion (LOESS|LOWESS)\n\nPopular family of methods called local regression that helps fitting non-linear functions just focusing locally on the data.\n\nLOESS and LOWESS (locally weighted scatterplot smoothing) are two strongly related non-parametric regression methods that combine multiple regression models in a k-nearest-neighbor-based meta-model.\n\n“LOESS” is a later generalization of LOWESS; although it is not a true initialism, it may be understood as standing for “LOcal regrESSion”.\n\nlocally weighted polynomial regression\n\nloess gives a better appearance, but is O(n^2) in memory, so does not work for larger datasets.\n\n## Procedure\n\nA linear function is fitted only on a local set of point delimited by a region. The polynomial is fitted using weighted least squares. The weights are given by the heights of the kernel (the weighting function) giving:\n\n• more weight to points near the target point (x) whose response is being estimated\n• and less weight to points further away.\n\nWe obtain then a fitted polynomial model but retains only the point of the model at the target point (x). The target point then moves away on the x axis and the procedure repeats and that traces out the orange curve.\n\nwhere:\n\n• The orange curve is the fitted function.\n• The points in the region are the orange one.\n• The kernel is a Gaussian distribution."
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https://gmatclub.com/forum/an-alpha-series-is-defined-by-the-rule-a-n-a-n-1-k-where-k-i-249619.html | [
"GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 16 Oct 2019, 21:14",
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"An \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\nHide Tags\n\nMath Expert",
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"V\nJoined: 02 Sep 2009\nPosts: 58385\nAn \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink]\n\nShow Tags",
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"based on 108 sessions\n\nHideShow timer Statistics\n\nAn \"alpha series\" is defined by the rule: $$A_n = A_{n - 1}*k$$, where k is a constant. If the 1st term of a particular \"alpha series\" is 64 and the 25th term is 192, what is the 9th term?\n\nA. $$\\sqrt{3}$$\n\nB. $$\\sqrt{3^9}$$\n\nC. $$64*\\sqrt{3^9}$$\n\nD. $$64*\\sqrt{3}$$\n\nE. $$64*\\sqrt{3^9}$$\n\n_________________\nRetired Moderator",
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"V\nStatus: Long way to go!\nJoined: 10 Oct 2016\nPosts: 1333\nLocation: Viet Nam\nRe: An \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink]\n\nShow Tags\n\n5\nBunuel wrote:\nAn \"alpha series\" is defined by the rule: $$A_n = A_{n - 1}*k$$, where k is a constant. If the 1st term of a particular \"alpha series\" is 64 and the 25th term is 192, what is the 9th term?\n\nA. $$\\sqrt{3}$$\n\nB. $$\\sqrt{3^9}$$\n\nC. $$64*\\sqrt{3^9}$$\n\nD. $$64*\\sqrt{3}$$\n\nE. $$64*\\sqrt{3^9}$$\n\n$$A_1=64$$\n$$A_2=A_1 * k$$\n$$A_3=A_2 * k$$\n$$A_4=A_3 * k$$\n\n$$...$$\n\n$$A_n=A_{n-1} * k$$\n________________________\n$$A_n=64 * k^{n-1}$$\n\nWe have $$A_{25}=64 * k^{24}=192 \\implies k^{24} = 3 \\implies k = (3)^{\\frac{1}{24}}$$\n\nHence $$A_9=64 * k^8 = 64 * \\Big ((3)^{\\frac{1}{24}} \\Big ) ^ 8 = 64 * (3)^{\\frac{8}{24}} = 64 * (3)^{\\frac{1}{3}} = 64 * \\sqrt{3}$$\n\n_________________\nGeneral Discussion\nManager",
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"B\nJoined: 04 May 2014\nPosts: 152\nLocation: India\nWE: Sales (Mutual Funds and Brokerage)\nAn \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink]\n\nShow Tags\n\n$$A_1$$=64\n\n$$A_2$$=$$A_1$$*k=64*k\n\n$$A_3$$=$$A_2$$*k=64*K*K=64*$$k^2$$\n\n$$A_4$$=$$A_3$$*k=64*$$k^3$$\n\nwe see a relationship between $$A_n$$ and k which is k is $$K^{n-1}$$ for any $$A_n$$\n\nwith above information\n\n$$A_{25}$$=64*$$k^{24}$$\n\nNow $$A_{25}$$ is given=192\n\ntherefore 192=65*$$k^{24}$$=$$\\frac{192}{65}$$=$$k^{24}$$\n\nor 3=$$k^{24}$$ or k=$$3^{1/24}$$\n\n$$A_9$$=64*$$K^8$$=64*{3^1/24}^8=64*3√3\nVeritas Prep GMAT Instructor",
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"V\nJoined: 16 Oct 2010\nPosts: 9705\nLocation: Pune, India\nAn \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink]\n\nShow Tags\n\n1\nBunuel wrote:\nAn \"alpha series\" is defined by the rule: $$A_n = A_{n - 1}*k$$, where k is a constant. If the 1st term of a particular \"alpha series\" is 64 and the 25th term is 192, what is the 9th term?\n\nA. $$\\sqrt{3}$$\n\nB. $$\\sqrt{3^9}$$\n\nC. $$64*\\sqrt{3^9}$$\n\nD. $$64*\\sqrt{3}$$\n\nE. $$64*\\sqrt{3^9}$$\n\nAs per the given rule for alpha series, it is a geometric progression.\n\n1st term = 64\n25th term = $$192 = 64 * k^{24}$$\n$$k^{24} = 3$$\n\nWe need to find the 9th term i.e. $$64*k^8$$\n\n$$k^8$$ is cube root of $$k^{24}$$ so it is cube root of 3. So 9th term is 64*cube root of 3\n\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\nNon-Human User",
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"Joined: 09 Sep 2013\nPosts: 13205\nRe: An \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink]\n\nShow Tags\n\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________",
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"Re: An \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink] 22 Oct 2018, 07:57\nDisplay posts from previous: Sort by\n\nAn \"alpha series\" is defined by the rule: A_n = A_{n - 1}*k, where k i\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89640313,"math_prob":0.99898887,"size":723,"snap":"2019-43-2019-47","text_gpt3_token_len":201,"char_repetition_ratio":0.09596662,"word_repetition_ratio":0.2,"special_character_ratio":0.25726143,"punctuation_ratio":0.15862069,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99995446,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T04:14:01Z\",\"WARC-Record-ID\":\"<urn:uuid:7b4402aa-c85c-40c2-a66e-d98d0a2738f4>\",\"Content-Length\":\"821240\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d204732-618b-4671-8047-3294f87e062f>\",\"WARC-Concurrent-To\":\"<urn:uuid:1e96a9cf-655c-494f-b30a-3d657775f5e4>\",\"WARC-IP-Address\":\"198.11.238.98\",\"WARC-Target-URI\":\"https://gmatclub.com/forum/an-alpha-series-is-defined-by-the-rule-a-n-a-n-1-k-where-k-i-249619.html\",\"WARC-Payload-Digest\":\"sha1:DHKIAIESHX7VWOAWP2C4IPKB2FFXFZQA\",\"WARC-Block-Digest\":\"sha1:RBBR6F5TIYMFMN45YABYPQKWRDPI735D\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986672548.33_warc_CC-MAIN-20191017022259-20191017045759-00158.warc.gz\"}"} |
https://answers.everydaycalculation.com/simplify-fraction/6-13 | [
"Solutions by everydaycalculation.com\n\n## Reduce 6/13 to lowest terms\n\n6/13 is already in the simplest form. It can be written as 0.461538 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 6 and 13 is 1\n2. Divide both the numerator and denominator by the GCD\n6 ÷ 1/13 ÷ 1\n3. Reduced fraction: 6/13\nTherefore, 6/13 simplified to lowest terms is 6/13.\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87781316,"math_prob":0.8828549,"size":347,"snap":"2021-43-2021-49","text_gpt3_token_len":89,"char_repetition_ratio":0.10204082,"word_repetition_ratio":0.0,"special_character_ratio":0.27665707,"punctuation_ratio":0.10606061,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95465267,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T15:45:15Z\",\"WARC-Record-ID\":\"<urn:uuid:888a14f8-dea5-4267-a9f6-81cca96d9cdf>\",\"Content-Length\":\"6406\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23838e0f-91ce-4a3a-a0c6-720580aac2ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:683a710b-072a-42dc-9285-b02d4162e952>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/simplify-fraction/6-13\",\"WARC-Payload-Digest\":\"sha1:UG25WHRU5CVVLIHXLSH26HKDMTXWNP74\",\"WARC-Block-Digest\":\"sha1:73WLV3O2TKZNZPF7DVR4MA4QJTFDPVJN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585270.40_warc_CC-MAIN-20211019140046-20211019170046-00607.warc.gz\"}"} |
https://ch.mathworks.com/matlabcentral/cody/problems/9-who-has-the-most-change/solutions/1021553 | [
"Cody\n\nProblem 9. Who Has the Most Change?\n\nSolution 1021553\n\nSubmitted on 18 Oct 2016 by francescoferrari\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\nTest Suite\n\nTest Status Code Input and Output\n1 Pass\na = [1 2 1 15]; b = 1; assert(isequal(most_change(a),b))\n\nb = 1\n\n2 Pass\na = [ 1 2 1 15; 0 8 5 9]; b = 2; assert(isequal(most_change(a),b))\n\nb = 2\n\n3 Pass\na = [ 1 22 1 15; 12 3 13 7; 10 8 23 99]; b = 3; assert(isequal(most_change(a),b))\n\nb = 3\n\n4 Pass\na = [ 1 0 0 0; 0 0 0 24]; b = 1; assert(isequal(most_change(a),b))\n\nb = 1\n\n5 Pass\na = [ 0 1 2 1; 0 2 1 1]; c = 1; assert(isequal(most_change(a),c))\n\nb = 1\n\n6 Pass\n% There is a lot of confusion about this problem. Watch this. a = [0 1 0 0; 0 0 1 0]; c = 2; assert(isequal(most_change(a),c)) % Now go back and read the problem description carefully.\n\nb = 2\n\n7 Pass\na = [ 2 1 1 1; 1 2 1 1; 1 1 2 1; 1 1 1 2; 4 0 0 0]; c = 5; assert(isequal(most_change(a),c))\n\nb = 5"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59379137,"math_prob":0.99934274,"size":1097,"snap":"2019-26-2019-30","text_gpt3_token_len":452,"char_repetition_ratio":0.16651419,"word_repetition_ratio":0.04526749,"special_character_ratio":0.4685506,"punctuation_ratio":0.13653137,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97487056,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-25T01:52:00Z\",\"WARC-Record-ID\":\"<urn:uuid:efe690cc-1c09-4b5a-a56b-b28ec141aeb6>\",\"Content-Length\":\"74418\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3a05956-4707-47c3-b694-57cc8714f02e>\",\"WARC-Concurrent-To\":\"<urn:uuid:41ca70a7-8650-464a-93b1-c1248bd96def>\",\"WARC-IP-Address\":\"104.118.213.50\",\"WARC-Target-URI\":\"https://ch.mathworks.com/matlabcentral/cody/problems/9-who-has-the-most-change/solutions/1021553\",\"WARC-Payload-Digest\":\"sha1:W4M4AODQP7W6D44TO3MB2DZTR7SHKOEE\",\"WARC-Block-Digest\":\"sha1:XJEX2TRIH5YGIUXXATYUELPM24TBB5VU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999783.49_warc_CC-MAIN-20190625011649-20190625033649-00042.warc.gz\"}"} |
https://werk.statt.codes/posts/2020-10-10-vienna-elections-2020-age-profile/ | [
"# Vienna Elections 2020: Age profile of electoral candidates\n\nAn empirical look at candidates’ age.\nAustria\nelections\nOCR\nregex\nreactable\ngt\nAuthor\n\nRoland Schmidt\n\nPublished\n\n10 Oct 2020\n\n# 1 Setup\n\nCode\n``````library(tidyverse)\nlibrary(here)\nlibrary(extrafont)\nloadfonts(device = \"win\", quiet = T)\nlibrary(hrbrthemes)\nhrbrthemes::update_geom_font_defaults(\nfamily = \"Roboto Condensed\",\nsize = 3.5,\ncolor = \"grey50\"\n)\nlibrary(scales)\nlibrary(knitr)\nlibrary(paletteer)\nlibrary(ggtext)\nlibrary(glue)\nlibrary(pdftools)\nlibrary(svglite)\nlibrary(tictoc)\nlibrary(tidytext)\nlibrary(gt)\nlibrary(reactable)\nlibrary(reactablefmtr)\nlibrary(ggforce)\nlibrary(ggiraph)\nlibrary(htmltools)``````\nCode: Define rmarkdown options\nCode\n``````knit_hooks\\$set(wrap = function(before, options, envir) {\nif (before) {\npaste0(\"<\", options\\$wrap, \">\")\n} else {\npaste0(\"</\", options\\$wrap, \">\")\n}\n})\n\nknitr::opts_chunk\\$set(\nfig.align = \"left\",\nmessage = FALSE,\nwarning = FALSE,\ndev = \"svglite\",\n# dev.args = list(type = \"CairoPNG\"),\ndpi = 300,\nout.width = \"100%\"\n)\noptions(width = 180, dplyr.width = 150)``````\nCode: Define plot theme, party colors, caption\nCode\n``````plot_bg_color <- \"white\"\n\ncaption_table <- \"Source:\\ndata: https://www.wien.gv.at/politik/wahlen/grbv/2020/ analysis: Roland Schmidt | @zoowalk | https://werk.statt.codes\"\n\ntheme_post <- function() {\nhrbrthemes::theme_ipsum_rc() +\ntheme(\nplot.background = element_rect(fill = plot_bg_color, color=NA),\npanel.background = element_rect(fill = plot_bg_color, color=NA),\n#panel.border = element_rect(colour = plot_bg_color, fill=NA),\n#plot.border = element_rect(colour = plot_bg_color, fill=NA),\nplot.margin = ggplot2::margin(l = 0,\nt = 0.25,\nunit = \"cm\"),\nplot.title = element_markdown(\ncolor = \"grey20\",\nface = \"bold\",\nmargin = ggplot2::margin(l = 0, unit = \"cm\"),\nsize = 11\n),\nplot.title.position = \"plot\",\nplot.subtitle = element_text(\ncolor = \"grey50\",\nmargin = ggplot2::margin(t = 0.2, b = 0.3, unit = \"cm\"),\nsize = 10\n),\nplot.caption = element_text(\ncolor = \"grey50\",\nsize = 8,\nhjust = c(0)\n),\nplot.caption.position = \"panel\",\naxis.title.x = element_text(\nangle = 0,\ncolor = \"grey50\",\nhjust = 1\n),\naxis.text.x = element_text(\nsize = 9,\ncolor = \"grey50\"\n),\naxis.title.y = element_blank(),\naxis.text.y = element_text(\nsize = 9,\ncolor = \"grey50\"\n),\npanel.grid.minor.x = element_blank(),\npanel.grid.major.x = element_blank(),\npanel.grid.minor.y = element_blank(),\npanel.spacing = unit(0.25, \"cm\"),\npanel.spacing.y = unit(0.25, \"cm\"),\nstrip.text = element_text(\nangle = 0,\nsize = 9,\nvjust = 1,\nface = \"bold\"\n),\nlegend.title = element_text(\ncolor = \"grey30\",\nface = \"bold\",\nvjust = 1,\nsize = 7\n),\nlegend.text = element_text(\nsize = 7,\ncolor = \"grey30\"\n),\nlegend.justification = \"left\",\nlegend.box = \"horizontal\", # arrangement of multiple legends\nlegend.direction = \"vertical\",\nlegend.margin = ggplot2::margin(l = 0, t = 0, unit = \"cm\"),\nlegend.spacing.y = unit(0.07, units = \"cm\"),\nlegend.text.align = 0,\nlegend.box.just = \"top\",\nlegend.key.height = ggplot2::unit(0.2, \"line\"),\nlegend.key.width = ggplot2::unit(0.5, \"line\"),\ntext = element_text(size = 5)\n)\n}\n\ndata_date <- format(Sys.Date(), \"%d %b %Y\")\n\nmy_caption <- glue::glue(\"data: https://www.wien.gv.at/politik/wahlen/grbv/2020/\\nanalysis: Roland Schmidt | @zoowalk | https://werk.statt.codes\")``````\n\n# 2 Context\n\nElections in Vienna are today and while glancing through the electoral lists I couldn’t help but paying attention to candidates’ birth years. Maybe that’s an age thing… This got me thinking that I haven’t seen any more systematic analysis of parties’/candidates’ age profile. So as a modest contribution to this end, here are my two cents. Again, I’ll focus mainly on the pertaining steps in R and related number crunching. Due to a lack of time and not being an expert on Vienna’s electoral system, I’ll be brief when it comes to substantive matters. But the presented results hopefully provide sufficient material to dig into.\n\nAs always, if you see any glaring error or have any constructive comment, feel free to let me now (best via twitter DM).\n\n# 3 Data\n\nAgain, as so often, the trickiest part is to get the data ‘liberated’ from the format it is provided in. The entire list of candidates is published in this pdf. Note that there are three lists: One for the city council (‘Gemeinderat’; composed on the basis of the results in 18 multi-member electoral districts), one for the 23 district councils (‘Bezirksrat’; one in each district), and the ‘city election proposal’ (‘Stadtwahlvorschlag’; admittedly a somewhat clumsy translation). The latter doesn’t constitute a body in itself, but serves to allocate mandates which remained unassigned after counting the votes for the city council (‘zweites Ermittlungsverfahren’/similar to the d’Hondt procedure).\n\nWhen it comes to extracting the data from the linked pdf, a difficulty may arise due to the two-column format of the document. Hence, simple row-wise extraction doesn’t help much since it would put candidates together which could be from different parties. Similarly, simply isolating the two columns and extracting candidates would also not do the trick since breaks betwee parties, districts etc run over two, and not one column. To illustrate this, I resorted to cutting edge technology and drew the two arrows below:\n\nLuckily, the `tabulizer` package is not only very powerful when it comes to extracting text/data from a pdf, it is also sophisticated enough to take into consideration the text flow highlighted above. I am not familiar with the underlying heuristic, but I assume it is contingent on consistently formatted section headings. Hence, empowered with this tool, retrieving the text becomes rather effortless. The subsequent steps are a battery of regular expressions to extract the specific data we are interested in. To see the code, unfold the snippets below.\n\nCode: Extract text from pdf\nCode\n``````df_raw <- tabulizer::extract_text(file=here::here(\"_blog_data\",\n\"vienna_elections_2020\",\n\"amtsblatt2020.pdf\"),\npages=c(1:115),\nencoding=\"UTF-8\") %>% #capital letters of UTF!\nenframe(name=NULL,\nvalue=\"text_raw\")``````\nCode: Extract data of interest (regex)\nCode\n``````df_clean <- df_raw %>%\nungroup() %>%\nmutate(text_split=str_split(text_raw, regex(\"\\r\\n\\\\s*(?=\\\\d+\\\\.)\"))) %>%\nunnest_longer(text_split) %>%\nmutate(text_split=text_split %>% str_squish() %>% str_trim()) %>%\nmutate(text_split=str_split(text_split, \".(?=Zustellungsbevollmächtigte(r)? Vertreter(in)?)\")) %>%\nunnest_longer(text_split)\n\n# get Listenplatz ------------------------------------------------------------\n\ndf_clean <- df_clean %>%\nmutate(listenplatz=str_extract(text_split, regex(\"^\\\\d+\\\\.?\\\\s+(?!Bezirk)\")) %>%\nstr_extract(., \"\\\\d*\") %>% as.numeric())\n\n# get elections -----------------------------------------------------------\n\ndf_clean <- df_clean %>%\nmutate(election=text_raw %>% str_extract(., regex(\"(?<=[A-Z]\\\\.)\\\\s*[A-z]+wahl(en)?\", dotall = T)) %>%\nstr_trim(., side=c(\"both\"))) %>%\ntidyr::fill(election, .direction=\"down\")\n\n# electoral district --------------------------------------------------------------\ndf_clean <- df_clean %>%\nmutate(wahlkreis=case_when(election==\"Bezirksvertretungswahlen\" ~ str_extract(text_split, \"\\\\d{1,2}\\\\. Bezirk\"),\nelection==\"Gemeinderatswahl\"~ str_extract(text_split, regex(\"Wahlkreis.*?(?=[:upper:]{2,}?)\",\ndotall = T,\nmultiline = T)),\nTRUE ~ as.character(\"missing\"))) %>%\nmutate(wahlkreis=str_trim(wahlkreis, side=c(\"both\"))) %>%\ntidyr::fill(wahlkreis, .direction=\"down\") %>%\nmutate(wahlkreis=str_remove(wahlkreis, \"Wahlkreis \") %>%\nstr_remove(., regex(\"\\\\(.*\\\\)\")) %>%\nstr_trim(., side=c(\"both\")))\n\n# other -------------------------------------------------------------------\n\ndf_clean <- df_clean %>%\nmutate(page=text_raw %>% str_extract(., regex(\"Seite \\\\d+\")) %>% str_extract(., \"\\\\d+\") %>%\nas.numeric()) %>%\nmutate(name=str_extract(text_split, regex(\"(?<=\\\\d\\\\.\\\\s?).*?(?=,\\\\s?\\\\d{4},)\")) %>%\nstr_trim(., side=c(\"both\"))) %>%\nmutate(first_name=text_split %>% str_extract(., regex(\"[:alpha:]*(?=,\\\\s?\\\\d+)\"))) %>%\nmutate(year_birth=text_split %>% str_extract(., regex(\"\\\\d{4}\")) %>%\nas.numeric()) %>%\nmutate(year_interval=cut(year_birth, seq(1930, 2005, 5))) %>%\nmutate(plz=text_split %>% str_extract(., regex(\"\\\\d{4}\\\\s(?=Wien)\")) %>%\nstr_trim(., side=c(\"both\")))\n\n# get party ---------------------------------------------------------------\n\ndf_clean <- df_clean %>%\nmutate(party=text_split %>% str_extract(., regex(\"(?<=^Zustellung)[:alpha:]*\\$\"))) %>%\nTRUE ~ NA_character_)) %>%\ntidyr::fill(party, .direction = \"down\") %>%\nmutate(party=party %>%\nas_factor() %>%\nfct_relevel(., sort) %>%\nfct_relevel(., \"SPÖ\", \"FPÖ\", \"GRÜNE\", \"ÖVP\", \"NEOS\"))\n\n# wrap up -----------------------------------------------------------------\n\ndf_clean <- df_clean %>%\nmutate(wahlkreis_plz=str_extract(wahlkreis, regex(\"\\\\d+\")) %>%\nas.numeric()+100) %>%\nmutate(wahlkreis_plz=wahlkreis_plz %>% as.character() %>% paste0(., \"0\")) %>%\nmutate(wahlkreis_plz=case_when(str_detect(wahlkreis, \"Zentrum\") ~ \"1010, 1040, 1050, 1060\",\nstr_detect(wahlkreis, \"Innen\") ~ \"1070, 1080, 1090\",\nstr_detect(wahlkreis, \"Landstraße\") ~ \"1030\",\nstr_detect(wahlkreis, \"Favoriten\") ~ \"1100\",\nstr_detect(wahlkreis, \"Simmering\") ~ \"1110\",\nstr_detect(wahlkreis, \"Meidling\") ~ \"1120\",\nstr_detect(wahlkreis, \"Hietzing\") ~ \"1130\",\nstr_detect(wahlkreis, \"Penzing\") ~ \"1140\",\nstr_detect(wahlkreis, \"Rudolf\") ~ \"1150\",\nstr_detect(wahlkreis, \"Ottakring\") ~ \"1160\",\nstr_detect(wahlkreis, \"Hernals\") ~ \"1170\",\nstr_detect(wahlkreis, \"Währing\") ~ \"1180\",\nstr_detect(wahlkreis, \"Döbling\") ~ \"1190\",\nstr_detect(wahlkreis, \"Brigittenau\") ~ \"1200\",\nstr_detect(wahlkreis, \"Floridsdorf\") ~ \"1210\",\nstr_detect(wahlkreis, \"Liesing\") ~ \"1230\",\nTRUE ~ as.character(wahlkreis_plz))) %>%\nmutate(residence=case_when(\nstr_detect(wahlkreis_plz, plz) ~ \"inside\",\n!str_detect(wahlkreis_plz, plz) ~ \"outside\",\nTRUE ~ as.character(\"missing\"))) #%>%\n\ndf_clean <- df_clean %>%\nselect(-text_raw) %>%\nfilter(!is.na(listenplatz)) ``````\n\nAfter these few steps we have a searchable/sortable table of all candidates (or better candidatures since one person can be candidate on multiple districts/lists). There have been 8,983 candidatures by 5,038 individuals.\n\nThe table essentially provides all necessary data for the subsequent analysis. While the pdf does not include the exact birth date of each candidate, it provides us with their birth years which we can take as a proxy for age. Note that I also extracted candidates’ residence zip code to see how often place of residence and candidature actually overlap (see below).\n\n## 3.1 Result\n\nCode: Table with all candidates\nCode\n``````tb_all <- reactable(df_clean %>%\nselect(election, wahlkreis, party, name, listenplatz, year_birth, plz),\ncolumns=list(election=colDef(name=\"Wahl\", width=130),\nwahlkreis=colDef(name=\"Wahlbezirk\", width=100),\nparty=colDef(name=\"Partei\", width=50),\nname=colDef(name=\"KandidatIn\"),\nlistenplatz=colDef(name=\"Listenplatz\",\nwidth=70,\nalign=\"center\"),\nyear_birth=colDef(name=\"Geburtsjahr\", width=90),\nplz=colDef(name=\"PLZ Wohnort\", width=90)),\nbordered=F,\ncompact = TRUE,\nhighlight = TRUE,\nstyle = list(fontSize = \"10px\"),\nfilterable = TRUE,\ndefaultPageSize = 23,\ntheme = reactablefmtr::nytimes()) %>%\nadd_title(title= \"WIEN-WAHL 2020: Liste aller KandidatInnen\") %>%\n\n# WIEN-WAHL 2020: Liste aller KandidatInnen\n\nSource: data: https://www.wien.gv.at/politik/wahlen/grbv/2020/ analysis: Roland Schmidt | @zoowalk | https://werk.statt.codes\n\n# 4 Analysis\n\n## 4.1 Oldest and youngest candidates\n\nLet’s now look at the overall youngest and oldest candidates. We could retrieve this information already from the main table provided above (sort column birth year). Here, however, let’s nest candidates’ different candidatures for the sake of clarity.\n\n### 4.1.1 Youngest candidates\n\nCode: Youngest candidates\nCode\n``````df_main <- df_clean %>%\ndistinct(name, year_birth, party) %>%\nslice_max(.,order_by=year_birth, n=10) %>%\narrange(name, desc(year_birth)) %>%\nmutate(index=dplyr::min_rank(year_birth)) %>%\nmutate(index_name=paste(index, \". \",name), .before=1) %>%\nselect(-index)\n\ntb_young <- reactable(df_main,\ncolumns=list(index_name=colDef(name=\"KandidatIn\",\nwidth=130),\nname=colDef(show=F),\nyear_birth=colDef(name=\"Geburtsjahr\",\nwidth=70),\nparty=colDef(name=\"Partei\",\nwidth=50)),\npagination = FALSE,\nonClick = \"expand\",\nbordered=F,\ncompact = TRUE,\nhighlight = TRUE,\nrowStyle = list(cursor = \"pointer\"),\nstyle = list(fontSize = \"10px\"),\ntheme = reactableTheme(\nborderWidth = 1,\nborderColor = \"#7f7f7f\",\nbackgroundColor = plot_bg_color,\nfilterInputStyle = list(\ncolor=\"green\",\nbackgroundColor = plot_bg_color)),\n\ndetails=function(index){\ndf_nested <- df_clean %>%\nslice_max(.,order_by=year_birth, n=10) %>%\nselect(name, election, wahlkreis, listenplatz) %>%\nfilter(name==df_main\\$name[index]) %>%\nselect(-name)\n\ntbl_nested <- reactable(df_nested,\ncolumns = list(\nelection=colDef(name=\"Wahl\"),\nwahlkreis=colDef(name=\"Wahlbezirk\"),\nlistenplatz=colDef(name=\"Listenplatz\")\n),\noutlined = TRUE,\nhighlight = TRUE,\nfullWidth = TRUE,\ntheme = reactableTheme(\nbackgroundColor = \"#ab8cab\"))\n\nhtmltools::div(style = list(margin = \"12px 45px\"), tbl_nested)}\n\n) %>%\nadd_source(source=\"Geburtsjahr lt. Wahlvorschlag als Basis. Top 10.\")``````\n\n# WIEN-WAHL 2020: Jüngesten KandidatInnen\n\nGeburtsjahr lt. Wahlvorschlag als Basis. Top 10.\n\nAs becomes clear from the table, there are overall 16 candidates who were all born in 2002.\n\n### 4.1.2 Oldest candidates\n\nCode: oldest candidates\nCode\n``````df_main <- df_clean %>%\ndistinct(name, year_birth, party) %>%\nslice_min(., order_by=year_birth, n=10) %>%\narrange(year_birth, name) %>%\nmutate(index=dplyr::min_rank(year_birth)) %>%\nmutate(index_name=paste(index, \". \", name), .before=1) %>%\nselect(-index)\n\ntb_old <- reactable(df_main,\ncolumns=list(index_name=colDef(name=\"KandidatIn\",\nwidth=130),\nname=colDef(show=F),\nyear_birth=colDef(name=\"Geburtsjahr\",\nwidth=70),\nparty=colDef(name=\"Partei\",\nwidth=50)),\npagination = FALSE,\nonClick = \"expand\",\nbordered=F,\ncompact = TRUE,\nhighlight = TRUE,\nrowStyle = list(cursor = \"pointer\"),\nstyle = list(fontSize = \"10px\"),\ntheme = reactableTheme(\nborderWidth = 1,\nborderColor = \"#7f7f7f\",\nbackgroundColor = plot_bg_color,\nfilterInputStyle = list(\ncolor=\"green\",\nbackgroundColor = plot_bg_color)),\n\ndetails=function(index){\ndf_nested <- df_clean %>%\nslice_min(.,order_by=year_birth, n=10) %>%\nselect(name, election, wahlkreis, listenplatz) %>%\nfilter(name==df_main\\$name[index]) %>%\nselect(-name)\n\ntbl_nested <- reactable(df_nested,\ncolumns = list(\nelection=colDef(name=\"Wahl\"),\nwahlkreis=colDef(name=\"Wahlbezirk\"),\nlistenplatz=colDef(name=\"Listenplatz\")\n),\noutlined = TRUE,\nhighlight = TRUE,\nfullWidth = T,\ntheme = reactableTheme(\nbackgroundColor = \"#ab8cab\"))\n\nhtmltools::div(style = list(margin = \"12px 45px\"), tbl_nested)}\n\n) %>%\nadd_source(source=\"Geburtsjahr lt. Wahlvorschlag als Basis. Top 10.\")``````\n\n# WIEN-WAHL 2020: Ältesten KandidatInnen\n\nGeburtsjahr lt. Wahlvorschlag als Basis. Top 10.\n\nThe oldest candidate is Waschiczek Wolfgang, who was born in 1928. Not bad.\n\n## 4.2 Avgerage birth year per election and party\n\nLet’s now look at the average year of birth of parties’ candidates on each of the different electoral levels. The table below provides the median, mean and standard deviation for each party. The thin white line in the density plots on the right indicates the median.\n\nCode: Median age of lists\nCode\n``````# summarize data (median, mean, sd)\ndf_list_age <- df_clean %>%\ngroup_by(election, party) %>%\nsummarize(year_median=median(year_birth, na.rm = T),\nyear_mean=mean(year_birth, na.rm=T),\nyear_sd=sd(year_birth, na.rm=T)) %>%\ngroup_by(election) %>%\narrange(desc(year_median), .by_group=T) #order as gt table\n\n#create graphs for table\n\n## define function creating plot\nfn_plot <- function(data){\ndata %>%\nggplot()+\nggridges::geom_density_ridges(aes(x=year_birth,\ny=0),\nfill=\"firebrick\",\ncolor=plot_bg_color,\nquantile_lines=T,\nquantiles=2,\npanel_scaling = F,\nlinewidth=12)+\nscale_x_continuous(limits=c(min(df_clean\\$year_birth),\nmax(df_clean\\$year_birth)),\nexpand=expansion(mult=0))+\nscale_y_discrete(expand=expansion(mult=0))+\ntheme(\nplot.background = element_rect(fill = plot_bg_color, color=NA),\npanel.background = element_rect(fill = plot_bg_color, color=NA),\nplot.margin = ggplot2::margin(0, unit=\"cm\"),\naxis.text = element_blank(),\naxis.title = element_blank()\n)\n\n}\n\n## apply function, dataframe with plots\nbox_plot <- df_clean %>%\nselect(election, party, year_birth) %>%\ngroup_by(election, party) %>%\nmutate(year_median=median(year_birth, na.rm = T)) %>%\n#ungroup() %>%\nnest(year_birth_nest=c(year_birth)) %>%\nmutate(plot=map(year_birth_nest, fn_plot)) %>%\ngroup_by(election) %>%\narrange(desc(year_median), .by_group=T) #order as gt table\n\n#create gt table & insert df with plots;\ntb_list_age <- df_list_age %>%\nmutate(year_mean=round(year_mean, digits = 2)) %>%\nselect(election, party, contains(\"year\")) %>%\ngroup_by(election) %>%\narrange(desc(year_median), .by_group=T) %>% #order as plots\nmutate(index=min_rank(-year_median), .before=1) %>%\nmutate(index_party=paste0(index,\". \", party)) %>%\nungroup() %>%\nselect(-index, -party) %>%\nmutate(boxplot=NA) %>%\ngt(groupname_col = \"election\", rowname_col = \"index_party\") %>%\ntab_header(title=md(\"**WIEN-WAHLEN 2020:<br>Durchschnittliches Geburtsjahr der KandidatInnen**\")) %>%\ngt::cols_label(index_party=\"Partei\",\nyear_median=\"Median\",\nyear_mean=\"arith. Mittel\",\nyear_sd=\"Std. Abw.\",\nboxplot=\"Verteilung\") %>%\ngt::fmt_number(columns=c(year_sd),\ndecimals=2,\nsuffixing=F) %>%\ngt::text_transform(\nlocations=cells_body(c(boxplot)),\nfn=function(x){\nmap(box_plot\\$plot, ggplot_image, height=px(11), aspect_ratio=4)}\n) %>%\ncols_width(\nc(boxplot) ~ px(200),\nc(index_party) ~ px(100)\n) %>%\ntable.background.color = plot_bg_color,\ntable.width=pct(100),\ntable.font.size = \"11px\",\nrow_group.font.weight = \"bold\",\n Median arith. Mittel Std. Abw. WIEN-WAHLEN 2020:Durchschnittliches Geburtsjahr der KandidatInnen 1991.5 1992.12 2.85",
null,
"1988.0 1985.63 10.57",
null,
"1986.0 1984.50 5.83",
null,
"1983.0 1978.02 16.36",
null,
"1981.0 1975.00 26.75",
null,
"1977.0 1977.59 13.20",
null,
"1976.0 1975.66 12.42",
null,
"1975.0 1975.26 16.06",
null,
"1973.0 1973.43 12.83",
null,
"1970.0 1971.81 14.96"
] | [
null,
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null,
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null,
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null,
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",
null,
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9OeDgB2F8EdAAAAtpU4jq8eiulUKvlPPy329hZ6e4t9fa1iMe3hAGBXE9wBAABgq1s7FBOHYWl4uHD+fKGvr9DbW52YcCgGALYOwR0AAAC2nPWHYuozM/lz54p9fYW+vpWhoajZTHs6AODGBHcAAABIXxzHwdVDMe1yOXnS6eqhmEIh7ekAgK9EcAcAAIB0rB2Kidrt0tBQ/vz5Yl9fsbe3OjXlUAwAbEeCOwAAANwia4U9iOPKlSvJGnuxv7908WLUbqc9HQDwTQnuAAAAsFnWn2JvLi7mP/202NdX7OsrXrjQqVTSng4A6DLBHQAAALonjuOrp9g7lUpyhL3Y31/s728sLKQ9HACwuQR3AAAA+AbiOI6i1VPsrVZxYGC1sPf1OcUOALuN4A4AAAA3Z+0UexyG5dHRQm9vssNevnQpDsO0pwMAUiO4AwAAwJe47mGna1diVoaHo2Yz7ekAgK1CcAcAAIDrXSvsQVCdnCz29hYHBooDA6WhoU6tlu5sAMCWJbgDAADAhsJem54u9vUVBwaK/f2loaF2uZzubADAdiG4AwAAsButL+z12dlCb+/KhQvFgYGVwcH2ykq6swEA25TgDgAAwK5wXWEv9vUVL1xYGRxcGRhoFYvpzgYA7AyCOwAAADvThisxU1OFvr6VCxdWLlxYGRqyww4AbAbBHQAAgB0hjuMoWi3scVy5fLnY378yOLgyOLgyNNSpVNKeDwDY+QR3AAAAtqc4juM409MTBEEchuVLl1YuXCheuLBy4ULp4sWwXk97PgBg1xHcAQAA2B7iKMpkMkEmEwRB2Gisbq8PDq4MDVUuXYra7bQHBAB2O8EdAACALWr9EfZWPl8cGFgZGioND68MDtay2TiK0h0PAOA6gjsAAABbw8Yj7LWpqaSwJ5G9ubiY9nwAAF9CcAcAACAdcRhmenrWTsSULl4sJXn94sXyyEinVkt7QACAmyO4AwAAcEusX2APgsb8/MrgYOnixZXh4dLwsBMxAMAOILgDAACwKdYvsEetVmlkpDQ4WBoZWRkeLo+MtEultAcEAOgywR0AAIBu2LjAXp+ZWRkcLI2MlC5eLI2M1LLZOAzTHRAAYLMJ7gAAANy8jXm9XSqVLl5M2nr54sXy6KgL7ADALiS4AwAA8OXiMFzL61GzWRodLV28WB4dLY+MlEZHm0tLQRynOyEAQOoEdwAAAK63/vx63OlULl8uXbxYvnSpPDpaHh2tTU97wCkAwGcJ7gAAALvdhrwehrVstjQyUh4dLY2OlkdHqxMTzq8DAHwVgjsAAMDusv44TByG1StXSiMj5UuXymNjlUuXqpOTUbud7oQAANuU4A4AALBzbXy0adRqVS5fLo+MlMfGymNjlfFx2+sAAF0kuAMAAOwQyV31TE9P8mm7XK6MjVXGx8tjY5WxsfLYWH1mxu11AIDNI7gDAABsQ3Ech2Fm7961T+uzs6WRkcrly5XLlyvj45Xx8VahkOqIAAC7juAOAACw1cVhGGQya6vrnWo12VuvXr5cHh+vXr5czWajVivdIQEAENwBAAC2kDiKgji+dnW93a5NTSXH1itXrlQnJirj483l5SCO050TAIDPEtwBAADScV1bj6OoPjt7ra1fuVKdmKjPzXmoKQDAdiG4AwAAbLrrHmcaR1Fjfj5p69WJicrERHVioj49HXU6qY4JAMA3IrgDAAB0TxzHUZTp6QkymeQLUbtdy2YrV67Ustnq5GQ1m61NTtZnZrR1AICdR3AHAAD4Oq47CBMEQXtlpToxsVbVq1NTtWy2ubiYrLcDALDjCe4AAABfKIrijWE9arVq09PVycna1NTqa3q6ls12qtUUxwQAIHWCOwAAQBDcaGM9arcbc3PVycnazEx9ero2M5O09ebSUhDHKY4KAMDWJLgDAAC7SxyGQSaz9vzSIAiiZrM+O1udmqrPzNSmp9fem0tLrsEAAPDVCe4AAMAO9Nl19SAIWvl8bWamPjNTn5urz87WZ2aST1vFoo11AAC+OcEdAADYrm5Y1Tvlcn1urjY93ZifXw3rc3P12dnG/HzUbqc1KgAAu4HgDgAAbGFxHEfRdRdggiBoFYuN+fn67Gwjl2vMz9fn5xtzc/W5ufr8fFirpTUsAAC7nOAOAACk6nOSetRsNhYWks30xsLC2ns9l2suLNhVBwBgCxLcAQCAzbV6+KWnJ8hk1n+9U602FxcbuVxjYaG5uNhYWGguLTUXFhq5XCOXa1cq7qoDALC9CO4AAMA3EodhEASf7elxGLby+aSeNxcXG0lMX1xsLi42l5aai4tho5HSyAAAsCkEdwAA4HMlMf2z916CIOhUq63l5cbiYmt5ubm01Fz3nnylXS5bUQcAYFcR3AEAYDeKoyiIoqCn57MlPYjjdrncKhSaS0utfL61vNwsFFqFQmt5ubm83Fxebi0vt/J5V9QBAOA6gjsAAOwcqxk9k8ns2XOjb8ftcrldLDbz+dbycqtYbOXzrUJh9YNisVUotAuFVrG4utgOAADcDMEdAAC2sDhOnjj6uQ09CKJ2u1Mur+byYrG1stK++moVi+1ice29U6nEUXSL/wYAALB7CO4AAHCrJPU8CIIbPWL02k+FYada7ZTLq+m8VOqUy+1yuX310w3vpZJHjwIAwBYhuAMAwM2I4ziOg+SVPEr0c7p58sOdWq1TrXaq1U6p1C6XO5VKp1rtVCrtcjnJ6J99DxsNzxoFAIDtSHAHAGDX2NjKV3P5l4na7ajR6NRqnVotyeVhtdqp1Vbf6/V2uZw09NX35INqtVOthvW6dA4AALuH4A4AwFa1doAladZfuk5+3W+HYdhoRM1mWK936vWwVuvUamGtFjYaYb2efJAU8w2f1mpJKE/ew1ot6nQ27W8IAADsKII7AAA3b/2qeBCsRvBMJpPJfPUgfu0f1ulErVbUbq/28UYjeUWNRthshs1m8sUo+frVT9d/JazXr3169VtCOQAAcIsJ7gAA21McB0EQryXvboTvDf/4TifqdOJ2O0pezWbYakWtVpQU8OTjz77a7eSD1Z9pNq99nPxu8sW1jN5sxu322nNEAQAAtjXBHQDg61rb8g6C1ZvgiW707iCOo04n7nSS3r22A74hbbfbUat1rYknn3Y6UfLDrdaNv7vx16//evKVTsflcQAAgJsluAMAO8gNC/hXezDmjf95V3t31G7H7fbaivfavva1Zr0uf4frM3erlfzu53167Y/Y+K04DLv3nwsAAAC3guAOAKQhiuL1Wfwmm3gchtfuliQnvNcf+F7XxDccP1l3FGXti9d+YP3yePKy5Q0AAMDNENwBgJsQR9HqczK/ciWPoyha91jLsFbr1GphoxHW62G9vuFbaw/MTD69GtOvJfWrh79tfwMAALAFCe4AsMskR1ei6CsV8zhejePVaqda7VQqnWq1U6utRvN6vVOthvV6WKt16vWwWg3r9fUxfTWpt9u36u8GAAAAaRLcAWA7i+PVlfOeni9I51GrtdrKK5V2qdSpVJJP25VKp1JZjenrX5VKp1oNa7Ww2XRTBQAAAL4iwR0AtpY4ipL188yePTf8gaSet0ul9spKe2WlXS53yuV2ubz2wXXvnUol6nRu8d8CAAAAdiHBHQBuhdU99M854RK120lAb+XzrWJxNaYXi+1SqZVU9ZWV1S+WSlGrdevnBwAAAL6U4A4A30xy1CUIbrCQHsedSqVVKDTz+VY+3yoUWoVCu1hsFQqt5L1QSGJ62GikMDkAAADQVYI7AHy+z4/pUbvdLhabS0vNxcXm8nIrn19f1VuFQiufb5dKcRimMTcAAACQAsEdgF3s83t6u1xuLS83FhaSnt5cWmotLzeXl1vLy82lpWY+36lWPU0UAAAAWE9wB2BHS5L6Zy6nR81mY3Gxkcs1crnVLfWlpebiYmNxsbm42Mrno3Y7rZEBAACAbUpwB2DbS862XLelHjWbjYWF+tzcalVfXGwsLKx9YD8dAAAA6DrBHYDt4QZVPY5bhUJ9bq4+O7sW1tdekjoAAABwiwnuAGwhcRQFcXzdrnqrUKjPztZnZupzc/W5ucbcXH1+vjE/38jlHH4BAAAAtg7BHYAUfDasR61WfXa2ms3WZ2bqMzO12dnG/Hx9draRy0WtVoqjAgAAAHxFgjsAm+izYT2s12szM7XJyVoS1pP36elWoeACDAAAALCtCe4AdEMcx1GU6ekJMpnkC1GrVZuerl65UpueTl716enazEy7VBLWAQAAgB1JcAfgJkVRvHFpvVUoVCcmqhMTtampajZby2ZrU1ONxUVhHQAAANhVBHcAPt/Gth5HUWN+vnL5ci2brWaz1cnJWjZby2Y7tVq6YwIAAABsBYI7AEEQXL0Js7a3HseNXK48Pp6srlcnJipXrtSnp6N2O9UpAQAAALYuwR1gN4rDcP299fbKSmV8vDw+Xr1ypXLlSvXKlWo2GzWb6Q4JAAAAsL0I7gA73cbV9bjTqU5Oli9dqoyPJ229cuVKe2Ul3RkBAAAAdgDBHWBn2ZjXO5VK+dKl8thYZXw8edWmp+MwTHdGAAAAgB1JcAfYzqIoDoJMT0/yWatQKI+MlC9dKo+PVy5dKo+NNZeXgzhOd0YAAACAXUJwB9g+Nm6vt/L50sWL5UuXyqOj5bGx8thYu1hMd0AAAACA3UxwB9i64jBcy+vtUql08eLqAvulS+XR0Za8DgAAALCVCO4AW0UcRZlMJshkgiCIO53y+HhpaKg0MlIaGSmPjDQWFhyHAQAAANjKBHeAlGy8D9PI5VYGB0sXL5YuXiyNjFSvXPFoUwAAAIDtRXAHuEXiKAquPuA0arVKIyOlwcHSyEgS2dulUtoDAgAAAPCNCO4AmyUOw0xPT3Iipl0qrVy4sDI4uDI0tDI0ZIEdAAAAYOcR3AG6Zv0zThu5XHFgYGVwsDQ0tDI4WJ+fd4EdAAAAYGcT3AG+vmuFPY6rk5PrC3urWEx7OgAAAABuKcEd4CZcV9gLvb3F/v6VCxdKw8OdWi3t6QAAAABIk+AO8EXWX4mpz84Wzp8vDgwU+/tXBgc7lUq6swEAAACwpQjuABvEUZTJZJInnTaXlpId9mJ//8rAgCsxAAAAAHwBwR3Y9eI4jqJkjb1TqRTOn08Ke3FgoJHLpT0cAAAAANuG4A7sRmuHYuIwLI2MFD79tNDbW+jtrU5MBHGc9nQAAAAAbEuCO7A7xHEcx5meniAImouL+U8+KfT2Fs6fXxkcDBuNtIcDAAAAYCcQ3IEda22NPWq3VwYG8p9+Wjh/vtDb25ifT3s0AAAAAHYgwR3YQdatsTdyueWPPy6cO1fo7S0ND0ftdtrDAQAAALDDCe7A9nbtGnsUlYaH8598kj93Ln/unDV2AAAAAG4xwR3YftYie6dazZ87l//kk/wnnxQHBsJaLe3RAAAAANi9BHdgO4jjOIqSyN7I5ZY//DCJ7OVLl+IoSns4AAAAAAgCwR3YsuI4Dq4eZC+Pja1G9nPn6rOzaY8GAAAAADcguANbSBxFmUwmyGTiKFoZHFz+4IPljz/OnzvXLhbTHg0AAAAAvoTgDqRsLbJH7Xbh/Pnljz7Kf/RR4fz5joPsAAAAAGwrgjuQgjiKklsxUbudP3du+YMPlj/8sNDXFzWbaY8GAAAAAF+T4A7cItcie6uV/+STpQ8+WP7ww2J/f9RqpT0aAAAAAHSB4A5sovXnYvLnzi29997yBx8U+/ujdjvt0QAAAACgywR3oNviOAiCtZvsS++/v/zBB4XeXpvsAAAAAOxsgjvQDXEcx3GmpyeI4+LAwOI77yy9917h/Pmw0Uh7MgAAAAC4RQR34OuLwzCzZ08QBJXx8YV33ll+//3lDz9sl8tpzwUAAAAAKRDcgZuzFtkbudzCW28tvffe0vvvN5eW0p4LAAAAAFImuANfLo6iTE9PEATtcnnpnXcW33136b33qtns6rl2AAAAAEBwBz7X1bPsUbud//jjxbNnF999tzQ8HEdR2pMBAAAAwFYkuAMbbDjL/tZbC++8k//oI88+BQAAAIAvJbgD6y7GrKwsvP324l+kjhQAABVuSURBVNmzi++808jl0p4LAAAAALYTwR12q6sXY+IwzJ87t3j27MLbb5eGhlyMAQAAAICvR3CH3WXtYkxtenrhzTcX33ln6f33O9Vq2nMBAAAAwLYnuMMusPb402Zz4ezZxbffXnj77drUVNpjAQAAAMCOIrjDjrW2zF6dnMy99trCW28tf/xx1GqlPRcAAAAA7EyCO+ws65bZF999d+GttxbeessyOwAAAADcAoI77ATrL7PPv/bawptvLn/0UdRspj0XAAAAAOwigjtsY3EUZXp6onZ7+f33c2+8sfDWW9XJybSHAgAAAIBdSnCHbSaJ7EEQNHK53Guv5V5/femDD8J6Pe25AAAAAGC3E9xhe1g9GhPHxf7+pLOXRkaCOE57LgAAAABgleAOW9jVJ6CG9frCm2/mXn899+abrXw+7bEAAAAAgBsQ3GHLWTsaU5+bm3/11dxrry1/9FHUbqc9FwAAAADwRQR32CquHY3p65v/619zr71WHhtzNAYAAAAAtgvBHVKW7LOHjcbCW2/l/vpXR2MAAAAAYJsS3CEFcRxngiDIZFr5/NyZM/Ovvrr0/vtRs5n2XAAAAADA1ye4w62zdpy9evny3CuvzP/1rysDA3EUpT0XAAAAANAFgjtsurXj7IVPP51/9dX5M2eq2WzaQwEAAAAAXSa4w2ZJOnvUbi+ePTt/5kzu9deby8tpDwUAAAAAbBbBHbosuRvTqVTmX311/tVXF8+e7dRqaQ8FAAAAAGw6wR26IY6DIAgymebS0txLL82dOZP/6KOo00l7LAAAAADg1hHc4RuIoiCTCTKZ2tTU7EsvzZ85U+zv9xBUAAAAANidBHe4acnRmCAISiMjSWcvj42tLrkDAAAAALuV4A5fVfIQ1CAIiv39cy+9NPfKK7WpqbSHAgAAAAC2CsEdvsRqZ4/j/CefJPvsjVwu7aEAAAAAgC1HcIcbSzp7HIZL7747+/LL86++2srn0x4KAAAAANi6BHfYYLWzdzoLb789+5e/5F5/vb2ykvZQAAAAAMA2ILhDEFzt7FG7nXv99bmXX869/nqnUkl7KAAAAABgOxHc2dVWO3urlXvttdm//GXhzTc7tVraQwEAAAAA25Lgzm50XWfPvflmqLMDAAAAAN+M4M4ucq2zv/767F/+knvjDZ0dAAAAAOgWwZ2d79p99uRuzBtvuBsDAAAAAHSd4M6OtdbZF954Y/bFF3Ovv66zAwAAAACbR3Bnp0k6e9zp5N58c7WzVyppDwUAAAAA7HyCOzvEamcPw8WzZ2deeCH317+2y+W0hwIAAAAAdhHBne0tjqJMT08Qx/mPP55+/vn5l19uFYtpDwUAAAAA7EaCO9vSamcPgmJ//8xzz8299FJjYSHtoQAAAACAXU1wZzuJ4zgTBEEmUx4dnXnuudkXX6xNT6c9FAAAAABAEAjubA9xHARBkMnUstmZ06dnXnihMj6e9kwAAAAAABsI7mxpyemYxsLC9OnTM88/XxoeXo3vAAAAAABbjODOVhSHYWbPnnaxOP3cc7PPP1/o7Y2jKO2hAAAAAAC+iODOFpJ09rDRmHvppZnnnlt6772o00l7KAAAAACAr0RwJ31JZ4/DMPfmmzOnT+defz2s19MeCgAAAADg5gjupCa5zx4EQf7TT2eefXb25ZfbxWLaQwEAAAAAfE2CO7dc8tTTTKZy6dL0s8/OPP98fW4u7ZkAAAAAAL4pwZ1bJ1lpr+dy06dOzZw+Xb50Ke2JAAAAAAC6RnBn0yUn2tvl8uzzz0+fPl349NM4itIeCgAAAACgywR3NkvS2aN2e/7VV2dOn154662o3U57KAAAAACAzSK4021xHGQyQRznP/lk+tSpuZdfbpfLac8EAAAAALDpBHe6ZO1RqJcvT508OXP6tEehAgAAAAC7iuDON5U8CrVVKEyfOjX97LMrw8Or8R0AAAAAYDcR3PmaVk+0N5tzL7889eyzS+++G4dh2kMBAAAAAKRGcOcmRVHQ0xPE8fKHH06dPDl/5kynWk17JgAAAACA9AnufFXJ6ZjKxMTUU09NP/dcY34+7YkAAAAAALYQwZ0vkZyOaReLU88+O33y5MrQkBPtAAAAAACfJbhzY8k+e9Ruz585M33q1OLZs1Gnk/ZQAAAAAABbl+DORsn2eiZT7Oubevrp2b/8pV0qpT0TAAAAAMA2ILizKllpr+dyU08/PX3yZHVyMu2JAAAAAAC2E8F9t0tOtIeNxuwLL0ydPJn/+OM4itIeCgAAAABg+xHcd6k4jjOZTBDHyx9+OPXMM3NnzoS1WtpDAQAAAABsY4L7rpOcjqlls1NPPz196lR9bi7tiQAAAAAAdgLBfbdITsd0qtWZ556beuaZQm/v6vNRAQAAAADoBsF9h1s7HbP07rtTzzwz/9e/ho1G2kMBAAAAAOxAgvuOtXY6Jvvkk9OnTjVyubQnAgAAAADYyQT3nSY5HRPWajPPPZd9+mmnYwAAAAAAbg3BfYdYOx2z/NFH2aeemj9zJqzX0x4KAAAAAGAXEdy3veR0TGN2Nvvkk1OnTtVnZtKeCAAAAABgNxLct6vV0zGNxuyLL049/fTyxx87HQMAAAAAkCLBfbtJqnomUzh/PvvUU3N/+UunVkt7JgAAAAAABPftI1lpby4tZZ98cuqZZ6qTk2lPBAAAAADANYL7VpecaI/DcO6VV6aeemrx3XfjMEx7KAAAAAAArie4b11Jaq+MjU0+8cTMc8+1CoW0JwIAAAAA4HMJ7ltOcjqmU6lMnzqVffrplcFBT0MFAAAAANj6BPct42pVX/7ww+yJE3Ovvho1m+lOBAAAAADAVye4py9Zaa/nctkTJ6ZPnqxNT6c9EQAAAAAAN01wT01yoj1qt+defjn75JPLH3wQR1HaQwEAAAAA8DUJ7ilIUnt5dDR74sT06dPtlZW0JwIAAAAA4JsS3G+d1aehVqvTp05ln3xyZXAw7YkAAAAAAOgawX3zJU9DzWTyn3wyeeLE/CuvhI1G2jMBAAAAANBlgvsmSlbam8vL2SefnHrqqWo2m/ZEAAAAAABsFsF9E8RxkMnEYZh7/fXsiRMLb78dh2HaMwEAAAAAsLkE925KnoZazWYnjx+fPnWqubSU9kQAAAAAANwignsXJJ09arVmXnghe+JE/ty51bvtAAAAAADsGoL7N5Kk9tLw8OTx47MvvNAul9OeCAAAAACAdAjuX0fyNNROpTJ18mT2qadKQ0NpTwQAAAAAQMoE95uRHIrJZPLnzk0+8cT8K6+EjUbaMwEAAAAAsCUI7l9JstLeKhazTz6ZffLJ6sRE2hMBAAAAALC1CO5fJI7jTCYTxPHi2bOTTzyx8MYbUaeT9lAAAAAAAGxFgvuNJSvtjVwue/z41DPP1Ofm0p4IAAAAAIAtTXDfKIqCnp44DOfPnJk8cWLp3XfjKEp7JgAAAAAAtgHBfVUcRZmenurU1OTx49MnTzaXl9OeCAAAAACA7WS3B/eks0ft9uyLL2ZPnFj++OMgjtMeCgAAAACA7Wf3BvcktVfGxiaPH58+fbq9spL2RAAAAAAAbGO7LrgnnT1sNGZOn548caLY32+lHQAAAACAb24XBfcktZeGhiaOH599/vlOtZr2RAAAAAAA7Bw7P7jHYZjZs6dTrU6fPDl54kRpeDjtiQAAAAAA2IF2cnBPVtqLfX0Tjz8+9/LLYb2e9kQAAAAAAOxYOzC4Jyvt7VJp6umnsydOlMfG0p4IAAAAAICdbwcF9+TZp5lM/pNPJh9/fO7MmajVSnsmAAAAAAB2i50Q3JOV9laxmH3qqeyJE9WJibQnAgAAAABg19nOwT1ZaQ+Cpfffnzx+PPfaa1G7ne5EAAAAAADsWtsyuCcr7c18PvvEE9mnnqpNTaU9EQAAAAAAu922Cu5XV9oXz56dPH4898YbcRimOxEAAAAAACS2R3BPVtobi4vJSnt9djbtiQAAAAAAYIMtHdzjOM5kMnEU5d58M3v8+MLbb1tpBwAAAABga9qiwX31SnsuN/H441NPP93I5dKeCAAAAAAAvsjWCu7XVtpfe23y+PHFd9+10g4AAAAAwLawVYL76pX2ubnJxx+feuaZxsJC2hMBAAAAAMBNSDm4r660h+HcK69kn3hi6f334yhKdyQAAAAAAPgaUgvuyUp7fWZm8s9/njp5srm0lNYkAAAAAADwzd3y4B7HQSYTh+HsSy9ln3hi+cMPrbQDAAAAALAD3Lrgnqy0V7PZyccfnzp5spXP37I/GgAAAAAANtumB/c4ijI9PVG7PffSS5NPPLH80UdBHG/2HwoAAAAAALfYJgb3JLVXJycnH398+uTJVrG4eX8WAAAAAACkq/vBfW2lffaFFyafeCJ/7pyVdgAAAAAAdrxuBvfVlfYrVyb+/Ofp06fbVtoBAAAAANg1uhDc11baZ55/Pnv8eP78eSvtAAAAAADsNt8ouCepvTI+PvHnP8+cPt0ulbo1FgAAAAAAbC9fJ7ivrrQ3m9PPPZd94olCX5+VdgAAAAAAdrmbC+5Jai+Pjk4+/vjMc8+1y+VNGgsAAAAAALaXrxTck84eNhozzz47eeJEcWDASjsAAAAAAKz3JcE9Se2lixcn//znmeef71Srt2YsAAAAAADYXm4c3FdX2uv16VOnJp94YmVw8BaPBQAAAAAA28uNg3uxvz974sTsCy90arVbPBAAAAAAAGxHNw7u5/73/67Pzt7iUQAAAAAAYPvqSXsAAAAAAADYCQR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAAAALpAcAcAAAAAgC4Q3AEAAAAAoAsEdwAAAAAA6ALBHQAAAAAAukBwBwAAAACALhDcAQAAAACgCwR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAAAALpAcAcAAAAAgC4Q3AEAAAAAoAsEdwAAAAAA6ALBHQAAAAAAukBwBwAAAACALhDcAQAAAACgCwR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAAAALpAcAcAAAAAgC4Q3AEAAAAAoAsEdwAAAAAA6ALBHQAAAAAAukBwBwAAAACALhDcAQAAAACgCwR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAAAALpAcAcAAAAAgC4Q3AEAAAAAoAsEdwAAAAAA6ALBHQAAAAAAukBwBwAAAACALhDcAQAAAACgCwR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAAAALpAcAcAAAAAgC4Q3AEAAAAAoAsEdwAAAAAA6ALBHQAAAAAAukBwBwAAAACALhDcAQAAAACgCwR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAAAALpAcAcAAAAAgC4Q3AEAAAAAoAsEdwAAAAAA6ALBHQAAAAAAukBwBwAAAACALhDcAQAAAACgCwR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAAAALpAcAcAAAAAgC4Q3AEAAAAAoAsEdwAAAAAA6ALBHQAAAAAAukBwBwAAAACALhDcAQAAAACgCwR3AAAAAADoAsEdAAAAAAC6QHAHAAAAAIAuENwBAAAAAKALBHcAAAAAAOgCwR0AAAD+/3bumCXqMIDjOGpeWHQGQtAiingQvhh3oTfQ4qxr0Gir0BsId+dmoVEXcbmGG5yENo27lqZqim9dB5/P+PDn4bf+v8MDABAQ3AEAAAAAICC4AwAAAABAQHAHAAAAAICA4A4AAAAAAAHBHQAAAAAAAoI7AAAAAAAEBHcAAAAAAAgI7gAAAAAAEBDcAQAAAAAgILgDAAAAAEBAcAcAAAAAgIDgDgAAAAAAAcEdAAAAAAACgjsAAAAAAAQEdwAAAAAACAjuAAAAAAAQENwBAAAAACAguAMAAAAAQEBwBwAAAACAgOAOAAAAAAABwR0AAAAAAAKCOwAAAAAABAR3AAAAAAAICO4AAAAAABAQ3AEAAAAAICC4AwAAAABAQHAHAAAAAICA4A4AAAAAAAHBHQAAAAAAAoI7AAAAAAAEBHcAAAAAAAgI7gAAAAAAEBDcAQAAAAAgILgDAAAAAEBAcAcAAAAAgIDgDgAAAAAAAcEdAAAAAAACgjsAAAAAAAQEdwAAAAAACAjuAAAAAAAQENwBAAAAACAguAMAAAAAQEBwBwAAAACAgOAOAAAAAAABwR0AAAAAAAKCOwAAAAAABAR3AAAAAAAICO4AAAAAABAQ3AEAAAAAICC4AwAAAABAQHAHAAAAAICA4A4AAAAAAAHBHQAAAAAAAoI7AAAAAAAEBHcAAAAAAAgI7gAAAAAAEBDcAQAAAAAgILgDAAAAAEBAcAcAAAAAgIDgDgAAAAAAAcEdAAAAAAACgjsAAAAAAAQEdwAAAAAACAjuAAAAAAAQENwBAAAAACAguAMAAAAAQEBwBwAAAACAgOAOAAAAAAABwR0AAAAAAAKCOwAAAAAABAR3AAAAAAAICO4AAAAAABAQ3AEAAAAAICC4AwAAAABAQHAHAAAAAICA4A4AAAAAAAHBHQAAAAAAAoI7AAAAAAAEBHcAAAAAAAgI7gAAAAAAEBDcAQAAAAAgILgDAAAAAEBAcAcAAAAAgIDgDgAAAAAAAcEdAAAAAAACgjsAAAAAAAQEdwAAAAAACAjuAAAAAAAQENwBAAAAACAguAMAAAAAQEBwBwAAAACAgOAOAAAAAAABwR0AAAAAAAKCOwAAAAAABAR3AAAAAAAICO4AAAAAABAQ3AEAAAAAICC4AwAAAABAQHAHAAAAAICA4A4AAAAAAAHBHQAAAAAAAoI7AAAAAAAEBHcAAAAAAAgI7gAAAAAAEBDcAQAAAAAgILgDAAAAAEBAcAcAAAAAgIDgDgAAAAAAgaXZbPbr6dfr6+nDw79fA/yZTxcXH8/P3x8dPVlbm/cWABbe8cnJq52d1/v78x4CwML7fHn54ezs7eHhi42N5MKllZXl1dXp/f1sOk0uBGCBvDs9fT4cvjk4mPeQH57t7i4PBj8dPvr9p6PR398DZO6vrm7G46ej0XA4nPcWABbel9vbl9vb63t78x4CwMKbTiY34/Hjra31zc15bwFg4U3u7r4NBv/5r4onZQAAAAAAICC4AwAAAABA4DvcQ4kDW7m/vwAAAABJRU5ErkJggg==",
null,
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",
null,
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",
null,
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",
null,
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https://www.projecteuclid.org/euclid.aos/1473685266 | [
"The Annals of Statistics\n\nEfficient estimation in semivarying coefficient models for longitudinal/clustered data\n\nAbstract\n\nIn semivarying coefficient modeling of longitudinal/clustered data, of primary interest is usually the parametric component which involves unknown constant coefficients. First, we study semiparametric efficiency bound for estimation of the constant coefficients in a general setup. It can be achieved by spline regression using the true within-subject covariance matrices, which are often unavailable in reality. Thus, we propose an estimator when the covariance matrices are unknown and depend only on the index variable. First, we estimate the covariance matrices using residuals obtained from a preliminary estimation based on working independence and both spline and local linear regression. Then, using the covariance matrix estimates, we employ spline regression again to obtain our final estimator. It achieves the semiparametric efficiency bound under normality assumption and has the smallest asymptotic covariance matrix among a class of estimators even when normality is violated. Our theoretical results hold either when the number of within-subject observations diverges or when it is uniformly bounded. In addition, using the local linear estimator of the nonparametric component is superior to using the spline estimator in terms of numerical performance. The proposed method is compared with the working independence estimator and some existing method via simulations and application to a real data example.\n\nArticle information\n\nSource\nAnn. Statist., Volume 44, Number 5 (2016), 1988-2017.\n\nDates\nRevised: September 2015\nFirst available in Project Euclid: 12 September 2016\n\nPermanent link to this document\nhttps://projecteuclid.org/euclid.aos/1473685266\n\nDigital Object Identifier\ndoi:10.1214/15-AOS1385\n\nMathematical Reviews number (MathSciNet)\nMR3546441\n\nZentralblatt MATH identifier\n1349.62128\n\nSubjects\nPrimary: 62G08: Nonparametric regression\n\nCitation\n\nCheng, Ming-Yen; Honda, Toshio; Li, Jialiang. Efficient estimation in semivarying coefficient models for longitudinal/clustered data. Ann. Statist. 44 (2016), no. 5, 1988--2017. doi:10.1214/15-AOS1385. https://projecteuclid.org/euclid.aos/1473685266\n\nSupplemental materials\n\n• Additional simulation results and technical material. Additional simulation results, proofs of the propositions and lemmas, and theory for the case of uniformly bounded cluster size and general link function."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74061924,"math_prob":0.78499925,"size":2902,"snap":"2019-43-2019-47","text_gpt3_token_len":647,"char_repetition_ratio":0.109040715,"word_repetition_ratio":0.032432433,"special_character_ratio":0.21192281,"punctuation_ratio":0.14705883,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98017514,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T05:20:09Z\",\"WARC-Record-ID\":\"<urn:uuid:3b555359-c3e6-4767-9ab8-80c55a6a4cda>\",\"Content-Length\":\"45600\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe34d1e7-7c02-400e-aa53-02c2a852abe6>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c476bc2-1c5e-4c6f-856d-dbf60738af98>\",\"WARC-IP-Address\":\"132.236.27.47\",\"WARC-Target-URI\":\"https://www.projecteuclid.org/euclid.aos/1473685266\",\"WARC-Payload-Digest\":\"sha1:IESEVMN7AKKKML6ZRZRKME4QVAD3TBNR\",\"WARC-Block-Digest\":\"sha1:CS2N2GXHLXPKDJXI5EYLOHWD4FJYG6BK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987798619.84_warc_CC-MAIN-20191022030805-20191022054305-00127.warc.gz\"}"} |
https://math.stackexchange.com/questions/2750601/projective-objects-in-a-presheaf-topos-is-a-retract-of-representables | [
"# Projective objects in a presheaf topos is a retract of representables\n\nAn object $P$ in a category is projective if the Hom-set $Hom(P,-)$ preserves epi. Considering the presheaf topos $\\hat{C}=\\mathbf{Sets}^{C^{op}}$, how can I show that an object in this category is projective then the object is a retract of a coproduct of representables?\n\nI know that a representable is projective and the coproduct of representables is projective. I also think of that a presheaf is a colimit of representables. But I'm not sure how to relate these two to solve the question.\n\n• This isn't true for this sense of \"projective\"-the coproduct of representables has no reason to be a retract of a representable in general. The sense of \"projective\" which is relevant here is \"small-projective\": an object $x$ such that maps out of $x$ commute with arbitrary small colimits. – Kevin Carlson Apr 23 '18 at 20:16\n• @KevinCarlson Sorry that should be a retract of a coproduct of representables. And this is exercise IV.15(d) from MacLane and Moerdijk. Sheaves in geometry and logic: A first introduction to topos theory. – user301513 Apr 23 '18 at 20:37\n\nIn a category with the relevant coproducts, every colimit can be written as the coequalizer of a coproduct. In particular, every presheaf $P$ has the form of a coequalizer\n$$\\coprod_i U_i \\rightrightarrows \\coprod_j V_j \\xrightarrow{\\rho} P$$\nwhere the $U_i$ and $V_j$ are represenetables.\nCoequalizers are epic, so we can apply the given property of $P$:\n$$\\hom\\left(P, \\coprod_j V_j \\right) \\xrightarrow{\\rho_*} \\hom(P, P)$$\nis surjective. In particular, there is a map $\\lambda : P \\to \\coprod_j V_j$ such that $\\rho \\circ \\lambda = 1_P$."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91371816,"math_prob":0.9660171,"size":492,"snap":"2019-26-2019-30","text_gpt3_token_len":129,"char_repetition_ratio":0.15163934,"word_repetition_ratio":0.0,"special_character_ratio":0.22560975,"punctuation_ratio":0.07368421,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971831,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T04:28:43Z\",\"WARC-Record-ID\":\"<urn:uuid:374a15b7-7c94-40f5-997f-960cc7a49737>\",\"Content-Length\":\"139139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc9f5654-bb7a-4667-bfa9-6be692d571b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:38bfb8c7-8188-4a99-9b6b-f877823dfd1b>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2750601/projective-objects-in-a-presheaf-topos-is-a-retract-of-representables\",\"WARC-Payload-Digest\":\"sha1:4G7PONABTRUM3E6I74L52UEQQCN2WQ4B\",\"WARC-Block-Digest\":\"sha1:KLMLGRMCDT7IGWCHRJG3E47FYNHJNLRY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527474.85_warc_CC-MAIN-20190722030952-20190722052952-00045.warc.gz\"}"} |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/245/6/ | [
"## Defining parameters\n\n Level: $$N$$ = $$245 = 5 \\cdot 7^{2}$$ Weight: $$k$$ = $$6$$ Nonzero newspaces: $$12$$ Sturm bound: $$28224$$ Trace bound: $$2$$\n\n## Dimensions\n\nThe following table gives the dimensions of various subspaces of $$M_{6}(\\Gamma_1(245))$$.\n\nTotal New Old\nModular forms 12000 10489 1511\nCusp forms 11520 10199 1321\nEisenstein series 480 290 190\n\n## Trace form\n\n $$10199 q - 28 q^{2} - 70 q^{3} + 174 q^{4} - 176 q^{5} - 710 q^{6} - 268 q^{7} - 858 q^{8} + 3013 q^{9} + O(q^{10})$$ $$10199 q - 28 q^{2} - 70 q^{3} + 174 q^{4} - 176 q^{5} - 710 q^{6} - 268 q^{7} - 858 q^{8} + 3013 q^{9} + 2655 q^{10} + 1094 q^{11} - 9146 q^{12} - 9084 q^{13} - 3516 q^{14} + 2975 q^{15} + 12454 q^{16} + 8000 q^{17} + 14708 q^{18} - 11414 q^{19} - 9067 q^{20} - 2370 q^{21} + 394 q^{22} + 18654 q^{23} + 37494 q^{24} + 12652 q^{25} + 590 q^{26} - 44818 q^{27} - 46420 q^{28} - 67292 q^{29} - 92747 q^{30} - 24030 q^{31} + 96586 q^{32} + 200158 q^{33} + 268770 q^{34} + 69663 q^{35} + 339782 q^{36} + 116036 q^{37} - 143578 q^{38} - 316744 q^{39} - 451999 q^{40} - 334652 q^{41} - 446754 q^{42} - 232230 q^{43} - 187058 q^{44} + 138583 q^{45} + 564626 q^{46} + 504194 q^{47} + 1068496 q^{48} + 403016 q^{49} + 263258 q^{50} + 326770 q^{51} + 252758 q^{52} + 13136 q^{53} - 638330 q^{54} - 72862 q^{55} - 809688 q^{56} - 500530 q^{57} - 1022350 q^{58} - 674602 q^{59} - 1190735 q^{60} - 393742 q^{61} - 165018 q^{62} + 577860 q^{63} + 457414 q^{64} + 311143 q^{65} + 1333034 q^{66} + 507818 q^{67} + 664678 q^{68} + 524622 q^{69} + 713835 q^{70} + 615358 q^{71} + 1456122 q^{72} + 904920 q^{73} + 739618 q^{74} - 120811 q^{75} - 654134 q^{76} - 195930 q^{77} - 1689902 q^{78} - 1065522 q^{79} - 715408 q^{80} - 175487 q^{81} - 894184 q^{82} - 1353582 q^{83} - 3003156 q^{84} - 109119 q^{85} - 1127212 q^{86} - 1146274 q^{87} - 2060820 q^{88} - 712596 q^{89} - 1418212 q^{90} - 237746 q^{91} - 1458960 q^{92} + 393318 q^{93} + 601788 q^{94} + 389605 q^{95} + 5427332 q^{96} + 2631178 q^{97} + 5712912 q^{98} + 2785820 q^{99} + O(q^{100})$$\n\n## Decomposition of $$S_{6}^{\\mathrm{new}}(\\Gamma_1(245))$$\n\nWe only show spaces with even parity, since no modular forms exist when this condition is not satisfied. Within each space $$S_k^{\\mathrm{new}}(N, \\chi)$$ we list the newforms together with their dimension.\n\nLabel $$\\chi$$ Newforms Dimension $$\\chi$$ degree\n245.6.a $$\\chi_{245}(1, \\cdot)$$ 245.6.a.a 1 1\n245.6.a.b 1\n245.6.a.c 2\n245.6.a.d 3\n245.6.a.e 4\n245.6.a.f 5\n245.6.a.g 5\n245.6.a.h 6\n245.6.a.i 6\n245.6.a.j 8\n245.6.a.k 8\n245.6.a.l 10\n245.6.a.m 10\n245.6.b $$\\chi_{245}(99, \\cdot)$$ 245.6.b.a 2 1\n245.6.b.b 2\n245.6.b.c 12\n245.6.b.d 14\n245.6.b.e 18\n245.6.b.f 18\n245.6.b.g 32\n245.6.e $$\\chi_{245}(116, \\cdot)$$ n/a 132 2\n245.6.f $$\\chi_{245}(48, \\cdot)$$ n/a 192 2\n245.6.j $$\\chi_{245}(79, \\cdot)$$ n/a 192 2\n245.6.k $$\\chi_{245}(36, \\cdot)$$ n/a 552 6\n245.6.l $$\\chi_{245}(68, \\cdot)$$ n/a 384 4\n245.6.p $$\\chi_{245}(29, \\cdot)$$ n/a 828 6\n245.6.q $$\\chi_{245}(11, \\cdot)$$ n/a 1128 12\n245.6.s $$\\chi_{245}(13, \\cdot)$$ n/a 1656 12\n245.6.t $$\\chi_{245}(4, \\cdot)$$ n/a 1656 12\n245.6.x $$\\chi_{245}(3, \\cdot)$$ n/a 3312 24\n\n\"n/a\" means that newforms for that character have not been added to the database yet\n\n## Decomposition of $$S_{6}^{\\mathrm{old}}(\\Gamma_1(245))$$ into lower level spaces\n\n$$S_{6}^{\\mathrm{old}}(\\Gamma_1(245)) \\cong$$ $$S_{6}^{\\mathrm{new}}(\\Gamma_1(5))$$$$^{\\oplus 3}$$$$\\oplus$$$$S_{6}^{\\mathrm{new}}(\\Gamma_1(7))$$$$^{\\oplus 4}$$$$\\oplus$$$$S_{6}^{\\mathrm{new}}(\\Gamma_1(35))$$$$^{\\oplus 2}$$$$\\oplus$$$$S_{6}^{\\mathrm{new}}(\\Gamma_1(49))$$$$^{\\oplus 2}$$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.50245786,"math_prob":1.000008,"size":2040,"snap":"2021-43-2021-49","text_gpt3_token_len":893,"char_repetition_ratio":0.2308448,"word_repetition_ratio":0.0,"special_character_ratio":0.6039216,"punctuation_ratio":0.20948617,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999658,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T12:42:56Z\",\"WARC-Record-ID\":\"<urn:uuid:463e8482-32ca-45ad-a7d0-564021b57335>\",\"Content-Length\":\"144687\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27e85fda-4caa-43b5-8b77-614d69f0727c>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ebef472-2871-4a5d-b7a0-19e9af27ffb8>\",\"WARC-IP-Address\":\"35.241.19.59\",\"WARC-Target-URI\":\"https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/245/6/\",\"WARC-Payload-Digest\":\"sha1:JUPYABDVR7Q3NZRMRVMWBUIQXBGEOVDB\",\"WARC-Block-Digest\":\"sha1:JOD7KCSYIPGYYDKFYIDYGXO4OIGVTOC3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363292.82_warc_CC-MAIN-20211206103243-20211206133243-00522.warc.gz\"}"} |
https://homework.zookal.com/questions-and-answers/here-is-the-code-to-complete-import-javautilscanner-fixme1-add-505520797 | [
"1. Engineering\n2. Computer Science\n3. here is the code to complete import javautilscanner fixme1 add...\n\n# Question: here is the code to complete import javautilscanner fixme1 add...\n\n###### Question details",
null,
"Here is the code to complete:\n\nimport java.util.Scanner;\n//Fixme(1) add a statement to import the Random class\n\npublic class RandomTest {\npublic static void main(String[] args) {\nScanner scnr = new Scanner(System.in);\n\nSystem.out.println(\"Enter the seed: \");\nint seed = scnr.nextInt();\n\n//Fixme(2) create a Random object with the user entered seed\n\n//Fixme(3) Generate and print out a random integer in the range of 0 to 100.\n\n//Fixme(4) Generate and print out a random integer in the range of -50 to 50.\n\n//Fixme(5) Generate and print out a random decimal number in the range of 0 to 1.0.\n\n//Fixme(6) Generate and print out a random decimal number in the range of 0 to 100.0.\n}\n}\n\n###### Solution by an expert tutor",
null,
""
] | [
null,
"https://homework-api-assets-production.s3.ap-southeast-2.amazonaws.com/uploads/store/505520797/1601595111502e331d678d8b54ddee4c37b14b74e8.png",
null,
"https://homework.zookal.com/images/blurredbg-mobile.png",
null
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https://lbs-to-kg.appspot.com/33.6-lbs-to-kg.html | [
"Pounds To Kg\n\n# 33.6 lbs to kg33.6 Pounds to Kilograms\n\nlbs\n=\nkg\n\n## How to convert 33.6 pounds to kilograms?\n\n 33.6 lbs * 0.45359237 kg = 15.240703632 kg 1 lbs\nA common question is How many pound in 33.6 kilogram? And the answer is 74.0753200941 lbs in 33.6 kg. Likewise the question how many kilogram in 33.6 pound has the answer of 15.240703632 kg in 33.6 lbs.\n\n## How much are 33.6 pounds in kilograms?\n\n33.6 pounds equal 15.240703632 kilograms (33.6lbs = 15.240703632kg). Converting 33.6 lb to kg is easy. Simply use our calculator above, or apply the formula to change the length 33.6 lbs to kg.\n\n## Convert 33.6 lbs to common mass\n\nUnitMass\nMicrogram15240703632.0 µg\nMilligram15240703.632 mg\nGram15240.703632 g\nOunce537.6 oz\nPound33.6 lbs\nKilogram15.240703632 kg\nStone2.4 st\nUS ton0.0168 ton\nTonne0.0152407036 t\nImperial ton0.015 Long tons\n\n## What is 33.6 pounds in kg?\n\nTo convert 33.6 lbs to kg multiply the mass in pounds by 0.45359237. The 33.6 lbs in kg formula is [kg] = 33.6 * 0.45359237. Thus, for 33.6 pounds in kilogram we get 15.240703632 kg.\n\n## 33.6 Pound Conversion Table",
null,
"## Alternative spelling\n\n33.6 lbs to kg, 33.6 lbs in kg, 33.6 lb to Kilogram, 33.6 lb in Kilogram, 33.6 Pounds to Kilograms, 33.6 Pounds in Kilograms, 33.6 Pounds to Kilogram, 33.6 Pounds in Kilogram, 33.6 lb to kg, 33.6 lb in kg, 33.6 lb to Kilograms, 33.6 lb in Kilograms, 33.6 lbs to Kilograms, 33.6 lbs in Kilograms, 33.6 lbs to Kilogram, 33.6 lbs in Kilogram, 33.6 Pound to Kilograms, 33.6 Pound in Kilograms"
] | [
null,
"https://lbs-to-kg.appspot.com/image/33.6.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8043011,"math_prob":0.9782608,"size":1149,"snap":"2022-27-2022-33","text_gpt3_token_len":416,"char_repetition_ratio":0.279476,"word_repetition_ratio":0.019607844,"special_character_ratio":0.4273281,"punctuation_ratio":0.23125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9578571,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T09:28:55Z\",\"WARC-Record-ID\":\"<urn:uuid:901cc734-a667-4d83-a82d-ddd597b1dfec>\",\"Content-Length\":\"28406\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41e7fd00-e4e3-42ca-8f7e-e57cd0f3c2f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:3646832c-562c-4491-a1ef-4efbe0d59aff>\",\"WARC-IP-Address\":\"142.250.81.212\",\"WARC-Target-URI\":\"https://lbs-to-kg.appspot.com/33.6-lbs-to-kg.html\",\"WARC-Payload-Digest\":\"sha1:2TI4NULH4L52H4EOTPP4A3IWLXOFPY2Q\",\"WARC-Block-Digest\":\"sha1:F3WSXWFO7D2C6QU4XVW2O2EWL6WJEPPQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571246.56_warc_CC-MAIN-20220811073058-20220811103058-00647.warc.gz\"}"} |
http://allaboutasp.net/2012/11/write-a-program-for-selection-sort/ | [
"Selection sort is a sorting algorithm, specifically an in-place comparison sort. It has O(n2) time complexity, making it inefficient on large lists, and generally performs worse than the similar insertion sort. Selection sort is noted for its simplicity, and it has performance advantages over more complicated algorithms in certain situations, particularly where auxiliary memory is limited.\n\nThe algorithm divides the input list into two parts: the sublist of items already sorted, which is built up from left to right at the front (left) of the list, and the sublist of items remaining to be sorted that occupy the rest of the list. Initially, the sorted sublist is empty and the unsorted sublist is the entire input list. The algorithm proceeds by finding the smallest (or largest, depending on sorting order) element in the unsorted sublist, exchanging it with the leftmost unsorted element (putting it in sorted order), and moving the sublist boundaries one element to the right\n\nCode\n\npublic static int[] SelectionSort(int[] input)\n{\nStreamWriter sw = new StreamWriter(@\"Sorting.txt\", true);\nsw.WriteLine(\"\\n\\t\\t Selecton Sorting \\n\");\nsw.WriteLine(System.DateTime.Now + \"\\t input Array \\t Total Element \\t\"\n+ input.Length); sw.WriteLine(string.Join(\",\", input)); int i, j; i = j = 0; int temp = 0; int iMin = 0; for (i = 0; i < input.Length; i++) { for (j = i + 1; j < input.Length; j++) { if (input[j] < input[iMin]) { iMin = j; } } if (iMin != i && iMin < input.Length) { temp = input[iMin]; input[iMin] = input[i]; input[i] = temp; } } sw.WriteLine(System.DateTime.Now + \"\\t Sorted Array \\n\"\n+ string.Join(\",\", input)); sw.Close(); return input; }\nTest Case\n\nprivate static void SelectionSortTest() {\nList<int> input = new List<int>();\nRandom rnd = new Random(1);\nint maxnumber = 1200;\n\ninput.Clear();\nfor (int i = 1; i <= maxnumber; i++)\n{\n}\nSorting.SelectionSort(input.ToArray());\n\ninput.Clear();\nmaxnumber = 100000;\nfor (int i = 1; i <= maxnumber; i++)\n{"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6417455,"math_prob":0.9806119,"size":2295,"snap":"2019-43-2019-47","text_gpt3_token_len":567,"char_repetition_ratio":0.13574858,"word_repetition_ratio":0.10557185,"special_character_ratio":0.28845316,"punctuation_ratio":0.22052401,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939898,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T02:16:12Z\",\"WARC-Record-ID\":\"<urn:uuid:5ef0ee90-d03e-4238-b11b-969ccb38d5e5>\",\"Content-Length\":\"30830\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f26fb5ea-0758-49a8-ad67-d4161f784ce6>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ed02837-ea4d-417d-ad4d-2edd01b7ae6b>\",\"WARC-IP-Address\":\"50.62.26.129\",\"WARC-Target-URI\":\"http://allaboutasp.net/2012/11/write-a-program-for-selection-sort/\",\"WARC-Payload-Digest\":\"sha1:XUC7G3YPH7UQIB5LVEGTXWAUBQUJREO4\",\"WARC-Block-Digest\":\"sha1:TRLPGQ72LOUO73ZN5YL32GYAITH2JAI3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986661296.12_warc_CC-MAIN-20191016014439-20191016041939-00427.warc.gz\"}"} |
https://github.com/parrt/dtreeviz | [
"Skip to content\n{{ message }}\n\n# parrt / dtreeviz\n\nA python library for decision tree visualization and model interpretation.\n\nSwitch branches/tags\nCould not load branches\nNothing to show\n\n## Latest commit\n\n* Display just a range of tree levels\n\n* Edit docs for depth_range_to_display parameter\nd116cc4\n\n## Files\n\nFailed to load latest commit information.\nType\nName\nCommit time\nSep 1, 2020\nJan 10, 2019\nApr 5, 2021\nAug 13, 2018\nApr 5, 2021\nApr 5, 2021\n\n# dtreeviz : Decision Tree Visualization\n\n## Description\n\nA python library for decision tree visualization and model interpretation. Currently supports scikit-learn, XGBoost, Spark MLlib, and LightGBM trees. With 1.3, we now provide one- and two-dimensional feature space illustrations for classifiers (any model that can answer predict_probab()); see below.\n\nAuthors:\n\nSee How to visualize decision trees for deeper discussion of our decision tree visualization library and the visual design decisions we made.\n\n### Feedback\n\nWe welcome info from users on how they use dtreeviz, what features they'd like, etc... via email (to parrt) or via an issue.\n\n## Quick start\n\nJump right into the examples using this Colab notebook\n\nTake a look in notebooks! Here we have a specific notebook for all supported ML libraries and more.\n\n## Discussion\n\nDecision trees are the fundamental building block of gradient boosting machines and Random Forests(tm), probably the two most popular machine learning models for structured data. Visualizing decision trees is a tremendous aid when learning how these models work and when interpreting models. Unfortunately, current visualization packages are rudimentary and not immediately helpful to the novice. For example, we couldn't find a library that visualizes how decision nodes split up the feature space. It is also uncommon for libraries to support visualizing a specific feature vector as it weaves down through a tree's decision nodes; we could only find one image showing this.\n\nSo, we've created a general package for decision tree visualization and model interpretation, which we'll be using heavily in an upcoming machine learning book (written with Jeremy Howard).\n\nThe visualizations are inspired by an educational animation by R2D3; A visual introduction to machine learning. With dtreeviz, you can visualize how the feature space is split up at decision nodes, how the training samples get distributed in leaf nodes, how the tree makes predictions for a specific observation and more. These operations are critical to for understanding how classification or regression decision trees work. If you're not familiar with decision trees, check out fast.ai's Introduction to Machine Learning for Coders MOOC.\n\n## Install\n\nInstall anaconda3 on your system, if not already done.\n\nYou might verify that you do not have conda-installed graphviz-related packages installed because dtreeviz needs the pip versions; you can remove them from conda space by doing:\n\nconda uninstall python-graphviz\nconda uninstall graphviz\n\nTo install (Python >=3.6 only), do this (from Anaconda Prompt on Windows!):\n\npip install dtreeviz # install dtreeviz for sklearn\npip install dtreeviz[xgboost] # install XGBoost related dependency\npip install dtreeviz[pyspark] # install pyspark related dependency\npip install dtreeviz[lightgbm] # install LightGBM related dependency\n\nThis should also pull in the graphviz Python library (>=0.9), which we are using for platform specific stuff.\n\nLimitations. Only svg files can be generated at this time, which reduces dependencies and dramatically simplifies install process.\n\nPlease email Terence with any helpful notes on making dtreeviz work (better) on other platforms. Thanks!\n\nFor your specific platform, please see the following subsections.\n\n### Mac\n\nMake sure to have the latest XCode installed and command-line tools installed. You can run xcode-select --install from the command-line to install those if XCode is already installed. You also have to sign the XCode license agreement, which you can do with sudo xcodebuild -license from command-line. The brew install shown next needs to build graphviz, so you need XCode set up properly.\n\nYou need the graphviz binary for dot. Make sure you have latest version (verified on 10.13, 10.14):\n\nbrew reinstall graphviz\n\nJust to be sure, remove dot from any anaconda installation, for example:\n\nrm ~/anaconda3/bin/dot\n\nFrom command line, this command\n\ndot -Tsvg\n\nshould work, in the sense that it just stares at you without giving an error. You can hit control-C to escape back to the shell. Make sure that you are using the right dot as installed by brew:\n\n$which dot /usr/local/bin/dot$ ls -l $(which dot) lrwxr-xr-x 1 parrt wheel 33 May 26 11:04 /usr/local/bin/dot@ -> ../Cellar/graphviz/2.40.1/bin/dot$\n\nLimitations. Jupyter notebook has a bug where they do not show .svg files correctly, but Juypter Lab has no problem.\n\n### Linux (Ubuntu 18.04)\n\nTo get the dot binary do:\n\nsudo apt install graphviz\n\nLimitations. The view() method works to pop up a new window and images appear inline for jupyter notebook but not jupyter lab (It gets an error parsing the SVG XML.) The notebook images also have a font substitution from the Arial we use and so some text overlaps. Only .svg files can be generated on this platform.\n\n### Windows 10\n\n(Make sure to pip install graphviz, which is common to all platforms, and make sure to do this from Anaconda Prompt on Windows!)\n\nDownload graphviz-2.38.msi and update your Path environment variable. Add C:\\Program Files (x86)\\Graphviz2.38\\bin to User path and C:\\Program Files (x86)\\Graphviz2.38\\bin\\dot.exe to System Path. It's windows so you might need a reboot after updating that environment variable. You should see this from the Anaconda Prompt:\n\n(base) C:\\Users\\Terence Parr>where dot\nC:\\Program Files (x86)\\Graphviz2.38\\bin\\dot.exe\n\n\n(Do not use conda install -c conda-forge python-graphviz as you get an old version of graphviz python library.)\n\nVerify from the Anaconda Prompt that this works (capital -V not lowercase -v):\n\ndot -V\n\n\nIf it doesn't work, you have a Path problem. I found the following test programs useful. The first one sees if Python can find dot:\n\nimport os\nimport subprocess\nproc = subprocess.Popen(['dot','-V'])\nprint( os.getenv('Path') )\n\nThe following version does the same thing except uses graphviz Python libraries backend support utilities, which is what we use in dtreeviz:\n\nimport graphviz.backend as be\ncmd = [\"dot\", \"-V\"]\nstdout, stderr = be.run(cmd, capture_output=True, check=True, quiet=False)\nprint( stderr )\n\nIf you are having issues with run command you can try copying the following files from: https://github.com/xflr6/graphviz/tree/master/graphviz.\n\nPlace them in the AppData\\Local\\Continuum\\anaconda3\\Lib\\site-packages\\graphviz folder.\n\nClean out the pycache directory too.\n\nJupyter Lab and Jupyter notebook both show the inline .svg images well.\n\n### Verify graphviz installation\n\nTry making text file t.dot with content digraph T { A -> B } (paste that into a text editor, for example) and then running this from the command line:\n\ndot -Tsvg -o t.svg t.dot\n\n\nThat should give a simple t.svg file that opens properly. If you get errors from dot, it will not work from the dtreeviz python code. If it can't find dot then you didn't update your PATH environment variable or there is some other install issue with graphviz.\n\n### Limitations\n\nFinally, don't use IE to view .svg files. Use Edge as they look much better. I suspect that IE is displaying them as a rasterized not vector images. Only .svg files can be generated on this platform.\n\n## Usage\n\ndtree: Main function to create decision tree visualization. Given a decision tree regressor or classifier, creates and returns a tree visualization using the graphviz (DOT) language.\n\n### Required libraries\n\nBasic libraries and imports that will (might) be needed to generate the sample visualizations shown in examples below.\n\nfrom sklearn.datasets import *\nfrom sklearn import tree\nfrom dtreeviz.trees import *\n\n### Regression decision tree\n\nThe default orientation of tree is top down but you can change it to left to right using orientation=\"LR\". view() gives a pop up window with rendered graphviz object.\n\nregr = tree.DecisionTreeRegressor(max_depth=2)\nboston = load_boston()\nregr.fit(boston.data, boston.target)\n\nviz = dtreeviz(regr,\nboston.data,\nboston.target,\ntarget_name='price',\nfeature_names=boston.feature_names)\n\nviz.view()",
null,
"### Classification decision tree\n\nAn additional argument of class_names giving a mapping of class value with class name is required for classification trees.\n\nclassifier = tree.DecisionTreeClassifier(max_depth=2) # limit depth of tree\niris = load_iris()\nclassifier.fit(iris.data, iris.target)\n\nviz = dtreeviz(classifier,\niris.data,\niris.target,\ntarget_name='variety',\nfeature_names=iris.feature_names,\nclass_names=[\"setosa\", \"versicolor\", \"virginica\"] # need class_names for classifier\n)\n\nviz.view()",
null,
"### Prediction path\n\nHighlights the decision nodes in which the feature value of single observation passed in argument X falls. Gives feature values of the observation and highlights features which are used by tree to traverse path.\n\nregr = tree.DecisionTreeRegressor(max_depth=2) # limit depth of tree\ndiabetes = load_diabetes()\nregr.fit(diabetes.data, diabetes.target)\nX = diabetes.data[np.random.randint(0, len(diabetes.data)),:] # random sample from training\n\nviz = dtreeviz(regr,\ndiabetes.data,\ndiabetes.target,\ntarget_name='value',\norientation ='LR', # left-right orientation\nfeature_names=diabetes.feature_names,\nX=X) # need to give single observation for prediction\n\nviz.view()",
null,
"If you want to visualize just the prediction path, you need to set parameter show_just_path=True\n\ndtreeviz(regr,\ndiabetes.data,\ndiabetes.target,\ntarget_name='value',\norientation ='TD', # top-down orientation\nfeature_names=diabetes.feature_names,\nX=X, # need to give single observation for prediction\nshow_just_path=True\n)",
null,
"#### Explain prediction path\n\nThese visualizations are useful to explain to somebody, without machine learning skills, why your model made that specific prediction.\nIn case of explanation_type=plain_english, it searches in prediction path and find feature value ranges.\n\nX = dataset[features].iloc\nprint(X)\nPclass 3.0\nAge 4.0\nFare 16.7\nSex_label 0.0\nCabin_label 145.0\nEmbarked_label 2.0\n\nprint(explain_prediction_path(tree_classifier, X, feature_names=features, explanation_type=\"plain_english\"))\n2.5 <= Pclass\nAge < 36.5\nFare < 23.35\nSex_label < 0.5\n\n\nIn case of explanation_type=sklearn_default (available only for scikit-learn), we can visualize the features' importance involved in prediction path only. Features' importance is calculated based on mean decrease in impurity.\nCheck Beware Default Random Forest Importances article for a comparison between features' importance based on mean decrease in impurity vs permutation importance.\n\nexplain_prediction_path(tree_classifier, X, feature_names=features, explanation_type=\"sklearn_default\")",
null,
"### Decision tree without scatterplot or histograms for decision nodes\n\nSimple tree without histograms or scatterplots for decision nodes. Use argument fancy=False\n\nclassifier = tree.DecisionTreeClassifier(max_depth=4) # limit depth of tree\ncancer = load_breast_cancer()\nclassifier.fit(cancer.data, cancer.target)\n\nviz = dtreeviz(classifier,\ncancer.data,\ncancer.target,\ntarget_name='cancer',\nfeature_names=cancer.feature_names,\nclass_names=[\"malignant\", \"benign\"],\nfancy=False ) # fance=False to remove histograms/scatterplots from decision nodes\n\nviz.view()",
null,
"For more examples and different implementations, please see the jupyter notebook full of examples.\n\n### Regression univariate feature-target space",
null,
"import pandas as pd\nimport matplotlib.pyplot as plt\nfrom sklearn.tree import DecisionTreeRegressor\nfrom dtreeviz.trees import *\n\ndf_cars = pd.read_csv(\"cars.csv\")\nX, y = df_cars[['WGT']], df_cars['MPG']\n\ndt = DecisionTreeRegressor(max_depth=3, criterion=\"mae\")\ndt.fit(X, y)\n\nfig = plt.figure()\nax = fig.gca()\nrtreeviz_univar(dt, X, y, 'WGT', 'MPG', ax=ax)\nplt.show()\n\n### Regression bivariate feature-target space",
null,
"from mpl_toolkits.mplot3d import Axes3D\nfrom sklearn.tree import DecisionTreeRegressor\nfrom dtreeviz.trees import *\n\ndf_cars = pd.read_csv(\"cars.csv\")\nX = df_cars[['WGT','ENG']]\ny = df_cars['MPG']\n\ndt = DecisionTreeRegressor(max_depth=3, criterion=\"mae\")\ndt.fit(X, y)\n\nfigsize = (6,5)\nfig = plt.figure(figsize=figsize)\nax = fig.add_subplot(111, projection='3d')\n\nt = rtreeviz_bivar_3D(dt,\nX, y,\nfeature_names=['Vehicle Weight', 'Horse Power'],\ntarget_name='MPG',\nfontsize=14,\nelev=20,\nazim=25,\ndist=8.2,\nshow={'splits','title'},\nax=ax)\nplt.show()\n\n### Regression bivariate feature-target space heatmap",
null,
"from sklearn.tree import DecisionTreeRegressor\nfrom dtreeviz.trees import *\n\ndf_cars = pd.read_csv(\"cars.csv\")\nX = df_cars[['WGT','ENG']]\ny = df_cars['MPG']\n\ndt = DecisionTreeRegressor(max_depth=3, criterion=\"mae\")\ndt.fit(X, y)\n\nt = rtreeviz_bivar_heatmap(dt,\nX, y,\nfeature_names=['Vehicle Weight', 'Horse Power'],\nfontsize=14)\n\nplt.show()\n\n### Classification univariate feature-target space",
null,
"from sklearn.tree import DecisionTreeClassifier\nfrom dtreeviz.trees import *\n\nknow = pd.read_csv(\"knowledge.csv\")\nclass_names = ['very_low', 'Low', 'Middle', 'High']\nknow['UNS'] = know['UNS'].map({n: i for i, n in enumerate(class_names)})\n\nX = know[['PEG']]\ny = know['UNS']\n\ndt = DecisionTreeClassifier(max_depth=3)\ndt.fit(X, y)\n\nct = ctreeviz_univar(dt, X, y,\nfeature_names = ['PEG'],\nclass_names=class_names,\ntarget_name='Knowledge',\nnbins=40, gtype='strip',\nshow={'splits','title'})\nplt.tight_layout()\nplt.show()\n\n### Classification bivariate feature-target space",
null,
"from sklearn.tree import DecisionTreeClassifier\nfrom dtreeviz.trees import *\n\nknow = pd.read_csv(\"knowledge.csv\")\nprint(know)\nclass_names = ['very_low', 'Low', 'Middle', 'High']\nknow['UNS'] = know['UNS'].map({n: i for i, n in enumerate(class_names)})\n\nX = know[['PEG','LPR']]\ny = know['UNS']\n\ndt = DecisionTreeClassifier(max_depth=3)\ndt.fit(X, y)\n\nct = ctreeviz_bivar(dt, X, y,\nfeature_names = ['PEG','LPR'],\nclass_names=class_names,\ntarget_name='Knowledge')\nplt.tight_layout()\nplt.show()\n\n### Leaf node purity\n\nLeaf purity affects prediction confidence.\nFor classification leaf purity is calculated based on majority target class (gini, entropy) and for regression is calculated based on target variance values.\nLeaves with low variance among the target values (regression) or an overwhelming majority target class (classification) are much more reliable predictors. When we have a decision tree with a high depth, it can be difficult to get an overview about all leaves purities. That's why we created a specialized visualization only for leaves purities.\n\ndisplay_type can take values 'plot' (default), 'hist' or 'text'\n\nviz_leaf_criterion(tree_classifier, display_type = \"plot\")",
null,
"### Leaf node samples\n\nIt's also important to take a look at the number of samples from leaves. For example, we can have a leaf with a good purity but very few samples, which is a sign of overfitting. The ideal scenario would be to have a leaf with good purity which is based on a significant number of samples.\n\ndisplay_type can take values 'plot' (default), 'hist' or 'text'\n\nviz_leaf_samples(tree_classifier, dataset[features], display_type='plot')",
null,
"#### Leaf node samples for classification\n\nThis is a specialized visualization for classification. It helps also to see the distribution of target class values from leaf samples.\n\nctreeviz_leaf_samples(tree_classifier, dataset[features], dataset[target])",
null,
"### Leaf plots\n\nVisualize leaf target distribution for regression decision trees.\n\nviz_leaf_target(tree_regressor, dataset[features_reg], dataset[target_reg], features_reg, target_reg)",
null,
"## Classification boundaries in feature space\n\nWith 1.3, we have introduced method clfviz() that illustrates one and two-dimensional feature space for classifiers, including colors the represent probabilities, decision boundaries, and misclassified entities. This method works with any model that answers method predict_proba() (and we also support Keras), so any model from scikit-learn should work. If you let us know about incompatibilities, we can support more models. There are lots of options would you can check out in the api documentation. See classifier-decision-boundaries.ipynb and classifier-boundary-animations.ipynb.\n\nclfviz(rf, X, y, feature_names=['x1', 'x2'], markers=['o','X','s','D'], target_name='smiley')",
null,
"clfviz(rf,x,y,feature_names=['f27'],target_name='cancer')",
null,
"clfviz(rf,x,y,\nfeature_names=['x2'],\ntarget_name = 'smiley',\ncolors={'scatter_marker_alpha':.2})",
null,
"Sometimes it's helpful to see animations that change some of the hyper parameters. If you look in notebook classifier-boundary-animations.ipynb, you will see code that generates animations such as the following (animated png files):\n\n## Visualization methods setup\n\nStarting with dtreeviz 1.0 version, we refactored the concept of ShadowDecTree. If we want to add a new ML library in dtreeviz, we just need to add a new implementation of ShadowDecTree API, like ShadowSKDTree, ShadowXGBDTree or ShadowSparkTree.\n\nInitializing a ShadowSKDTree object:\n\nsk_dtree = ShadowSKDTree(tree_classifier, dataset[features], dataset[target], features, target, [0, 1])\n\n\nOnce we have the object initialized, we can used it to create all the visualizations, like :\n\ndtreeviz(sk_dtree)\n\nviz_leaf_samples(sk_dtree)\n\nviz_leaf_criterion(sk_dtree)\n\n\nIn this way, we reduced substantially the list of parameters required for each visualization and it's also more efficient in terms of computing power.\n\nYou can check the notebooks section for more examples of using ShadowSKDTree, ShadowXGBDTree or ShadowSparkTree.\n\n## Install dtreeviz locally\n\nMake sure to follow the install guidelines above.\n\nTo push the dtreeviz library to your local egg cache (force updates) during development, do this (from anaconda prompt on Windows):\n\npython setup.py install -f\n\nE.g., on Terence's box, it add /Users/parrt/anaconda3/lib/python3.6/site-packages/dtreeviz-0.3-py3.6.egg.\n\n## Customize colors\n\nEach function has an optional parameter colors which allows passing a dictionary of colors which is used in the plot. For an example of each parameter have a look at this notebook.\n\n### Example\n\ndtreeviz.trees.dtreeviz(regr,\nboston.data,\nboston.target,\ntarget_name='price',\nfeature_names=boston.feature_names,\ncolors={'scatter_marker': '#00ff00'})\n\nwould paint the scatter (dots) in red.",
null,
"### Parameters\n\nThe colors are defined in colors.py, all options and default parameters are shown below.\n\n COLORS = {'scatter_edge': GREY,\n'scatter_marker': BLUE,\n'split_line': GREY,\n'mean_line': '#f46d43',\n'axis_label': GREY,\n'title': GREY,\n'legend_title': GREY,\n'legend_edge': GREY,\n'edge': GREY,\n'color_map_min': '#c7e9b4',\n'color_map_max': '#081d58',\n'classes': color_blind_friendly_colors,\n'rect_edge': GREY,\n'text': GREY,\n'highlight': HIGHLIGHT_COLOR,\n'wedge': WEDGE_COLOR,\n'text_wedge': WEDGE_COLOR,\n'arrow': GREY,\n'node_label': GREY,\n'tick_label': GREY,\n'leaf_label': GREY,\n'pie': GREY,\n}\n\nThe color needs be in a format matplotlib can interpret, e.g. a html hex like '#eeefff' .\n\nclasses needs to be a list of lists of colors with a minimum length of your number of colors. The index is the number of classes and the list with this index needs to have the same amount of colors.\n\n## Authors\n\nSee also the list of contributors who participated in this project.\n\n## License\n\nThis project is licensed under the terms of the MIT license, see LICENSE.\n\n## Deploy\n\n\\$ python setup.py sdist upload\n\n\n## About\n\nA python library for decision tree visualization and model interpretation.\n\n## Packages 0\n\nNo packages published"
] | [
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null,
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null,
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null,
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null,
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null,
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https://exceptionshub.com/php-how-can-i-make-timediff-function-to-convert-negative-time-to-positive.html | [
"Home » Php » php – How can I make timediff function to convert negative time to positive?\n\n# php – How can I make timediff function to convert negative time to positive?\n\nQuestions:\n\nI have query where I have to find the difference between two times , it works well when the first time is greater than the second time .\n\nBut when the first one is smaller than the second one it returns a negative time; Just wondering is there any way to make it positive?\n\nBelow is the query that returns negative time.\n\n``````SELECT TIMEDIFF(\"13:10:11\", \"13:20:10\");\n``````\n\noutput\n\n``````-00:09:59\n``````\n\nexpected output\n\n``````00:09:59\n``````\n\nIf what you want is to get the absolute time difference between two times, you could use a `CASE` expression to swap the times so that the first was always greater than the second e.g.\n\n``````SELECT CASE WHEN time1 < time2 THEN TIMEDIFF(time2, time1)\nELSE TIMEDIFF(time1, time2)\nEND AS delta\n``````\n\n`````` SELECT DATEDIFF(MINUTE, '11:10:10' , '11:20:00') AS MinuteDiff"
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https://www.esaim-cocv.org/articles/cocv/ref/2021/01/cocv200005/cocv200005.html | [
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http://pomucoco.com/Info/single-101.shtml.htm | [
"",
null,
"负责办理传统民事法律事务，以稳中求精为原则，根据当事人的委托，处理非诉讼法律事务，代理委托人参加各类诉讼和仲裁活动，并在诉讼和仲裁相关的调解、调查等方面提供法律服务。\n\n提供：人身损害诉讼、医疗纠纷诉讼、交通事故诉讼、婚姻家庭诉讼、产品质量法律服务等。\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`"
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http://www.sport-tx.com/datasheet/MC1413/Z2xqbG2WZw==.html | [
"< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\"> < id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\"> < id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">< id=\"6e2kg\">",
null,
"型号： MC1413 厂商：",
null,
"ON Semiconductor 分类： 半导体 描述： peripheral driver arrays PDF： 预览 报错 收藏 赞\n\nDatasheet下载地址\n\nMC1413的详细信息\n\n•",
null,
"• 亲，您要的详细信息都在下载文档里了！\n\nOrder this document by MC1413/D\nThe seven NPN Darlington connected transistors in these arrays are well\nsuited for driving lamps, relays, or printer hammers in a variety of industrial\nand consumer applications. Their high breakdown voltage and internal\nsuppression diodes insure freedom from problems associated with inductive\nloads. Peak inrush currents to 500 mA permit them to drive incandescent\nlamps.\nPERIPHERAL\nDRIVER ARRAYS\nSEMICONDUCTOR\nTECHNICAL DATA\nThe MC1413, B with a 2.7 k? series input resistor is well suited for\nsystems utilizing a 5.0 V TTL or CMOS Logic. The MC1416, B uses a series\n10.5 k? resistor and is useful in 8.0 to 18 V MOS systems.\n16\n1\nP SUFFIX\nPLASTIC PACKAGE\nCASE 648\nORDERING INFORMATION\nOperating\nTemperature Range\nPlastic DIP\nSOIC\nMC1413P (ULN2003A)\nMC1416P (ULN2004A)\nMC1413D\nMC1416D\nT\n= –20° to +85°C\n= –40° to +85°C\nA\n16\n1\nMC1413BP\nMC1416BP\nMC1413BD\nMC1416BD\nD SUFFIX\nPLASTIC PACKAGE\nCASE 751B\nT\nA\n(SO–16)\nPIN CONNECTIONS\nRepresentative Schematic Diagrams\n1\n2\n3\n4\n5\n6\n16\n15\n14\n13\n12\n11\n10\n9\n1/7 MC1413, B\n2.7 k\n5.0 k\nPin 9\n3.0 k\n1/7 MC1416, B\n10.5 k\n5.0 k\nPin 9\n7\n8\n3.0 k\n(Top View)\n?\nMotorola, Inc. 1996\nRev 0",
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"http://p.datasheet.21ic.com/source3/img/partimg/2014101904/35/6be9a99d-f27b-4b38-9f35-4ad14666ab56.jpeg",
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"http://www.sport-tx.com/logo/112ONSEMI.GIF",
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"http://www.sport-tx.com/datasheet/qrcode.html",
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https://pyquil-docs.rigetti.com/en/v2.19.0/migration2-qc.html | [
"# New in Forest 2 - QuantumComputer¶\n\nPyQuil is for constructing and running quantum programs on real quantum computers. With the release of pyQuil 2, we have changed parts of the API to better reflect that focus. Instead of swapping between a QVMConnection and a QPUConnection, you will primarily deal with a QuantumComputer with consistent API and behavior regardless of\n\n• QVM / QPU\n• Presence of noise model\n• Device topology\n\n## Running a program¶\n\nLet’s show how you can run a simple program on a QuantumComputer first we start with the relevant imports.\n\n:\n\nfrom pyquil import Program\nfrom pyquil.gates import *\n\n\nWe’ll write a function that takes a list of qubits and returns a pyQuil Program that constructs an entangled “GHZ” state. This is a generalization of the two-qubit Bell state.\n\n:\n\ndef ghz_state(qubits):\n\"\"\"Create a GHZ state on the given list of qubits by applying\na Hadamard gate to the first qubit followed by a chain of CNOTs\n\"\"\"\nprogram = Program()\nprogram += H(qubits)\nfor q1, q2 in zip(qubits, qubits[1:]):\nprogram += CNOT(q1, q2)\nreturn program\n\n\nFor example, creating a GHZ state on qubits 1, 2, and 3 would look like:\n\n:\n\nprogram = ghz_state(qubits=[0, 1, 2])\nprint(program)\n\nH 0\nCNOT 0 1\nCNOT 1 2\n\n\n\n## Debugging with WavefunctionSimulator¶\n\nWe can check that this program gives us the desired wavefunction by using WavefunctionSimulator.wavefunction()\n\n:\n\nfrom pyquil.api import WavefunctionSimulator\nwfn = WavefunctionSimulator().wavefunction(program)\nprint(wfn)\n\n(0.7071067812+0j)|000> + (0.7071067812+0j)|111>\n\n\nWe can’t get the wavefunction from a real quantum computer though, so instead we’ll sample bitstrings. We expect to always measure the bitstring 000 or the bitstring 111 based on the definition of a GHZ state and confirmed by our wavefunction simulation.\n\n## get_qc¶\n\nWe’ll construct a QuantumComputer via the helper method get_qc. You may be tempted to use the QuantumComputer constructor directly. Please refer to the advanced documentation to see how to do that. Our program uses 3 qubits, so we’ll ask for a 3-qubit QVM.\n\n:\n\nfrom pyquil import get_qc\nqc = get_qc('3q-qvm')\nqc\n\n:\n\nQuantumComputer[name=\"3q-qvm\"]\n\n\nWe can do a quick check to make sure it has 3 qubits\n\n:\n\nqc.qubits()\n\n:\n\n[0, 1, 2]\n\n\n## Sampling with run_and_measure¶\n\nQuantumComputer.run_and_measure will run a given program (that does not have explicit MEASURE instructions) and then measure all qubits present in the quantum computer.\n\n:\n\nbitstrings = qc.run_and_measure(program, trials=10)\nbitstrings\n\n:\n\n{0: array([1, 0, 0, 1, 1, 1, 1, 0, 1, 0]),\n1: array([1, 0, 0, 1, 1, 1, 1, 0, 1, 0]),\n2: array([1, 0, 0, 1, 1, 1, 1, 0, 1, 0])}\n\n\nLet’s programatically verify that we always measure 000 or 111 by “summing” each bitstring and checking if it’s eather 0 (for 000) or 3 (for 111)\n\n:\n\nimport numpy as np\nbitstring_array = np.vstack([bitstrings[q] for q in qc.qubits()]).T\nsums = np.sum(bitstring_array, axis=1)\nsums\n\n:\n\narray([3, 0, 0, 3, 3, 3, 3, 0, 3, 0])\n\n:\n\nsample_is_ghz = np.logical_or(sums == 0, sums == 3)\nsample_is_ghz\n\n:\n\narray([ True, True, True, True, True, True, True, True, True,\nTrue])\n\n:\n\nnp.all(sample_is_ghz)\n\n:\n\nTrue\n\n\n## Change alert: run_and_measure will return a dictionary of 1d bitstrings.¶\n\nNot a 2d array. To demonstrate why, consider a lattice whose qubits are not contiguously indexed from 0.\n\n:\n\n# TODO: we need a lattice that is not zero-indexed\n# qc = get_qc('Aspen-0-3Q-B')\n# qc.run_and_measure(ghz_state(qubits=[1,2,3]))\n\n\n## Change alert: All qubits are measured¶\n\nPyQuil 1.x’s run_and_measure would only measure qubits used in the given program. Now all qubits (per qc.qubits()) are measured. This is easier to reason about and reflects the reality of running on a QPU. When accounting for noise or when running QCVV tasks, you may be interested in the measurement results of qubits that weren’t even used in your program!\n\n:\n\nqc = get_qc('4q-qvm')\nbitstrings = qc.run_and_measure(Program(X(0), X(1), X(2)), trials=10)\nbitstrings\n\n:\n\n{0: array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1]),\n1: array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1]),\n2: array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1]),\n3: array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0])}\n\n\nYou can drop qubits you’re not interested in by indexing into the returned dictionary\n\n:\n\n# Stacking everything\nnp.vstack([bitstrings[q] for q in qc.qubits()]).T\n\n:\n\narray([[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0],\n[1, 1, 1, 0]])\n\n:\n\n# Stacking what you want (contrast with above)\nqubits = [0, 1, 2]\nnp.vstack([bitstrings[q] for q in qubits]).T\n\n:\n\narray([[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1],\n[1, 1, 1]])\n\n\n## Change alert: run_and_measure works with noise models now.¶\n\nIn pyQuil 1.x, run_and_measure would not work with noise models. Now noise models are supported. Pre-configured noise models can be used via get_qc('xxxx-noisy-qvm').\n\nAs a consequence, run_and_measure for large numbers of trials will be slower in Pyquil 2.\n\n:\n\nqc = get_qc('3q-noisy-qvm')\nbitstrings = qc.run_and_measure(program, trials=10)\nbitstrings\n\n:\n\n{0: array([0, 1, 1, 0, 0, 1, 1, 0, 1, 1]),\n1: array([0, 1, 1, 0, 0, 1, 1, 0, 1, 1]),\n2: array([0, 1, 1, 0, 0, 1, 1, 0, 1, 1])}\n\n:\n\nbitstring_array = np.vstack([bitstrings[q] for q in qc.qubits()]).T\nsums = np.sum(bitstring_array, axis=1)\nsums\n\n:\n\narray([0, 3, 3, 0, 0, 3, 3, 0, 3, 3])\n\n:\n\n# Noise means now we measure things other than 000 or 111\nnp.all(np.logical_or(sums == 0, sums == 3))\n\n:\n\nTrue\n\n\n## list_quantum_computers¶\n\nYou can find all possible arguments to get_qc with list_quantum_computers\n\n:\n\nfrom pyquil import list_quantum_computers\n# TODO: unauthenticated endpoint\n# list_quantum_computers()\n\n\n## QuantumComputers have a topology¶\n\nAn important restriction when running on a real quantum computer is the mapping of qubits to the supported two-qubit gates. The QVM is designed to provide increasing levels of “realism” to guarantee that if your program executes successfully on get_qc(\"Aspen-xxx-noisy-qvm\") then it will execute successfully on get_qc(\"Aspen-xxx\")*\n\n* guarantee not currently guaranteed. This is a work in progress.\n\n## Inspecting the topology¶\n\nYou can access a topology by qc.qubit_topology(), which will return a NetworkX representation of qubit connectivity. You can access the full set of supported instructions by qc.get_isa(). For example, we include a generic QVM named \"9q-square-qvm\" that has a square topology.\n\n:\n\nqc = get_qc('9q-square-qvm')\n%matplotlib inline\nimport networkx as nx\nnx.draw(qc.qubit_topology())\nfrom matplotlib import pyplot as plt\n_ = plt.title('9q-square-qvm', fontsize=18)",
null,
"## What If I don’t want a topology?¶\n\nWavefunctionSimulator still has no notion of qubit connectivity, so feel free to use that for simulating quantum algorithms that you aren’t concerned about running on an actual QPU.\n\nAbove we used get_qc(\"3q-qvm\"), \"4q-qvm\", and indeed you can do any \"{n}q-qvm\" (subject to computational resource constraints). These QVM’s are constructed with a topology! It just happens to be fully connected\n\n:\n\nnx.draw(get_qc('5q-qvm').qubit_topology())\n_ = plt.title('5q-qvm is fully connected', fontsize=16)",
null,
"## Heirarchy of realism¶\n\n• WavefunctionSimulator to debug algorithm\n• get_qc(\"5q-qvm\") to debug sampling\n• get_qc(\"9q-square-qvm\") to debug mapping to a lattice\n• get_qc(\"9q-square-noisy-qvm\") to debug generic noise characteristics\n• get_qc(\"Aspen-0-16Q-A-qvm\") to debug mapping to a real lattice\n• get_qc(\"Aspen-0-16Q-A-noisy-qvm\") to debug noise characteristics of a real device\n• get_qc(\"Aspen-0-16Q-A\") to run on a real device\n\n## “What is a QuantumComputer?” Advanced Edition¶\n\nA QuantumComputer is a wrapper around three constituent parts, each of which has a programatic interface that must be respected by all classes that implement the interface. By having clear interfaces we can write backend-agnostic methods on QuantumComputer and mix-and-match backing objects.\n\nThe following diagram shows the three objects that must be provided when constructing a QuantumComputer “by hand”. The abstract classes are backed in grey with example implementing classes listed below. Please consult the api reference for details on each interface.",
null,
"As an example, let’s construct a 5-qubit QVM with one central node and only even numbered qubits.\n\n:\n\ntopology = nx.from_edgelist([\n(10, 2),\n(10, 4),\n(10, 6),\n(10, 8),\n])\nfrom pyquil.device import NxDevice\ndevice = NxDevice(topology)\n\nfrom pyquil.api._qac import AbstractCompiler\nclass MyLazyCompiler(AbstractCompiler):\ndef quil_to_native_quil(self, program, *, protoquil=None):\nreturn program\n\ndef native_quil_to_executable(self, nq_program):\nreturn nq_program\n\nfrom pyquil.api import QuantumComputer, QVM, ForestConnection\nmy_qc = QuantumComputer(\nname='my-qvm',\nqam=QVM(connection=ForestConnection()),\ndevice=device,\ncompiler=MyLazyCompiler(),\n)\n\nnx.draw(my_qc.qubit_topology())",
null,
":\n\nmy_qc.run_and_measure(Program(X(10)), trials=5)\n\n:\n\n{2: array([0, 0, 0, 0, 0]),\n4: array([0, 0, 0, 0, 0]),\n6: array([0, 0, 0, 0, 0]),\n8: array([0, 0, 0, 0, 0]),\n10: array([1, 1, 1, 1, 1])}"
] | [
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"https://pyquil-docs.rigetti.com/en/v2.19.0/_images/migration2-qc_36_0.png",
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"https://pyquil-docs.rigetti.com/en/v2.19.0/_images/pyquil-objects.png",
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"https://pyquil-docs.rigetti.com/en/v2.19.0/_images/migration2-qc_39_0.png",
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https://kmusi.pugliasrl.it/rate-of-change-of-acceleration-unit.html | [
"### Go math grade 7 practice and skills fluency workbook answer key\n\nStart studying Motion, Velocity, Speed, Acceleration. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Dec 11, 2013 · If we look at the rate of change of velocity, the second derivative (by time) of the object’s displacement (i.e. the rate of change of the rate of change of its displacement), then we have calculated the object’s acceleration. If we now look at the rate of change of acceleration, the third derivative of the object’s displacement (i.e. the ...\n\n### How to lock honeywell 6000 thermostat\n\nderiving the above equation w.r.t time will give you the rate of change of acceleration. That will be dependent on two things, da/dt = (1/m)dF/dt - F.(1/m^2)dm/dt As rate of change of mass can be assumed 0, the equation reduces to da/dt= (1/m)dF/dt\nThis precalculus video tutorial explains how to calculate the average rate of change of a function over an interval. This video contains plenty of examples ... 14 Slope and Rate of Change Rate of Change Quiz 3 Practice Test Practice Test How are you doing? Get this page from your teacher Math Journal Math Journal Journal entry based on criteria on handout and question jointly chosen. Self-Assessment Self-Assessment On the next page, complete the self-assessment assignment. Unit Test Unit Test\n\n### Where is my phone icon on my samsung tablet\n\nAcceleration: It is the rate of change of velocity of the object in respect to the time frame. It's SI unit is meter per second squared (m/s 2), as the velocity in meters per second changes by the acceleration value, every second. It is a vector quantity as it contains both magnitude and direction.\nDec 19, 2016 · Velocity , Both , Or Acceleration?? 1. A rate of change 2. Unit m/s 3. Unit m/s^2 4. Can be positive or negative 5. Change in velocity over change in time 6. Change in displacement over change in time Sep 30, 2016 · The rate of change of acceleration is called jolt or jerk. Jerk is a vector quantity as it has both direction and magnitude. Jerk is expressed in meters per second cube. (m/sec^3) Jerk is the time derivative of acceleration.\n\n### Webassign answers pdf statistics\n\nCHA-3.A.2 The derivative can be used to express information about rates of change in applied contexts. CHA-3.A.3 The unit for f' (x) is the unit for f divided by the unit for x. CHA-3.B.1 The derivative can be used to solve rectilinear motion problems involving position, speed, velocity, and acceleration.\nIts SI unit is m/s 2. Acceleration due to gravity is a vector, which means it has both a magnitude and a direction. The acceleration due to gravity at the surface of Earth is represented by the letter g. It has a standard value defined as 9.80665 m/s 2 (32.1740 ft/s 2). However, the actual acceleration of a body in free fall varies with location. Tangential Acceleration The tangential accelerationis the rate of change of the magnitude of the velocity Angular acceleration: rate of change of angular velocity with time 2 00 2 t lim lim tt vdd aR RRR t t dt dt ωω θ α Δ→ Δ→ ΔΔ == == = ΔΔ α= dω dt = d2θ dt2 (units: rad⋅s-2)\n\n### Magpul pmag 10 pack dark earth\n\nThough the St. Lucie Unit 1 and Unit 2 TS states that the power rate-of-change trip may be bypassed below 10-4% and above 15% of RATED THERMAL POWER, the power rate-of-change trip is automatically bypassed below 1 o-4% and above 15% power and is automatically restored upon re-entering this power range.\nRate means per unit of time. 5 miles per hour is a rate because it is expressed as distance per unit of time. Acceleration is a simple rate of change -- the rate changes. Hmm, maybe I look too deeply. Acceleration (a) is defined as the rate of change of velocity of an object with time. Here velocity is the rate with an object moves from one place to another. Velocity is a vector quantity, which means it has both a magnitude, called speed, and a direction. This makes acceleration also a vector quantity.\n\n### Juniper syslog examples\n\nRate of Change Definition Basically, the ratio of the change in the output value and change in the input value of a function is called as rate of change. Rate of change is the ratio that shows the relationship between the two variables in equation.\nTIPS4RM: MCV4U: Unit 1 – Rates of Change 2008 12 1.1.4: Frayer Model Solutions (Teacher) Definition Instantaneous Rate of Change is the measure of the rate of change for a continuous function at point on the function. Characteristics • The rate can be represented as the slope of the tangent line to a curve at a particular point Mar 07, 2001 · The paradigm shift rate (i.e., the overall rate of technical progress) is currently doubling (approximately) every decade; that is, paradigm shift times are halving every decade (and the rate of acceleration is itself growing exponentially).\n\n### Mikhaila peterson eggs\n\nAcceleration and Speed are both kinds of rates. and for a quantity to be a rate, such as speed, the change in something is inseparable from the change in something else. You cannot define a rate without expressing the change in two related quantities.\nDec 17, 1991 · According to the invention, an acceleration rate-of-change limitation is provided to ensure that the driving system does not experience any sudden changes as a result of the acceleration changes a 1. Such an acceleration rate-of-change limitation can be achieved for example by filtering the acceleration values to round off the edges of a particular acceleration pulse so that the respective new acceleration value is reached gradually. Acceleration is the rate of change of velocity, hence the units. If you were to plot a graph of velocity against time the gradient would be the So even if something is travelling with a constant speed, if its direction changes then it is accelerating. This effect is illustrated in circular motion, but an explaintion...\n\n### Vuetify container fixed width\n\nChecklist for Unit 1a: 1.1 Slope and Rate of Change 1.2 Grade, Angle of Elevation 1.3 Rate of Change\nDefine acceleration. acceleration synonyms, acceleration pronunciation, acceleration translation, English dictionary definition of acceleration. n. 1. a. The act 2. Abbr. a Physics The rate of change of velocity with respect to time. American Heritage® Dictionary of the English Language, Fifth Edition.\n\n### Oracion de la salud de mi familia\n\n• Nonprofit board of directors voting rights\n• #### Winaero tweaker reddit\n\n• Physical science chapter 13 forces in fluids test answers\n\n• #### Balsa electric rc plane kits\n\n• Kubota la450s specs\n• #### Davinci resolve generate optimized media error\n\n• Lobo how can i tell her about you lyrics chords\n• #### Machines a sous gratuites 3d tropezia palace\n\n• Virtual learning expectations\n\n• #### Hot rod cars\n\nDeposit check with paypal\n\n### Moze deathless bloodletter build\n\nacceleration unit — noun a unit for measuring acceleration • Hypernyms: ↑unit of measurement, ↑unit • Hyponyms: ↑gal … Useful english dictionary. measurement — I (New American Roget s College Thesaurus) Determination of size Nouns 1. measurement, measure, admeasurement, mensuration...\n14 Slope and Rate of Change Rate of Change Quiz 3 Practice Test Practice Test How are you doing? Get this page from your teacher Math Journal Math Journal Journal entry based on criteria on handout and question jointly chosen. Self-Assessment Self-Assessment On the next page, complete the self-assessment assignment. Unit Test Unit Test"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88737184,"math_prob":0.9015865,"size":10925,"snap":"2021-04-2021-17","text_gpt3_token_len":2568,"char_repetition_ratio":0.19183224,"word_repetition_ratio":0.23159592,"special_character_ratio":0.22517163,"punctuation_ratio":0.107893504,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99442685,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T14:11:31Z\",\"WARC-Record-ID\":\"<urn:uuid:d9021241-6a57-4ffc-9be0-d22533252191>\",\"Content-Length\":\"37444\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ca765fc-31d0-41bb-a50e-4711377e6af2>\",\"WARC-Concurrent-To\":\"<urn:uuid:869188f3-6236-466c-bf13-6439593e276b>\",\"WARC-IP-Address\":\"104.21.14.73\",\"WARC-Target-URI\":\"https://kmusi.pugliasrl.it/rate-of-change-of-acceleration-unit.html\",\"WARC-Payload-Digest\":\"sha1:QPAJUSVTQXZZ2MPYI4A3MJU4NVWL6J4X\",\"WARC-Block-Digest\":\"sha1:LZZHCGTHOAJQYHOZCLUYTLROFHWIQ63I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038492417.61_warc_CC-MAIN-20210418133614-20210418163614-00438.warc.gz\"}"} |
http://www.ferinheighttocelsius.com/fahrenheit-to-celsius-formula/ | [
"# Fahrenheit to Celsius formula\n\n## Conversion between temperature scales\n\nUse the following equations to calculate temperature in degrees Celsius (t[C]) from degrees Fahrenheit (t[F]) and vice versa. More generally, the National Institute of Standards and Technology (NIST) website is one of the most reliable resources for information about systems of units and unit conversion.\n\n### Fahrenheit to Celsius\n\n``````t[C] = (5/9)*(t[F] - 32)\n``````\n\n### Celsius to Fahrenheit\n\n``````t[F] = (9/5)*t[C] + 32\n``````"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78386915,"math_prob":0.9652864,"size":583,"snap":"2019-43-2019-47","text_gpt3_token_len":131,"char_repetition_ratio":0.12435233,"word_repetition_ratio":0.0,"special_character_ratio":0.22298457,"punctuation_ratio":0.05154639,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9943369,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-16T21:17:54Z\",\"WARC-Record-ID\":\"<urn:uuid:e8fc46de-64a3-499c-a3d3-437e2e18b9c2>\",\"Content-Length\":\"10943\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb43f23d-87eb-491f-9722-ec083066bb01>\",\"WARC-Concurrent-To\":\"<urn:uuid:9380b8fe-0fe5-4a0c-85fb-b762e676f411>\",\"WARC-IP-Address\":\"104.28.20.191\",\"WARC-Target-URI\":\"http://www.ferinheighttocelsius.com/fahrenheit-to-celsius-formula/\",\"WARC-Payload-Digest\":\"sha1:TWY65ICZBZNQTKKFS5XRYOYVYACDRGGT\",\"WARC-Block-Digest\":\"sha1:SBHFYBY7BI5N5FZRROB2QX52I6ZFJVEU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668765.35_warc_CC-MAIN-20191116204950-20191116232950-00339.warc.gz\"}"} |
https://pijaeducation.com/arduino/how-to-take-output-from-arduino-uno/blink-leds-in-stack-form-using-for-loop/ | [
"# Blink LEDs in Stack Form Using for loop\n\nTo blink LEDs in stack form using for loop we first know about “Stack (click) and basic program of LEDs stack pattern (previous topic)“.\n\n## Required hardware or components for Lighting up LED in Stack form using Arduino\n\nS.N.ComponentQuantity\n1.Arduino Uno1\n3.LED4\n4.Resistor 220 or 280 ohm4\n\n## CIRCUIT DIAGRAM : BLINK LEDS IN STACK FORM USING FOR LOOP\n\n### CONNECTION TABLE\n\nS.N.ArduinoLED\n1.GNDCathode (-)\n2.2LED 1 - Anode (+)\n3.3LED 2 - Anode (+)\n4.4LED 3 - Anode (+)\n5.5LED 4 - Anode (+)\n\n## ARDUINO PROGRAMMING CODE : BLINK LEDS IN STACK FORM USING FOR LOOP\n\n```int i;\nvoid setup() {\nfor (i = 2 ; i <= 5; i++) {\npinMode(i, OUTPUT);\n}\n}\n\nvoid loop() {\nfor (i = 2; i <= 5 ; i++) {\ndigitalWrite(i, HIGH);\ndelay(500);\n}\nfor (i = 5; i >= 2 ; i--) {\ndigitalWrite(i, LOW);\ndelay(500);\n}\n}\n```\n\n## CODE EXPLANATION\n\nStep-1: We define an integer variable i.\n\n`int i;`\n\nStep-2: In this block we use a for-loop because we want to do the same work again and again i.e. We want to set the mode of the pin number as output in an order from number 2 to 5. Loop will run 4 times as the loop runs in the first cycle it assigns pin 2 as an output then in the second cycle it assigns pin 3 as an output and same in third cycle pin 4 and in forth cycle pin 5.\n\n```void setup() {\nfor (i = 2 ; i <= 5; i++) {\npinMode(i, OUTPUT);\n}\n}```\n\nStep-3:\n\nIn first for-loop we want to set the state of pin number 2, 3, 4 & 5 as HIGH after a delay of 0.5 second next to each other to turn ON led connected on pin number 2, 3, 4 & 5.\n\nIn the second for-loop we want to set the state of pin number 5, 4, 3, & 2 as LOW after a delay of 0.5 second next to each other to turn OFF led connected on pin number 5, 4, 3 & 2.\n\n```void loop() {\nfor (i = 2; i <= 5 ; i++) {\ndigitalWrite(i, HIGH);\ndelay(500);\n}\nfor (i = 5; i >= 2 ; i--) {\ndigitalWrite(i, LOW);\ndelay(500);\n}\n}```\n\nAnd the void loop() gets executed continuously and follows the same pattern.\n\nNow let’s try a new pattern\n\nSubscribe\nNotify of",
null,
""
] | [
null,
"https://secure.gravatar.com/avatar/",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.68433464,"math_prob":0.9639393,"size":2010,"snap":"2023-40-2023-50","text_gpt3_token_len":669,"char_repetition_ratio":0.09770688,"word_repetition_ratio":0.29787233,"special_character_ratio":0.36666667,"punctuation_ratio":0.1487069,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9808456,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T14:41:06Z\",\"WARC-Record-ID\":\"<urn:uuid:9e90de8d-3d8c-4cda-aa41-be345ef6ca1c>\",\"Content-Length\":\"268223\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:059e9fd9-b37f-405e-a172-54b4d823d0b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc1140ed-3b58-4b9f-8890-fbdb842f1f82>\",\"WARC-IP-Address\":\"134.119.218.58\",\"WARC-Target-URI\":\"https://pijaeducation.com/arduino/how-to-take-output-from-arduino-uno/blink-leds-in-stack-form-using-for-loop/\",\"WARC-Payload-Digest\":\"sha1:XFXTF654SDRSQIF4RS5UYS2ZEFCMPEBA\",\"WARC-Block-Digest\":\"sha1:M6HFRYE4FALORSVKW2XD7HL2TOKYXPRT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506646.94_warc_CC-MAIN-20230924123403-20230924153403-00527.warc.gz\"}"} |
https://kr.mathworks.com/matlabcentral/cody/problems/10-determine-whether-a-vector-is-monotonically-increasing/solutions/1544827 | [
"Cody\n\n# Problem 10. Determine whether a vector is monotonically increasing\n\nSolution 1544827\n\nSubmitted on 31 May 2018 by Danuanping\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = [0 1 2 3 4]; assert(isequal(mono_increase(x),true));\n\nans = logical 1\n\n2 Pass\nx = ; assert(isequal(mono_increase(x),true));\n\nans = logical 1\n\n3 Pass\nx = [0 0 0 0 0]; assert(isequal(mono_increase(x),false));\n\nans = logical 0\n\n4 Pass\nx = [0 1 2 3 -4]; assert(isequal(mono_increase(x),false));\n\nans = logical 0\n\n5 Pass\nx = [-3 -4 2 3 4]; assert(isequal(mono_increase(x),false));\n\nans = logical 0\n\n6 Pass\nx = 1:.1:10; assert(isequal(mono_increase(x),true));\n\nans = logical 1\n\n7 Pass\nx = cumsum(rand(1,100)); x(5) = -1; assert(isequal(mono_increase(x),false));\n\nans = logical 0\n\n8 Pass\nx = cumsum(rand(1,50)); assert(isequal(mono_increase(x),true));\n\nans = logical 1\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5579031,"math_prob":0.9991184,"size":1083,"snap":"2020-45-2020-50","text_gpt3_token_len":362,"char_repetition_ratio":0.17145506,"word_repetition_ratio":0.13017751,"special_character_ratio":0.37026778,"punctuation_ratio":0.15625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99763966,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T21:43:13Z\",\"WARC-Record-ID\":\"<urn:uuid:679811d4-00e2-47a4-b0fb-6dc22a3bc90d>\",\"Content-Length\":\"83588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c764e697-1a77-4e2d-aef7-984dcf24bab3>\",\"WARC-Concurrent-To\":\"<urn:uuid:2b584141-26d5-46dc-8a2c-0915d24fa4d6>\",\"WARC-IP-Address\":\"23.56.12.57\",\"WARC-Target-URI\":\"https://kr.mathworks.com/matlabcentral/cody/problems/10-determine-whether-a-vector-is-monotonically-increasing/solutions/1544827\",\"WARC-Payload-Digest\":\"sha1:C37GC7AFD6CB6CN5QSXT2D3R3USE3S7O\",\"WARC-Block-Digest\":\"sha1:6IWJUQNG6MHMW6UWJ6M52PGMRUHXQE6Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141194171.48_warc_CC-MAIN-20201127191451-20201127221451-00695.warc.gz\"}"} |
https://blog.vedicmathsindia.org/complex-mergers/ | [
"SHARE\n\nThis post is, once again, a summary of a longer post authored by me on my own blog. My blog covers a lot of areas, including Vedic Mathematics. If you are interested in reading my thoughts on other topics, please feel free to visit my blog and post comments on the other articles you find there also! Thank you!\n\nIn the previous lesson, we saw how it is sometimes possible to apply the merger methodology to equations that have more than 3 terms originally, bringing them to a form that ultimately makes it possible to solve them using the merger formulae we derived earlier. In this lesson, we will look at a different type of equation whose terms can be merged down to a form where we can apply simple formulae to solve the equation.\n\nConsider the equation below:\n\n4/(x + 3) – 16/(4x + 13) = 1/(2x + 5) – 1/(2x + 9)\n\nThis does not seem to be a prime candidate for a merger of any sort. Now, look at the series of manipulations we can perform on the above equation:\n\n4/(x + 3) – 16/(4x + 13) = 1/(2x + 5) – 1/(2x + 9) becomes\n[4*(4x + 13) – 16*(x + 3)]/[(x + 3)*(4x + 13)] = [(2x + 9) – (2x + 5)]/[(2x + 5)*(2x + 9)] becomes\n4/[(x + 3)*(4x + 13)] = 4/[(2x + 5)*(2x + 9)]\n\nNow, we see that the numerator is the same on both sides of the equation. We can then cancel out the common numerator, and invert the two terms to get a simple equation as below:\n\n(x + 3)*(4x + 13) = (2x + 5)*(2x + 9)\n\nWe now recognize this to be a generalized form of the third type of equation we derived a formula for in the lesson on solving equations using the Paravartya Yojayet sutra. Applying the formula we derived then to the equation above, we can easily find the solution for the equation to be x = -2.\n\nThe operation we performed on the given equation to reduce it to a form that is easily amenable to solution using the formula we derived earlier is called a complex merger.\n\nHow do we identify when an equation is capable of being solved using complex merger? There are three tests that will tell us when the conditions are correct for the application of a complex merger operation on a given equation. Consider a general equation as below:\n\na/(bx + c) – d/(ex + f) = g/(hx + j) – k/(mx + n)\n\nThe first condition the equation has to satisfy is as below:\n\na/b = d/e and g/h = k/m\n\nThus, we have to make sure that the ratio of the numerator to the coefficient of the unknown is the same in both terms on the left hand side of the equation as well as in both terms on the right hand side of the equation. The ratios on the left and right hand side don’t have to be the same, just the ratios of the two terms on either side should be the same.\n\nWhat does this first condition ensure? Consider the equality a/b = d/e. It can be rewritten as ae = bd. On the left hand side of the given equation, when we take the LCM of (bx + c) and (ex + f), and then cross-multiply to combine the two terms into one, the coefficients of the unknown in the numerator will be ae and -bd. If ae = bd, then these coefficients will cancel out, leaving the numerator on the left hand side free of unknown terms. The condition g/h = k/m similarly ensures that the numerator on the right hand side contains no unknown terms either.\n\nThe second condition is that af – dc = gn – kj. The astute reader will have already surmised that the actual numerator on the left hand side of the equation (when we take the LCM of the two denominators and cross-multiply) is af – dc after the cancellation of the coefficients of the unknown term (according to the first condition). Similarly, gn – kj is the final numerator on the right hand side of the equation when that side is expanded out by cross-multiplication, after cancellation of the coefficients of the unknown terms.\n\nWhen this second condition is satisfied, it enables us to cancel out the numerators on either side of the equation and invert the equation into the form below:\n\n(bx + c)*(ex + f) = (hx + j)*(mx + n)\n\nThe third condition comes into play at this point. To ensure that this equation can be solved using the formula derived earlier, the third condition stipulates that be = hm. When this condition is satisfied, the two quadratic terms on either side of the equation cancel out, leaving us a linear equation that can be solved using the formula derived using the Paravartya Yojayet sutra. The final solution is then derived as x = (jn – cf)/(bf + ce – hn – mj)\n\nIf this third condition is not satisfied, all is not lost though. Because the first two conditions are satisfied, we have still managed to reduce a potentially cubic equation to just a quadratic equation which is easy to solve using the quadratic formula. Thus, if the first two conditions are satisfied, we should go ahead and take advantage of the complex merger to reduce the equation to a quadratic even if we are not able to reduce it to a linear form.\n\nWhat happens if the second condition is not satisfied either? We are then left with an equation as below:\n\n(af – dc)/[(bx + c)*(ex + f)] = (gn – kj)/[(hx + j)*(mx + n)]\n\nWhen we cross-multiply, we get:\n\n(af – dc)*(hx + j)*(mx + n) = (gn – kj)*(bx + c)*(ex + f)\n\nIf, at this point, we find that (af – dc)*hm = (gn – kj)*be, then the quadratic terms on the two sides of the equation will cancel out, and we will be left with a linear equation that can be solved by transposing and adjusting appropriately."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.927439,"math_prob":0.99977404,"size":5876,"snap":"2020-34-2020-40","text_gpt3_token_len":1436,"char_repetition_ratio":0.15888965,"word_repetition_ratio":0.07520143,"special_character_ratio":0.2593601,"punctuation_ratio":0.070940174,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99983954,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-23T00:07:53Z\",\"WARC-Record-ID\":\"<urn:uuid:797d8665-7d92-45ae-a68d-74c4ca84a853>\",\"Content-Length\":\"128728\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:86e04599-cd6e-44a1-a9fc-7e62f16d0038>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d040e3e-aaa6-4494-882f-21f63b6b5dbd>\",\"WARC-IP-Address\":\"157.245.110.226\",\"WARC-Target-URI\":\"https://blog.vedicmathsindia.org/complex-mergers/\",\"WARC-Payload-Digest\":\"sha1:KIAY64P7TQSXXVLKO4NIBCQGSK3PT5UO\",\"WARC-Block-Digest\":\"sha1:DZKUW4PBZH6GA5PCGU5WM4DE6COLCHKB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400208095.31_warc_CC-MAIN-20200922224013-20200923014013-00461.warc.gz\"}"} |
https://openstax.org/books/university-physics-volume-1/pages/16-2-mathematics-of-waves | [
"University Physics Volume 1\n\n# 16.2Mathematics of Waves\n\nUniversity Physics Volume 116.2 Mathematics of Waves\n\n### Learning Objectives\n\nBy the end of this section, you will be able to:\n\n• Model a wave, moving with a constant wave velocity, with a mathematical expression\n• Calculate the velocity and acceleration of the medium\n• Show how the velocity of the medium differs from the wave velocity (propagation velocity)\n\nIn the previous section, we described periodic waves by their characteristics of wavelength, period, amplitude, and wave speed of the wave. Waves can also be described by the motion of the particles of the medium through which the waves move. The position of particles of the medium can be mathematically modeled as wave functions, which can be used to find the position, velocity, and acceleration of the particles of the medium of the wave at any time.\n\n### Pulses\n\nA pulse can be described as wave consisting of a single disturbance that moves through the medium with a constant amplitude. The pulse moves as a pattern that maintains its shape as it propagates with a constant wave speed. Because the wave speed is constant, the distance the pulse moves in a time $ΔtΔt$ is equal to $Δx=vΔtΔx=vΔt$ (Figure 16.8).\n\nFigure 16.8 The pulse at time $t=0t=0$ is centered on $x=0x=0$ with amplitude A. The pulse moves as a pattern with a constant shape, with a constant maximum value A. The velocity is constant and the pulse moves a distance $Δx=vΔtΔx=vΔt$ in a time $Δt.Δt.$ The distance traveled is measured with any convenient point on the pulse. In this figure, the crest is used.\n\n### Modeling a One-Dimensional Sinusoidal Wave using a Wave Function\n\nConsider a string kept at a constant tension $FTFT$ where one end is fixed and the free end is oscillated between $y=+Ay=+A$ and $y=−Ay=−A$ by a mechanical device at a constant frequency. Figure 16.9 shows snapshots of the wave at an interval of an eighth of a period, beginning after one period $(t=T).(t=T).$\n\nFigure 16.9 Snapshots of a transverse wave moving through a string under tension, beginning at time $t=Tt=T$ and taken at intervals of $18T.18T.$ Colored dots are used to highlight points on the string. Points that are a wavelength apart in the x-direction are highlighted with the same color dots.\n\nNotice that each select point on the string (marked by colored dots) oscillates up and down in simple harmonic motion, between $y=+Ay=+A$ and $y=−A,y=−A,$ with a period T. The wave on the string is sinusoidal and is translating in the positive x-direction as time progresses.\n\nAt this point, it is useful to recall from your study of algebra that if f(x) is some function, then $f(x−d)f(x−d)$ is the same function translated in the positive x-direction by a distance d. The function $f(x+d)f(x+d)$ is the same function translated in the negative x-direction by a distance d. We want to define a wave function that will give the y-position of each segment of the string for every position x along the string for every time t.\n\nLooking at the first snapshot in Figure 16.9, the y-position of the string between $x=0x=0$ and $x=λx=λ$ can be modeled as a sine function. This wave propagates down the string one wavelength in one period, as seen in the last snapshot. The wave therefore moves with a constant wave speed of $v=λ/T.v=λ/T.$\n\nRecall that a sine function is a function of the angle $θθ$, oscillating between $+1+1$ and $−1−1$, and repeating every $2π2π$ radians (Figure 16.10). However, the y-position of the medium, or the wave function, oscillates between $+A+A$ and $−A−A$, and repeats every wavelength $λλ$.\n\nFigure 16.10 A sine function oscillates between $+1+1$ and $−1−1$ every $2π2π$ radians.\n\nTo construct our model of the wave using a periodic function, consider the ratio of the angle and the position,\n\n$θx=2πλ,θ=2πλx.θx=2πλ,θ=2πλx.$\n\nUsing $θ=2πλxθ=2πλx$ and multiplying the sine function by the amplitude A, we can now model the y-position of the string as a function of the position x:\n\n$y(x)=Asin(2πλx).y(x)=Asin(2πλx).$\n\nThe wave on the string travels in the positive x-direction with a constant velocity v, and moves a distance vt in a time t. The wave function can now be defined by\n\n$y(x,t)=Asin(2πλ(x−vt)).y(x,t)=Asin(2πλ(x−vt)).$\n\nIt is often convenient to rewrite this wave function in a more compact form. Multiplying through by the ratio $2πλ2πλ$ leads to the equation\n\n$y(x,t)=Asin(2πλx−2πλvt).y(x,t)=Asin(2πλx−2πλvt).$\n\nThe value $2πλ2πλ$ is defined as the wave number. The symbol for the wave number is k and has units of inverse meters, $m−1:m−1:$\n\n$k≡2πλk≡2πλ$\n16.2\n\nRecall from Oscillations that the angular frequency is defined as $ω≡2πT.ω≡2πT.$ The second term of the wave function becomes\n\n$2πλvt=2πλ(λT)t=2πTt=ωt.2πλvt=2πλ(λT)t=2πTt=ωt.$\n\nThe wave function for a simple harmonic wave on a string reduces to\n\n$y(x,t)=Asin(kx∓ωt),y(x,t)=Asin(kx∓ωt),$\n\nwhere A is the amplitude, $k=2πλk=2πλ$ is the wave number, $ω=2πTω=2πT$ is the angular frequency, the minus sign is for waves moving in the positive x-direction, and the plus sign is for waves moving in the negative x-direction. The velocity of the wave is equal to\n\n$v=λT=λT(2π2π)=ωk.v=λT=λT(2π2π)=ωk.$\n16.3\n\nThink back to our discussion of a mass on a spring, when the position of the mass was modeled as $x(t)=Acos(ωt+ϕ).x(t)=Acos(ωt+ϕ).$ The angle $ϕϕ$ is a phase shift, added to allow for the fact that the mass may have initial conditions other than $x=+Ax=+A$ and $v=0.v=0.$ For similar reasons, the initial phase is added to the wave function. The wave function modeling a sinusoidal wave, allowing for an initial phase shift $ϕ,ϕ,$ is\n\n$y(x,t)=Asin(kx∓ωt+ϕ)y(x,t)=Asin(kx∓ωt+ϕ)$\n16.4\n\nThe value\n\n$(kx∓ωt+ϕ)(kx∓ωt+ϕ)$\n16.5\n\nis known as the phase of the wave, where $ϕϕ$ is the initial phase of the wave function. Whether the temporal term $ωtωt$ is negative or positive depends on the direction of the wave. First consider the minus sign for a wave with an initial phase equal to zero $(ϕ=0).(ϕ=0).$ The phase of the wave would be $(kx−ωt).(kx−ωt).$ Consider following a point on a wave, such as a crest. A crest will occur when $sin(kx−ωt)=1.00sin(kx−ωt)=1.00$, that is, when $kx−ωt=nπ+π2,kx−ωt=nπ+π2,$ for any integral value of n. For instance, one particular crest occurs at $kx−ωt=π2.kx−ωt=π2.$ As the wave moves, time increases and x must also increase to keep the phase equal to $π2.π2.$ Therefore, the minus sign is for a wave moving in the positive x-direction. Using the plus sign, $kx+ωt=π2.kx+ωt=π2.$ As time increases, x must decrease to keep the phase equal to $π2.π2.$ The plus sign is used for waves moving in the negative x-direction. In summary, $y(x,t)=Asin(kx−ωt+ϕ)y(x,t)=Asin(kx−ωt+ϕ)$ models a wave moving in the positive x-direction and $y(x,t)=Asin(kx+ωt+ϕ)y(x,t)=Asin(kx+ωt+ϕ)$ models a wave moving in the negative x-direction.\n\nEquation 16.4 is known as a simple harmonic wave function. A wave function is any function such that $f(x,t)=f(x−vt).f(x,t)=f(x−vt).$ Later in this chapter, we will see that it is a solution to the linear wave equation. Note that $y(x,t)=Acos(kx+ωt+ϕ′)y(x,t)=Acos(kx+ωt+ϕ′)$ works equally well because it corresponds to a different phase shift $ϕ′=ϕ−π2.ϕ′=ϕ−π2.$\n\n### Problem-Solving Strategy\n\n#### Finding the Characteristics of a Sinusoidal Wave\n\n1. To find the amplitude, wavelength, period, and frequency of a sinusoidal wave, write down the wave function in the form $y(x,t)=Asin(kx−ωt+ϕ).y(x,t)=Asin(kx−ωt+ϕ).$\n2. The amplitude can be read straight from the equation and is equal to A.\n3. The period of the wave can be derived from the angular frequency $(T=2πω).(T=2πω).$\n4. The frequency can be found using $f=1T.f=1T.$\n5. The wavelength can be found using the wave number $(λ=2πk).(λ=2πk).$\n\n### Example 16.3\n\n#### Characteristics of a Traveling Wave on a String\n\nA transverse wave on a taut string is modeled with the wave function\n$y(x,t)=Asin(kx−wt)=0.2msin(6.28m−1x−1.57s−1t).y(x,t)=Asin(kx−wt)=0.2msin(6.28m−1x−1.57s−1t).$\n\nFind the amplitude, wavelength, period, and speed of the wave.\n\n#### Strategy\n\nAll these characteristics of the wave can be found from the constants included in the equation or from simple combinations of these constants.\n\n#### Solution\n\n1. The amplitude, wave number, and angular frequency can be read directly from the wave equation:\n$y(x,t)=Asin(kx−wt)=0.2msin(6.28m−1x−1.57s−1t).y(x,t)=Asin(kx−wt)=0.2msin(6.28m−1x−1.57s−1t).$\n$(A=0.2m;k=6.28m−1;ω=1.57s−1)(A=0.2m;k=6.28m−1;ω=1.57s−1)$\n2. The wave number can be used to find the wavelength:\n$k=2πλ.λ=2πk=2π6.28m−1=1.0m.k=2πλ.λ=2πk=2π6.28m−1=1.0m.$\n3. The period of the wave can be found using the angular frequency:\n$ω=2πT.T=2πω=2π1.57s−1=4s.ω=2πT.T=2πω=2π1.57s−1=4s.$\n4. The speed of the wave can be found using the wave number and the angular frequency. The direction of the wave can be determined by considering the sign of $kx∓ωtkx∓ωt$: A negative sign suggests that the wave is moving in the positive x-direction:\n$|v|=ωk=1.57s−16.28m−1=0.25m/s.|v|=ωk=1.57s−16.28m−1=0.25m/s.$\n\n#### Significance\n\nAll of the characteristics of the wave are contained in the wave function. Note that the wave speed is the speed of the wave in the direction parallel to the motion of the wave. Plotting the height of the medium y versus the position x for two times $t=0.00st=0.00s$ and $t=0.80st=0.80s$ can provide a graphical visualization of the wave (Figure 16.11).\nFigure 16.11 A graph of height of the wave y as a function of position x for snapshots of the wave at two times. The dotted line represents the wave at time $t=0.00st=0.00s$ and the solid line represents the wave at $t=0.80s.t=0.80s.$ Since the wave velocity is constant, the distance the wave travels is the wave velocity times the time interval. The black dots indicate the points used to measure the displacement of the wave. The medium moves up and down, whereas the wave moves to the right.\n\nThere is a second velocity to the motion. In this example, the wave is transverse, moving horizontally as the medium oscillates up and down perpendicular to the direction of motion. The graph in Figure 16.12 shows the motion of the medium at point $x=0.60mx=0.60m$ as a function of time. Notice that the medium of the wave oscillates up and down between $y=+0.20my=+0.20m$ and $y=−0.20my=−0.20m$ every period of 4.0 seconds.\n\nFigure 16.12 A graph of height of the wave y as a function of time t for the position $x=0.6m.x=0.6m.$ The medium oscillates between $y=+0.20my=+0.20m$ and $y=−0.20my=−0.20m$ every period. The period represented picks two convenient points in the oscillations to measure the period. The period can be measured between any two adjacent points with the same amplitude and the same velocity, $(∂y/∂t).(∂y/∂t).$ The velocity can be found by looking at the slope tangent to the point on a y-versus-t plot. Notice that at times $t=3.00st=3.00s$ and $t=7.00s,t=7.00s,$ the heights and the velocities are the same and the period of the oscillation is 4.00 s.\n\nThe wave function above is derived using a sine function. Can a cosine function be used instead?\n\n### Velocity and Acceleration of the Medium\n\nAs seen in Example 16.4, the wave speed is constant and represents the speed of the wave as it propagates through the medium, not the speed of the particles that make up the medium. The particles of the medium oscillate around an equilibrium position as the wave propagates through the medium. In the case of the transverse wave propagating in the x-direction, the particles oscillate up and down in the y-direction, perpendicular to the motion of the wave. The velocity of the particles of the medium is not constant, which means there is an acceleration. The velocity of the medium, which is perpendicular to the wave velocity in a transverse wave, can be found by taking the partial derivative of the position equation with respect to time. The partial derivative is found by taking the derivative of the function, treating all variables as constants, except for the variable in question. In the case of the partial derivative with respect to time t, the position x is treated as a constant. Although this may sound strange if you haven’t seen it before, the object of this exercise is to find the transverse velocity at a point, so in this sense, the x-position is not changing. We have\n\n$y(x,t)=Asin(kx−ωt+ϕ)vy(x,t)=∂y(x,t)∂t=∂∂t(Asin(kx−ωt+ϕ))=−Aωcos(kx−ωt+ϕ)=−vymaxcos(kx−ωt+ϕ).y(x,t)=Asin(kx−ωt+ϕ)vy(x,t)=∂y(x,t)∂t=∂∂t(Asin(kx−ωt+ϕ))=−Aωcos(kx−ωt+ϕ)=−vymaxcos(kx−ωt+ϕ).$\n\nThe magnitude of the maximum velocity of the medium is $|vymax|=Aω|vymax|=Aω$. This may look familiar from the Oscillations and a mass on a spring.\n\nWe can find the acceleration of the medium by taking the partial derivative of the velocity equation with respect to time,\n\n$ay(x,t)=∂vy∂t=∂∂t(−Aωcos(kx−ωt+ϕ))=−Aω2sin(kx−ωt+ϕ)=−aymaxsin(kx−ωt+ϕ).ay(x,t)=∂vy∂t=∂∂t(−Aωcos(kx−ωt+ϕ))=−Aω2sin(kx−ωt+ϕ)=−aymaxsin(kx−ωt+ϕ).$\n\nThe magnitude of the maximum acceleration is $|aymax|=Aω2.|aymax|=Aω2.$ The particles of the medium, or the mass elements, oscillate in simple harmonic motion for a mechanical wave.\n\n### The Linear Wave Equation\n\nWe have just determined the velocity of the medium at a position x by taking the partial derivative, with respect to time, of the position y. For a transverse wave, this velocity is perpendicular to the direction of propagation of the wave. We found the acceleration by taking the partial derivative, with respect to time, of the velocity, which is the second time derivative of the position:\n\n$ay(x,t)=∂2y(x,t)∂t2=∂2∂t2(Asin(kx−ωt+ϕ))=−Aω2sin(kx−ωt+ϕ).ay(x,t)=∂2y(x,t)∂t2=∂2∂t2(Asin(kx−ωt+ϕ))=−Aω2sin(kx−ωt+ϕ).$\n\nNow consider the partial derivatives with respect to the other variable, the position x, holding the time constant. The first derivative is the slope of the wave at a point x at a time t,\n\n$slope=∂y(x,t)∂x=∂∂x(Asin(kx−ωt+ϕ))=Akcos(kx−ωt+ϕ).slope=∂y(x,t)∂x=∂∂x(Asin(kx−ωt+ϕ))=Akcos(kx−ωt+ϕ).$\n\nThe second partial derivative expresses how the slope of the wave changes with respect to position—in other words, the curvature of the wave, where\n\n$curvature=∂2y(x,t)∂x2=∂2∂x2(Asin(kx−ωt+ϕ))=−Ak2sin(kx−ωt+ϕ).curvature=∂2y(x,t)∂x2=∂2∂x2(Asin(kx−ωt+ϕ))=−Ak2sin(kx−ωt+ϕ).$\n\nThe ratio of the acceleration and the curvature leads to a very important relationship in physics known as the linear wave equation. Taking the ratio and using the equation $v=ω/kv=ω/k$ yields the linear wave equation (also known simply as the wave equation or the equation of a vibrating string),\n\n$∂2y(x,t)∂t2∂2y(x,t)∂x2=−Aω2sin(kx−ωt+ϕ)−Ak2sin(kx−ωt+ϕ)=ω2k2=v2,∂2y(x,t)∂t2∂2y(x,t)∂x2=−Aω2sin(kx−ωt+ϕ)−Ak2sin(kx−ωt+ϕ)=ω2k2=v2,$\n$∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2.∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2.$\n16.6\n\nEquation 16.6 is the linear wave equation, which is one of the most important equations in physics and engineering. We derived it here for a transverse wave, but it is equally important when investigating longitudinal waves. This relationship was also derived using a sinusoidal wave, but it successfully describes any wave or pulse that has the form $y(x,t)=f(x∓vt).y(x,t)=f(x∓vt).$ These waves result due to a linear restoring force of the medium—thus, the name linear wave equation. Any wave function that satisfies this equation is a linear wave function.\n\nAn interesting aspect of the linear wave equation is that if two wave functions are individually solutions to the linear wave equation, then the sum of the two linear wave functions is also a solution to the wave equation. Consider two transverse waves that propagate along the x-axis, occupying the same medium. Assume that the individual waves can be modeled with the wave functions $y1(x,t)=f(x∓vt)y1(x,t)=f(x∓vt)$ and $y2(x,t)=g(x∓vt),y2(x,t)=g(x∓vt),$ which are solutions to the linear wave equations and are therefore linear wave functions. The sum of the wave functions is the wave function\n\n$y1(x,t)+y2(x,t)=f(x∓vt)+g(x∓vt).y1(x,t)+y2(x,t)=f(x∓vt)+g(x∓vt).$\n\nConsider the linear wave equation:\n\n$∂2(f+g)∂x2=1v2∂2(f+g)∂t2∂2f∂x2+∂2g∂x2=1v2[∂2f∂t2+∂2g∂t2].∂2(f+g)∂x2=1v2∂2(f+g)∂t2∂2f∂x2+∂2g∂x2=1v2[∂2f∂t2+∂2g∂t2].$\n\nThis has shown that if two linear wave functions are added algebraically, the resulting wave function is also linear. This wave function models the displacement of the medium of the resulting wave at each position along the x-axis. If two linear waves occupy the same medium, they are said to interfere. If these waves can be modeled with a linear wave function, these wave functions add to form the wave equation of the wave resulting from the interference of the individual waves. The displacement of the medium at every point of the resulting wave is the algebraic sum of the displacements due to the individual waves.\n\nTaking this analysis a step further, if wave functions $y1(x,t)=f(x∓vt)y1(x,t)=f(x∓vt)$ and $y2(x,t)=g(x∓vt)y2(x,t)=g(x∓vt)$ are solutions to the linear wave equation, then $Ay1(x,t)+By2(x,y),Ay1(x,t)+By2(x,y),$ where A and B are constants, is also a solution to the linear wave equation. This property is known as the principle of superposition. Interference and superposition are covered in more detail in Interference of Waves.\n\n### Example 16.4\n\n#### Interference of Waves on a String\n\nConsider a very long string held taut by two students, one on each end. Student A oscillates the end of the string producing a wave modeled with the wave function $y1(x,t)=Asin(kx−ωt)y1(x,t)=Asin(kx−ωt)$ and student B oscillates the string producing at twice the frequency, moving in the opposite direction. Both waves move at the same speed $v=ωk.v=ωk.$ The two waves interfere to form a resulting wave whose wave function is $yR(x,t)=y1(x,t)+y2(x,t).yR(x,t)=y1(x,t)+y2(x,t).$ Find the velocity of the resulting wave using the linear wave equation $∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2.∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2.$\n\n#### Strategy\n\nFirst, write the wave function for the wave created by the second student. Note that the angular frequency of the second wave is twice the frequency of the first wave $(2ω)(2ω)$, and since the velocity of the two waves are the same, the wave number of the second wave is twice that of the first wave $(2k).(2k).$ Next, write the wave equation for the resulting wave function, which is the sum of the two individual wave functions. Then find the second partial derivative with respect to position and the second partial derivative with respect to time. Use the linear wave equation to find the velocity of the resulting wave.\n\n#### Solution\n\n1. Write the wave function of the second wave: $y2(x,t)=Asin(2kx+2ωt).y2(x,t)=Asin(2kx+2ωt).$\n2. Write the resulting wave function:\n$yR(x,t)=y1(x,t)+y(x,t)=Asin(kx−ωt)+Asin(2kx+2ωt).yR(x,t)=y1(x,t)+y(x,t)=Asin(kx−ωt)+Asin(2kx+2ωt).$\n3. Find the partial derivatives:\n$∂yR(x,t)∂x=−Akcos(kx−ωt)+2Akcos(2kx+2ωt),∂2yR(x,t)∂x2=−Ak2sin(kx−ωt)−4Ak2sin(2kx+2ωt),∂yR(x,t)∂t=−Aωcos(kx−ωt)+2Aωcos(2kx+2ωt),∂2yR(x,t)∂t2=−Aω2sin(kx−ωt)−4Aω2sin(2kx+2ωt).∂yR(x,t)∂x=−Akcos(kx−ωt)+2Akcos(2kx+2ωt),∂2yR(x,t)∂x2=−Ak2sin(kx−ωt)−4Ak2sin(2kx+2ωt),∂yR(x,t)∂t=−Aωcos(kx−ωt)+2Aωcos(2kx+2ωt),∂2yR(x,t)∂t2=−Aω2sin(kx−ωt)−4Aω2sin(2kx+2ωt).$\n4. Use the wave equation to find the velocity of the resulting wave:\n\n$∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2,−Ak2sin(kx−ωt)−4Ak2sin(2kx+2ωt)=1v2(−Aω2sin(kx−ωt)−4Aω2sin(2kx+2ωt)),k2(−Asin(kx−ωt)−4Asin(2kx+2ωt))=ω2v2(−Asin(kx−ωt)−4Asin(2kx+2ωt)),k2=ω2v2,|v|=ωk.∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2,−Ak2sin(kx−ωt)−4Ak2sin(2kx+2ωt)=1v2(−Aω2sin(kx−ωt)−4Aω2sin(2kx+2ωt)),k2(−Asin(kx−ωt)−4Asin(2kx+2ωt))=ω2v2(−Asin(kx−ωt)−4Asin(2kx+2ωt)),k2=ω2v2,|v|=ωk.$\n\n#### Significance\n\nThe speed of the resulting wave is equal to the speed of the original waves $(v=ωk).(v=ωk).$ We will show in the next section that the speed of a simple harmonic wave on a string depends on the tension in the string and the mass per length of the string. For this reason, it is not surprising that the component waves as well as the resultant wave all travel at the same speed.\n\nThe wave equation $∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2∂2y(x,t)∂x2=1v2∂2y(x,t)∂t2$ works for any wave of the form $y(x,t)=f(x∓vt).y(x,t)=f(x∓vt).$ In the previous section, we stated that a cosine function could also be used to model a simple harmonic mechanical wave. Check if the wave\n\n$y(x,t)=0.50mcos(0.20πm−1x−4.00πs−1t+π10)y(x,t)=0.50mcos(0.20πm−1x−4.00πs−1t+π10)$\n\nis a solution to the wave equation.\n\nAny disturbance that complies with the wave equation can propagate as a wave moving along the x-axis with a wave speed v. It works equally well for waves on a string, sound waves, and electromagnetic waves. This equation is extremely useful. For example, it can be used to show that electromagnetic waves move at the speed of light."
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https://www.geeksforgeeks.org/java-data-types-question-3/?ref=rp | [
"# Java | Data Types | Question 3\n\nPredict the output of the following program.\n\n `class` `Test ` `{ ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``Double object = ``new` `Double(``\"2.4\"``); ` ` ``int` `a = object.intValue(); ` ` ``byte` `b = object.byteValue(); ` ` ``float` `d = object.floatValue(); ` ` ``double` `c = object.doubleValue(); ` ` ` ` ``System.out.println(a + b + c + d ); ` ` ` ` ``} ` `} `\n\n(A) 8\n(B) 8.8\n(C) 8.800000095367432"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73299295,"math_prob":0.9189357,"size":1505,"snap":"2020-34-2020-40","text_gpt3_token_len":400,"char_repetition_ratio":0.24050634,"word_repetition_ratio":0.20070423,"special_character_ratio":0.2963455,"punctuation_ratio":0.09338521,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99652946,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-26T10:00:05Z\",\"WARC-Record-ID\":\"<urn:uuid:61fd6254-99df-4951-8d10-0f3549dac4e1>\",\"Content-Length\":\"91685\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0965dda-50ed-405e-aeac-b352bf54a5b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:a392b4e7-269a-49e5-a844-80d2fe2c8e58>\",\"WARC-IP-Address\":\"23.56.172.242\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/java-data-types-question-3/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:WRRXR22TRPQWQ7ZEH63M6L6LNO2FB4MO\",\"WARC-Block-Digest\":\"sha1:CVGIMQ7IWYKWA4QQSKMF4AQM3V4KKBUC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400238038.76_warc_CC-MAIN-20200926071311-20200926101311-00123.warc.gz\"}"} |
https://origin.geeksforgeeks.org/category/algorithm/bit-magic/ | [
"# Category Archives: Bit Magic\n\nGiven a positive integer N, the task is to find the two smallest integers A and B such that the Bitwise XOR of A and… Read More\nGiven two integers N and K, the task is to find the number of N-length arrays that satisfies the following conditions: The sum of the… Read More\nGiven a positive integer N, the task is to count the number of N-digit numbers such that the count of distinct odd and distinct even… Read More\nGiven two binary arrays, A[] and B[] of size N respectively, the task is to find the number of elements in array A[] that will… Read More\nGiven a binary string str[] of size N and an integer M. This binary string can be modified by flipping all the 0’s to 1… Read More\nGiven a binary tree, the task is to compress all the nodes on the same vertical line into a single node such that if the… Read More\nGiven an integer N, the task is to reduce N to 0 in the minimum number of operations using the following operations any number of… Read More\nGiven a binary string S, the task is to find that the decimal representation of the given binary string is divisible by integer K or… Read More\nGiven two binary strings s1[] and s2[] of the same length N, the task is to find the minimum number of operations to make them… Read More\nGiven two positive integers N and K, the task is to find the number of arrays of size N such that each array element lies… Read More\nGiven an array arr[] consisting of N positive integers and a positive integer X, the task is to find the number of pairs (i, j)… Read More\nGiven a positive integer N, the task is to print the minimum count of consecutive numbers less than N such that the bitwise AND of… Read More\nGiven an integer N, the task is to print the count of all the non-negative integers K less than or equal to N, such that… Read More\nGiven a sorted array arr[] of the first N Natural Numbers and an integer X, the task is to print the last remaining element after… Read More\nGiven a positive integer N, the task is to count the number of N-digit numbers such that the bitwise AND of adjacent digits equals 0.… Read More"
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http://codeforces.com/blog/entry/73244 | [
"Codeforces celebrates 10 years! We are pleased to announce the crowdfunding-campaign. Congratulate us by the link https://codeforces.com/10years. ×\n\n### neal's blog\n\nBy neal, 5 weeks ago,",
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"Given the numbers\n\n$1, 2, \\dots, 20$\n\n, you want to choose some subset of these numbers such that no number you choose is divisible by any other number you choose. What's the largest subset you can get?\n\nOne simple idea is to choose all the primes: 2, 3, 5, 7, 11, 13, 17, and 19, leading to a subset of size 8. But you can do better! Try to figure out the solution yourself before reading below.\n\nFor 1292F - Nora's Toy Boxes from Round 614 Div. 1, after you make several observations and pull out a sick counting DP, you get a solution that is\n\n$\\displaystyle \\mathcal{O} \\left (N^2 \\cdot 2^S \\right )$\n\n, where $S$ is the number of values in our array $A$ that are 1) not divisible by any other value in $A$ and 2) have at least two other values in $A$ divisible by it. (These are the only values we should consider using as $a_i$ in the operation described in the problem.) So what is that time complexity really?\n\nDon't worry if you haven't actually read or tried to solve the problem. In the rest of this post I'll focus on this question: out of all subsets of\n\n$1, 2, \\dots, N$\n\n, what is the maximum possible value of $S$? I'll show that we can attain a precise bound of\n\n$\\displaystyle \\frac{N}{6}$\n\n.\n\nLet's call the numbers that satisfy our constraints above special. First, since each of our special numbers needs to have two other multiples in $A$, any special number $x$ must satisfy\n\n$3x \\leq N$\n\n, so\n\n$\\displaystyle x \\leq \\left \\lfloor \\frac{N}{3} \\right \\rfloor$\n\n. So we should just find the maximum subset of\n\n$\\displaystyle 1, 2, \\dots, \\left \\lfloor \\frac{N}{3} \\right \\rfloor$\n\nsuch that no number in our subset divides another, and then we can add every number greater than\n\n$\\displaystyle \\left \\lfloor \\frac{N}{3} \\right \\rfloor$\n\nto ensure that we hit the second requirement of at least two multiples for each.\n\nWith\n\n$N = 60$\n\nin our case, this leads to the problem described above: given the numbers\n\n$1, 2, \\dots, 20$\n\n, what is the largest subset of these numbers we can choose such that no number we choose is divisible by another number we choose? The answer turns out to be 10. First we provide a simple construction: choose\n\n$11, 12, \\dots, 20$\n\n. (We can also generalize this: for\n\n$1, 2, \\dots, K$\n\n$\\displaystyle \\left \\lceil \\frac{K}{2} \\right \\rceil$\n\n.)\n\nNow we need to prove that this is the maximum possible. We can do this by building the following chains of numbers:\n\n• 1 — 2 — 4 — 8 — 16\n• 3 — 6 — 12\n• 5 — 10 — 20\n• 7 — 14\n• 9 — 18\n• 11\n• 13\n• 15\n• 17\n• 19\n\nEvery number occurs in exactly one chain, and we can never take two numbers from the same chain. So our subset can't be larger than the number of chains, which is again exactly 10. (In general it's the number of odd numbers up to $K$, which is again\n\n$\\displaystyle \\left \\lceil \\frac{K}{2} \\right \\rceil$\n\n.)\n\nSo in the general case, the maximum value of $S$ is exactly\n\n$\\displaystyle \\left \\lceil \\frac{\\left \\lfloor \\frac{N}{3} \\right \\rfloor}{2} \\right \\rceil = \\left \\lfloor \\frac{N + 3}{6} \\right \\rfloor$\n\n, meaning the algorithm has a complexity of\n\n$\\displaystyle \\mathcal{O} \\left (N^2 \\cdot 2^{\\frac{N}{6}} \\right )$\n\n.\n\nThanks scott_wu for discussing and thinking through this problem.",
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https://puzzling.stackexchange.com/questions/12119/coin-flipping-game-with-the-devil | [
"# Coin Flipping Game with the Devil\n\nYou die, and awake in Hell. Satan awaits you, and has prepared a curious game. He has arranged $n$ quarters in a line, going in the east/west direction. He placed the coins at the ends tails up, and all others heads up, like so: $$\\text{T H H H }\\cdots \\text{ H H H T}$$\n\nSatan explains the rules.\n\n• Once a day, a coin is removed from the east end, and placed on the west end.\n• If the coin was initially tails-up, then you get to choose whether the coin is placed heads up or down.\n• If it was initially heads-up, then Satan gets to make this choice.\n• If the coins are all heads up at the end of a day, you get to leave Hell.\n\nSatan will of course try his hardest to make sure you never leave.\n\nFor example, when $n=5$, we start with $\\text{T H H } \\color{red}{\\text H} \\color{green}{\\text{ T}}$. The first day is your choice; if you choose heads, the arrangement becomes $\\color{green}{\\text{H }}\\text{T H H } \\color{red}{\\text H}$. The next three days, however, will be Satan's choice. He may fight back on the second day by choosing tails, resulting in $\\color{red}{\\text{T } }\\color{green}{\\text{H }}\\text{T H H }$.\n\nIs there a strategy that eventually guarantees your salvation? Or can Satan conspire to keep you in Hell forever?\n\nAddendum: To give credit where it is due, this puzzle from The Puzzle Toad, (under the name \"Zeroise Me\") which is a superb collection of similarly clever and enjoyable conundrums.\n\n• All the answers give are quite fun but the devil would win, you're in hell, he's the devil, he does not play by the rules, he can do what ever he wants in hell. Apr 15, 2015 at 8:55\n• The Devil went down to Georgia / lookin' for a soul to claim / so he sat me down at his table / to play a combinatorial game Apr 15, 2015 at 9:23\n• If hell is a place where we get to play logic puzzles with the devil for eternity, why try to leave? Apr 15, 2015 at 12:16\n• The game seems fair until you n days in you find out that the last coin is tails on both sides. Apr 15, 2015 at 14:40\n• @RemcoGerlich This game only occupies a minute or so of each day, presumedly the rest of the day is not as intellectually stimulating. Apr 15, 2015 at 15:30\n\nSatan should stick to fiddling. You will win, and here is a simple proof.\n\nConsider the game $n$ turns at a time. After each cycle of $n$ turns, all the coins are in their original position (though not necessarily flipped the same way).\n\nReplace $H$ with $0$ and $T$ with $1$.\n\nIn each cycle, you flip all $1$'s to $0$'s, until Satan flips a $0$ to a $1$. Once Satan makes a flip, you stop and leave the rest of this cycle's coins alone.\n\nSatan must always make a flip during a cycle. If not, then you have just flipped all the coins to $0$, and you win.\n\nRead the sequence of coins as a binary number. Each cycle's play starts at the ones place and progresses to the largest place. Satan makes the last flip in each cycle, and that flip flips a $0$ to a $1$. Therefore, after each cycle, the number gets larger.\n\nBut it can't get larger forever. After at most $2^n$ cycles, it reaches $111...1$. Put on your smuggest face and flip all the coins for a well-deserved win.\n\n• What a nice, simple proof! Unfortunately for Satan, fiddling didn't turn out so well either.\n– xnor\nApr 15, 2015 at 9:16\n• Since we have now proven you win from every configuration, we can also prove you can win in $2^n$ steps, after all there are only $2^n$ configurations. Apr 15, 2015 at 14:20\n• Although $2^n$ is indeed an upper bound, I contend that your solution is not optimal in terms of fewest moves.\n– user88\nApr 15, 2015 at 23:04\n• @JoeZ. I agree. As far as I can tell, the optimal solution is as of yet an open problem. One way to optimize this strategy is to rotate the viewing window on certain turns, which can increase the current number. The dynamics quickly become difficult to describe. Apr 16, 2015 at 0:24\n• I described them in my solution, although I still haven't made a program to count them yet.\n– user88\nApr 16, 2015 at 0:25\n\nCharlie always wins.\n\nLet's split the chain of coins up into two sections: the tails-chain and the heads-chain. The tails-chain is Charlie's long chain of tails. I will use a vertical bar to denote where the chains split. It starts out as the two Ts in TTH...H, written TT|H...H. The heads chain is everything bounded by the two H coins of Satan's.\n\nA postulate: Tails in a chain will always connect, and Satan will never get them back. This is pretty easy to prove. Once the tails have joined Charlie's chain, Charlie just has to choose not to flip them to heads again.\n\nA postulate: Satan must have at least one tails in his chain. If Satan at any point has all heads in his chain, then the chain must pass through the state H...H|T...T, which means he loses. Thus, there must be at least one tails in his chain.\n\nFinal postulate: Charlie can force the tails in Satan's chain one step further towards its 'tail' with each loop through. (Note that there is a special case for this, detailed below.)\n\nThis leads to the win condition. In order to win, Charlie needs to force Satan's chain to spit out tails. Those tails will connect with Charlie's chain, and Satan will lose them. Repeat this process indefinitely, and Charlie wins.\n\nTo do this, Charlie flips any tails resting between two heads to heads. That means the soonest Satan can put a new tails in is the next coin. This forces Satan's tails coin one coin further down the heads-chain, until it contacts the tails-chain and becomes part of it.\n\nAt that point, he loses another heads, as he has to have a tails in his chain. Satan will eventually only have one heads, and will lose.\n\nAn example chain: HHTH|TT -> HTHH|TT -> THHH|TT. At this point, Charlie's chain is actually three tails long: TTT|HHH. Satan can't help but move HTH|TTT -> THH|TTT, which means Charlie's chain becomes four tails long. At this point, Satan's loss is inevitable.\n\nAnother example chain: HHHTH|TT -> HTHHH|TT. At this point Satan might flip the fourth coin again resulting in HTHTH|TT, but if Charlie leaves this alone and flips the second tail back when he gets around to it, Satan will have no choice but to flip the third coin from the left, resulting in HTTHH|TT. Then, on the next go-around, Charlie increases the length of the chain by 1 again. Consequently, Satan will eventually lose all his open coins.\n\nAfter there are enough coins, Satan may be able to introduce multiple tails. This is non-trivial, but results in a loss for Satan anyway. If Satan introduces multiple tails, do nothing. Then, start pushing coins from the front of the chain toward the back.\n\nFor instance, let's say you have TT|HTHHH. Then, Satan is a jerk and creates HTH|TT|HT. Now, instead of flipping your lingering tails coin, do nothing with it: TT|HTHTH. Now, keep flipping your tails to heads until Satan flips a coin. This leads to the following chain:\n\nH|TT|HTHT -> HH|TT|HTH -> HHH|TT|HT -> HHHH|TT|H -> THHHH|TT -> TTT|HHHH\n\n\nIn other words, up until the point where Satan flips a coin, you keep flipping tails to heads. As an example with more coins:\n\nTT|HTHHHTH -> H|TT|HTHHHT -> HH|TT|HTHHH -> THH|TT|HTHH -> TT|HTHHTHH\nTT|HTHHTHH -> HH|TT|HTHHT -> HHH|TT|HTHH -> THHH|TT|HTH -> TT|HTHTHHH\n\n\nIt's pretty clear that eventually you'll end up in the configuration:\n\nTT|HTTTHHH\n\n\nYou can keep pushing Satan's tokens further down the chain. Once you do, you're in control of them, and Satan needs to add another tails further up the chain otherwise you'll bump the last heads into your chain.\n\nTo illustrate, let's say Satan doesn't add a new tails. Then you have: TT|THHHHHH -> TTT|HHHHHH\n\nHowever, if he adds another one, it will become TT|HTTTHTH, which you can easily force into TT|HTTTTHH on your next turn. From there, Satan has nowhere to go, and loses a coin.\n\nTherefore, no matter what Satan does, Charlie always wins.\n\nHere are a few extraneous notes.\n\nSome raw experimentation shows that the number of 'cycles' it takes to flip $n$ coins over is $F_n$, the $n$th Fibonacci number. (Here, I'm defining a cycle as the number of moves until the tails-chain is at the front again.)\n\nBecause the tails-chain gets shifted backward one step for every coin assimilated, the total number of moves ends up being $n*F_n-(n-1)=n(F_n-1)+1$.\n\nI'd gander that if there were even 20 coins, it would take $20*6765-19=135281$ days for you to actually escape. Several centuries would have passed before you ever left, for any reasonable number of coins. What a jerk, that Satan.\n\n• I've moved the comments here to chat, as there's valuable content but it's mostly chat content.\n– user20\nApr 15, 2015 at 5:46\n• One question - who's Charlie? Apr 16, 2015 at 19:15\n• @mdc Just a name I gave our poor tortured friend.\n– user20\nApr 16, 2015 at 19:16\n\nFor $n = 2$, the solution is trivial - you flip both coins heads-up for the first two days, and leave.\n\nFor $n = 3$, you leave the first coin alone, leaving Satan with TTH. But no matter whether Satan flips this coin or leaves it alone, you can flip all the T's in a row afterwards, so you win.\n\nFor $n = 4$, you leave the first coin alone, leaving Satan with TTHH. Regardless of how Satan flips these two coins, in all four cases you're left with a string of 2, 3, or 4 consecutive T's, so you can flip all of them and win.\n\n(In the case of flipping only the second one, you're left with THTT, but just leave those two alone and no matter what Satan flips, you can flip all of them the second time around.)\n\n$n = 5$ is the first interesting case. Then, if Satan flips only the second H out of three after you leave the first coin alone, you're left with HTHTT. If you leave this alone, then Satan will also leave it alone, and you'll be stuck in a loop forever. So the only thing to do is to either flip the solitary T, or flip one or both of the two consecutive T's.\n\n1. If you flip one of the two consecutive T's, you've left Satan with either HTHTH or THHTH depending on which one you flipped. Satan can just leave this alone, which means you're stuck in an infinite loop until you decide to flip one of the others.\n\n• If you flip the first one, Satan's left with HHHTH. If he leaves this alone, then you flip the T and win. So he has to flip this H, leaving you with THHHT, back where you started.\n\n• If you flip the second one, Satan's left with HHTHH. But he can just leave the first H alone and flip the second one, leaving you back where you started as well.\n\n2. If you flip the solitary T, then Satan's left with HHTTH. If he leaves this H alone, then you flip the T and win. So he has to flip the T, leaving you with THHTT. Just leave these two alone, leaving Satan with TTTHH and the same quandary he had in the $n = 4$ case.\n\nIn general, you want to leave Satan with a situation where all the T's are in a single bunch and a single H follows them. Then, he's forced to increase the length of the T chain by 1 (and thereby effectively reduce the number of coins in the game inductively), and if you can keep doing that, then by induction you win.\n\nFor $n = 6$, Satan can flip one, the other, or both of the two coins not next to the existing T's, leaving you with one of HHTHTT, HTHHTT, or HTTHTT once you've gone a full revolution. In any case where Satan flips the coin right before the leftmost H, you leave them alone (and your first two T's) the first time around, and flip them all back to heads the second round, leaving Satan with HHHTTH and you winning.\n\nSo Satan has to flip the first coin only where he's left with HHTTHH and has the choice of flipping the T, which is the only nontrivial variation from $n = 5$.\n\nLet's suppose he does this. Then you're left with HTHHTT, which, from above, is one of the winning strategies. So you win as well.\n\nWe still haven't deduced a general strategy yet, so let's continue with $n = 7$.\n\nSatan can leave you with HTHTHTT this time, which circumvents the flipping-the-one-next-to-last rule described above since the first flipped T prevents Satan from being forced to flip the H immediately before the string of two T's. But in this case, if you flip the first solitary T the second time around, then Satan has to flip the next H, because if he leaves it alone, you can just flip the next T as well, leaving Satan with HHHHTTH and starting the $n = 6$ case. So he has to flip that H, leaving you with HTTHHTT the second time around. Then you simply flip the left sequence of T's and force the $n = 6$ case again.\n\nBut there is one wrinkle here. What if Satan flips to HHTHHTT, and you flip that T only for Satan to flip the H immediately afterwards (resulting in HTHHHTT) and then Satan flips the second H only, resulting in HTHTTHT on your turn? If you flip the T now, and then the next solitary T right afterward, Satan will just flip the next T again, returning you to HHTHHTT.\n\nSo, you just leave that second T alone for two days, and the HTHTHTT case plays out exactly as described in the first place.\n\nAt this point, we finally notice a pattern. You will always flip the series of T's immediately after your longest chain (which starts at length 2 and keeps growing until it's the entire string of coins), which forces Satan to flip the H immediately to the left of them in order to ward off defeat. And if Satan flips any of the H's beforehand, leave the rest of the T's alone and let it go around one more revolution.\n\nThis will always result in the set of stray T's moving left towards the end of your longest chain and merging together with other stray sets, or a set of new T's being created that will have to move left anyway, possibly creating an even longer chain in the process.\n\nThe longest Satan can drag the game out for when your longest chain is $m$ T's long (without making a longer chain in the process) is until the chain of $m$ T's is followed by a series of chains of $m$ T's each separated by a single H. At that point, any head Satan flips will automatically create a longer chain, so you can flip all the chains of $m$ T's, leaving Satan with the coin right before the last chain of $m$ T's which he must then extend to $m+1$.\n\nAnd this way, you can always defeat Satan and escape from Hell.\n\nBelow is a program-like algorithm that you can follow, assuming that instead of moving coins to the left, you keep track of a cursor that moves to the left one space every day.\n\n• Every day, the cursor moves left one space. If the cursor reaches the leftmost coin, it jumps back to the right.\n• If you are on your rightmost set of consecutive T's, pass.\n• If Satan has not flipped a coin yet since the cursor last jumped to the right and you are currently on a T, flip it. Otherwise, pass.\n• If Satan flips the leftmost coin, move that coin to the rightmost position, keeping the cursor on it.\n\nAn example of the longest match for $n = 11$, to show how this algorithm might work:\n\nHHHHHHHHHTT You leave the two T's alone, Satan flips coin 8.\nHHHHHHHTHTT You flip the first T, forcing Satan to flip coin 7.\nHHHHHHTHHTT You flip the first T, forcing Satan to flip coin 6.\n...\nHTHHHHHHHTT Satan flips coin 8 again, and you leave everything alone.\nHTHHHHHTHTT You flip coin 8, and keep moving left until Satan reaches coin 4.\nHTHTHHHHHTT Satan flips coin 8 again. This flipping coin 8 and moving to the left continues until all the even-numbered slots are occupied.\nHTHTHTHTHTT You start flipping all the T's from coin 8 onward. If Satan flips any coins before that, stop and let the cursor wrap around again.\nHTTHHHHTHTT This might happen. Then, Satan flips coin 8 and starts the above process all over.\n...\nHTTHTTHHHTT This might happen again.\nHTTHTTHTHTT Satan flips coin 8 one last time, but he cannot dodge anymore. You leave the coins alone for one more iteration.\nTHHHHHHHHTT You flip all groups of coins, leaving Satan with no choice but to flip coin 1.\n\nHHHHHHHHTTT The frame of reference of the coins move forward one space.\nHHHHHHTHTTT Satan starts with coin 7 this time.\n...\nHTHTHTHHTTT Chains of length 1.\nHTHTHTTHTTT Satan flips coin 7. You leave all the other coins alone.\nHTTHHHHHTTT Satan is forced to do this again.\nHTTHHHTHTTT Same process as before.\n...\nHTTHTTHHTTT Chains of length 2. You flip the first chain, Satan flips coin 4.\nHTTTHHHHTTT First chain of length 3. Satan flips coin 7 again, and the process continues.\nHTTTHHTHTTT\nHTTTHTHHTTT\nHTTTHTTHTTT In 3 moves, Satan exhausts the chain of length 3. You flip all sequences of coins to the left of your first chain.\nTHHHHHHHTTT Again, Satan has no choice but to flip coin 1. The chain extends by length 1 again.\n...\nTHHHHHHTTTT ???\n......\nTTTTTTTTTTT Profit!\n\n\nIt turns out that the number of moves required to force a win grows exponentially compared to the number of coins in a row.\n\n• The two possibilities after HTHTT in scenario 1 are THHTH and HTHTH. Your solution flips the coins, but doesn't ever move them. Apr 15, 2015 at 3:50\n• I'm in the middle of editing to clarify that.\n– user88\nApr 15, 2015 at 3:56\n• Also, the reason my solution doesn't move the coins is because it deals in complete revolutions.\n– user88\nApr 15, 2015 at 4:14\n• @JoeZ. I think I fully understand what you are doing, but your second to last section, where you explain the strategy, could use some expanding. This description, as it stands, does not cover all cases (you turn the first series of heads you see.. then what?). I get that you can piece together what to do from your entire post's discussion, but it would be nicer if you at some point wrote a simple algorithm that clearly describe's Charlie's decisions in all cases, as if programming a Turing machine. Apr 15, 2015 at 6:15\n• Yes it is, and I can't think of any way Satan could beat that. But you should convince Charlie beyond all doubt that your strategy works, for all $n$; false hope is a dangerous thing in Hell, so Charlie won't accept any strategy without a proof. Apr 15, 2015 at 7:09\n\nTo see why, let's consider 8 coins to see the pattern. The goal is to gain control, or in other words to get as many tails in a row as possible so that you can get all heads in the end. First, keep tails on the first move to keep the two tails together:\n\n00: TTHHHHHH //Position after first move\n\n\nLet's call this the base position. Below are the states each time the coins return to the base position if you and satan both flip optimally for your goals (assuming he wants to keep you there and you want to get out):\n\nTTHHHHHH //Position after first move\nTTHHHHTH //satan flips the 2nd coin to tails.\n-he doesn't flip the first because that would give you 3 tails in a row\n-If he doesn't flip any you could flip your tails to heads at the end\nTTHHHTHH //You flip the 2nd coin to tails and he flips the 3rd to heads\nTTHHHTTH //satan flips the 2nd coin to tails and you leave the 3rd alone\nTTHHTHHH //You flip 2 and 3 to heads and he flips 4 to tails\nTTHHTHTH\nTTHHTTHH\nTTHHTTTH\nTTHTHHHH\nTTHTHHTH\nTTHTHTHH\nTTHTHTTH\nTTHTTHHH\nTTHTTHTH\nTTHTTTHH\nTTHTTTTH\nTTTHHHHH //Yay, 3 tails in a row!\nTTTHHHTH\nTTTHHTHH\nTTTHHTTH\nTTTHTHHH\nTTTHTHTH\nTTTHTTHH\nTTTHTTTH\nTTTTHHHH //Yay, 4 tails in a row!\nTTTTHHTH\nTTTTHTHH\nTTTTHTTH\nTTTTTHHH //Yay, 5 tails in a row!\nTTTTTHTH\nTTTTTTHH //Yay, 6 tails in a row!\nTTTTTTTH //Yay, 7 tails in a row...almost free!\nHHHHHHHT //Yay, one more day until you win the game and go free!\n\n\nThe next day, you get up early and excitedly flip the easternmost coin from tales to heads as you move it to the west side of the pile.\n\nSatan sighs and says in a worn-out tone \"Congratulations, you've won the game and you are now free to go.\"\n\nThis has been the news you have been waiting for! You eagerly await the Devil's instructions on how to leave this God-forsaken place.\n\nHowever, after about 15 seconds of awkward silence where you nervously shift between looking at the burning embers and him, you finally ask \"Ok, and how do I go about leaving?\"\n\nThe devil looks up at the cave-like ceiling miles above and says \"That's a good question. To tell you the truth I've been thinking about that a lot myself, but I haven't found a way yet. Don't worry though, when I find one I'll be sure and tell you.\"\n\nYou let the words sink in for a bit. \"So I'm free to go, but there's no way out? Curse you, what was this game all about?!!?\"\n\nThe Devil sighs tiredly and says, \"Oh nothing really. I just made it up as a way to pass the time...\"\n\n• Good point. Prisoner is he. Apr 16, 2015 at 20:09\n\nYou can force a win.\n\nYour strategy is to aim for the string $\\phi\\gamma$ where $\\gamma$ is comprised only of $T$ and $\\phi$ is comprised only of $H$ and it is your turn next..\n\nClearly, a configuration of $H^iT^j$ wins for us... we just make all $T$ into $H$ and win - regardless of $i$, $j$. Call this configuration $win$.\n\nThe strategy in general is, We always play $T$ unless it's a $T$ that is not part of our $win$ T string ($\\gamma$) in which case we play $H$ (in order to establish $\\phi$).\n\nHere is a demonstration: Say, $n=5$: the string, $S$, is $THHHT$.\n\nSo your first choice will always be to choose $T$: $TTHHH$.\n\nOur job is to keep the string of $T$s as long as possible.\n\nThere are now 8 ways that Satan can play his moves:\n\n$HHHTT$\n$TTHTT$\n$THTTT$\n$HTTTT$\n$HHTTT$\n$HTHTT$\n$THHTT$\n$TTTTT$\n\nI'll remove the ones that are an instance of $win$.\n\n$TTHTT$\n$THTTT$\n$HTHTT$\n$THHTT$\n\nSo Satan has 4 ways which don't lose automatically.\n\nNow in all these cases, it's our choice. We want to create or facilitate $win$. We leave the $T$ as they are. I will number them now and remove duplicates.\n\n$1.\\ TTTTH$\n$2.\\ TTHTH$\n$3.\\ TTTHH$\n\nNow it's Satan's turn. In case 1. Satan has lost. He cannot avoid establishing $win$ for us. Leaving us only with these options.\n\n$1.\\ TTHTH$\n$2.\\ TTTHH$\n\nLet's focus on case $2. TTTHH$. There are 4 ways he can play this one.\n\n$2.1\\ HHTTT$\n$2.2\\ HTTTT$\n$2.3\\ THTTT$\n$2.4\\ TTTTT$\n\nThese are all now lost for Satan. $2.1$,$2.2$ and $2.4$ are an instance of $win$ and $2.3$ after our moves gives $TTTTH$ where Satan cannot avoid establishing $win$ for us.\n\nSo, the only complicated case is $1. TTHTH$. This is only case of a broken string of $H$ or $T$ when considered East to West overflow.\n\nSatan has 2 ways to play the next one:\n\n$1.1\\ HTTHT$\n$1.2\\ TTTHT$\n\nIn case $1.2$ we win again, by choosing $T$. For case, $1.1\\ HTTHT$ which in general will be the from of the only tricky position. We have 2 ways to play. If we play $T$ nothing changes. This is the only case where we must play $H$. And the result is: $HHTTH$\n\nNow Satan has 2 ways to play.\n\n$1.\\ HHHTT$ - this is $win$.\n$2.\\ THHTT$ - we choose $T$ -> $TTTHH$ - now Satan is lost as before. What a loser :D\n\n• Best way to explain it, +1, easyer to follow than the others :) Apr 15, 2015 at 7:26\n\nThe base of this solution is the binary interpretation strategy as shown in Lopsy's answer, but with an improvement to $2^n$ moves instead of $n2^n$\n\nWith $T \\to 1$ and $H \\to 0$, interpret the sequence of coins as a binary number starting from a position called the mark (denoted by inserting an M) with the string left of the mark appended to the right end. The value of the mark at that position is the binary value of that sequence. So if $M_1$ is the mark $0M1100$ its value would be $V(M_1)=11000b=24$\n\n### Choosing the mark\n\nThe correct position of the mark is the one which yields the highest value for the given sequence. For example, $01101 \\to 0M1101$ and the value at that mark is binary value of $11010b=26$. There can be multiple positions which are candidates for the mark (e.g. $M101010$, $10M1010$, $1010M10$). In such cases the mark is chosen by:\n\n• If there are no 1s to the left of the leftmost candidate, then the leftmost candidate\n• otherwise the rightmost candidate.\n\nFor example $0M10010$ or $10110M1$ would be correct marks.\n\nIf after a turn the mark is between different coins than it was before the turn, we will say that the mark \"jumped\". This can happen either when a coin is flipped changing the set of candidate marks, or when the coins are rotated without flipping and a 1 appears on the left forcing a switch from the leftmost candidate to the rightmost candidate.\n\n### The Player's strategy:\n\n• If there are no 1s to the left of the mark, flip $1\\to 0$\n• otherwise do not flip $1\\to 1$\n\n### Observation\n\nThe position of the mark and P's strategy are both completely determined by the current sequence of coins. If the opponent (S) can cause the same sequence to appear twice, then S has a strategy that can repeat that loop forever because P will respond to it exactly the same the second time. By the contrapositive, if we can prove P's stategy must win, then no sequence repeats, so the total number of moves to win is $< 2^n$.\n\n### Lemma 1:\n\nThe mark can only jump when S flips. Consider the other cases\n\n1. No flip. When there is no flip, the set of mark candidates won't change, so the only way the mark can jump is if the mark started as the leftmost candidate and then a 1 was placed to it's left forcing a switch to the rightmost candidate.\n• 1a. P does not flip. If there are no 1's left of the mark, P's strategy says flip, so this case can't occur.\n• 1b. S does not flip. It's S's move and he doesn't flip, so the coin placed on the left is a 0 and there is no jump.\n2. P does flip. Let $M_1$ be the position of the mark prior to the move, which has value $V(M_1)$. Note that unless there are no 1s, meaning P just won, any valid mark candidate has to have a 1 on its left. Since P just flipped, all bits left of $M_1$ are 0, so no position left of $M_1$ is a mark candidate. Let $M_2$ be some position to the right of $M_1$, with value $V(M_2)$. Since $M_1$ was the mark prior to the move, $V(M_2) <= V(M_1)$. The move reduces the value of each mark based on the significance of the bit that was flipped $1\\to 0$. Denote the values of $M_1$ and $M_2$ after the move as $V'(M_1)$ and $V'(M_2)$ respectively. Since $M_2$ is right of $M_1$, the flipped bit is more significant to $M_2$ than it is to $M_1$, so $V(M_2)-V'(M_2) > V(M_1)-V'(M_1)$ and thus $V'(M_2) < V'(M_1)'$. So, no position right of $M_1$ can have as high a value as a mark. Therefore $M_1$ is still the correct mark.\n\n### Lemma 2: If S doesn't flip during any full cycle of the mark (from left back to left again), then P wins.\n\nBy P's strategy, if S does not flip for a full cycle, P will have flipped all coins to 0, winning.\n\n### Lemma 3: When S flips a coin, the marked value after the flip will be higher than the marked value after the last time S flipped\n\n1. S's first flip Trivially satisfied\n2. The last flipper was S. S can only flip $0\\to 1$ so S's move increase the value, which was unchanged since his last flip since P has not flipped since then.\n3. The last flipper was P. By P's strategy and Lemma 1, P has not flipped any coins to the right of the mark since S's last flip. Therefore the total decrease in value from P's flips (since S's last flip) is less than the increase in value from S's flip (since it is a more significant bit).\n\n### Conclusion\n\nCombining the above, S must keep making flips every cycle to avoid losing, causing the marked value to increase with each flip. Eventually the value reaches $2^n-1$ and the sequence is all 1s, allowing P to win by flipping all coins. Using our original observation, this means P wins in $<2^n$ steps.\n\nWe could improve the upper bound slightly given this starting point by eliminating sequences which can be shown to be not encountered. In particular we can see we will never encounter any sequence without one of (first and last position are 1) or (at least 1 run of 1s of length 2) until we get $0...01$ right before P wins. This only eliminates around $F(n-1) + F(n-3) - 2$, which is very small compared to $2^n$ as n grows.\n\n### Speculation on lower bound\n\nThe following strategy for S might be generally effective\n\n• if there are no 1s left of the mark and flipping will not jump the mark then flip,\n• otherwise do not flip\n\nIf this strategy can force us to count through all values which wouldn't cause a jump in the mark, then we could use n-nacci numbers to place a lower bound on the number of non-mark jumping values by playing it safe and skipping all values which have a run of 1s as long as the existing longest run of 1s. The would prove that for any $k < 2$, winning takes $\\omega(k^n)$ (i.e. that for some constant c > 0 and and sufficiently large n, the number of moves exceeds $ck^n$).\n\nOriginally, this question was asked and solved by Gera Weiss in a paper called \"A Combinatorial Game Approach to State Nullification by Hybrid Feedback\". Here's a link:\n\nhttp://www.cs.bgu.ac.il/~geraw/Publications_files/CombinatorialGameApproachToStateNullification.pdf\n\n# OPTIMAL STRATEGY to escape in less than 2^n days\n\nYou can escape in LESS than 2n days\n\n1. It was already proven that you can ALWAYS escape in 2n \"cycles\" (each cycle has n days) which means escape is always possible. Proof for that is in \"The Puzzle Toad\" link from original problem, even if that is not optimal play (uses more days than needed, ~n*2n days ).\n2. Optimal play require that you never repeat same position (arrangement of the coins) - because that would mean Devil has won since it will just repeat. But in #1 we proved that you can always escape, therefore an optimal win strategy must exist.\n3. Since there are at max 2n different positions, it means optimal play will free the prisoner in max 2n days from any starting position.\n4. But since we start with THH..HHT position, it will require less than 2n days.\n\n• define how to generate optimal play strategy for n coins\n• find an exact formula for max needed days with n coins starting in THH..HHT position, as days= f(n), where f(n) < 2n\n\nI proved above that it is always possible to escape in less than 2^n days with optimal play. Here I will also show 'how' to do it (strategy), and prove that it is optimal and always possible.\n\n## Strategy\n\n1. Generate ordered list of 'win/optimal positions' ( arrangements of n H/T coins ):\n• starts with HH..HH at position #1, which is target win/optimal position (oPos)\n• assuming previous oPos is in form xCCC (where CCC=CC..CC= all coins except leftmost 'x'), then next oPos is either CCCT ( remove 'x' and set 'T' Tail as rightmost ) or CCCH ( if CCCT already exists as previous oPos )\n• continue until you reach 'THH..HHT' position\n2. Start playing:\n• find current position in list, and play so that it turns into next (lower numbered) position\n• that is optimal move for both players - for devil, it guarantee maximal days given that you play optimally, and for you it guarantee minimal number of days given that devil plays optimally\n• therefore, after optimal list is generated in step '1', position of 'THH..HHT' is also optimal number of days needed\n\nExample list for 4 coins:\n\n 1: HHHH\n2: HHHT\n3: HHTT\n4: HTTT\n5: TTTT\n6: TTTH\n7: TTHT\n8: THTT\n9: HTTH\n10: TTHH\n11: THHT ***\n12: HHTH\n13: HTHT\n14: THTH\n15: HTHH\n16: THHH\n\n\n## Proof\n\nIf we have list of previous 'win positions' ( those that guarantee win for devil's opponent ), next win position is one that will necessarily result in one of those existing (previous) optimal positions after single move.\n\nIf we mark previous win position as xCCC, it is obvious that new position CCCT will result in that previous xCCC in one move ( it is your move since CCCT ends up in 'T', and you would turn it into xCCC, whatever 'x' was ).\n\nWhen you can not add CCCT (because it already exists in list), you can always add CCCH. Reason for that is that existence of CCCT in list imply existence of both TCCC and HCCC in the list - we already have xCCC as last position, but we also must have yCCC ( where y is complement of x, so if x=heads, y=tails and reverse). That is due to fact that already having CCCT somewhere in the list means that position immediately before it was also in the form of yCCC , and since both that yCCC and last position xCCC are on the list (and we can not have two same positions on the list), it means that x<>y ... so one of them is H and other is T , and we are certain we have both TCCC and HCCC as previous win positions.\n\nWhen we know that both TCCC and HCCC are previous win position, we can add CCCH as new win position, even if that is position where devil will play, because whatever devil play it will end at other win position. It is in devils interest to play xCCC ( last/previous position) instead of yCCC (which would be earlier win position, and would allow you to escape earlier), but regardless of what devil play it ends up in previous win position, so this CCCH can be added as new latest win position.\n\nThis proves that , for every last win position, it is always possible to generate single new win position (that new win position would need one more day for win). Since there are 2^n possible positions ( arrangements of Heads/Tails on n coins ), ordered list of win positions will have exactly 2^n positions.\n\nOptimal days ( needed for you to escape if both you and devil play optimally) for any starting coin arrangement is equal to position of that starting coin arrangement in optimal list. For 4 coins example above, \"THHT\" is at #11 position , which means you will go free \"on 11th day\" (or reduce one if you count how many days passed instead, so \"in 10 days\" )\n\nIt also means that, regardless of what is initial coin arrangement, it is possible to escape in 2^n days as worst case. And since \"THH..HHT\" will not be last position in list, it is possible to escape in less than 2^n days if optimally played.\n\n## Formula for days=f(n)\n\nAs I mentioned before, it is possible to express optimal number of days ( days needed for you to escape if both you and devil play optimally ) as a function of number of coins, when your starting position is THH..HHT.\n\nObviously, formula for different starting positions would be different. Trivial example is for HH..HT starting position - it will get you free on 2nd day ( in 1 day, after your first move), so f'(n)=2 , regardless of n. It is not guaranteed that all starting position have formula that can return number of days, but THH..HHT has such formula. That formula is without proof here, but it was not part of the question anyway:\n\ndays(n)= 2^n - A014739[n-2]\n\n$$days(n)= 2^n - \\frac{1}{2}\\left[(3+\\sqrt5)*(\\frac{1+\\sqrt5}{2})^{n-2}+(3-\\sqrt5)*(\\frac{1-\\sqrt5}{2})^{n-2}\\right]+2$$\n\nWhere A014739 is integer sequence #A014739 in Sloane index for expansion of (1+x^2)/(1-2*x+x^3). This formula returns \"on which day\" player will go free, assuming first day is #1. If you need \"in how many days\" or \"moves\", then subtract 1.\n\nUpper bound with optimal approach for any starting position is 2^n, and above days(n) formula is obviously days(n)<=2^n. As a comparison, formula for non-optimal \"cycle\" solution would have upper bound as upCycle(n)= n * 2^n for any starting position, or cycleDays(n)~[2^(n-1)-2]*(n-1) for starting \"THHT\".\n\nExample for different number of coins:\n\n• n= 4 coins : upCycle= 64 ; 2^n= 16 ; optimal days(n)= 11\n• n=20 coins : upCycle= 20,971,520; 2^n= 1,048,576 ; optimal days(n)= 1,033,451\n• Nice strategy! The explanation is a bit hard to follow, but I guess it's still correct. Anyway, as bobble mentioned, this needs to be edited into your previous answer (or edit the previous answer into this one, doesn't really matter here). Jul 22, 2021 at 10:28\n\nFor n=5 I see no way out.\n\nI'm solving from the end state, not from the beginning.\n\nSatan could always make a T at west out of the H at east. That means, what you need as last state before winning is one of these states:\n\nHHHT\n\nHHHTT\n\nHHTTT\n\nHTTTT\n\nFor all these, it's impossible that the H at west was an H at east, because satan would not let it unflipped and make you win.\n\nSo, the H at west must have been a T at east. That means HHHHT must have been HHHTT, HHHTT must have been HHTTT and so on.\n\nThis only leaves TTTTT as entry state to this set of states. But satan can prevent TTTTT by not flipping TTTTH to TTTTT. But if he does so, he has to give you HTTTT and you would win. So, TTTTH (which forces satan to one of two winning nodes) must be on your route to win.\n\nSo, how to get to TTTTH?\n\nGoing reverse here again:\n\nTTTTH => TTTHH Impossible, satan doesn't make you win.\n\nTTTTH => TTTHT Yes, it was yours.\n\nThis goes for every of the T at west. It was never satans H.\n\nSo here's the rest.\n\nTTTHT => TTHTT Yes, it was yours.\n\nTTHTT => THTTT Yes, it was yours.\n\nTHTTT => HTTTT Yes, it was yours."
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https://www.sharecodepoint.com/2017/10/cse205-data-structure-and-algorithm_3.html | [
"# CSE205 : Data Structure And Algorithm important MCQ Questions for Exam\n\n1. Which of the following is not the required condition for binary search\n(A) There must be mechanism to delete and/or insert elements in list\n(B) There should be the direct access to the middle element in any sublist.\n(C) The list must be sorted\n(D) none of above\n\n2 Big O notation is used when function g(n) defines ________ for function f(n)\n(A) Upper bound (B) Lower bound\n(C) Both Upper and Lower (D) None\n\n3. The worst case occur in linear search algorithm\n(A) When Item is not in the array at all\n(B) When Item is the last element in the array\n(C) When Item is somewhere in the middle of the array\n(D) When Item is the last element in the array or is not there at all\n\n4. Consider the following three claims\nI (n + r)m= O(nm), where k and m are constants\nII 2(n + 1000)= O(2n)\nIII n+logn= O(n)\nWhich of these claims are correct?\n(A) I and III (B) I and II\n(C) II and III (D) I, II and III\n\n5. The complexity of searching an element using Binary search Algorithm is\n(A) O(n) (B) O(log n)\n(C) O(n2) (D) O(n log n)\n\n6. Time complexities of three algorithms are given. Which should execute the\nfastest for large values of N?\n(A) O(N2) (B) O(NN)\n(C) O(Nlog N) (D) O(2N)\n\n7. The complexity of Bubble sort algorithm is\n(A) O(n) (B) O(log n)\n(C) O(n2) (D) O(n log n)\n\n8. The time required to insert a node x at last position in aunsorted linked list having n nodes\n(A) O (n) (B) O (log n)\n(C) O (1) (D) O (n log n)\n\n9. Two main measures for the efficiency of an algorithm are\n(A) Processor and memory (B) Complexity and capacity\n(C) Time and space (D) Data and space\n\n10. Number of comparisons required to sort letters in ALGORITHM\n(A) 36 (B) 8\n(C) 45 (D) 9\n\n11. The space factor when determining the efficiency of algorithm is measured by\n(A) Counting the maximum memory needed by the algorithm\n(B) Counting the minimum memory needed by the algorithm\n(C) Counting the average memory needed by the algorithm\n(D) Counting the maximum disk space needed by the algorithm\n\n12. Linked List are known as\n(A). Self Calling Function (B). Self Referential Function\n(C). Self Calling Structures (D). Self Referential Structures\n\n13. The operation of processing each element in the list is known as\n(A) Sorting (B) Merging\n(C) Inserting (D) Traversal\n\n14. Linked lists are best suited\n(A) for relatively permanent collections of data\n(B) for the size of the structure and the data in the structure are constantly changing\n(C) for both of above situation\n(D) for none of above situation\n\n15. Which of the following operations is not O(1) for an array of sorted data. You may assume that array elements are distinct.\n(A) Find the ith largest element\n(B) Find the ith smallest element\n(C) Delete an element\n(D) All of the above\n\n16. Which of the given options provides the increasing order of asymptotic\ncomplexity of functions f1, f2, f3 and f4?\nf1(n) = 2n\nf2(n) = n(3/2)\nf3(n) = n log n\nf4(n) = n (log n)\nA) f3, f2, f1, f4 B) f3, f2, f4, f1\nC) f2, f3, f1, f4 D) f2, f3, f4, f1\n\n17. Which of the following data structures are not indexed structures?\n(A) linear arrays (B) linked lists\n(C) both of above (D) none of above\n\n18. Consider the following for loops are part of a program:\nfor(i = 0; i< n; i++){}\nfor(i = 0; i< n; i += 2){}\nfor(i = 1; i< n; i *= 2){}\nIf n is the size of input(positive), what will be time complexity?\n(A) O(n) (B) O(n2) (C) O(n3) (D) O(logn)\n\n19. Sparse matrix has\n(A) Many zero entries (B) Many non zero entries\n(C) Many dimensions (D) All diagonal elements zero\n\n20. Which of the following is not a limitation of binary search algorithm?\n(A) must use a sorted array\n(B) there must be a mechanism to access middle element directly\n(C) requirement of sorted array is expensive when a lot of insertion and deletions\nare needed\n(D) binary search algorithm is not efficient when the data elements are more than\n1000.\n\n21. The time factor when determining the efficiency of algorithm is measured by\n(A) Counting microseconds\n(B) Counting the number of key operations\n(C) Counting the number of statements\n(D) Counting the kilobytes of algorithm\n\n22. The situation when in a linked list START=NULL is\n(A) underflow (B) overflow\n(C) housefull (D) saturated\n\n23 The OS of a computer may periodically collect all the free memory space to\nform contiguous block of free space. This is called\n(A) Garbage Management (B) Garbage collection\n(C) Waste Collection (D) Waste Management\n\n24 If LOCP is NULL for a given LOC and to delete item at LOC following is used\n(C) LINK [LOC] = START (D) START = NULL\n\n25 Which instructions are used to get a free node from AVAIL list\n(A) NEW = AVAIL, AVAIL = LINK [AVAIL]\n(B) AVAIL = NEW, LINK [AVAIL] = AVAIL\n(C) AVAIL = NEW, AVAIL = LINK [AVAIL]\n(D) NEW = AVAIL, LINK [AVAIL] = AVAIL\n\n26. Calculate number of elements in A( -10:5)\n\n27 Consider 25 X 4 matrix array SCORE. LetBase(SCORE) = 325 and w = 4. Tell address of SCORE[12,3] if arrays are stored in column major order? [2 marks]\n\n28. Consider 8 X 4 matrix array SCORE. Suppose Base(SCORE) = 18 and w = 4.\nWhat will be the address of SCORE[12,3] if arrays are stored in row major order?\n[2 marks]\n\n### Sharecodepoint\n\nSharecodepoint is the junction where every essential thing is shared for college students in the well-defined packets of codes. We are focused on providing you the best material package like Question papers, MCQ'S, and one NIGHT STUDY MATERIAL. facebook twitter youtube instagram\n\n1.",
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https://foreach.id/EN/fluids/flowratevol/milliliter%7Cday-to-cubic_inch%7Chour.html | [
"# Convert milliliter/day to inch³/hour (mL/d to in³/h)\n\nBatch Convert\n• milliliter/day [mL/d]\n• inch³/hour [in³/h]\nCopy\n_\nCopy\n• milliliter/day [mL/d]\n• inch³/hour [in³/h]\n\n## Milliliter/day to Inch³/hour (mL/d to in³/h)\n\n### Milliliter/day (Symbol or Abbreviation: mL/d)\n\nMilliliter/day is one of volumetric flow rate units. Milliliter/day abbreviated or symbolized by mL/d. The value of 1 milliliter/day is equal to 1.1574e-11 meter³/second. In its relation with inch³/hour, 1 milliliter/day is equal to 0.0025427 inch³/hour.\n\n#### Relation with other units\n\n1 milliliter/day equals to 1.1574e-11 meter³/second\n\n1 milliliter/day equals to 0.000001 meter³/day\n\n1 milliliter/day equals to 4.1667e-8 meter³/hour\n\n1 milliliter/day equals to 6.9444e-10 meter³/minute\n\n1 milliliter/day equals to 1 centimeter³/day\n\n1 milliliter/day equals to 0.041667 centimeter³/hour\n\n1 milliliter/day equals to 0.00069444 centimeter³/minute\n\n1 milliliter/day equals to 0.000011574 centimeter³/second\n\n1 milliliter/day equals to 0.001 liter/day\n\n1 milliliter/day equals to 0.000041667 liter/hour\n\n1 milliliter/day equals to 6.9444e-7 liter/minute\n\n1 milliliter/day equals to 1.1574e-8 liter/second\n\n1 milliliter/day equals to 0.041667 milliliter/hour\n\n1 milliliter/day equals to 0.00069444 milliliter/minute\n\n1 milliliter/day equals to 0.000011574 milliliter/second\n\n1 milliliter/day equals to 0.00026417 gallon (US)/day\n\n1 milliliter/day equals to 0.000011007 gallon (US)/hour\n\n1 milliliter/day equals to 1.8345e-7 gallon (US)/minute\n\n1 milliliter/day equals to 3.0575e-9 gallon (US)/second\n\n1 milliliter/day equals to 0.00021997 gallon (UK)/day\n\n1 milliliter/day equals to 0.0000091654 gallon (UK)/hour\n\n1 milliliter/day equals to 1.5276e-7 gallon (UK)/minute\n\n1 milliliter/day equals to 2.5459e-9 gallon (UK)/second\n\n1 milliliter/day equals to 6.2898e-9 kilobarrel (US)/day\n\n1 milliliter/day equals to 0.0000062898 barrel (US)/day\n\n1 milliliter/day equals to 2.6208e-7 barrel (US)/hour\n\n1 milliliter/day equals to 4.3679e-9 barrel (US)/minute\n\n1 milliliter/day equals to 7.2799e-11 barrel (US)/second\n\n1 milliliter/day equals to 2.9591e-7 acre foot/year\n\n1 milliliter/day equals to 8.1071e-10 acre foot/day\n\n1 milliliter/day equals to 3.378e-11 acre foot/hour\n\n1 milliliter/day equals to 3.5315e-7 100 foot³/day\n\n1 milliliter/day equals to 1.4714e-8 100 foot³/hour\n\n1 milliliter/day equals to 2.4524e-10 100 foot³/minute\n\n1 milliliter/day equals to 0.0014089 ounce/hour\n\n1 milliliter/day equals to 0.000023482 ounce/minute\n\n1 milliliter/day equals to 3.9137e-7 ounce/second\n\n1 milliliter/day equals to 0.0014665 ounce (UK)/hour\n\n1 milliliter/day equals to 0.000024441 ounce (UK)/minute\n\n1 milliliter/day equals to 4.0735e-7 ounce (UK)/second\n\n1 milliliter/day equals to 5.4498e-8 yard³/hour\n\n1 milliliter/day equals to 9.083e-10 yard³/minute\n\n1 milliliter/day equals to 1.5138e-11 yard³/second\n\n1 milliliter/day equals to 0.0000014714 foot³/hour\n\n1 milliliter/day equals to 2.4524e-8 foot³/minute\n\n1 milliliter/day equals to 4.0873e-10 foot³/second\n\n1 milliliter/day equals to 0.0025427 inch³/hour\n\n1 milliliter/day equals to 0.000042378 inch³/minute\n\n1 milliliter/day equals to 7.0629e-7 inch³/second\n\n1 milliliter/day equals to 1.8865e-8 pound/second (Gasoline at 15.5%b0C)\n\n1 milliliter/day equals to 0.0000011319 pound/minute (Gasoline at 15.5%b0C)\n\n1 milliliter/day equals to 0.000067914 pound/hour (Gasoline at 15.5%b0C)\n\n1 milliliter/day equals to 0.0016299 pound/day (Gasoline at 15.5%b0C)\n\n1 milliliter/day equals to 8.557e-9 kilogram/second (Gasoline at 15.5%b0C)\n\n1 milliliter/day equals to 5.1342e-7 kilogram/minute (Gasoline at 15.5%b0C)\n\n1 milliliter/day equals to 0.000030805 kilogram/hour (Gasoline at 15.5%b0C)\n\n1 milliliter/day equals to 0.00073933 kilogram/day (Gasoline at 15.5%b0C)\n\n### Inch³/hour (Symbol or Abbreviation: in³/h)\n\nInch³/hour is one of volumetric flow rate units. Inch³/hour abbreviated or symbolized by in³/h. The value of 1 inch³/hour is equal to 4.552e-9 meter³/second. In its relation with milliliter/day, 1 inch³/hour is equal to 393.29 milliliter/day.\n\n#### Relation with other units\n\n1 inch³/hour equals to 4.552e-9 meter³/second\n\n1 inch³/hour equals to 0.00039329 meter³/day\n\n1 inch³/hour equals to 0.000016387 meter³/hour\n\n1 inch³/hour equals to 2.7312e-7 meter³/minute\n\n1 inch³/hour equals to 393.29 centimeter³/day\n\n1 inch³/hour equals to 16.387 centimeter³/hour\n\n1 inch³/hour equals to 0.27312 centimeter³/minute\n\n1 inch³/hour equals to 0.004552 centimeter³/second\n\n1 inch³/hour equals to 0.39329 liter/day\n\n1 inch³/hour equals to 0.016387 liter/hour\n\n1 inch³/hour equals to 0.00027312 liter/minute\n\n1 inch³/hour equals to 0.000004552 liter/second\n\n1 inch³/hour equals to 393.29 milliliter/day\n\n1 inch³/hour equals to 16.387 milliliter/hour\n\n1 inch³/hour equals to 0.27312 milliliter/minute\n\n1 inch³/hour equals to 0.004552 milliliter/second\n\n1 inch³/hour equals to 0.1039 gallon (US)/day\n\n1 inch³/hour equals to 0.004329 gallon (US)/hour\n\n1 inch³/hour equals to 0.00007215 gallon (US)/minute\n\n1 inch³/hour equals to 0.0000012025 gallon (US)/second\n\n1 inch³/hour equals to 0.086512 gallon (UK)/day\n\n1 inch³/hour equals to 0.0036047 gallon (UK)/hour\n\n1 inch³/hour equals to 0.000060078 gallon (UK)/minute\n\n1 inch³/hour equals to 0.0000010013 gallon (UK)/second\n\n1 inch³/hour equals to 0.0000024737 kilobarrel (US)/day\n\n1 inch³/hour equals to 0.0024737 barrel (US)/day\n\n1 inch³/hour equals to 0.00010307 barrel (US)/hour\n\n1 inch³/hour equals to 0.0000017179 barrel (US)/minute\n\n1 inch³/hour equals to 2.8631e-8 barrel (US)/second\n\n1 inch³/hour equals to 0.00011638 acre foot/year\n\n1 inch³/hour equals to 3.1884e-7 acre foot/day\n\n1 inch³/hour equals to 1.3285e-8 acre foot/hour\n\n1 inch³/hour equals to 0.00013889 100 foot³/day\n\n1 inch³/hour equals to 0.000005787 100 foot³/hour\n\n1 inch³/hour equals to 9.6451e-8 100 foot³/minute\n\n1 inch³/hour equals to 0.55411 ounce/hour\n\n1 inch³/hour equals to 0.0092352 ounce/minute\n\n1 inch³/hour equals to 0.00015392 ounce/second\n\n1 inch³/hour equals to 0.57674 ounce (UK)/hour\n\n1 inch³/hour equals to 0.0096124 ounce (UK)/minute\n\n1 inch³/hour equals to 0.00016021 ounce (UK)/second\n\n1 inch³/hour equals to 0.000021433 yard³/hour\n\n1 inch³/hour equals to 3.5722e-7 yard³/minute\n\n1 inch³/hour equals to 5.9537e-9 yard³/second\n\n1 inch³/hour equals to 0.0005787 foot³/hour\n\n1 inch³/hour equals to 0.0000096451 foot³/minute\n\n1 inch³/hour equals to 1.6075e-7 foot³/second\n\n1 inch³/hour equals to 0.016667 inch³/minute\n\n1 inch³/hour equals to 0.00027778 inch³/second\n\n1 inch³/hour equals to 0.0000074194 pound/second (Gasoline at 15.5%b0C)\n\n1 inch³/hour equals to 0.00044517 pound/minute (Gasoline at 15.5%b0C)\n\n1 inch³/hour equals to 0.02671 pound/hour (Gasoline at 15.5%b0C)\n\n1 inch³/hour equals to 0.64104 pound/day (Gasoline at 15.5%b0C)\n\n1 inch³/hour equals to 0.0000033654 kilogram/second (Gasoline at 15.5%b0C)\n\n1 inch³/hour equals to 0.00020192 kilogram/minute (Gasoline at 15.5%b0C)\n\n1 inch³/hour equals to 0.012115 kilogram/hour (Gasoline at 15.5%b0C)\n\n1 inch³/hour equals to 0.29077 kilogram/day (Gasoline at 15.5%b0C)\n\n### How to convert Milliliter/day to Inch³/hour (mL/d to in³/h):\n\n#### Conversion Table for Milliliter/day to Inch³/hour (mL/d to in³/h)\n\nmilliliter/day (mL/d) inch³/hour (in³/h)\n0.01 mL/d 0.000025427 in³/h\n0.1 mL/d 0.00025427 in³/h\n1 mL/d 0.0025427 in³/h\n2 mL/d 0.0050853 in³/h\n3 mL/d 0.007628 in³/h\n4 mL/d 0.010171 in³/h\n5 mL/d 0.012713 in³/h\n6 mL/d 0.015256 in³/h\n7 mL/d 0.017799 in³/h\n8 mL/d 0.020341 in³/h\n9 mL/d 0.022884 in³/h\n10 mL/d 0.025427 in³/h\n20 mL/d 0.050853 in³/h\n25 mL/d 0.063566 in³/h\n50 mL/d 0.12713 in³/h\n75 mL/d 0.1907 in³/h\n100 mL/d 0.25427 in³/h\n250 mL/d 0.63566 in³/h\n500 mL/d 1.2713 in³/h\n750 mL/d 1.907 in³/h\n1,000 mL/d 2.5427 in³/h\n100,000 mL/d 254.27 in³/h\n1,000,000,000 mL/d 2,542,700 in³/h\n1,000,000,000,000 mL/d 2,542,700,000 in³/h\n\n#### Conversion Table for Inch³/hour to Milliliter/day (in³/h to mL/d)\n\ninch³/hour (in³/h) milliliter/day (mL/d)\n0.01 in³/h 3.9329 mL/d\n0.1 in³/h 39.329 mL/d\n1 in³/h 393.29 mL/d\n2 in³/h 786.58 mL/d\n3 in³/h 1,179.9 mL/d\n4 in³/h 1,573.2 mL/d\n5 in³/h 1,966.4 mL/d\n6 in³/h 2,359.7 mL/d\n7 in³/h 2,753 mL/d\n8 in³/h 3,146.3 mL/d\n9 in³/h 3,539.6 mL/d\n10 in³/h 3,932.9 mL/d\n20 in³/h 7,865.8 mL/d\n25 in³/h 9,832.2 mL/d\n50 in³/h 19,664 mL/d\n75 in³/h 29,497 mL/d\n100 in³/h 39,329 mL/d\n250 in³/h 98,322 mL/d\n500 in³/h 196,640 mL/d\n750 in³/h 294,970 mL/d\n1,000 in³/h 393,290 mL/d\n100,000 in³/h 39,329,000 mL/d\n1,000,000,000 in³/h 393,290,000,000 mL/d\n1,000,000,000,000 in³/h 393,290,000,000,000 mL/d\n\n#### Steps to Convert Milliliter/day to Inch³/hour (mL/d to in³/h)\n\n1. Example: Convert 600 milliliter/day to inch³/hour (600 mL/d to in³/h).\n2. 1 milliliter/day is equivalent to 0.0025427 inch³/hour (1 mL/d is equivalent to 0.0025427 in³/h).\n3. 600 milliliter/day (mL/d) is equivalent to 600 times 0.0025427 inch³/hour (in³/h).\n4. Retrieved 600 milliliter/day is equivalent to 1.5256 inch³/hour (600 mL/d is equivalent to 1.5256 in³/h).\n\n▸▸"
] | [
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http://dan.saigon.ro/gui-python-with-tkinter-part-4-the-grid-geometry-manager/ | [
"# GUI Python with Tkinter – Part 4 – The grid() geometry manager\n\nThe grid geometry manager organizes the container frame into a table with rows and columns. Each cell in the table can hold one widget. Widgets can span multiple cells.\n\nExample:\n\nLet’s create a login window using the `grid()` geometry manager. I placed all the code in a file named `test04.py`.\n\n``````from tkinter import *\nroot = Tk()\n\nlabel1 = Label(root, text = \"Username\")\nlabel1.grid(row = 0, sticky = W)\nlabel2 = Label(root, text = \"Password\")\nlabel2.grid(row = 1, sticky = W)\nentry1 = Entry(root)\nentry1.grid(row = 0, column = 1, sticky = E)\nentry2 = Entry(root)\nentry2.grid(row = 1, column = 1, sticky = E)\nbutton1 = Button(root, text = \"Login\")\nbutton1.grid(row = 2, column = 1, sticky = E)\n\nroot.mainloop()``````\n\nWe aligned the position of each widget using the `sticky` option. The `sticky` option decides how the widget is expanded. The sticky option can be specified using one of the cardinal points.\n\nNote:\nNot specifying any value, defaults to stickiness to the center of the widget in the cell.\n\nIf you run the file in the terminal you will get something like this:",
null,
"The `sticky = W` makes the labels stick on the left side.\nThe width and the height of each column is decided automatically. You can specify the `width` for widgets if you want total control.\nThe `sticky = NSEW` argument makes the widget expandable and fills the entire cell of the grid.\n\nYour widgets can span across multiple cells in the grid. For that you can use the options `rowspan` and `columnspan`.\n\nThe `padx` and `pady` options allow you to set a padding around a widget. The `ipadx` and `ipady` options are used for internal padding. The default value of an external and internal padding is 0.\n\nLet’s use these options in an example. I created another file named `test05.py` and I placed all the code for this example in it.\n\n``````from tkinter import *\nroot = Tk()\n\nroot.title('Motorcycles')\nLabel(root, text = \"Brand:\").grid(row = 0, column = 0, sticky = E)\nEntry(root, width = 35).grid(row = 0, column = 1, padx = 2, pady = 2, sticky = 'we', columnspan = 4)\nLabel(root, text=\"Model:\").grid(row = 1, column = 0, sticky = E)\nEntry(root).grid(row = 1, column = 1, padx = 2, pady = 2, sticky = 'we', columnspan = 4)\nButton(root, text = \"Find\").grid(row = 0, column = 5, sticky = E + W, padx = 2, pady = 2)\nButton(root, text = \"Options\").grid(row = 1, column = 5, sticky = E + W, padx = 2)\nButton(root, text = \"Custom\").grid(row = 2, column = 5, sticky = E + W, padx = 2)\nButton(root, text = \"Reset\").grid(row = 3, column = 5, sticky = E + W, padx = 2)\nCheckbutton(root, text = \"Retro\").grid(row = 2, column = 1, columnspan = 2, sticky = W)\nCheckbutton(root, text = \"Accesories\").grid(row = 3, column = 1, columnspan = 2, sticky = W)\nCheckbutton(root, text = \"Delivered\").grid(row = 4, column = 1, columnspan = 2, sticky = W)\nLabel(root, text = \"Color:\").grid(row = 2, column = 3, sticky = W, padx = 8)\nRadiobutton(root, text = \"Red\", value = 1).grid(row = 3, column = 3, sticky = W)\nRadiobutton(root, text = \"Blue\", value = 2).grid(row = 3, column = 3, sticky = E)\n\nroot.mainloop()``````\n\nIf you run the file in the terminal you should get something like shown below:",
null,
"Other `grid()` options commonly used are `widget.grid_forget()` and `widget.grid_remove()` methods. There are many more! For a complete `grid` reference, type the following command in the Python shell:\n\n``````>>> import tkinter\n>>> help(tkinter.Grid)``````\n\nYou can configure a grid’s column and row sizes to override the default customization given automatically. The height of all the grid rows (and columns) is automatically adjusted to be the height of its tallest cell.\n\nTo override this automatic sizing of columns and rows you can use something like shown below:\n\n``````widget_name.columnconfigure(n, option = value, ...)\nwidgen_name.rowconfigure(N, option = value, ...)``````\n\nUse these to configure the options for a given widget, `widget_name` , in either the nth column or the nth row. You can specify `minsize`, `pad`, and `weight`.\n\nExample:\n\n``````w.columnconfigure(0, weight = 2)\nw.columnconfigure(1, weight = 4)``````\n\nTwo-sixths of the extra space is allocated to the first column and four-sixths to the second column.\n\nNote:\nThe `columnconfigure()` and rowconfigure() methods are often used for dynamic resizing of widgets."
] | [
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https://www.chegg.com/homework-help/introductory-chemistry-4th-edition-chapter-13-problem-53p-solution-9780321687937 | [
"",
null,
"Introductory Chemistry (4th Edition) Edit edition Problem 53P from Chapter 13\n\nWe have solutions for your book!\nChapter: Problem:\nStep-by-step solution:\n100%(3 ratings)\nfor this solution\nChapter: Problem:\n• Step 1 of 4\n\nSince 1 L",
null,
"contains = 1000 mL",
null,
"So 4.8 L",
null,
"will have",
null,
"4800 mL of",
null,
"solution contains 4800 mL of silver. Because,",
null,
"Given that the density of solution 1.01g/mL.\n\nTherefore, mass of 4800 mL of silver solution",
null,
"",
null,
"silver\n\n• Chapter , Problem is solved.\nCorresponding Textbook",
null,
"Introductory Chemistry | 4th Edition\n9780321687937ISBN-13: 0321687930ISBN: Nivaldo J. TroAuthors:\nAlternate ISBN: 9780321730107, 9780321730183, 9780321747808, 9780321778529, 9780321785978, 9780321811325, 9780321830500"
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https://www.hindawi.com/journals/ahep/2014/794626/ | [
"/ / Article\nSpecial Issue\n\n## Black Holes Physics\n\nView this Special Issue\n\nResearch Article | Open Access\n\nVolume 2014 |Article ID 794626 | 8 pages | https://doi.org/10.1155/2014/794626\n\n# Researching on Hawking Effect in a Kerr Space Time via Open Quantum System Approach\n\nAccepted30 Jan 2014\nPublished09 Mar 2014\n\n#### Abstract\n\nIt has been proposed that Hawking radiation from a Schwarzschild or a de Sitter spacetime can be understood as the manifestation of thermalization phenomena in the framework of an open quantum system. Through examining the time evolution of a detector interacting with vacuum massless scalar fields, it is found that the detector would spontaneously excite with a probability the same as the thermal radiation at Hawking temperature. Following the proposals, the Hawking effect in a Kerr space time is investigated in the framework of an open quantum systems. It is shown that Hawking effect of the Kerr space time can also be understood as the the manifestation of thermalization phenomena via open quantum system approach. Furthermore, it is found that near horizon local conformal symmetry plays the key role in the quantum effect of the Kerr space time.\n\n#### 1. Introduction\n\nHawking radiation arising from the quantization of matter field in a curved background space-time with the event horizon is a prominent quantum effect. The research to understand Hawking radiation, which is related to general relativity, quantum theory, and thermodynamics, has attracted widespread interest in the physics community. Since Hawking’s original derivation of black hole thermal radiation , several alternative methods have been proposed, such as Damour-Ruffini method [4, 5], the tunneling method , and gravitational anomaly method [12, 13].\n\nHowever, from a physical viewpoint, the black hole thermodynamics system should be more like a nonequilibrium system rather than an equilibrium system. Hawking effect should be investigated in the frame of nonequilibrium statistics physics. In quantum mechanics and nonequilibrium statistics physics, the open quantum theory system has gotten a lot of successful development . The quantum dynamics of an open quantum system characterized by the effects of decoherence and dissipation cannot be represented in terms of a unitary time evolution. It has been applied to quantum information science, modern quantum optics, atomic and many-body systems, soft condensed matter physics, and biophysics. Recently, in the paradigm of open quantum system, based on , Yu and Zhang proposed a new insight to understand Hawking radiation in a Schwarzschild space time . Through examining the time evolution of a detector interacting with vacuum massless scalar field, they got a conclusion that the detector in both Unruh and Hartle-Hawking vacua would spontaneously excite with a nonvanishing probability the same as Hawking thermal radiation from the black hole. This new approach has been extended to understand the Gibbons-Hawking effect of de Sitter space-time . However, there remain some challenges to study Hawking radiation from a generic Kerr space time under the manifestation of thermalization phenomena in an open quantum system.\n\nOur motivation comes from the fact that the near-horizon geometry plays the key role to the character of a black hole space time . In 1998, Strominger discussed the near-horizon asymptotic symmetry in a Kerr black hole and found that there was a holographic duality between extremal and near-extremal Kerr black hole and a 2-dimensional conformal field theory. In [23, 24], it was shown that around the horizon of a Kerr space time, the scalar field theory can be reduced to a 2-dimensional effective field theory. Now, the thermal radiation of scalar particles from a Kerr black hole can be derived on the basis of a conformal symmetry arising from the near-horizon geometry .\n\nUsing the near-horizon geometry and open quantum system approach, Hawking effect in a Kerr space time will be investigated. We will examine the time evolution of a static detector (modeled by a two-level atom) outside a Kerr space time immersed in a vacuum massless scalar field. The dynamics of the detector can be obtained from the complete time evolution describing the total system (detector plus external field) by integrating over the field degrees of freedom. Our results show that the detector would spontaneously excite in the Unruh vacuum state with a probability the same as the thermal radiation at Hawking temperature, indicating that Hawking radiation from a Kerr space time can be understood as the manifestation of thermalization phenomena in the framework of open quantum systems. The conformal invariance of the wave equation near the horizon is at the key point of Hawking quantum effect in the Kerr space time.\n\nThe organization of our paper is as follows. In Section 2, we will review the basic formulae, including the master equation describing the system of the detector plus external vacuum scalar field in the weak-coupling limit and the reduced dynamical equation for the finite time evolution of the detector. In Section 3, the dimensional reduction technique is used to investigate the massless scalar field in a Kerr space time, and the Wightman function is obtained. In Section 4, applying the method and results of the preceding sections to calculate the probability of a spontaneous transition of the detector from the ground state to the excited state outside a Kerr space time. Finally, some discussions and conclusions will be given in Section 5.\n\n#### 2. Review of the Open Quantum System Approach\n\nIn this section, we will review the open quantum system approach to get the master equation describing the combined system , where a static detector (two-level atom) as an open system which is coupled to another quantum system of a vacuum massless scalar field in a Kerr space time. Our derivation mostly follows the works in . Here, we will consider the evolution of the static detector in the proper time and assume the combined system () to be initially prepared in a factorized state, with the detector keeping static in the exterior region of the Kerr black hole and the field keeping in vacuum state. The static detector is a two-level simplest quantum system whose Hilbert space is spanned over just two states, an excited state , and a ground state . The Hilbert space of such a system is equivalent to that of a spin- system. So the states of the detector can be represented by a density matrix, which is Hermitian , and normalized with . For simplicity, the Hamiltonian of the detector may be taken as where is the Pauli matrix and is the energy level spacing. The standard Hamiltonian of massless, free scalar field in a Kerr space time can be denoted as , which will be discussed in detail in Section 3. The interaction Hamiltonian of the detector with the scalar field can be denoted as Therefore, the Hilbert space of the total system is given by the tensor product space . The total Hamiltonian can be taken as where is the coupling constant and, and denote the identity operators in and , respectively.\n\nNow in order to get the reduced dynamics of the subsystem , we assume that the interaction between the detector and the scalar field is weak as is small and the finite time evolution describing the dynamics of the detector takes the form of a one-parameter semigroup of completely positive map.\n\nInitially, the complete system is described by the total density matrix , where is the initial reduced density matrix of the detector and is the Kerr space-time vacuum state of field . In the frame of the atom, the evolution in the proper time of the total density of the complete system satisfies which is often referred to the von Neumann or Liouville-von Neumann equation, where represents the Liouville operator associated with as follows: The dynamics of the detector can be obtained by summing over the degrees of freedom of the field ; that is, by applying to with the trace projection operator as follows: In the limit of weak coupling, we can find that the reduced density obeys an equation in the Kossakowski-Lindblad form as follows: where The matrix and the effective Hamiltonian are determined by the Fourier transform and Hilbert transform of the vacuum field correlation function (the Wightman function) as follows: and they are defined as The coefficients of the Kossakowski matrix can be written as with The effective Hamiltonian contains a correction term, the so-called Lamb shift, and one can find that it can be obtained by replacing in with a renormalized energy level spacing as follows: where a suitable subtraction is assumed in the definition of to remove the logarithmic divergence which would otherwise be presented.\n\nTo facilitate the discussion of the properties of solutions for (7) and (8), let us express the density matrix in terms of the Pauli matrices as follows: Substituting (14) into (8), the Bloch vector of components satisfies where denotes a constant vector . The exact form of the matrix reads Equation (15) can be solved exactly and its solution is where the matrix is defined by series expansion as usual. However, obeys a cubic eigenvalue equation, so powers of higher than 2 can always be written in terms of combinations of , , and . Actually, three eigenvalues of are , . We can write where Equation (19) reveals that a freely falling atom in a Kerr space time is subjected to the effects of decoherence and dissipation by the exponentially decaying factors including the real parts of the eigenvalues of and oscillating terms associated with the imaginary part. These nonunitary effects can be analyzed by examining the evolution behavior in time of suitable atom observable. For any observable of the atom represented by a Hermitian operator , the behavior of its mean value is determined by Let the observable be an admissible atom state , the probability , that the atom evolves to the expected state represented by density matrix from an initial one , should be If initially the atom is in the ground state, its Bloch vector is , and the final state is the excited state given by the Bloch vector , according to (17)–(22), we have\n\nThe probability per unit time of the transition from the ground state to the excited state, in the limit of infinitely slow switching on and off the atom-field interaction, that is, the spontaneous excitation rate, can be calculated by taking the time derivative of at as\n\n#### 3. Scalar Wave Equation Near the Event Horizon in a Kerr Space-Time\n\n##### 3.1. Dimensional Reduction Near the Horizon\n\nIn order to find out how the reduced density evolves with proper time from (7), we will investigate the scalar wave equation of the Kerr space time. In Boyer-Lindquist coordinates, the stationary Kerr space time can be written as where , and . The parameters and represent the mass and the angular momentum per unit mass of the black hole, respectively. The event horizon of the Kerr black hole is located at . The line element in (25) is stationary and axisymmetric, with and as the corresponding Killing vector fields.\n\nAnd then, we will show that the scalar field theory in the background (25) can be reduced to a 2-dimensional field theory in the near-horizon region with the dimensional reduction technique. This technique firstly has been employed for the Kerr black hole by Murata and Soda and developed with a more general technique by Iso et al. .\n\nThe action for the scalar field in a Kerr space time is where the first term is the kinetic term and the second term represents the mass, potential, and interaction terms.\n\nBy substituting (25) into (26), we obtain Now, we transform the radial coordinate into the tortoise coordinate defined by After the transformation, the action (27) can be written as Now we consider this action in the region near the horizon. Since at , we only retain dominant terms in (29). We have where we have ignored by using at . Because the theory becomes high-energy case near the horizon and the kinetic term dominates, we can ignore all the terms in . After this analysis, we return to the expression written in terms of . So, we have\n\nFollowing Murata and Soda’s method , we transform the coordinates to the corotating coordinate system. They employed a locally corotating coordinate system, and we will use a globally corotating coordinate system as Under these new coordinates, we can rewrite the action (31) as so the angular terms disappear completely. Using the spherical harmonics expansion , we obtain the effective 2-dimensional action where we have used the orthonormal condition for the spherical harmonics as follows:\n\nFrom the action (34), it is obvious to find that can be considered as a -dimensional massless scalar field in the backgrounds of the dilaton . The effective 2-dimensional metric and the dilaton can be written as\n\nSo far, we have reduced the 4-dimensional field theory to a 2-dimensional case. This is consistent with . This 2-dimensional metric tells us that, near the horizon, the geometry of a Kerr space time can be regarded as a Rindler space time when . In the extremal case , the near horizon geometry reduces to which is consistent with [19, 23]. The same as the Schwarzschild space time , we will define two vacuum states by using the two natural notions of time translation of this effective 2-dimensional metric, namely, the Killing time and the proper time as measured by a congruence of freely falling observers.\n\n##### 3.2. The Boulware Vacuum\n\nUsing the tortoise coordinate in (28), the effective 2-dimensional metric (36) can be changed into\n\nWe can see that the part of the metric has the form of Minkowski metric. Now in this 2-dimensional space time, the wave equation of can be written as\n\nIts standard ingoing and outgoing orthonormal mode solutions are where are null coordinates. These modes are positive frequency modes with respect to the killing vector field for , and they satisfy\n\nIt is obvious that the wave equation (39) is manifestly invariant under the infinite-dimensional group of conformal transformation in two dimensions , . In the following, we will show how this conformal symmetry does play the key role in the quantum effect of a Kerr space time.\n\nNear the event horizon, we only consider the outgoing modes along the rays constant. Quantizing the field in the exterior of the black hole, we can expand it as follows where and are the annihilation and creation operators acting on the vacuum state, which corresponds to the Boulware vacuum. The Fock vacuum state can be defined as . So, with the proper prescription, the Wightman function of state can be written as where , .\n\n##### 3.3. The Unruh Vacuum\n\nIn order to define the Unruh vacuum state, one can write the Kerr line element in the near-horizon region in terms of Kruskal-like coordinates defined as where , and is the surface gravity of the event horizon. The effective 2-dimensional space time (36) becomes where , which is a finite constant near the event horizon . The part of the metric also has the form of Minkowski metric. The interval of time then corresponds to the interval of proper time of a radial freely falling observer crossing the horizon. The 2-dimensional space time in the coordinates is well-behaved near the event horizon. One can obtain the wave equation of as\n\nSimilar to previous proceeding, we can obtain the out-going wave solution as . These modes are positive frequency modes with respect to the freely falling observer for , satisfying As pointed in previous subsection, there are conformal transformations from to , as (45).\n\nSubsequently, quantizing the field in the exterior of the black hole, we can expand it as follows: where and are the annihilation and creation operators acting on the vacuum state, which can be defined as . This vacuum state is just the so-called Unruh vacuum defined in the maximally extended geometry. So, with the proper prescription, the Wightman function of the state can be written as where , .\n\n#### 4. Probability of Spontaneous Transition of the Detector in a Kerr Space-Time\n\nIn what follows, we will calculate the spontaneous excitation rate in the two vacuum states with the open quantum system approach.\n\n##### 4.1. The State-Boulware Vacuum\n\nFirstly, let us turn to the state-Boulware vacuum case. Thinking of the relation between the proper time and the coordinate time, the Fourier transform of the Wightman function (44) with respect to the proper time can be expressed as\n\nAccording to (12), we have\n\nSo the spontaneous excitation rate can be obtained as\n\nTherefore, no spontaneous excitation would ever occur in the state-Boulware vacuum. In fact, the Boulware vacuum corresponds to our familiar notion of a vacuum state. This result is consistent with the conclusion in .\n\n##### 4.2. The State-Unruh Vacuum\n\nUsing the Wightman function (50) and the relation between the proper time and coordinate time (51), the Fourier transform can be given as where .\n\nAccording to (12), we have\n\nUsing (23) and (24), we have and the spontaneous excitation rate is which reveals that, the ground state detector in the vacuum would spontaneously excite with an excitation rate that one would expect in the case of a flux of thermal radiation at the temperature It is obvious that there is a near horizon conformal symmetry from to as (45), which is just the reason of the nonvanishing spontaneously excitation. This proposal is similar to . In fact, the effective temperature in (59) approaches to Hawking temperature as . This suggests that the thermal radiation emanating from the horizon of a Kerr black hole is just Hawking radiation, which is in agreement with [16, 17].\n\n#### 5. Conclusions and Discussions\n\nIn brief, we have investigated the Hawking radiation from a Kerr space time through examining the evolution of a detector (modeled by a two-level atom) interacting with the vacuum massless scalar field in the framework of open quantum systems.\n\nFirst of all, using the dimensional reduction technique, the 4-dimensional spherically nonsymmetric Kerr metric can be regarded as a 2-dimensional effective spherically symmetric metric near the event horizon. So we can construct two conformal vacuum states in this 2-dimensional effective space time: one is the state-Boulware vacuum, the other is the state-Unruh vacuum. Then we give the Wightman functions of the two vacuum states, respectively.\n\nOn the basis of these, we have calculated the time evolution of the detector in the two vacuum states. It is found that the detector in the state-Unruh vacuum would spontaneously excite with a nonvanishing probability the same as thermal radiation at Hawking temperature from a Kerr black hole. Hawking-Unruh effect of a Kerr space time can be understood as a manifestation of thermalization phenomena in an open quantum system. Meanwhile, it is also found that the probability of spontaneous transition of the detector would be vanishing in the state-Boulware vacuum. It suggests that near horizon conformal symmetry plays the key role in the full quantum phenomena in the Kerr space time.\n\n#### Conflict of Interests\n\nThe authors declare that there is no conflict of interests regarding the publication of this paper.\n\n#### Acknowledgments\n\nThe authors would like to thank Dr. Kui Xiao and Dr. Shi-Wei Zhou for their helpful discussions and suggestions. This research is supported by the National Natural Science Foundation of China under Grant no. 11365008.\n\n1. S. W. 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View at: Publisher Site | Google Scholar\n12. S. P. Robinson and F. Wilczek, “Relationship between Hawking radiation and gravitational anomalies,” Physical Review Letters, vol. 95, no. 1, Article ID 011303, 4 pages, 2005. View at: Publisher Site | Google Scholar | MathSciNet\n13. K. Xiao, W. Liu, and H. B. Zhang, “Anomalies of the Achucarro-Ortiz black hole,” Physics Letters B, vol. 647, no. 5-6, pp. 482–485, 2007. View at: Publisher Site | Google Scholar\n14. H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, Oxford, UK, 2002. View at: MathSciNet\n15. F. Benatti and R. Floreanini, “Entanglement generation in uniformly accelerating atoms: reexamination of the Unruh effect,” Physical Review A, vol. 70, no. 1, Article ID 012112, 12 pages, 2004. View at: Publisher Site | Google Scholar\n16. H. Yu and J. 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Piani, “Environment induced entanglement in Markovian dissipative dynamics,” Physical Review Letters, vol. 91, no. 7, Article ID 070402, 4 pages, 2003. View at: Google Scholar\n31. V. Gorini, A. Kossakowski, and E. C. G. Surdarshan, “Completely positive dynamical semigroups of N level systems,” Journal of Mathematical Physics, vol. 17, p. 821, 1976. View at: Publisher Site | Google Scholar\n32. G. Lindblad, “On the generators of quantum dynamical semigroups,” Communications in Mathematical Physics, vol. 48, no. 2, pp. 119–130, 1976. View at: Publisher Site | Google Scholar\n33. N. D. Birrell and P. C. W. Davies, Quantum Fields in Curved Space, vol. 7, Cambridge University Press, Cambridge, UK, 1982. View at: MathSciNet\n\nWe are committed to sharing findings related to COVID-19 as quickly and safely as possible. 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http://onnocenter.or.id/wiki/index.php?title=Keras:_5_Step_Life-Cycle_for_Long_Short-Term_Memory_Model&diff=56807&oldid=0 | [
"# Difference between revisions of \"Keras: 5 Step Life-Cycle for Long Short-Term Memory Model\"\n\nDeep learning neural networks are very easy to create and evaluate in Python with Keras, but you must follow a strict model life-cycle.\n\nIn this post, you will discover the step-by-step life-cycle for creating, training, and evaluating Long Short-Term Memory (LSTM) Recurrent Neural Networks in Keras and how to make predictions with a trained model.\n\nAfter reading this post, you will know:\n\n• How to define, compile, fit, and evaluate an LSTM in Keras.\n• How to select standard defaults for regression and classification sequence prediction problems.\n• How to tie it all together to develop and run your first LSTM recurrent neural network in Keras.\n\nDiscover how to develop LSTMs such as stacked, bidirectional, CNN-LSTM, Encoder-Decoder seq2seq and more in my new book, with 14 step-by-step tutorials and full code.\n\nOverview\n\nBelow is an overview of the 5 steps in the LSTM model life-cycle in Keras that we are going to look at.\n\n• Define Network\n• Compile Network\n• Fit Network\n• Evaluate Network\n• Make Predictions\n\nEnvironment\n\nThis tutorial assumes you have a Python SciPy environment installed. You can use either Python 2 or 3 with this example.\n\nThis tutorial assumes you have Keras v2.0 or higher installed with either the TensorFlow or Theano backend.\n\nThis tutorial also assumes you have scikit-learn, Pandas, NumPy, and Matplotlib installed.\n\nNext, let’s take a look at a standard time series forecasting problem that we can use as context for this experiment.\n\nIf you need help setting up your Python environment, see this post:\n\n``` How to Setup a Python Environment for Machine Learning and Deep Learning with Anaconda\n```\n\n## Step 1. Define Network\n\nThe first step is to define your network.\n\nNeural networks are defined in Keras as a sequence of layers. The container for these layers is the Sequential class.\n\nThe first step is to create an instance of the Sequential class. Then you can create your layers and add them in the order that they should be connected. The LSTM recurrent layer comprised of memory units is called LSTM(). A fully connected layer that often follows LSTM layers and is used for outputting a prediction is called Dense().\n\nFor example, we can do this in two steps:\n\n```model = Sequential()\n```\n\nBut we can also do this in one step by creating an array of layers and passing it to the constructor of the Sequential.\n\n```layers = [LSTM(2), Dense(1)]\nmodel = Sequential(layers)\n```\n\nThe first layer in the network must define the number of inputs to expect. Input must be three-dimensional, comprised of samples, timesteps, and features.\n\n• Samples. These are the rows in your data.\n• Timesteps. These are the past observations for a feature, such as lag variables.\n• Features. These are columns in your data.\n\nAssuming your data is loaded as a NumPy array, you can convert a 2D dataset to a 3D dataset using the reshape() function in NumPy. If you would like columns to become timesteps for one feature, you can use:\n\n```data = data.reshape((data.shape, data.shape, 1))\n```\n\nIf you would like columns in your 2D data to become features with one timestep, you can use:\n\n```data = data.reshape((data.shape, 1, data.shape))\n```\n\nYou can specify the input_shape argument that expects a tuple containing the number of timesteps and the number of features. For example, if we had two timesteps and one feature for a univariate time series with two lag observations per row, it would be specified as follows:\n\n```model = Sequential()\n```\n\nLSTM layers can be stacked by adding them to the Sequential model. Importantly, when stacking LSTM layers, we must output a sequence rather than a single value for each input so that the subsequent LSTM layer can have the required 3D input. We can do this by setting the return_sequences argument to True. For example:\n\n```model = Sequential()\n```\n\nThink of a Sequential model as a pipeline with your raw data fed in at in end and predictions that come out at the other.\n\nThis is a helpful container in Keras as concerns that were traditionally associated with a layer can also be split out and added as separate layers, clearly showing their role in the transform of data from input to prediction.\n\nFor example, activation functions that transform a summed signal from each neuron in a layer can be extracted and added to the Sequential as a layer-like object called Activation.\n\n```model = Sequential()\n```\n\nThe choice of activation function is most important for the output layer as it will define the format that predictions will take.\n\nFor example, below are some common predictive modeling problem types and the structure and standard activation function that you can use in the output layer:\n\n``` Regression: Linear activation function, or ‘linear’, and the number of neurons matching the number of outputs.\nBinary Classification (2 class): Logistic activation function, or ‘sigmoid’, and one neuron the output layer.\nMulticlass Classification (>2 class): Softmax activation function, or ‘softmax’, and one output neuron per class value, assuming a one-hot encoded output pattern.\n```\n\n## Step 2. Compile Network\n\nOnce we have defined our network, we must compile it.\n\nCompilation is an efficiency step. It transforms the simple sequence of layers that we defined into a highly efficient series of matrix transforms in a format intended to be executed on your GPU or CPU, depending on how Keras is configured.\n\nThink of compilation as a precompute step for your network. It is always required after defining a model.\n\nCompilation requires a number of parameters to be specified, specifically tailored to training your network. Specifically, the optimization algorithm to use to train the network and the loss function used to evaluate the network that is minimized by the optimization algorithm.\n\nFor example, below is a case of compiling a defined model and specifying the stochastic gradient descent (sgd) optimization algorithm and the mean squared error (mean_squared_error) loss function, intended for a regression type problem.\n\n```model.compile(optimizer='sgd', loss='mean_squared_error')\n```\n\nAlternately, the optimizer can be created and configured before being provided as an argument to the compilation step.\n\n```algorithm = SGD(lr=0.1, momentum=0.3)\nmodel.compile(optimizer=algorithm, loss='mean_squared_error')\n```\n\nThe type of predictive modeling problem imposes constraints on the type of loss function that can be used.\n\nFor example, below are some standard loss functions for different predictive model types:\n\n• Regression: Mean Squared Error or ‘mean_squared_error’.\n• Binary Classification (2 class): Logarithmic Loss, also called cross entropy or ‘binary_crossentropy‘.\n• Multiclass Classification (>2 class): Multiclass Logarithmic Loss or ‘categorical_crossentropy‘.\n\nThe most common optimization algorithm is stochastic gradient descent, but Keras also supports a suite of other state-of-the-art optimization algorithms that work well with little or no configuration.\n\nPerhaps the most commonly used optimization algorithms because of their generally better performance are:\n\n• Stochastic Gradient Descent, or ‘sgd‘, that requires the tuning of a learning rate and momentum.\n• RMSprop, or ‘rmsprop‘, that requires the tuning of learning rate.\n\nFinally, you can also specify metrics to collect while fitting your model in addition to the loss function. Generally, the most useful additional metric to collect is accuracy for classification problems. The metrics to collect are specified by name in an array.\n\nFor example:\n\n```model.compile(optimizer='sgd', loss='mean_squared_error', metrics=['accuracy'])\n```\n\n## Step 3. Fit Network\n\nOnce the network is compiled, it can be fit, which means adapt the weights on a training dataset.\n\nFitting the network requires the training data to be specified, both a matrix of input patterns, X, and an array of matching output patterns, y.\n\nThe network is trained using the backpropagation algorithm and optimized according to the optimization algorithm and loss function specified when compiling the model.\n\nThe backpropagation algorithm requires that the network be trained for a specified number of epochs or exposures to the training dataset.\n\nEach epoch can be partitioned into groups of input-output pattern pairs called batches. This defines the number of patterns that the network is exposed to before the weights are updated within an epoch. It is also an efficiency optimization, ensuring that not too many input patterns are loaded into memory at a time.\n\nA minimal example of fitting a network is as follows:\n\n```history = model.fit(X, y, batch_size=10, epochs=100)\n```\n\nOnce fit, a history object is returned that provides a summary of the performance of the model during training. This includes both the loss and any additional metrics specified when compiling the model, recorded each epoch.\n\nTraining can take a long time, from seconds to hours to days depending on the size of the network and the size of the training data.\n\nBy default, a progress bar is displayed on the command line for each epoch. This may create too much noise for you, or may cause problems for your environment, such as if you are in an interactive notebook or IDE.\n\nYou can reduce the amount of information displayed to just the loss each epoch by setting the verbose argument to 2. You can turn off all output by setting verbose to 1. For example:\n\n```history = model.fit(X, y, batch_size=10, epochs=100, verbose=0)\n```\n\n## Step 4. Evaluate Network\n\nOnce the network is trained, it can be evaluated.\n\nThe network can be evaluated on the training data, but this will not provide a useful indication of the performance of the network as a predictive model, as it has seen all of this data before.\n\nWe can evaluate the performance of the network on a separate dataset, unseen during testing. This will provide an estimate of the performance of the network at making predictions for unseen data in the future.\n\nThe model evaluates the loss across all of the test patterns, as well as any other metrics specified when the model was compiled, like classification accuracy. A list of evaluation metrics is returned.\n\nFor example, for a model compiled with the accuracy metric, we could evaluate it on a new dataset as follows:\n\n```loss, accuracy = model.evaluate(X, y)\n```\n\nAs with fitting the network, verbose output is provided to give an idea of the progress of evaluating the model. We can turn this off by setting the verbose argument to 0.\n\n```loss, accuracy = model.evaluate(X, y, verbose=0)\n```\n\n## Step 5. Make Predictions\n\nOnce we are satisfied with the performance of our fit model, we can use it to make predictions on new data.\n\nThis is as easy as calling the predict() function on the model with an array of new input patterns.\n\nFor example:\n\n```predictions = model.predict(X)\n```\n\nThe predictions will be returned in the format provided by the output layer of the network.\n\nIn the case of a regression problem, these predictions may be in the format of the problem directly, provided by a linear activation function.\n\nFor a binary classification problem, the predictions may be an array of probabilities for the first class that can be converted to a 1 or 0 by rounding.\n\nFor a multiclass classification problem, the results may be in the form of an array of probabilities (assuming a one hot encoded output variable) that may need to be converted to a single class output prediction using the argmax() NumPy function.\n\nAlternately, for classification problems, we can use the predict_classes() function that will automatically convert uncrisp predictions to crisp integer class values.\n\n```predictions = model.predict_classes(X)\n```\n\nAs with fitting and evaluating the network, verbose output is provided to given an idea of the progress of the model making predictions. We can turn this off by setting the verbose argument to 0.\n\n```predictions = model.predict(X, verbose=0)\n```\n\n## End-to-End Worked Example\n\nLet’s tie all of this together with a small worked example.\n\nThis example will use a simple problem of learning a sequence of 10 numbers. We will show the network a number, such as 0.0 and expect it to predict 0.1. Then show it 0.1 and expect it to predict 0.2, and so on to 0.9.\n\n``` Define Network: We will construct an LSTM neural network with a 1 input timestep and 1 input feature in the visible layer, 10 memory units in the LSTM hidden layer, and 1 neuron in the fully connected output layer with a linear (default) activation function.\nCompile Network: We will use the efficient ADAM optimization algorithm with default configuration and the mean squared error loss function because it is a regression problem.\nFit Network: We will fit the network for 1,000 epochs and use a batch size equal to the number of patterns in the training set. We will also turn off all verbose output.\nEvaluate Network. We will evaluate the network on the training dataset. Typically we would evaluate the model on a test or validation set.\nMake Predictions. We will make predictions for the training input data. Again, typically we would make predictions on data where we do not know the right answer.\n```\n\nThe complete code listing is provided below.\n\n```# Example of LSTM to learn a sequence\nfrom pandas import DataFrame\nfrom pandas import concat\nfrom keras.models import Sequential\nfrom keras.layers import Dense\nfrom keras.layers import LSTM\n# create sequence\nlength = 10\nsequence = [i/float(length) for i in range(length)]\nprint(sequence)\n# create X/y pairs\ndf = DataFrame(sequence)\ndf = concat([df.shift(1), df], axis=1)\ndf.dropna(inplace=True)\n# convert to LSTM friendly format\nvalues = df.values\nX, y = values[:, 0], values[:, 1]\nX = X.reshape(len(X), 1, 1)\n# 1. define network\nmodel = Sequential()\n# 2. compile network\n# 3. fit network\nhistory = model.fit(X, y, epochs=1000, batch_size=len(X), verbose=0)\n# 4. evaluate network\nloss = model.evaluate(X, y, verbose=0)\nprint(loss)\n# 5. make predictions\npredictions = model.predict(X, verbose=0)\nprint(predictions[:, 0])\n```\n\nRunning this example produces the following output, showing the raw input sequence of 10 numbers, the mean squared error loss of the network when making predictions for the entire sequence, and the predictions for each input pattern.\n\nOutputs were spaced out for readability.\n\nWe can see the sequence is learned well, especially if we round predictions to the first decimal place.\n\n```[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]\n\n4.54527471447e-05\n\n[ 0.11612834 0.20493418 0.29793766 0.39445466 0.49376178 0.59512401\n0.69782174 0.80117452 0.90455914]\n```\n\n``` Keras documentation for Sequential Models.\nKeras documentation for LSTM Layers.\nKeras documentation for optimization algorithms.\nKeras documentation for loss functions.\n```\n\n## Summary\n\nIn this post, you discovered the 5-step life-cycle of an LSTM recurrent neural network using the Keras library.\n\nSpecifically, you learned:\n\n• How to define, compile, fit, evaluate, and make predictions for an LSTM network in Keras.\n• How to select activation functions and output layer configurations for classification and regression problems.\n• How to develop and run your first LSTM model in Keras."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8435343,"math_prob":0.9430391,"size":15864,"snap":"2019-51-2020-05","text_gpt3_token_len":3440,"char_repetition_ratio":0.14079446,"word_repetition_ratio":0.040194884,"special_character_ratio":0.22144479,"punctuation_ratio":0.13083802,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959251,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-25T05:13:17Z\",\"WARC-Record-ID\":\"<urn:uuid:35c981a3-d889-4fbd-be77-19896c54c029>\",\"Content-Length\":\"33799\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:703dc821-bef0-4c11-b69e-e0392d4df119>\",\"WARC-Concurrent-To\":\"<urn:uuid:080c5cc1-c204-4f48-ac03-a8028bb35210>\",\"WARC-IP-Address\":\"103.23.22.141\",\"WARC-Target-URI\":\"http://onnocenter.or.id/wiki/index.php?title=Keras:_5_Step_Life-Cycle_for_Long_Short-Term_Memory_Model&diff=56807&oldid=0\",\"WARC-Payload-Digest\":\"sha1:ALETGBIND2JOQ4Z5MWGXC2XTXWWTPVES\",\"WARC-Block-Digest\":\"sha1:GDLQBW4GT3XIJIOX5PFFY2VHM5EXIQX2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251669967.70_warc_CC-MAIN-20200125041318-20200125070318-00385.warc.gz\"}"} |
https://www.gcflcm.com/gcf-of-65-and-80 | [
"# What is the Greatest Common Factor of 65 and 80?\n\nGreatest common factor (GCF) of 65 and 80 is 5.\n\nGCF(65,80) = 5\n\nWe will now calculate the prime factors of 65 and 80, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 65 and 80.\n\nGCF Calculator and\nand\n\n## How to find the GCF of 65 and 80?\n\nWe will first find the prime factorization of 65 and 80. After we will calculate the factors of 65 and 80 and find the biggest common factor number .\n\n### Step-1: Prime Factorization of 65\n\nPrime factors of 65 are 5, 13. Prime factorization of 65 in exponential form is:\n\n65 = 51 × 131\n\n### Step-2: Prime Factorization of 80\n\nPrime factors of 80 are 2, 5. Prime factorization of 80 in exponential form is:\n\n80 = 24 × 51\n\n### Step-3: Factors of 65\n\nList of positive integer factors of 65 that divides 65 without a remainder.\n\n1, 5, 13\n\n### Step-4: Factors of 80\n\nList of positive integer factors of 80 that divides 65 without a remainder.\n\n1, 2, 4, 5, 8, 10, 16, 20, 40\n\n#### Final Step: Biggest Common Factor Number\n\nWe found the factors and prime factorization of 65 and 80. The biggest common factor number is the GCF number.\nSo the greatest common factor 65 and 80 is 5.\n\nAlso check out the Least Common Multiple of 65 and 80"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89372104,"math_prob":0.9684924,"size":1228,"snap":"2022-40-2023-06","text_gpt3_token_len":363,"char_repetition_ratio":0.21895425,"word_repetition_ratio":0.06837607,"special_character_ratio":0.32410422,"punctuation_ratio":0.125,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9998091,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T15:46:27Z\",\"WARC-Record-ID\":\"<urn:uuid:f2010250-89e7-4b54-b0b0-a9225856767e>\",\"Content-Length\":\"20837\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81a60cf3-9826-42fe-a78b-8d399b647501>\",\"WARC-Concurrent-To\":\"<urn:uuid:332fa5b4-d3da-405e-aa2a-6ac8c69f3a91>\",\"WARC-IP-Address\":\"34.133.163.157\",\"WARC-Target-URI\":\"https://www.gcflcm.com/gcf-of-65-and-80\",\"WARC-Payload-Digest\":\"sha1:CMFCP3WKP66MUMNPL54Q3I2L34MQO4QS\",\"WARC-Block-Digest\":\"sha1:QRMABJDL2MEKW32NREUHUAXY5G44WBDG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499946.80_warc_CC-MAIN-20230201144459-20230201174459-00178.warc.gz\"}"} |
https://www.examsolutions.net/tutorials/equation-circle/ | [
"In this tutorial you are shown how the equation of a circle is derived when you know the centre and radius.\n\nFormulae to remember:\n\n• (x – x1)2 + (y – y1)2 = r2 where the centre has coordinates (x1 , y1) and radius = r2\n\nYou will also be taken through this example.\n\nExample:\n\n1. Find the equation of a circle with centre (4, -2) and radius 3.\nCartesian Equation of a Circle | ExamSolutions - youtube Video"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87397176,"math_prob":0.99781257,"size":337,"snap":"2019-43-2019-47","text_gpt3_token_len":94,"char_repetition_ratio":0.13513513,"word_repetition_ratio":0.03076923,"special_character_ratio":0.30267063,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99931335,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T23:22:48Z\",\"WARC-Record-ID\":\"<urn:uuid:2cb8e793-2b11-49d1-a482-a3cedecc095d>\",\"Content-Length\":\"842341\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b4c58a0-e28a-48b8-8ee4-55c4ecefea16>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ecf0a3c-3b11-494a-94ee-c10078a5e31b>\",\"WARC-IP-Address\":\"5.134.13.111\",\"WARC-Target-URI\":\"https://www.examsolutions.net/tutorials/equation-circle/\",\"WARC-Payload-Digest\":\"sha1:D4SLXFHDU3EW2SAEMQVKIT5ZKMLAST2L\",\"WARC-Block-Digest\":\"sha1:A25JUP7OJMET3JUA46EIEXWRFIBJ2742\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660829.5_warc_CC-MAIN-20191015231925-20191016015425-00084.warc.gz\"}"} |
https://www.colorhexa.com/5b6730 | [
"# #5b6730 Color Information\n\nIn a RGB color space, hex #5b6730 is composed of 35.7% red, 40.4% green and 18.8% blue. Whereas in a CMYK color space, it is composed of 11.7% cyan, 0% magenta, 53.4% yellow and 59.6% black. It has a hue angle of 73.1 degrees, a saturation of 36.4% and a lightness of 29.6%. #5b6730 color hex could be obtained by blending #b6ce60 with #000000. Closest websafe color is: #666633.\n\n• R 36\n• G 40\n• B 19\nRGB color chart\n• C 12\n• M 0\n• Y 53\n• K 60\nCMYK color chart\n\n#5b6730 color description : Very dark desaturated yellow.\n\n# #5b6730 Color Conversion\n\nThe hexadecimal color #5b6730 has RGB values of R:91, G:103, B:48 and CMYK values of C:0.12, M:0, Y:0.53, K:0.6. Its decimal value is 5990192.\n\nHex triplet RGB Decimal 5b6730 `#5b6730` 91, 103, 48 `rgb(91,103,48)` 35.7, 40.4, 18.8 `rgb(35.7%,40.4%,18.8%)` 12, 0, 53, 60 73.1°, 36.4, 29.6 `hsl(73.1,36.4%,29.6%)` 73.1°, 53.4, 40.4 666633 `#666633`\nCIE-LAB 41.435, -13.921, 29.227 9.698, 12.138, 4.628 0.366, 0.459, 12.138 41.435, 32.373, 115.469 41.435, -4.962, 33.86 34.84, -11.282, 16.512 01011011, 01100111, 00110000\n\n# Color Schemes with #5b6730\n\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #3c3067\n``#3c3067` `rgb(60,48,103)``\nComplementary Color\n• #675830\n``#675830` `rgb(103,88,48)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #406730\n``#406730` `rgb(64,103,48)``\nAnalogous Color\n• #573067\n``#573067` `rgb(87,48,103)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #304067\n``#304067` `rgb(48,64,103)``\nSplit Complementary Color\n• #67305b\n``#67305b` `rgb(103,48,91)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #305b67\n``#305b67` `rgb(48,91,103)``\n• #673c30\n``#673c30` `rgb(103,60,48)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #305b67\n``#305b67` `rgb(48,91,103)``\n• #3c3067\n``#3c3067` `rgb(60,48,103)``\n• #2d3318\n``#2d3318` `rgb(45,51,24)``\n• #3c4420\n``#3c4420` `rgb(60,68,32)``\n• #4c5628\n``#4c5628` `rgb(76,86,40)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #6a7838\n``#6a7838` `rgb(106,120,56)``\n• #7a8a40\n``#7a8a40` `rgb(122,138,64)``\n• #899b48\n``#899b48` `rgb(137,155,72)``\nMonochromatic Color\n\n# Alternatives to #5b6730\n\nBelow, you can see some colors close to #5b6730. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #676530\n``#676530` `rgb(103,101,48)``\n• #646730\n``#646730` `rgb(100,103,48)``\n• #606730\n``#606730` `rgb(96,103,48)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #566730\n``#566730` `rgb(86,103,48)``\n• #526730\n``#526730` `rgb(82,103,48)``\n• #4d6730\n``#4d6730` `rgb(77,103,48)``\nSimilar Colors\n\n# #5b6730 Preview\n\nThis text has a font color of #5b6730.\n\n``<span style=\"color:#5b6730;\">Text here</span>``\n#5b6730 background color\n\nThis paragraph has a background color of #5b6730.\n\n``<p style=\"background-color:#5b6730;\">Content here</p>``\n#5b6730 border color\n\nThis element has a border color of #5b6730.\n\n``<div style=\"border:1px solid #5b6730;\">Content here</div>``\nCSS codes\n``.text {color:#5b6730;}``\n``.background {background-color:#5b6730;}``\n``.border {border:1px solid #5b6730;}``\n\n# Shades and Tints of #5b6730\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #080904 is the darkest color, while #fdfdfb is the lightest one.\n\n• #080904\n``#080904` `rgb(8,9,4)``\n• #14170b\n``#14170b` `rgb(20,23,11)``\n• #202411\n``#202411` `rgb(32,36,17)``\n• #2c3117\n``#2c3117` `rgb(44,49,23)``\n• #383f1d\n``#383f1d` `rgb(56,63,29)``\n• #434c24\n``#434c24` `rgb(67,76,36)``\n• #4f5a2a\n``#4f5a2a` `rgb(79,90,42)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #677436\n``#677436` `rgb(103,116,54)``\n• #73823c\n``#73823c` `rgb(115,130,60)``\n• #7e8f43\n``#7e8f43` `rgb(126,143,67)``\n• #8a9d49\n``#8a9d49` `rgb(138,157,73)``\n• #96aa4f\n``#96aa4f` `rgb(150,170,79)``\n• #9fb25a\n``#9fb25a` `rgb(159,178,90)``\n• #a7b968\n``#a7b968` `rgb(167,185,104)``\n• #afbf75\n``#afbf75` `rgb(175,191,117)``\n• #b6c583\n``#b6c583` `rgb(182,197,131)``\n• #becb90\n``#becb90` `rgb(190,203,144)``\n• #c6d19d\n``#c6d19d` `rgb(198,209,157)``\n• #ced8ab\n``#ced8ab` `rgb(206,216,171)``\n• #d6deb8\n``#d6deb8` `rgb(214,222,184)``\n• #dde4c5\n``#dde4c5` `rgb(221,228,197)``\n``#e5ead3` `rgb(229,234,211)``\n• #edf1e0\n``#edf1e0` `rgb(237,241,224)``\n• #f5f7ee\n``#f5f7ee` `rgb(245,247,238)``\n• #fdfdfb\n``#fdfdfb` `rgb(253,253,251)``\nTint Color Variation\n\n# Tones of #5b6730\n\nA tone is produced by adding gray to any pure hue. In this case, #4e5047 is the less saturated color, while #759502 is the most saturated one.\n\n• #4e5047\n``#4e5047` `rgb(78,80,71)``\n• #515641\n``#515641` `rgb(81,86,65)``\n• #545b3c\n``#545b3c` `rgb(84,91,60)``\n• #586136\n``#586136` `rgb(88,97,54)``\n• #5b6730\n``#5b6730` `rgb(91,103,48)``\n• #5e6d2a\n``#5e6d2a` `rgb(94,109,42)``\n• #627324\n``#627324` `rgb(98,115,36)``\n• #65781f\n``#65781f` `rgb(101,120,31)``\n• #687e19\n``#687e19` `rgb(104,126,25)``\n• #6b8413\n``#6b8413` `rgb(107,132,19)``\n• #6f8a0d\n``#6f8a0d` `rgb(111,138,13)``\n• #729007\n``#729007` `rgb(114,144,7)``\n• #759502\n``#759502` `rgb(117,149,2)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #5b6730 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.52689457,"math_prob":0.70625544,"size":3708,"snap":"2019-51-2020-05","text_gpt3_token_len":1663,"char_repetition_ratio":0.12284017,"word_repetition_ratio":0.011070111,"special_character_ratio":0.5666127,"punctuation_ratio":0.23756906,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9906249,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T13:56:40Z\",\"WARC-Record-ID\":\"<urn:uuid:40b0177e-b90d-4e81-8cf9-941e15740d74>\",\"Content-Length\":\"36287\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e3b1e864-75e0-47b0-b2ba-0a721ef8ac81>\",\"WARC-Concurrent-To\":\"<urn:uuid:7d1180ba-7070-4126-94b4-4198527981ec>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/5b6730\",\"WARC-Payload-Digest\":\"sha1:KMR4KEFW7J7G43EDU5MCCYR56CC2YTUI\",\"WARC-Block-Digest\":\"sha1:SRIHQYT67E7CRCJDDIIEGJHZQBGWXS6Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594603.8_warc_CC-MAIN-20200119122744-20200119150744-00453.warc.gz\"}"} |
https://ktbssolutions.com/2nd-puc-physics-question-bank-chapter-11/ | [
"# 2nd PUC Physics Question Bank Chapter 11 Dual Nature of Radiation and Matter\n\nYou can Download Chapter 11 Dual Nature of Radiation and Matter Questions and Answers, Notes, 2nd PUC Physics Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.\n\n## Karnataka 2nd PUC Physics Question Bank Chapter 11 Dual Nature of Radiation and Matter\n\n### 2nd PUC Physics Dual Nature of Radiation and Matter NCERT Text Book Questions and Answers\n\nQuestion 1.\nFind the\n(a) Maximum frequency, and\n(b) Minimum wavelength of X-rays produced by 30 kV electrons.\nAnswer:\nV=30kV = 30×103 V\n(a) The maximum frequency of X-rays produced is given by,",
null,
"Question 2.\nThe work function of caesium metal is 2.14 eV. When light of frequency 6 x 1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the\n(a) Maximum kinetic energy of the emitted electrons,\n(b) Stopping potential, and\n(c) Maximum speed of the emitted photoelectrons?\nAnswer:\nυ=6×1014Hz\nω =2.14eV = 2.14 x 1.6 x 10-19J\n(a) The maximum kinetic energy of the emitted electrons is given by",
null,
"",
null,
"Question 3.\nThe photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photo electrons emitted?\nAnswer:\nV0 = 1.5 V\n∴ Maximum kinetic energy of the emitted electrons,\n$$\\frac { 1 }{ 2 } m{ V }_{ max }^{ 2 }=e{ V }_{ 0 }=1.5j=15eV$$",
null,
"Question 4.\nMonochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 raW.\n\n(a) Find the energy and momentum of each photon in the light beam,\n\n(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and\n\n(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?\nAnswer:\nλ= 632.8 nm= 632 x 10-9m\npower, P = 9.42 mW = 9.42 x 10-3 W\n(a) Energy of each photon in the light beam,",
null,
"",
null,
"Question 5.\nThe energy flux of sunlight reaching the surface of the earth is 1.388 x 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.\nAnswer:\nEnergy flux of sunlight reaching the surface of the earth,",
null,
"Question 6.\nIn an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10-15 V s. Calculate the value of Planck’s constant.\nAnswer:\nSlope of the cut off voltage versus frequency of incident light graph\n= 4.12 x 10-15 Vs\nbut the slope of the graph = $$\\frac { h }{ e }$$\nh = e x slope of the graph\n= 1.6 x 1019 x 4.12 x 10-15 = 6.592 x10-34Js",
null,
"Question 7.\nA 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.\n(a) What is the energy per photon associated with the sodium light?\n(b) At what rate are the photons delivered to the sphere?\nAnswer:\nPower of the lamp, P = 100 W\nλ =589nm=589×10-9m\n(a) Therefore, energy of a photon of sodium lamp light,",
null,
"Question 8.\nThe threshold frequency for a certain metal is 3.3 x 1014 Hz. If light of frequency\n8.2 x 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.\nAnswer:",
null,
"Question 9.\nThe work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?\nAnswer:\nX, = 330 nm = 330 x 10-9m\n∴ Energy of a photon of incident light,",
null,
"Since the energy of the photon of incident light is less than the work function for the metal, photoelectric emission will not take place.",
null,
"Question 10.\nLight of frequency 7.21 x 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?\nAnswer:",
null,
"Question 11.\nLight of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material from which the emitter is made.\nAnswer:",
null,
"Question 12.\nCalculate the\n(a) Momentum, and\n(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.\nAnswer:\nV = 56 V\n(a) When an electron is accelerated through a potential difference V, it acquires kinetic energy, which is given by, $$\\frac { 1 }{ 2 }$$mV2 – eV Therefore, momentum of electron.",
null,
"Question 13.\nWhat is the\n(a) Momentum,\n(b) Speed, and\n(c) de Broglie wavelength of an electron .with kinetic energy of 120 eV.\nAnswer:\nV = 120eV\n(a) When an electron is accelerated through a potential difference V, it acquires kinetic energy which is given by",
null,
"Question 14.\nThe wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which\n(a) An electron, and\n(b) A neutron, would have the same de Broglie wavelength.\nAnswer:\nλ = 589 nm= 589 x 109 m.\nThe de-Broglie wavelength of a particle having energy E is given by,",
null,
"Question 15.\nWhat is the de Broglie wavelength of\n(a) A bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,\n(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and\n(c) A dust particle of mass 1.0 x 10 9 kg drifting with a speed of 2.2 m/s?\nAnswer:\nThe de-Broglie wavelength of a particle moving with a velocity V is given by",
null,
"Question 16.\nAn electron and a photon each have a wavelength of 1.00 nm. Find\n(a) Their momenta,\n(b) The energy of the photon, and\n(c) The kinetic energy of electron.\nAnswer:\nh = 6.63 × 10-34 Js\nλ= 1 nm= 10-9 m\n(a) The electron and the photon will possess the same momentum, which is given by",
null,
"",
null,
"Question 17.\n(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 x 10-10m?\n(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.\nAnswer:",
null,
"Question 18.\nShow that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).\nAnswer:\nConsider an electromagnetic radiation of wavelength λ and frequency υ. Then,",
null,
"But\n$$\\frac { h\\upsilon }{ C } =P………………(2)\\quad$$\nP = momentum of the electromagnetic radiation. From equations (1) and (2), we have\n$$\\lambda =\\frac { h }{ P }$$\nHence the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength associated with its quantum.",
null,
"Question 19.\nWhat is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root- mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)\nAnswer:\nK= 1.38 x 10-23Jkg-1\nT = 300 K\nNitrogen is a diatomic gas\n∴ Mass of the nitrogen gas molecule,\nm= 2 x 14.0076 × 1027amu\n= 2 x 14.0076 x 1027 = 4.65 ×10-26 kg.\nThe root mean square speed of the nitrogen gas molecule is given by",
null,
"### 2nd PUC Physics Dual Nature of Radiation and Matter Additional Exercises\n\nQuestion 20.\n(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of, the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 x 1011 C kg-1\n\n(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?\nAnswer:\n(a) V = 500 V\ne/m= 1.76×1011 C kg-1\nDue to the applied potential difference between the emitter and the collector, an emitted electron gains kinetic energy, which is given by",
null,
"Using the same formula as in (a), the speed of electrons at 10 mV comes out to be greater than the speed of light. Since no material particle can move with the speed of light, the result is clearly wrong.\n\nIn fact, equation (1) is valid for kinetic energy of the electron, so long as V/C < < 1. At 10 mV, the electrons attain a speed comparable to the speed of light i.e. become relativistic.\n\nIn the relativistic domain:- The mass of the electron having rest mass m() is",
null,
"Question 21.\n(a) A monoenergetic electron beam with electron speed of 5.20 x 106 m s-1 is subject to a magnetic field of 1.30 x 10-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 x 10nC kg-1.\n\n(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified ?\n\n[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]\nAnswer:\n(a) V = 5.2×106 ms1\nB = 1.3 × 10-4\ne/m=1.76 x 1011Ckg-1\n\nWhen a magnetic field is applied normally to the path of electron, the force due to the field makes the electron to move along the circular path. The radius of the circular path is given by,",
null,
"The 10 MeV electron attain a speed comparable to the speed of light i.e. becomes relativistic. In the relativistic domain, the mass of an electron, having rest mass M0 is given by",
null,
"Question 22.\nAn electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~10-2 mm of Hg). A magnetic field of 2.83 x 10-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.\nAnswer:\nV = 100 V\nB = 2.83 ×10-4 T\nr = 12 cm = 12 x 10-2 m\nDue to the applied potential difference between the emitter and the collector, an emitted electron gains kinetic energy which is given by,",
null,
"When a magnetic field is applied normal to the path of electron, the force due to the filed makes the electron to move along the circular path. The radius of the circular path is given by,",
null,
"Question 23.\n(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?\n(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?\nAnswer:\nλ= 0.45 A = 0.45 x 10-10 m\nh = 6.62×10-34 Js\nC = 3 x 108ms-1\n(a) The maximum energy of the photon is given by,\n$$E=\\frac { hC }{ \\lambda }$$",
null,
"(b) To produce electrons of energy 25.81 keV, accelerating potential of 25.81 kV i.e. order of 26 kV is required.",
null,
"Question 24.\nIn an accelerator experiment on high- energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ -rays of equal energy. What is the wavelength ‘ associated with each y-ray? (IBeV – 109 eV)\nAnswer:",
null,
"Question 25.\nEstimating the following two numbers should be interesting. The first number will tell you why radio engineers do not, need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.\n\n(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.\n\n(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~10-10 W m-2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 x 1014Hz.\nAnswer:\n(a) Power of energy transmitted per second,\nE = 10 kW = 104 W = 104 Js-1\nλ = 500 m\nEnergy of one photon,",
null,
"(b) Minimum intensity of white light = 1010 Wm2\nArea of the pupil = 0.4cm2 = 0.4 x 104 m2\n∴ Light energy falling on pupil per second\n= 10-10 x 0.4 x 10-4= 0.4 x 10-14W\nFrequency of white light, u = 6x 1014 Hz\n∴ Energy of the photon of the white light,\nhV= 6.62×10-34 x 6 x 1014J\n∴ Number of photons entering the pupil of our eye per second",
null,
"",
null,
"Question 26.\nUltraviolet light of wavelength 2271 A from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~ 10s W m-2) red light of wavelength 6328 A produced by a He-Ne laser?\nAnswer:",
null,
"",
null,
"Since wavelength 6.328 Å produced by He-Ne laser and incident on the photocell is greater than,λo the photo cell will not respond.\n\nQuestion 27.\nMonochromatic radiation of wavelength nm ( 1 nm = 10-9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.\nAnswer:",
null,
"",
null,
"Question 28.\nA mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photocell, the following lines from a mercury source were used:\nλ1= 3650 Å, λ2= 4047 A, λ3= 4358 A, λ4= 5461 A,λ5= 6907 A,\nThe stopping voltages, respectively, were measured to be:\nV01 = 1.28 V, F02 = 0.95 V, Vn = 0.74 V, V04 = 0.16 V, Fos = 0 V\nDetermine the value of Planck’s constant h, the threshold frequency and work function for the material.\n[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 x 10-19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.\nAnswer:\n(a) Let the respective frequencies of the five spectral lines of mercury be υ1, υ2, υ3, υ4 and υ5. Then",
null,
"It represents the equation of straight line, whose slope is hυ0/e and makes an intercept role on negative V0 – axis. The plot of graph between V (along x-axis) and V0 (along y-axis) for the given data for the five spectral lines will be shown as in figure.",
null,
"",
null,
"Question 29.\nThe work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?\nAnswer:\nλ. = 3300Å= 3300 ×10-10m\nEnergy of a photon of incident light,",
null,
"Since the work functions of Mo and Ni are greater than the energy of the photon of incident light, photoelectric emission will not occur for these metals.\n\nIf the laser is brought nearer, the intensity of incident radiation on the metal will increase. It will increase the photoemission in case of Na and K. However, no photo electric emission will .take place in case of Mo and N.",
null,
"Question 30.\nLight of intensity 10-5 W m-2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?\nAnswer:\nSince size of the atom is about 10-10m, the effective area of an atom may be considered as 10-20 m2.\nArea of each layer in sodium = 2cm2 = 2 x 10-4m2\nNumber of atoms in 5 layers of sodium\n$$\\frac { 2\\times { 10 }^{ -4 }\\times 5 }{ { 10 }^{ -20 } }$$\nIf we assume that sodium has one conduction electron per atom, then number of electrons in the 5 layers of sodium = $${ 10 }^{ 17 }$$\nEnergy incident per second on the surface of the photocell\n= 10-5 x 2 x 10-4\n= 2×10-9\nSince incident energy is observed by 5 layers of sodium, according to wave picture of radiation, the electrons present in all the 5 layers of sodium will share the incident equally.\n∴ Energy received by any one electron in the 5 layers of sodium",
null,
"Since work function of sodium is 2 eV, an electron will be ejected as soon as it gathers energy equal to 2 eV.\n∴ Time required for photoelectric emission\n= $$\\frac { 2 }{ 1.25\\times { 10 }^{ 7 } }$$ = 1.6 x 107s ≈ 0.5 year\nIt is contrary to the observed fact that there is no time lag between the incidence of light and the emission of photoelectrons.\n\nQuestion 31.\nCrystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 A, which is of the order of inter-atomic spacing in the lattice) (me=9.11 ×10-31 kg).\nAnswer:\nFor x- ray photon of wavelength of 1Å",
null,
"Question 32.\n(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn = 1.675 x 10-27 kg)\n\n(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.\nAnswer:",
null,
"Since the interatomic spacing (≈10-10m) is about\n100 times greater than the de=Broglie wavelength of 150 eV neutrons, they are not suitable for crystal diffraction experiments.\n\n(b)T=27+273=300k\nEnergy of the neutron at temperature, T,",
null,
"Thus, thermal neutrons have wavelength of the order of interatomic spacing\n(≈10-10m) and therefore they are suitable for diffraction experiments. Fast neutrons will possess wavelength quite small as compared to interatomic spacing. Hence for neutron diffraction experiments, fast neutron beam needs to be thermalised.\n\nQuestion 33.\nAn electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?\nAnswer:\nHere, V = 50 kV\nTherefore, energy of electrons,\nE = 50keV = 50 x 103 x 1.6 x 1019 =80 x 10-15",
null,
"The resolving power of a microscope is inversely proportional to the wave length of the radiation used. Since wavelength of the yellow light is 5,990 A i.e. 5.99 x 10-7 m, it follows that the electron microscope will possess resolving power 10 times as large as that of the optical microscope.",
null,
"Question 34.\nThe wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10-15m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)\nAnswer:\nAs the quark structure of protons and neutrons appears a the length-scale of 10-15m, the electron beam used should be of the wavelength of this order i.e.\nλ= 10-15m\nTherefore, momentum of the electron beam,",
null,
"The electron having momentum of this order possesses speed comparable to the speed of light. Therefore, its energy can be found by using the relativistic formula for die total energy of the electron i.e.",
null,
"Question 35.\nFind the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.\nAnswer:\nMass of the He-atom,\n$$m=\\frac { atomic\\quad weight\\quad of\\quad He }{ Avogadro\\quad Number } \\frac { 4 }{ 6.02\\times { 10 }^{ 23 } }$$\n= 6.645 × 10-24 g = .6645 ×10-27 kg\nBy proceeding as in solved problem No.2.09,\nfind the de-Broglie wavelength of the He-atom at 27°C.\nIt can be obtained that de-Broglie wavelength of the He-atom,\nλ= 0.73 x 10-10 m = 0.73 Å\nIf V is volume of the gas and dj the mean separation between two atoms in the gas, then V = d3 N,\nWhere N is Avogadro’s number\nFrom the perfect gas equation",
null,
"i.e the separation between the two He-atoms in the gas is much larger than the Broglie wavelength of the atom.\n\nQuestion 36.\nCompute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10-10 [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishability has many fundamental implications which you will explore in more advanced Physics courses.]\nAnswer:\nIt cap be obtained that de-Broglie wavelength of the electron,\nλ – 62.3 x 10-10 m = 62.3 A\nMean separation between two electrons in the metal,",
null,
"i.e de-Broglie wavelength of the electron is much larger than the separtation between two electrons in the metal.",
null,
"Question 37.\nAnswer the following questions:\n\n(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (-1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?\n(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?\n(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?\n(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?\n(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:\n$$\\frac { h }{ \\lambda }$$\nBut while the value of x is physically significant, the value of v (and therefore, the value of the phase speed v X) has no physical significance. Why?\nAnswer:\n(a) Quarks having the fractional charges are thought to be confined within a proton or a neutron only. These quarks are bound by forces, which grow stronger, if they are tried to be pulled apart. Thus, the quarks always remain together. It follows that though fractional v charges exist in nature, the observable charges are always integral multiples of e.\n\n(b) The motion of an electron inside electric and magnetic fields is governed by\neV = $$\\frac { 1 }{ 2 }$$-mv2 = eV; eE = ma and B eV =$$\\frac { m{ v }^{ 2 } }{ r }$$ where the symbols have their usual meanings.\nIn all the three relations, e and m occur separately. Further, the value of a physical variable involved in any of these reations can be found by substituting the value of e/m.\n\n(c) The gases keep on getting ionised due to energetic rays [X-rays, Cosmic rays, etc.,] passing through them.\n\nHowever, at atmospheric pressure, the positive and negative ions are too close and recombine to form neutral gas atoms. Due to this, the gases behave as insulatiors at ordinary pressure.\n\nAt low pressures, the gas ions do not recombine as they are appreciably apart. Due to the presence of free ions, the gases start conducting at low pressures.\n\n(d) Work function of a metal is the minimum energy required to knock out an electron (from the conduction band). Since electrons are present in a continuous band of levels, different electrons require different amounts of energy to get out of the atom. For this reason electrons knocked off by a monochromatic radiation posses different energies.\n\n(e) Whereas energy E (= hv) of a particle is arbitrary to within an additive constant, its momentum p (= h/ λ) is not. So, for the matter wave associated with an electron the wavelength λ is physically significant and its frequency v, is not so likewise, the phase speed v λ, is also not physically significant.\n\n### 2nd PUC Physics Dual Nature of Radiation and Matter Additional Questions and Answers\n\nQuestion 1.\nAn electron and an alpha particle have the same de-broglie wavelength associated with them. How are their kinetic energies related to each other?\nAnswer:",
null,
"Question 2.\nIf the frequency of the incident radiation is equal to the threshold frequency, what will be the value of the stopping potential?\nAnswer:\nFrom Einstein’s photoelectric equation,\nhv = hv0 + eV0\nv = v0\nhv0=hv0 + eV0\nv0=o\n∴ Stopping potential = 0\n\nQuestion 3.\nIt is harder to remove a free electron from copper than from sodium. Which metal has greater work function? Which has higher threshed wavelength?\nAnswer:\nAs it is harder to remove an electron from copper than sodium, the work function of copper is greater",
null,
"As threshold wavelength is inversely proportional to the work function, its value will be more for sodium.\n\nQuestion 4.\nAn electromagnetic wave of wavelength X is incident on a photosensitive surface of negligible work function. If the phot- electrons emitted from this surface have de-Broglie wavelength λ1 prove that $$\\lambda =\\frac { 2mc }{ h } { \\lambda }_{ 1 }^{ 2 }$$\nAnswer:\nSince photosensitive surface of negligible work function, K.E of emitted electron = energy of incident photon",
null,
"Question 5.\nFind the momentum of an x-ray photon of wavelength.\nAnswer:\nLet p be the momentum of the x-ray photon of wavelength x, then energy of x-ray photon,",
null,
"Question 6.\nThe work function of a substance is 4eV. The longest wavelength of light that can cause photoelectric emission from this substance is approximately\n(A) 540 nm\n(B) 400 nm\n(C) 310 nm\n(D) 220 nm\nAnswer:\n(C) 310 nm",
null,
"Question 7.\nThe mass of a photon at rest is\n(A) zero\n(B) 1.67 x 10-35kg\n(C) 1 amu\n(D) 9 x 1031 kg\nAnswer:\n(A) zero\n\nQuestion 8.\nWhich one among the following shows particle nature of light?\n(A) Photoelectric effect\n(B) Interference\n(C) Refraction\n(D) Polarisation\nAnswer:\n(A) Photoelectric effect",
null,
"Question 9.\nPlanck’s constant has the dimensions of\n(A) linear momentum\n(B) angular momentum\n(C) energy\n(D) power\nAnswer:",
null,
"Question 10.\nThe de-Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it\n(A) $$\\frac { h }{ mv }$$\n(B) hmv\n(C) $$\\frac { mh }{ v }$$\n(D) $$\\frac { m }{ hv }$$\nAnswer:",
null,
""
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https://meanderingpi.com/microbit-analog-referencing/ | [
"# micro:bit Internal Analog Reference",
null,
"For the last few days I have been playing with the micro:bits builtin capability to measure analog voltages on P0, P1 & P2 (you can set up some of the other IO lines as analog inputs but this is a bit of an involved process and best avoided if possible).\n\n** WARNING ** – YOU MUST ENSURE THAT ANY VOLTAGE APPLIED TO THE MICRO:BIT DOES NOT EXCEED THE SUPPLY RAIL [TYPICALLY 3V] DOING SO WILL RESULT IN DAMAGE TO YOUR MICRO:BIT\n\nBy default the analog converter in the micro:bit is referenced to the micro:bit supply rail and is configured for 10 bit operation. Hence the following code;\n\n```from microbit import *\nprint(a)\ndisplay.scroll(str(a))```\n\nwill return 1023 if you connect the 3V & P0 pads together, irrespective of battery condition of if you power via the micro USB connector. For many applications this is fine and using the micro:bits 3V supply as the source voltage for any analog experiments is the best way to minimize risk of damage the the micro:bit.\n\nHowever if you want to measure a voltage not derived from the micro:bits supply things can become a bit more problematic. It’s fine if you can power the micro:bit from a constant source – for instance a phone mains power adapter, but not so good if you are running from a battery. This because as the ADC (analog to digit converter) reference voltage will drop as the battery discharges resulting in a change to the count value returned. For instance assuming a new set of batteries provides 3V, a 1.5V source on P0 would return 512, but if the battery voltage drops to 2.5V the reading would change to 614 counts. To get around this the ADC in the micro:bit can be configured to reference an internal fixed 1.2V voltage reference. This means the count value will not vary with a changing supply voltage.",
null,
"In it’s basic setup this would mean a voltage of 1.2V on P0 would give a result of 1023. However 0-1.2V is quite a small range so the ADC circuit includes a configurable ‘pre-scaler’ to help us. Configuring this to 3:1 we get full scale ADC range of 0-3.6V. It is important to note however that the rule of keeping the maximum voltage applied to a micor:bit input pin at or below the supply rail still applies.\n\nThe following code will setup the ADC to use the internal reference on P0 together with a 3:1 pre-scaler.\n\n```Code to demo using the micro:bit internal ADC reference\n# based on material and help from Carlos Pereira Atencio https://gist.github.com/microbit-carlos/\n# David Saul 2017. Released to the public domain.\n\nfrom microbit import *\n\n# functions used to modify ADC setup\n@micropython.asm_thumb\nldr(r0, [r0, 0])\n\n@micropython.asm_thumb\ndef reg_bit_clear(r0, r1):\nldr(r2, [r0, 0])\nbic(r2, r1)\nstr(r2, [r0, 0])\n\n# change ADC config for pin0\n\n# mbed configured: 10bit, input 1/3, ref 1/3 vdd, enable ADC in AIN4-P0.03-pin0\n#Set ref voltage to internal ref 1.2V (band gap), set bits 5-6 to 0x00\n\n#variable definition\nvt = 0 # raw / scaled analog value\nvf = \"0000\" # measured voltage as a formatted string\ncal = 352 # calibration variable\n# cal is calculated as (Vin / raw count)*10000\nwhile True\nprint ('raw count = ',vt)\n\n# the following lines of code scale the measured value and derive a\n# string (vf) that equates to it - there are simpler ways to do this\n# but this method avoids the use of floating variables which are very\n# inefficient on cards like the micro:bit\n\nvt = int((vt * cal)/1000)\nvs = str(vt)\nif vt > 1000:\nvf = vs[0:2]+'.'+vs[2:3]\nelif vt > 100 :\nvf = vs[0:1]+'.'+vs[1:3]\nelif vt > 10 :\nvf = '0.'+vs\nelif vt > 1 :\nvf = '0.0'+vs\nelse :\nvf = '0.000'\n\nvf = vf+\"V\"\nprint ('Scaled', vf)\ndisplay.scroll(vf)\n\nsleep(2000)```"
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https://www.gsp.com/cgi-bin/man.cgi?section=3&topic=zunmql.f | [
"",
null,
"",
null,
"Quick Navigator\n\n Search Site Miscellaneous Server Agreement Year 2038 Credits",
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"",
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"",
null,
"zunmql.f(3) LAPACK zunmql.f(3)\n\nzunmql.f -\n\n# SYNOPSIS\n\n## Functions/Subroutines\n\nsubroutine zunmql (SIDE, TRANS, M, N, K, A, LDA, TAU, C, LDC, WORK, LWORK, INFO)\n\nZUNMQL\n\n# Function/Subroutine Documentation\n\n## subroutine zunmql (characterSIDE, characterTRANS, integerM, integerN, integerK, complex*16, dimension( lda, * )A, integerLDA, complex*16, dimension( * )TAU, complex*16, dimension( ldc, * )C, integerLDC, complex*16, dimension( * )WORK, integerLWORK, integerINFO)\n\nZUNMQL\nPurpose:\n``` ZUNMQL overwrites the general complex M-by-N matrix C with\n\nSIDE = 'L' SIDE = 'R'\nTRANS = 'N': Q * C C * Q\nTRANS = 'C': Q**H * C C * Q**H\n\nwhere Q is a complex unitary matrix defined as the product of k\nelementary reflectors\n\nQ = H(k) . . . H(2) H(1)\n\nas returned by ZGEQLF. Q is of order M if SIDE = 'L' and of order N\nif SIDE = 'R'.\n```\nParameters:\nSIDE\n``` SIDE is CHARACTER*1\n= 'L': apply Q or Q**H from the Left;\n= 'R': apply Q or Q**H from the Right.\n```\nTRANS\n``` TRANS is CHARACTER*1\n= 'N': No transpose, apply Q;\n= 'C': Transpose, apply Q**H.\n```\nM\n``` M is INTEGER\nThe number of rows of the matrix C. M >= 0.\n```\nN\n``` N is INTEGER\nThe number of columns of the matrix C. N >= 0.\n```\nK\n``` K is INTEGER\nThe number of elementary reflectors whose product defines\nthe matrix Q.\nIf SIDE = 'L', M >= K >= 0;\nif SIDE = 'R', N >= K >= 0.\n```\nA\n``` A is COMPLEX*16 array, dimension (LDA,K)\nThe i-th column must contain the vector which defines the\nelementary reflector H(i), for i = 1,2,...,k, as returned by\nZGEQLF in the last k columns of its array argument A.\n```\nLDA\n``` LDA is INTEGER\nThe leading dimension of the array A.\nIf SIDE = 'L', LDA >= max(1,M);\nif SIDE = 'R', LDA >= max(1,N).\n```\nTAU\n``` TAU is COMPLEX*16 array, dimension (K)\nTAU(i) must contain the scalar factor of the elementary\nreflector H(i), as returned by ZGEQLF.\n```\nC\n``` C is COMPLEX*16 array, dimension (LDC,N)\nOn entry, the M-by-N matrix C.\nOn exit, C is overwritten by Q*C or Q**H*C or C*Q**H or C*Q.\n```\nLDC\n``` LDC is INTEGER\nThe leading dimension of the array C. LDC >= max(1,M).\n```\nWORK\n``` WORK is COMPLEX*16 array, dimension (MAX(1,LWORK))\nOn exit, if INFO = 0, WORK(1) returns the optimal LWORK.\n```\nLWORK\n``` LWORK is INTEGER\nThe dimension of the array WORK.\nIf SIDE = 'L', LWORK >= max(1,N);\nif SIDE = 'R', LWORK >= max(1,M).\nFor optimum performance LWORK >= N*NB if SIDE = 'L', and\nLWORK >= M*NB if SIDE = 'R', where NB is the optimal\nblocksize.\n\nIf LWORK = -1, then a workspace query is assumed; the routine\nonly calculates the optimal size of the WORK array, returns\nthis value as the first entry of the WORK array, and no error\nmessage related to LWORK is issued by XERBLA.\n```\nINFO\n``` INFO is INTEGER\n= 0: successful exit\n< 0: if INFO = -i, the i-th argument had an illegal value\n```\nAuthor:\nUniv. of Tennessee\nUniv. of California Berkeley\nNAG Ltd.\nDate:\nNovember 2011\nDefinition at line 169 of file zunmql.f.\n\n# Author\n\nGenerated automatically by Doxygen for LAPACK from the source code.\n Sat Nov 16 2013 Version 3.4.2\n\nSearch for or go to Top of page | Section 3 | Main Index",
null,
"Visit the GSP FreeBSD Man Page Interface.\nOutput converted with ManDoc."
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https://tex.stackexchange.com/questions/109856/how-to-format-maxwell-equations | [
"# How to format Maxwell equations?\n\nIs there a variant of split environment from amsmath package which would allow for two aligned equations in a row? I want to format two pairs of Maxwell equations in two rows and put single equation number between the rows. Unfortunetely, split allows only one alignment tab & in a row, and I need three as in the following example:\n\n\\documentclass{article}\n\\usepackage{amsmath}\n\\DeclareMathOperator{\\Div}{div}\n\\DeclareMathOperator{\\Rot}{rot}\n\\begin{document}\n\\begin{align}\n\\label{18.1:1}\n%\\begin{split}\n\\Rot\\vec{E} &=-\\frac{1}{c}\\parder{\\vec{B}}{t}\n,\n&\n\\Div\\vec{B} &=0,\n\\\\\n\\label{18.1:2}\n\\Rot\\vec{B} &=\\frac{1}{c}\\parder{\\vec{E}}{t}\n+\\frac{4\\pi}{c}\\,\\vec{j}\n,\n&\n\\Div\\vec{E} &=4\\pi\\rho_{\\varepsilon }\n%\\end{split}\n\\end{align}\n\\end{document}\n\n\nHere every row has its own number.\n\n• Your code is not compilable: \\parder is nowhere defined. Also, you may want to use a more descriptive name for your labels, and it's best to precede punctuation marks such commas and full stops by a \\,. Apr 21, 2013 at 10:02\n• The \\parder is defined as: \\newcommand{\\parder}{\\frac{\\partial {#1}}{\\partial {#2}}}. Apr 21, 2013 at 10:18\n• @m0nhawk, thank you, indeed, i've missed the definition of parder. Apr 22, 2013 at 6:25\n• @Jubobs, Sorry, I didn't ask how to format punctuation marks or how to choose names for labels. Apr 23, 2013 at 4:03\n• Sorry, I didn't mean to offend you. That was just a side note that I thought could be useful. Apr 23, 2013 at 8:27\n\nI think you are after aligned:",
null,
"\\documentclass{article}\n\n\\usepackage{amsmath}\n\n\\DeclareMathOperator{\\Div}{div}\n\\DeclareMathOperator{\\Rot}{rot}\n\\newcommand{\\parder}{\\frac{\\partial {#1}}{\\partial {#2}}}\n\n\\begin{document}\n\n\\begin{align}\n\\label{18.1:1}\n\\begin{aligned}\n\\Rot\\vec{E} &=-\\frac{1}{c}\\parder{\\vec{B}}{t},&\n\\Div\\vec{B} &=0,\n\\\\\n\\Rot\\vec{B} &=\\frac{1}{c}\\parder{\\vec{E}}{t}\n+\\frac{4\\pi}{c}\\,\\vec{j},&\n\\Div\\vec{E} &=4\\pi\\rho_{\\varepsilon}.\n\\end{aligned}\n\\end{align}\n\n\\end{document}\n\n\nJust use split:\n\n\\documentclass{article}\n\\usepackage{amsmath}\n\n\\DeclareMathOperator{\\Div}{div}\n\\DeclareMathOperator{\\Rot}{rot}\n\n\\newcommand\\parder{\\frac{\\partial #1}{\\partial #2}}\n\n\\begin{document}\n\\begin{align}\n\\label{18.1:1}\n\\begin{split}\n\\Rot\\vec{E} &=-\\frac{1}{c}\\parder{\\vec{B}}{t},\n\\\\\n\\Div\\vec{B} &=0,\n\\end{split}\n\\\\[2ex]\n\\label{18.1:2}\n\\begin{split}\n\\Rot\\vec{B} &=\\frac{1}{c}\\parder{\\vec{E}}{t}+\\frac{4\\pi}{c}\\,\\vec{j},\n\\\\\n\\Div\\vec{E} &=4\\pi\\rho_{\\varepsilon }\n\\end{split}\n\\end{align}\n\\end{document}",
null,
"• Interesting solution. Apr 22, 2013 at 6:26"
] | [
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https://math.stackexchange.com/questions/327201/are-there-infinite-sets-of-axioms | [
"# Are there infinite sets of axioms?\n\nI'm reading Behnke's Fundamentals of mathematics:\n\nIf the number of axioms is finite, we can reduce the concept of a consequence to that of a tautology.\n\nI got curious on this: Are there infinite sets of axioms? The only thing I could think about is the possible existence of unknown axioms and perhaps some belief that this number of axioms is infinite.\n\n• Might be a useful post <math.stackexchange.com/questions/309147/…> – MITjanitor Mar 11 '13 at 5:53\n• Due to the presence of axiom schemata, $\\sf ZFC$ has an infinite number of axioms. However, more can be said, namely, that it is not finitely axiomatizable. This means that it is impossible to replace the infinitely many axioms of $\\sf ZFC$ by a finite list. – Haskell Curry Mar 11 '13 at 10:39\n• @HaskellCurry Propositional calculus can get founded with axioms and a rule of substitution. Even in that system though we can always get more axioms if we have countably infinite variables. – Doug Spoonwood Oct 17 '14 at 0:29\n\n## 7 Answers\n\nIn first order Peano axioms the principal of mathematical induction is not one axiom, but a \"template\" called an axiom scheme. For every possible expression (or \"predicate\") with a free variable, $P(n)$, we have the axiom:\n\n$$(P(0) \\land \\left(\\forall n: P(n)\\implies P(n+1)\\right))\\implies \\\\\\forall n: P(n)$$\n\nSo, if $P(x)$ is the predicate, $x\\cdot 0 = 1$ then we'd have the messy axiom:\n\n$$(0\\cdot 0=1 \\land \\left(\\forall n: n\\cdot 0 =1\\implies (n+1)\\cdot 0=1\\right))\\implies \\\\\\forall n: n\\cdot 0 = 1$$\n\nOur inclination is to think of this as a single axiom when preceded by \"$\\forall P$\", but in first-order theory, there is only one \"type.\" In first-order number theory, that type is \"natural number.\" So there is no room in the language for the concept of $\\forall P$. In second order theory, we can say $\\forall P$.\n\nIn set theory, you have a similar rule, the \"axiom of specification\" which lets you construct a set from any predicate, $P(x,y)$, with two free variables:\n\n$$\\forall S:\\exists T: \\forall x: (x\\in T\\iff (x\\in S\\land P(x,S)))$$\n\n(The axiom lets you do more, but this is a simple case.)\n\nwhich essentially means that there exists a set:\n\n$$\\{x\\in S: P(x,S)\\}$$\n\nAgain, there is no such object inside set theory as a \"predicate.\"\n\nFor most human axiom systems, even when the axioms are infinite, we have a level of verifiability. We usually desire an ability to verify a proof using mechanistic means, and therefore, given any step in a proof, we desire the ability to verify the step in a finite amount of time.\n\nMany important theories, most significantly first-order Peano arithmetic, and ZFC, the most commonly used axiomatic set theory, have an infinite number of axioms. So does the theory of algebraically closed fields.\n\n• Specifically, in Peano, mathematical induction really describes and infinite set of axioms. – Thomas Andrews Mar 11 '13 at 5:54\n• I remember of this answer I thought that there's an axiom for every number, but I'm not sure. – Billy Rubina Mar 11 '13 at 5:55\n• I don't think that's true, @GustavoBandeira. – Thomas Andrews Mar 11 '13 at 6:02\n• Among the commonly encountered, there are also the theories of vector spaces over a given (infinite) field, of fields of characteristic zero... – tomasz Mar 11 '13 at 6:18\n• More interestingly, $\\sf ZFC$ is not finitely axiomatizable, as first demonstrated by Richard Montague using the Reflection Principle. – Haskell Curry Mar 11 '13 at 10:35\n\nAre there infinite sets of axioms? Yes!\n\nAs a trivial (and somewhat meaningless) example, consider the first-order language whose only non-logical symbols are the constant symbol $0$ and the unary function symbol $S$. Then the following would comprise an infinite set of axioms:\n\n• $0 \\neq S(0)$;\n• $S(0) \\neq S(S(0))$;\n• $S(S(0)) \\neq S(S(S(0)))$;\n$\\vdots$\n• $S ( S ( S ( \\cdots ( 0 ) \\cdots ) ) ) \\neq S ( S ( \\cdots ( 0 ) \\cdots ) )$\n$\\vdots$\n\nFor a more meaningful example we have, as others have pointed out, Peano Arithmetic. More than just being an infinite list of axioms, this theory necessarily has an infinite set of axioms. By a result of C. Ryll-Nardzewski, (assuming it is consistent) Peano Arithmetic is not finitely axiomatisable: there is no finite set of axioms (in the same language) that has the same collection of formal theorems. (A similar result holds for ZF(C).)\n\nPerhaps surprisingly even the classical (Łukasiewicz's) axiomatization of propositional logic has an infinite number of axioms. The axioms are all substitution instances of\n\n• $(p \\to (q \\to p))$\n• $((p \\to (q \\to r)) \\to ((p \\to q) \\to (p \\to r)))$\n• $((\\neg p \\to \\neg q) \\to (q \\to p))$\n\nso we have an infinite number of axioms.\n\nUsually the important thing is not if the set of axioms is finite or infinite, but if it is decidable. We can only verify proofs in theories with decidable sets of axioms. If a set of axioms is undecidable, we can't verify a proof, because we can't tell if a formula appearing in it is an axiom or not. (If a set of axioms is only semi-decidable, we're able to verify correct proofs, but we're not able to refute incorrect ones.)\n\nFor example, if I construct a theory with the set of axioms given as\n\n$T(\\pi)$ is an axiom if $\\pi$ is a representation of a terminating program.\n\nThen I can \"prove\" that some program $p$ is terminating in a one-line proof by simply stating $T(p)$. But of course such theory has no real use because nobody can verify such a proof.\n\n• Your semi-decidable set of axioms, however, proves the same theorems as the decidable one where the axioms are those $$\\underbrace{x=x\\land \\cdots\\land x=x}_{n\\text{ conjuncts}}\\land T(m)$$ where $m$ is the representation of a program that terminates in $n$ steps. (Every semi-dedicable set of axioms is equivalent to a decidable one in essentially this way). – Henning Makholm Mar 12 '13 at 21:58\n• You can have a recursively enumerable axiom set - it just requires all proofs to include the amount of time before enumerating the axiom. The above is recursively enumerable, so we can prove things in it. The axiom system $S(\\pi)$ meaning that $\\pi$ does not terminate is not recursively enumerable, and therefore we can't write checkable proofs in that theory. – Thomas Andrews Aug 11 '13 at 18:07\n• Lukasiewicz's axiomitization didn't actually use axiom schema. It had 3 axioms and the rule of uniform substitution, which consists of a rule of inference. So actually, the system you've referenced only has 3 axioms. However, even with each axiom as finite, the possible axiomizations of classical propositional logic are infinite since (p→(q→p)) can get replaced by any theorem of the system {(p→(q→p))} under condensed detachment, and since {(p→(q→p))} under condensed detachment has an infinity of theorems. For example, (p→(q→p)) could get replaced by (a→(b→(c→(d→(p→(q→p))))). – Doug Spoonwood May 7 '14 at 2:19\n\nAn example of an infinite set of axioms used for practical use is the foundation of non-standard analysis.\n\nCreate a first-order language whose symbols include:\n\n• Every element of $\\mathbb{R}$\n• Every element of $\\mathcal{P}(\\mathbb{R})$\n• Every element of $\\mathcal{P}(\\mathcal{P}(\\mathbb{R}))$\n• :\n\nNow select a set of axioms to be the set of all true statements in this language.\n\nThis language and set of axioms is enough to do lots of things; e.g. any theorem you've learned in calculus can be stated in this language and is automatically a theorem.\n\nBy general properties of first-order logic, this set of axioms is consistent and thus there exists a non-standard model! By construction, every theorem you know about 'standard' things is automatically true for non-standard things as well! In this formulation, the transfer principle from non-standard analysis -- every 'standard' object has a nonstandard analog, and every statement of 'standard' analysis is true if and only if its translation to the non-standard model is true -- becomes an utter triviality.\n\nWhile one can surely attempt to select a smaller set of axioms, there is no actual reason to; for the specific application we have to apply logic in this instance, it is by far easier and more convenient to take all of them.\n\nAs others have noted, first order PA has an infinite number of axioms. But it is worth stressing that this is not just a superficial feature of the usual mode of presenting the induction axioms via a schema or template. It is provable that there is no finitely axiomatized theory in the language of PA whose theorems coincide with those of PA.\n\n• Someone did mention this in another answer. ;-) (BTW: I just pre-ordered a coy of your Introduction to Gödel's Theorems text earlier today. Looking forward to thumbing through it!) – user642796 Mar 11 '13 at 9:25\n• Ooops, sorry @ArthurFischer: very careless reading by me. Enjoy Gödel! – Peter Smith Mar 11 '13 at 14:21\n\nJust to make the matter maybe a bit more clear giving another example:\n\nSimilar to what André Nicolas stated, $ACF_0$ (the theory of algebraically closed fields of charateristic $0$) is axiomatized by $ACF$(the theory of algebraically closed fields) + $\\{\\neg\\phi_p$ : $p$ is prime$\\}$, where $\\phi_p$ says \"I have characteristic $p$.\" $ACF$ is already not finitely axiomatizable itself but I am sure observing that $\\{\\neg\\phi_p$ : $p$ is prime$\\}$ is an infinite set of axioms is way more obvious."
] | [
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https://isabelle.in.tum.de/repos/isabelle/file/d8205bb279a7/doc-src/Ref/substitution.tex | [
"doc-src/Ref/substitution.tex\n author lcp Wed, 10 Nov 1993 05:00:57 +0100 changeset 104 d8205bb279a7 child 148 67d046de093e permissions -rw-r--r--\nInitial revision\n\n%% $Id$\n\\chapter{Substitution Tactics} \\label{substitution}\n\\index{substitution|(}\nReplacing equals by equals is a basic form of reasoning. Isabelle supports\nseveral kinds of equality reasoning. {\\bf Substitution} means to replace\nfree occurrences of~$t$ by~$u$ in a subgoal. This is easily done, given an\nequality $t=u$, provided the logic possesses the appropriate rule ---\nunless you want to substitute even in the assumptions. The tactic\n\\ttindex{hyp_subst_tac} performs substitution in the assumptions, but it\nworks via object-level implication, and therefore must be specially set up\nfor each suitable object-logic.\n\nSubstitution should not be confused with object-level {\\bf rewriting}.\nGiven equalities of the form $t=u$, rewriting replaces instances of~$t$ by\ncorresponding instances of~$u$, and continues until it reaches a normal\nform. Substitution handles one-off' replacements by particular\nequalities, while rewriting handles general equalities.\nChapter~\\ref{simp-chap} discusses Isabelle's rewriting tactics.\n\n\\section{Simple substitution}\n\\index{substitution!simple}\nMany logics include a substitution rule of the form\\indexbold{*subst}\n$$\\List{\\Var{a}=\\Var{b}; \\Var{P}(\\Var{a})} \\Imp \\Var{P}(\\Var{b}) \\eqno(subst)$$\nIn backward proof, this may seem difficult to use: the conclusion\n$\\Var{P}(\\Var{b})$ admits far too many unifiers. But, if the theorem {\\tt\neqth} asserts $t=u$, then \\hbox{\\tt eqth RS subst} is the derived rule\n$\\Var{P}(t) \\Imp \\Var{P}(u).$\nProvided $u$ is not an unknown, resolution with this rule is\nwell-behaved.\\footnote{Unifying $\\Var{P}(u)$ with a formula~$Q$\nexpresses~$Q$ in terms of its dependence upon~$u$. There are still $2^k$\nunifiers, if $Q$ has $k$ occurrences of~$u$, but Isabelle ensures that\nthe first unifier includes all the occurrences.} To replace $u$ by~$t$ in\nsubgoal~$i$, use\n\\begin{ttbox}\nresolve_tac [eqth RS subst] $$i$$ {\\it.}\n\\end{ttbox}\nTo replace $t$ by~$u$ in\nsubgoal~$i$, use\n\\begin{ttbox}\nresolve_tac [eqth RS ssubst] $$i$$ {\\it,}\n\\end{ttbox}\nwhere \\ttindexbold{ssubst} is the swapped' substitution rule\n$$\\List{\\Var{a}=\\Var{b}; \\Var{P}(\\Var{b})} \\Imp \\Var{P}(\\Var{a}). \\eqno(ssubst)$$\nIf \\ttindex{sym} denotes the symmetry rule\n$$\\Var{a}=\\Var{b}\\Imp\\Var{b}=\\Var{a}$$, then {\\tt ssubst} is just\n\\hbox{\\tt sym RS subst}. Many logics with equality include the rules {\\tt\nsubst} and {\\tt ssubst}, as well as {\\tt refl}, {\\tt sym} and {\\tt trans}\n(for the usual equality laws).\n\nElim-resolution is well-behaved with assumptions of the form $t=u$.\nTo replace $u$ by~$t$ or $t$ by~$u$ in subgoal~$i$, use\n\\begin{ttbox}\neresolve_tac [subst] $$i$$ {\\it or} eresolve_tac [ssubst] $$i$$ {\\it.}\n\\end{ttbox}\n\n\\section{Substitution in the hypotheses}\n\\index{substitution!in hypotheses}\nSubstitution rules, like other rules of natural deduction, do not affect\nthe assumptions. This can be inconvenient. Consider proving the subgoal\n$\\List{c=a; c=b} \\Imp a=b.$\nCalling \\hbox{\\tt eresolve_tac [ssubst] $$i$$} simply discards the\nassumption~$c=a$, since $c$ does not occur in~$a=b$. Of course, we can\nwork out a solution. First apply \\hbox{\\tt eresolve_tac [subst] $$i$$},\nreplacing~$a$ by~$c$:\n$\\List{c=b} \\Imp c=b$\nEquality reasoning can be difficult, but this trivial proof requires\nnothing more sophisticated than substitution in the assumptions.\nObject-logics that include the rule~$(subst)$ provide a tactic for this\npurpose:\n\\begin{ttbox}\nhyp_subst_tac : int -> tactic\n\\end{ttbox}\n\\begin{description}\n\\item[\\ttindexbold{hyp_subst_tac} {\\it i}]\nselects an equality assumption of the form $t=u$ or $u=t$, where $t$ is a\nfree variable or parameter. Deleting this assumption, it replaces $t$\nby~$u$ throughout subgoal~$i$, including the other assumptions.\n\\end{description}\nThe term being replaced must be a free variable or parameter. Substitution\nfor constants is usually unhelpful, since they may appear in other\ntheorems. For instance, the best way to use the assumption $0=1$ is to\ncontradict a theorem that states $0\\not=1$, rather than to replace 0 by~1\nin the subgoal!\n\nReplacing a free variable causes similar problems if they appear in the\npremises of a rule being derived --- the substitution affects object-level\nassumptions, not meta-level assumptions. For instance, replacing~$a$\nby~$b$ could make the premise~$P(a)$ worthless. To avoid this problem, call\n\\ttindex{cut_facts_tac} to insert the atomic premises as object-level\nassumptions.\n\n\\section{Setting up {\\tt hyp_subst_tac}}\nMany Isabelle object-logics, such as {\\tt FOL}, {\\tt HOL} and their\ndescendants, come with {\\tt hyp_subst_tac} already defined. A few others,\nsuch as {\\tt CTT}, do not support this tactic because they lack the\nrule~$(subst)$. When defining a new logic that includes a substitution\nrule and implication, you must set up {\\tt hyp_subst_tac} yourself. It\nis packaged as the \\ML{} functor \\ttindex{HypsubstFun}, which takes the\nargument signature~\\ttindexbold{HYPSUBST_DATA}:\n\\begin{ttbox}\nsignature HYPSUBST_DATA =\nsig\nval subst : thm\nval sym : thm\nval rev_cut_eq : thm\nval imp_intr : thm\nval rev_mp : thm\nval dest_eq : term -> term*term\nend;\n\\end{ttbox}\nThus, the functor requires the following items:\n\\begin{description}\n\\item[\\ttindexbold{subst}] should be the substitution rule\n$\\List{\\Var{a}=\\Var{b};\\; \\Var{P}(\\Var{a})} \\Imp \\Var{P}(\\Var{b})$.\n\n\\item[\\ttindexbold{sym}] should be the symmetry rule\n$\\Var{a}=\\Var{b}\\Imp\\Var{b}=\\Var{a}$.\n\n\\item[\\ttindexbold{rev_cut_eq}] should have the form\n$\\List{\\Var{a}=\\Var{b};\\; \\Var{a}=\\Var{b}\\Imp\\Var{R}} \\Imp \\Var{R}$.\n\n\\item[\\ttindexbold{imp_intr}] should be the implies introduction\nrule $(\\Var{P}\\Imp\\Var{Q})\\Imp \\Var{P}\\imp\\Var{Q}$.\n\n\\item[\\ttindexbold{rev_mp}] should be the reversed'' implies elimination\nrule $\\List{\\Var{P}; \\;\\Var{P}\\imp\\Var{Q}} \\Imp \\Var{Q}$.\n\n\\item[\\ttindexbold{dest_eq}] should return the pair~$(t,u)$ when\napplied to the \\ML{} term that represents~$t=u$. For other terms, it\nshould raise exception~\\ttindex{Match}.\n\\end{description}\nThe functor resides in {\\tt Provers/hypsubst.ML} on the Isabelle\ndistribution directory. It is not sensitive to the precise formalization\nof the object-logic. It is not concerned with the names of the equality\nand implication symbols, or the types of formula and terms. Coding the\nfunction {\\tt dest_eq} requires knowledge of Isabelle's representation of\nterms. For {\\tt FOL} it is defined by\n\\begin{ttbox}\nfun dest_eq (Const(\"Trueprop\",_) $(Const(\"op =\",_)$t$u)) = (t,u); \\end{ttbox} Here {\\tt Trueprop} is the coercion from type'~$o$to type~$prop$, while \\hbox{\\tt op =} is the internal name of the infix operator~{\\tt=}. Pattern-matching expresses the function concisely, using wildcards~({\\tt_}) to hide the types. Given a subgoal of the form $\\List{P@1; \\cdots ; t=u; \\cdots ; P@n} \\Imp Q,$ \\ttindexbold{hyp_subst_tac} locates a suitable equality assumption and moves it to the last position using elim-resolution on {\\tt rev_cut_eq} (possibly re-orienting it using~{\\tt sym}): $\\List{P@1; \\cdots ; P@n; t=u} \\Imp Q$ Using$n$calls of \\hbox{\\tt eresolve_tac [rev_mp]}, it creates the subgoal $\\List{t=u} \\Imp P@1\\imp \\cdots \\imp P@n \\imp Q$ By \\hbox{\\tt eresolve_tac [ssubst]}, it replaces~$t$by~$u$throughout: $P'@1\\imp \\cdots \\imp P'@n \\imp Q'$ Finally, using$n$calls of \\hbox{\\tt resolve_tac [imp_intr]}, it restores$P'@1$, \\ldots,$P'@n\\$ as assumptions:\n$\\List{P'@n; \\cdots ; P'@1} \\Imp Q'$\n\n\\index{substitution|)}"
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http://docs.h2o.ai/h2o/latest-stable/h2o-docs/data-science/algo-params/tweedie_variance_power.html | [
"# tweedie_variance_power¶\n\n• Available in: GLM, GAM\n\n• Hyperparameter: yes\n\n## Description¶\n\nWhen family=tweedie, this option can be used to specify the power for the tweedie variance. This option defaults to 0.\n\nTweedie distributions are a family of distributions that include gamma, normal, Poisson and their combinations. This distribution is especially useful for modeling positive continuous variables with exact zeros. The variance of the Tweedie distribution is proportional to the $$p$$-th power of the mean $$var(y_i) = \\phi\\mu{^p_i}$$.\n\nThe Tweedie distribution is parametrized by variance power $$p$$. It is defined for all $$p$$ values except in the (0,1) interval and has the following distributions as special cases:\n\n• $$p = 0$$: Normal\n\n• $$p = 1$$: Poisson\n\n• $$p \\in (1,2)$$: Compound Poisson, non-negative with mass at zero\n\n• $$p = 2$$: Gamma\n\n• $$p = 3$$: Gaussian\n\n• $$p > 2$$: Stable, with support on the positive reals\n\nThe following table shows the acceptable relationships between family functions, tweedie variance powers, and tweedie link powers.\n\nFamily Function\n\nTweedie Variance Power\n\nPoisson\n\n1\n\n0, 1-vpow, 1\n\nGamma\n\n2\n\n0, 1-vpow, 2\n\nGaussian\n\n3\n\n1, 1-vpow\n\n## Example¶\n\nlibrary(h2o)\nh2o.init()\n\n# import the auto dataset:\n# this dataset looks at features of motor insurance policies and predicts the aggregate claim loss\n# the original dataset can be found at https://cran.r-project.org/web/packages/HDtweedie/HDtweedie.pdf\nauto <- h2o.importFile(\"https://s3.amazonaws.com/h2o-public-test-data/smalldata/glm_test/auto.csv\")\n\n# set the predictor names and the response column name\npredictors <- colnames(auto)[-1]\n# The response is aggregate claim loss (in $1000s) response <- \"y\" # split into train and validation sets auto_splits <- h2o.splitFrame(data = auto, ratios = 0.8) train <- auto_splits[] valid <- auto_splits[] # try using the tweedie_variance_power parameter: # train your model, where you specify tweedie_variance_power auto_glm <- h2o.glm(x = predictors, y = response, training_frame = train, validation_frame = valid, family = 'tweedie', tweedie_variance_power = 1) # print the mse for validation set print(h2o.mse(auto_glm, valid = TRUE)) # grid over tweedie_variance_power # select the values for tweedie_variance_power to grid over hyper_params <- list( tweedie_variance_power = c(0, 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3, 5, 7) ) # this example uses cartesian grid search because the search space is small # and we want to see the performance of all models. For a larger search space use # random grid search instead: {'strategy': \"RandomDiscrete\"} # build grid search with previously selected hyperparameters grid <- h2o.grid(x = predictors, y = response, training_frame = train, validation_frame = valid, family = 'tweedie', algorithm = \"glm\", grid_id = \"auto_grid\", hyper_params = hyper_params, search_criteria = list(strategy = \"Cartesian\")) # Sort the grid models by mse sorted_grid <- h2o.getGrid(\"auto_grid\", sort_by = \"mse\", decreasing = FALSE) sorted_grid # print the mse for the validation data print(h2o.mse(auto_glm, valid = TRUE)) import h2o from h2o.estimators.glm import H2OGeneralizedLinearEstimator h2o.init() # import the auto dataset: # this dataset looks at features of motor insurance policies and predicts the aggregate claim loss # the original dataset can be found at https://cran.r-project.org/web/packages/HDtweedie/HDtweedie.pdf auto = h2o.import_file(\"https://s3.amazonaws.com/h2o-public-test-data/smalldata/glm_test/auto.csv\") # set the predictor names and the response column name predictors = auto.names predictors.remove('y') # The response is aggregate claim loss (in$1000s)\nresponse = \"y\"\n\n# split into train and validation sets\ntrain, valid = auto.split_frame(ratios = [.8])\n\n# try using the tweedie_variance_power parameter:\n# initialize the estimator then train the model\nauto_glm = H2OGeneralizedLinearEstimator(family = 'tweedie', tweedie_variance_power = 1)\nauto_glm.train(x = predictors, y = response, training_frame = train, validation_frame = valid)\n\n# print the mse for the validation data\nprint(auto_glm.mse(valid=True))\n\n# grid over tweedie_variance_power\n# import Grid Search\nfrom h2o.grid.grid_search import H2OGridSearch\n\n# select the values for tweedie_variance_power to grid over\nhyper_params = {'tweedie_variance_power': [0, 1, 1.1, 1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9,2,\n2.1, 2.2,2.3,2.4,2.5,2.6,2.7,2.8,2.9,3, 5, 7]}\n\n# this example uses cartesian grid search because the search space is small\n# and we want to see the performance of all models. For a larger search space use\n# random grid search instead: {'strategy': \"RandomDiscrete\"}\n# initialize the GLM estimator\nauto_glm_2 = H2OGeneralizedLinearEstimator(family = 'tweedie')\n\n# build grid search with previously made GLM and hyperparameters\ngrid = H2OGridSearch(model = auto_glm_2, hyper_params = hyper_params,\nsearch_criteria = {'strategy': \"Cartesian\"})\n\n# train using the grid\ngrid.train(x = predictors, y = response, training_frame = train, validation_frame = valid)\n\n# sort the grid models by mse\nsorted_grid = grid.get_grid(sort_by='mse', decreasing=False)\nprint(sorted_grid)"
] | [
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http://eu.swi-prolog.org/pack/file_details/pljulia/prolog/julia.pl | [
"Did you know ... Search Documentation:",
null,
"Pack pljulia -- prolog/julia.pl\n\nWhen loaded, this module starts an embedded instance of Julia. Expressions may be executed (either discarding the return value or printing it in julia) or evaluated, unifying certain Julia return types with Prolog terms.\n\nSee `library(dcg/julia)` for description of valid expression terms.\n\nThe type `val` denotes values that can be exchanged with Julia by direct conversion, rather than via strings. It is a union:\n\n```val == integer | float | string | bool | symbol | void | arr_int64(_) | arr_float64(_)\n| tuple() | tuple(A) | tuple(A,B) | tuple(A,B,C) | ...\n\nbool ---> true; false.\nsymbol ---> :atom.\nvoid ---> nothing.\narr_int64(D) ---> int64(shape(D), nested(D,integer)).\narr_float64(D) ---> float64(shape(D), nested(D,float)).\ntuple() ---> #()\ntuple(A) ---> #(A)\ntuple(A,B) ---> #(A,B)```\n\nThese types are mapped to and from Julia types as follows:\n\n```Int64 <--> integer\nFloat64 <--> float\nString <--> string\nSymbol <--> symbol\nBool <--> bool\nVoid <--> void\nTuple{...} <--> tuple(...)```\n\nA D-dimensional arrays is represented using a `shape(D)`, which is list of exactly D integers, in REVERSE ORDER from the Julia form, and a D-level nested list containing the values, where the first dimension is the length of the outer list, the second dimension is the length of each 1st level nested list, and so on.\n\nThese arrays can be passed to jl_call/N, returned from jl_eval, and also passed to the Julia DCG via the higher-level term level evaluators/executors.\n\njl_exec(+S:text) is det\nExecute Julia expression S, which may be string, atom, codes or chars. If any Julia exceptions are raised, the Julia function `showerror` is used to print the details, and a `julia_error` exception is raised.\njl_eval(+S:text, -X:val) is det\nEvaluate Julia expression S, which may be string, atom, codes or chars. The following Julia return types are recognised and are converted to Prolog values and unified with X: Any other return type raises an `unsupported_julia_return_type_error` exception. If any Julia exceptions are raised, the Julia function `showerror` is used to print the details, and a Prolog `julia_error` exception is raised.\njl_call(+F:text, +Arg:val, -Result:val) is det\njl_call(+F:text, +Arg1:val, +Arg2:val, -Result:val) is det\njl_call(+F:text, +Arg1:val, +Arg2:val, +Arg3:val, -Result:val) is det\nEvaluates F to get a function and applies it to given arguments. The arguments are converted to Julia values directly in C code, which should be faster than conversion via DCG formatting and parsing in Julia.\njl_call_(+F:text, +Arg:val) is det\njl_call_(+F:text, +Arg1:val, +Arg2:val) is det\njl_call_(+F:text, +Arg1:val, +Arg2:val, +Arg3:val) is det\nSame as jl_call/3, jl_call/4 etc but ignoring the return value. This can be useful when the return value is not supported by pljulia.\njl_apply(+F:text, +Args:list(val), -Result:val) is det\nEvaluates Julia expression F to a function and applies it to any number of arguments in Args, unifying the return value with Result.\njl_apply_(+F:text, +Args:list(val)) is det\nEvaluates Julia expression F to a function and applies it to any number of arguments in Args and discarding the return value.\n!(+E:expr) is det\nExecute Julia expression E using jl_exec/1. No output is printed. E is converted to a string using term_jlstring/2. Use `debug(julia)` to see the Julia expression before execution.\n?(+E:expr) is det\nExecute Julia expression term E using (!)/1 and display the result using Julia.\n?>(+E:expr, -X:val) is det\nEvaluate the Julia expression term E using jl_eval/2 and unify result with X. E is converted to a string using term_jlstring/2. Use `debug(julia)` to see the Julia expression before execution.\n<?(-X:val, +E:expr) is det\nEvalute Julia expression E and unify result with X. `X <? E` is the same as `E ?> X`, but may be more convenient for interactive use.\n??(-X:val, +E:expr) is det\nEvalute Julia expression E as with (<?)/2, but instead of converting the result into a Prolog representation, store it in a freshly allocated Julia workspace variable and unify X with a term containing the name of the new variable. This can be used in subsequent Julia expressions, but it cannot be passed to jl_apply/4 and jl_apply_/3. When X goes out of scope, garbage collection on the Prolog side releases the value in the Julia workspace so that it becomes available for garbage collection on the Julia side.\n\n## Undocumented predicates\n\nThe following predicates are exported, but not or incorrectly documented.\n\njl_call(Arg1, Arg2, Arg3, Arg4)\njl_call_(Arg1, Arg2, Arg3)\njl_call(Arg1, Arg2, Arg3, Arg4, Arg5)\njl_call_(Arg1, Arg2, Arg3, Arg4)\nterm_jlstring(Arg1, Arg2)"
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"http://eu.swi-prolog.org/icons/swipl.png",
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https://retrobasic.allbasic.info/index.php?topic=825.msg6575 | [
"###",
null,
"Author Topic: TriQuad Remake (Read 1234 times)\n\n#### B+\n\n• Guest",
null,
"« on: July 21, 2019, 01:51:05 am »\nThis is total overhaul of Rick's TriQuad clone done in Naalaa some time ago. This game allows user to choose from 3 to 9 square pieces per side, uses a different method to build the puzzle and uses a different method to detect/check solutions since it is possible by random luck to have multiple solutions to a puzzle.\n\nQB64v1.3\nCode: [Select]\n`OPTION _EXPLICIT_TITLE \"TriQuad Puzzle\" 'B+ start 2019-07-17 trans to QB64 from:' TriQuad.bas SmallBASIC 0.12.8 [B+=MGA] 2017-03-26' inspired by rick3137's recent post at Naalaa of cute puzzle' 2019-07 Complete remake for N X N puzzles, not just 3 X 3's.RANDOMIZE TIMERCONST xmax = 1000, margin = 50 'screen size, margin that should allow a line above and below the puzzle displayCONST topLeftB1X = margin, topLeftB2X = xmax / 2 + .5 * margin, topY = margin'these have to be decided from user input from Intro screenDIM SHARED ymax, N, Nm1, NxNm1, sq, sq2, sq4ymax = 500 'for starters in intro screen have resizing in pixels including ymaxREDIM SHARED B1(2, 2), B2(2, 2) ' B1() box container for scrambled pieces of C(), B2 box container to build solutionREDIM SHARED C(8, 3) '9 squares 4 colored triangles, C() contains the solution as created by code, may not be the only one!DIM mx, my, mb, bx, by, holdF, ky AS STRING, again AS STRINGSCREEN _NEWIMAGE(xmax, ymax, 32)_SCREENMOVE 300, 40introrestart:assignColorsholdF = N * NWHILE 1 CLS showB (1) showB (2) WHILE _MOUSEINPUT: WEND mx = _MOUSEX: my = _MOUSEY: mb = _MOUSEBUTTON(1) IF mb THEN DO WHILE mb WHILE _MOUSEINPUT: WEND mx = _MOUSEX: my = _MOUSEY: mb = _MOUSEBUTTON(1) LOOP IF topY <= my AND my <= topY + N * sq THEN by = INT((my - topY) / sq) 'LOCATE 1, 1: PRINT SPACE\\$(20) 'LOCATE 1, 1: PRINT \"bY = \"; by IF topLeftB1X <= mx AND mx <= topLeftB1X + N * sq THEN 'mx in b1 bx = INT((mx - topLeftB1X) / sq) 'LOCATE 2, 1: PRINT SPACE\\$(20) 'LOCATE 2, 1: PRINT \"bX = \"; bx IF holdF < N * N THEN 'trying to put the piece on hold here? IF B1(bx, by) = N * N THEN B1(bx, by) = holdF: holdF = N * N END IF ELSEIF holdF = N * N THEN IF B1(bx, by) < N * N THEN holdF = B1(bx, by): B1(bx, by) = N * N END IF END IF ELSEIF topLeftB2X <= mx AND mx <= topLeftB2X + N * sq THEN 'mx in b2 bx = INT((mx - topLeftB2X) / sq) 'LOCATE 2, 1: PRINT SPACE\\$(20) 'LOCATE 2, 1: PRINT \"bX = \"; bx IF holdF < N * N THEN IF B2(bx, by) = N * N THEN B2(bx, by) = holdF: holdF = N * N END IF ELSEIF holdF = N * N THEN IF B2(bx, by) < N * N THEN holdF = B2(bx, by): B2(bx, by) = N * N END IF END IF 'my out of range END IF END IF END IF IF solved THEN COLOR hue(9) LOCATE 2, 1: centerPrint \"Congratulations puzzle solved!\" _DISPLAY _DELAY 3 EXIT WHILE END IF ky = INKEY\\$ IF LEN(ky) THEN IF ky = \"q\" THEN showSolution COLOR hue(9) LOCATE 2, 1: centerPrint \"Here is solution (for 10 secs), Goodbye!\" _DISPLAY _DELAY 10 SYSTEM END IF END IF _DISPLAY _LIMIT 100WENDCOLOR hue(9): LOCATE 2, 1: centerPrint SPACE\\$(50): LOCATE 2, 1centerPrint \"Press enter to play again, any + enter ends... \"_DISPLAYagain = INKEY\\$WHILE LEN(again) = 0: again = INKEY\\$: _LIMIT 200: WENDIF ASC(again) = 13 THEN GOTO restart ELSE SYSTEMFUNCTION solved 'since it is possible that a different tile combination could be a valid solution we have to check points DIM x, y 'first check that there is a puzzle piece in every slot of b2 FOR y = 0 TO Nm1 FOR x = 0 TO Nm1 IF B2(x, y) = N * N THEN EXIT FUNCTION NEXT NEXT 'check left and right triangle matches in b2 FOR y = 0 TO Nm1 FOR x = 0 TO Nm1 - 1 IF POINT(topLeftB2X + x * sq + sq2 + sq4, topY + y * sq + sq2) <> POINT(topLeftB2X + (x + 1) * sq + sq4, topY + y * sq + sq2) THEN EXIT FUNCTION NEXT NEXT 'check to and bottom triangle matches in b2 FOR y = 0 TO Nm1 - 1 FOR x = 0 TO Nm1 'the color of tri4 in piece below = color tri1 of piece above IF POINT(topLeftB2X + x * sq + sq2, topY + y * sq + sq2 + sq4) <> POINT(topLeftB2X + x * sq + sq2, topY + (y + 1) * sq + sq4) THEN EXIT FUNCTION NEXT NEXT 'if made it this far then solved solved = -1END FUNCTIONSUB showSolution DIM x, y, index FOR y = 0 TO Nm1 FOR x = 0 TO Nm1 drawSquare index, x * sq + topLeftB2X, y * sq + topY index = index + 1 NEXT NEXTEND SUBSUB showB (board) DIM x, y, index FOR y = 0 TO Nm1 FOR x = 0 TO Nm1 IF board = 1 THEN index = B1(x, y) drawSquare index, x * sq + topLeftB1X, y * sq + topY ELSE index = B2(x, y) drawSquare index, x * sq + topLeftB2X, y * sq + topY END IF NEXT NEXTEND SUBSUB drawSquare (index, x, y) LINE (x, y)-STEP(sq, sq), &HFF000000, BF LINE (x, y)-STEP(sq, sq), &HFFFFFFFF, B IF index < N * N THEN LINE (x, y)-STEP(sq, sq), &HFFFFFFFF LINE (x + sq, y)-STEP(-sq, sq), &HFFFFFFFF PAINT (x + sq2 + sq4, y + sq2), hue(C(index, 0)), &HFFFFFFFF PAINT (x + sq2, y + sq2 + sq4), hue(C(index, 1)), &HFFFFFFFF PAINT (x + sq4, y + sq2), hue(C(index, 2)), &HFFFFFFFF PAINT (x + sq2, y + sq4), hue(C(index, 3)), &HFFFFFFFF END IFEND SUBSUB assignColors () 'the pieces are indexed 0 to N X N -1 (NxNm1) ' y(index) = int(index/N) : x(index) = index mod N ' index(x, y) = (y - 1) * N + x DIM i, j, x, y 'first assign a random color rc to every triangle FOR i = 0 TO NxNm1 'piece index FOR j = 0 TO 3 'tri color index for piece C(i, j) = rand(1, 9) NEXT NEXT 'next match c0 to c3 of square to right FOR y = 0 TO Nm1 FOR x = 0 TO Nm1 - 1 'the color of tri3 of next square piece to right = color of tri0 to left of it C(y * N + x + 1, 2) = C(y * N + x, 0) NEXT NEXT FOR y = 0 TO Nm1 - 1 FOR x = 0 TO Nm1 'the color of tri4 in piece below = color tri1 of piece above C((y + 1) * N + x, 3) = C(y * N + x, 1) NEXT NEXT ' C() now contains one solution for puzzle, may not be the only one ' scramble pieces to box1 DIM t(0 TO NxNm1), index 'temp array FOR i = 0 TO NxNm1: t(i) = i: NEXT FOR i = NxNm1 TO 1 STEP -1: SWAP t(i), t(rand(0, i)): NEXT FOR y = 0 TO Nm1 FOR x = 0 TO Nm1 B1(x, y) = t(index) index = index + 1 B2(x, y) = N * N 'PRINT B1(x, y), B2(x, y) NEXT NEXTEND SUBFUNCTION hue~& (n) SELECT CASE n CASE 0: hue~& = &HFF000000 CASE 1: hue~& = &HFFA80062 CASE 2: hue~& = &HFF000050 CASE 3: hue~& = &HFFE3333C CASE 4: hue~& = &HFFFF0000 CASE 5: hue~& = &HFF008000 CASE 6: hue~& = &HFF0000FF CASE 7: hue~& = &HFFFF64FF CASE 8: hue~& = &HFFFFFF00 CASE 9: hue~& = &HFF00EEEE CASE 10: hue~& = &HFF663311 END SELECTEND FUNCTIONFUNCTION rand% (n1, n2) DIM hi, lo IF n1 > n2 THEN hi = n1: lo = n2 ELSE hi = n2: lo = n1 rand% = (RND * (hi - lo + 1)) \\ 1 + loEND FUNCTIONSUB intro 'use intro to select number of pieces DIM test AS INTEGER CLS: COLOR hue(8): LOCATE 3, 1 centerPrint \"TriQuad Instructions:\": PRINT: COLOR hue(9) centerPrint \"This puzzle has two boxes that contain up to N x N square pieces of 4 colored triangles.\" centerPrint \"The object is to match up the triangle edges from left Box to fill the Box on the right.\": PRINT centerPrint \"You may move any square piece to an empty space on either board by:\" centerPrint \"1st clicking the piece to disappear it,\" centerPrint \"then clicking any empty space for it to reappear.\": PRINT centerPrint \"You may press q to quit and see the solution displayed.\": PRINT centerPrint \"Hint: the colors without matching\" centerPrint \"complement, are edge pieces.\": PRINT centerPrint \"Good luck!\": COLOR hue(5) LOCATE CSRLIN + 2, 1: centerPrint \"Press number key for square pieces per side (3 to 9, 1 to quit)...\" WHILE test < 3 OR test > 9 test = VAL(INKEY\\$) IF test = 1 THEN SYSTEM WEND N = test ' pieces per side of 2 boards Nm1 = N - 1 ' FOR loops NxNm1 = N * N - 1 ' FOR loop of piece index 'sizing sq = (xmax / 2 - 1.5 * margin) / N 'square piece side size sq2 = sq / 2: sq4 = sq / 4 ymax = sq * N + 2 * margin REDIM B1(Nm1, Nm1), B2(Nm1, Nm1), C(NxNm1, 3) SCREEN _NEWIMAGE(xmax, ymax, 32) '_SCREENMOVE 300, 40 'need again? 'PRINT ymaxEND SUBSUB centerPrint (s\\$) LOCATE CSRLIN, (xmax / 8 - LEN(s\\$)) / 2: PRINT s\\$END SUB`\n« Last Edit: July 21, 2019, 02:08:23 am by B+ »\n\n#### Rick3137\n\n• Guest",
null,
"« Reply #1 on: July 21, 2019, 08:47:24 pm »\nAwesome!!\n\nMuch better than the one I made. Thanks for sharing.\n\nI noticed a slight problem with qb64, when I was downloading. I have a habit of copy and paste to notepad or wordpad when I get a short program from a forum. This does not work with qb64.\nI had to paste straight to the qb64 editor. There is some problem somewhere with the text format that causes errors.\n\n#### B+\n\n• Guest",
null,
""
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https://fdocument.org/document/atomic-theory-amp-structure-weebly.html | [
"of 15 /15\n\nothers\n• Category\n\n## Documents\n\n• view\n\n2\n\n0\n\nEmbed Size (px)\n\n### Transcript of Atomic Theory & Structure - Weebly",
null,
"",
null,
"",
null,
"c = λνE = hν\n\nc = speed of light (2.998 x 108 m/s)\n\nλ = wavelength (m) ν = frequency (Hz) h = Planck’s constant\n\n(6.626 x 10-34 J·s)",
null,
"Photons of light shining on a metal causes electrons to be emitted from the surface.\n\nWhat energy does the photon need to cause this to happen?",
null,
"When an atom absorbs energy, it is able to emit photons of light with specific energies.",
null,
"Only orbits, corresponding to these energies, exist for the electron in a hydrogen atom.\n\nElectrons in these allowed energy state do not radiate energy, and as a result, does not spiral in to the radius.\n\nAs an electron changes energy states, it emits or absorbs energy as a photon with an energy equal to the difference in the two energy states.",
null,
"Bohr just assumed that electrons would not spiral into the nucleus.\n\nHe didn’t explain WHY this doesn’t happen!\n\nBohr’s model says that all electrons in the same principal energy level have the same energy.\n\nPhotoelectron spectroscopy (PES) proves that this isn’t true!",
null,
"It is impossible for us to know the exact momentum (energy) and exact location (orbital) of an electron at the same time!\n\nThis opened the door for the Quantum Mechanical Model.",
null,
"Describes the probability of finding an electron with a given energy in a certain region of space in an atom.",
null,
"Principal QN (n) = 1, 2, 3, …\n\nHigher n Larger orbital (further from nucleus) with higher energy\n\nAngular momentum QN (l) = 0, 1, 2, …, (n-1)\n\nShape of orbital (0 = s, 1 = p, 2 = d, 3 = f)\n\nMagnetic QN (ml) = -l, …, -1, 0, 1, …, l\n\nOrientation of orbital around nucleus\n\nSpin QN (ms) = +1/2, -1/2\n\nSpin of electron",
null,
"No two electrons can have the same set of 4 quantum numbers\n\nMaximum of 2 electrons in the same orbital\n\nElectrons in the same orbital must have opposite spins",
null,
"",
null,
"For degenerate orbitals, the lowest energy exists when there are the maximum number of electrons with the same spin.",
null,
"Shows the noble-gas core of an atom by a bracketed symbol and includes the valence electrons.",
null,
""
] | [
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"data:image/gif;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABCAQAAAC1HAwCAAAAC0lEQVR42mM8Uw8AAh0BTZud3BwAAAAASUVORK5CYII=",
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null
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http://jsoc.stanford.edu/cvs/JSOC/proj/sharp/apps/sw_functions.c.diff?r1=1.28&r2=1.29 | [
"",
null,
"Return to sw_functions.c CVS log",
null,
"",
null,
"Up to [Development] / JSOC / proj / sharp / apps\n\n### Diff for /JSOC/proj/sharp/apps/sw_functions.c between version 1.28 and 1.29\n\nversion 1.28, 2014/03/13 18:40:43 version 1.29, 2014/05/16 21:55:54\n Line 17\n Line 17\nMEANPOT Mean photospheric excess magnetic energy density in ergs per cubic centimeter MEANPOT Mean photospheric excess magnetic energy density in ergs per cubic centimeter\nTOTPOT Total photospheric magnetic energy density in ergs per cubic centimeter TOTPOT Total photospheric magnetic energy density in ergs per cubic centimeter\nMEANSHR Mean shear angle (measured using Btotal) in degrees MEANSHR Mean shear angle (measured using Btotal) in degrees\nR_VALUE Karel Schrijver's R parameter\n\nThe indices are calculated on the pixels in which the conf_disambig segment is greater than 70 and The indices are calculated on the pixels in which the conf_disambig segment is greater than 70 and\npixels in which the bitmap segment is greater than 30. These ranges are selected because the CCD pixels in which the bitmap segment is greater than 30. These ranges are selected because the CCD\n Line 1035 int computeShearAngle(float *bx_err, flo\n Line 1036 int computeShearAngle(float *bx_err, flo\n} }\n\n/*===========================================*/ /*===========================================*/\n/* Example function 15: R parameter as defined in Schrijver, 2007 */\n//\n// Note that there is a restriction on the function fsample()\n// If the following occurs:\n// nx_out > floor((ny_in-1)/scale + 1)\n// ny_out > floor((ny_in-1)/scale + 1),\n// where n*_out are the dimensions of the output array and n*_in\n// are the dimensions of the input array, fsample() will usually result\n// in a segfault (though not always, depending on how the segfault was accessed.)\n\nint computeR(float *bz_err, float *los, int *dims, float *Rparam, float cdelt1, int computeR(float *bz_err, float *los, int *dims, float *Rparam, float cdelt1,\nfloat *rim, float *p1p0, float *p1n0, float *p1p, float *p1n, float *p1, float *rim, float *p1p0, float *p1n0, float *p1p, float *p1n, float *p1,\nfloat *pmap, int nx1, int ny1) float *pmap, int nx1, int ny1)\n{ {\n\nint nx = dims; int nx = dims;\nint ny = dims; int ny = dims;\nint i = 0; int i = 0;\n Line 1058 int computeR(float *bz_err, float *los,\n Line 1068 int computeR(float *bz_err, float *los,\ninit_fresize_gaussian(&fresgauss,sigma,20,1); init_fresize_gaussian(&fresgauss,sigma,20,1);\n\nfsample(los, rim, nx, ny, nx, nx1, ny1, nx1, scale, 0, 0, 0.0); fsample(los, rim, nx, ny, nx, nx1, ny1, nx1, scale, 0, 0, 0.0);\n\nfor (i = 0; i < nx1; i++) for (i = 0; i < nx1; i++)\n{ {\nfor (j = 0; j < ny1; j++) for (j = 0; j < ny1; j++)\n\nLegend:\n Removed from v.1.28 changed lines Added in v.1.29"
] | [
null,
"http://jsoc.stanford.edu/icons/small/back.gif",
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"http://jsoc.stanford.edu/icons/small/text.gif",
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"http://jsoc.stanford.edu/icons/small/dir.gif",
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https://www.scholars.northwestern.edu/en/publications/dsubpsub-convergence-and-regularity-theorems-for-entropy-and-scal | [
"# dp –convergence and –regularity theorems for entropy and scalar curvature lower bounds\n\nMan Chun Lee, Aaron Naber, Robin Tonra Neumayer\n\nResearch output: Contribution to journalArticlepeer-review\n\n1 Scopus citations\n\n## Abstract\n\nConsider a sequence of Riemannian manifolds.Min; gi/whose scalar curvatures and entropies are bounded from below by small constants Ri; The goal of this paper is to understand notions of convergence and the structure of limits for such spaces. As a first issue, even in the seemingly rigid case ! 0, we will construct examples showing that from the Gromov–Hausdorff or intrinsic flat points of view, such a sequence may converge wildly, in particular to metric spaces with varying dimensions and topologies and at best a Finsler-type structure. On the other hand, we will see that these classical notions of convergence are the incorrect ones to consider. Indeed, even a metric space is the wrong underlying category to be working on. Instead, we will introduce a weaker notion of convergence called dp –convergence, which is valid for a class of rectifiable Riemannian spaces. These rectifiable spaces will have a well-behaved topology, measure theory and analysis. This includes the existence of gradients of functions and absolutely continuous curves, though potentially there will be no reasonably associated distance function. Under this dp notion of closeness, a space with almost nonnegative scalar curvature and small entropy bounds must in fact always be close to Euclidean space, and this will constitute our –regularity theorem. In particular, any sequence.Min; gi/with lower scalar curvature and entropies tending to zero must dp –converge to Euclidean space. More generally, we have a compactness theorem saying that sequences of Riemannian manifolds.Min; gi/with small lower scalar curvature and entropy bounds Ri; must dp –converge to such a rectifiable Riemannian space X. In the context of the examples from the first paragraph, it may be that the distance functions of Mi are degenerating, even though in a well-defined sense the analysis cannot be. Applications for manifolds with small scalar and entropy lower bounds include an L1 –Sobolev embedding and a priori Lp scalar curvature bounds for p < 1.\n\nOriginal language English (US) 227-350 124 Geometry and Topology 27 1 https://doi.org/10.2140/GT.2023.27.227 Published - 2023\n\n## ASJC Scopus subject areas\n\n• Geometry and Topology\n\n## Fingerprint\n\nDive into the research topics of 'dp –convergence and –regularity theorems for entropy and scalar curvature lower bounds'. Together they form a unique fingerprint."
] | [
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https://linuxhint.com/sql-rank-over/ | [
"SQL Standard\n\n# SQL Rank Over\n\nRanking in SQL allows you to calculate the rank of each record within a partition within a result set.\n\nThe rank() will return a 1-based index for each record in an ordered partition in Standard SQL. Remember that the function will assign the same rank value to the partitions with similar values.\n\nThe number of previous rank values increments each consequent row rank value. If you wish to increment the rank value by 1, use the dense_rank() function.\n\n## Rank() Function Syntax\n\nThe syntax for the rank function is as shown below:\n\nRANK() OVER (\n\n[PARTITION BY expression, ]\n\nORDER BY expression (ASC | DESC) );\n\n);\n\nTo illustrate how to use this function, consider the example below:\n\ncreate table users(\n\nid serial primary key,\n\nfirst_name varchar(100),\n\nlast_name varchar(100),\n\nstate varchar(25),\n\nactive bool\n\n);\n\ninsert into users(first_name, last_name, state, active) values (\n\n'Mary', 'Smith', 'New York', TRUE);\n\ninsert into users(first_name, last_name, state, active) values (\n\ninsert into users(first_name, last_name, state, active) values (\n\n'Taylor', 'Moore', 'Utah', TRUE);\n\ninsert into users(first_name, last_name, state, active) values (\n\n'Susan', 'Wilson', 'Washington', TRUE);\n\ninsert into users(first_name, last_name, state, active) values (\n\n'Mary', 'Smith', 'New York', TRUE);\n\ninsert into users(first_name, last_name, state, active) values (\n\n'Taylor', 'Moore', 'Utah', TRUE);\n\nThe above queries create and insert sample data into the table.\n\nTo assign a rank to the records in the result set, we can use the rank() function as illustrated below.\n\nselect id, first_name, last_name, state, active,\n\nrank () over (partition by active order by id) rank_value\n\nfrom users;\n\nIn the above query, we partition the data by the active column. The column contains Boolean values. We then rank over each item in the partitions.\n\nThe resulting set is as shown:",
null,
"Note that the result contains two partitions, one containing false values and the other containing true values.\n\nIn the “false’ partition, the function assigns the rank values. The same case for the “true” partition. Note that the function starts the rank value from 1 in a new partition.\n\n## Closing\n\nThis tutorial showed you how to perform row ranking by partitioning using the rank() function. Check the documentation for your database engine to learn more.",
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""
] | [
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http://gooddealvn.com/addition-and-subtraction-worksheets-pdf/ | [
"# Addition And Subtraction Worksheets Pdf\n\nAddition And Subtraction Worksheets Pdf addition and subtraction worksheets pdf halloween add or subtract worksheet 3 free. addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents free. addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents template. Addition And Subtraction Worksheets Pdf addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents download. addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents templates. addition and subtraction worksheets pdf math addition subtraction worksheets dongola. Addition And Subtraction Worksheets Pdf addition and subtraction worksheets pdf subtraction worksheets addition with regrouping worksheets sample templates.",
null,
"Addition And Subtraction Worksheets Pdf Halloween Add Or Subtract Worksheet 3 Free",
null,
"Addition And Subtraction Worksheets Pdf 17 Sample Addition Subtraction Worksheets Free Pdf Documents Free",
null,
"Addition And Subtraction Worksheets Pdf 17 Sample Addition Subtraction Worksheets Free Pdf Documents Template",
null,
"Addition And Subtraction Worksheets Pdf 17 Sample Addition Subtraction Worksheets Free Pdf Documents Download",
null,
"Addition And Subtraction Worksheets Pdf 17 Sample Addition Subtraction Worksheets Free Pdf Documents Templates",
null,
"Addition And Subtraction Worksheets Pdf Math Addition Subtraction Worksheets Dongola",
null,
"Addition And Subtraction Worksheets Pdf Subtraction Worksheets Addition With Regrouping Worksheets Sample Templates\n\nAddition And Subtraction Worksheets Pdf addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents free. addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents template. addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents download. addition and subtraction worksheets pdf 17 sample addition subtraction worksheets free pdf documents templates. addition and subtraction worksheets pdf math addition subtraction worksheets dongola.\n\n### Related Post to Addition And Subtraction Worksheets Pdf\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] | [
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"http://gooddealvn.com/wp-content/uploads/2018/11/addition-and-subtraction-worksheets-pdf-halloween-add-or-subtract-worksheet-3-free.jpg",
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"http://gooddealvn.com/wp-content/uploads/2018/11/addition-and-subtraction-worksheets-pdf-17-sample-addition-subtraction-worksheets-free-pdf-documents-free.jpg",
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https://stanford.library.sydney.edu.au/archives/win2015/entries/spacetime-supertasks/ | [
"First published Tue Jun 29, 1999; substantive revision Wed Nov 11, 2009\n\nSupertasks have posed problems for philosophy since the time of Zeno of Elea. The term ‘supertask’ is new but it designates an idea already present in the formulation of the old motion paradoxes of Zeno, namely the idea of an infinite number of actions performed in a finite amount of time. The main problem lies in deciding what follows from the performance of a supertask. Some philosophers have claimed that what follows is a contradiction and that supertasks are, therefore, logically impossible. Others have denied this conclusion, and hold that the study of supertasks can help us improve our understanding of the physical world, or even our theories about it.\n\n## 1. What is a Supertask\n\n### 1.1 Definitions\n\nA supertask may be defined as an infinite sequence of actions or operations carried out in a finite interval of time. The terms ‘action’ and ‘operation’ must not be understood in their usual sense, which involves a human agent. Human agency may be involved but it is not necessary. To show this, let us see how actions can be characterised precisely without any references to man. We will assume that at each instant of time the state of the world relevant to a specific action can be described by a set S of sentences. Now an action or operation applied to a state of the world results in a change in that state, that is, in the set S corresponding to it. Consequently, an arbitrary action a will be defined (Allis and Koetsier 1995) as a change in the state of the world by which the latter changes from state S before the change to state a(S) after it. This means that an action has a beginning and an end, but does not entail that there is a finite lapse of time between them. For instance, take the case of a lamp that is on at t = 0 and remains so until t = 1, an instant at which it suddenly goes off. Before t =1 the state of the lamp (which is the only relevant portion of the world here) can be described by the sentence ‘lamp on’, and after t =1 by the sentence ‘lamp off’, without there being a finite lapse of time between the beginning and the end of the action. Some authors have objected to this consequence of the definition of action, and they might be right if we were dealing with the general philosophical problem of change. But we need not be concerned with those objections at this stage, since in the greatest majority of the relevant supertasks instantaneous actions (i.e., actions without any duration) can be replaced by actions lasting a finite amount of time without affecting the analysis at any fundamental point.\n\nThere is a particular type of supertask called hypertasks. A hypertask is a non-numerable infinite sequence of actions or operations carried out in a finite interval of time. Therefore, a supertask which is not a hypertask will be a numerable infinite sequence of actions or operations carried out in a finite interval of time. Finally, a task can be defined as a finite sequence of actions or operations carried out in a finite interval of time.\n\n### 1.2 The Philosophical Problem of Supertasks\n\nTo gain a better insight into the fundamental nature of the philosophical problem posed by supertasks, consider the distinction between tasks in general (finite sequences of actions of the type (a1, a2, a3, … , an)) and one particular type of supertasks, namely those consisting of an infinite sequence of actions of the type (a1, a2, a3, … , an, … ) and thus having the same type of order as the natural order of positive integers: 1, 2, 3, … , n, … (it is customary to denote this type of order with letter ‘w’ and so the related supertasks can be called supertasks of type w).\n\nIn the case of a task T = (a1, a2, a3, … , an) it is natural to say that T is applicable in state S if:\n\na1 is applicable to S,\na2 is applicable to a1(S),\na3 is applicable to a2(a1(S)),\n… , and,\nan is applicable to an−1(an−2(… (a2(a1(S)))… )).\n\nThe successive states of the world relevant to task T can be defined by means of the finite sequence of sets of sentences:\n\nS, a1(S), a2(a1(S)), a3(a2(a1(S))), …, an(an−1(an−2 (… (a2(a1(S)))…))),\n\nwhose last term will therefore describe the relevant state of the world after the performance of T. Or, equivalently, the state resulting from applying T to S will be T(S) =\n\nan(an−1(an−2 (… (a2(a1(S)))… ))).\n\nNow take the case of a supertask T = (a1, a2, a3, …, an, …). Let us give the name Tn to the task which consists in performing the first n actions of T. That is, Tn = (a1, a2, a3, …, an). Now it is natural to say that T is applicable in state S if Tn is applicable in S for each natural number n, and, obviously,\n\nTn(S) = an(an−1(an−2 (…(a2(a1(S)))…))).\n\nThe successive states of the world relevant to supertask T can be described by means of the infinite sequence of sets of sentences:\n\nS, T1(S), T2(S), …, Tn(S), …\n\nA difficulty arises, however, when we want to specify the set of sentences which describe the relevant state of the world after the performance of supertask T, because the infinite sequence above lacks a final term. Put equivalently, it is difficult to specify the relevant state of the world resulting from the application of supertask T to S because there seems to be no final state resulting from such an application. This inherent difficulty is increased by the fact that, by definition, supertask T is performed in a finite time, and so there must exist one first instant of time t* at which it can be said that the performance happened. Now notice that the world must naturally be in a certain specific state at t*, which is the state resulting from the application of T, but that, nevertheless, we have serious trouble to specify this state, as we have just seen.\n\n### 1.3 Supertask: A Fuzzy Concept\n\nSince we have defined supertasks in terms of actions and actions in terms of changes in the state of the world, there is a basic indeterminacy regarding what type of processes taking place in time should be considered supertasks, which is linked to the basic indeterminacy that there is regarding which type of sets of sentences are to be allowed in descriptions of the state of the world and which are not. For this reason, there are some processes that would be regarded as supertasks by virtually every philosopher and some about which opinions differ. For an instance of the first sort of process, take the one known as ‘Thomson's lamp’. Thomson's lamp is basically a device consisting of a lamp and a switch set on an electrical circuit. The switch can be in one of just two positions and the lamp has got to be lit — when the switch is in position ‘on’ — or else dim — when the switch is in position ‘off’. Assume that initially (at t = 12 A.M., say) the lamp is dim and that it is thenceforth subject to the following infinite sequence of actions: when half of the time remaining until t* = 1 P.M. has gone by, we carry out the action a1 of turning the switch into position ‘on’ and, as a result, the lamp is lit (a1 is thus performed at t = 1/2 P.M.); when half the time between the performance of a1 and t* = 1 P.M. has gone by, we carry out action a2 of turning to the switch into position ‘off’ and, as a result, the lamp is dim (a2 is thus performed at t = 1/2 + 1/4 P.M.); when half the time between the performance of a2 and t* = 1 P.M. has gone by, we carry out the action of turning the switch into position ‘on’ and, as a result, the lamp is lit (a3 is thus performed at t = 1/2 + 1/4 + 1/8), and so on. When we get to instant t* = 1 P.M. we will have carried out an infinite sequence of actions, that is, a supertask T = (a1, a2, a3, … , an, … ). If, for the sake of simplicity, we are only concerned about the evolution of the lamp (not the switch) the state of the world relevant to the description of our supertask admits of only two descriptions, one through the unitary set of sentences {lamp lit} and the other through the set {lamp dim}.\n\nAs an instance of the second sort of processes we referred to above, those about which no consensus has been reached as to whether they are supertasks, we can take the process which is described in one of the forms of Zeno's dichotomy paradox. Suppose that initially (at t = 12 A.M., say) Achilles is at point A (x = 0) and moving in a straight line, with a constant velocity v = 1 km/h, towards point B (x = 1), which is 1 km. away from A. Assume, in addition, that Achilles does not modify his velocity at any point. In that case, we can view Achilles's run as the performance of a supertask, in the following way: when half the time until t* = 1 P.M. has gone by, Achilles will have carried out the action a1 of going from point x = 0 to point x = 1/2 (a1 is thus performed in the interval of time between t =12 A.M. and t = 1/2 P.M.), when half the time from the end of the performance of a1 until t* = 1 P.M. will have elapsed, Achilles will have carried out the action a2 of going from point x = 1/2 to point x = 1/2 + 1/4 (a2 is thus performed in the interval of time between t = 1/2 P.M. and t = 1/2 + 1/4 P.M.), when half the time from the end of the performance of a2 until t* = 1 P.M. will have elapsed, Achilles will have carried out the action a3 of going from point x = 1/2 + 1/4 to point x = 1/2 + 1/4 + 1/8 (a3 is thus performed in the interval of time between t = 1/2 + 1/4 P.M. and t = 1/2 + 1/4 + 1/8 P.M.), and so on. When we get to instant t* = 1 P.M., Achilles will have carried out an infinite sequence of actions, that is, a supertask T = (a1, a2, a3, … , an, … ), provided we allow the state of the world relevant for the description of T to be specified, at any arbitrary instant, by a single sentence: the one which specifies Achilles's position at that instant. Several philosophers have objected to this conclusion, arguing that, in contrast to Thomson's lamp, Achilles's run does not involve an infinity of actions (acts) but of pseudo-acts. In their view, the analysis presented above for Achilles's run is nothing but the breakdown of one process into a numerable infinity of subprocesses, which does not make it into a supertask. In Allis and Koetsier's words, such philosophers believe that a set of position sentences is not always to be admitted as a description of the state of the world relevant to a certain action. In their opinion, a relevant description of a state of the world should normally include a different type of sentences (as is the case with Thomson's lamp) or, in any case, more than simply position sentences.\n\n## 2. On The Conceptual Possibility Of Supertasks\n\nIn section 1.2 I have pointed out and illustrated the fundamental philosophical problem posed by supertasks. Obviously, one will only consider it a problem if one deems the concepts employed in its formulation acceptable. In fact, some philosophers reject them, because they regard the very notion of supertask as problematic, as leading to contradictions or at least to insurmountable conceptual difficulties. Among these philosophers the first well-known one is Zeno of Elea.\n\nConsider the dichotomy paradox in the formulation of it given in section 1.3. According to Zeno, Achilles would never get to point B (x = 1) because he would first have to reach the mid point of his run (x = 1/2), after that he would have to get to the mid point of the span which he has left (x = 1/2 + 1/4), and then to the mid point of the span which is left (x = 1/2 + 1/4 + 1/8), and so on. Whatever the mid point Achilles may reach in his journey, there will always exist another one (the mid point of the stretch that is left for him to cover) that he has not reached yet. In other words, Achilles will never be able to reach point B and finish his run. According to Owen (Owen 1957–58), in this as well as in his other paradoxes, Zeno was concerned to show that the Universe is a simple, global entity which is not comprised of different parts. He tried to demonstrate that if we take to making divisions and subdivisions we will obtain absurd results (as in the dichotomy case) and that we must not yield to the temptation of breaking up the world. Now the notion of supertask entails precisely that, division into parts, as it involves breaking up a time interval into successive intervals. Therefore, supertasks are not feasible in the Zenonian world and, since they lead to absurd results, the notion of supertask itself is conceptually objectionable.\n\nIn stark contrast to Zeno, the dichotomy paradox is standardly solved by saying that the successive distances covered by Achilles as he progressively reaches the mid points of the spans he has left to go through — 1/2, 1/4, 1/8, 1/16, … — form an infinite series 1/2 + 1/4 + 1/8 + 1/16 + … whose sum is 1. Consequently, Achilles will indeed reach point B (x = 1) at t* = 1 P.M. (which is to be expected if he travels with velocity v = 1 km/h, as has been assumed). Then there is no problem whatsoever in splitting up his run into smaller sub-runs and, so, no inherent problem about the notion of supertask. An objection can be made, however, to this standard solution to the paradox: it tells us where Achilles is at each instant but it does not explain where Zeno's argument breaks down. Importantly, there is another objection to the standard solution, which hinges on the fact that, when it is claimed that the infinite series 1/2 + 1/4 + 1/8 + 1/16 + … adds up to 1, this is substantiated by the assertion that the sequence of partial sums 1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, … has limit 1, that is, that the difference between the successive terms of the sequence and number 1 becomes progressively smaller than any positive integer, no matter how small. But it might be countered that this is just a patch up: the infinite series 1/2 + 1/4 + 1/8 + … seems to involve infinite sums and thus the performance of a supertask, and the proponent of the standard solution is in fact presupposing that supertasks are feasible just in order to justifiy that they are. To this the latter might reply that the assertion that the sum of the series is 1 presupposes no infinite sum, since, by definition, the sum of a series is the limit to which its partial (and so finite) sums approach. His opponent can now express his disagreement with the response that the one who supports the standard solution is deducing a matter of fact (that Achilles is at x = 1 at t* = 1 P.M.) from a definition pertaining to the arithmetic of infinite series, and that it is blatantly unacceptable to deduce empirical propositions from mere definitions.\n\n### 2.3 On Thomson's Impossibility Arguments\n\nThomson (1954–55) put forward one more argument against the logical possibility of his lamp. Let us assign to the lamp the value 0 when it is dim and the value 1 when it is lit. Then lighting the lamp means adding one unity (going from 0 to 1) and dimming it means subtracting one unity (going from 1 to 0). It thus seems that the final state of the lamp at t* = 1 P.M., after an infinite, and alternating, sequence of lightings (additions of one unity) and dimmings (subtractions of one unity), should be described by the infinite series 1−1+1−1+1… If we accept the conventional mathematical definition of the sum of a series, this series has no sum, because the partial sums 1, 1−1, 1−1+1, 1−1+1−1, … , etc. take on the values 1 and 0 alternatively, without ever approaching a definite limit that could be taken to be the proper sum of the series. But in that case it seems that the final state of the lamp can neither be dim (0) nor lit (1), which contradicts our assumption that the lamp is at all times either dim or lit. Benacerraf's (1962) reply was that even though the first, second, third, … , n-th partial sum of the series 1−1+1−1+1… does yield the state of the lamp after one, two, three, … , n actions ai (of lighting or dimming), it does not follow from this that the final state of the lamp after the infinite sequence of actions ai must of necessity be given by the sum of the series, that is, by the limit to which its partial sums progressively approach. The reason is that a property shared by the partial sums of a series does not have to be shared by the limit to which those partial sums tend. For instance, the partial sums of the series 0.3 + 0.03 + 0.003 + 0.0003 + … are 0.3, 0.3 + 0.03 = 0.33, 0.3 + 0.03 + 0.003 = 0.333,… , all of them, clearly, numbers less than 1/3; however, the limit to which those partial sums tend (that is, the sum of the original series) is 0.3333… , which is precisely the number 1/3.\n\n### 2.4 On Black's Impossibility Argument\n\nAnother one of the classical arguments against the logical possibility of supertasks comes from Black (1950–51) and is constructed around the functioning of an infinity machine of his own invention. An infinity machine is a machine that can carry out an infinite number of actions in a finite time. Black's aim is to show that an infinity machine is a logical impossibility. Consider the case of one such machine whose sole task is to carry a ball from point A (x = 0) to point B (x = 1) and viceversa. Assume, in addition, that initially (at t = 12 A.M., say) the ball is at A and that the machine carries out the following infinite sequence of operations: when half the time until t* = 1 P.M. has gone by, it does the action a1 of taking the ball from position A to position B (a1 is thus carried out at t = 1/2 P.M.); when half the time between the performance of a1 and t* = 1 P.M. has gone by, it does the action a2 of taking the ball from position B to position A (a2 is thus carried out at t = 1/2 + 1/4 P.M.); when half the time between the performance of a2 and t* = 1 P.M. has gone by, the machine does the action a3 of taking the ball from position A to position B (a3 is thus performed in t = 1/2 + 1/4 + 1/8 P.M.), and so on. When we get to instant t* = 1 P.M. the machine will have carried out an infinite sequence of actions, that is, a supertask T = (a1, a2, a3, … , an, … ). The parallelism with Thomson's lamp is clear when it is realised that the ball in position A corresponds to the dim lamp and the ball in position B corresponds to the lit lamp. Nevertheless, Black believes that by assuming that at each instant the ball is either in A or else in B (and note that assuming this means that the machine transports the ball from A to B and viceversa instantaneously, but we need not be worried by this, since all that we are concerned with now is logical or conceptual possibility, not physical possibility), he can deduce, by a totally different route from Thomson's based on the symmetrical functioning of his machine, a contradiction regarding its state at t* = 1 P.M.. However, Benacerraf's criticisms also applies to Black's argument. In effect, the latter hinges on how the machine works, and as this has only been specified for instants of time previous to t* = 1 P.M., it follows that what can be logically inferred from the functioning of the machine is only applicable to those instants previous to t* = 1 P.M.. Black seeks to deduce a contradiction at t* = 1 P.M. but he fails at the same point as Thomson: whatever happens to the ball at t* = 1 P.M. cannot be a logical consequence of what has happened to it before. Of course, one can always specify the functioning of the machine for instants t greater than or equal to 1 P.M. by saying that at all those instants the machine will not perform any actions at all, but that is not going to help Black. His argument is fallacious because he seeks to reach a logical conclusion regarding instant t* = 1 P.M. from information relative to times previous to that instant.. In the standard argument against Zeno's dichotomy one could similarly specify Achilles's position at t* = 1 P.M. saying, for instance, that he is at B (x = 1), but there is no way that this is going to get us a valid argument out of a fallacious one, which seeks to deduce logically where Achilles will be at t* = 1 P.M. from information previous to that instant of time.\n\n### 2.5 Benacerraf's Critique and the Dichotomy Arguments\n\nThe cases dealt with above are examples of how Benacerraf's strategy can be used against supposed demonstrations of the logical impossibility of supertasks. We have seen that the strategy is based on the idea that\n\n(I) the state of a system at an instant t* is not a logical consequence of which states he has been in before t* (where by ‘state’ I mean ‘relevant state of the world’, see section 1.1)\n\nand occasionally on the idea that\n\n(II) the properties shared by the partial sums of a series do not have to be shared by the limit to which those partial sums tend.\n\nSince the partial sums of a series make up a succession (of partial sums), (II) may be rewritten as follows:\n\n(III) the properties shared by the terms of a succession do not have to be shared by the limit to which that succession tends.\n\nIf we keep (I), (II) and (III) well in mind, it is easy not to yield to the perplexing implications of certain supertasks dealt with in the literature. And if we do not yield to the perplexing results, we will also not fall into the trap of considering supertasks conceptually impossible. (III), for instance, may be used to show that it is not impossible for Achilles to perform the supertasks of the inverse and the direct dichotomy of Zeno. Take the case of the direct dichotomy: the limit of the corresponding succession of instants of time t1, t2, t3, … at which each one of Achilles's successive sub-runs is finished can be the instant at which Achilles's supertask has been accomplished, even if such a supertask is not achieved at any one of the instants in the infinite succession t1, t2, t3, … (all of this in perfect agreement with (III)).\n\n## 3. On The Physical Possibility Of Supertasks\n\nWe have gone through several arguments for the conceptual impossibility of supertasks and counterarguments to these. Those who hold that supertasks are conceptually possible may however not agree as to whether they are also physically possible. In general, when this issue is discussed in the literature, by physical possibility is meant possibility in relation with certain broad physical principles, laws or ‘circumstances’ which seems to operate in the real world, at least as far as we know. But it is a well-known fact that authors do not always agree about which those principles, laws or circumstances are.\n\n### 3.1 Kinematical Impossibility\n\nIn our model of Thomson's lamp we are assuming that at each moment the switch can be in just one of two set positions (‘off’, ‘on’). If there is a fixed distance d between them, then clearly, since the switch swings an infinite number of times from the one to the other from t = 12 A.M. until t* = 1 P.M., it will have covered an infinite distance in one hour. For this to happen it is thus necessary for the speed with which the switch moves to increase unboundedly during this time span. Grünbaum has taken this requirement to be physically impossible to fulfil. Grünbaum (1970) believes that there is a sort of physical impossibility of a purely kinematical nature (kinematical impossibility) and describes it in more precise terms by saying that a supertask is kinematically impossible if:\n\na) At least one of the moving bodies travels at an unboundedly increasing speed,\n\nb) For some instant of time t*, the position of at least one of the moving bodies does not approach any defined limit as we get arbitrarily closer in time to t*.\n\nIt is clear then that the Thomson's lamp supertask, in the version presented so far, is kinematically (and eo ipso physically) impossible, since not only does the moving switch have to travel at a speed that increases unboundedly but also—because it oscillates between two set positions which are a constant distance d apart—its position does not approach any definite limit as we get closer to instant t* = 1 P.M., at which the supertask is accomplished. Nevertheless, Grünbaum has also shown models of Thomson's lamp which are kinematically possible. Take a look at Figure 1, in which the switch (in position ‘on’ there) is simply a segment AB of the circuit connecting generator G with lamp L. The circuit segment AB can shift any distance upwards so as to open the circuit in order for L to be dimmed. Imagine we push the switch successively upwards and downwards in the way illustrated in Figure 2, so that it always has the same velocity v = 1.",
null,
"Figure 1",
null,
"Figure 2\n\nThe procedure is the following. Initially (t = 0) the switch is in position AB′ (lamp dim) a height of 0.2 above the circuit and moving downwards (at v = 1). At t = 0.2 it will be in position AB (lamp lit) and will begin moving upwards (v = 1). At t = 0.2 + 0.01 it will be in position AB″ (lamp dim) and will begin moving downwards (v = 1). At t = 0.2 + 0.01 + 0.01 = 0.2 + 0.02 it will be in position AB (lamp lit) and will begin moving upwards (v = 1). At t = 0.2 + 0.02 + 0.001 it will be in position A′′′B′′′ (lamp dim), and so on. Obviously, between t = 0 and t* = 0.2 + 0.02 + 0.002 + … = 0.2222… = 2/9, the lamp is in the states ‘dim’ and ‘lit’ an infinite number of times, and so a supertask is accomplished. But this supertask is not kinematically impossible, because it has been so designed that the switch always moves with velocity v = 1 — and, therefore, condition (a) for kinematical impossibility is not fulfilled — and that, additionally, as we get closer to the limit time t* = 2/9 (the only one which could cause us any trouble) the switch approaches more and more a well-defined limit position, position AB (lamp lit)—and, therefore, condition (b) for kinematical impossibility is not fulfilled either. Once the kinematical possibility has been established, what is the state of the lamp at t* = 2/9? What has been said so far does not enable us to give a determinate answer to this question (just as the obvious kinematical possibility of Achilles's supertask in the dichotomy paradox does not suffice to determine where Achilles will be at t* = 1 P.M.), but there exists a ‘natural’ result. It seems intuitively acceptable that the position the switch will occupy at t* = 2/9 will be position AB, and so the lamp will be lit at that instant. There is no mysterious asymmetry about this result. Figure 3 shows a model of Thomson's lamp with a switch that works according to exactly the same principles as before, but which will yield the ‘natural’ result that the lamp will in the end be dim at t* = 2/9. In effect, the switch will now finally end up in the ‘natural’ position AB at t* = 2/9 and will thereby bring about an electrical short-circuit that will make all the current in the generator pass through the cable on which the switch is set, leaving nothing for the more resistant path where the lamp is.",
null,
"Figure 3\n\nThere are some who believe that the very fact that there exist Thomson's lamps yielding an intuitive result of ‘lamp lit’ when the supertask is accomplished but also other lamps whose intuitive result is ‘lamp dim’ brings up back to the contradiction which Thomson thought to have found originally. But we have nothing of that sort. What we do have is different physical models with different end-results. This does not contradict but rather corroborates the results obtained by Benacerraf: the final state is not logically determined by the previous sequence of states and operations. This logical indeterminacy can indeed become physical determinacy, at least sometimes, depending on what model of Thomson's lamp is employed.\n\nA conspicuous instance of a supertask which is kinematically impossible is the one performed by Black's infinity machine, whose task it is to transport a ball from position A (x = 0) to position B (x = 1) and from B to A an infinite number of times in one hour. As with the switch in our first model of Thomson's lamp, it is obvious that the speed of the ball increases unboundedly (and so condition a) for impossibility is met), while at the same time, as we approach t* = 1 P.M., its position does not tend to any defined limit, due to the fact that it must oscillate continuously between two set positions A and B one unity distance apart from each other (and so also condition b) for impossibility is met).\n\n### 3.2 The Principle Of Continuity and the Solution to the Philosophical Problem of Supertasks\n\nSince Benacerraf's critique, we know that there is no logical connection between the position of Achilles at t* = 1 P.M. and his positions at instants previous to t* = 1 P.M. Sainsbury (1988) has tried to bridge the gap opened by Benacerraf. He claims that this can be achieved by drawing a distinction between abstract space of a mathematical kind and physical space. No distinction between mathematical and physical space has to be made, however, to attain that goal; one need only appeal to a single principle of physical nature, which is, moreover, simple and general, namely, that the trajectories of material bodies are continuous lines. To put it more graphically, what this means is that we can draw those trajectories without lifting our pen off the paper. More precisely, that the trajectory of a material body is a continuous line means that, whatever the instant t, the limit to which the position occupied by the body tends as time approaches t coincides precisely with the position of the body at t. Moreover, the principle of continuity is highly plausible as a physical hypothesis: the trajectories of all physical bodies in the real world are in fact continuous. What matters is that we realise that, aided by this principle, we can now finally demonstrate that after the accomplishment of the dichotomy supertask, that is, at t* = 1 P.M., Achilles will be in point B (x = 1). We know, in fact, that as the time Achilles has spent running gets closer and closer to t* = 1 P.M., his position will approach point x = 1 more and more, or, equivalently, we know that the limit to which the position occupied by Achilles tends as time get progressively closer to t* = 1 P.M. is point B (x = 1). As Achilles's trajectory must be continuous, by the definition of continuity (applied to instant t = t* = 1 P.M.) we obtain that the limit to which the position occupied by Achilles tends as time approaches t* = 1 P.M. coincides with Achilles's position at t* = 1 P.M. Since we also know that this limit is point B (x = 1), it finally follows that Achilles's position at t* = 1 P.M. is point B (x = 1). Now is when we can spot the flaw in the standard argument against Zeno mentioned in section 2.1, which was grounded on the observation that the sequence of distances covered by Achilles (1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, … ) has 1 as its limit. This alone does not suffice to conclude that Achilles will reach point x = 1, unless it is assumed that if the distances run by Achilles have 1 as their limit, then Achilles will as a matter of fact reach x = 1, but assuming this entails using the principle of continuity. This principle affords us a rigorous demonstration of what, in any event, was already plausible and intuitively ‘natural’: that after having performed the infinite sequence of actions (a1, a2, a3, … , an, … ) Achilles will have reached point B (x = 1). In addition, now it is easy to show how, with a switch like the one in Figure 2, Thomson's lamp in Figure 1 will reach t* = 2/9 with its switch in position AB and will therefore be lit. We have in fact already pointed out (3.1) that in this case, as we get closer to the limit time t* = 2/9, the switch indefinitely approaches a well-defined limit position—position AB. Due to the fact that the principle of continuity applies to the switch, because it is a physical body, this well-defined limit position must coincide precisely with the position of the switch at t* = 2/9. Therefore, at t* = 2/9 the latter will be in positon AB and, consequently, the lamp will be lit. By the same token, it can also be shown that the lamp in Figure 3 will be dim at time t* = 2/9.\n\n### 3.4 On Burke's Impossibility Argument\n\nBurke (2000b) proposed a new argument for the physical impossibility of supertasks based on the idea that their execution would entail the possibility of violating Newton's first law: a body cannot alter its state of motion without some external force acting on it. He imagines a passive Achilles being moved successively by an infinite numerable set of machines Mn. My presentation differs in secondary details from Burke's to concentrate on the essential issues. For each positive integer n, Mn will only act on Achilles in an interval of time of the magnitude 1/n − 1/(n+1), and will proceed as follows: if at some moment Achilles is at point xn = 1/n2 then it will take him to the point xn+1 = 1/(n+1)2 in a time 1/n − 1/(n+1) and (vice versa) if at any time Achilles is at xn+1 = 1/(n+1)2 it will take him and transfer him to xn = 1/n2 in a time 1/n − 1/(n+1). Now suppose that Achilles places himself at x1 = 1 in t = 1. Presumably the machines Mn will act successively on him, collectively performing a supertask that will bring the warrior in t = 2 to rest at x = 0 (as, under the principle of continuity, Achilles' path must be continuous). Is this true? Burke appeals to the temporal symmetry of the physical laws involved to argue that, if this is the case, then the machines Mn must also be capable of taking Achilles to x1 = 1 in t = 1 if he is initially at rest at x = 0 in t = 0 (executing the reverse process to the previous one). But, says Burke, if Achilles is at rest at x = 0 in t = 0, the machines Mn will be incapable of taking him to x1 = 1 in t = 1 for the simple reason that each machine operates at a finite distance to the right of x = 0. So, none of them is capable of setting him in motion, which means that Achilles remains at x = 0 even after t = 0 (Newton's first law). This means that the reverse process is not physically possible and, for reasons of temporal symmetry, the direct one is not possible either. And as the direct process is a standard supertask, its physical impossibility justifies the physical impossibility of supertasks in general.\n\nThe argument is original and refuting it involves an idea important in the context of other problems associated with supertasks, as we shall see later on. The key lies in recognizing that the fact that no Mn can take Achilles away from x = 0 is not enough to conclude that, in the absence of any other possible influence, Achilles cannot leave x = 0. We might consider the set of Mn as a new machine, ‘super M’, which obviously can exert a non-null force on Achilles at points arbitrarily close to x = 0 (unlike any individual machine). The Mn are individually incapable of setting somebody (Achilles) to their left in motion, but it does not follow that they are also incapable of doing so collectively. Burke's argument fails here. Furthermore (as Pérez Laraudogoitia argues in (2006a) and is immediately verifiable) we can design machines M*n capable of guaranteeing that, by handling Achilles exactly as the Mn would, an Achilles at rest at x = 0 in t = 0 would be taken by them to x1 = 1 in t = 1. It is enough to program each M*n thus: whatever the point at which Achilles is at in the instant tn+1 = 1/(n+1), it shall take him to xn = 1/n2 in the instant tn = 1/n.\n\n## 4. The Physics of Supertasks\n\nAs we do not know exactly what laws of nature there are, it goes without saying that the question whether a particular supertask is physically possible (that is, compatible with those laws) cannot be given a definitive answer in general. What we have done in 3 above is rather to set out necessary conditions for physical possibility which are plausible (such as the principle of continuity) and sufficient conditions for physical impossibility which are likewise plausible (such as Grünbaum's criterion of kinematical impossibility). In this section we shall look into a problem related to the one just dealt with, but one to which a definitive answer can be given: the problem of deciding whether a certain supertask is possible within the framework of a given physical theory, that is, whether it is compatible with the principles of that theory. These are two distinct problems. At this stage our object are theories whereas in 3 above we were concerned with the real world. What we are after is supertasks formulated within the defined framework of a given physical theory which can tell us something exciting and/or new about that theory. We will discover that this search will lead us right into the heart of basic theoretical problems.\n\n### 4.1 A New Form of Indeterminism: Spontaneous Self-Excitation\n\nClassical dynamics is a theory that studies the motion of physical bodies which interact among themselves in various ways. The vast majority of interesting examples of supertasks within this theory have been elaborated under the assumption that the particles involved only interact with one another by means of elastic collisions, that is, collisions in which no energy is dissipated. We shall see here that supertasks of type w* give rise to a new form of indeterministic behaviour of dynamical systems. The most simple type of case (Pérez Laraudogoitia 1996) is illustrated by the particle system represented in Figure 4 at three distinct moments. It consists of an infinite set of identical point particles P1, P2, P3, … , Pn, … arranged in a straight line. Take the situation depicted in Figure 4A first. In it P1 is at one unity distance from the coordinate origin O, P2 at a distance 1/2 of O, P3 at a distance 1/3 of O and so on. In addition, let it be that all the particles are at rest, except for P1, which is approaching O with velocity v = 1. Suppose that all this takes place at t = 0. Now we will employ",
null,
"Figure 4\n\nThe previous model of supertask in the form of spontaneous self-excitation is valid in relativistic classical dynamics as well as non-relativistic classical dynamics and can also be extended—though not in a completely obvious way, see Pérez Laraudogoitia (2001)—to the Newtonian theory of universal gravitation. The core idea behind indeterministic behaviour in all these cases is that the configuration of a physical system consisting of a denumerable infinite number of parts can be such that the solutions to the dynamic equations—in principle, one for each one of the parts—turn out to be coupled. A particular case of this situation (but probably the most important case, as it is the one that can be generalised more straightforwardly) is that in which the connection between solutions stems from the fact that the dynamic equations themselves are coupled as a result of the configuration of the system. Norton (1999) has availed himself of precisely this possibility, thus introducing a model of spontaneous self-excitation in quantum mechanics. Even though the indeterminism vanishes in this case when the normalizability of the state vector is imposed, this does not make his proposal any less interesting: after all, the free particle solutions to Schrodinger's equation are not normalizable either.\n\nFor reference purposes, we shall call the type of spontaneous self-excitation we have considered until now a type I self-excitation, which is characterized by the fact that the internal energy of an isolated system of particles changes (increases) suddenly and unpredictably. But the physics of supertasks allows, in the specific case of non-relativistic classical dynamics, a qualitatively different variety of self-excitation that we shall call a type II self-excitation: this is the self-excitation of a void (Pérez Laraudogoitia, 1998). To see how it works, we should remember once again the configuration of Figure 4B, which we modify trivially in two steps: a) by ensuring all the particles Pi are at rest in xi = (1/(i+1)) − 1 = − i/(i+1), which means taking them en bloc one unit to the left and b) by simultaneously placing a numerable infinity of particles P*i (identical to Pi) at rest at x*i = i/(i+1). The resulting configuration is symmetrical with respect to the origin of coordinates O, with the asterisk particles to the right and the non-asterisk ones to the left. We already know that both the set of particles to the left of the origin and the set of particles to the right may spontaneously and unpredictably self-excite provoking the successive movement of the particles at any velocity (for the sake of brevity we shall then say they self-excite at the particular velocity in question). Suppose now that: 1) in t = 1 the particles on the left self-excite at velocity v = 2 (the excitation will extinguish in t = 1+1), 2) in t = 1+1 the particles to the right self-excite at velocity v = 4 (the excitation will extinguish in t = 1+1+(1/2)), 3) in t = 1+1+(1/2)) the particles on the left self-excite at velocity v = 8 (the excitation will extinguish in t = 1+1+(1/2)+(1/4)), 4) in t = 1+1+(1/2)+(1/4) the particles on the right self-excite at at velocity v = 16 (the excitation will extinguish in t = 1+1+(1/2)+(1/4)+(1/8)), and so on successively. In t = 1+1+1 = 3 a numerable infinity of self-excitations of type I will have occurred. During this (finite) time, any one particle will have oscillated an infinite number of times between its initial position and the position of the particle to its right that was initially closest to it. As there is a finite distance between these two positions that will have been covered an infinite number of times in a finite time, the particle in question essentially evolves like Black's particle analyzed in section 2.4. It will therefore disappear in t = 3. In other words, in t = 3 the infinity of particles Pi, P*i will have disappeared leaving a void. By now performing the temporal reversion of this whole process of disappearance, what occurs is in fact a complex process of self-excitation of the void by means of which, spontaneously and unpredictably, there emerges a numerable infinity of identical particles. This is an example of type II self-excitation. The disappearance in t = 3 of the infinity of particles Pi, P*i certainly violates the Postulate of Permanence introduced in section 3.3 (and also the necessary conditions for kinematic possibility of Grünbaum, as seen in section 3.1) but what is important here is that it does not violate any of the postulates of classical mechanics. In particular, the standard formulation of the principle of conservation of mass: ‘Particle World lines do not have beginning or end points and mass is constant along a World line’ (Earman 1986) is not contradicted, because none of the World lines of particles Pi, P*i genuinely has an end point.\n\nUntil now we have only seen examples of self-excitations in systems of infinite mass, but this is not an essential condition. Taking advantage of the structural similarity between the equations of the dynamics of translation and the dynamics of rotation, Pérez Laraudogoitia (2007a) proposed an analogous model to the one in Figure 4, but in the dynamics of rotation, in which, instead of infinite particles of equal mass placed in a line, infinite rectangular-shaped thin rigid rods of equal momentum of inertia are included in one another. Despite having a finite mass, this system may self-excite as simply as the one in Figure 4B, although its spatial extension must be infinite to maintain the condition of equal momentums of inertia. Atkinson (2007) proved that the spatially limited system of particles in Figure 4B can also spontaneously self-excite in classical mechanics when the mass of Pn is mn = (2/(n+1)) − (2/(n+2)), which obviously corresponds to a total mass of unit value. Finally, Pérez Laraudogoitia (2007b) showed that, for this same system but in the relativistic case, the spontaneous self-excitation is possible providing qn+1 = (kqn+k)/(1+k+k2kqn) (where qn = mn+1/mn, with 0 < k < 1), and which also corresponds to a case of total finite mass if q1 < k.\n\n### 4.2 A New Form of Indeterminism: Global Interaction\n\nA simple modification of the initial conditions illustrated in Figure 4A leads to a qualitatively different type of supertask that is interesting in itself. Now suppose that all the particles P1, P2, P3, … are at rest at the beginning, while a new particle P0 (identical to them) approaches the origin O at unit velocity but from the left of O. Alper & Bridger (1998) were the first to consider this situation, arguing that it is incompatible with Newtonian mechanics. Their argument went like this: P0 cannot hit Pi because in order to do so P0 must have passed through Pi+1, the particle lying immediately to the left of Pi. Therefore P0 cannot hit any particle. On the other hand, if P0 does not hit any particle, then its motion is undisturbed, and it arrives at, for example, position x = 1/2 at a certain instant of time. But there is a particle, namely P2, at x = 1/2. P0 must hit P2, contradicting the conclusion that P0 hits none of the particles. As Pérez Laraudogoitia (2005) underscored, this argument is fallacious. To see why, consider the innocent example of the collision between two identical solid spheres (of, let us say, unit ratio), X and Y. Let us suppose that Y is mentally divided into an infinite number of concentric spheres Yi in the following way: Y1 is a solid sphere of radius 1/2 and Yi (i > 1) is a hollow sphere of interal radius (i−1)/i and external radius i/(i+1). Clearly, although X collides with Y it does not collide with any of the Yi (Yi is pushed at most by Yi+1 and Yi−1) and this despite the fact that Y is the set of Yis. This may seem enigmatic at first sight, largely because we tend to pass inadvertently from\n\n(A) If X collides with Y then at least one division into parts of Y exists such that X collides with one of the parts\n\nwhich is true, to\n\n(B) If X collides with Y then for all divisions into parts of Y, X collides with one of the parts\n\nwhich is false. In the process, one goes fallaciously from an existential to a universal statement, and after wrongly settling on (B), one finds it impossible for X to collide with Y given the division of Y into Yi parts, none of which collides with X. One then commits a fallacy of composition by assuming here that the whole must have the properties of its parts: if none of the Yi parts collides with X then Y cannot collide with X. Of course Y collides with X, and as Y is the set of Yis it turns out that X collides with the set of Yis, but does not collide with any Yi separately. One can say that X collides collectively, but not distributively, with the set of Yis and, in more physical terms, that X interacts (collides) globally with the Yis. We thus recover the idea of collective action seen in the analysis of Burke's argument concerning impossibility in 3.4. The fact that the Yis were obtained by a process of mental division of Y is irrelevant. One may convert them, if one so desires, into ‘physical’ parts by supposing that Yi is made from matter with density i/(i+1). The conclusion, then, is that a physical system S1 can collide (globally) with another system S2 without colliding with any of the physical parts of which S2 is made. Now the error in Alper and Bridger's argument becomes clear: the particle P0 collides (globally) with the set of the Pi (i > 0) without doing so with any separately, which means that their statement ‘if P0 does not hit any particle, then its motion is undisturbed’ is not justified. This of course means that the conclusion (based on the foregoing statement) that their system is incompatible with Newtonian mechanics is not justified either.\n\nIndeed, the global collision of P0 with the set of the Pi leads to a new form of indeterminism. If, as in a typical elastic collision, we impose the classical laws of the conservation of energy and linear momentum, then one possible evolution from the instant of the global collision is one in which P0 excites the infinite system of the Pi (i > 0) at unit velocity (remember the terminology of section 4.1) before coming to rest and P1 finally moving away from origin O at that unit velocity. But there is also the possibility of P0 rebounding backwards after its global collision to provoke the excitation of the system of the Pi at not one but two different velocities. For instance, readers may verify that the laws of conservation referred to are observed in Newtonian mechanics if, in the final state, P0 moves away from the origin O to the left at velocity 1/4 whereas finally P1 moves away to the right at velocity (5+51/2)/8 and P2 at velocity (5−51/2)/8 while the other Pi come to rest. It is not difficult to verify that the global collision of P0 could also excite the system of the Pi with any number greater than 2 of different velocities. Having seen in section 4.1 that the self-excitation of a dynamical system entailed a new source of indeterminism, we now verify that global interaction provides another, different source, although linked to the previous one by the fact that it causes excitations that, in absence of external influences, could be spontaneous.\n\nPeijnenburg & Atkinson (2008 and 2010) also analyzed the problem proposed by Alper & Bridger (1998), proposing a different solution, although still closely linked to the previous one. They reject the latter's path to contradiction at the same point at which Pérez Laraudogoitia (2005) does, namely: ‘if P0 does not hit any particle, then its motion is undisturbed’ is not justified. But their rejection has another formulation. They propose distinguishing between ‘collision’ as ‘ making contact’ (when two bodies X and Y collide, then X and Y come in contact with one another, in the sense that at least one point of X occupies the same location in space as does one point of Y) and ‘collision’ as ‘zero distance’. Although both notions coincide in topologically closed bodies, this is not the case when topologically open bodies are involved, like the set of particles Pi (i > 0). In the instant P0 arrives at the origin O it does not collide with the other particles in the first sense of ‘making contact’. But under the definition of ‘ collision’ as ‘zero distance ’, P0 does indeed collide with the set of stationary balls at the origin O. Although Peijnenburg & Atkinson's treatment might seem equivalent to Pérez Laraudogoitia's, it is in fact less general, as there are cases of global interaction that it can't account for. Simplifying the model proposed in Pérez Laraudogoitia (2006b), let us assume that for t < 0 we have, in the plane z = 0, a rigid disc D of radius r = 2 at rest and centred on the origin, and above it, in the plane z = 1, we have in t = 0 a denumerable infinite set of rigid rings Ri of equal mass, null thickness, radii ri = i/(i+1) and velocities vi = i moving towards D. In no instant subsequent to t = 0 may D still be in the plane z = 0, because in that case it would have been ‘pierced through’ by an infinite number of rigid rings. This means that, in t = 0, D interacts globally with the infinite set of Ri in Pérez Laraudogoitia's sense, despite the fact that, the distance between D and the Ri being equal in this instant to the unit, D cannot collide with the Ri in any of the senses considered by Peijnenburg & Atkinson. Peijnenburg & Atkinson's analysis fails when faced with situations like this, demonstrating that it is not sufficiently general.\n\n### 4.3 Energy and Momentum Conservation in Supertasks\n\nIn sections 4.1 and 4.2 we considered the problem of indeterminism in supertasks formulated principally in the sphere of classical mechanics or relativistic mechanics. This section provides a brief review of the closely related issue of energy and momentum conservation in supertasks generated from initial states topologically similar to the ones in Figure 4A in the following precise sense: P1 may have any velocity (towards the left) while the rest of the particles will be at rest to the left of P1 in arbitrary positions on P1's line of advance (but with Pj to the left of Pi if j > i) and such that these positions have a point of accumulation O.\n\nFrom 4.1 it obviously follows that when the total mass is infinite both energy conservation and momentum conservation may be violated (in classical and relativistic mechanics alike). They are violated, for instance, when all the particles are identical. Atkinson (2007) proposed the first model with total finite mass in Newtonian mechanics in which the energy was not conserved, although momentum was, and showed that in this mechanics momentum will be conserved provided that: C1) total mass is finite, C2) if j > i the mass of Pj is less than those of the Pi, and C3) each particle Pn+1 undergoes in the process exactly two collisions (one with each of its immediate neighbours) while P1 undergoes just one. He also demonstrated that, in relativistic mechanics and under these same three conditions, energy conservation will be violated if and only if momentum conservation is violated (in marked contrast to the classical case), and that such violations will in fact take place unless the masses of the particles Pn decrease very rapidly with n (Atkinson 2008 shows that violation will occur, for example, if mn = 2n·m0). Finally, Atkinson & Johnson (forthcoming) have shown that the temporal reversal of the state resulting from a process performed under conditions C1), C2) and C3) lead in turn to an indeterministic process: there is an arbitrary parameter in the general solution to the equations of the particles' movement that corresponds to the injection of an arbitrary amount of energy (classically), or energy-momentum (relativistically), into the system at the point of accumulation of the locations of these particles. This last, highly general result poses the question of whether in the classical (or relativistic) dynamics of supertasks there is an intrinsic connection between indeterminism and the non-conservation of energy. Pérez Laraudogoitia (2009a) proves there isn't by showing a specific example of supertask (although qualitatively different from the ones considered by Atkinson & Johnson) in which the energy is conserved despite the evolution of the physical system analized being indeterministic. This latter model is also of interest because it remains valid when the rigid bodies assumed in its description are replaced, whether classically or relativistically, by deformable solids. Furthermore, as the total mass may be as small as one wishes without affecting the description of the evolution, it might be argued that even the inclusion of gravitation will not affect the results. This is probably the clearest example of how supertasks are possible in the context of specific physical theories that also make realistic assumptions.\n\n### 4.4 Actions without Interaction\n\nThe configuration of identical particles introduced in section 4.1, and the resulting possibility of spontaneous self-excitation, has inspired a new type of problem that can be formulated in terms of supertasks (although not only in terms of supertasks), that is qualitatively different to the ones considered until now. It is a commonplace that, when we act physcally on a system (which entails exerting influence in some way on its evolution), we do so by interacting with it. However, Pérez Laraudogoitia (2009b) has shown that a numerable infinity of actors (Gods, or machines suitably programmed) may act physically on an infinite set of particles provoking its self-excitation without interacting with the set in question. Let us consider the system of identical particles in Figure 4A, with the sole difference that P1 is also at rest (in general, Pi is at rest at xi = 1/i). We know that it may spontaneously self-excite at any moment at any velocity, although we shall assume that in t = 0 all the Pi are at rest. Let God1, God2, God3, … be a numerable totality of Gods (machines) such that Godn has the capacity to control particle Pn and only particle Pn. Taking t* > 0, let us suppose that the Gods now decide to pursue the following course of action: Godn shall prevent, at any 0 < t < t* + (1/n), Pn from abandoning point xn = 1/n, but if, from t* + (1/n), Pn does not acquire a unit velocity towards the right by other means (although even for a limited time) then it (Godn) shall provide it (n = 1, 2, 3, …). The interesting, and surprising, thing in this case is that the Gods will manage to make the system of particles self-excite at t* at unit velocity and they will do so without interacting with it, meaning we will indeed have a case of action without interaction that does, however, alter the dynamic state of the particles (all at rest in t = 0). This is easy to see in two steps:\n\n(I) For every n, Pn will acquire from t* + (1/n) unit velocity to the right (although, for n > 1, for a limited time). This is so because, should there be no other process providing it with unit velocity from t* + (1/n), we know that Godn will provide it.\n\n(II) For every n, Pn will acquire from t* + (1/n) unit velocity to the right as a result of receiving the impact of Pn+1. This is so because, according to (I), Pn+1 acquired from t* + (1/(n+1)) unit velocity to the right which led it to collide with Pn at t* + (1/(n+1)) + ((1/n) − (1/(n+1))) = t* + (1/n), transmitting that velocity to it from t* + (1/n), which implies that Godn does not in fact interact with Pn.\n\nFrom (II) it follows that the set of particles will self-excite precisely at t*, that no Godn interacts with any particle, but that the set of Gods caused this self-excitation by acting physically without interaction on the set in question on the basis of the policy referred to earlier. Indeed, if only the particles existed (without Gods), then from the configuration in t = 0 an infinity of different possibilities of evolution would be open to them, an infinity that the Gods have reduced significantly by preventing any self-excitation before t* and provoking one in t*: this is precisely how they acted on them. From this Pérez Laraudogoitia (2009b) concludes that indeterminism is a necessary condition for actions without interaction.\n\nBenardete (1964) formulates what has since then come to be known as the spaceship paradox in the following terms: ‘Indeed one minute will suffice to enable us to exhaust an infinite world. We have only to launch a rocket and travel one thousand miles into outer space during the first 1/2 minute, another one thousand miles after that during the next 1/4 minute, still another one thousand miles during the next 1/8 minute, &c. ad infinitum . At the end of the minute we shall have succeeded in travelling an infinite distance’ (p. 149). The situation described here has all the characteristic ingredients of a classical supertask and the whiff of paradox arises when we wonder, like Benardete: ‘At the end of the minute we find ourselves an infinite distance from Earth … Where in the World are we?’ The question only makes sense in a Newtonian universe, where there is no limit to the velocity that a material body can achieve. Indeed, Pérez Laraudogoitia (1997) constructed a model of the spaceship paradox in classical particle mechanics in which a numerable infinity of identical particles, distributed along the positive X axis with unbounded velocities, collide with each other in an orderly fashion in a finite time, each transmitting to the following one progressively increasing velocities in accordance with a process essentially identical to the one imagined by Benardete. Oppy (2006) rejected the possibility of this kind of process, reasoning that there is no possible World in which the speed of the spaceship is infinite. However, his criticism is not convincing to the extent that in Benardete's scenario only the average velocity (and average acceleration) needs to be infinite, not the instantaneous velocity (or acceleration). In any case the air of paradox disappears if we apply coherently the principle of continuity seen in section 3.2. From that, it follows that at the end of the minute Benardete's spaceship will not be anywhere, as we saw occurring in section 4.1 with Black's particle introduced previously in section 2.4.\n\nThe temporal reversion of this curious evolution of a flight to infinity is an unpredictable process of arrival from infinity, which would appear to sanction an odd asymmetry between the two situations: the escape is deterministic while the arrival from infinity is not. However, Pérez Laraudogoitia (forthcoming) has shown that this asymmetry is much less profound than it appears. By elaborating an infinite sequence of suitably programmed finite tasks (which recall, although with far greater complexity, the ones taken on by the Gods in section 4.4 to achieve actions without interaction), it is possible to design supertasks capable of ‘bringing’ a spaceship from infinity in a perfectly planned and deterministic way. The intriguing possibility of manipulating the conditions at infinity to predict what we are going to ‘take’ from there (consistently, certainly, with the principle of mass conservation announced in section 4.1) is a further example of the new conceptual possibilities opened up by the supertask concept.\n\nHowever, to date we have only discussed models of supertasks in classical mechanics, relativistic mechanics and quantum mechanics. The physics of supertasks in the context of the other major closed physical theory, general relativity, will be the topic of the following two subsections.\n\nWithin relativity theory, supertasks have been approached from a radically different perspective from the one adopted here so far. This new perspective is inherently interesting, since it links the problem of supertasks up with the relativistic analysis of the structure of space-time. To get an insight into the nature of that connection, let us first notice that, according to the theory of relativity, the duration of a process will not be the same in different reference systems but will rather vary according to the reference system within which it is measured. This leaves open the possibility that a process which lasts an infinite amount of time when measured within reference system O may last a finite time when measured within a different reference system O′.\n\nThe supertask literature has needed to exploit space-times with sufficiently complicated structure that global reference systems cannot be defined in them. In these and other cases, the time of a process can be represented by its ‘proper time’. If we represent a process by its world-line in space-time, the proper time of the process is the time read by a good clock that moves with the process along its world-line. A familiar example of its use is the problem of the twins in special relativity. One twin stays home on earth and grows old. Forty years of proper time, for example, elapses along his world-line. The travelling twin accelerates off into space and returns to find his sibling forty years older. But much less time — say only a year of proper time — will have elapsed along the travelling twin world-line if he has accelerated to sufficiently great speeds.\n\nIf we take this into account it is easily seen that the definition of supertask that we have been using is ambiguous. In section 1 above we defined a supertask as an infinite sequence of actions or operations carried out in a finite interval of time. But we have not specified in whose proper time we measure the finite interval of time. Do we take the proper time of the process under consideration? Or do we take the proper time of some observer who watches the process? It turns out that relativity theory allows the former to be infinite while the latter is finite. This fact opens new possibilities for supertasks. Relativity theory thus forces us to disambiguate our definition of supertask, and there is actually one natural way to do it. We can use Black's idea — presented in 2.4 — of an infinity machine, a device capable of performing a supertask, to redefine a supertask as an infinite sequence of actions or operations carried out by an infinity machine in a finite interval of the machine's own proper time measured within the reference system associated to the machine. This redefinition of the notion of supertask does not change anything that has been said until now; our whole discussion remains unaffected so long as ‘finite interval of time’ is read as ‘finite interval of the machine's proper time’. This notion of supertask, disambiguated so as to accord with relativity theory, will be denoted by the expression ‘supertask-1’. Thus:\n\nSupertask-1: an infinite sequence of actions or operations carried out by an infinity machine in a finite interval of the machine's proper time.\n\nHowever we might also imagine a machine that carries out an infinite sequence of actions or operations in an infinite machine proper time, but that the entire process can be seen by an observer in a finite amount of the observer's proper time.\n\nIt is convenient at this stage to introduce a contrasting notion:\n\nSupertask-2: an infinite sequence of actions or operations carried out by a machine in a finite interval of an observer's proper time.\n\nWhile we did not take relativity theory into account, the notions of supertask-1 and supertask-2 coincided. The duration of an interval of time between two given events is the same for all observers. However in relativistic spacetimes this is no longer so and the two notions of supertasks become distinct. Even though all supertasks-1 are also supertasks-2, there may in principle be supertasks-2 which are not supertasks-1. For instance, it could just so happen that there is a machine (not necessarily an infinity machine) which carries out an infinite number of actions in an interval of its own proper time of infinite duration, but in an interval of some observer's proper time of finite duration. Such a machine would have performed a supertask-2 but not a supertask-1.\n\n### 4.7 Bifurcated Supertasks and the Solution to the Philosophical Problem of Supertasks\n\nWhat shape does the philosophical problem posed by supertasks — introduced in Section 1.2 — take on now? Remember that the problem lay in specifying the set of sentences which describe the state of the world after the supertask has been performed. The problem will now be to specify the set of sentences which describe the relevant state of the world after the bifurcated supertask has been performed. Before this can done, of course, the question needs to be answered whether a bifurcated supertask is physically possible. Given that we agree that compatibility with relativity theory is a necessary and sufficient condition of physical possibility, we can reply in the affirmative.\n\nPitowsky (1990) first showed how this compatibility might arise. He considered a Minkowski spacetime, the spacetime of special relativity. He showed that an observer O* who can maintain a sufficient increase in his acceleration will find that only a finite amount of proper time elapses along his world-line in the course of the complete history of the universe, while other unaccelerated observers would find an infinite proper time elapsing on theirs.\n\nLet us suppose that some machine M accomplishes a bifurcated supertask in such a way that the infinite sequence of actions involved happens in a finite interval of an observer O's proper time. If we imagine such an observer at some event on his world-line, all those events from which he can retrieve information are in the ‘past light cone’ of the observer. That is, the observer can receive signals travelling at or less than the speed of light from any event in his past light cone. The philosophical problem posed by the bifurcated supertask accomplished by M has a particularly simple solution when the infinite sequence of actions carried out by M is fully contained within the past light cone of an event on observer O's world-line. In such a case the relevant state of the world after the bifurcated supertask has been performed is M's state, and this, in principle, can be specified, since O has causal access to it. Unfortunately, a situation of this type does not arise in the simple bifurcated supertask devised by Pitowsky (1990). In his supertask, while the accelerated observer O* will have a finite upper bound on the proper time elapsed on his world-line, there will be no event on his world-line from which he can look back and see an infinity of time elapsed along the world-line of some unaccelerated observer.\n\nTo find a spacetime in which the philosophical problem posed by bifurcated supertasks admits of the simple solution that has just been mentioned, we will move from the flat spacetime of special relativity to the curved spacetimes of general relativity. One type of spacetime in the latter class that admits of this simple solution has been dubbed Malament-Hogarth spacetime, from the names of the first scholars to use them (Hogarth 1992). An example of such a spacetime is an electrically charged black hole (the Reissner-Nordstroem spacetime). A well known property of black holes is that, in the view of those who remain outside, unfortunates who fall in appear to freeze in time as they approach the event horizon of the black hole. Indeed those who remain outside could spend an infinite lifetime with the unfortunate who fell in frozen near the event horizon. If we just redescribe this process from the point of view of the observer who does fall in to the black hole, we discover that we have a bifurcated supertask. The observer falling in perceives no slowing down of time in his own processes. He sees himself reaching the event horizon quite quickly. But if he looks back at those who remain behind, he sees their processes sped up indefinitely. By the time he reaches the event horizon, those who remain outside will have completed infinite proper time on their world-lines. Of course, the cost is high. The observer who flings himself into a black hole will be torn apart by tidal forces and whatever remains after this would be unable to return to the world in which he started.\n\n## 5. What Supertasks Entail for the Philosophy of Mathematics\n\n### 5.1 A Critique of Intuitionism\n\nAs Benacerraf and Putnam (1964) have observed, the acknowledgement that supertasks are possible has a profound influence on the philosophy of mathematics: the notion of truth (in arithmetic, say) would no longer be doubtful, in the sense of dependent on the particular axiomatisation used. The example mentioned earlier in connection with Goldbach's conjecture can indeed be reproduced and generalised to all other mathematical statements involving numbers (although, depending on the complexity of the statement, we might need to use several infinity machines instead of just one), and so, consequently, supertasks will enable us to decide on the truth or falsity of any arithmetical statement; our conclusion will no longer depend on provability in some formal system or constructibility in a more or less strict intuitionistic sense. This conclusion seems to lead to a Platonist philosophy of mathematics.\n\nHowever, the situation here is more subtle than the previous comments suggest. Above I introduced a supertask of type w that can settle the truth or falsity of Goldbach's conjecture, but the reference (essential in it) to time contrasts with the lack of specification regarding how to make the necessary computations.When one tries to make up for this omission one discovers that the defence of Platonism is more debatable than it seems at first sight. Davies (2001) has proposed a model of an infinite machine (an infinite machine is a computer which can carry out an infinite number of computations within a finite length of time) based on the Newtonian dynamics of continuous media which reveals the nature of the difficulty. One cannot attempt to decide on mathematical questions such as Goldbach's conjecture by using a mechanical computer which carries out operations at an increasing speed, as if it were a Turing machine. The reason is that the different configurations the computer adopts at increasingly short intervals of time eventually (if the conjecture is true) lead to a paradox of the type of Thomson's lamp, where (if we do not assume continuity in the sense of section 3.2) the final state of the computer is indeterminate, which makes it useless for our purpose. Davies's clever solution consists in assuming an infinite machine capable of building a replica of itself that has twice its own memory but is smaller and works at greater speed. The replica can in its turn build another (even smaller and quicker) replica of itself and so on. With the details Davies gives about the working of his infinite machines, it is clear that they will in no case lead to an indeterminacy paradox (since each replica carries out only a finite part of the task).The problem is that to settle questions like Goldbach's conjecture (if, as I said above, it is true) a numerable (actual) infinity of replicating machines is required, and this will surely be rejected by anyone who, like intuitionists, has a strong dislike of the actual infinity. In more abstract words, the mathematical theory that models the computation process presupposes a Platonic conception of the infinite and thus begs the question by, circularly, supporting Platonism.\n\n### 5.2 The Importance of the Malament-Hogarth Spacetime\n\nAt first sight, the intuitionistic criticism of the possibility of supertasks is less effective in the case of bifurcated supertasks, because in this latter case it is not required that there is any sort of device capable of carrying out an infinite number of actions or operations in a finite time (measured in the reference system associated to the device in question, which is the natural reference system to consider). In contrast, from the possibility of bifurcated supertasks in Malament-Hogarth space-times strong arguments seem to follow against an intuitionistic philosophy of mathematics. But, again, one must be very cautious at this point, as we were at the end of our previous section 5.1. The mathematical theory which models a bifurcated supertask is general relativity, and this theory, as it fully embraces classical mathematical analysis, entails a strong commitment to the—intuitionistically unacceptable—objective status of the set of all natural numbers. It is difficult to believe, therefore, that a radical constructivist lets himself be influenced by the current literature on bifurcated supertasks. This does not make that literature less interesting, as, in establishing unthought-of connections between computability and the structure of space-time, it enriches (as does the existing literature on supertasks in general) the set of consequences that can be derived from our most interesting physical theories.\n\n## Bibliography\n\n• Allis, V. and Koetsier, T., 1991, ‘On Some Paradoxes of the Infinite’, British Journal for the Philosophy of Science, 42: 187–194.\n• Allis, V. and Koetsier, T., 1995, ‘On Some Paradoxes of the Infinite II’, British Journal for the Philosophy of Science, 46: 235–247.\n• Alper, J.S. and Bridger, M., 1998, ‘Newtonian Supertasks: A Critical Analysis’ Synthese, 114: 355–369.\n• Alper, J.S., Bridger, M., Earman , J. and Norton , J.D., 2000, ‘What is a Newtonian System? The Failure of Energy Conservation and Determinism in Supertasks’ Synthese, 124: 281–293.\n• Aristotle, Physics, (W. Charlton, trans.), Oxford: Oxford University Press, 1970.\n• Atkinson, D., 2007, ‘Losing Energy in Classical, Relativistic, and Quantum Mechanics’, Studies in History and Philosophy of Modern Physics, 38: 170–180.\n• Atkinson, D., 2008, ‘A Relativistic Zeno Effect’, Synthese, 160: 77–86.\n• Atkinson, D. and Johnson, P.W., 2009, ‘Nonconservation of Energy and Loss of Determinism. I. Infinitely many colliding balls’, Foundations of Physics, 39: 937–957. [Atkinson and Johnson 2009 available online].\n• Benacerraf, P., 1962, ‘Tasks, Super-Tasks, and Modern Eleatics’, Journal of Philosophy, LIX: 765–784; reprinted in Salmon .\n• Benacerraf, P. and Putnam, H., 1964, Introduction, Philosophy of Mathematics: Selected Readings, P. Benacerraf and H. Putnam (eds.), 2nd edition, Cambridge: Cambridge University Press, pp. 1–27.\n• Benardete, J.A., 1964, Infinity: An Essay in Metaphysics, Oxford: Clarendon Press.\n• Berresford, G. C., 1981, ‘A Note on Thomson's Lamp “Paradox”’, Analysis, 41: 1–7.\n• Black, M., 1950–1, ‘Achilles and the Tortoise’, Analysis, XI: 91–101; reprinted in Salmon .\n• Black, M., 1954, Problems of Analysis, Ithaca: Cornell University Press.\n• Black, R., 2002, ‘Solution to a small infinite puzzle’, Analysis, 62(4): 345–346.\n• Bokulich, A., 2003, ‘Quantum measurement and supertasks’, International Studies in the Philosophy of Science, 17(2): 127–136.\n• Bostock, D., 1972–73, ‘Aristotle, Zeno and the Potential Infinite’, Proceedings of the Aristotelian Society, 73: 37–51.\n• Burke, M.B., 1984, ‘The Infinitistic Thesis’, The Southern Journal of Philosophy, 22: 295–305.\n• Burke, M.B., 2000a, ‘The Impossibility of Superfeats’, The Southern Journal of Philosophy, XXXVIII: 207–220.\n• Burke, M.B., 2000b, ‘The Staccato Run: a Contemporary Issue in the Zenonian Tradition’, The Modern Schoolman, LXXVIII: 1–8.\n• Bridger, M., and Alper, J. S., 1999, ‘On the Dynamics of Perez-Laraudogoitia's Supertask’, Synthese, 119: 325–337.\n• Chihara, C., 1965, ‘On the Possibility of Completing an Infinite Task’, Philosophical Review, LXXIV: 74–87.\n• Cooke, M.C., 2003, ‘Infinite Sequences: Finitist Consequence’, British Journal for the Philosophy of Science, 54: 591–599.\n• Cotogno, P., 2003, ‘Hypercomputation and the Physical Church-Turing Thesis’, British Journal for the Philosophy of Science, 54: 181–223.\n• Davies, E.B., 2001, ‘Building Infinite Machines’, British Journal for the Philosophy of Science, 52: 671–682.\n• Earman, J., 1986, A Primer on Determinism, Dordrecht: Reidel.\n• Earman, J., 1994, Bangs, Crunches, Shrieks, and Whimpers: Singularities and Acausalities in Relativistic Spacetimes, New York: Oxford University Press.\n• Earman, J., and Norton, J.D., 1993, ‘Forever Is a Day: Supertasks in Pitowsky and Malament-Hogarth Spacetimes’, Philosophy of Science, 60: 22–42.\n• Earman, J., and Norton, J. D., 1996, ‘Infinite Pains: The Trouble with Supertasks’, in Benacerraf and His Critics, A. Morton and S. Stich (eds.), Oxford: Blackwell, pp. 231–261.\n• Earman, J., and Norton, J. D., 1998, ‘Comments on Laraudogoitia's “Classical Particle Dynamics, Indeterminism and a Supertask”’, British Journal for the Philosophy of Science, 49: 123–133.\n• Faris, J. A., 1996, The Paradoxes of Zeno, Aldershot: Avebury.\n• Forrest, P., 1999, ‘Supertasks and material objects’, Logique et Analyse, 166–167: 441–446.\n• Friedman, K.S., 2002, ‘A small infinite puzzle’, Analysis, 62(4): 344–345.\n• Gale, R. M. (ed.), 1968, The Philosophy of Time, London: MacMillan.\n• Glazebrook, T., 2001, ‘Zeno Against Mathematical Physics’, Journal of the History of Ideas, 62: 193–210.\n• Groarke, L., 1982, ‘Zeno's Dichotomy: Undermining the Modern Response’, Auslegung, 9: 67–75.\n• Grünbaum, A. 1950–52 ‘Messrs. Black and Taylor on Temporal Paradoxes’, Analysis, 11–12: 144–148.\n• Grünbaum, A., 1967, Modern Science and Zeno's Paradoxes, Middletown, CT: Wesleyan University Press.\n• Grünbaum, A., 1968, ‘Are “Infinity Machines” Paradoxical?’, Science, CLIX: 396–406.\n• Grünbaum, A., 1969, ‘Can an Infinitude of Operations be Performed in a Finite Time?’, British Journal for the Philosophy of Science, 20: 203–218.\n• Grünbaum, A., 1970, ‘Modern Science and Zeno's Paradoxes of Motion’, in Salmon , pp. 200–250.\n• Hawthorne, J., 2000, “Before-Effect and Zeno Causality”, Noˆs, 34(4): 622–633.\n• Hogarth, M. L., 1992, ‘Does General Relativity Allow an Observer to View an Eternity in a Finite Time?’, Foundations of Physics Letters, 5: 173–181.\n• Hogarth, M. L., 1994, ‘Non-Turing Computers and Non-Turing Computability’, in PSA 1994, D. Hull, M. Forbes and R.M. Burian (eds.), 1, East Lansing: Philosophy of Science Association, pp. 126–138.\n• Holgate, P., 1994, ‘Discussion: Mathematical Notes on Ross's Paradox’, British Journal for the Philosophy of Science, 45: 302–304.\n• Koetsier, T. and Allis, V., 1997, ‘Assaying Supertasks’, Logique & Analyse, 159: 291–313.\n• McLaughlin, W. I., 1998, ‘Thomson's Lamp is Dysfunctional’, Synthese, 116: 281–301.\n• Moore, A. W., 1989–90, ‘A Problem for Intuitionism: The Apparent Possibility of Performing Infinitely Many Tasks in a Finite Time’, Proceedings of the Aristotelian Society, 90: 17–34.\n• Moore, A. W., 1990, The Infinite, London: Routledge.\n• Norton, J. D., 1999, ‘A Quantum Mechanical Supertask’, Foundations of Physics, 29: 1265–1302.\n• Oppy, G., 2006, Philosophical Perspectives on Infinity, Cambridge: Cambridge University Press.\n• Owen, G. E. L., 1957–58, ‘Zeno and the Mathematicians’, Proceedings of the Aristotelian Society, LVIII: 199–222, reprinted in Salmon .\n• Parsons, J., 2004, ‘The Eleatic hangover cure’, Analysis, 64: 364–366.\n• Parsons, J., 2006, ‘Topological drinking problems’, Analysis, 66: 149–154.\n• Peijnenburg, J. and Atkinson, D., 2008, ‘Achilles, the Tortoise and Colliding Balls’, History of Philosophy Quarterly, 25: 187–201.\n• Peijnenburg, J. and Atkinson, D., 2010, ‘Lamps, cubes, balls and walls: Zeno problems and solutions’, Philosophical Studies, 150: 49–59.\n• Pérez Laraudogoitia, J., 1996, ‘A Beautiful Supertask’, Mind, 105: 81–83.\n• –––, 1997, ‘Classical Particle Dynamics, Indeterminism and a Supertask’, British Journal for the Philosophy of Science, 48: 49–54.\n• –––, 1998, ‘Infinity Machines and Creation Ex Nihilo’, Synthese, 115: 259–265.\n• –––, 2001, ‘Indeterminism, classical gravitation and non-collision singularities’, International Studies in the Philosophy of Science, 15(3): 269–274.\n• –––, 2005, ‘An Interesting Fallacy Concerning Dynamical Supertasks’, British Journal for the Philosophy of Science, 56: 321–334.\n• –––, 2006a, ‘A Look at the Staccato Run’, Synthese, 148: 433–441.\n• –––, 2006b, ‘Global Interaction in Classical Mechanics’, International Studies in the Philosophy of Science, 20: 173–183.\n• –––, 2007a, ‘Avoiding Infinite Masses’, Synthese, 156: 21–31.\n• –––, 2007b, ‘Supertasks, dynamical attractors and indeterminism’, Studies in History and Philosophy of Modern Physics, 38: 724–731.\n• –––, 2009a, ‘On Energy Conservation in Infinite Systems’, in Energy Conservation: New Research, G. Spadoni (ed.), New York: Nova Science Publishers, pp. 3–12.\n• –––, 2009b, ‘Physical Action Without Interaction’, Erkenntnis, 70: 365–377.\n• –––, forthcoming, ‘The Inverse Spaceship Paradox’, Synthese.\n• Pitowsky, I., 1990, ‘The Physical Church Thesis and Physical Computational Complexity’, Iyyun, 39: 81–99.\n• Priest, G., 1982, ‘To Be and Not to Be: Dialectical Tense Logic’, Studia Logica, XLI: 157–176.\n• Ray, C., 1991, Time, Space and Philosophy, London: Routledge.\n• Sainsbury, R. M., 1988, Paradoxes, Cambridge: Cambridge University Press.\n• Salmon, W., (ed.), 1970, Zeno's Paradoxes, Indianapolis: Bobbs-Merril.\n• Salmon, W., 1980, Space, Time and Motion: A Philosophical Introduction, Minneapolis: University of Minnesota Press.\n• Smith, J. W., 1986, Reason, Science and Paradox, London: Croom Helm.\n• Sorabji, R., 1983, Time, Creation and the Continuum, London: Gerald Duckworth and Co. Ltd.\n• Svozil, K., 1993, Randomness and Undecidability in Physics, Singapore: World Scientific.\n• Taylor, R., 1951–52, ‘Mr. Black on Temporal Paradoxes’, Analysis, XII: 38–44.\n• Taylor, R., 1952–53, ‘Mr. Wisdom on Temporal Paradoxes’, Analysis, XIII: 15–17.\n• Thomson, J., 1954–55, ‘Tasks and Super-Tasks’, Analysis, XV: 1–13; reprinted in Salmon .\n• Thomson, J., 1967, ‘Infinity in Mathematics and Logic’, in Encyclopedia of Philosophy, P. Edwards (ed.), 4, New York: MacMillan, pp. 183–90.\n• Thomson, J., 1970, ‘Comments on Professor Benacerraf's Paper’, in Salmon , pp. 130–38.\n• Van Bendegem, J. P., 1994, ‘Ross' Paradox is an Impossible Super Task’, British Journal for the Philosophy of Science, 45: 743–48.\n• Van Bendegem, J. P., 1995–1997, ‘In Defence of Discrete Space and Time’, Logique & Analyse, 150–151–152: 127–150.\n• Vlastos, G., 1966, ‘Zeno's Race Course’, Journal of the History of Philosophy, IV: 95–108.\n• Vlastos, G., 1967, ‘Zeno of Elea’, in Encyclopedia of Philosophy, P. Edwards (ed.), 8, New York: MacMillan, pp. 369–79.\n• Watling, J., 1953, ‘The Sum of an Infinite Series’, Analysis, XIII: 39–46.\n• Wedeking, G. A., 1968, ‘On a Finitist “Solution” to Some Zenonian Paradoxes’, Mind, 77: 420–26.\n• Weyl, H., 1949, Philosophy of Mathematics and Natural Science, Princeton: Princeton University Press.\n• Whitrow, G. J., 1961, The Natural Philosophy of Time, Edinburgh: Thomas Nelson & Sons.\n• Wisdom, J. O., 1951–52, ‘Achilles on a Physical Racecourse’, Analysis, XII: 67–72, reprinted in Salmon .",
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-1-foundations-for-algebra-chapter-review-1-7-the-distributive-property-page-71/82 | [
"## Algebra 1\n\n$y-3$\nThe distributive property states that for any real numbers $a, b$, and $c$, $a(b+c) = a\\cdot b + a\\cdot c$ Use the formula above to obtain: $=(6y-5y)-3 \\\\=(6-5)y - 3 \\\\=1(y)-3 \\\\=y-3$"
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https://www.mathway.com/examples/calculus/exponential-and-logarithmic-functions/exponential-equations?id=208 | [
"# Calculus Examples\n\nStep 1\nTake the natural logarithm of both sides of the equation to remove the variable from the exponent.\nStep 2\nExpand by moving outside the logarithm.\nStep 3\nSimplify the left side.\nStep 3.1\nApply the distributive property.\nStep 4\nMove all the terms containing a logarithm to the left side of the equation.\nStep 5\nMove all terms not containing to the right side of the equation.\nStep 5.1\nSubtract from both sides of the equation.\nStep 5.2\nAdd to both sides of the equation.\nStep 6\nDivide each term in by and simplify.\nStep 6.1\nDivide each term in by .\nStep 6.2\nSimplify the left side.\nStep 6.2.1\nCancel the common factor of .\nStep 6.2.1.1\nCancel the common factor.\nStep 6.2.1.2\nDivide by .\nStep 6.3\nSimplify the right side.\nStep 6.3.1\nCancel the common factor of .\nStep 6.3.1.1\nCancel the common factor.\nStep 6.3.1.2\nDivide by .\nStep 7\nThe result can be shown in multiple forms.\nExact Form:\nDecimal Form:"
] | [
null
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https://en.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:log-intro/v/plotting-exponential-logarithm | [
"If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## Algebra 2\n\n### Course: Algebra 2>Unit 8\n\nLesson 1: Introduction to logarithms\n\n# Relationship between exponentials & logarithms: graphs\n\nGiven a few points on the graph of an exponential function, Sal plots the corresponding points on the graph of the corresponding logarithmic function. Created by Sal Khan.\n\n## Want to join the conversation?\n\n• Hi,\n\nThis might be a silly question but how is y = log_b (X) the inverse of y=b^x? Isn't the inverse of y=b^x this log_b (y) = X?\n\nThanks",
null,
"•",
null,
"I guess if you look at the function as y = f(x) = b^x. The inverse of the function would be x = f(y) = b^y. So, the logarithmic form of the inverse function is log_b(x) = y.\nEy that's just my take, might be wrong too lol.\n• How did he know to use points 1,4, and 16 in the second chart? I thought that, given the question, he would have taken the x values from the existing points and plugged them into the 2nd function.",
null,
"•",
null,
"",
null,
"For anyone who watches this video and has a similar question, think about it this way:\ny = b^x can be rewritten as log_b(y) = x.\n\nSo for b = 4 (you can deduce that from the graph) and y = 1, 4, 16 (also deduced from the graph) you have\n\nlog_4(1) = 0\nlog_4(4) = 1\nlog_4(16) = 2\n\nWhich represent points (1, 0), (4, 1) and (16,2) the inverse of the points on the given graph!\n• at around , why is sal taking the y values of the first table and using them as the x values of the second table?",
null,
"• I had a little trouble on this too, but here's what I figured: The purpose of an exponential function is to find what value a number equals when it is raised to the power of x. However, the *purpose of a logarithm is to find the exponent (x value)* that gives you a certain amount when you raise a base by the unknown x value. Though it might sound complicated, if you try to define the purpose yourself, you'll end up with the same answer. For this reason, they are like the inverse functions of each other, just like multiplication and division.\n\nIn other words, the logarithm tries to lead you to the exponent needed to reach the value, while the exponential graph tries to lead you to the value given by the exponent's use.\n\nTherefore, they are inverse operations as they undo each other. Furthermore, they reflect over the y=x line, and their coordinates are switched. If you check at , you'll see Sal agrees. If you check the background at this time, you'll also see how when the b= 4, everything fits together perfectly.\n\nHope this helps.\n• Sorry about this but I have two questions. The first might be silly, but here goes: How does one solve for x when it is an exponent such as 2^x=1/64. I've been having to guess this whole time and try for answers on my calculator, because solving for x has not been working. I think I might be doing it wrong...\nSecond is, I didn't understand this video too well. I don't understand what the graph has\nto do with any of this, even though I've been okay with everything pertaining to logarithms so far. Any answer is much appreciated. Thank you!",
null,
"• When the variable is in the exponent, you need to use logarithms of whatever the base of the exponent is.\nFor 2^x = 1 / 64, the base is 2. Therefore, we'll be taking log base 2 of each side of the equation.\nBut before doing that, it's usually easiest to express both sides of the equation using the same base.\nSo, 2^x = 1 / 64 = 1 / 2^6 = 2^(-6)\nlog_2 ( 2^x ) = log_2 ( 2^-6 )\nx = - 6\nHope this helps.\n(I'll let someone else pick up on your question about the graph.)\n• At , how do we know that b raised to the power 1 is equal to four. Is it a definite variable with its value",
null,
"• Hi, can you please explain how did you know that y=log_b(x) is inverse of y=b^x. a detailed answer would be appreciated. Thanks.",
null,
"• The actual definition of a logarithm that proves this is several years ahead of the math you're having now, so it wouldn't mean much to you at this stage.\n\nBut, the short answer is that the logarithm was invented to be the inverse operation of an exponential function. In other words, there is a difficult mathematical process that was worked out over the course of about 80 years by Nicholas Mercator, Leonhard Euler and others. The term \"log\" is just a shorthand way of expressing this difficult bit of math.\n\nSo, a log is just a quick way of writing the inverse function of an exponential function. Thus, by definition, the log must be the inverse function of the exponential function.\n\nSo, the reason you are not given exactly what a log is or how to compute it by hand at this level of study is that it is very difficult math. Since we have calculators that can do this math for us, it is possible for people to use logs without having to master very advanced and difficult computations.\n\nBut, just for reference, here is one way to write the actual definition of a logarithm\nlog_b (x)= lim n→0 [x^(n) - 1] / [b^(n) - 1]\nwhere b is the base of the logarithm.\n• I'm sorry but I'm confused by this video. I've understood everything so far but I got lost. How does this work?",
null,
"• To solve for other points, couldn't I figure out that b=4 from the pattern in y=b^x and go from there?",
null,
"• From this set it is obvious that b=4, but the point of the exercise is to highlight the fact that the logarithmic function is an inverse of the exponential function, meaning that you still could have mapped the second set of points simply by reflecting over the line y = x, even if the values present did not provide a simple/obvious value for b.\n• I've seen a notation such as 10 log (25/5) = 10 log (5). I'm confused about the \"10\" and the \"log\" placement. Is this the same as saying log_10 5? Another (very similar) notation is 20 log (0.1/2)=20 log (0.05). Again, why put the 10 and the 20 before the \"log\"? Thank You!\n(1 vote)",
null,
"• My book has the following example:\nYou can choose a prize of either a \\$20,000 car or one penney on the first day, double that (2cents) on the 2nd day, and so on for a month. On what day would you receive more than the value of the car?\n\\$20,000 is 2,000,000 cents. One day you would receive 1 cent, or 2^0 cents. On day 2, you would receive 2 cents or 2^1, and so on. So, on day n you would receive 2^(n-1) cents.\nHere is the solution:\n2^(n-1)>2X10^6\nlog 2^(n-1)>log(2X10^6)\n(n-1 log 2>log 2+log10^6\n(n-1)log2>log2 +6\nn-1>(log2+6)/log2\nn>(0.301+6)/(0.301)+1\nn>21.93\nThe answer is : On day 22\n\nNow based on the problem above the book asks:\nSuppose that you receive triple the amount each day. On what day would you receive at least 1 millon dollars?\n\nI solved it the following way but I could not get the right answer which is day 18.\n3^(n-1)>10^6\nlog 2(n-1)>log10^6\n(n-1)log3 >lo 10^6\n(n-1)log3>6\nn-1 >6/log3\nn >6/log3 +1\nI got 13.58, but the book says 18 days",
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https://www.thermal-engineering.org/what-is-reynolds-number-definition/ | [
"# What is Reynolds Number – Definition\n\nThe Reynolds number is the ratio of inertial forces to viscous forces and is a convenient parameter for predicting if a flow condition will be laminar or turbulent. Thermal Engineering\n\n## Reynolds Number\n\nThe Reynolds number is the ratio of inertial forces to viscous forces and is a convenient parameter for predicting if a flow condition will be laminar or turbulent. It can be interpreted that when the viscous forces are dominant (slow flow, low Re) they are sufficient enough to keep all the fluid particles in line, then the flow is laminar. Even very low Re indicates viscous creeping motion, where inertia effects are negligible. When the inertial forces dominate over the viscous forces (when the fluid is flowing faster and Re is larger) then the flow is turbulent.",
null,
"It is a dimensionless number comprised of the physical characteristics of the flow. An increasing Reynolds number indicates an increasing turbulence of flow.\n\nwhere:\nV is the flow velocity,\nD is a characteristic linear dimension, (travelled length of the fluid; hydraulic diameter etc.)\nρ fluid density (kg/m3),\nμ dynamic viscosity (Pa.s),\nν kinematic viscosity (m2/s); ν = μ / ρ.\n\n[/su_accordion]\n\n## Laminar vs. Turbulent Flow\n\nLaminar flow:\n\n• Re < 2000\n• ‘low’ velocity\n• Fluid particles move in straight lines\n• Layers of water flow over one another at different speeds with virtually no mixing between layers.\n• The flow velocity profile for laminar flow in circular pipes is parabolic in shape, with a maximum flow in the center of the pipe and a minimum flow at the pipe walls.\n• The average flow velocity is approximately one half of the maximum velocity.\n• Simple mathematical analysis is possible.\n• Rare in practice in water systems.\n\nTurbulent Flow:\n\n• Re > 4000\n• ‘high’ velocity\n• The flow is characterized by the irregular movement of particles of the fluid.\n• Average motion is in the direction of the flow\n• The flow velocity profile for turbulent flow is fairly flat across the center section of a pipe and drops rapidly extremely close to the walls.\n• The average flow velocity is approximately equal to the velocity at the center of the pipe.\n• Mathematical analysis is very difficult.\n• Most common type of flow.\n\nClassification of Flow Regimes\nFrom a practical engineering point of view the flow regime can be categorized according to several criteria.\n\nAll fluid flow is classified into one of two broad categories or regimes. These two flow regimes are:\n\n• Single-phase Fluid Flow\n• Multi-phase Fluid Flow (or Two-phase Fluid Flow)\n\nThis is a basic classification. All of the fluid flow equations (e.g. Bernoulli’s Equation) and relationships that were discussed in this section (Fluid Dynamics) were derived for the flow of a single phase of fluid whether liquid or vapor. Solution of multi-phase fluid flow is very complex and difficult and therefore it is usually in advanced courses of fluid dynamics.",
null,
"Another usually more common classification of flow regimes is according to the shape and type of streamlines. All fluid flow is classified into one of two broad categories. The fluid flow can be either laminar or turbulent and therefore these two categories are:\n\n• Laminar Flow\n• Turbulent Flow\n\nLaminar flow is characterized by smooth or in regular paths of particles of the fluid. Therefore the laminar flow is also referred to as streamline or viscous flow. In contrast to laminar flow, turbulent flow is characterized by the irregular movement of particles of the fluid. The turbulent fluid does not flow in parallel layers, the lateral mixing is very high, and there is a disruption between the layers. Most industrial flows, especially those in nuclear engineering are turbulent.\n\nThe flow regime can be also classified according to the geometry of a conduit or flow area. From this point of view, we distinguish:\n\n• Internal Flow\n• External Flow\n\nInternal flow is a flow for which the fluid is confined by a surface. Detailed knowledge of behaviour of internal flow regimes is of importance in engineering, because circular pipes can withstand high pressures and hence are used to convey liquids. On the other hand, external flow is such a flow in which boundary layers develop freely, without constraints imposed by adjacent surfaces. Detailed knowledge of behaviour of external flow regimes is of importance especially in aeronautics and aerodynamics.\n\nTable from Life in Moving Fluids\n\n## Reynolds Number Regimes",
null,
"Laminar flow. For practical purposes, if the Reynolds number is less than 2000, the flow is laminar. The accepted transition Reynolds number for flow in a circular pipe is Red,crit = 2300.\n\nTransitional flow. At Reynolds numbers between about 2000 and 4000 the flow is unstable as a result of the onset of turbulence. These flows are sometimes referred to as transitional flows.\n\nTurbulent flow. If the Reynolds number is greater than 3500, the flow is turbulent. Most fluid systems in nuclear facilities operate with turbulent flow.\n\n## Reynolds Number and Internal Flow\n\nThe internal flow (e.g. flow in a pipe) configuration is a convenient geometry for heating and cooling fluids used in energy conversion technologies such as nuclear power plants.\n\nIn general, this flow regime is of importance in engineering, because circular pipes can withstand high pressures and hence are used to convey liquids. Non-circular ducts are used to transport low-pressure gases, such as air in cooling and heating systems.\n\nFor internal flow regime an entrance region is typical. In this region a nearly inviscid upstream flow converges and enters the tube. To characterize this region the hydrodynamic entrance length is introduced and is approximately equal to:",
null,
"The maximum hydrodynamic entrance length, at ReD,crit = 2300 (laminar flow), is Le = 138d, where D is the diameter of the pipe. This is the longest development length possible. In turbulent flow, the boundary layers grow faster, and Le is relatively shorter. For any given problem, Le / D has to be checked to see if Le is negligible when compared to the pipe length. At a finite distance from the entrance, the entrance effects may be neglected, because the boundary layers merge and the inviscid core disappears. The tube flow is then fully developed.\n\nFlow Velocity Profile - Power-law velocity profile",
null,
"Source: U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2 and 3. June 1992.\n\n## Power-law velocity profile – Turbulent velocity profile",
null,
"The velocity profile in turbulent flow is flatter in the central part of the pipe (i.e. in the turbulent core) than in laminar flow. The flow velocity drops rapidly extremely close to the walls. This is due to the diffusivity of the turbulent flow.\n\nIn case of turbulent pipe flow, there are many empirical velocity profiles. The simplest and the best known is the power-law velocity profile:",
null,
"where the exponent n is a constant whose value depends on the Reynolds number. This dependency is empirical and it is shown at the picture. In short, the value n increases with increasing Reynolds number. The one-seventh power-law velocity profile approximates many industrial flows.\n\n## Hydraulic Diameter\n\nSince the characteristic dimension of a circular pipe is an ordinary diameter D and especially reactors contains non-circular channels, the characteristic dimension must be generalized.\n\nFor these purposes the Reynolds number is defined as:",
null,
"where Dh is the hydraulic diameter:",
null,
"",
null,
"The hydraulic diameter, Dh, is a commonly used term when handling flow in non-circular tubes and channels. The hydraulic diameter transforms non-circular ducts into pipes of equivalent diameter. Using this term, one can calculate many things in the same way as for a round tube. In this equation A is the cross-sectional area, and P is the wetted perimeter of the cross-section. The wetted perimeter for a channel is the total perimeter of all channel walls that are in contact with the flow.\n\nExample: Reynolds number for a primary piping and a fuel bundle\n`It is an illustrative example, following data do not correspond to any reactor design.`\n\nPressurized water reactors are cooled and moderated by high-pressure liquid water (e.g. 16MPa). At this pressure water boils at approximately 350°C (662°F). Inlet temperature of the water is about 290°C (⍴ ~ 720 kg/m3). The water (coolant) is heated in the reactor core to approximately 325°C (⍴ ~ 654 kg/m3) as the water flows through the core.\n\nThe primary circuit of typical PWRs is divided into 4 independent loops (piping diameter ~ 700mm), each loop comprises a steam generator and one main coolant pump. Inside the reactor pressure vessel (RPV), the coolant first flows down outside the reactor core (through the downcomer). From the bottom of the pressure vessel, the flow is reversed up through the core, where the coolant temperature increases as it passes through the fuel rods and the assemblies formed by them.\n\nAssume:\n\n• the primary piping flow velocity is constant and equal to 17 m/s,\n• the core flow velocity is constant and equal to 5 m/s,\n• the hydraulic diameter of the fuel channel, Dh, is equal to 1 cm\n• the kinematic viscosity of the water at 290°C is equal to 0.12 x 10-6 m2/s\n\nDetermine\n\n• the flow regime and the Reynolds number inside the fuel channel\n• the flow regime and the Reynolds number inside the primary piping\n\nThe Reynolds number inside the primary piping is equal to:\n\nReD = 17 [m/s] x 0.7 [m] / 0.12×10-6 [m2/s] = 99 000 000\n\nThis fully satisfies the turbulent conditions.\n\nThe Reynolds number inside the fuel channel is equal to:\n\nReDH = 5 [m/s] x 0.01 [m] / 0.12×10-6 [m2/s] = 416 600\n\nThis also fully satisfies the turbulent conditions.\n\n## Reynolds Number and External Flow\n\nThe Reynolds number describes naturally the external flow as well. In general, when a fluid flows over a stationary surface, e.g. the flat plate, the bed of a river, or the wall of a pipe, the fluid touching the surface is brought to rest by the shear stress to at the wall. The region in which flow adjusts from zero velocity at the wall to a maximum in the main stream of the flow is termed the boundary layer.\n\nBasic characteristics of all laminar and turbulent boundary layers are shown in the developing flow over a flat plate. The stages of the formation of the boundary layer are shown in the figure below:",
null,
"Boundary layers may be either laminar, or turbulent depending on the value of the Reynolds number.\n\nAlso here the Reynolds number represents the ratio of inertia forces to viscous forces and is a convenient parameter for predicting if a flow condition will be laminar or turbulent. It is defined as:",
null,
"in which V is the mean flow velocity, D a characteristic linear dimension, ρ fluid density, μ dynamic viscosity, and ν kinematic viscosity.\n\nFor lower Reynolds numbers, the boundary layer is laminar and the streamwise velocity changes uniformly as one moves away from the wall, as shown on the left side of the figure. As the Reynolds number increases (with x) the flow becomes unstable and finally for higher Reynolds numbers, the boundary layer is turbulent and the streamwise velocity is characterized by unsteady (changing with time) swirling flows inside the boundary layer.\n\nTransition from laminar to turbulent boundary layer occurs when Reynolds number at x exceeds Rex ~ 500,000. Transition may occur earlier, but it is dependent especially on the surface roughness. The turbulent boundary layer thickens more rapidly than the laminar boundary layer as a result of increased shear stress at the body surface.\n\nThe external flow reacts to the edge of the boundary layer just as it would to the physical surface of an object. So the boundary layer gives any object an “effective” shape which is usually slightly different from the physical shape. We define the thickness of the boundary layer as the distance from the wall to the point where the velocity is 99% of the “free stream” velocity.\n\nTo make things more confusing, the boundary layer may lift off or “separate” from the body and create an effective shape much different from the physical shape. This happens because the flow in the boundary has very low energy (relative to the free stream) and is more easily driven by changes in pressure.\n\nSpecial reference: Schlichting Herrmann, Gersten Klaus. Boundary-Layer Theory, Springer-Verlag Berlin Heidelberg, 2000, ISBN: 978-3-540-66270-9\n\nExample: Transition Layer\nA long thin flat plate is placed parallel to a 1 m/s stream of water at 20°C. Assume that kinematic viscosity of water at 20°C is equal to 1×10-6 m2/s.\n\nAt what distance x from the leading edge will be the transition from laminar to turbulent boundary layer (i.e. find Rex ~ 500,000).\n\nSolution:\n\nIn order to locate the transition from laminar to turbulent boundary layer, we have to find x at which Rex ~ 500,000.\n\nx = 500 000 x 1×10-6 [m2/s] / 1 [m/s] = 0.5 m\n\nReferences:\nReactor Physics and Thermal Hydraulics:\n1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).\n2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.\n3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.\n4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317\n5. Todreas Neil E., Kazimi Mujid S. Nuclear Systems Volume I: Thermal Hydraulic Fundamentals, Second Edition. CRC Press; 2 edition, 2012, ISBN: 978-0415802871\n6. Zohuri B., McDaniel P. Thermodynamics in Nuclear Power Plant Systems. Springer; 2015, ISBN: 978-3-319-13419-2\n7. Moran Michal J., Shapiro Howard N. Fundamentals of Engineering Thermodynamics, Fifth Edition, John Wiley & Sons, 2006, ISBN: 978-0-470-03037-0\n8. Kleinstreuer C. Modern Fluid Dynamics. Springer, 2010, ISBN 978-1-4020-8670-0.\n9. U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2 and 3. June 1992.\n10. White Frank M., Fluid Mechanics, McGraw-Hill Education, 7th edition, February, 2010, ISBN: 978-0077422417"
] | [
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null
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https://statnmap.com:443/2020-07-31-buffer-area-for-nearest-neighbour/ | [
"",
null,
"# Buffer areas for nearest neighbours or national marine areas delineation\n\nA few weeks ago, Christina Buelow asked on Twitter how she could create the polygons of coastal waters using the land polygons, and making sure there is no overlap in buffer areas. This would make it possible to know which land area is closest to any place in the sea. Let’s explore what solution emerged from the discussions.\n\n## TL;DR\n\n``````# Define where to save the dataset\nextraWD <- tempdir()\n# Get some data available to anyone\nif (!file.exists(file.path(extraWD, \"departement.zip\"))) {\ngithubURL <- \"https://github.com/statnmap/blog_tips/raw/master/2018-07-14-introduction-to-mapping-with-sf-and-co/data/departement.zip\"\nunzip(file.path(extraWD, \"departement.zip\"), exdir = extraWD)\n}``````\n• Reduce the dataset to a small region\n``````library(sf)\n# remotes::install_github(\"statnmap/cartomisc\")\nlibrary(cartomisc)\nlibrary(dplyr)\nlibrary(ggplot2)\n\ndepartements_l93 <- read_sf(dsn = extraWD, layer = \"DEPARTEMENT\")\n\n# Reduce dataset\nbretagne_l93 <- departements_l93 %>%\nfilter(NOM_REG == \"BRETAGNE\")``````\n• Calculate the regional buffer area using `regional_seas()`\n``````bretagne_regional_2km_l93 <- regional_seas(\nx = bretagne_l93,\ngroup = \"NOM_DEPT\",\ndist = units::set_units(30, km), # buffer distance\ndensity = units::set_units(0.5, 1/km) # density of points (the higher, the more precise the region attribution)\n)``````\n• Plot the data\n``````ggplot() +\ngeom_sf(data = bretagne_regional_2km_l93,\naes(colour = NOM_DEPT, fill = NOM_DEPT),\nalpha = 0.25) +\ngeom_sf(data = bretagne_l93,\naes(fill = NOM_DEPT),\ncolour = \"grey20\",\nalpha = 0.5) +\nscale_fill_viridis_d() +\nscale_color_viridis_d() +\ntheme_bw()``````",
null,
"If you want to know the long story, you can continue reading this blog post, otherwise, you have everything you need to start on your own data !\n\n## Create buffer area around adjacent polygons\n\nAs always when trying to figure out a solution to a specific problem, let’s start with a reproducible example, commonly called “reprex” among R users. This means, we create the minimal code that allows anyone to execute the code, face the same problem and show everything we tested. In the present case, I couldn’t find enough time to reach a solution during the discussions, so I prepared a gist showing what I tried, what point I reached, and more importantly what I plan to achieve.\n\n``````library(sf)\nlibrary(ggplot2)\nlibrary(dplyr)\n\nmy_blue <- \"#1e73be\"``````\n\n``````# Define where to save the dataset\nextraWD <- \".\"\n# Get some data available to anyone\nif (!file.exists(file.path(extraWD, \"departement.zip\"))) {\ngithubURL <- \"https://github.com/statnmap/blog_tips/raw/master/2018-07-14-introduction-to-mapping-with-sf-and-co/data/departement.zip\"\nunzip(file.path(extraWD, \"departement.zip\"), exdir = extraWD)\n}``````\n\nI reduce the dataset to better see what I am doing. Here, I filter the Brittany area.\nAs you can see, there are 4 departments in this region, thus 4 spatial features.\n\n``````departements_l93 <- read_sf(dsn = extraWD, layer = \"DEPARTEMENT\")\n\n# Reduce dataset\nbretagne_l93 <- departements_l93 %>%\nfilter(NOM_REG == \"BRETAGNE\")\n\nbretagne_l93``````\n``````## Simple feature collection with 4 features and 11 fields\n## Geometry type: MULTIPOLYGON\n## Dimension: XY\n## Bounding box: xmin: 99217.1 ymin: 6704195 xmax: 400572.3 ymax: 6881964\n## Projected CRS: RGF93_Lambert_93\n## # A tibble: 4 x 12\n## ID_GEOFLA CODE_DEPT NOM_DEPT CODE_CHF NOM_CHF X_CHF_LIEU Y_CHF_LIEU X_CENTROID\n## * <chr> <chr> <chr> <chr> <chr> <int> <int> <int>\n## 1 DEPARTEM… 35 ILLE-ET… 238 RENNES 351870 6789646 326237\n## 2 DEPARTEM… 22 COTES-D… 278 SAINT-… 274947 6839206 260216\n## 3 DEPARTEM… 29 FINISTE… 232 QUIMPER 171326 6789950 115669\n## 4 DEPARTEM… 56 MORBIHAN 260 VANNES 267889 6744074 256790\n## # … with 4 more variables: Y_CENTROID <int>, CODE_REG <chr>, NOM_REG <chr>,\n## # geometry <MULTIPOLYGON [m]>``````\n\nNow, I can plot the output of buffer areas around separate polygons.\n\n``````# Create buffer\nbretagne_buffer_l93 <- bretagne_l93 %>%\nst_buffer(\ndist =\nunits::set_units(10, km)\n)\n\nggplot() +\ngeom_sf(data = bretagne_buffer_l93, fill = my_blue, alpha = 0.2, colour = my_blue) +\ngeom_sf(data = bretagne_l93, fill = NA) +\ntheme_bw()``````",
null,
"You can see there are overlapping areas.\nThis is where Voronoï polygons come to help us. From here, I will take some ideas that Christina gathered during her Twitter discussions. Her complete code for coastal buffers is here.\n\n``````# Create a merged region entity\nbretagne_union_l93 <- bretagne_l93 %>%\nsummarise()\n\nggplot() +\ngeom_sf(data = bretagne_union_l93, colour = my_blue, fill = NA)``````",
null,
"``````# Create a doughnut for regional seas areas, 30km off coasts\nbretagne_seas_l93 <- bretagne_union_l93 %>%\nst_buffer(\ndist = units::set_units(30, km)\n) %>%\nst_cast()\n\nggplot() +\ngeom_sf(data = bretagne_seas_l93, colour = my_blue, fill = NA)``````",
null,
"``````# Remove inside terrestrial parts\nbretagne_donut_l93 <- bretagne_seas_l93 %>%\nst_difference(bretagne_union_l93) %>%\nst_cast()\n\nggplot() +\ngeom_sf(data = bretagne_donut_l93, colour = my_blue, fill = my_blue, alpha = 0.2)``````",
null,
"## Let’s divide this area in multiples polygons\n\nWe use the coastlines as a basis.\n\n``````# density <- units::set_units(0.1, 1/km)\n\n# First merge everything and transform as lines\nbretagne_points_l93 <- bretagne_union_l93 %>%\n# transform polygons as multi-lines\nst_cast(\"MULTILINESTRING\") %>%\n# transform multi-lines as separate unique lines\nst_cast(\"LINESTRING\") %>%\n# transform as points, 0.1 per km (= 1 every 10 km)\n# Choose density according to precision needed\nst_line_sample(density = units::set_units(0.1, 1/km)) %>%\nst_sf() #%>%\n# remove empty geometry (small islands where sample is not possible with this density)\n# filter(!st_is_empty(geometry))\n\nggplot() +\ngeom_sf(data = bretagne_points_l93, colour = my_blue, fill = NA)``````",
null,
"``````# Create voronoi polygons around points\nbretagne_voronoi_l93 <- bretagne_points_l93 %>%\nst_combine() %>%\nst_voronoi() %>%\nst_cast()\n\nggplot() +\ngeom_sf(data = bretagne_voronoi_l93, colour = my_blue, fill = NA)``````",
null,
"## Join voronoi and donut polygon\n\n``````bretagne_voronoi_donut_l93 <- bretagne_voronoi_l93 %>%\nst_intersection(bretagne_donut_l93) %>%\nst_cast() %>%\nst_sf()\n\nggplot() +\ngeom_sf(data = bretagne_voronoi_donut_l93, colour = my_blue, fill = NA)``````",
null,
"# Join voronoi-donut polygons to original dataset to intersect with regions\n\nWe realise sort of a `left_join` based on spatial attributes using `st_join()`. This will keep the original geometry of our voronoi-donut polygons, while retrieving variables of joined terrestrial dataset when there is intersection.\n\n``````# Equivalent to left_join\nbretagne_vd_region_l93 <- bretagne_voronoi_donut_l93 %>%\nst_join(bretagne_l93)\n\nggplot() +\ngeom_sf(data = bretagne_vd_region_l93,\naes(colour = NOM_DEPT, fill = NOM_DEPT),\nalpha = 0.25)``````",
null,
"``````# Interactive view\nmapview::mapview(bretagne_vd_region_l93, zcol = \"NOM_DEPT\")``````\n\nIf we zoom on the map using `mapview()`, you can see that polygons between two regions sees a double attribution. This is because voronoi intersect with both regions. If you want to reduce the size of the overlapping polygons, you can increase the density of points used to create the voronoi polygons in `st_line_sample()`. However, we will always have the double attribution. Remember that `st_join()` works like `left_join()`. If there are duplicate keys, entities are doubled.\nTo avoid this, we will have to choose between one region or the other. First, we give a number to the voronoi polygons before joining to be able to identify those that are doubled. Then, we will choose to keep only the first of them in the list. Note that `summarise_all` does not exists for {sf} objects. I cannot use `group_by(id_v) + summarise_all(first)` as for classical datasets, neither the new `across()` function. I use `distinct()` with parameter `keep_all = TRUE` so that it keeps all variables with the first entity for each voronoi polygon.\n\n``````bretagne_vd_region_distinct_l93 <- bretagne_voronoi_donut_l93 %>%\nmutate(id_v = 1:n()) %>%\nst_join(bretagne_l93) %>%\n# group_by(id_v) %>%\n# summarise(across(.cols = everything(), .fns = first)) # Not on {sf}\n# summarise_all(.funs = first) # Not on {sf}\ndistinct(id_v, .keep_all = TRUE)\n\nggplot() +\ngeom_sf(data = bretagne_vd_region_distinct_l93,\naes(colour = NOM_DEPT, fill = NOM_DEPT),\nalpha = 0.25)``````",
null,
"``````# Interactive view\nmapview::mapview(bretagne_vd_region_distinct_l93, zcol = \"NOM_DEPT\")``````"
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