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2200	https://www.sigmaaldrich.com/US/en/product/saj/010310?srsltid=AfmBOoq7Q9OLmt5F2oEyUD08ebpcKWf-4NTTrU6IZ9jCAqK6bZjQj496	Acetic acid = 99.7 64-19-7 Skip to Content Products Cart 0 IL EN Products Products Applications Services Documents Support Analytical ChemistryCell Culture & AnalysisChemistry & BiochemicalsClinical & DiagnosticsFiltration Greener Alternative Products Industrial MicrobiologyLab AutomationLabwareMaterials ScienceMolecular Biology & Functional GenomicsmAbs Development & ManufacturingmRNA Development & ManufacturingPharma & Biopharma ManufacturingProtein BiologyWater Purification Analytical ChemistryCell Culture & AnalysisChemistry & SynthesisClinical & DiagnosticsEnvironmental & Cannabis TestingFood & Beverage Testing & ManufacturingGenomicsMaterials Science & EngineeringMicrobiological TestingmAbs Development & ManufacturingmRNA Development & ManufacturingPharma & Biopharma ManufacturingProtein BiologyResearch & Disease AreasWater Purification Contract ManufacturingContract TestingCustom ProductsDigital Solutions for Life ScienceIVD Development & ManufacturingProduct ServicesSupport Safety Data Sheets (SDS) Certificates of Analysis (COA) Certificates of Origin (COO) Certificates of Quality (COQ) Customer Support Contact Us Get Site Smart FAQ Quality & Regulatory Calculators & Apps Webinars Login Order Lookup Quick Order Cart 0 Products Acids 010310 Key Documents SDS COA COO View All Documentation Skip To Safety Information Documentation Peer Reviewed Papers Questions & Answers Reviews 01-0310 Share Acetic acid ≥99.7% No rating value Same page link. (0) Write a review Ask a question Synonym(s): Glacial acetic acid Sign Into View Organizational & Contract Pricing Select a Size Change View About This Item Linear Formula: CH 3 CO 2 H CAS Number: 64-19-7 Molecular Weight: 60.05 Beilstein: 506007 MDL number: MFCD00036152 UNSPSC Code: 51113400 PubChem Substance ID: 329747182 Pricing and availability is not currently available.Contact Technical Service Technical Service Need help? Our team of experienced scientists is here for you. 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Visit the Document Library Customers Also Viewed Slide 1 of 5 1 of 2 Sigma-Aldrich L092003 Ethyl acetate Quick View Sigma-Aldrich 8.18969 Isoamyl alcohol Quick View DENWAT Pure Water Density Standard Quick View Sigma-Aldrich A5376 Acetylsalicylic acid Quick View Supelco PX1000 o-Phosphoric Acid 85% W/W Quick View Peer Reviewed Papers Lysine acetylation in sexual stage malaria parasites is a target for antimalarial small molecules. Katharine Trenholme et al. Antimicrobial agents and chemotherapy, 58(7), 3666-3678 (2014-04-16) Therapies to prevent transmission of malaria parasites to the mosquito vector are a vital part of the global malaria elimination agenda. Primaquine is currently the only drug with such activity; however, its use is limited by side effects. The development Fibrosis, vascular activation, and immune abnormalities resembling systemic sclerosis in bleomycin-treated Fli-1-haploinsufficient mice. Takashi Taniguchi et al. Arthritis & rheumatology (Hoboken, N.J.), 67(2), 517-526 (2014-11-12) Fli-1, a potential predisposing factor for systemic sclerosis (SSc), is constitutively down-regulated in the lesional skin of patients with SSc by an epigenetic mechanism. To investigate the impact of Fli-1 deficiency on the induction of an SSc phenotype in various Netrin-1 controls sympathetic arterial innervation. Isabelle Brunet et al. The Journal of clinical investigation, 124(7), 3230-3240 (2014-06-18) Autonomic sympathetic nerves innervate peripheral resistance arteries, thereby regulating vascular tone and controlling blood supply to organs. Despite the fundamental importance of blood flow control, how sympathetic arterial innervation develops remains largely unknown. Here, we identified the axon guidance cue Functional diversification of duplicated chalcone synthase genes in anthocyanin biosynthesis of Gerbera hybrida. Xianbao Deng et al. The New phytologist, 201(4), 1469-1483 (2013-11-26) • Chalcone synthase (CHS) is the key enzyme in the first committed step of the flavonoid biosynthetic pathway and catalyzes the stepwise condensation of 4-coumaroyl-CoA and malonyl-CoA to naringenin chalcone. In plants, CHS is often encoded by a small family Phosphorylation of eIF4E promotes EMT and metastasis via translational control of SNAIL and MMP-3. N Robichaud et al. Oncogene, 34(16), 2032-2042 (2014-06-10) The progression of cancers from primary tumors to invasive and metastatic stages accounts for the overwhelming majority of cancer deaths. Understanding the molecular events which promote metastasis is thus critical in the clinic. Translational control is emerging as an important View All Related Papers Questions Be the first to ask a question Reviews ★★★★★ No rating value Be the first to review this product . 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2201	https://www.jove.com/science-education/v/12713/work-energy-theorem-for-rotational-motion	Skip to content Sign In EN - EnglishCN - 简体中文DE - DeutschES - EspañolKR - 한국어IT - ItalianoFR - FrançaisPT - Português do BrasilPL - PolskiHE - עִבְרִיתRU - РусскийJA - 日本語TR - TürkçeAR - العربية Sign In Start Free Trial #### JoVE Journal Peer reviewed scientific video journal ##### Cancer Research ##### Developmental Biology ##### View All #### JoVE Encyclopedia of Experiments Video encyclopedia of advanced research methods ##### Biological Techniques ##### Biology ##### Cancer Research ##### Microbiology #### JoVE Visualize Visualizing science through experiment videos EDUCATION #### JoVE Core Video textbooks for undergraduate courses ##### Analytical Chemistry ##### Anatomy and Physiology ##### Biology ##### Cell Biology ##### Chemistry ##### Civil Engineering ##### Electrical Engineering ##### View All #### JoVE Science Education Visual demonstrations of key scientific experiments ##### Advanced Biology ##### Basic Biology ##### Chemistry ##### View All #### JoVE Lab Manual Videos of experiments for undergraduate lab courses ##### Biology ##### Chemistry BUSINESS #### JoVE Business Video textbooks for business education ##### Finance ##### Marketing ##### Microeconomics OTHERS #### JoVE Quiz Interactive video based quizzes for formative assessments Authors Teaching Faculty Librarians K12 Schools Products RESEARCH JoVE Journal Peer reviewed scientific video journal JoVE Encyclopedia of Experiments Video encyclopedia of advanced research methods JoVE Visualize Visualizing science through experiment videos EDUCATION JoVE Core Video textbooks for undergraduates JoVE Science Education Visual demonstrations of key scientific experiments JoVE Lab Manual Videos of experiments for undergraduate lab courses BUSINESS JoVE Business Video textbooks for business education OTHERS JoVE Quiz Interactive video based quizzes for formative assessments Solutions Authors Teaching Faculty Librarians K12 Schools Language English EN English CN 简体中文 DE Deutsch ES Español KR 한국어 IT Italiano FR Français PT Português do Brasil PL Polski HE עִבְרִית RU Русский JA 日本語 TR Türkçe AR العربية Menu JoVE Journal Behavior Biochemistry Bioengineering Biology Cancer Research Chemistry Developmental Biology Engineering Environment Genetics Immunology and Infection Medicine Neuroscience Menu JoVE Encyclopedia of Experiments Biological Techniques Biology Cancer Research Immunology Neuroscience Microbiology Menu JoVE Core Analytical Chemistry Anatomy and Physiology Biology Cell Biology Chemistry Civil Engineering Electrical Engineering Introduction to Psychology Mechanical Engineering Medical-Surgical Nursing View All Menu JoVE Science Education Advanced Biology Basic Biology Chemistry Clinical Skills Engineering Environmental Sciences Physics Psychology View All Menu JoVE Lab Manual Biology Chemistry Menu JoVE Business Finance Marketing Microeconomics Start Free Trial Home JoVE Core Physics Work-Energy Theorem for Rotational Motion JoVE Core Physics A subscription to JoVE is required to view this content. Sign in or start your free trial. JoVE Core Physics Work-Energy Theorem for Rotational Motion Work-Energy Theorem for Rotational Motion 6,380 Views 01:11 min May 22, 2025 Overview The work-energy theorem for rotational motion is analogous to the work-energy theorem in translational motion. It states that the net work done by an external force to rotate a rigid body equals the change in the object's rotational kinetic energy. The power delivered is simply the time derivative of the work done; therefore, power is the dot product of torque and angular velocity. This relation is analogous to power in translational motion, which is given by the dot product of force and velocity. It is assumed that frictional force is absent here; however, this is not always the case for a real system. For example, an airplane's engine does work to set the propeller into a spinning motion. However, air friction and the friction between the mechanical parts of the engine lead to inevitable losses, as the work done by the engine translates into the change in rotational kinetic energy of the propeller. Therefore, even after the propeller gains the final desired angular velocity (and hence the desired rotational kinetic energy), the engine still needs to work to balance the opposing forces, which could otherwise slow down the spinning propeller. This text is adapted from Openstax, University Physics Volume 1, Section 10.8: Work and Power for Rotational Motion. Transcript Consider a rigid body, for instance a fidget spinner. If an external force is applied on it such that it rotates from θ1 to θ2, about a fixed axis, then the total work done on the body equals the integration of the product of the net torque on it and its angular displacement. Since the net torque on any rigid body is equal to the moment of inertia times angular acceleration, substituting the value of torque in the work expression and using the chain rule, dω/dt can be expressed as dθ/dt multiplied by dω/dθ. Now, dθ/dt is the angular velocity, ω. Therefore, the integrand in the equation equals Iω. Integrating within the limits of initial and final angular velocities, the net work done by the external force to rotate the rigid object about a fixed axis equals the change in the object’s rotational kinetic energy. This is the expression of the work-energy theorem for a rotating rigid body. 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2202	https://ocw.mit.edu/courses/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/0d0ec730e8e02ed0399d39ff05054a72_MIT8_333F13_Lec1.pdf	I. Thermodynamics I.A Fundamental deﬁnitions • Thermodynamics is a phenomenological description of equilibrium properties of macro scopic systems. ⋆ As a phenomenological description, it is based on a number of empirical observations which are summarized by the laws of thermodynamics. A coherent logical and mathe matical structure is then constructed on the basis of these observations, which leads to a variety of useful concepts, and to testable relationships among various quantities. The laws of thermodynamics can only be justiﬁed by a more fundamental (microscopic) theory of nature. For example, statistical mechanics attempts to obtain these laws starting from classical or quantum mechanical equations for the evolution of collections of particles. ⋆ A system under study is said to be in equilibrium when its properties do not change appreciably with time over the intervals of interest (observation times). The dependence on the observation time makes the concept of equilibrium subjective. For example, window glass is in equilibrium as a solid over many decades, but ﬂows like a ﬂuid over time scales of millennia. At the other extreme, it is perfectly legitimate to consider the equilibrium between matter and radiation in the early universe during the ﬁrst minutes of the big bang. ⋆ The macroscopic system in equilibrium is characterized by a number of thermodynamic coordinates or state functions. Some common examples of such coordinates are pressure and volume (for a ﬂuid), surface tension and area (for a ﬁlm), tension and length (for a wire), electric ﬁeld and polarization (for a dielectric), · · ·. A closed system is an ide alization similar to a point particle in mechanics in that it is assumed to be completely isolated by adiabatic walls that don’t allow any exchange of heat with the surroundings. By contrast, diathermic walls allow heat exchange for an open system. In addition to the above mechanical coordinates, the laws of thermodynamics imply the existence of other equilibrium state functions as described in the following sections. 1 I.B The zeroth law The zeroth law of thermodynamics describes the transitive nature of thermal equilib rium. It states: • If two systems, A and B, are separately in equilibrium with a third system C, then they are also in equilibrium with one another. Despite its apparent simplicity, the zeroth law has the consequence of implying the existence of an important state function, the empirical temperature Θ, such that systems in equilibrium are at the same temperature. Proof: Let the equilibrium state of systems A, B, and C be described by the coordinates {A1, A2, · · ·}, {B1, B2, · · ·}, and {C1, C2, · · ·} respectively. The assumption that A and C are in equilibrium implies a constraint between the coordinates of A and C, i.e. a change in A1 must be accompanied by some changes in {A2, · · · ; C1, C2, · · ·} to maintain equilibrium of A and C. Denote this constraint by fAC(A1, A2, · · · ; C1, C2, · · ·) = 0. (I.1) The equilibrium of B and C implies a similar constraint fBC(B1, B2, · · · ; C1, C2, · · ·) = 0. (I.2) Each of the above equations can be solved for C1 to yield C1 =FAC(A1, A2, · · · ; C2, · · ·), (I.3) C1 =FBC(B1, B2, · · · ; C2, · · ·). Thus if C is separately in equilibrium with A and B we must have FAC(A1, A2, · · · ; C2, · · ·) = FBC(B1, B2, · · · ; C2, · · ·). (I.4) However, according to the zeroth law there is also equilibrium between A and B, implying the constraint fAB(A1, A2, · · · ; B1, B2, · · ·) = 0. (I.5) Therefore it must be possible to simplify eq.(I.4) by cancelling the coordinates of C. Hence, the condition (I.5) for equilibrium of A and B must be expressible as ΘA(A1, A2, · · ·) = ΘB(B1, B2, · · ·), (I.6) 2 i.e., equilibrium is characterized by a function Θ of thermodynamic coordinates. This function speciﬁes the equation of state, and isotherms of A are described by the condition ΘA(A1, A2, · · ·) = Θ. Example: Consider three systems: (A) a wire of length L with tension F, (B) a param agnet of magnetization M in a magnetic ﬁeld B, and (C) a gas of volume V at pressure P. Observations indicate that when these systems are in equilibrium, the following constraints are satisﬁed between their coordinates: a P + (V −b)(L −L0) −c[F −K(L −L0)] = 0, V 2 a (I.7) P + (V −b)M −dB = 0. V 2 Clearly these constraints can be organized into three empirical temperature functions as a F B Θ ∝ P + (V −b) = c −K = d . (I.8) V 2 L −L0 M These are the well known equations of state describing:  P  + a/V 2)(V −b = NkBT (van der Waals gas)   M = (Nµ2 BB)/(3kBT) (Curie paramagnet) . (I.9)    F = (K + DT)(L −L0) (Hook ′ s law for rubber) The ideal gas temperature scale: As the above example indicates, the zeroth law merely states the presence of isotherms. In order to set up a practical temperature scale at this stage, a reference system is necessary. The ideal gas occupies an important place in thermodynamics and provides the necessary reference. Empirical observations indicate that the product of pressure and volume is constant along the isotherms of any gas that is suﬃciently dilute. The ideal gas refers to this dilute limit of real gases, and the ideal gas temperature is proportional to the product. The constant of proportionality is determined by reference to the temperature of the triple point of the ice–water–gas system, which was set to 273.16 degrees Kelvin (0K) by the 10th General Conference on Weights and Measures in 1954. Using a dilute gas (i.e. as P → 0) as thermometer, the temperature of a system can be obtained from T(oK) ≡ 273.16 × lim (PV )system/ lim (PV )ice−water−gas . (I.10) P →0 P →0 3 P I.C The First law We now consider transformations between diﬀerent equilibrium states. Such transfor mations can be achieved by applying work or heat to the system. The ﬁrst law states that both work and heat are forms of energy, and that the total energy is conserved. We shall use the following formulation: • The amount of work required to change the state of an otherwise adiabatically isolated system depends only on the initial and ﬁnal states, and not on the means by which the work is performed, or on the intermediate stages through which the system passes. As a consequence, we conclude the existence of another state function, the internal energy, E(X). Up to a constant, E(X) can be obtained from the amount of work ΔW needed for an adiabatic transformation from an initial state Xi to a ﬁnal state Xf , using ΔW = E(Xf ) −E(Xi). (I.11) In a generic (non–adiabatic) transformation, the amount of work does not equal to the change in the internal energy. The diﬀerence ΔQ = ΔE − ΔW is deﬁned as the heat intake of the system from its surroundings. Clearly in such transformations, ΔQ and ΔW are not separately functions of state, in that they depend on external factors such as the means of applying work, and not only on the ﬁnal states. To emphasize this, for a diﬀerential transformation we write ¯ = ¯ (I.12) dQ dE −dW, where dE = dQ and ¯ ∂iEdXi can be obtained by diﬀerentiation, while ¯ dW generally i can not. Also note the convention that the signs of work and heat are chosen to indicate the energy added to the system, and not vice versa. A quasi-static transformation is one that is performed suﬃciently slowly so that the system is always in equilibrium. Thus at any stage of the process, the thermodynamic coordinates of the system exist and can in principle be computed. For such transformations, the work done on the system (equal in magnitude but opposite in sign to the work done by the system) can be related to changes in these coordinates. Typically one can divide the state functions {X} into a set of generalized displacements {x}, and their conjugate generalized forces {J}, such that for an inﬁnitesimal quasi-static transformation X ¯ dW = Jidxi. (I.13) i 4 ¯ Table provides some common examples of such coordinates. Note that the displacement is usually an extensive quantity, i.e. proportional to system size, while the forces are intensive and independent of size. Also note that pressure is by convention calculated from the force exerted by the system on the walls, as opposed to the force on a spring which is exerted in the opposite direction. This is the origin of the negative sign that usually accompanies hydrostatic work. System Force Displacement Wire Tension F Length L Film Surface Tension S Area A Fluid Pressure −P Volume V Magnet Magnetic Field H Magnetization M Dielectric Electric Field E Polarization P Chemical Reaction Chemical Potential µ Particle Number N Table 1: Generalized Forces and Displacements Joule’s Free Expansion Experiment: Another important property of the ideal gas is the behavior of its internal energy. Observations indicate that if such a gas expands adiabatically (but not necessarily quasi-statically), from a volume Vi to Vf, the initial and ﬁnal temperatures are the same. Since the transformation is adiabatic (ΔQ = 0) and there is no external work done on the system (ΔW = 0), the internal energy of the gas is unchanged. Since the pressure and volume of the gas change in the process, but its temperature does not, we conclude that the internal energy depends only on temperature, i.e. E(V, T) = E(T). This property of the ideal gas is in fact a consequence of the form of its equation of state as will be proved in test 1 review problems. Response functions are the usual method for characterizing the macroscopic behav ior of a system. They are experimentally measured from the changes of thermodynamic coordinates with external probes. Some common response functions are: Heat Capacities are obtained from the change in temperature upon addition of heat to the system. Since heat is not a function of state, the path by which it is supplied must also be speciﬁed. For example, for a gas we can calculate the heat capacities at constant volume or pressure, denoted by CV = dQ/dT\|V and CP = dQ/dT\|P respectively. The latter is ¯ ¯ larger since some of the heat is used up in the work done in changes of volume: dQ ¯ dE −dW dE + PdV ∂E CV = = = = , dT dT dT ∂T V V V V (I.14) dQ ¯ ¯ dE −dW dE + PdV ∂E ∂V CP = = = = + P . dT dT dT ∂T ∂T P P P P P 5 Force Constants measure the (inﬁnitesimal) ratio of displacement to force and are gener alizations of the spring constant. Examples include the isothermal compressibility of a gas κT = − ∂V/∂P\|T /V , and the susceptibility of a magnet χT = ∂M/∂B\|T /V . From the equation of state of an ideal gas PV ∝ T, we obtain κT = 1/P. Thermal Responses probe the change in the thermodynamic coordinates with temperature. For example, the expansivity of a gas is given by αP = ∂V/∂T\|P /V , which equals 1/T for the ideal gas. Since the internal energy of an ideal gas depends only on its temperature, ∂E/∂T\|V = ∂E/∂T\|P = dE/dT, and eq.(I.14) simpliﬁes to ∂V PV CP −CV = P = PV αP = ≡ NkB. (I.15) ∂T T P The last equality follows from extensivity: for a given amount of ideal gas, the constant PV/T is proportional to N, the number of particles in the gas; the ratio is Boltzmann’s constant with a value of kB ≈ 1.4 × 10−23J0K−1 . 6 MIT OpenCourseWare 8.333 Statistical Mechanics I: Statistical Mechanics of Particles Fall 2013 For information about citing these materials or our Terms of Use, visit:
2203	https://biostatistics.letgen.org/mikes-biostatistics-book/linear-regression/relationship-between-the-slope-and-the-correlation/	17.2 – Relationship between the slope and the correlation 17.2 – Relationship between the slope and the correlation Introduction Coefficient of determination and the Product moment correlation Questions Chapter 17 contents Introduction Product moment correlation is used to indicate the direction and the strength of the linear association between two ratio-scale variables; the slope tells you the rate of change between the two variables. When the correlation is negative, the slope will be negative; when correlation is positive, so too will the slope. As you might suspect, there is a mathematical relationship between the product moment correlation, r, and the regression slope, b1. We haven’t spent much time explaining the equations presented in this text, but correlation and linear regression are such important tools it’s worth a closer look. Recall from our discussion in Chapter 16.1, the equation of the correlation is where the numerator is termed the covariance between X and Y and the denominator contains the standard deviations of X and Y variables. We can say the at the covariance is standardized by the variability in X and Y. In contrast, the regression slope is equal to the covariance divided by the variance in X. Thus, with a little algebra, we can see that the slope and correlation are equal to each other as This should drive home the following statistical reasoning point. You can always calculate a slope from a correlation, but recall that correlation analysis is intended as a test of the hypothesis of a linear association between variables for which cause and effect model — though perhaps reasonable — should be implied. Just because it is mathematically possible does not mean the analysis is correct for the problem. Coefficient of determination and the Product moment correlation , the coefficient of determination, was introduced in the last chapter. It’s the ratio of variation of the data explained by the linear regression model divided by the total variation in the data. Values range from 0% to 100% — it’s a measure of fit, how well the data are described by a line. Note that part of our description for , the product moment correlation — strength of the linear association — is simply another way to describe model fit. Thus, Of course, we lose the direction information by squaring the correlation. Questions If the correlation is 0.6, , and , what is the slope estimate? Chapter 17 contents Introduction Simple Linear Regression Relationship between the slope and the correlation Estimation of linear regression coefficients OLS, RMA, and smoothing functions Testing regression coefficients ANCOVA – analysis of covariance Regression model fit Assumptions and model diagnostics for Simple Linear Regression References and suggested readings (Ch17 & 18)
2204	https://www.scribd.com/document/434045629/Hamming-Distance-Wikipedia-1	Hamming Distance - Wikipedia \| PDF \| Encodings \| Notation Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 345 views 23 pages Hamming Distance - Wikipedia The Hamming distance is a measure of the minimum number of substitutions required to change one string into another. It is used in coding theory and information theory to detect and correct … Full description Uploaded by Er Mandar Kulkarni AI-enhanced title and description Go to previous items Go to next items Download Save Save Hamming Distance - Wikipedia (1) For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Hamming Distance - Wikipedia (1) For Later You are on page 1/ 23 Search Fullscreen Hamming distance This articleincludes alist of references,but its sources remain unclear because it has insuﬃcient inline citations . Please help to improve this article by introducing more precise citations. (May 2015) (Learn how and when to remove this templatemessage) adDownload to read ad-free 3-bit binary cube for ﬁnding Hamming distance Two example distances:100 → 011 has distance 3;010 → 111 has distance 2 The minimum distance between any two vertices is the Hamming distance between the two binary strings.4-bit binary tesseract for ﬁnding Hamming Two example distances: adDownload to read ad-free In information theory, the Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are differe nt.In other words, it measures the minimum number of substitutions required to change one string into the other, or the minimum number of errors that could have transformed one string into the other. In a more gener al context, the Hamming distance is one of sever al string metrics distance.0100 → 1001 has distance 3;0110 → 1110 has distance 1 adDownload to read ad-free for measuring the edit distance between two sequences. It is named after the American mathematician Richard Hamming.A major application is in coding theory,more speciﬁcally to block codes, in which the equal-length strings are vectors over a ﬁnite ﬁeld. Examples The Hamming distance between:" ka rol in " and " ka thr in " is 3." k a r ol in " and " k e r st in " is 3. 10 1 1 1 01 and 10 0 1 0 01 is 2. adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like BCS502 Module 2 Word No ratings yet BCS502 Module 2 Word 43 pages Hamming Distance Ppt2 No ratings yet Hamming Distance Ppt2 11 pages Hamming Code No ratings yet Hamming Code 8 pages GEMATMW Coding Theory Notes 100% (1) GEMATMW Coding Theory Notes 8 pages Approximate Matching No ratings yet Approximate Matching 16 pages Hamming Distance Explained No ratings yet Hamming Distance Explained 3 pages Hamming Algos No ratings yet Hamming Algos 1 page Block Coding and Error Detection No ratings yet Block Coding and Error Detection 2 pages Lecture 12 Distance Metrics Different Distance Metrics in Machine Learning No ratings yet Lecture 12 Distance Metrics Different Distance Metrics in Machine Learning 12 pages Binary Codes & Hamming Distance No ratings yet Binary Codes & Hamming Distance 5 pages Computer Networking Ind Assignment No ratings yet Computer Networking Ind Assignment 3 pages Coding Theory No ratings yet Coding Theory 4 pages Clustering Part4 No ratings yet Clustering Part4 79 pages Hamming Distance No ratings yet Hamming Distance 4 pages BioInfor Assignment No ratings yet BioInfor Assignment 4 pages Hamming Distance in Data Science No ratings yet Hamming Distance in Data Science 5 pages CN Assignment No2 No ratings yet CN Assignment No2 22 pages Coding Theory Tohoku June 09 No ratings yet Coding Theory Tohoku June 09 68 pages Linear Codes & Hamming Distance No ratings yet Linear Codes & Hamming Distance 30 pages Search Algorithms No ratings yet Search Algorithms 14 pages Hamming Codes No ratings yet Hamming Codes 7 pages Encoding Function Error Analysis No ratings yet Encoding Function Error Analysis 8 pages Hamming Code Design & Analysis No ratings yet Hamming Code Design & Analysis 7 pages Hamming Code No ratings yet Hamming Code 9 pages Lecture 10 No ratings yet Lecture 10 9 pages Hamming Codes: Error Correction Guide No ratings yet Hamming Codes: Error Correction Guide 31 pages Shreve S.E. 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2205	https://www.cut-the-knot.org/pythagoras/two_coins.shtml	Two coins puzzle Site What's new Content page Front page Index page About Privacy policy Help with math Subjects Arithmetic Algebra Geometry Probability Trigonometry Visual illusions Articles Cut the knot! What is what? Inventor's paradox Math as language Problem solving Collections Outline mathematics Book reviews Interactive activities Did you know? Eye opener Analogue gadgets Proofs in mathematics Things impossible Index/Glossary Simple math Fast Arithmetic Tips Stories for young Word problems Games and puzzles Our logo Make an identity Elementary geometry Two coins puzzle Place two identical coins side by side and roll one along the circumference of another without slipping. (Does it remind you of the way to draw epicycloids?). Two very related questions may now be asked: What will be the position of the rolling coin when it gets to the other side of the coin at rest, i.e., after traversing only one half of the circumference of the latter? Will it be upside down or the right side up? How many revolutions will the rolling coin undergo before it returns to its original position? The puzzle is actually an old-timer but never fails to surprise people who never saw it before. Even with the solution suggested by the picture on the right, it takes some experimenting before finally but still with the feeling of disbelief the right answer is accepted: The coin will be the right side up. It will take two revolutions for the rolling coin to get back to its starting position. The puzzle illustrates how shaky our intuition about motion (and, by extension, intuition in general) may be. From observing drawing of the epicycloid it becomes apparent that the point on the rolling coin nearest to the other coin (i.e. the point where the two coins touch) will be the farthest from the latter after the rolling coin traverses a half-circumference. It's not yet a proof but an experiment that might help our intuition to set things straight. The ultimate proof comes from the following observation. The rolling coin actually participates in two separate motions not unlike the moon relative to the earth: It rotates around its own center. It rotates around the center of the other coin. To answer the first question, note that in both motions the moving coin makes a half-turn. A half-turn around a point can be described as a reflection in that point. Choose three points A,B,C on the rolling coin. When reflected in a point O, A is carried into A' while, B and C are carried into B' and C', respectively. Another reflection in a point O', moves A', B', C' into A", B", and C", respectively. Now note, that O is the middle point of the segments AA' and BB' whereas O' is the middle point of two segments A'A" and B'B". Therefore, in both triangles AA'A" and BB'B", the line OO' connects midpoints of two sides thus being parallel and equal to half size of the third side. From here, AA" and BB" are equal and parallel (both, of course, are equal and parallel to CC" as well.) Since points A,B,C have been chosen arbitrary on the moving coin, we can now say that the whole coin was translated along a vector AA". Maxim R. made this remark: The main point is that there is no slipping, so the point of contact will travel the distance Pir on both circles. If we label the rightmost point of the right coin A and rotate the right coin until it's on the left of the other coin, the point of contact will be A. That means the coin is the same side up as at the start. References J.R. Newman, The World of Mathematics, v. 3, Simon & Schuster Books, 2001 \|Contact\|\|Front page\|\|Contents\|\|Geometry\|\|Up\| Copyright © 1996-2018 Alexander Bogomolny 73256491
2206	https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOorBhGG2UMkyJCsjLJk0brQ7L24-giev3CEw-N-leQWomt-niPTq	Art of Problem Solving AM-GM Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki AM-GM Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AM-GM Inequality In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same. In symbols, the inequality states that for any real numbers , with equality if and only if . The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests. Contents [hide] 1 Proofs 2 Generalizations 2.1 Weighted AM-GM Inequality 2.2 Mean Inequality Chain 2.3 Power Mean Inequality 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Proofs Main article: Proofs of AM-GM All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for . Generalizations The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM. Weighted AM-GM Inequality The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article. Mean Inequality Chain Main article: Mean Inequality Chain The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain. Power Mean Inequality Main article: Power Mean Inequality The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality. Problems Introductory For nonnegative real numbers , demonstrate that if then . (Solution) Find the maximum of for all positive . (Solution) Intermediate Find the minimum value of for . (Source) Olympiad Let , , and be positive real numbers. Prove that (Source) See Also Proofs of AM-GM Mean Inequality Chain Power Mean Inequality Cauchy-Schwarz Inequality Inequality Retrieved from " Categories: Algebra Inequalities Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2207	https://www.reddit.com/r/learnmath/comments/jqo1jn/is_there_an_easier_way_to_work_on_the_derivative/	Is there an easier way to work on the derivative of x^x? : r/learnmath Skip to main contentIs there an easier way to work on the derivative of x^x? : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 403K Members Online •5 yr. ago BlizzardAk Is there an easier way to work on the derivative of x^x? RESOLVED The question is in the title, and every way I see x^x, or even n^x where we differentiate for x just seems extremely overcomplicated. I've been primarily making it; y = n^x to ln(y) = x ln(n) But it doesn't feel like a definitive way of solving for any given function, and my professor wants us to use the form of making it e^f(x) which is just, way worst to work and more complex. Edit: Thanks for the answers, I think this is probably the worst thing I've had to learn in Calc 1. By far. Read more Share Related Answers Section Related Answers Best ways to find derivative of x^x Derivative of x ln x techniques How to differentiate natural log functions Derivative of e^x proof methods Effective strategies for mastering algebra New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of November 9, 2020 Reddit reReddit: Top posts of November 2020 Reddit reReddit: Top posts of 2020 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
2208	https://mathsdemo.cf.ac.uk/maths/resources/Probability_Answers.pdf	Charles M. Grinstead and J. Laurie Snell: INTRODUCTION to PROBABILITY Published by AMS Solutions to the exercises SECTION 1.1 1. As n increases, the proportion of heads gets closer to 1/2, but the diﬀerence between the number of heads and half the number of ﬂips tends to increase (although it will occasionally be 0). 2. n must be approximately 100. 3. (b) If one simulates a suﬃciently large number of rolls, one should be able to conclude that the gamblers were correct. 4. Player one has a probability of about .83 of winning. 5. The smallest n should be about 150. 7. The graph of winnings for betting on a color is much smoother (i.e. has smaller ﬂuctuations) than the graph for betting on a number. 8. For two tosses both probabilities are 1/2. For four tosses they are both 6/16. They are, in fact, the same for any even number of tosses. (This is not at all obvious; see Chapter 12 for a discussion of this and related topics.) 9. Each time you win, you either win an amount that you have already lost or one of the original numbers 1,2,3,4, and hence your net winning is just the sum of these four numbers. This is not a foolproof system, since you may reach a point where you have to bet more money than you have. If you and the bank had unlimited resources it would be foolproof. 10. You are very likely to win 5 dollars, but we shall see that this is still an unfair game, so we might say that Thackeray was right. 11. For two tosses, the probabilities that Peter wins 0 and 2 are 1/2 and 1/4, respectively. For four tosses, the probabilities that Peter wins 0, 2, and 4 are 3/8, 1/4, and 1/16, respectively. 13. Your simulation should result in about 25 days in a year having more than 60 percent boys in the large hospital and about 55 days in a year having more than 60 percent boys in the small hospital. 14. About 1/2 the time you win 2, 1/4 of the time you win 4, 1/8 of the time you win 8, etc. If you add up all of these potential winnings, weighted by their probabilities, you get ∞, so it would seem that you should be willing to pay quite a lot to play this game. Few are willing to pay more than $10. 15. In about 25 percent of the games the player will have a streak of ﬁve. 16. In the case of having children until they have a boy, they should have about 200,000 children. In the case that they have children until they have both a boy and a girl, they should have about 300,000 children, or about 100,000 more. SECTION 1.2 1. P({a, b, c}) = 1 P({a}) = 1/2 P({a, b}) = 5/6 P({b}) = 1/3 1 P({b, c}) = 1/2 P({c}) = 1/6 P({a, c}) = 2/3 P(φ) = 0 2. (a) Ω= {A elected, B elected}. (b) Ω= {Head,Tail}. (c) Ω= {(Jan., Mon.), (Jan., Tue.),. . . ,(Jan., Sun.),. . . ,(Dec., Sun.)}. (d) Ω= {Student 1,. . . ,Student 10}. (e) Ω= {A, B, C, D, F}. 3. (b), (d) 4. (a) In three tosses of a coin the ﬁrst outcome is a head. (b) In three tosses of a coin the same side turns up on each toss. (c) In three tosses of a coin exactly one tail turns up. (d) In three tosses of a coin at least one tail turns up. 5. (a) 1/2 (b) 1/4 (c) 3/8 (d) 7/8 6. 4 7. 7. 11/12 8. Art 1 4, Psychology 1 2, Geology 1 4. 9. 3/4, 1 10. 1 2. 11. 1 : 12, 1 : 3, 1 : 35 12. 3 4. 13. 11:4 14. (a) mY (2) = 1/5, mY (3) = 1/5, mY (4) = 2/5, mY (5) = 1/5 (b) mZ(0) = 1/5, mZ(1) = 3/5, mZ(4) = 1/5 15. Let the sample space be: ω1 = {A, A} ω4 = {B, A} ω7 = {C, A} ω2 = {A, B} ω5 = {B, B} ω8 = {C, B} ω3 = {A, C} ω6 = {B, C} ω9 = {C, C} where the ﬁrst grade is John’s and the second is Mary’s. You are given that P(ω4) + P(ω5) + P(ω6) = .3, P(ω2) + P(ω5) + P(ω8) = .4, P(ω5) + P(ω6) + P(ω8) = .1. Adding the ﬁrst two equations and subtracting the third, we obtain the desired probability as P(ω2) + P(ω4) + P(ω5) = .6. 16. 10 per cent. An example: 10 lost eye, ear, hand, and leg; 15 eye, ear, and hand; 20 eye, ear, and leg; 25 eye, hand, and leg; 30 ear, hand, and leg. 2 17. The sample space for a sequence of m experiments is the set of m-tuples of S’s and F’s, where S represents a success and F a failure. The probability assigned to a sample point with k successes and m −k failures is 1 n kn −1 n m−k . (a) Let k = 0 in the above expression. (b) If m = n log 2, then lim n→∞ 1 −1 n m = lim n→∞ 1 −1 n nlog 2 = lim n→∞( 1 −1 n nlog 2 = e−1log 2 = 1 2 . (c) Probably, since 6 log 2 ≈4.159 and 36 log 2 ≈24.953. 18. (a) The right-hand side is the sum of the probabilities of all outcomes occurring in the left-hand side plus some more because of duplication. (b) 1 ≥P(A ∪B) = P(A) + P(B) −P(A ∩B). 19. The left-side is the sum of the probabilities of all elements in one of the three sets. For the right side, if an outcome is in all three sets its probability is added three times, then subtracted three times, then added once, so in the ﬁnal sum it is counted just once. An element that is in exactly two sets is added twice, then subtracted once, and so it is counted correctly. Finally, an element in exactly one set is counted only once by the right side. 20. We would have to have the same probability assigned to all outcomes. If this probability is 0, the sum of the probabilities would be 0 so that P(Ω) = 0 instead of 1 as it should be. If this common probability is a > 0, then the sum of all the probabilities of the ﬁrst n outcomes would be na and for large enough n this would be greater than 1, contradicting the requirement that the sum of the probabilities for all possible outcomes should be 1. 21. 7/212 22. Ω= {1, 2, 3, . . .} and the distribution is m(n) = (5/6)n−1(1/6). Now if 0 < x < 1, then ∞ X n=0 xj = 1 1 −x . Hence (1/6) ∞ X n=1 (5/6)n−1 = (1/6) · 1 1 −5/6 = 1. 23. We have ∞ X n=0 m(ωn) = ∞ X n=0 r(1 −r)n = r 1 −(1 −r) = 1 . 24. He just meant that if you pick a month at random within a complete 400-year cycle of the calendar the thirteenth of the month is more likely to fall on Friday than on any other day. 3 25. They call it a fallacy because if the subjects are thinking about probabilities they should realize that P(Linda is bank teller and in feminist movement) ≤P(Linda is bank teller). One explanation is that the subjects are not thinking about probability as a measure of likelihood. For another explanation see Exercise 52 of Section 4.1. 26. The probability that the two cards are of the same rank is 52·3 52·51 = 1 17. Thus 2x + 1 17 = 1 giving x = 8 17 27. Px = P(male lives to age x) = number of male survivors at age x 100, 000 . Qx = P(female lives to age x) = number of female survivors at age x 100, 000 . 28. (a) 1 3 (b) P3(N) = [ N 3 ] N , where [ N 3 ] is the greatest integer in N 3 . Note that N 3 −1 ≤ N 3 ≤N 3 . From this we see that P3 = lim N→∞P3(N) = 1 3 . (c) If A is a ﬁnite set with K elements then A(N) N ≤K N , so lim N→∞ A(N) N = 0 . On the other hand, if A is the set of all positive integers, then lim N→∞ A(N) N = lim N→∞ N N = 1 . (d) Let Nk = 10k −1. Then the integers between Nk−1 + 1 and Nk have exactly k digits. Thus, if k is odd, then A(Nk) = (Nk −Nk−1) + (Nk−2 −Nk−3) + . . . , while if k is even, then A(Nk) = (Nk−1 −Nk−2) + (Nk−3 −Nk−4) + . . . . Thus, if k is odd, then A(Nk) ≥(Nk −Nk−1) = 9 · 10k−1 , while if k is even, then A(Nk) ≤Nk−1 < 10k−1 . So, if k is odd, then A(Nk) Nk ≥9 · 10k−1 10k −1 > 9 10 , 4 while if k is even, then A(Nk) Nk < 10k−1 10k −1 < 2 10 . Therefore, lim N→∞ A(N) N does not exist. 29. (Solution by Richard Beigel) (a) In order to emerge from the interchange going west, the car must go straight at the ﬁrst point of decision, then make 4n + 1 right turns, and ﬁnally go straight a second time. The probability P(r) of this occurring is P(r) = ∞ X n=0 (1 −r)2r4n+1 = r(1 −r)2 1 −r4 = 1 1 + r2 − 1 1 + r , if 0 ≤r < 1, but P(1) = 0. So P(1/2) = 2/15. (b) Using standard methods from calculus, one can show that P(r) attains a maximum at the value r = 1 + √ 5 2 − s 1 + √ 5 2 ≈.346 . At this value of r, P(r) ≈.15. 30. In order to depart to the east, one must make 4n + 3 right-hand turns in succession, and then go straight. The probability is P(r) = ∞ X n=0 (1 −r)r4n+3 = r3 (1 + r)(1 + r2) , if 0 ≤r < 1. This function increases on the interval [0, 1), so the maximum value of P(r), if a maximum exists, must occur at r = 1. Unfortunately, if r = 1, then the car never leaves the interchange, so no maximum exists. 31. (a) Assuming that each student gives any given tire as an answer with probability 1/4, then prob-ability that they both give the same answer is 1/4. (b) In this case, they will both answer ‘right front’ with probability (.58)2, etc. Thus, the probability that they both give the same answer is 39.8%. SECTION 2.1 The problems in this section are all computer programs. SECTION 2.2 1. (a) f(ω) = 1/8 on [2, 10] (b) P([a, b]) = b−a 8 . 5 2. (a) c = 1/48. (b) P(E) = 1 96(b2 −a2). (c) P(X > 5) = 75 96, P(x < 7) = 45 96 . (d) P(x2 −12x + 35 > 0) = P(x −5 > 0, x −7 > 0) + P(x −5 < 0, x −7 < 0) = P(x > 7) + P(x < 5) = 3 4 . 3. (a) C = 1 log 5 ≈.621 (b) P([a, b]) = (.621) log(b/a) (c) P(x > 5) = log 2 log 5 ≈.431 P(x < 7) = log(7/2) log 5 ≈.778 P(x2 −12x + 35 > 0) = log(25/7) log 5 ≈.791 . 4. (a) .04, (b) .36, (c) .25, (d) .09. 5. (a) 1 −1 e1 ≈.632 (b) 1 −1 e3 ≈.950 (c) 1 −1 e1 ≈.632 (d) 1 6. (a) e−.01T , (b) T = 100log(2) = 69.3. 7. (a) 1/3, (b) 1/2, (c) 1/2, (d) 1/3 8. (a) 1/8, (b) 1 2(1 + log(2)), (c) .75, (d) .25, (e) 3/4, (f) 1/4, (g) 1/8, (h) π/8, (i) π/4. 12. 1/4. 13. 2 log 2 −1. 14. (a) 13/24, (b) 1/48. 15. Yes. 16. Consider the circumference to be the interval [0, 1], as in the hint. Let A = 0. There are two cases to consider; 0 < B < 1/2, and 1/2 < B < 1. In the ﬁrst case, C must lie between 1/2 and B + 1/2, for otherwise there would be a gap of length greater than 1/2, corresponding to a semicircle containing none of the points. Similarly, if 1/2 < B < 1, it can be seen that C must lie between B −1/2 and 1/2. The probability of one of these two cases occurring is 1/4. SECTION 3.1 1. 24 2. 1/12 3. 232 6 4. At this writing, 37 Presidents have died. The probability that no two people from a group of 37 (all of whom are dead) died on the same day is about .15. Thus, the probability that at least two died on the same day is .85. Yes; Jeﬀerson, Adams, and Monroe (all signers of the Declaration of Independence) died on July 4. 5. 9, 6. 6. Since we do not get a diﬀerent situation if we rotate the table we can consider one person’s position as ﬁxed, and then there are (n −1)! possible arrangements for the other n −1 people. 7. 5! 55 . 8. Each subset S corresponds to a unique r-tuple of 0’s and 1’s, where a 1 in the i’th location means that i is an element of S. Since each location has two possibilities, there are 2r r-tuples, and hence there are 2r subsets. 10. 1/13 11. 3n −2 n3 , 7 27, 28 1000. 12. (a) 30 · 15 · 9 = 4050 (b) 4050 · (3 · 2 · 1) = 24300 (c) 148824 13. (a) 263 × 103 (b) 6 3 × 263 × 103 14. (a) 5 · 4 · 3 · 2 · 1 = 120 (b) 60 15. 3 1 × (2n −2) 3n . 16. The number of possible four-digit phone numbers is 104 = 10, 000, but there are more than this number of people in Atlanta who have a telephone. 17. 1 −12 · 11 · . . . · (12 −n + 1) 12n , if n ≤12, and 1, if n > 12. 18. 36 20. Think of the person on your right at lunch and at dinner as determining a permutation. Do the same for the person on your left at lunch and at dinner. We have two examples of the problem of a random permutation having no ﬁxed point. The probability of no match for large n for each random permutation would be approximately e−1 and if they were independent the probability of no match in either would be e−2. They are not quite independent but for large n they are close enough to being independent to make this a good estimate. 21. They are the same. 22. If x1, x2, . . . , x15, x16 are the number observed with a maximum of 56, and we assume there are N counterfeits then P(x1, x2, . . . , x15, x16) = ( 1 N )16 for any N ≥56 and 0 for any N < 56. Thus, this probability is greatest when N = 56. Your program should verify that Watson’s guess is much better. 23. (a) 1 n, 1 n (b) She will get the best candidate if the second best candidate is in the ﬁrst half and the best candidate is in the secon half. The probability that this happens is greater than 1/4. 7 SECTION 3.2 1. (a) 20 (b) .0064 (c) 21 (d) 1 (e) .0256 (f) 15 (g) 10 2. 10 5 = 252 3. 9 7 = 36 5. .998, .965, .729 6. If Charles has the ability, the probability that he wins is b(10, .75, 7) + b(10, .75, 8) + b(10, .75, 9) + b(10, .75, 10) = .776. If Charles is guessing, the probability that Ruth wins is 1 −b(10, .5, 7) −b(10, .5, 8) −b(10, .5, 9) −b(10, .5, 10) = .828. 7. b(n, p, j) b(n, p, j −1) = n j pjqn−j n j −1 pj−1qn−j+1 = n! j!(n −j)! (n −j + 1)!(j −1)! n! p q = (n −j + 1) j p q . But (n −j + 1) j p q ≥1 if and only if j ≤p(n + 1), and so j = [p(n + 1)] gives b(n, p, j) its largest value. If p(n + 1) is an integer there will be two possible values of j, namely j = p(n + 1) and j = p(n + 1) −1. 8. b(30, 1 6, 5) = 30 5 1 6 55 6 25 = .1921. The most probable number of times is 5. 9. n = 15, r = 7 10. 11 64 ≈.172 11. Eight pieces of each kind of pie. 12. (a) 4/ 52 5 ≈.0000015 (b) 36/ 52 5 ≈.000014 (c) 624/ 52 5 ≈.00024 (d) 3744/ 52 5 ≈.0014 (e) 5108/ 52 5 ≈.0020 (f) 10200/ 52 5 ≈.0039 8 13. The number of subsets of 2n objects of size j is 2n j . 2n i 2n i −1 = 2n −i + 1 i ≥1 ⇒i ≤n + 1 2. Thus i = n makes 2n i maximum. 14. By Stirling’s formula, n! ∼ √ 2πn(nn)e−n. Thus, b(2n, 1 2, n) = 20 n 1 22n = 2n! (n!)2 · 1 22n ∼1 22n √ 2π2n(2n)2ne−2n 2πn(n2n)e−2n = 1 √πn. 15. .3443, .441, .181, .027. 16. There are 8 3 ways of winning three games. After winning, there are 5 3 ways of losing three games. After losing, there is only one way of tying two games. Thus the the total number of ways to win three games, lose three games, and tie two is 8 3 5 3 = 560. 17. There are n a ways of putting a diﬀerent objects into the 1st box, and then n−a b ways of putting b diﬀerent objects into the 2nd and then one way to put the remaining objects into the 3rd box. Thus the total number of ways is n a n −a b = n! a!b!(n −a −b)!. 18. P(no student gets 2 or fewer correct) = b(340, 7/128, 0) ≈4.96 · 10−9; P(no student gets 0 correct) = b(340, 1/1024, 0) ≈.717. So Prosser is right to expect at least one student with 2 or fewer correct, but Crowell is wrong to expect at least one student with none correct. 19. (a) 4 1 13 10 52 10 = 7.23 × 10−8. (b) 4 1 3 2 13 4 13 3 13 3 52 10 = .044. (c) 4! 13 4 13 3 13 2 13 1 52 13 = .315. 20. (a) 13 6 / 52 6 ≈.000084 9 (b) 4 3 4 2 4 1 / 52 6 ≈.0000047 (c) 4 2 13 3 13 3 / 52 6 21. 3(25) −3 = 93 (We subtract 3 because the three pure colors are each counted twice.) 22. 8 2 = 28 23. To make the boxes, you need n + 1 bars, 2 on the ends and n −1 for the divisions. The n −1 bars and the r objects occupy n−1+r places. You can choose any n−1 of these n−1+r places for the bars and use the remaining r places for the objects. Thus the number of ways this can be done is n −1 + r n −1 = n −1 + r r . 24. 19 10 / 29 20 ≈.009 25. (a) 6! 10 6 /106 ≈.1512 (b) 10 6 / 15 6 ≈.042 26. (a) pq, qp, p2, q2 27. Ask John to make 42 trials and if he gets 27 or more correct accept his claim. Then the probability of a type I error is X k≥27 b(42, .5, k) = .044, and the probability of a type II error is 1 − X k≥27 b(42, .75, k) = .042. 28. n = 114, m = 81 29. b(n, p, m) = n m pm(1 −p)n−m. Taking the derivative with respect to p and setting this equal to 0 we obtain m(1 −p) = p(n −m) and so p = m/n. 30. (a) p(.5) = .5, p(.6) = .71, p(.7) = .87 (b) Mets have a 95.2% chance of winning in a 7-game series. 31. .999996. 32. If u = 1, you only need to be sure to send at least one to each side. If u = 0, it doesn’t matter what you do. Let v = 1 −u and q = 1 −p. If 0 < v < 1, let x be the nearest integer to n 2 −1 2 log(p/q) log v . 33. By Stirling’s formula, 2n n 2 4n 2n = (2n!)2(2n!)2 n!4(4n)! ∼ ( √ 4πn(2n)2ne−2n)4 ( √ 2πn(nn)e−n)4p 2π(4n)(4n)4ne−4n = r 2 πn. 10 34. Let Ei be the event that you do not get the ith player’s picture. Then for any k of these events P(Ei1 ∩Ei2 ∩. . . ∩Eik) = n −k n m . You have n k ways of choosing k diﬀerent Ei’s. Thus the result follows from Theorem 9. 35. Consider an urn with n red balls and n blue balls inside. The left side of the identity 2n n = n X j=0 n j 2 = n X j=0 n j n n −j counts the number of ways to choose n balls out of the 2n balls in the urn. The right hand counts the same thing but breaks the counting into the sum of the cases where there are exactly j red balls and n −j blue balls. 36. (a) n j (b) 1 − n −j j / n j 38. Consider the Pascal triangle (mod 3) for example. 0 1 1 1 1 2 1 2 1 3 1 0 0 1 4 1 1 0 1 1 5 1 2 1 1 2 1 6 1 0 0 2 0 0 1 7 1 1 0 2 2 0 1 1 8 1 2 1 2 1 2 1 2 1 9 1 0 0 0 0 0 0 0 0 1 10 1 1 0 0 0 0 0 0 0 1 1 11 1 2 1 0 0 0 0 0 0 1 2 1 12 1 0 0 1 0 0 0 0 0 1 0 0 1 13 1 1 0 1 1 0 0 0 0 1 1 0 1 1 14 1 2 1 1 2 1 0 0 0 1 2 1 1 2 1 15 1 0 0 2 0 0 1 0 0 1 0 0 2 0 0 1 16 1 1 0 2 2 0 1 1 0 1 1 0 2 2 0 1 1 17 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 18 1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 1 Note ﬁrst that the entries in the third row are 0 for 0 < j < 3. Lucas notes that this will be true for any p. To see this assume that 0 < j < p. Note that p j = p(p −1) · · · p −j + 1 j(j −1) · · · 1 is an integer. Since p is prime and 0 < j < p, p is not divisible by any of the terms of j!, and so (p −1)! must be divisible by j!. Thus for 0 < j < p we have p j = 0 mod p. Let us call the triangle of the ﬁrst three rows a basic triangle. The fact that the third row is 1 0 0 1 11 produces two more basic triangles in the next three rows and an inverted triangle of 0’s between these two basic triangles. This leads to the 6’th row 1 0 0 2 0 0 1 . This produces a basic triangle, a basic triangle multiplied by 2 (mod 3), and then another basic triangle in the next three rows. Again these triangles are separated by inverted 0 triangles. We can continue this way to construct the entire Pascal triangle as a bunch of multiples of basic triangles separated by inverted 0 triangles. We need only know what the mutiples are. The multiples in row np occur at positions 0, p, 2p, ..., np. Looking at the triangle we see that the multiple at position (mp, jp) is the sum of the multiples at positions (j −1)p and jp in the (m−1)p’th row. Thus these multiples satisfy the same recursion relation n j = n −1 j −1 + n −1 j that determined the Pascal triangle. Therefore the multiple at position (mp, jp) in the triangle is m j . Suppose we want to determine the value in the Pascal triangle mod p at the position (n, j). Let n = sp + s0 and j = rp + r0, where s0 and r0 are < p. Then the point (n, j) is at position (s0, r0) in a basic triangle multiplied by s r . Thus n j = s r s0 r0 . But now we can repeat this process with the pair (s, r) and continue until s < p. This gives us the result: n j = k Y i=0 si rj (mod p) , where s = s0 + s1p1 + s2p2 + · · · + skpk , j = r0 + r1p1 + r2p2 + · · · + rkpk . If rj > sj for some j then the result is 0 since, in this case, the pair (sj, rj) lies in one of the inverted 0 triangles. If we consider the row pk −1 then for all k, sk = p−1 and rk ≤p−1 so the product will be positive resulting in no zeros in the rows pk −1. In particular for p = 2 the rows pk −1 will consist of all 1’s. 39. b(2n, 1 2, n) = 2−2n 2n! n!n! = 2n(2n −1) · · · 2 · 1 2n · 2(n −1) · · · 2 · 2n · 2(n −1) · · · 2 = (2n −1)(2n −3) · · · 1 2n(2n −2) · · · 2 . SECTION 3.3 3. (a) 96.99% (b) 55.16% 12 SECTION 4.1 2. (a) 1/2 (b) 1/4 (c) 1/2 (d) 0 (e) 1/2 3. (a) 1/2 (b) 2/3 (c) 0 (d) 1/4 4. (a) 1/2 (b) 4/13 (c) 1/13 5. (a) (1) and (2) (b) (1) 6. 3/10 7. (a) P(A ∩B) = P(A ∩C) = P(B ∩C) = 1 4 , P(A)P(B) = P(A)P(C) = P(B)P(C) = 1 4 , P(A ∩B ∩C) = 1 4 ̸= P(A)P(B)P(C) = 1 8 . (b) P(A ∩C) = P(A)P(C) = 1 4, so C and A are independent, P(C ∩B) = P(B)P(C) = 1 4, so C and B are independent, P(C ∩(A ∩B)) = 1 4 ̸= P(C)P(A ∩B) = 1 8 , so C and A ∩B are not independent. 8. P(A ∩B ∩C) = P({a}) = 1 8 , P(A) = P(B) = P(C) = 1 2 . Thus while P(A ∩B ∩C) = P(A)P(B)P(C) = 1 8 , P(A ∩B) = P(A ∩C) = P(B ∩C) = 5 16 , P(A)P(B) = P(A)P(C) = P(B)P(C) = 1 4 . Therefore no two of these events are independent. 9. (a) 1/3 (b) 1/2 10. It is probably a reasonable estimate. One might refer to earlier life tables to see how much this number has changed over the last four or ﬁve censuses. 12. .0481 13. 1/2 13 14. 1/8 15. (a) 48 11 4 2 52 13 − 48 13 ≈.307 . (b) 48 11 3 1 51 12 ≈.328 . 16. P(A)P(B\|A)P(C\|A ∩B) = P(A) · P(A ∩B) P(A) · P(A ∩B ∩C) P(A ∩B) = P(A ∩B ∩C) . 17. (a) P(A ∩˜ B) = P(A) −P(A ∩B) = P(A) −P(A)P(B) = P(A)(1 −P(B)) = P(A)P( ˜ B) . (b) Use (a), replacing A by ˜ B and B by A. 18. P(D1\|+) = 4/9 P(D2\|+) = 1/3 P(D3\|+) = 2/9 19. .273. 20. It can be shown that after n draws, P(k white balls, n+1−k black balls in the urn) = 1/(n + 1) for any 0 ≤k ≤n. Thus you are equally likely to have any proportion of white balls after n draws. In fact, the fraction of white balls will tend to a limit but this limit is a random number. This rather suprising fact is a consequence of the fact that the Polya Urn model is mathematicallly exactly the same as the following apparently very diﬀerent model. You have a coin where the probability of heads p is chosen by rnd. Once this random p is chosen, the coin is tosses n times. If a head turns up you say you have a white ball, and if a tail turns up you have a black ball. Then the probability for any particular sequence of colors for the balls is exactly the same as the probability of this sequence occurring in the Polya urn model. In the coin model it is obvious that the proportion of heads will tend to a limit which is again a random number since it just depends upon what kind of a coin was chosen by rnd. This random coin model will be discussed more in the next section. 21. No. 22. 1/2 23. Put one white ball in one urn and all the rest in the other urn. This gives a probability of nearly 3/4, in particular greater than 1/2, for obtaining a white ball which is what you would have with an equal number of balls in each urn. Thus the best choice must have more white balls in one urn than the other. In the urn with more white balls, the best we can do is to have probability 1 of getting a white ball if this urn is chosen. In the urn with less white balls than black, the best we can do is to have one less white ball than black and then to have as many white balls as possible. Our solution is thus best for the urn with more white balls than black and also for the urn with more black balls than white. Therefore our solution is the best we can do. 24. P(A head on the jth trial and a total of k heads in n trials) = 1 2 nn −1 k −1 . P(Exactly k heads in n trials)= 1 2 nn k . 14 Thus P(Head on j’th trial \| k heads in n trials) = n −1 k −1 n k = k n . 25. We must have p n j pkqn−k = p n −1 k −1 pk−1qn−k . This will be true if and only if np = k. Thus p must equal k/n. 26. P(A\|B) = P(B\|A) implies that P(A) = P(B) . Thus, since since P(A ∩B) > 0, 1 = P(A ∪B) = P(A) + P(B) −P(A ∩B) < 2P(A) , and P(A) > 1 2. 27. (a) P(Pickwick has no umbrella, given that it rains)= 2 9 . (b) P(It does not rain, given that he brings his umbrella)= 5 12 . 28. The most obvious objection is the assumption that all of the events in question are independent. A more subtle objection is that, since Los Angeles is so large, it is reasonable to ask for the probability that there is a second couple with the same discription, given that there is one such couple. This probability is not so small. (See Exercise 23 of Section 9.3.) 29. P(Accepted by Dartmouth \| Accepted by Harvard) = 2 3 . The events ‘Accepted by Dartmouth’ and ‘Accepted by Harvard’ are not independent. 30. Neither has a convincing argument based upon comparing grouped data only. You need more information. For example, suppose that each defective bulb cost $10 whether a regular or a softglow bulb. Then in making 3000 bulbs the loss to A is $130 and to B is $110 so B has a smaller loss than A. But suppose that a defective regular bulb results in a loss of $20 and a defective softglow bulb in a loss of $10. Now making 3000 bulbs A has a loss of 2 × $20 + $11 × $10 = $150 while B has a loss of 5 × $20 + 6 × $10 = $160. Thus A has a smaller loss than B. Paradoxes caused by comparing percentages when data are grouped are called Simpson paradoxes . An interesting real life example can be found in Parsani, and Purvis, W.W. Norton l978. In a study of the admission to graduate school at the University of California, Berkeley in 1973 it was found that about 44% of the men who applied were admitted, but 35% of the women who applied were admitted. Suspecting sex bias, an attempt was made to locate where it occurred by examining the individual departments. But within individual departments there did not seem to be any bias. Indeed the only department with a signiﬁcant diﬀerence was one that favored women. Again more information is needed. In this case the explantion lay in the fact that the women applied to majors that were diﬃcult to get into (lower acceptance rate) and men applied generally to the majors that were easy to get in (high acceptance rate). In each of these examples you are interested in comparing one trait in the presence of a second confounding trait. In the ﬁrst example it was good or bad bulb confounded by the type of bulb and in the second example it was sex confounded by the major. Freedman et al. discuss how to control the confounding trait to make a more valid comparison. 31. The probability of a 60 year old male living to 80 is .41, and for a female it is .62. 15 32. (a) pq (b) 1 −(1 −p)(1 −q) (c) .958 33. You have to make a lot of calculations, all of which are like this: P( ˜ A1 ∩A2 ∩A3) = P(A2)P(A3) −P(A1)P(A2)P(A3) = P(A2)P(A3)(1 −P(A1)) = P( ˜ A1)P(A2)P(A3). 34. PXj = 1 0 1 4 3 4 . They are not independent. For example, if we know that X1 = 1, X2 = 1, and X3 = 1, then it must be the case that X4 = 1. 35. The random variables X1 and X2 have the same distributions, and in each case the range values are the integers between 1 and 10. The probability for each value is 1/10. They are independent. If the ﬁrst number is not replaced, the two distributions are the same as before but the two random variables are not independent. 36. p = 1 2 3 4 5 6 11 36 9 36 7 36 5 36 3 36 1 36 . 37. P(max(X, Y ) = a) = P(X = a, Y ≤a) + P(X ≤a, Y = a) −P(X = a, Y = a). P(min(X, Y ) = a) = P(X = a, Y > a) + P(X > a, Y = a) + P(X = a, Y = a). Thus P(max(X, Y ) = a) + P(min(X, Y ) = a) = P(X = a) + P(Y = a) and so u = t + s −r. 38. (a) pX = 0 1 2 1 4 1 2 1 4 , pY = 0 1 1 2 1 2 . (b) pZ = 0 1 2 3 1 8 3 8 3 8 1 8 . (c) pW = −1 0 1 2 1 8 3 8 3 8 1 8 . 39. (a) 1/9 (b) 1/4 (c) No (d) pZ = −2 −1 0 1 2 4 1 6 1 6 1 6 1 6 1 6 1 6 40. p = 1/2 pX = 0 1 1/2 1/2 pY = 3 4 5 1/4 3/8 3/8 Independent p = 2/3 pX = 0 1 17/81 64/81 pY = 3 4 5 1/3 10/27 8/27 Not independent 42. Let u = N −r and v = N −s be the number of games that A and B, respectively, must win to win the series. Then the series will surely be over in u + v −1 games, so Fermat extended the game to assure this many plays. The player with the most points in the extended game wins. Therefore, P(r, s) = P(u, v) = u+v−1 X j=u u + v −1 j pjqu+v−1−j. 16 An alternative formula can be derived without extending the game. To win the series in u + j games, A must win u −1 games among the ﬁrst u + j −1 games and then win the (u + j)th game with j ≤v −1. Thus, P(r, s) = P(u, v) = v−1 X j=0 u + j −1 j puqj. 43. .710. 44. (a) First convention. If you serve N + 1 times, then your opponent must serve N times. The total number of points played is 2N + 1, so one of you must have won at least N + 1 points. That is a contradition, since the game is over when a player has won N points. Second convention. If you serve N + 1 times, then except for the ﬁrst time, before each time you serve, you have won a point. Thus at the (N + 1)st time you serve you have already won N points. The game should have already ended. Therefore, you serve at most N times. Before each serve of your apponent he won the previous point. Thus, as he serves for the Nth time he has already won N points. Therefore, your opponent serves at most N −1 times. (b) Since the total number of points for the two players is 2N −1, and one player has already got N points before the game is extended, the other can get at most N −1 points in the extended game and hence not change the winner. (c) For the extended game, probabilistically, the two methods are the same: in one, we have N Bernoulli trials with probability p for success and in the other we have N −1 trials with probability ¯ p for success, and in either method you win if you win the most points. 45. (a) The probability that the ﬁrst player wins under either service convention is equal to the proba-bility that if a coin has probability p of coming up heads, and the coin is tossed 2N + 1 times, then it comes up heads more often than tails. This probability is clearly greater than .5 if and only if p > .5. (b) If the ﬁrst team is serving on a given play, it will win the next point if and only if one of the following sequences of plays occurs (where ‘W’ means that the team that is serving wins the play, and ‘L’ means that the team that is serving loses the play): W, LLW, LLLLW, . . . . The probability that this happens is equal to p + q2p + q4p + . . . , which equals p 1 −q2 = 1 1 + q . Now, consider the game where a ‘new play’ is deﬁned to be a sequence of plays that ends with a point being scored. Then the service convention is that at the beginning of a new play, the team that won the last new play serves. This is the same convention as the second convention in the preceding problem. ¿From part a), we know that the ﬁrst team to serve under the second service convention will win the game more than half the time if and only if p > .5. In the present case, we use the new value of p, which is 1/(1 + q). This is easily seen to be greater than .5 as long as q < 1. Thus, as long as p > 0, the ﬁrst team to serve will win the game more than half the time. 46. P(X = i) = P(Y = i) = 4 i 5 −i 48 52 5 , 17 P(X = i, Y = j) = 4 i 4 j 44 5 −i −j 52 5 if i ≤4, j ≤4, and i + j ≤5, P(X = i, Y = j) = 0, otherwise. 47. (a) P(Y1 = r, Y2 = s) = P(Φ1(X1) = r, Φ2(X2) = s) = X Φ1(a)=r Φ2(b)=s P(X1 = a, X2 = b) . (b) If X1, X2 are independent, then P(Y1 = r, Y2 = s) = X Φ1(a)=r Φ2(b)=s P(X1 = a, X2 = b) = X Φ1(a)=r Φ2(b)=s P(X1 = a)P(X2 = b) = X Φ1(a)=r P(X1 = a) X Φ2(b)=s P(X2 = b) = P(Φ1(X1) = r)P(Φ2(X2) = s) = P(Y1 = r)P(Y2 = s) , so Y1 and Y2 are independent. 48. X ω∈Ω mE(ω) = 1 P(E) X ω∈Ω P(ω ∩E) = 1 P(E)P(E) = 1 . 49. P(both coins turn up using (a)) = 1 2p2 1 + 1 2p2 2. P(both coins turn up heads using (b)) = p1p2. Since (p1 −p2)2 = p2 1 −2p1p2 + p2 2 > 0, we see that p1p2 < 1 2p2 1 + 1 2p2 2, and so (a) is better. 50. For any sequence B1, . . . , Bn with Bk = Ak or ˜ Ak P(B1 ∩· · · ∩Bn) = P(B1)P(B2) · · · P(Bn) > 0 . Thus there is at least one sample point ω in each of the sets B1 ∩B2 ∩· · · ∩Bn. Since there are 2n such subsets, there must be at least this many sample points. 51. P(A) = P(A\|C)P(C) + P(A\| ˜ C)P( ˜ C) ≥P(B\|C)P(C) + P(B\| ˜ C)P( ˜ C) = P(B) . 52. P(coin not found in the i′th box) = P(coin not in i′th box) + P(coin in i′th box but not found) = 1 −pi + (1 −ai)pi = 1 −aipi . Thus P(coin is in j′th box \| not found in i′th box) = pj/(1 −aipi) if j ̸= i. 18 P(coin is in the i′th box \| not found in the i′th box) = 1 − X j̸=i P(coin is in j′th box \| not found in i′th box) = 1 − 1 −pi (1 −aipi) = (1 −ai)pi. 1 −aipi . 53. We assume that John and Mary sign up for two courses. Their cards are dropped, one of the cards gets stepped on, and only one course can be read on this card. Call card I the card that was not stepped on and on which the registrar can read government 35 and mathematics 23; call card II the card that was stepped on and on which he can just read mathematics 23. There are four possibilities for these two cards. They are: Card I Card II Prob. Cond. Prob. Mary(gov,math) John(gov, math) .0015 .224 Mary(gov,math) John(other,math) .0025 .373 John(gov,math) Mary(gov,math) .0015 .224 John(gov,math) Mary(other,math) .0012 .179 In the third column we have written the probability that each case will occur. For example, for the ﬁrst one we compute the probability that the students will take the appropriate courses: .5 × .1 × .3 × .2 = .0030 and then we multiply by 1/2, the probability that it was John’s card that was stepped on. Now to get the conditional probabilities we must renormalize these probabilities so that they add up to one. In this way we obtain the results in the last column. From this we see that the probability that card I is Mary’s is .597 and that card I is John’s is .403, so it is more likely that that the card on which the registrar sees Mathematics 23 and Government 35 is Mary’s. 54. (a) Let A = {(a, •):a ∈a subset A of {♣, ♦, ♥, ♠}, • ∈{2, 3, . . . , J, Q, K, A}} be a suit event and B = {(•, b):b ∈a subset B of {2, 3, . . . , J, Q, K, A}, • ∈{♣, ♦, ♥, ♠}} be a rank event. Then P(A) = size of A 4 P(B) = size of B 13 P(A ∩B) = P{(a, b):a ∈A, b ∈B} = (size of A)(size of B) 52 = P(A)P(B) . (b) The possible sizes of a rank event are (4i + 3j), where i = 0,. . . ,12 and j= 0 or 1, and the possible sizes of a suit event are (13m + 12n), where m = 0,1,2,3, and n = 0,1. By the hint we must have (4i + 3j)(13m + 12n) 51 an integer. This means that either 4i + 3 or m + 12n must be divisible by 17. We show that this is not possible. Assume for example that 4i + 3j = 17k where k is 1 or 2. Since 19 17k = 16k + k, when 17k is divided by 4 we would get a remainder of k, that is, 1 or 2. But when 4i + 3j is divided by 4 we get a remainder of 3j, which is 0 or 3 depending on the value of j. Therefore we cannot have 4i + 3j = 17k, for k= 1 or 2 and j = 0 or 1. Similarly, assume that 13m + 12n = 17k with k = 1 or 2. Since 17k = 13k + 4k, when 17k is divided by 13 we get a remainder of 4 or 8 depending on the value of k, but when 13m + 12n is divided by 13 we get a remainder of 0 or 12 depending on the value of n. Thus we cannot have 17k = 13m + 12n for k = 1 or 2 and n = 0 or 1. (c) A = {(♠, 2), (♠, 3), (♠, 4)} B = {(♠, 4), . . . , (♠, 7), (♥, 1), . . . , (♥, K)}. (d) Let a be the size of A, b the size of B, and c the size of A ∩B. Then A and B independent implies that c 53 = a 53 · b 53. Thus ab = 53c. But, since 53 is prime, this means that either a or b must be 53, which means that either A or B must be trivial. 55. P(R1) = 4 52 5 = 1.54 × 10−6. P(R2 ∩R1) = 4 · 3 52 5 47 5 . Thus P(R2 \| R1) = 3 47 5 = 1.96 × 10−6. Since P(R2\|R1) > P(R1), a royal ﬂush is attractive. P(player 2 has a full house) = 13 · 12 4 3 4 2 52 5 . P(player 1 has a ﬂush and player 2 has a full house) = 4 · 8 · 7 4 3 4 2 + 4 · 8 · 5 4 3 · 3 2 + 4 · 5 · 8 · 3 3 4 2 + 4 · 5 · 4 3 3 3 2 52 5 47 5 . Taking the ratio of these last two quantities gives: P(player 1 has a royal ﬂush \| player 2 has a full house) = 1.479 × 10−6. Since this probability is less than the probability that player 1 has a royal ﬂush (1.54 × 10−6), a full house repels a royal ﬂush. 56. P(B\|A) > P(B) ⇔P(B ∩A) > P(A)P(B) ⇔P(A\|B) = P(A ∩B) P(B) > P(A) . 20 57. P(B\|A) ≤P(B) and P(B\|A) ≥P(A) ⇔P(B ∩A) ≤P(A)P(B) and P(B ∩A) ≥P(A)P(B) ⇔P(A ∩B) = P(A)P(B) . 58. P(A\|B) > P(A) ⇔P(A ∩B) > P(A)P(B) ⇔P(A ∩B) −P(A)P(A ∩B) > P(A)P(B) −P(A)P(B ∩A) ⇔P(A ∩B)P( ˜ A) > P(A)P(B∩˜ A) ⇔P(A ∩B) P(A) > P(B ∩˜ A) P( ˜ A) ⇔P(B\|A) > P(B\| ˜ A) . 59. Since A attracts B, P(B\|A) > P(A) and P(B ∩A) > P(A)P(B) , and so P(A) −P(B ∩A) < P(A) −P(A)P(B) . Therefore, P( ˜ B ∩A) < P(A)P( ˜ B) , P( ˜ B\|A) < P( ˜ B) , and A repels ˜ B. 60. If A attracts B and C, and A repels B ∩C then we have P(A ∩B) > P(A)P(B) , P(A ∩C) > P(A)P(C) , P(B ∩C ∩A) < P(B ∩C)P(A) . Thus P(A ∩(B ∪C)) = P((A ∩B) ∪(A ∩C)) = P(A ∩B) + P(A ∩C) −P(A ∩B ∩C) > P(A)P(B) + P(A)P(C) −P(B ∩C)P(A) = P(A)(P(B) + P(C) −P(B ∩C)) = P(A)P(B ∪C) . Therefore P(B ∪C\|A) > P(B ∪C) . Here is an example in which A attracts B and C and repels B ∪C. Let Ω= {a, b, c, d} , p(a) = .2, p(b) = .25, p(c) = .25, p(d) = .3 . Let A = {a, d}, B = {b, d}, C = {c, d} . Then P(B\|A) = .6 > P(B) = .55 , 21 P(C\|A) = .6 > P(C) = .55 , and P(B ∪C\|A) = .6 < P(B ∪C) = .8 . 61. Assume that A attracts B1, but A does not repel any of the Bj’s. Then P(A ∩B1) > P(A)P(B1), and P(A ∩Bj) ≥P(A)P(Bj), 1 ≤j ≤n. Then P(A) = P(A ∩Ω) = P(A ∩(B1 ∪. . . ∪Bn)) = P(A ∩B1) + · · · + P(A ∩Bn) > P(A)P(B1) + · · · + P(A)P(Bn) = P(A) P(B1) + · · · + P(Bn) = P(A) , which is a contradiction. SECTION 4.2 1. (a) 2/3 (b) 1/3 (c) 1/2 (d) 1/2 2. (a) 1 −e−.09 = .086 (b) 1 −e−.5 = .393 (c) 1 (d) (1 −e−1)/(1 −e−2) = .731 3. (a) .01 (b) e−.01 T where T is the time after 20 hours. (c) e−.2 ≈.819 (d) 1 −e−.01 ≈.010 4. (a) 1/2 (b) 1/4 (c) 3/4 (d) 1/2 5. (a) 1 (b) 1 (c) 1/2 (d) π/8 22 (e) 1/2 6. (a) f(x) = 4/3, if 1/4 < x < 1, 0, otherwise. (b) f(t) = e−t/(e−1 −e−10), if 1 < t < 10, 0, otherwise. (c) f(x, y) = π/50, if (x, y) is in upper half of the target, 0, otherwise. (d) f(x, y) = 2, if x > y, 0, otherwise. 7. P(X > 1 3, Y > 2 3) = Z 1 1 3 Z 1 2 3 dydx = 2 9 . But P(X > 1 3)P(Y > 2 3) = 2 3 · 1 3 , so X and Y are independent. 8. a and b; c and d. 10. (b) Let Z be a number chosen with uniform density on the interval [0, a]. We ﬁnd the density for max(Z, a −Z). This density is nonzero only on the interval [a/2,a]. For x in this interval: P(max(Z, a −Z)) ≤x) = P(a −Z ≥Z, a −Z ≤x) + P(Z > a −Z, Z ≤x) = P(Z ≥a −x, Z ≤a 2) + P(Z > a 2, Z ≤x) = 2x −a a . Taking a = 1, we see that the density for the length of the largest stick from the ﬁrst cut is uniform on the interval [ 1 2, 1]. Assume that the length of this longest piece is X. Let Y be position on the interval [0, X] of the second cut. Then we obtain a triangle if max(Y, X −Y ) ≤1 2. But by our ﬁrst computation with a = X and x = 1 2, we see that P(max(Y, X −Y )) ≤1 2 = 1 −X X . Thus P(triangle) = 2 Z 1 1 2 1 −x x dx = 2 log 2 −1 . 11. If you have drawn n times (total number of balls in the urn is now n + 2) and gotten j black balls, (total number of black balls is now j + 1), then the probability of getting a black ball next time is (j + 1)/(n + 2). Thus at each time the conditional probability for the next outcome is the same in the two models. This means that the models are determined by the same probability distribution, so either model can be used in making predictions. Now in the coin model, it is clear that the proportion of heads will tend to the unknown bias p in the long run. Since the value of p was assumed to be unformly distributed, this limiting value has a random value between 0 and 1. Since this is true in the coin model, it is also true in the Polya Urn model for the proportion of black balls.(See Exercise 20 of Section 4.1.) 12. A new beta density with α = 6 and β = 9. It will be successful next time with probability .4. SECTION 4.3 23 1. 2/3 2. . Let M be the event that the hand has an ace. Let N be the event that the hand has at least two aces. Then P(N\|M) = P(N ∩M) P(M) = P(N) P(M) = 4 2 48 11 + 4 3 48 10 + 4 4 48 9 52 13 − 48 13 ≈.3696 . Let S be the event that the hand has the ace of hearts. Then P(N\|S) = P(S ∩T) P(S) = 3 1 48 11 + 3 2 48 10 + 3 3 48 9 52 13 − 51 13 ≈.5612 . 3. (a) Consider a tree where the ﬁrst branching corresponds to the number of aces held by the player, and the second branching corresponds to whether the player answers ‘ace of hearts’ or anything else, when asked to name an ace in his hand. Then there are four branches, corresponding to the numbers 1, 2, 3, and 4, and each of these except the ﬁrst splits into two branches. Thus, there are seven paths in this tree, four of which correspond to the answer ‘ace of hearts.’ The conditional probability that he has a second ace, given that he has answered ‘ace of hearts,’ is therefore 48 12 + 1 2 3 1 48 11 + 1 3 3 2 48 10 + 1 4 3 3 48 9 .52 13 ! 51 12 .52 13 ! ≈.6962 . (b) This answer is the same as the second answer in Exercise 2, namely .5612. 5. Let x = 2k. It is easy to check that if k ≥1, then px/2 px/2 + px = 3 4 . If x = 1, then px/2 px/2 + px = 0 . Thus, you should switch if and only if your envelope contains 1. SECTION 5.1 1. (a), (c), (d) 2. Yes. 3. Assume that X is uniformly distributed, and let the countable set of values be {ω1, ω2, . . .}. Let p be the probability assigned to each outcome by the distribution function f of X. If p > 0, then ∞ X i=1 f(ωi) = ∞ X i=1 p , 24 and this last sum does not converge. If p = 0, then ∞ X i=1 f(ωi) = 0 . So, in both cases, we arrive at a contradiction, since for a distribution function, we must have ∞ X i=1 f(ωi) = 1 . 4. One way to help decide whether the given experiment will result in a random subset is to ask whether one might reasonably expect any of the possible subsets to occur if the experiment is performed. Since some close friends almost always eat lunch together, the ﬁrst experiment will almost never give a subset which has as a member exactly one of a pair of close friends. The second experiment will always give the same subset when performed on the same student body. In addition, Social Security numbers are assigned based upon geography, among other things. Thus, depending upon the use we are going to make of this subset, this experiment may or may not give an adequate subset. Finally, the third experiment will probably return each subset with approximately the same probability. (A tenth-ﬂoor window would be much better.) 5. (b) Ask the Registrar to sort by using the sixth, seventh, and ninth digits in the Social Security numbers. (c) Shuﬄe the cards 20 times and then take the top 100 cards. (Can you think of a method of shuﬄing 3000 cards? 6. The distribution function of Y is given by f(x) = (k −x + 1)n −(k −x)n kn , for 1 ≤x ≤k. The numerator counts the number of n-tuples, all of whose entries are at least x, and subtracts the number of n-tuples, all of whose entries are at least x + 1. 7. (a) pj(n) = 1 6 5 6 n−1 for j = 0, 1, 2, . . . . (b) P(T > 3) = (5 6)3 = 125 216 . (c) P(T > 6 \| T > 3) = (5 6)3 = 125 216 . 8. 1 8 . 9. (a) 1000 (b) 100 10 N−100 90 N 100 (c) N = 999 or N = 1000 10. (a) N k N−k n1−k N−n1 n2−k N n1 N n2 (b) Let pN denote the probability that X = n12, given that the population size is N. Then pN equals the expression in the answer to part (a), with k replaced by n12. After some gruesome algebra, one obtains pN+1 pN = N 2 −(n1 + n2 −2)N + (n1 −1)(n2 −1) N 2 −(n1 + n2 −n12 −2)N + (n12 + 1 −n1 −n2 . 25 Let aN and bN denote the numerator and denominator of this expression. We want the smallest value of N for which the expression is less than or equal to 1, or equivalently, we want the smallest value of N for which aN ≤bN. If we solve this inequality for N, we obtain N = ln1n2 n12 m −1 . 13. .7408, .2222, .0370 14. (a) e−10 ≈4.54 × 10−5 (b) We need e−n/1000 = 1/2, or n = 1000 · log2 ≈694. 16. P(miss 0 calls) + P(miss 1 call) = .0498+.1494 = .1992. 17. 649741 18. (a) m = 600 × 1 500 so P(no raisins) = e−m = .301. (b) m = 400 × 1 500, so P(exactly two chocolate chips) = m2e−m 2! = .144. (c) m = 1000 × 1 500, so P(at least two bits) = 1 −P(0 bits) −P(1 bit) = 1 −.1353 −.2707 = .594. 19. The probability of at least one call in a given day with n hands of bridge can be estimated by 1 −e−n·(6.3×10−12). To have an average of one per year we would want this to be equal to 1 365. This would require that n be about 400,000,000 and that the players play on the average 8,700 hands a day. Very unlikely! It’s much more likely that someone is playing a practical joke. 20. e−5 ≈.00674 21. (a) b(32, j, 1/80) = 32 j 1 80 j79 80 32−j (b) Use λ = 32/80 = 2/5. The approximate probability that a given student is called on j times is e−2/5(2/5)j/j! . Thus, the approximate probability that a given student is called on more than twice is 1 −e−2/5 (2/5)0 0! + (2/5)1 1! + (2/5)2 2! ≈.0079 . 22. .0077 23. P(outcome is j + 1)/P(outcome is j) = mj+1e−m (j + 1)! .mje−m j! = m j + 1. Thus when j +1 ≤m, the probability is increasing, and when j +1 ≥m it is decreasing. Therefore, j = m is a maximum value. If m is an integer, then the ratio will be one for j = m −1, and so both j = m −1 and j = m will be maximum values. (cf. Exercise 7 of Chapter 3, Section 2) 24. The probability that Kemeny receives no mail on a given weekday can be estimated by e−10 = 4.54×10−5. Thus in ten years, the probability that at least one day brings no mail can be estimated by 1 −e−3000·4.54×10−5 = .127. Thus he ﬁnds that the probability is .127, which is inconclusive. 25. Without paying the meter Prosser pays 2 · 5e−5 1! + (5 · 2)52e−5 2! + · · · (5 · n)5ne−5 n! + · · · = 25 −15e−5 = $24.90. 26 He is better oﬀputting a dime in the meter each time for a total cost of $10. 26. number observed expected 0 229 227 1 211 211 2 93 99 3 35 31 4 7 9 5 1 1 27. m = 100 × (.001) = .1. Thus P(at least one accident) = 1 −e−.1 = .0952. 28. .9084 29. Here m = 500 × (1/500) = 1, and so P(at least one fake) = 1 −e−1 = .632. If the king tests two coins from each of 250 boxes, then m =250 × 2 500 = 1, and so the answer is again .632. 30. P(win ≥3 times) ≈.5071, expected winnings ≈−2.703 31. The expected number of deaths per corps per year is 1 · 91 280 + 2 · 32 280 + 3 · 11 280 + 4 · 2 280 = .70. The expected number of corps with x deaths would then be 280· (.70)xe−(.70) x! . From this we obtain the following comparison: Number of deaths Corps with x deaths Expected number of Corps 0 144 139.0 1 91 97.3 2 32 34.1 3 11 7.9 ≥4 2 1.6 The ﬁt is quite good. 32. P(X + Y = j) = j X k=0 P(X = k)P(Y = j −k) = j X k=0 mke−m k! · ¯ mj−ke−¯ m (j −k)! = e−(m+ ¯ m) j X k=0 j!mk ¯ mj−k k!(j −k)! 1 j! = e−(m+ ¯ m) 1 j! j X k=0 j k mk ¯ mj−k = (m + ¯ m)j j! e−(m+ ¯ m). Thus, X + Y has a Poisson density with mean m + ¯ m. 33. Poisson with mean 3. 34. .168 35. (a) In order to have d defective items in s items, you must choose d items out of D defective ones and the rest from S −D good ones. The total number of sample points is the number of ways to choose s out of S. 27 (b) Since min(D,s) X j=0 P(X = j) = 1, we get min(D,s) X j=0 D j s −D s −j = S s . 36. D = 20. This illustrates the general fact that the maximum probability is achieved when d D = s S . 37. The maximum likelihood principle gives an estimate of 1250 moose. 38. With replacement: P(X = 1) ≈.396 Without replacement: P(X = 1) ≈.440 40. (a) 4! 2 13 4 13 4 13 3 13 2 52 13 = .2155. (b) 4! 2 13 5 13 3 13 3 13 2 52 13 = .1552. 42. pX1 = 0 4 1 3 2 3 , pX2 = 3 1 , pX3 = 2 6 2 3 1 3 , pX4 = 1 5 1 2 1 2 . If your friend chooses die 1, choose die 4; if she chooses die 2, choose die 1; if she chooses die 3, choose die 2; if she chooses die 4, choose die 3. Then P(X1 < X4) = P(X2 < X1) = P(X3 < X2) = P(X4 < X3) = 2 3. Thus you are assured of winning with probability 2/3. 43. If the traits were independent, then the probability that we would obtain a data set that diﬀers from the expected data set by as much as the actual data set diﬀers is approximately .00151. Thus, we should reject the hypothesis that the two traits are independent. 44. The value of χ2 corresponding to the data is v = 9931.6, which is much greater than v0, so the hypothesis that the chosen numbers are uniformly distributed should be rejected. SECTION 5.2 1. (a) f(x) = 1 on [2, 3]; F(x) = x −2 on [2, 3]. (b) f(x) = 1 3x−2/3 on [0, 1]; F(x) = x1/3 on [0, 1]. 2. (a) F(x) = 2 −1 x, f(x) = 1 x2 on [ 1 2, 1]. (b) F(x) = ex −1, f(x) = ex on [0, log 2]. 28 5. (a) F(x) = 2x, f(x) = 2 on [0, 1 2]. (b) F(x) = 2√x, f(x) = 1 √x on [0, 1 4]. 7. Using Corollary 5.2, we see that the expression √ rnd will simulate the given random variable. 9. (a) F(y) = ( y2 2 , 0 ≤y ≤1; 1 −(2−y)2 2 , 1 ≤y ≤2, f(y) = y, 0 ≤y ≤1; 2 −y 1 ≤y ≤2. (b) F(y) = 2y −y2, f(y) = 2 −2y, 0 ≤y ≤1. 10. (a) F(x) = x2 and f(x) = 2x on [0, 1]. (b) F(x) = 2x −x2 and f(x) = 2 −2x on [0, 1]. 12. (a) 1/2 (b) 1 (c) .2 13. (a) F(r) = √r , f(r) = 1 2√r , on [0,1]. (b) F(s) = 1 − √ 1 −4s , f(s) = 2 √1 −4s ,on [0, 1 4]. (c) F(t) = t 1 + t , f(t) = 1 (1 + t)2 , on [0, ∞]. 14. (a) 3/4 (b) π/16 15. F(d) = 1 −(1 −2d)2, f(d) = 4(1 −2d) on [0, 1 2]. 16. (a) c = 6 (b) F(x) = 3x2 −2x3 (c) .156 17. (a) f(x) = π 2 sin(πx), 0 ≤x ≤1; 0, otherwise. (b) sin2( π 8 ) = .146. 18. FW (w) = a > 0 : FX( w−b a ); a = 0 : 1, w ≥b; 0, otherwise; a < 0 : 1 −FX( w−b a ). 19. a ̸= 0 : fW (w) = 1 \|a\|fX( w−b a ), a = 0: fW (w) = 0 if w ̸= 0. 20. (a) 1 d −c (b) c c −d 21. P(Y ≤y) = P(F(X) ≤y) = P(X ≤F −1(y)) = F(F −1(y)) = y on [0, 1]. 22. (a) a + b 2 (b) µ (c) 1 λ log 2 29 23. The mean of the uniform density is (a + b)/2. The mean of the normal density is µ. The mean of the exponential density is 1/λ. 24. The mode of the uniform density is any number in [0, 1]. The mode of the normal is µ. The mode of the exponential is 0. 25. (a) .9773, (b) .159, (c) .0228, (d) .6827. 26. 13.4% are likely to be rejected. For 1% rejection rate, let σ = .0012. 27. A: 15.9%, B: 34.13%, C: 34.13%, D: 13.59%, F: 2.28%. 28. 2.4% 29. e−2, e−2. 30. The car will last for 4 years with probability 1/e ≈.368. 31. 1 2. 34. P(X < Y ) = Z ∞ x=0 Z ∞ y=x f(x)g(y)dxdy = Z ∞ x=0 f(x)(1 −G(x))dx. Thus P(X < Y ) = Z ∞ 0 λe−λx · e−µxdx = λ λ + µ. Therefore, the probability that a 100 watt bulb will outlast a 60 watt bulb is (1/200) 1/200+1/100 = 1/3. 35. P(size increases) = P(Xj < Yj) = λ/(λ + µ). P(size decreases) = 1 −P(size increases) = µ/(λ + µ). 36. Exponential with parameter λ/r. 37. FY (y) = 1 √2πy e−log2(y) 2 , for y > 0. 38. P(Y1 ≤y1, Y2 ≤y2) = P(X1 ≤Φ−1 1 (y1), X2 ≤Φ−1 2 (y2)) = P(X1 ≤Φ−1 1 (y1))P(X2 ≤Φ−1 2 (y2)) = P(Y1 ≤y1)P(Y2 ≤y2), so Y1 and Y2 are independent. SECTION 6.1 1. -1/9 2. -1/2 3. 5′ 10.1” 4. -1/19 5. -1/19 6. Let U and V be independent identically distributed random variables with the density: pU = 1 2 3 4 5 6 1 6 1 6 1 6 1 6 1 6 1 6 . 30 Then XY = (U + V )(U −V ) = U 2 −V 2, so E(XY ) = E(U 2) −E(V 2) = 0. Since E(Y ) = E(U) −E(V ) = 0, we have E(XY ) = E(X)E(Y ) = 0. But X and Y are not independent. For example, if we know that X = 12, then we know that Y = 0. 7. Since X and Y each take on only two values, we may choose a, b, c, d so that U = X + a b , V = Y + c d take only values 0 and 1. If E(XY ) = E(X)E(Y ) then E(UV ) = E(U)E(V ). If U and V are independent, so are X and Y . Thus it is suﬃcient to prove independence for U and V taking on values 0 and 1 with E(UV ) = E(U)E(V ).Now E(UV ) = P(U = 1, V = 1) = E(U)E(V ) = P(U = 1)P(V = 1), and P(U = 1, V = 0) = P(U = 1) −P(U = 1, V = 1) = P(U = 1)(1 −P(V = 1)) = P(U = 1)P(V = 0). Similarly, P(U = 0, V = 1) = P(U = 0)P(V = 1) P(U = 0, V = 0) = P(U = 0)P(V = 0). Thus U and V are independent, and hence X and Y are also. 8. The expected number of boys and the expected number of girls are both 7 8. 9. The second bet is a fair bet so has expected winning 0. Thus your expected winning for the two bets is the same as the original bet which was -7/498 = -.0141414... On the other hand, you bet 1 dollar with probability 1/3 and 2 dollars with probability 2/3. Thus the expected amount you bet is 1 2 3 dollars and your expected winning per dollar bet is -.0141414/1.666667 = -.0085 which makes this option a better bet in terms of the amount won per dollar bet. However, the amount of time to make the second bet is negligible, so in terms of the expected winning per time to make one play the answer would still be -.0141414. 11. The roller has expected winning -.0141; the pass bettor has expected winning -.0136. 12. 0 13. 45 14. E(Xj) = 1 N . For j ̸= k, E(XjEk) = 1 N(N −1) . Thus Xj and Xk are not independent. 15. E(X) = 1 5, so this is a favorable game. 16. (a) E(X) = (1 + 2 + 3 + 4 + 5 + 6) 6 = 31 2. 31 (b) The large sums are much less likely to occur than small sums. For example P(total = 21) = (1/6)6 = 2.14 × 10−5 and P(total = 0) = (5/6)6 = .335. 17. pk = p( k−1 times z }\| { S · · · S F) = pk−1(1 −p) = pk−1q, k = 1, 2, 3, . . . . ∞ X k=1 pk = q ∞ X k=0 pk = q 1 1 −p = 1 . E(X) = q ∞ X k=1 kpk−1 = q (1 −p)2 = 1 q . (See Example 6.4.) 18. 7/2 19. E(X) = 4 4 4 4 (3 −3) + 3 2 4 3 (3 −2) + 3 3 4 3 (0 −3) + 3 1 4 2 (3 −1) + 3 2 4 2 (0 −2) + 3 0 4 1 (3 −0) + 3 1 4 1 (0 −1) = 0 . 20. (a) E(X) = 1 2 · 2 + 1 2 2 · 22+ 1 2 3 ·23 · · · = 1 + 1 + 1 + · · · = ∞, and so E(X) does not exist. This means that if we could play the game, it would be favorable now matter how much we pay to play it. However, we cannot realize this game, since it requires arbitrarly large amounts of money. (b) E(X) = 1 2 · 2 + 1 2 2 · 22 + · · · + 1 2 10 · 210 + 210 1 211 + 1 212 + · · · = 10 + 1 2 + 1 22 + · · · = 11 . (d) If the utility of n dollars is √n, then the expected utility of the payment is given by ∞ X i=1 1 2i √ 2i = 1 √ 2 −1 . If the uility of n dollars is log n, then the expected utility of the payment is given by ∞ X i=1 1 2i log(2i) = 2 log 2 . 22. The expected number of days in a year with more than 60 percent boys for the large hospital is 365 · k=45 X k=28 b(45, .5, k) = 24.67 . For the small hospital it is 365 · k=15 X k=10 b(15, .5, k) = 55.1 . 32 23. 10 25. (b) Let S be the number of stars and C the number of circles left in the deck. Guess star if S > C and guess circle if S < C. If S = C toss a coin. (d) Consider the recursion relation: h(S, C) = max(S, C) S + C + S S + C h(S −1, C) + C S + C h(S, C −1) and h(0, 0) = h(−1, 0) = h(0, −1) = 0. In this equation the ﬁrst term represents your expected winning on the current guess and the next two terms represent your expected total winning on the remaining guesses. The value of h(10, 10) is 12.34. 26. (a) Let L be the horizontal line passing through S −C. If the random walk is below L, then there are more stars than circles in the remaining deck, and so, using the optimal strategy, you guess star. If you are right, the graph goes up. If the walk is above L, then there are more circles than stars, and you guess circle. If you are right, the graph goes down. Since S ≥C, the graph ends at (S + C, S −C). Let a be the number of times the graph goes up under L, b the number of times it goes down under L, c the number of times it goes down above L, and d the number of times it goes up above L. Then a + b + c + d = S + C, a −b = S −C, c −d = 0. Thus 2a −S + C + 2c = S + C, and this implies a + c = S, i.e., we have S correct guesses. (b) We arrive at (x, x) if S −x stars turn up and C −x circles turn up in S + C −2x guesses. The probability of this happening is S S−x C C−x S+C S+C−2x = S x C x S+C 2x . (c) The number of correct guesses equals the number of correct guesses when the graph is under or above L plus the number of correct guesses when the graph hits L. Thus the expected number of correct guesses is: S + C X x=1 S x C x S+C 2x · 1 2 . 27. (a) 4 (b) 4 + 4 X x=1 4 x 4 x 8 x = 5.79 . 28. (a) Assume that n = 2k −1. Choose the middle number of the numbers from 1 to 2k −1, and then continue to choose the middle number until you guess the number correctly. If you have not yet succeeded after k −1 guesses you will be down to a single number and will be sure to get it on the kth question. This strategy obviously works just as well if n < 2k−1. (b) Whenever you make a guess and are wrong the search is narrowed to a new and smaller interval [a, b]. The probability that you guess correctly on a question when the interval is [a, b] is P(correct) = (b −a) n · 1 (b −a) = 1 n . Thus the probabililty that you guess the number on the kth question is a(k)/n where a(k) is the number of possible subintervals for the kth question. The probability of guessing the number correctly for a strategy with at most k guesses is P k a(k) n . 33 Thus any strategy that makes this sum as large as possible is optimal. We can at most double the number of intervals on each question.Thus any strategy that achieves this is optimal.The resulting probability of guessing the number in k questions is Pk−1 j=0 2k n = 2k −1 n . If n ≥2k −1 we can achieve this optimal strategy by continuing to bisect the numbers between 1 and 2k −1. 29. If you have no ten-cards and the dealer has an ace, then in the remaining 49 cards there are 16 ten cards. Thus the expected payoﬀof your insurance bet is: 2 · 16 49 −1 · 33 49 = −1 49 . If you are playing two hands and do not have any ten-cards then there are 16 ten-cards in the remaining 47 cards and your expected payoﬀon an insurance bet is: 2 · 16 47 −1 · 31 47 = 1 47 . Thus in the ﬁrst case the insurance bet is unfavorable and in the second it is favorable. 30. (a) P(Xk = j)= P(j −1 boxes have old pictures and the jth box has a new picture) = k −1 n jn −k + 1 n , and so Xk has a geometric distribution with p = (n −k + 1)/n. (c) The expected time for getting the ﬁrst half of the players is E(X1) + · · · + E(Xn) = 2n 2n −1 + 1 + 2n 2n −2 + 1 + · · · + 2n 2n −n + 1 = 2n 1 2n + 1 2n −1 + · · · 1 n + 1 . The expected time for getting the second half of the players is: E(Xn+1) + · · · + E(X2n) = 2n 2n −(n + 1) −1 + · · · + 2n 2n −2n + 1 = 2n 1 n + 1 n −1 + · · · + 1 1 . (d) 1 + 1 2 + · · · + 1 n ∼log n + .5772 + 1 2n . 1 + 1 2 + · · · + 1 n + 1 n + 1 + · · · + 1 2n ∼log 2n + .5772 + 1 4n . 2n 1 2n + · · · + 1 n + 1 ∼2n log 2n + 1 4n −log n −1 2n . = 2n log 2 −1 4n = 2nlog 2 −1 2 2n 1 + 1 2 + · · · + 1 n ∼2n log n + .5772 + 1 2n . 34 31. (a) 1 −(1 −p)k . (b) N k · (k + 1)(1 −(1 −p)k) + (1 −p)k . (c) If p is small, then (1 −p)k ∼1 −kp, so the expected number in (b) is ∼N[kp + 1 k], which will be minimized when k = 1/√p. 32. Your estimate should be near e = 2.718... 33. We begin by noting that P(X ≥j + 1) = P((t1 + t2 + · · · + tj) ≤n) . Now consider the j numbers a1, a2, · · · , aj deﬁned by a1 = t1 a2 = t1 + t2 a3 = t1 + t2 + t3 . . . . . . . . . aj = t1 + t2 + · · · + tj . The sequence a1, a2, · · · , aj is a monotone increasing sequence with distinct values and with succes-sive diﬀerences between 1 and n. There is a one-to-one correspondence between the set of all such sequences and the set of possible sequences t1, t2, · · · , tj. Each such possible sequence occurs with probability 1/nj. In fact, there are n possible values for t1 and hence for a1. For each of these there are n possible values for a2 corresponding to the n possible values of t2. Continuing in this way we see that there are nj possible values for the sequence a1, a2, · · · , aj. On the other hand, in order to have t1 + t2 + · · · + tj ≤n the values of a1, a2, · · · , aj must be distinct numbers lying between 1 to n and arranged in order. The number of ways that we can do this is n j . Thus we have P(t1 + t2 + · · · + tj ≤n) = P(X ≥j + 1) = n j 1 nj . E(X) = P(X = 1) + P(X = 2) + P(X = 3) · · · + P(X = 2) + P(X = 3) · · · + P(X = 3) · · · . . If we sum this by rows we see that E(X) = n−1 X j=0 P(X ≥j + 1) . Thus, E(X) = n X j=1 n j 1 n j = 1 + 1 n n . The limit of this last expression as n →∞is e = 2.718... . There is an interesting connection between this problem and the exponential density discussed in Section 2.2 (Example 2.17). Assume that the experiment starts at time 1 and the time between occurrences is equally likely to be any value between 1 and n. You start observing at time n. Let T be the length of time that you wait. This is the amount by which t1 +t2 +· · ·+tj is greater than n. Now imagine a sequence of plays of a game in which you pay n/2 dollars for each play and for 35 the j’th play you receive the reward tj. You play until the ﬁrst time your total reward is greater than n. Then X is the number of times you play and your total reward is n+T. This is a perfectly fair game and your expected net winning should be 0. But the expected total reward is n + E(T). Your expected payment for play is n 2 E(X). Thus by fairness, we have n + E(T) = (n/2)E(X) . Therefore, E(T) = n 2 E(X) −n . We have seen that for large n, E(X) ∼e. Thus for large n, E(waiting time) = E(T) ∼n(e 2 −1) = .718n . Since the average time between occurrences is n/2 we have another example of the paradox where we have to wait on the average longer than 1/2 the average time time between occurrences. 34. (a) We prove ﬁrst that for Bernoulli trials the probability that the kth failure precedes the rth success is f(k, p, r) = r + k −1 k pr−1qk · p . To prove this, we note that for the kth failure to precede the rth success we must have r −1 successes and k failures in the ﬁrst r + k −1 trials and then have a success. The probability that this happens is f(k, p, r). Now consider a Bernoulli trials process where success is getting a match from the right pocket. In order to have r matches in the left pocket when the right pocket has none we must have N −r failures before the (N + 1)st success. Thus the probability that there are r matches in the left pocket when the right pocket has none is f(N −r, 1 2, N + 1) . The same argument applies for the probability that there are r matches in the right pocket when the left pocket has none. Thus pr = 2f(N −r, 1 2, N + 1) = 2N −r N 1 2 2N−r . (c) (N −r 2)pr+1 = (N −r 2) 2N −r −1 N 1 2 2N−r−1 = (2N −r) 2N −r −1 N 1 2 2N−r = (N −r) 2N −r N 1 2 2N−r = (N −r)pr . (d) N X r=0 pr = 1 . (e) N X r=0 (N −r)pr = N X r=0 1 2(2N + 1)pr+1 − N X r=0 1 2(r + 1)pr+1 . 36 Thus N −E = 1 2(2N + 1)(1 −p0) −1 2E , and E = p0(2N + 1) −1 . But p0 = 2N N 1 2 2N ∼ 1 √ πN , so E ∼2 r N π . Using this asymptotic expression leads to an estimate of 133 for the number of matches needed in each pocket to make E = 13. It is easy to make an exact calculation with the computer, and this gives 153 matches. 35. One can make a conditionally convergent series like the alternating harmonic series sum to anything one pleases by properly rearranging the series. For example, for the order given we have E = ∞ X n=0 (−1)n+1 2n n · 1 2n = ∞ X n=0 (−1)n+1 1 n = log 2 . But we can rearrange the terms to add up to a negative value by choosing negative terms until they add up to more than the ﬁrst positive term, then choosing this positive term, then more negative terms until they add up to more than the second positive term, then choosing this positive term, etc. 36. c k c + d 37. (a) Under option (a), if red turns up, you win 1 franc, if black turns up, you lose 1 franc, and if 0 turns up, you lose 1/2 franc. Thus, the expected winnings are 1 18 37 + (−1) 18 37 + −1 2 1 37 ≈−.0135 . (b) Under option (b), if red turns up, you win 1 franc, if black turns up, you lose 1 franc, and if 0 comes up, followed by black or 0, you lose 1 franc. Thus, the expected winnings are 1 18 37 + (−1) 18 37 + (−1) 1 37 19 37 ≈−.0139 . (c) 38. (Solution by Victor Hern´ andez) Let pij be the probability that book i is above book j. Then the average depth of book j is dj = X i̸=j pij , where the top book is considered to be at depth 0. Now if book i is above book j, then the relative order of books i and j is changed after a call if and only if book j is consulted. Hence, pij = pij(1 −pj) + pjipi = pij(1 −pj) + (1 −pij)pi = pi + pij(1 −pi −pj) . 37 Thus, we have pij = pi pi + pj , and dj = X k̸=j pk pk + pj . If pi ≥pj, then pk pk + pi ≤ pk pk + pj for k ̸= i, j, and pj pi + pj ≤ pi pi + pj . Since each term in the sum for di is less than or equal to the corresponding term in the sum for dj, we have di ≤dj. 39. (Solution by Peter Montgomery) The probability that book 1 is in the right place is the probability that the last phone call referenced book 1, namely p1. The probability that book 2 is in the right place, given that book 1 is in the right place, is p2 + p2p1 + p2p2 1 + . . . = p2 (1 −p1) . Continuing, we ﬁnd that P = p1 p2 (1 −p1) p3 (1 −p1 −p2) · · · pn (1 −p1 −p2 −. . . −pn−1 . Now let q be a real number between 0 and 1, let p1 = 1 −q , p2 = q −q2 , and so on, and ﬁnally let pn = qn−1 . Then P = (1 −q)n−1 , so P can be made arbitrarily close to 1. 40. If a1, a2, . . . is the sequence, then the event {a1 < a2 < . . . < ak} occurs with probability 1/k!, since there are k! diﬀerent orderings of k real numbers, and all of them are equally likely to occur in this experiment. Therefore, P(X > k) = 1 k! . Now let pk = P(X = k) . Then E(X) = p1 + 2p2 + 3p3 + . . . = (p1 + p2 + . . .) + (p2 + p3 + . . .) + . . . = P(X > 0) + P(X > 1) + . . . = 1 + 1 1! + 1 2! + . . . = e . 38 SECTION 6.2 1. E(X) = 0, V (X) = 2 3, σ = D(X) = r 2 3 . 2. E(X) = 4 3, V (X) = 17 9 , σ = D(X) = √ 17 3 . 3. E(X) = −1 19 , E(Y ) = −1 19 , V (X) = 33.21, V (Y ) = .99 . 4. (a) 10015, (b) 310, (c) −100, (d) 15, (e) √ 15. 5. (a) E(F) = 62, V (F) = 1.2 . (b) E(T) = 0, V (T) = 1.2 . (c) E(C) = 50 3 , V (C) = 10 27 . 7. V (X) = 3 4 , D(X) = √ 3 2 . 8. E(S2400) = 960, V (S2400) = 576, σ = D(S2400) = 24. 9. V (X) = 3 4 , D(X) = 2 √ 5 3 . 10. (a) V (X + c) = V (X), so D(X + c) = D(X). (b) V (cX) = c2V (X), so D(cX) = \|c\|X. 11. E(X) = (1 + 2 + · · · + n)/n = (n + 1)/2. V (X) = (12 + 22 + · · · + n2)/n −(E(X))2 = (n + 1)(2n + 1)/6 −(n + 1)2/4 = (n + 1)(n −1)/12. 12. E X −µ/σ = (1/σ)(E(X) −µ) = 0, V X −µ/σ 2 = (1/σ2)E(X −µ)2 = σ2/σ2 = 1. 13. Let X1, . . . , Xn be identically distributed random variables such that P(Xi = 1) = P(Xi = −1) = 1 2. Then E(Xi) = 0, and V (Xi) = 1. Thus Wn = Pn j=1 Xi. Therefore E(Wn) = Pn i=1 E(Xi) = 0, and V (Wn) = Pn i=1 V (Xi) = n. 14. Let X be the number of boys and Y be the number of girls. Then E(X) = E(Y ) = 7 8 , and V (X) = 7 64 , V (Y ) = 71 64 . 15. (a) PXi = 0 1 n−1 n 1 n . Therefore, E(Xi)2 = 1/n for i ̸= j. (b) PXiXj = 0 1 1 − 1 n(n−1) 1 n(n−1) for i ̸= j . Therefore, E(XiXj) = 1 n(n −1) . 39 (c) E(Sn)2 = X i E(Xi)2 + X i X j̸=i E(XiXj) = n · 1 n + n(n −1) · 1 n(n −1) = 2 . (d) V (Sn) = E(Sn)2 −E(Sn)2 = 2 −(n · (1/n))2 = 1 . 16. (a) For p = .5: k 1 2 3 10 .656 .979 .998 N 30 .638 .957 .999 50 .678 .967 .997 For p = .2: k 1 2 3 10 .772 .967 .994 N 30 .749 .964 .997 50 .629 .951 .997 (b) Use Exercise 12 and the fact that E(Sn) = np and V (Sn) = npq. The two examples in (a) suggests that the probability that the outcome is within k standard deviations is approximately the same for diﬀerent values of p. We shall see in Chapter 9 that the Central Limit Theorem explains why this is true. 18. (a) E(¯ x) = 1 n n X i=1 E(xi) = 1 n · nµ = µ . (b) We have E (¯ x −µ)2 = V (¯ x) , which was shown to equal σ2/n in Theorem 6.9. (c) We have from the hint: n X i=1 (xi −¯ x)2 = n X i=1 (xi −µ)2 −n(¯ x −µ)2 . Thus, E(s2) = 1 nE n X i=1 (xi −¯ x)2 ! = 1 n E n X i=1 (xi −µ)2 ! −nE(¯ x −µ)2 = 1 n(nσ2 −nσ2 n ) = n −1 n σ2 , where we have used the deﬁnition of the variance and part (b) to obtain the penultimate expression. 40 (d) Since the expectation operator is linear, and the ‘new’ s2 is n/(n −1) times the ‘old’ s2, the new s2 has expectation n n −1 n −1 n σ2 = σ2 . 19. Let X1, X2 be independent random variables with pX1 = pX2 = −1 1 1 2 1 2 . Then pX1+X2 = −2 0 2 1 4 1 2 1 4 . Then ¯ σX1 = ¯ σX2 = 1, ¯ σX1+X2 = 1 . Therefore V (X1 + X2) = 1 ̸= V (X1) + V (X2) = 2 , and ¯ σX1+X2 = 1 ̸= ¯ σX1 + ¯ σX2 = 2 . 20. (a) E(¯ µ) = µ (b) w = V (X2) V (X1) + V (X2) 21. f ′(x) = − X ω 2(X(ω) −x)p(ω) = −2 X ω X(ω)p(ω) + 2x X ω p(ω) = −2µ + 2x . Thus x = µ is a critical point. Since f ′′(x) ≡2, we see that x = µ is the minimum point. 22. X, Y, X + Y , and X −Y have the same distribution, so they have the same mean and variance. Thus E(X) = E(Y ) = E(X) + E(Y ). This implies that E(X) = E(Y ) = 0 . It also implies that E(X + Y )2 = E(X −Y )2 = E(X2) = E(Y 2) . Thus E(XY ) = 0 and E(Y 2) = E(X2) = 0. Therefore, P(X = Y = 0) = 1. 23. If X and Y are independent, then Cov(X, Y ) = E(X −E(X)) · E(Y −E(Y )) = 0 . Let U have distribution pU = 0 π/2 π 3π/2 1/4 1/4 1/4 1/4 . Then let X = cos(U) and Y = sin(U). X and Y have distributions pX = 1 0 −1 0 1/4 1/4 1/4 1/4 , 41 pY = 0 1 0 −1 1/4 1/4 1/4 1/4 . Thus E(X) = E(Y ) = 0 and E(XY ) = 0, so Cov(X, Y ) = 0. However, since sin2(x) + cos2(x) = 1, X and Y are dependent. 24. Consider the variance of Sn: V (Sn) = n X i=1 pi(1 −pi) = n X i=1 pi − n X i=1 p2 i , with the constraint n X i=1 pi = np . Assume that we have values of pi that satisfy the constraint and that m of the values of pi are equal to x and one is equal to y with x > y. By rearranging the terms if necessary we can assume that the ﬁrst m are equal to x and the (m + 1)st is equal to y. We shall show that we can increase the variance by making these m + 1 values equal. To do this we deﬁne ¯ pi = pi −ϵ m, for i = 1 to m and ¯ pm+1 = pm+1 + ϵ , where ϵ = m m + 1(x −y) . Then the new ¯ pis satisfy the constraint, and the diﬀerence between the new variance ¯ V and the old variance V is ¯ V −V = m(x −ϵ m)2 + (y + ϵ)2 −mx2 −y2 . After simplifying and substituting the value for ϵ this becomes ¯ V −V = m m + 1(x −y)2 . Since this value is positive we have increased the variance by making the ﬁrst m + 1 values equal. The same argument applies in case y > x. By induction we see that the variance is maximized by making all the values equal. (Note: Students who know about the technique of Lagrange multipliers will ﬁnd this easier to prove using that method.) 25. (a) The expected value of X is µ = E(X) = 5000 X i=1 iP(X = i) . The probability that a white ball is drawn is P(white ball is drawn) = n X i=1 P(X = i) i 5000 . Thus P(white ball is drawn) = µ 5000 . 42 (b) To have P(white,white) = P(white)2 we must have 5000 X i=1 ( i 5000)2P(X = i) = ( n X i=1 i 5000P(X = i))2 . But this would mean that E(X2) = E(X)2, or V (X) = 0. Thus we will have independence only if X takes on a speciﬁc value with probability 1. (c) From (b) we see that P(white,white) = 1 50002 E(X2) . Thus V (X) = (σ2 + µ2) 50002 . 26. (a) P(X = k) = pqk−1, k = 1, 2, . . . Thus by Example 3 we have E(Xj) = 1 p, V (Xj) = q p2 . (b) E(Tn) = n/p, V (Tn) = nq/p2. (c) E(Tn) = 2n, V (Tn) = 2n. 27. The number of boxes needed to get the j’th picture has a geometric distribution with p = (2n −k + 1) 2n . Thus V (Xj) = 2n(k −1) (2n −k + 1)2 . Therefore, for a team of 26 players the variance for the number of boxes needed to get the ﬁrst half of the pictures would be 13 X k=1 26(k −1) (26 −k + 1)2 = 7.01 , and to get the second half would be 26 X k=14 26(k −1) (26 −k + 1)2 = 979.23 . Note that the variance for the second half is much larger than that for the ﬁrst half. SECTION 6.3 1. (a) µ = 0, σ2 = 1/3 (b) µ = 0, σ2 = 1/2 (c) µ = 0, σ2 = 3/5 (d) µ = 0, σ2 = 3/5 2. (a) µ = 0, σ2 = 1/5 43 (b) µ = 0, σ2 = π2 −8 π2 (c) µ = 1/3, σ2 = 2/9 (d) µ = 1/2, σ2 = 3/20 3. µ = 40, σ2 = 800 4. (a) R 1 −1(ax + b)dx = 2b = 1, so b = 1 2. (b) ax + 1 2 ≥0, so when x = 1, a ≥−1 2, and when x = −1, a ≤1 2. Thus −1 2 ≤a ≤1 2. (c) µ = R 1 −1(ax2 + bx)dx = 2 3a. (d) E(X2) = R 1 −1(ax3 + bx2)dx = 2b 3 = 1/3 −(4/9)a2. Thus σ2(X) = 2 3b −4 9a2. 5. (d) a = −3/2, b = 0, c = 1 (e) a = 45 48, b = 0, c = 3 16 6. (a) µT = 1 3, σ2 T = 1 9. (b) µT = 1 3, σ2 T = 2 9. (c) µT = 1 2, σ2 T = 3 4. 7. f(a) = E(X −a)2 = Z (x −a)2f(x)dx . Thus f ′(a) = − Z 2(x −a)f(x)dx = −2 Z xf(x)dx + 2a Z f(x)dx = −2µ(X) + 2a . Since f ′′(a) = 2, f(a) achieves its minimum when a = µ(X). 8. a(σ2 + µ2) + bµ + c . 9. (a) 3µ, 3σ2 (b) E(A) = µ, V (A) = σ2 3 (c) E(S2) = 3σ2 + 9µ2, E(A2) = σ2 3 + µ2 10. (a) 1 3 (b) 2 3 (c) 1 3 (d) 2 3 (e) 7 6 11. In the case that X is uniformly distributed on [0, 100], one ﬁnds that E(\|X −b\|) = 1 200 b2 + (100 −b)2 , 44 which is minimized when b = 50. When fX(x) = 2x/10,000, one ﬁnds that E(\|X −b\|) = 200 3 −b + b3 15000 , which is minimized when b = 50 √ 2. 12. Z 1 0 Z 1 0 xydxdy = log(2) ≈.693 . 13. Integrating by parts, we have E(X) = Z ∞ 0 xdF(x) = −x(1 −F(x)) ∞ 0 + Z ∞ 0 (1 −F(x))dx = Z ∞ 0 (1 −F(x))dx . To justify this argment we have to show that a(1 −F(a)) approaches 0 as a tends to inﬁnity. To see this, we note that Z ∞ 0 xf(x)dx = Z a 0 xf(x)dx + Z ∞ a xf(x)dx ≥ Z a 0 xf(x)dx + Z a 0 af(x)dx = Z a 0 xf(x)dx + a(1 −F(a)) . Letting a tend to inﬁnity, we have that E(X) ≥E(X) + lim a→∞a(1 −F(a)) . Since both terms are non-negative, the only way this can happen is for the inequality to be an equality and the limit to be 0. To illustrate this with the exponential density, we have Z ∞ 0 (1 −F(x))dx = Z ∞ 0 e−λxdx = 1 λ = E(X) . 15. E(Y ) = 9.5, E(Z) = 10, E(\|X −Y \|) = 1/2, E(\|X −Z\|) = 1/2 . Z is better, since it has the same expected value as X and the variance is only slightly larger. 17. (a) Cov(X, Y ) = E(XY ) −µ(X)E(Y ) −E(X)µ(Y ) + µ(X)µ(Y ) = E(XY ) −µ(X)µ(Y ) = E(XY ) −E(X)E(Y ) . (b) If X and Y are independent, then E(XY ) = E(X)E(Y ), and so Cov(X, Y ) = 0. (c) V (X + Y ) = E(X + Y )2 −(E(X + Y ))2 = E(X2) + 2E(XY ) + E(Y 2) −E(X)2 −2E(X)E(Y ) −E(Y )2 = V (X) + V (Y ) + 2Cov(X, Y ) . 45 18. (a) 0 ≤V X σ(X) + Y σ(Y ) = V X σ(X) + V Y σ(Y ) + 2Cov X σ(X)), Y σ(Y ) = 1 + 1 + 2 cov(X, Y ) p V (X)V (Y ) = 2(1 + ρ(X, Y )) . (b) 0 ≤V X σ(X) − Y σ(Y ) = V X σ(X)) + V ( Y σ(Y ) −2Cov X σ(X)), − Y σ(Y ) = 1 + 1 −2Cov(X, Y ) σ(X)σ(Y ) = 2(1 −ρ(X, Y )) . (c) From (a), 1 + ρ(X, Y ) ≥0, so ρ(X, Y ) ≥−1. From (b), 1 −ρ(X, Y ) ≥0, so ρ(X, Y ) ≤1. Thus −1 ≤ρ(X, Y ) ≤1. 19. (a) 0 (b) 1 √ 2 (c) −1 √ 2 (d) 0 20. (a) fX(x) = 1 2π p 1 −ρ2 Z ∞ −∞ exp −(x2 −2ρxy + y2) 2(1 −ρ2) dy = 1 2π p 1 −ρ2 Z ∞ −∞ exp(−(y −ρx)2) · exp(−1 2x2)dy = 1 √ 2π · exp(−1 2x2) . Thus X has a standard normal density. By symmetry, Y also has a standard normal density. (b) E(XY ) = 1 2π p 1 −ρ2 Z ∞ −∞ Z ∞ −∞ xy · exp −(x2 −2ρxy + y2) 2(1 −ρ2) dxdy = 1 2π p 1 −ρ2 Z ∞ −∞ Z ∞ −∞ y · exp −(y −ρx)2 (2(1 −ρ2) dy x · exp(−1 2x2)dx = 1 √ 2π Z ∞ −∞ ρx2· exp(−1 2x2) = ρ . Now Cov(X, Y ) = E(XY ) −E(X)E(Y ) p V (X)V (Y ) . Since E(X) = E(Y ) = 0 and V (X) = V (Y ) = 1 , Cov(X, Y ) = E(XY ) = ρ. 46 21. We have fXY (x,y) fY (y) = 1 2π√ 1−ρ2 · exp −(x2−2ρxy+y2) 2(1−ρ2) √ 2π · exp(−y2 2 ) = 1 p 2π(1 −ρ2) · exp −(x −ρy)2 2(1 −ρ2) which is a normal density with mean ρy and variance 1 −ρ2. Thus, E(X\|Y = y) = Z ∞ −∞ x 1 p 2π(1 −ρ2) · exp −(x −ρy)2 2(1 −ρ2) dx = ρy Z ∞ −∞ 1 p 2π(1 −ρ2) · exp(−(x −ρy)2) = ρy < y, if 0 < ρ < 1; y, if ρ = 1. 22. We have fX(x) = Z 1 0 fX,Y (x, y)dy = Z 1 0 fX\|Y (x\|y)fY (y)dy = Z 1 x 1 y dy = −log x , if 0 < x ≤1. 24. (a) Since the father’s height is 72 inches, Y = (72 −68)/2.7 = 1.48. Therefore the density for X given Y is normal with mean .5 · 1.48 = .74 and variance 1 −.52 = .75. Thus the density for the son’s height, given that the father’s height is 72, is normal with mean 2.7 · .74 + 68 = 70 and variance (2.7)2 · .75 = 5.47. 26. (a) Let θ denote the angle that our path makes with the river bank, and assume without loss of generality that 0 ≤θ ≤π/2. Let X denote the distance from P to the river. Then X = sin(θ). Thus, the cumulative distribution function of X is given by FX(x) = P(X ≤x) = P sin(θ) ≤x = P(θ ≤arcsin x) = 2 π arcsin x . So, fX(x) = 2 π 1 √ 1 −x2 . Therefore, E(X) = Z 1 0 2 π x √ 1 −x2 dx = 2 π h (1 −x2)1/2i1 0 = 2 π . 47 (b) For a ﬁxed θ between 0 and π/2, let Aθ denote the set of angles α that you can choose at P and get back to the river by walking at most 1 mile in the direction α. If α = 0 corresponds to the direction directly towards the river from P, then Aθ = h θ −π 2 , π 2 −θ i . So the probability that you choose a good angle α, given that you are at P, is \|Aθ\| 2π = π −2θ 2π . This must be averaged over all θ ∈[0, π/2] to obtain the ﬁnal answer: 2 π Z π/2 0 1 2 −1 π θ dθ = 2 π hθ 2 −1 2π θ2iπ/2 0 = 1 4 . 27. Let Z represent the payment. Then P(Z = k\|X = x) = P(Y1 ≤x, Y2 ≤x, . . . , Yk ≤x, Yk+1 > x) = xk(1 −x) . Therefore, P(Z = k) = Z 1 0 xk(1 −x) dx = 1 k + 1xk+1 − 1 k + 2xk+2 1 0 = 1 k + 1 − 1 k + 2 = 1 (k + 1)(k + 2) . Thus, E(Z) = ∞ X k=0 k 1 (k + 1)(k + 2) , which diverges. Thus, you should be willing to pay any amount to play this game. SECTION 7.1 1. (a) .625 (b) .5 2. −2 −1 0 1 2 3 4 1 16 1 4 5 16 3 16 9 64 1 32 1 64 . 3. 0 1 2 3 4 1 64 3 32 17 64 3 8 1 4 48 5. (a) 3 4 5 6 1 12 4 12 4 12 3 12 (b) 1 2 3 4 1 12 4 12 4 12 3 12 6. (a) P(Tr = k) = r + k −1 k prqk . (b) P(Cr = k) = b(r, p, k) = r k pkqr−k . (c) E(Cr) = rp, V (Cr) = rpq . 7. (a) P(Y3 ≤j) = P(X1 ≤j, X2 ≤j, X3 ≤j) = P(X1 ≤j)3. Thus pY3 = 1 2 3 4 5 6 1 216 7 216 19 216 37 216 61 216 91 216 . This distribution is not bell-shaped. (b) In general, P(Yn ≤j) = P(X1 ≤j)3 = j n n . Therefore, P(Yn = j) = j n n − j −1 n n . This distribution is not bell-shaped for large n. 8. (b) .304 (c) .325 9. Let p1, . . . , p6 be the probabilities for one die and q1, . . . , q6 be the probabilities for the other die. Assume ﬁrst that all probabilities are positive. Then p1q1 > p1q6, since there is only one way to get a 2 and several ways to get a 7. Thus q1 > q6. In the same way q6q6 > q1p6 and so q6 > q1. This is a contradiction. If any of the sides has probability 0, then we can renumber them so that it is side 1. But then the probability of a 2 is 0 and so all sums would have to have probability 0, which is impossible. Here’s a fancy way to prove it. Deﬁne the polynomials p(x) = 5 X k=0 p(k+1)xk and q(x) = 5 X k=0 q(k+1)xk . Then we must have p(x)q(x) = 10 X k=0 xk 11 = (1 −x11) (1 −x) . The left side is the product of two ﬁfth degree polynomials. A ﬁfth degree polynomial must have a real root which will not be 0 if p1 > 0. Consider the right side as a polynomial. For x to be a non-zero root of this polynomial it would have to be a real eleventh root of unity other than 1, and there are no such roots. Hence again we have a contradiction. 49 10. Let n = rs. Then consider the following two distributions pX = 0 1 2 . . . r −1 1/r 1/r 1/r . . . 1/r , pY = 0 r 2r . . . (s −1)r 1/s 1/s 1/s . . . 1/s . If X and Y are independent, then X + Y takes on all possible values from 0 to n −1. Further, there is only one choice of X and Y that gives X + Y a particular value and the probability for this choice is 1/rs. Thus X + Y has a uniform distribution on the values from 0 to n −1. SECTION 7.2 2. (a) fZ(x) = x + 2 4 on [−2, 0] and 2 −x 4 on [0, 2]. 3. (a) fZ(x) = x3/24, if 0 ≤x ≤2; x −x3/24 −4/3, if 2 ≤x ≤4. (b) fZ(x) = (x3 −18x2 + 108x −216)/24, if 6 ≤x ≤8; (−x3 + 18x2 −84x + 40)/24, if 8 ≤x ≤10. (c) fZ(x) = x2/8, if 0 ≤x ≤2; 1/2 −(x −2)2/8, if 2 ≤x ≤4. 4. fZ(x) =    x2/2, if 0 ≤x ≤1; (−2x2 + 6x −3)/2, if 1 ≤x ≤2; (x −3)2/2, if 2 ≤x ≤3. 5. (a) fZ(x) = ( λµ µ+λeλx, x < 0; λµ µ+λe−µx, x ≥0. (b) fZ(x) = 1 −e−λx, 0 < x < 1; (eλ −1)e−λx, x ≥1. 6. Z is normally distributed with mean µ = µ1 + µ2 and variance σ2 = σ2 1 + σ2 2. 7. We ﬁrst ﬁnd the density for X2 when X has a general normal density fX(x) = 1 σ √ 2π e−(x−µ)2/2σ2dx . Then (see Theorem 1 of Chapter 5, Section 5.2 and the discussion following) we have f 2 X(x) = 1 σ √ 2π 1 2√x exp(−x/2σ2 −µ2/2σ2) exp(√xµ/σ2) + exp(−√xµ/σ2) . 50 Replacing the last two exponentials by their series representation, we have f 2 X(x) = e−µ/2σ2 ∞ X r=0 µ 2σ2 r 1 r!G(1/2σ2, r + 1/2, x) , where G(a, p, x) = ap Γ(p)e−axxp−1 is the gamma density. We now consider the original problem with X1 and X2 two random variables with normal density with parameters µ1, σ1 and µ2, σ2. This is too much generality for us, and we shall assume that the variances are equal, and then for simplicity we shall assume they are 1. Let c = q µ2 1 + µ2 2 . We introduce the new random variables Z1 = 1 c (µ1X1 + µ2X2) , Z2 = 1 c (µ2X1 −µ1X2) . Then Z1 is normal with mean c and variance 1 and Z2 is normal with mean 0 and variance 1. Thus, fZ2 1 = e−c2/2 ∞ X r=0 c2 2 r 1 r!G(1/2, r + 1/2, x) , and fZ2 2 = G(1/2, 1/2, x) . Convoluting these two densities and using the fact that the convolution of a gamma density G(a, p, x) and G(a, q, x) is a gamma density G(a, p + q, x) we ﬁnally obtain fZ2 1+Z2 2 = fX2 1+X2 2 = e−c2/2 ∞ X r=0 c2 2 r 1 r!G 1/2, r + 1, x . (This derivation is adapted from that of C.R. Rao in his book Advanced Statistical Methods in Biometric Research, Wiley, l952.) 8. fR2 = ( π/4, if 0 ≤x ≤1; (1/2) arcsin (2 −x)/x , if 1 ≤x ≤2. fR = ( (π/2)x, if 0 ≤x ≤1; x arcsin (2 −x2)/x2 , if 1 ≤x ≤ √ 2. 9. P(X10 > 22) = .341 by numerical integration. This also could be estimated by simulation. 10. P(min(X1, . . . , Xn > x) = (P(X1 > x))n = (e−x/µ)n = e−(n/µ)x. Thus fmin(X1,...,Xn) = n µe−(n/µ)x . This is the exponential density with mean µ/n. 11. 10 hours 51 12. By Exercise 10 the ﬁrst claim has the mean of µ/50. If µ is about 30 years, then µ/50 is about 7 months, which is practical. Once we have estimated µ/50, we have an estimate for µ. 13. Y1 = −log(X1) has an exponential density fY1 (x) = e−x. Thus Sn has the gamma density fSn(x) = xn−1e−x (n −1)! . Therefore fZn(x) = 1 (n −1)! log 1 x n−1 . 14. X3 = −X2 has density f−X2 (x) = eλx, −∞< x ≤0; 0, otherwise. Thus Z = X1 + X3 has density fZ(x) = Z ∞ 0 eλ(x−2y)dy = 1 2λeλx, x < 0; = Z ∞ x eλ(x−2y)dy = 1 2λeλx e−2λx = 1 2λe−λx, x ≥0. 19. The support of X + Y is [a + c, b + d]. 20. We prove it by induction. It is true for n = 1. Suppose that fSk has support on [kc, kb]. Then fSk+1 = fSk ∗fX has support on [ka + a, kb + b] = [(k+1)a,(k+1)b] (See Exercise 19 above.) 21. (a) fA(x) = 1 √ 2πn e−x2/(2n) . (b) fA(x) = nnxne−nx/(n −1)! . SECTION 8.1 1. 1/9 3. We shall see that Sn −n/2 tends to inﬁnity as n tends to inﬁnity. While the diﬀerence will be small compared to n/2, it will not tend to 0. On the other hand the diﬀerence Sn/n −1/2 does tend to 0. 4. You will lose on the average 1.41 percent of the money that you bet. Thus if you play a long time, you will lose a lot. The law of large numbers tells you that the probability that you will be ahead in the long run tends to 0. 5. k = 10 6. V Sn n −p = V Sn n = p(1 −p) n . Thus P Sn n −p ≥ϵ ≤p(1 −p) nϵ2 . 7. p(1 −p) = 1 4 − 1 4 −p + p2 = 1 4 −(1 2 −p)2 ≤1 4 . 52 Thus, max 0≤p≤1 p(1 −p) = 1 4. From Exercise 6 we have that P \|Sn n −p\| ≥ϵ ≤p(1 −p) nϵ2 ≤ 1 4nϵ2 . 8. No. 9. P(Sn ≥11) = P(Sn −E(Sn) ≥11 −E(Sn)) = P(Sn −E(Sn) ≥10) ≤V (Sn) 102 = .01. 10. P(X ≥k + 1) = P(X −E(X) ≥k + 1 −E(X)) = P(X −E(X) ≥k) ≤V (X) k2 = 1 k2 . 11. No, we cannot predict the proportion of heads that should turn up in the long run, since this will depend upon which of the two coins we pick. If you have observed a large number of trials then, by the Law of Large Numbers, the proportion of heads should be near the probability for the coin that you chose. Thus, in the long run, you will be able to tell which coin you have from the proportion of heads in your observations. To be 95 percent sure, if the proportion of heads is less than .625, predict p = 1/2; if it is greater than .625, predict p = 3/4. Then you will get the correct coin if the proportion of heads does not deviate from the probability of heads by more than .125. By Exercise 7, the probability of a deviation of this much is less than or equal to 1/(4n(.125)2). This will be less than or equal to .05 if n > 320. Thus with 321 tosses we can be 95 percent sure which coin we have. 12. P Sn n −Mn n > ϵ = P 1 n n X i=1 (Xi −mi) > ϵ ! = P n X i=1 (Xi −mi) > nϵ ! ≤ 1 n2ϵ2 n X i=1 σ2 k < nR n2ϵ2 = R nϵ2 . This last expression approaches 0 as n goes to ∞. 14. E(\|X −E(X)\|) = X ω \|X(ω) −E(X)\| ≥ X {ω:\|X(ω)−E(X)\|≥ϵ} \|X(ω) −E(X)\| ≥ϵP(\|X −E(X)\| ≥ϵ). 15. Take as Ωthe set of all sequences of 0’s and 1’s, with 1’s indicating heads and 0’s indicating tails. We cannot determine a probability distribution by simply assigning equal weights to all inﬁnite sequences, since these weights would have to be 0. Instead, we assign probabilities to ﬁnite 53 sequences in the usual way, and then probabilities of events that depend on inﬁnite sequences can be obtained as limits of these ﬁnite sequences. (See Exercise 28 of Chapter 1, Section 1.2.) 16. The exercise as stated in the text is incorrect. The following replacement exercise, sent to us by David Maslen, is correct: In this exercise, we shall construct an example of a sequence of random variables that satisﬁes the weak law of large numbers, but not the strong law. The distribution of Xi will have to depend on i, because otherwise both laws would be satisﬁed. As a preliminary, we need to prove a lemma, which is one of the Borel-Cantelli lemmas. Suppose we have an inﬁnite sequence of mutually independent events A1, A2, . . .. Let ai = Prob(Ai), and let r be a positive integer. (a) Find an expression of the probability that none of the Ai with i > r occur. (b) Use the fact that x −1 ≤e−x to show that Prob(No Ai with i > r occurs) ≤e−P∞ i=r ai . (c) Prove that if P∞ i=1 ai diverges, then Prob(inﬁnitely many Ai occur) = 1 . Now, let Xi be a sequence of mutually independent random variables such that for each positive integer i ≥2, Prob(Xi = i) = 1 2i log i , Prob(Xi = −i) = 1 2i log i , Prob(Xi = 0) = 1 − 1 i log i . When i = 1 we let Xi = 0 with probability 1. As usual we let Sn = X1 + · · · + Xn. Note that the mean of each Xi is 0. (d) Find the variance of Sn. (e) Show that the sequence {Xi} satisﬁes the weak law of large numbers, i.e. prove that for any ϵ > 0 Prob Sn n ≥ϵ ! →0 , as n tends to inﬁnity. We now show that {Xi} does not satisfy the strong law of large numbers. Suppose that Sn/n →0. Then because Xn n = Sn n −n −1 n Sn−1 n −1 , we know that Xn/n →0. From the deﬁnition of limits, we conclude that the inequality \|Xi\| ≥i/2 can only be true for ﬁnitely many i. (f) Let Ai be the event \|Xi\| ≥i/2. Find Prob(Ai). Show that ∞ X i=1 Prob(Ai) diverges (think Integral Test). (g) Prove that Ai occurs for inﬁnitely many i. (h) Prove that Prob Sn n →0 = 0 , and hence that the Strong Law of Large Numbers fails for the sequence {Xi}. 17. For x ∈[0, 1], let us toss a biased coin that comes up heads with probability x. Then E f(Sn) n →f(x). 54 But E f(Sn) n = n X k=0 f k n n k xk(1 −x)n−k. The right side is a polynomial, and the left side tends to f(x). Hence n X k=0 f k n n k xk(1 −x)n−k →f(x). This shows that we can obtain obtain any continuous function f(x) on [0,1] as a limit of polynomial functions. SECTION 8.2 1. (a) 1 (b) 1 (c) 100/243 (d) 1/12 2. (a) E(X) = 10, V (X) = 100/3. (b) P(\|X −10\| ≥2) = 4/5, P(\|X −10\| ≥5) = 1/2. P(\|X −10\| ≥9) = 1/10, P(\|X −10\| ≥20) = 0. 3. f(x) = n 1 −x/10, if 0 ≤x ≤10; 0 otherwise. g(x) = 100 3x2 . 4. (a) E(X) = 1/λ = 10, V(X) = (1/λ)2= 100. (b) For the ﬁrst three probabilities Chebyshev’s estimate is greater than 1, and so the best estimate is 1. For the last one Chebyshev’s estimate gives P(\|X −10\| ≥20) ≤.25. (c) Comparing these Chebyshev’s estimates with the exact values, we have: (1, .852), (1, .617), (1, .245), (.25, .0498). 5. (a) 1, 1/4, 1/9 (b) 1 vs. .3173, .25 vs. .0455, .11 vs. .0027 6. (a) 1, (b) 1/4, (c) 1/ 9, (d) 1/ 16. The exact values are (a) .3173, (b) .0455, (c) .0027, (d) 0. 7. (b) 1, 1, 100/243, 1/12 8. (a) P(\|X∗\| ≥a) = P X −µ σ ≥a = P(\|X −µ\| ≥aσ) ≤ σ2 σ2a2 = 1 a2 . 55 (b) P(\|X∗\| ≥2) = 1/4, P(\|X∗\| ≥5) = 1/25. P(\|X∗\| ≥9) = 1/81. 9. (a) 0 (b) 7/12 (c) 11/12 10. (a) P(65 ≤X ≤75) = P(65 −70 ≤X −70 ≤75 −70) = P(−5 ≤X −70 ≤5) = 1 −P(\|X −70\| ≥5) ≥1 −25/25 = 0. Thus Chebyshev’s estimate gives us a useless lower bound in this case. (b) E( ¯ X) = 70, V ( ¯ X) = 25/100 = .25. Thus P(65 ≤¯ X ≤75) = 1 −P(\| ¯ X −70\| ≥5) ≥1 −.25 25 = .99. Therefore, Chebyshev’s estimate gives a lower bound of .99. 11. (a) 0 (b) 7/12 12. (a) E(Y2) = 30, V (Y2) = 1 4. Thus P(25 ≤Y2 ≤35) ≥.99. (b) E(Y11) = 30, V (Y11) = 10 4 . Thus P(25 ≤Y11 ≤35) ≥.9. (c) E(Y101) = 30, V (Y101) = 100 4 . Thus P(25 ≤Y101 ≤35) ≥0. 13. (a) 2/3 (b) 2/3 (c) 2/3 17. E(X) = R ∞ −∞xp(x)dx. Since X is non-negative, we have E(X) ≥ Z x≥a xp(x)dx ≥aP(X ≥a) . 18. Since E(X) = 20 and X is non-negative, we have: 20 = Z ∞ 0 xp(x)dx ≥ Z ∞ a xp(x)dx ≥aP(X ≥a) . Therefore, P(X ≥a) ≤20 a . This is interesting only for a ≥20. (b) Now assume E(X) = 20 and V (X) = 25. Then E(X2) = V (X) + E(X)2 = 425 . Thus 425 = Z ∞ 0 x2p(x)dx ≥ Z ∞ a x2dx ≥a2P(X ≥a) . 56 Therefore P(X ≥a) ≤425 a2 . ¿From part (a) we also have that P(X ≥a) ≤20 a . Thus our best upper bounds are: P(X ≥a) ≤20 a if 20 ≤a ≤21.25 , and P(X ≥a) ≤425 a2 if a ≥21.25 . (c) Since X is non-negative and the density is symmetric with mean 20, we must have p(x) positive only on the interval [0,40]. Again by symmetry we have 10 = Z 40 20 xp(x) dx . Thus for a ≥20, 10 = Z 40 20 xp(x) dx ≥ Z 40 a xp(x) dx ≥aP(X ≥a) . Therefore, P(X ≥a) ≤10 a . Again by symmetry we have Z 40 20 (x −20)2p(x)dx = 12.5 . Then Z 40 20 x2p(x)dx −40 Z 40 20 xp(x)dx + 400 · 1 2 = 12.5 . ¿From this we obtain Z 40 20 x2p(x)dx = 12.5 + 40 · 10 −200 = 212.5 . Therefore, for a ≥20 we have 212.5 = Z 40 20 x2p(x)dx ≥ Z 40 a xp(x)dx ≥a2P(X ≥a) . Thus for a ≥20 we have P(X ≥a) ≤212.5 a2 . Combining our two estimates we have: P(X ≥a) ≤10 a if 20 ≤a ≤21.25 , and P(X ≥a) ≤212.5 a2 if 21.25 ≤a ≤40 . 57 SECTION 9.1(The answers to the problems in this chapter do not use the ‘1/2 correction mentioned in Section 9.1. 1. (a) .158655 (b) .6318 (c) .0035 (d) .9032 2. (a) .0564 (b) .0208 (c) 1.033 × 10−3 3. (a) P(June passes) ≈.985 (b) P(April passes) ≈.056 4. (a) P(499, 500 < S1,000,000 < 500, 500) ≥0 by Chebyschev. (b) P(499, 500 < S1,000,000 < 500, 500) ≈.6826 by the Central Limit Theorem. (a) P(499, 000 < S1,000,000 < 501, 000) ≥.75 by Chebyschev. (b) P(499, 000 < S1,000,000 < 501, 000) ≈.9545 by the Central Limit Theorem. (a) P(498, 500 < S1,000,000 < 501, 500) ≥.8889 by Chebyschev. (b) P(498, 500 < S1,000,000 < 501, 500) ≈.9973 by the Central Limit Theorem. 5. Since his batting average was .267, he must have had 80 hits. The probability that one would obtain 80 or fewer successes in 300 Bernoulli trials, with individual probability of success .3, is approximately .115. Thus, the low average is probably not due to bad luck (but a statistician would not reject the hypothesis that the player has a probability of success equal to .3). 6. We need to choose k so that P(S1000 ≤k) ≥.99. This is the same as P S∗ 1000 ≤k −500 15.81 ≥.99 . Thus we want k −500 15.81 = 2.33 . This will be true if k = 537. 7. .322 8. We want np + 2√npq = 108 and np −2√npq = 72. Adding and subtracting gives 2np = 180 and 4√npq = 36. Solving these two equations for n and p gives p = .1 and n = 900. 9. (a) 0 (b) 1 (Law of Large Numbers) (c) .977 (Central Limit Theorem) (d) 1 (Law of Large Numbers) 10. We want P(S10,000 ≤931) = P(S∗ 10,000 ≤931 −1000 30 ) = P(S∗ 10,000 ≤−2.3) ≈.011. 12. 13 58 13. P (S1900 ≥115) = P S∗ 1900 ≥ 115 −95 √ 1900 · .05 · .95 = P (S∗ 1900 ≥2.105) = .0176. 14. (a) 64 to 96 (b) 6400 16. n = 108, m = 77 17. We want 2√pq √n = .01. Replacing √pq by its upper bound 1 2, we have 1 √n = .01. Thus we would need n = 10,000. Recall that by Chebyshev’s inequality we would need 50,000. SECTION 9.2 1. (a) .4762 (b) .0477 2. .3174 3. (a) .5 (b) .9987 5. (a) P(S210 < 700) ≈.0757. (b) P(S189 ≥700) ≈.0528 (c) P(S179 < 700, S210 ≥700) = P(S179 < 700) −P(S179 < 700, S210 < 700) = P(S179 < 700) −P(S210 < 700) ≈.9993 −.0757 = .9236 . 6. (a) The expected loss is .2 cents and the variance of this loss is .36. (b) .2024 (c) .047 (d) .9994 (e) 54 7. (a) Expected value = 200, variance = 2 (b) .9973 8. P(S30 = 0) ≈ N(0) √ 30 · 1.5 = .0595. 9. P Sn n −µ ≥ϵ = P Sn −nµ ≥nϵ = P Sn −nµ √ nσ2 ≥ nϵ √ nσ2 . By the Central Limit Theorem, this probability is approximated by the area under the normal curve between √nϵ σ and inﬁnity, and this area approaches 0 as n tends to inﬁnity. 10. (a) The law of large numbers states that the average of Peter’s fortune will be close to 0. (b) The Central Limit Theorem states, for example, that with probability .95 Peter will not have won or lost more than $2 after the 10,000 plays. 11. Her expected loss is 60 dollars. The probability that she lost no money is about .0013. 12. Betting 1 dollar on red gives E(X) = −1 37 and Var(X)= .9787. 59 Betting 1 dollar on 17 gives E(X) = −1 37 and Var(X) = 34.08. Thus, by the Central Limit Theorem, if we bet 1 dollar on red for 100 plays, P(S100 > 20) = P(S∗ 100 > 2.295) ≈.011. If we bet 1 dollar on 17 for 100 plays, we have P(S100 > 20) = P(S∗ 100 > .389) ≈.3489. Thus, betting 1 dollar on 17 for 100 plays gives a higher probability of winning $20. (Of course, it also gives a higher probability of losing $20.) Similar calculations show that if we bet 1 dollar on red for 100 plays, the probability that we win any money is .3732. If we bet 1 dollar on 17 for 100 plays, the probability that we win any money is .4781. 13. p = .0056 SECTION 9.3 1. E(X∗) = 1 σ (E(X) −µ) = 1 σ (µ −µ) = 0 , σ2(X∗) = E X −µ σ 2 = 1 σ2 σ2 = 1 . 2. S∗ n = Sn −nµ √nσ = Sn √n = nAn √n = √nAn . 3. Tn = Y1 + Y2 + · · · + Yn = Sn −nµ σ . Since each Yj has mean 0 and variance 1, E(Tn) = 0 and V (Tn) = n. Thus T ∗ n = Tn √n = Sn −nµ σ√n = S∗ n . 4. For one uniform random variable on [0,20] the mean is 10 and the variance is 400/12. For the sum of 25 such random variables the mean is 250 and the standard deviation is p 25(400/12) = 28.87. Thus the normal density used to approximate the sum is: f(x) = 1 28.87 1 √ 2π exp −1 2 x −250 28.87 2! . For the standardized sum S∗the density for the normal approximation is the density with mean 0 and standard deviation 1: f(x) = 1 √ 2π e−x2/2 . The average of 25 numbers A25 has expected value 10 and standard deviation p (400/12)/25 = 1.155. Thus the normal density used to approximate the average is: f(x) = 1 1.155 1 √ 2π exp −1 2 x −10 1.155 2! . 10. By Chebyshev’s inequality we would need 10/n ≤.05 or n ≥200. By the Central Limit Theorem we would need 2 p 10/n ≈1, or n ≈40. To ﬁnd the variance necessary for 10 measurements to 60 suﬃce using Chebyshev’s inequality, we would need σ2/10 ≈.05, or σ2 ≈.5. Using the Central Limit Theorem we would need 2 p σ2/10 ≈1, or σ2 ≈2.5.(A larger variance is easier to obtain.) 11. (a) .5 (b) .148 (c) .018 13. .0013 14. (b) (20.53, 25.87) SECTION 10.1 1. In each case, to get g(t) just replace z by et in h(z). (a) h(z) = 1 2(1 + z) (b) h(z) = 6 X j=1 zj (c) h(z) = z3 (d) h(z) = 1 k + 1zn k X j=1 zj (e) h(z) = zn(pz + q)k (f) h(z) = 2 3 −z 2. (a) µ1 = 1/2, µ2 = 1/2, h′(1) = 1/2 = µ1, h′′(1) = 0 = µ2 −µ1. (b) µ1 = 7/2, µ2 = 91/6, h′(1) = 7/2, h′′(1) = 70/6 = µ2 −µ1. (c) µ1 = 3, µ2 = 9, h′(1) = 3, h′′(1) = 6 = µ2 −µ1. (d) µ1 = n + k/2, µ2 = n2 + nk + k(1 + 2k)/6, h′(1) = n + k/2 = µ1, h′′(1) = n2 + (k −1)n + k(k −1)/3 = µ2 −µ1. 3. (a) h(z) = 1 4 + 1 2z + 1 4z2 . (b) g(t) = h(et) = 1 4 + 1 2et + 1 4e2t . (c) g(t) = 1 4 + 1 2 ∞ X k=0 tk k! + 1 4 ∞ X k=0 2k k! tk = 1 + ∞ X k=1 1 2k! + 2k−2 k! tk = 1 + ∞ X k=1 µk k! tk . Thus µ0 = 1, and µk = 1 2 + 2k−2 for k ≥1. (d) p0 = 1 4, p1 = 1 2, p2 = 1 4 . 4. h(z) = 1 −3 2µ1 + 1 2µ2 + (2µ1 −µ2)z + (µ2 −µ1) 2 z2 . Thus p0 = 1 −3 2µ1 + 1 2µ2, p1 = 2µ1 −µ2, p3 = µ2 −µ1 2 . 61 5. (a) µ1(p) = µ1(p′) = 3, µ2(p) = µ2(p′) = 11 µ3(p) = 43, µ3(p′) = 47 µ4(p) = 171, µ4(p′) = 219 6. (a) p2 = 0 1 2 3 4 0 0 1/4 4/9 4/9 . (b) h(z) = z 3(1 + 2z), h2(z) = z2 9 (1 + 4z + 4z2) = h(z) 2. (c) hn(z) = z 3(1 + 2z) n . (d) µ1 = 5 3n, µ2 = 25 9 n2 + 2 9n. Thus the mean of pn = 5 3n, and the variance of pn = 2 9n. Since p has mean 5 3, we see that the mean of pn is n times the mean of p. (e) pn(j) > 0 for j = n, · · · , 2n. 7. (a) g−X(t) = g(−t) (b) gX+1(t) = etg(t) (c) g3X(t) = g(3t) (d) gaX+b = ebtg(at) 8. (a) h(z) = 1 3(z + 2) , (b) h2(z) = 1 3(z + 2) 2 , (c) hn(z) = 1 3(z + 2) n , (d) hn(z) = 1 3(z1/n + 2) n , (e) hn(z) = 1 z √nµ/σ 1 3(z1/√nσ + 2) n , g(t) = 1 3(et + 2) . g2(t) = 1 3(et + 2) 2 . gn(t) = 1 3(et + 2) n . gn(t) = 1 3(et/n + 2) n . gn(t) = e−t√nµ/σ1 3(et/√nσ + 2) n . Note: µ = 1/3 and σ = √ 2/3. 9. (a) hX(z) = 6 X j=1 ajzj, hY (z) = 6 X j=1 bjzj. (b) hZ(z) = 6 X j=1 ajzj 6 X j=1 bjzj . (c) Assume that hZ(z) = (z2 + · · · + z12)/11 . Then 6 X j=1 ajzj−1 6 X j=1 bjzj−1 = 1 + z + · · · z10 11 = z11 −1 11(z −1) . Either 6 X j=1 ajzj−1or 6 X j=1 bjzj−1 is a polynomial of degree 5 (i.e., either a6 ̸= 0 or b6 ̸= 0). Suppose that 6 X j=1 ajzj−1 is a polynomial of degree 5. Then it must have a real root, which is a real root of (z11 −1)/(z −1). However (z11 −1)/(z −1) has no real roots. This is because the only real root of z11 −1 is 1, which cannot be a real root of (z11 −1)/(z −1). Thus, we have a contradiction. 62 This means that you cannot load two dice in such a way that the probabilities for any sum from 2 to 12 are the same. (cf. Exercise 11 of Section 7.1). 10. h(1) = 1 −√1 −4pq 2q = 1 − p 1 −4p + 4p2 2q = 1 −\|2p −1\| 2q = q/p, if p ≤q, 1, if p ≥q. h′(z) =        4pqz 2qz p 1 −4pqz2 − 1 2qz2 (1 − p 1 −4pqz) , if p > q, −1 z2 1 − p 1 −z2 + 1 √ 1 −z2 . if p = q. Therefore, h′(1) = 1/(p −q), if p > q, ∞, if p = q. 11. Let pn = probability that the gambler is ruined at play n. Then pn = 0, if n is even, p1 = q, pn = p(p1pn−2 + p3pn−4 + · · · + pn−2p1), if n > 1 is odd. Thus h(z) = qz + pz h(x) 2 , so h(z) = 1 − p 1 −4pqz2 2pz . By Exercise 10 we have h(1) = q/p, if q ≤p, 1, if q ≥p, h′(1) = 1/(q −p), if q > p, ∞, if q = p. This says that when q > p, the gambler must be ruined, and the expected number of plays before ruin is 1/(q −p). When p > q, the gambler has a probability q/p of being ruined. When p = q, the gambler must be ruined eventually, but the expected number of plays until ruin is not ﬁnite. 12. dn dxn (1 −x)1/2 x=0 = (−1)n 1 2 1 2 −1 · · · 1 2 −n + 1 . Thus, (1 −x)1/2 = ∞ X n=0 (−1)n 1 2 1 2 −1 · · · 1 2 −n + 1 n! (−x)n = ∞ X n=0 1 2 1 2 −1 · · · 1 2 −n + 1 n! xn = ∞ X n=0 1/2 n xn. 63 Therefore, h(z) = 1 − p 1 −4pqz2 2pz = 1 − ∞ X n=0 1/2 n (−4pqz2)n 2pz = ∞ X n=1 1 2p(−1)n 1/2 n (4pq)nz2n−1 , and pT (n) =    1 2p(−1)k 1/2 k (4pq)k, if n = 2k −1 = odd, 0, if n = even. 13. (a) From the hint: hk(z) = hU 1(z) · · · hU k (z) = h(z) k . (b) hk(1) = h(1) k = (q/p)k if q ≤p, 1 if q ≥p. h′(1) = k/(q −p) if q > p, ∞ if q = p. Thus the gambler must be ruined if q ≥p. The expected number of plays in this case is k/(q −p) if q > p and ∞if q = p. When q < p he is ruined with probability (q/p)k. 14. gX∗(t) = E(eX∗t) = E(e X−µ σ t) = e−µ σ tE(e X σ t) = e−µ σ tgX t σ . SECTION 10.2 1. (a) d = 1 (b) d = 1 (c) d = 1 (d) d = 1 (e) d = 1/2 (f) d ≈.203 2. (a) .618, (b) .414, (c) 0 if t = 0, −1 + √ 5 2 =.618 if t ̸= 0 3. (a) 0 (b) 276.26 64 4. h(z) = E ZSn = X k E(ZSk\|N = k)P(N = k) = X k E(ZX1) k P(N = k) = X k (f(z))kP(N = k) = g(f(z)) . 5. Let Z be the number of oﬀspring of a single parent. Then the number of oﬀspring after two generations is SN = X1 + · · · + XN , where N = Z and Xi are independent with generating function f. Thus by Exercise 4, the generating function after two generations is h(z) = f(f(z)). 6. (a) f(z) = p + qz, g(z) = rz 1 −(1 −r)z . (c) If p < r, then the expected length of the busy period is ﬁnite. (If p = r, then the expected length of the busy period is inﬁnite, but with probability 1, the busy period will be of ﬁnite length.) 7. Let N be the time she needs to be served. Then the number of customers arriving during this time is X1 + · · · + XN, where Xi are identically distributed independent of N. P(X0 = 0) = p, P(Xi = 1) = q. Thus by Exercise 4, h(z) = g(f(z)). 8. The server ultimately has a time when he is not busy if the branching procees dies out. For this we need m ≤1 or h′(1) = 1. But h(z) = g(f(z)), so we need h′(1) = g′(1)f ′(1) = mean arrival rate · mean service time ≤1 . This means that we need q/r ≤1. 9. If there are k oﬀspring in the ﬁrst generation, then the expected total number of oﬀspring will be kN, where N is the expected total numer for a single oﬀspring. Thus we can compute the expected total number by counting the ﬁrst oﬀspring and then the expected number after the ﬁrst generation. This gives the formula N = 1 + X k kpk = 1 + mN . ¿From this it follows that N is ﬁnite if and only if m < 1, in which case N = 1/(1 −m). 10. You can work this by passing to the limit in the expressions given in Example 4, but it is easier to do it directly as follows: The generating function h1(z) = h(z) for the population after one generation is h(z) = ∞ X k=0 (1/2)j+1zj = 1/2 1 −(1/2)z = 1 (2 −z) . Then we can get the generating functions for future generations by using the relation hn+1(z) = hn(h(z)). For example, h2(z) = 1 (2 − 1 (2−z)) = 2 −x 3 −2x . Continuing in this way, we get h3(z) = 3 −2x 4 −3x . 65 These results suggest that the general case is hn(z) = n −x(n −1) (n + 1) −nx . It is easy to check that this satisﬁes the equation hn+1 = hn(h(z)), so by induction we see that our guess for hn(z) is correct. Then hn(z) = n −x(n −1) n + 1 1 1 − n n+1x = n −z(n −1) n + 1 ∞ X j=0 n n + 1 j zj . The constant term is pn(0) = n/(n + 1). Collecting the coeﬃcients of zj and simplifying gives p(n)(j) = 1 n(n + 1) n n + 1 j . (b) The probability that the population dies out at the nth generation is equal to the diﬀerence between the probability that it has died out by the nth generation and the probability that it has died out by the (n −1)st generation. This is: pn 0 −pn−1 0 = n n + 1 −n −1 n = 1 n(n + 1) . (c) The expected lifetime is X n n · P(population dies out on the nth generation) = ∞ X n=1 n n(n + 1) = ∞ X n=1 1 n + 1 = ∞. SECTION 10.3 1. (a) g(t) = 1 2t(e2t −1) (b) g(t) = e2t(2t −1) + 1 2t2 (c) g(t) = e2t −2t −1 2t2 (d) g(t) = e2t(ty −1) + 2et −t −1 t2 (e) (3/8) e2t(4t2 −4t + 2) −2 t3 2. (a) µ1 = 1 = g′(0), µ2 = 4 3 = g′′(0) . (b) µ1 = 4 3 = g′(0), µ2 = 2 = g′′(0) . (c) µ1 = 2 3 = g′(0), µ2 = 2 3 = g′′(0) . 66 (d) µ1 = 1 = g′(0), µ2 = 3 2 = g′′(0) . (e) µ1 = 3 2 = g′(0), µ2 = 12 5 = g′′(0) . 3. (a) g(t) = 2 2 −t (b) g(t) = 4 −3t 2(1 −t)(2 −t) (c) g(t) = 4 (2 −t)2 (d) g(t) = λ λ + t , t < λ . 4. (a) µ1 = 1 2 = g′(0), µ2 = 1 2 = g′′(0) . (b) µ1 = 3 4 = g′(0), µ2 = 5 4 = g′′(0) . (c) µ1 = 1 = g′(0), µ2 = 3 2 = g′′(0) . (d) µ1 = n λ = g′(0), µ2 = n(n + 1) λ2 = g′′(0) . 5. (a) k(τ) = 1 2iτ (e2iτ −1) (b) k(τ) = e2iτ(2iτ −1) + 1 −2τ 2 (c) k(τ) = e2iτ −2iτ −1 −2τ 2 (d) k(τ) = e2iτ(iτ −1) + 2eiτ −iτ −1 −τ 2 (e) k(τ) = (3/8) e2iτ(−4τ 2 −4iτ + 2 −iτ 3 6. fX(x) = 1 2π Z ∞ −∞ e−iτe\|τ\|dτ = 1 2π Z ∞ 0 e−iτe−τdτ + 1 2π Z 0 −∞ e−iτeτdτ = 1 2π Z ∞ 0 e−iτx + eiτx e−τdτ = 1 π Z ∞ 0 cos(τx)e−τdτ . Now to calculate this last integral: 1 π Z ∞ 0 cos(τx)e−τdτ = 1 π h −e−τcos(τx) ∞ 0 −x Z ∞ 0 e−τsin(τx)dτ i = 1 π h 1 −x(−e−τsin(τx)) ∞ 0 + x2 Z ∞ 0 e−τcos(τx) i = 1 π h 1 −x2 Z ∞ 0 e−τcos(τx)dτ i . Solving this for the integral we obtain: 1 π Z ∞ 0 cos(τx)e−τdτ = 1 π(1 + x2) . Thus, fX(x) = 1 π Z ∞ 0 cos(τx)e−τdτ = 1 π(1 + x2) . 67 7. (a) g(−t) = 1 −e−t t (b) etg(t) = e2t −et t (c) g(et) = e3t −1 3t (d) ebg(at) = eb(eat −1) at 8. (a) g(t) = eat −ebt t(b −a) . (b) g(t) = eat −ebt t(b −a) 2 . (c) g(t) = eat −ebt t(b −a) n . (d) g(t) = eat/n −ebt/n t(b −a) n . (e) g(t) = e−√nµ/σeat/√nσ −ebt/√nσ t(b −a) n . 9. (a) g(t) = et2+t (b) g(t) 2 (c) g(t) n (d) g(t/n) n (e) et2/2 10. (a) m = 0, σ2 = 2 . (b) gX1(t) = 1 1 −t2 , gSn(t) = 1 1 −t2 n , t < 1 , gAn(t) = 1 1 −( t n)2 n gS∗ n(t) = 1 1 −t2 2n n . (c) gS∗ n(t) →e−t2 2 as n →∞. (d) gAn(t) →1 as n →∞. SECTION 11.1 1. w(1) = (.5, .25, .25) w(2) = (.4375, .1875, .375) w(3) = (.40625, .203125, .390625) 2. P = 1 0 1 2 1 2 , P2 = 1 0 3 4 1 4 , P3 = 1 0 7 8 1 8 . Pn = 1 0 2n−1 2n 1 2n → 1 0 1 0 . 68 Whatever the President’s decision, in the long run each person will be told that he or she is going to run. 3. Pn = P for all n. 4. .7. 5. 1 6. w(1) = w(2) = w(3) = w(n) = (.25, .5, .25). 7. (a) Pn = P (b) Pn = P, if n is odd, I, if n is even. 8. P = 0 1 0 1 −p p 1 p 1 −p . 9. p2 + q2, q2, 0 1 0 p q 1 q p 11. .375 12. (a) P =     P SL UL NS P .64 .08 .08 .2 SL .16 .48 .16 .2 UL .2 .2 .4 .2 NS 0 0 0 1    . (b) .24. 19. (a) 5/6. (b) The ‘transition matrix’ is P = H T H 5/6 1/6 T 1/2 1/2 . (c) 9/10. (d) No. If it were a Markov chain, then the answer to (c) would be the same as the answer to (a). SECTION 11.2 1. a = 0 or b = 0 2. H is the absorbing state. Y and D are transient states. It is possible to go from each of these states to the absorbing state, in fact in one step. 3. Examples 11.10 and 11.11 4. N = Gg gg GG 2 0 gg 2 1 . 69 5. The transition matrix in canonical form is P =         GG, Gg GG, gg Gg, Gg Gg, gg GG, GG gg, gg GG, Gg 1/2 0 1/4 0 1/4 0 GG, gg 0 0 1 0 0 0 Gg, Gg 1/4 1/8 1/4 1/4 1/16 1/16 Gg, gg 0 0 1/4 1/2 0 1/4 GG, GG 0 0 0 0 1 0 gg, gg 0 0 0 0 0 1         . Thus Q =       GG, Gg GG, gg Gg, Gg Gg, gg GG, Gg 1/2 0 1/4 0 GG, gg 0 0 1 0 Gg, Gg 1/4 1/8 1/4 1/4 Gg, gg 0 0 1/4 1/2 .       , and N = (I −Q)−1 =     GG, Gg GG, gg Gg, Gg Gg, gg GG, Gg 8/3 1/6 4/3 2/3 GG, gg 4/3 4/3 8/3 4/3 Gg, Gg 4/3 1/3 8/3 4/3 Gg, gg 2/3 1/6 4/3 8/3    . From this we obtain t = Nc =     GG, Gg 29/6 GG, gg 20/3 Gg, Gg 17/3 Gg, gg 29/6    , and B = NR =     GG, GG gg, gg GG, Gg 3/4 1/4 GG, gg 1/2 1/2 Gg, Gg 1/2 1/2 Gg, gg 1/4 3/4    . 6. The canonical form of the transition matrix is P =   N S R N 0 1/2 1/2 S 1/4 1/2 1/4 R 0 0 1  , N = N S N 4/3 4/3 S 2/3 8/3 , t = Nc = N 8/3 S 10/3 , B = NR = N 1 S 1 . 70 Here is a typical interpretation for an entry of N. If it is snowing today, the expected number of nice days before the ﬁrst rainy day is 2/3. The entries of t give the expected number of days until the next rainy day. Starting with a nice day this is 8/3, and starting with a snowy day it is 10/3. The entries of B reﬂect the fact that we are certain to reach the absorbing state (rainy day) starting in either state N or state S. 7. N =   2.5 3 1.5 2 4 2 1.5 3 2.5   Nc =   7 8 7   B =   5/8 3/8 1/2 1/2 3/8 5/8   8. The transition matrix in canonical form is P =       1 2 3 0 4 1 0 2/3 0 1/3 0 2 1/3 0 2/3 0 0 3 0 1/3 0 0 2/3 0 0 0 0 1 0 4 0 0 0 0 1 cr       , N =   1 2 3 1 7/5 6/5 4/5 2 3/5 9/5 6/5 3 1/5 3/5 7/5  , B = NR =   0 4 1 7/15 8/15 2 3/15 12/15 3 1/15 14/15  , t = NC =   1 17/5 2 18/5 3 11/5  . 9. 2.08 12. P =           ABC AC BC A B C none ABC 5/18 5/18 4/18 0 0 4/18 0 AC 0 5/12 0 5/2 0 1/12 1/12 BC 0 0 10/18 0 5/18 2/18 1/18 A 0 0 0 1 0 0 0 B 0 0 0 0 1 0 0 C 0 0 0 0 0 1 0 none 0 0 0 0 0 0 1           N =   1.385 .659 .692 0 1.714 0 0 0 2.25   71 Nc =   2.736 1.714 2.25   B =   A B C none ABC .275 .192 .440 .093 AC .714 0 .143 .143 BC 0 .625 .25 .125   13. Using timid play, Smith’s fortune is a Markov chain with transition matrix P =               1 2 3 4 5 6 7 0 8 1 0 .4 0 0 0 0 0 .6 0 2 .6 0 .4 0 0 0 0 0 0 3 0 .6 0 .4 0 0 0 0 0 4 0 0 .6 0 .4 0 0 0 0 5 0 0 0 .6 0 .4 0 0 0 6 0 0 0 0 .6 0 .4 0 0 7 0 0 0 0 0 .6 0 .4 0 0 0 0 0 0 0 0 0 1 0 8 0 0 0 0 0 0 0 0 1               . For this matrix we have B =           0 8 1 .98 .02 2 .95 .05 3 .9 .1 4 .84 .16 5 .73 .27 6 .58 .42 7 .35 .65           . For bold strategy, Smith’s fortune is governed instead by the transition matrix P =       1 2 4 0 8 1 0 .4 0 .6 0 2 0 0 .4 .6 0 4 0 0 0 .6 .4 0 0 0 0 1 0 8 0 0 0 0 1       , with B =   0 8 1 .936 .064 2 .84 .16 4 .6 .4  . From this we see that the bold strategy gives him a probability .064 of getting out of jail while the timid strategy gives him a smaller probability .02. Be bold! 14. It is the same. 15. (a) P =       3 4 5 1 2 3 0 2/3 0 1/3 0 4 1/3 0 2/3 0 0 5 0 2/3 0 0 1/3 1 0 0 0 1 0 2 0 0 0 0 1       . 72 (b) N =   3 4 5 3 5/3 2 4/3 4 1 3 2 5 2/3 2 7/3  , t =   3 5 4 6 5 5  , B =   1 2 3 5/9 4/9 4 1/3 2/3 5 2/9 7/9  . (c) Thus when the score is deuce (state 4), the expected number of points to be played is 6, and the probability that B wins (ends in state 2) is 2/3. 16.         g, GG G, Gg g, Gg G, gg G, GG g, gg g, GG 0 1 0 0 0 0 G, Gg .25 .25 .25 0 .25 0 g, Gg 0 .25 .25 .25 0 .25 G, gg 0 0 1 0 0 0 G, GG 0 0 0 0 1 0 g, gg 0 0 0 0 0 1         , N =    1.667 2.667 1.333 .333 .667 2.667 1.333 .333 .333 1.333 2.667 .667 .333 1.333 2.667 1.667   , Nc =      6 5 5 6 ,      B =    .667 .333 .667 .333 .333 .667 .333 .667   . 17. For the color-blindness example, we have B =     G, GG g, gg g, GG 2/3 1/3 G, Gg 2/3 1/3 g, Gg 1/3 2/3 G, gg 1/3 2/3    , and for Example 9 of Section 11.1, we have B =     GG, GG gg, gg GG, Gg 3/4 1/4 GG, gg 1/2 1/2 Gg, Gg 1/2 1/2 Gg, gg 1/4 3/4    . 73 In each case the probability of ending up in a state with all G’s is proportional to the number of G’s in the starting state. The transition matrix for Example 9 is P =         GG, GG GG, Gg GG, gg Gg, Gg Gg, gg gg, gg GG, GG 1 0 0 0 0 0 GG, Gg 1/4 1/2 0 1/4 0 0 GG, gg 0 0 0 1 0 0 Gg, Gg 1/16 1/4 1/8 1/4 1/4 1/16 Gg, gg 0 0 0 1/4 1/2 1/4 gg, gg 0 0 0 0 0 1         . Imagine a game in which your fortune is the number of G’s in the state that you are in. This is a fair game. For example, when you are in state Gg,gg your fortune is 1. On the next step it becomes 2 with probability 1/4, 1 with probability 1/2, and 0 with probability 1/4. Thus, your expected fortune after the next step is equal to 1, which is equal to your current fortune. You can check that the same is true no matter what state you are in. Thus if you start in state Gg,gg, your expected ﬁnal fortune will be 1. But this means that your ﬁnal fortune must also have expected value 1. Since your ﬁnal fortune is either 4 if you end in GG, GG or 0 if you end in gg, gg, we see that the probability of your ending in GG, GG must be 1/4. 18. (a)       1 2 3 F G 1 r p 0 q 0 2 0 r p q 0 3 0 0 r q p F 0 0 0 1 0 G 0 0 0 0 1       (b) Expected time in second year = 1.09. Expected time in med school = 3.3 years. (c) Probability of an incoming student graduating = .67. 19. (a) P =     1 2 0 3 1 0 2/3 1/3 0 2 2/3 0 0 1/3 0 0 0 1 0 3 0 0 0 1    . . (b) N = 1 2 1 9/5 6/5 2 6/5 9/5 , B = 0 3 1 3/5 2/5 2 2/5 3/5 , t = 1 3 2 3 . (c) The game will last on the average 3 moves. (d) If Mary deals, the probability that John wins the game is 3/5. 20. Consider the Markov chain with state i (for 1 ≤i < k) the length of the current run, and k an absorbing state. Then when in state i < k, the chain goes to i + 1 with probability 1/m or to 1 74 with probability (m −1)/m. Thus, starting in state 1, in order to get to state j + 1 the chain must be in state j and then move to j + 1. This means that N1,j+1 = N1,j(1/m) , or N1,j = mN1,j+1 . This will be true also for j + 1 = k if we interpret N1,k as the number of times that the chain enters the state k, namely, 1. Thus, starting with N1,k = 1 and working backwards, we see that N1,j = mk−j for j = 1, · · · , k. Therefore, the expected number of experiments until a run of k occurs is 1 + m + m2 + · · · + mk−1 = mk −1 m −1 . (The initial 1 is to start the process oﬀ.) Putting m = 10 and k = 9 we see that the expected number of digits in the decimal expansion of π until the ﬁrst run of length 7 would be about 111 million if the expansion were random. Thus we should not be surprised to ﬁnd such a run in the ﬁrst 100,000,000 digits of π and indeed there are runs of length 9 among these digits. 21. The problem should assume that a fraction qi = 1 − X j qij > 0 of the pollution goes into the atmosphere and escapes. (a) We note that u gives the amount of pollution in each city from today’s emission, uQ the amount that comes from yesterday’s emission, uQ2 from two days ago, etc. Thus wn = u + uQ + · · · uQn−1 . (b) Form a Markov chain with Q-matrix Q and with one absorbing state to which the process moves with probability qi when in state i. Then I + Q + Q2 + · · · + Qn−1 →N , so w(n) →w = uN . (c) If we are given w as a goal, then we can achieve this by solving w = Nu for u, obtaining u = w(I −Q) . 22. (a) The total amount of goods that the ith industry needs to produce $1 worth of goods is x1q1i + x2q2i + · · · + xnqni . This is the i’th component of the vector xQ. (b) By part (a) the amounts the industries need to meet their internal demands is xQ. Thus to meet both internal and external demands, the companies must produce amounts given by a vector x satifying the equation x = xQ + d . (c) From Markov chain theory we can always solve the equation x = xQ + d 75 by writing it as x(I −Q) = d and then using the fact that (I −Q)N = I to obtain x = dN . (d) If the row sums of Q are all less than 1, this means that every industry makes a proﬁt. A company can rely directly or indirectly on a proﬁt-making company. If for any value of n, qn ij > 0, then i depends at least indirectly on j. Thus depending upon is equivalent in the Markov chain interpretation to being able to reach. Thus the demands can be met if every company is either proﬁt-making or depends upon a proﬁt-making industry. (e) Since x = dN, we see that xc = dNc = dt . 24. When the walk is in state i, it goes to i+1 with probability p and i−1 with probability q. Condition (a) just equates the probability of winning in terms of the current state to the probability after the next step. Clearly, if our fortune is 0, then the probability of winning is 0, and if it is T, then the probability is 1. Here is an instructive way (not the simplest way) to see that the values of w are uniquely determined by (a), (b), and (c). Let P be the transition matrix for our absorbing chain. Then these equations state that Pw = w . That is, the column vector w is a ﬁxed vector for P. Consider the transition matrix for an arbitrary Markov chain in canonical form and assume that we have a vector w such that w = Pw. Multiplying through by P, we see that P2w = w, and in general Pnw = w. But Pn → 0 B 0 I . Thus w = 0 B 0 I w . If we write w = wT wA , where T is the set of transient states and A the set of absorbing states, then by the argument above we have w = wT wA = BwA wA . Thus for an absorbing Markov chain, a ﬁxed column vector w is determined by its values on the absorbing states. Since in our example we know these values are (0,1), we know that w is completely determined. The solutions given clearly satisfy (b) and (c), and a direct calculation shows that they also satisfy (a). 26. Again, it is easy to check that the proposed solution f(x) = x(n−x) satisﬁes conditions (a) and (b). The hard part is to prove that these equations have a unique solution. As in the case of Exercise 23, it is most instructive to consider this problem more generally. We have a special case of the following situation. Consider an absorbing Markov chain with transition matrix P in canonical form and with transient states T and absorbing states A. Let f and g be column vectors that satisfy the following system of equations Q R 0 I fA 0 + gA 0 = fA 0 , 76 where gA is given and it is desired to determine fA. In our example, gA has all components equal to 1. To solve for fA we note that these equations are equivalent to QfA + gA = fA , or (I −Q)fA = gA . Solving for fA, we obtain fA = NgA . Thus fA is uniquely determined by gA. 27. Use the solution to Exercise 24 with w = f. 28. Using the program Absorbing Chain for the transition matrix corresponding to the pattern HTH, we ﬁnd that t =   HT 6 H 8 ∅ 10  . Thus E(T) = 10. For the pattern HHH the transition matrix is P =     HHH HH H ∅ HHH 1 0 0 0 HH .5 0 0 .5 H 0 .5 0 .5 ∅ 0 0 .5 .5    . Solving for t for this matrix gives t =   HH 8 H 12 ∅ 14  . Thus for this pattern E(T) = 14. 29. For the chain with pattern HTH we have already veriﬁed that the conjecture is correct starting in HT. Assume that we start in H. Then the ﬁrst player will win 8 with probability 1/4, so his expected winning is 2. Thus E(T\|H) = 10 −2 = 8, which is correct according to the results given in the solution to Exercise 28. The conjecture can be veriﬁed similarly for the chain HHH by comparing the results given by the conjecture with those given by the solution to Exercise 28. 30. T must be at least 3. Thus when you sum the terms P(T > n) = 2P(T = n + 1) + 8P(T = n + 3), the coeﬃcients of the 2 and the 8 just add up to 1 since they are all possible probabilies for T. Let T be an integer-valued random variable. We write E(T) = P(T = 1) + P(T = 2) + P(T = 3) + · · · + P(T = 2) + P(T = 3) + · · · + P(T = 3) + · · · If we add the terms by columns, we get the usual deﬁnition of expected value; if we add them by rows, we get the result that E(T) = ∞ X n=0 P(T > n) . 77 That the order does not matter follows from the fact that all the terms in the sum are positive. 31. You can easily check that the proportion of G’s in the state provides a harmonic function. Then by Exercise 27 the proportion at the starting state is equal to the expected value of the proportion in the ﬁnal aborbing state. But the proportion of 1s in the absorbing state GG, GG is 1. In the other absorbing state gg, gg it is 0. Thus the expected ﬁnal proportion is just the probability of ending up in state GG, GG. Therefore, the probability of ending up in GG, GG is the proportion of G genes in the starting state.(See Exercise 17.) 32. The states with all squares the same color are absorbing states. From any non-absorbing state it is possible to reach any absorbing state corresponding to a color still represented in the state. To see that the game is fair, consider the following argument. In order to decrease your fortune by 1 you must choose a red square and then choose a neighbor that is not red. With the same probability you could have chosen the neighbor and then the red square and your fortune would have been increased by 1. Since it is a fair game, if at any time a proportion p of the squares are red, for example, then p is also the probability that we end up with all red squares. 33. In each case Exercise 27 shows that f(i) = biNf(N) + (1 −biN)f(0) . Thus biN = f(i) −f(0) f(N) −f(0) . Substituting the values of f in the two cases gives the desired results. SECTION 11.3 1. (a), (f) 2. (a) P3 =   .5 .333 .167 .562 .250 .187 .375 .500 .125  . Since P3 has no zero entries, P is regular. (b) 1/6. (c) w = (1/2, 1/3, 1/6). 3. (a) a = 0 or b = 0 (b) a = b = 1 (c) (0 < a < 1 and 0 < b < 1) or (a = 1 and 0 < b < 1) or (0 < a < 1 and b = 1). 4. w = (b/(b + a), a/(b + a)). 5. (a) (2/3, 1/3) (b) (1/2, 1/2) (c) (2/7, 3/7, 2/7) 6. Let P = 0 1 1 0 . Then P2n+1 = P and P2n = I. Thus P is not regular. However, the average An = 1 n P + P2 + · · · + Pn of these matrices converges to 1/2 1/2 1/2 1/2 . The vector w = (1/2, 1/2) is a ﬁxed vector for P. Its components represent the average number of times the process will be in each state in the long run. 78 7. The ﬁxed vector is (1, 0) and the entries of this vector are not strictly positive, as required for the ﬁxed vector of an ergodic chain. 8. The vectors (1,0,0) and (0,0,1) are ﬁxed vectors, and so is any vector of the form a(1,0,0) + (1 −a)(0,0,1) for 0 < a < 1. Pn →   1 0 0 1/2 0 1/2 0 0 1  . 9. Let P =   p11 p12 p13 p21 p22 p23 p31 p32 p33  , with column sums equal to 1. Then (1/3, 1/3, 1/3)P = (1/3 3 X j=1 pj1, 1/3 3 X j=1 pj2, 1/3 3 X j=1 pj3) = (1/3, 1/3, 1/3) . The same argument shows that if P is an n × n transition matrix with columns that add to 1 then w = (1/n, · · · , 1/n) is a ﬁxed probability vector. For an ergodic chain this means the the average number of times in each state is 1/n. 10. In Example 11.10 of Section 11.1, the state GG is an absorbing state, and it is impossible to go from this state to either of the other two states. 11. In Example 11.11 of Section 11.1, the state (GG, GG) is absorbing, and the same reasoning as in the immediately preceding answer applies to show that this chain is not ergodic. 12. The chain is ergodic but not regular. Note that it is impossible to reach states 1 and 3 from state 0 in an even number of steps, and it is impossible to reach states 0, 2, and 4 from state 0 in an odd number of steps. 13. The ﬁxed vector is w = (a/(b + a), b/(b + a)). Thus in the long run a proportion b/(b + a) of the people will be told that the President will run. The fact that this is independent of the starting state means it is independent of the decision that the President actually makes. (See Exercise 2 of Section 11.1) 14. The ﬁxed vector is the common row of P. The chain is regular if and only if the entries of this vector are strictly positive. 15. It is clearly possible to go between any two states, so the chain is ergodic. From 0 it is possible to go to states 0, 2, and 4 only in an even number of steps, so the chain is not regular. For the general Erhrenfest Urn model the ﬁxed vector must statisfy the following equations: 1 nw1 = w0 , wj+1 j + 1 n + wj−1 n −j + 1 n = wj, if 0 < j < n, 1 nwn−1 = wn . It is easy to check that the binomial coeﬃcients satisfy these conditions. 79 16. P2 is strictly positive. The ﬁxed vector is w = (1/4, 1/2, 1/4). 17. Consider the Markov chain whose state is the value of Sn mod(7), that is, the remainder when Sn is divided by 7. Then the transition matrix for this chain is P =           0 1 2 3 4 5 6 0 0 1/6 1/6 1/6 1/6 1/6 1/6 1 1/6 0 1/6 1/6 1/6 1/6 1/6 2 1/6 1/6 0 1/6 1/6 1/6 1/6 3 1/6 1/6 1/6 0 1/6 1/6 1/6 4 1/6 1/6 1/6 1/6 0 1/6 1/6 5 1/6 1/6 1/6 1/6 1/6 0 1/6 6 1/6 1/6 1/6 1/6 1/6 1/6 0           . Since the column sums of this matrix are 1, the ﬁxed vector is w = (1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7) . 18. 2r + 1 (by this power there must have been a repetition of the pattern of positive numbers in the matrix so nothing new can occur). N(3) = 5. See Exercise 19. 19. (a) For the general chain it is possible to go from any state i to any other state j in r2−2r+2 steps. We show how this can be done starting in state 1. To return to 1, circle (1, 2, .., r −1, 1) r −2 times (r2 −3r + 2 steps) and (1, ..., r, 1) once (r steps). For k = 1, ..., r −1 to reach state k + 1, circle (1, 2, . . . , r, 1) r−k times (r2−rk steps) then (1, 2, . . . , r−1, 1) k−2 times (rk−2r−k+2 steps) and then move to k +1 in k steps.You have taken r2 −2r +2 steps in all. The argument is the same for any other starting state with everything translated the appropriate amount. (b) P =   0 ∗ 0 ∗ 0 ∗ ∗ 0 0  , P2 =   ∗ 0 ∗ ∗ ∗ 0 0 ∗ 0  , P3 =   ∗ ∗ 0 ∗ ∗ ∗ ∗ 0 ∗  , P4 =   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0  , P5 =   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗  . 20. The transition matrix is        0 1 2 . . . n −1 n 0 1 −p p 0 . . . 0 0 1 r(1 −p) pr + (1 −p)(1 −r) p(1 −r) . . . 0 0 2 0 r(1 −p) pr + (1 −p)(1 −r) . . . 0 0 . . . . . . . . . ... . . . n 0 0 . . . . . . r 1 −r        . This transition matrix has a property called reversibility which will be discussed in Section 11.5. For such a chain the ﬁxed vector w satisﬁes the condition wipij = wjpji . When this condition is satisﬁed, it is easy to determine the ﬁxed vector. For this example, re-versibility in this chain means that wip(1 −r) = wi+1r(1 −p) , 0 < i < n , 80 or wi+1 wi = p(1 −r) r(1 −p) , 0 < i < n . Thus wi = cρi, 0 < i < n, where ρ = p(1 −r) r(1 −p) . The values for w0 and wn are obtained from w0p = w1r(1 −p) and wnr = wn−1p(1 −r). The constant c is then chosen to make the probabilities add to 1. If the traﬃc intensity s = p/r is greater than 1 then ρ is greater than 1, and if it is less than 1 then ρ is less than one. Thus when the traﬃc intensity is greater than 1 the ﬁxed vector is large for large values of i, and when it is less than 1 the ﬁxed vector is small for large values of n. This means that the queue size will build up when the traﬃc intensity is greater than or equal to 1. The case ρ eaual to 1 is a border-line case, and in this case the equilibrium vector is constant for 0 < i < n. 22. (a) P(S = j) = P(customer does not ﬁnish in j −1 sec.,but does in jth sec. ) = (1 −r)j−1r , P(T = j) = P(no arrival in j −1 seconds,but arrival in jth second ) = (1 −p)j−1p . (b) S and T have geometric distributions, and so E(S) = 1/r and E(T) = 1/p. (c) The traﬃc intensity s greater than 1 means p is greater than r, or E(S) is greater than E(T). This means that the arrival rate is faster than the service rate, so the queue size builds up; s equal to 1 means p is equal to r or that E(S) is equal to E(T); s less than 1 means that p is less than r, or E(S) is less than E(T) and the queue size does not build up. 24. Fixed probability vector is (1/5, 4/5). Thus wP = w implies 1 5 × .5 + 4 5p = 1 5 , so p = .125. 25. To each Markov chain we can associate a directed graph, whose vertices are the states i of the chain, and whose edges are determined by the transition matrix: the graph has an edge from i to j if and only if pij > 0. Then to say that P is ergodic means that from any state i you can ﬁnd a path following the arrows until you reach any state j. If you cut out all the loops in this path you will then have a path that never interesects itself, but still leads from i to j. This path will have at most r −1 edges, since each edge leads to a diﬀerent state and none leads to i. Following this path requires at most r −1 steps. 26. Assume that P is ergodic. Let M be the maximum of the steps required to go between two states. Then it is possible to go from any state i to any state j in M steps. To see this, assume that it is possible to go from i to j in m steps with m < M. Then just stay in i for M −m steps before starting on your route to M. Thus P is regular. 27. If P is ergodic it is possible to go between any two states. The same will be true for the chain with transition matrix 1 2(I+P). But for this chain it is possible to remain in any state; therefore, by Exercise 26, this chain is regular. 28. Assume that wP = w. Then w(I + P)/2 = (w + w)/2 = w . 81 Conversely, if w(I + P)/2 = w then wP/2 = w/2, and wP = w. 29. (b) Since P has rational transition probabilities, when you solve for the ﬁxed vector you will get a vector a with rational components. We can multiply through by a suﬃciently large integer to obtain a ﬁxed vector u with integer components such that each component of u is an integer multiple of the corresponding component of a. Let a(n) be the vector resulting from the nth iteration. Let b(n) = a(n)P. Then a(n+1) is obtained by adding chips to b(n+1). We want to prove that a(n+1) ≥a(n). This is true for n = 0 by construction. Assume that it is true for n. Then multiplying the inequality by P gives that b(n+1) ≥b(n). Consider the component a(n+1) j . This is obtained by adding chips to b(n+1) j until we get a multiple of aj. Since b(n) j ≤b(n+1) j , any multiple of aj that could be obtained in this manner to deﬁne a(n+1) j could also have been obtained to deﬁne a(n) j by adding more chips if necessary. Since we take the smallest possible multiple aj, we must have a(n) j ≤an+1 j . Thus the results after each iteration are monotone increasing. On the other hand, they are always less than or equal to u. Since there are only a ﬁnite number of integers between components of a and u, the iteration will have to stop after a ﬁnite number of steps. 30. Assume that the tape can hold at most n units.Then the transition matrix is P =        0 1 2 3 . . . n −1 n 0 0 1 0 0 . . . 0 0 1 0 p q 0 . . . 0 0 2 0 p2 2 1 pq q2 . . . 0 0 . . . . . . . . . . . . . . . ... . . . . . . n pn n 1 pn−1q n 2 pn−2q2 n 3 pn−3q3 . . . n n−1 pqn−1 qn        . (Note that we assume that the request has no eﬀect when the tape is full). When the chain is in state i the expected value of the next position is 1 −ip. No matter how small p is, for large enough i this will be negative. Thus, no matter how small p is, we see that if the tape is big enough, there will be a tendancy to free up space when a large number of spaces are occupied. 31. If the maximum of a set of numbers is an average of other elements of the set, then each of the elements with positive weight in this average must also be maximum. By assumption, Px = x. This implies Pnx = x for all n. Assume that xi = M, where M is the maximum value for the xk’s, and let j be any other state. Then there is an n such that pn ij > 0. The ith row of the equation Pnx = x presents xi as an average of values of xk with positive weight,one of which is xj. Thus xj = M, and x is constant. 32. If 0 is the average of non-negative numbers, then any element of this average occurring with positive weight must be 0. Assume that w is a ﬁxed probability vector for an ergodic chain P. Then wP = w and wPn = w. Assume that wi = 0. For any j there is an n such that pn ij > 0. Thus the ith column of wPn presents wi as an average of other values of wk, with wj occurring with positive weight. Hence wj = 0. SECTION 11.4 1. 1/3 1/3 2. Pn →cw. Thus Pny →cwy = wyc, since wy is a number. 82 3. For regular chains, only the constant vectors are ﬁxed column vectors. 4. All vectors of the form ay + bz for a and b constants are ﬁxed vectors for the matrix P of Exercise 3. There are no other ﬁxed vectors. 6. Let y(n) = Pny. Then y(n+1) i = P j pijy(n) j . Thus y(n+1) i ≤P j pijMn = Mn. This means that Mn is a upper bound for y(n+1) i . Therefore, Mn+1 ≤Mn. A similar argument shows that mn+1 ≥mn. Hence Mn+1 −mn+1 ≤Mn −mn for any n ≥1. Thus if we can show that a subsequence of diﬀerences Mn −mn tends to 0, then the same will be true for the entire sequence of diﬀerences, since this sequence is monotone decreasing. 8. This game has the ﬂavor of Doeblin’s coupling idea. Once you and your friend happen to be looking at the same number, from that time on you will continue together. If this happens you will end up together and you can successfully state that she stopped the same place you did. To see how successful you will be, you will have to estimate the probability that the coupling takes place. It is easy to write a program using random permutations of 52 objects to simulate this process. If you do, you will ﬁnd that about 75 percent of the time the coupling is successful. SECTION 11.5 1. Z = 11/9 −2/9 −1/9 10/9 . and M = 0 2 4 0 . 2. P =   S A W S 1/2 1/2 0 A 1/4 1/2 1/4 W 0 1/3 2/3  . Z =   S A W S 1.333 0 −1 A −.222 .889 −.333 W −.889 −.444 1.667  , w = (2/9, 4/9, 3/9). (a) Mean recurrence time for S = 1/w1 = 4.5. (b) m31 = 10. 3. 2 4. Mean recurrence time for Yes is 1 + a/b, and for No it is 1 + b/a. For a = .5 and b = .75 this gives a mean recurrence time of 5/3 for Yes and 5/2 for No. 5. The ﬁxed vector is w = (1/6,1/6,1/6,1/6,1/6,1/6), so the mean recurrence time is 6 for each state. 6. P =   R N S R 1 0 0 N 1/2 0 1/2 S 1/4 1/4 1/2  , N = N S N 4/3 4/3 S 2/3 8/3 , 83 t = Nc = N 8/3 S 10/3 . This tells us that if it is nice today, then the expected number of days until rain is 8/3. If it is snowy today, then the expected number of days until rain is 10/3. 7. (a)         1 2 3 4 5 6 1 0 0 1 0 0 0 2 0 0 1 0 0 0 3 1/4 1/4 0 1/4 1/4 0 4 0 0 1/2 0 0 1/2 5 0 0 1/2 0 0 1/2 6 0 0 0 1/2 1/2 0         (b) The rat alternates between the sets {1, 2, 4, 5} and {3, 6}. (c) w = (1/12, 1/12, 4/12, 2/12, 2/12, 2/12). (d) m1,5 = 7 8. The mean recurrence time for state 0 is the average time between times that the server is busy. 9. (a) if n is odd, P is regular. If n is even, P is ergodic but not regular. (b) w = (1/n, · · · , 1/n). (c) From the program Ergodic we obtain M =       0 1 2 3 4 0 0 4 6 6 4 1 4 0 4 6 6 2 6 4 0 4 6 3 6 6 4 0 4 4 4 6 6 4 0       . This is consistent with the conjecture that mij = d(n −d), where d is the clockwise distance from i to j. 10. The transition matrix is: P =         0 1 2 3 4 5 0 0 1 0 0 0 0 1 1/2 0 1/2 0 0 0 2 0 1/2 0 1/2 0 0 3 0 0 1/2 0 1/2 0 4 0 0 0 1/2 0 1/2 5 0 0 0 0 1 0         , w = (.25, .125, .125, .125, .125, .25), and M =         0 1 2 3 4 5 0 0 1 4 9 16 25 1 9 0 3 8 15 24 2 16 7 0 5 12 21 3 21 12 5 0 7 16 4 24 15 8 3 0 9 5 25 16 9 4 1 0         . 84 Note that the entries of the ﬁrst passage matrix are all integers. They also form arithmetic progres-sions going down diagonals. General formuals for basic quantities for random walks can be found in Finite Markov Chains by John G. Kemeny and J. Laurie Snell (New York: Springer-Verlag, l976). 11. Yes, the reverse transition matrix is the same matrix. 12. P =   1/2 0 1/2 2/3 1/3 0 0 2/3 1/3  , w = (4/10, 3/10, 3/10), w1p12 = 0 ̸= w2p21. 13. Assume that w is a ﬁxed vector for P. Then X i wip∗ ij = X i wiwjpji wi = X i wjpji = wj, so w is a ﬁxed vector for P. Thus if w is the unique ﬁxed vector for P we must have w = w. 14. No. In the Land of Oz example we found the mean ﬁrst-passage matrix to be M =   R N S R 0 4 10/3 N 8/3 0 8/3 S 10/3 4 0  . Note, for example, that mNR = 8/3 ̸= mRN = 4. 15. If pij = pji then P has column sums 1. We have seen (Exercise 9 of Section 11.3) that in this case the ﬁxed vector is a constant vector. Thus for any two states si and sj, wi = wj and pij = pji. Thus wipij = wjpji, and the chain is reversible. 16. We ﬁrst show that (I + P + · · · + Pn−1)(I −P + W) = I −Pn + nW . Recall that PW = W so also PkW = W. Thus, just multiplying out the left side gives this equality. Now if we divide through by n and pass to the limit we have I + P + . . . + Pn−1 n (I −P + W) →W Multiplying both sides on the right by Z and recalling that WZ = W we see that I + P + . . . + Pn−1 n →W 17. We know that wZ = w. We also know that mki = (zii −zki)/wi and wi = 1/ri. Putting these in the relation ¯ mi = X k wkmki + wiri , we see that ¯ mi = X k wk zii −zki wi + 1 = zii wi X k wk −1 wi X k wkzki + 1 = zii wi −1 + 1 = zii wi . 85 18. Form a Markov chain whose states are the possible outcomes of a roll. After 100 rolls we may assume that the chain is in equilibrium. We want to ﬁnd the mean time in equilibrium to obtain snake eyes for the ﬁrst time. For this chain mki is the same as ri, since the starting state does not eﬀect the time to reach another state for the ﬁrst time. The ﬁxed vector has all entries equal to 1/36, so ri = 36. Using this fact, we obtain ¯ mi = X k wkmki + wiri = 35 + 1 = 36. We see that the expected time to obtain snake eyes is 36, so the second argument is correct. 19. Recall that mij = X j zjj −zij wj . Multiplying through by wj summing on j and, using the fact that Z has row sums 1, we obtain mij = X j zjj − X j zij = X j zjj −1 = K, which is independent of i. 20. Assume that you start in state a. Then the expected amount you win on the nth step is P j P n a,jfj. From this it follows that your expected winning on the ﬁrst n steps can be represented by the column vector g(n), with g(n) = (I + P + P2 + · · · + Pn)f. But since wf = 0 also Wf = 0. Thus we have g(n) = (I + (P −W) + (P2 −W) + · · · + (Pn −W))f Letting n →∞we obtain, g(n) →Zf. We used here that if P is the transition matrix for a regular chain then Z is equal to the inﬁnite series: Z = I + (P −W) + (P2 −W) + (P3 −W) · · · . 21. The transition matrix is P =     GO A B C GO 1/6 1/3 1/3 1/6 A 1/6 1/6 1/3 1/3 B 1/3 1/6 1/6 1/3 C 1/3 1/3 1/6 1/6    . Since the column sums are 1, the ﬁxed vector is w = (1/4, 1/4, 1/4, 1/4) . From this we see that wf = 0. From the result of Exercise 20 we see that your expected winning starting in GO is the ﬁrst component of the vector Zf where f =    15 −30 −5 20   . Using the program ergodic we ﬁnd that the long run expected winning starting in GO is 10.4. 86 22. PW = W follows from the fact the columns of W are constant and the row sums of P are 1. Similarly WW = W follows from the fact that W has row sums 1 and constant columns. Thus W2 = W. Multiplying both sides by W gives W3 = W2 = W. Continuing in this way we obtain Wk = W. 23. Assume that the chain is started in state si. Let X(n) j equal 1 if the chain is in state si on the nth step and 0 otherwise. Then S(n) j = X(0) j + X(1) j + X(2) j + . . . X(n) j . and E(X(n) j ) = P n ij. Thus E(S(n) j ) = n X h=0 p(n) ij . If now follows then from Exercise 16 that lim n→∞ E(S(n) j ) n = wj. 24. He got it! 87
2209	https://jamanetwork.com/journals/jamaneurology/fullarticle/795358	Published Time: 2008-04-01 Harrison’s Neurology in Clinical Medicine \| JAMA Neurology \| JAMA Network [Skip to Navigation] Our website uses cookies to enhance your experience. 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Arch Neurol. 2008;65(4):554. doi:10.1001/archneur.65.4.554 Manage citations: Select Format Download citation Copy citation Close Permissions View Metrics Arch Neurol Published Online: April 2008 2008;65;(4):554. doi:10.1001/archneur.65.4.554 related icon Related Articlesmultimedia icon Media Close Trending GLP-1RA Exposure and NAION Prevalence by Diabetes StatusJAMA Ophthalmology Research Letter July 24, 2025 Diverticulitis: A ReviewJAMA Review July 24, 2025 Modified Target Delineation and Radiotherapy for High-Grade GliomaJAMA Network Open Original Investigation July 2025 Close Harrison's Principles of Internal Medicine is a household name for medical students, residents, and physicians throughout the world. It has been translated into many languages and is now in its 16th edition. Since its first edition published in 1950, it has contained a section on neurology. Eminent neurologists served as section editors: Houston Merritt, MD, Raymond Adams, MD, and Joseph Martin, MD. Stephen Hauser, MD, currently is the spiritus rector. The publisher and editor decided to publish Harrison's Neurology in Clinical Medicine as a stand-alone volume with the aim “to provide expanded coverage of clinically important topics geared to the needs of the practicing internist, while retaining the focus on pathophysiology and therapy that has always been characteristic of Harrison’s.” The book is organized in 7 sections, each containing between 1 and 32 chapters, with a review and self-assessment appendix. Section 1 is an introduction to neurology that crisply summarizes principles of the pathogenesis of neurological diseases, the approach to the patient with neurological disorders, and neuroimaging. Section 2 provides a symptom-oriented overview of the clinical manifestations of neurological diseases. Section 3 deals with all major categories of diseases affecting the central nervous system, while section 4 covers diseases of the peripheral nerves and muscle. Section 5 reviews chronic fatigue syndrome, section 6 presents an overview on psychiatric disorders relevant to the internist and neurologist, and section 7 addresses alcoholism and drug dependence. The text is enhanced by many excellent tables; boxes outlining the diagnostic approach and reviewing current treatments; excellent flow diagrams encapsulating the diagnostic approach; numerous excellent illustrations using state-of-the-art graphics, often almost artistic in nature; and radiologic images. Each chapter contains recommendations for further readings. Much in line with the tradition of Harrison's Principles, the authors successfully emphasize the etiology and pathogenic mechanisms of neurological diseases and lay out the fundamentals of their treatment. While it is difficult to single out any chapter, those on dizziness, syncope, and vertigo; cerebral vascular diseases; Alzheimer disease and other dementias; Parkinson disease and other movement disorders; ataxias; paraneoplastic syndromes; multiple sclerosis; approach to the patient with peripheral neuropathy; and polymyositis, dermatomyositis, and inclusion body myositis impressed me most as exemplary didactic accounts. I have not detected significant flaws. I do not think that the target audience for this book is only practicing internists; it is equally suited to serve as an excellent reference text for neurologists whether in residency, subspecialty training, or practice. Get Access View Full TextDownload PDF Button To Top Select Your Interests Customize your JAMA Network experience by selecting one or more topics from the list below. 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2210	https://www.youtube.com/watch?v=oPTw52MPVvE	TMWYF: Vertex-magic total labelings of graphs (Dan McQuillan) Talk Math With Your Friends 2395 views 7 Mar 2022 A total labeling of a graph G is an assignment of the integers 1,2,...,v+e, to the v vertices and e edges of G. The weight of a vertex is the sum of its "label" with the sum of its incident edge labels. The total labeling is vertex-magic (or a VMTL) if the weight of each vertex is a constant, called the magic constant for the VMTL. The questions are easy to state and reasonable to solve. Many results and proofs are quite elegant. There are many remaining open problems that are accessible to undergraduate students. This talk will discuss several past undergraduate research successes and possible future projects.
2211	https://www.reddit.com/r/EnglishLearning/comments/j62ex3/food_vs_foods_which_one_is_correct/	food vs. foods - Which one is correct? : r/EnglishLearning Skip to main contentfood vs. foods - Which one is correct? : r/EnglishLearning Open menu Open navigationGo to Reddit Home r/EnglishLearning A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to EnglishLearning r/EnglishLearning r/EnglishLearning A place for learning English. 英語の学びのスペースです。 Un lugar para aprender Inglés. مكان لتعلم اللغة الإنجليزية. Un lieu pour apprendre l'Anglais. Ein Ort zum Englisch lernen. 613K Members Online •5 yr. ago heeeda food vs. foods - Which one is correct? Hi all Could you tell me which one of the below is correct? Love exploring new food and recipes Love exploring new foods and recipes From my understanding, food is an uncountable noun but we use 'foods' when we refer to a specific type of food. (e.g. Vegan foods, British foods.. etc.) When we say 'new food', are we referring to a specific type? I am not quite sure about this. Appreciate your help and thank you in advance. Read more Archived post. New comments cannot be posted and votes cannot be cast. Share Related Answers Section Related Answers Difference between food and foods grammar Is food a countable or uncountable noun Plural form of food When to use foods instead of food Usage of all food in sentences New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of October 6, 2020 Reddit reReddit: Top posts of October 2020 Reddit reReddit: Top posts of 2020 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
2212	https://www.effortlessmath.com/math-topics/reversing-derivatives-made-easy-power-rule-of-integration/?srsltid=AfmBOop9Xj4OLj7azAIhA4iQnGq5HEsdMa8pvxVFT80DAVbqcsDwNVM3	Reversing Derivatives Made Easy: Power Rule of Integration - Effortless Math: We Help Students Learn to LOVE Mathematics Effortless Math X +eBooks +ACCUPLACER Mathematics +ACT Mathematics +AFOQT Mathematics +ALEKS Tests +ASVAB Mathematics +ATI TEAS Math Tests +Common Core Math +CLEP +DAT Math Tests +FSA Tests +FTCE Math +GED Mathematics +Georgia Milestones Assessment +GRE Quantitative Reasoning +HiSET Math Exam +HSPT Math +ISEE Mathematics +PARCC Tests +Praxis Math +PSAT Math Tests +PSSA Tests +SAT Math Tests +SBAC Tests +SIFT Math +SSAT Math Tests +STAAR Tests +TABE Tests +TASC Math +TSI Mathematics +Worksheets +ACT Math Worksheets +Accuplacer Math Worksheets +AFOQT Math Worksheets +ALEKS Math Worksheets +ASVAB Math Worksheets +ATI TEAS 6 Math Worksheets +FTCE General Math Worksheets +GED Math Worksheets +3rd Grade Mathematics Worksheets +4th Grade Mathematics Worksheets +5th Grade Mathematics Worksheets +6th Grade Math Worksheets +7th Grade Mathematics Worksheets +8th Grade Mathematics Worksheets +9th Grade Math Worksheets +HiSET Math Worksheets +HSPT Math Worksheets +ISEE Middle-Level Math Worksheets +PERT Math Worksheets +Praxis Math Worksheets +PSAT Math Worksheets +SAT Math Worksheets +SIFT Math Worksheets +SSAT Middle Level Math Worksheets +7th Grade STAAR Math Worksheets +8th Grade STAAR Math Worksheets +THEA Math Worksheets +TABE Math Worksheets +TASC Math Worksheets +TSI Math Worksheets +Courses +AFOQT Math Course +ALEKS Math Course +ASVAB Math Course +ATI TEAS 6 Math Course +CHSPE Math Course +FTCE General Knowledge Course +GED Math Course +HiSET Math Course +HSPT Math Course +ISEE Upper Level Math Course +SHSAT Math Course +SSAT Upper-Level Math Course +PERT Math Course +Praxis Core Math Course +SIFT Math Course +8th Grade STAAR Math Course +TABE Math Course +TASC Math Course +TSI Math Course +Puzzles +Number Properties Puzzles +Algebra Puzzles +Geometry Puzzles +Intelligent Math Puzzles +Ratio, Proportion & Percentages Puzzles +Other Math Puzzles +Math Tips +Articles +Blog Reversing Derivatives Made Easy: Power Rule of Integration The power rule of integration is a fundamental technique in calculus for finding the integral of a function raised to a power. The power rule for integration is a fundamental and widely used tool in calculus. Its simplicity makes it a first-line method for integrating power functions, playing a crucial role in both theoretical and applied mathematics. Understanding and applying this rule correctly is essential for anyone studying calculus. Here’s a detailed explanation: Definition of the Power Rule for Integration The power rule states that for any real number n 𝑛 different from −1−1, the integral of x n 𝑥 𝑛 with respect to x 𝑥 is: ∫x n d x=x n+1 n+1+C∫𝑥 𝑛 𝑑 𝑥=𝑥 𝑛+1 𝑛+1+𝐶 where C 𝐶 is the constant of integration. Why the Exclusion of n=−1 𝑛=−1 The case where n=−1 𝑛=−1 is excluded because it leads to the function x−1 𝑥−1, which is 1 x 1 𝑥, and its integral is the natural logarithm function, not a power function. The integral of 1 x 1 𝑥 is ln\|x\|+C. Applying the Power Rule General Application: To integrate a function like x 3, you would apply the power rule as follows: ∫x 3 d x=x 3+1 3+1+C=x 4 4+C Negative Powers: It also applies to negative powers (except for −1). For instance: ∫x−2 d x=x−2+1−2+1+C=−1 x+C Importance in Calculus The power rule is a go-to technique for integrating polynomials and any function that can be expressed as a power of x. It simplifies the process of finding antiderivatives, which is crucial in solving problems involving areas under curves and in various physical applications. Limitations The power rule is not applicable to functions that cannot be expressed as x n. In such cases, other integration methods like substitution or integration by parts are required. For n=−1, a different approach (integration of 1 x) must be used. by: Effortless Math Team about 2 years ago (category: Articles) Effortless Math Team 4 weeks ago Effortless Math Team Related to This Article CalculusIntegralsReverse Power Rule More math articles Using Strip Diagrams to Represent Fractions How to Estimate Quotients Using Compatible Numbers for One-digit Divisors Full-Length ALEKS Math Practice Test Full-Length SAT Math Practice Test Algebra Puzzle – Critical Thinking 11 The Ultimate MCA Algebra 1 Course (+FREE Worksheets) Top 10 Free Websites for CLEP College Math Preparation How to Construct the Incircle of a Triangle 4th Grade MAP Math FREE Sample Practice Questions What if I Fail the CBEST Test? What people say about "Reversing Derivatives Made Easy: Power Rule of Integration - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. 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2213	https://sites.cs.ucsb.edu/~omer/DOWNLOADABLE/Fibonacci-run_Graphs_II_21.pdf	Discrete Applied Mathematics 300 (2021) 56–71 Contents lists available at ScienceDirect Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam Fibonacci-run graphs II: Degree sequences Ömer Eğecioğlu a, Vesna Iršič b,c,∗ a Department of Computer Science, University of California Santa Barbara, Santa Barbara, CA 93106, USA b Faculty of Mathematics and Physics, University of Ljubljana, Slovenia c Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia a r t i c l e i n f o Article history: Received 9 October 2020 Received in revised form 12 May 2021 Accepted 13 May 2021 Available online xxxx Keywords: Fibonacci cube Fibonacci-run graph Degree sequence Generating function a b s t r a c t Fibonacci cubes are induced subgraphs of hypercube graphs obtained by restricting the vertex set to those binary strings which do not contain consecutive 1s. This class of graphs has been studied extensively and generalized in many different directions. Induced subgraphs of the hypercube on binary strings with restricted runlengths as vertices define Fibonacci-run graphs. These graphs have the same number of vertices as Fibonacci cubes, but fewer edges and different graph theoretical properties. Basic properties of Fibonacci-run graphs are presented in a companion paper, while in this paper we consider the nature of the degree sequences of Fibonacci-run graphs. The generating function we obtain is a refinement of the generating function of the degree sequences, and has a number of corollaries, obtained as specializations. We also obtain several properties of Fibonacci-run graphs viewed as a partially ordered set, and discuss its embedding properties. © 2021 Elsevier B.V. All rights reserved. 1. Introduction The n-dimensional hypercube Qn is the graph with all binary strings of length n as vertices, where two vertices v1v2 . . . vn and u1u2 . . . un are adjacent if and only if vi ̸= ui for exactly one index i ∈[n]. We have \|V(Qn)\| = 2n, and \|E(Qn)\| = n2n−1. Fibonacci cubes Γn are a family of subgraphs of Qn, and were introduced by Hsu . The vertices of Γn are the Fibonacci strings of length n, Fn = {v1v2 . . . vn ∈{0, 1}n \| vi · vi+1 = 0 , i ∈[n −1]} , and two vertices are adjacent if and only if they differ in exactly one coordinate. Shortly, Γn is the subgraph of Qn, induced by the vertices that do not contain consecutive 1s. This family of graphs turned out to be interesting, and has been widely investigated, see for example [3,8,11,12,14]. Recall that a graph isomorphic to a Fibonacci cube is obtained by adding the string 00 to the end of every vertex. We call such binary strings extended Fibonacci strings. With this interpretation, we can set V(Γn) = {w00 \| w ∈Fn} , and make two vertices adjacent if and only if they differ in exactly one coordinate. Instead of considering extended Fibonacci strings as the vertex set, it is possible to consider run-constrained binary strings, which are used to define Fibonacci-run graphs introduced in . Run-constrained binary strings are strings of 0s ∗Corresponding author at: Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia. E-mail addresses: omer@cs.ucsb.edu (Ö. Eğecioğlu), vesna.irsic@fmf.uni-lj.si (V. Iršič). 0166-218X/© 2021 Elsevier B.V. All rights reserved. Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Fig. 1. Fibonacci-run graphs Rn for n ∈. and 1s, in which every run of 1s appearing in the word is immediately followed by a strictly longer run of 0s. Run-constrained strings, together with the null word λ and the singleton 0, are generated freely (as a monoid) by the letters from the infinite alphabet R = 0, 100, 11000, 1110000, . . . (1) This means that every run-constrained binary string can be written uniquely as a concatenation of zero or more strings from R. Note that run-constrained strings of length n ≥2 must end with 00. For n ≥0, the Fibonacci-run graph Rn, has the vertex set V(Rn) = {w00 \| w00 is a run-constrained binary string of length n + 2} , and edge set E(Rn) = {{u00, v00} \| H(u, v) = 1} , where H(u, v) is the Hamming distance between u, v ∈{0, 1}n, i.e. the number of coordinates in which u and v differ. We take R0 to be the graph with a single vertex corresponding to the label 00, which after the removal of the trailing pair of zeros, corresponds to the null word. Clearly, Rn is a subgraph of Qn+2. However it is more natural to see it as a subgraph of Qn after suppressing the trailing 00 in the vertex labels of Rn. In this way, we can view the vertices of Rn without the trailing pair of zeros as V(Rn) = {w \| w00 is a run-constrained binary string of length n + 2} . Note that here we use the same notation Rn, even though the obtained graph is just isomorphic to Rn. This is the same kind of convention as viewing Γn as a subgraph of Qn+2 if one thinks of the vertices as extended Fibonacci strings, or as a subgraph of Qn as usual by suppressing the trailing 00 of the vertex labels. The graphs R1 – R4 with this truncated labeling of run-constrained strings are shown in Fig. 1. Basic properties of Fibonacci-run graphs such as the number of vertices, the number of edges, diameter, the decom-position into lower dimensional Fibonacci-run graphs, Hamiltonicity and the nature of the asymptotic average degree are studied in . The rest of the paper is organized as follows. After the general preliminaries in Section 2, we consider a decomposition of run-constrained binary strings and prove a result on special collections of words in Section 3. In Section 4, we consider the problem of keeping track of both the up-degree and the down-degree of a run-constrained string. Calculation of the generating function of the up–down degree enumerator polynomials is presented in Section 5. The proof is divided into a number of subsections. In Section 6, we derive a number of consequences of the generating function obtained in Section 5. Among these is the generating function for the degree enumerator polynomials of Fibonacci-run graphs. Following this, in Section 7, we consider a number of parameters for Fibonacci-run graphs as partially ordered sets. These include the rank generating polynomial, enumeration of the maximal elements, and the calculation of the Möbius function. A combinatorial aspect of run-constrained strings, namely the generating function by inversions is presented in Section 8. Embedding and related results are in Section 9, followed by conjectures, questions and further directions in Section 10. 2. Preliminaries In this section, we present definitions and some known results which are needed in the paper. To avoid possible confusion that may arise due to the initial values, we reiterate that Fibonacci numbers are defined as f0 = 0, f1 = 1, and fn = fn−1 + fn−2 for n ≥2. The Hamming weight of a binary string u is the number of 1s in u, denoted by \|u\|1. If a, b are strings, then ab denotes the concatenation of these two strings in that order. Similarly, for a set of strings B, we set aB = {ab \| b ∈B}. The distance between vertices u and v in a graph G is denoted by dG(u, v). 57 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Fig. 2. The Hasse diagram of the Fibonacci-run graph R4 when viewed as a partially ordered set. The number of vertices and the number of edges of Rn for n ≥5 are given by \|V(Rn)\| = \|V(Γn)\| = fn+2 , \|E(Rn)\| = \|E(Γn)\| −\|E(Γn−4)\| = (3n + 4)fn−6 + (5n + 6)fn−5 , as proved in [5, Lemma 3.1] and [5, Corollary 4.3]. Fibonacci-run graphs can also be viewed as partially ordered sets whose structure is inherited from the Boolean algebra of subsets of [n]. The elements here correspond to all binary strings of length n and the covering relation is flipping a 0 to a 1. Therefore in Rn we have a natural distinction between up- and down-degree of a vertex, denoted by degup(v) and degdown(v). Here degup(v) is the number of vertices u in Rn obtained by changing a 0 to a 1, and degdown(v) is the number of vertices u in Rn obtained from v by changing a 1 to a 0. Clearly deg(v) = degup(v) + degdown(v) . Note that in Rn, degdown(v) is not necessarily equal to the Hamming weight of v because the vertices of the graph are restricted to be run-constrained binary strings. The degree sequences, i.e. the nature of the vertices of a given degree in a graph, has been well studied for Fibonacci cubes . Here we keep track of the degree sequences of our graphs Rn as the coefficients of a polynomial. This polynomial is called the degree enumerator polynomial of the graph denoted by gn(x). The coefficient of xi in the degree enumerator polynomial is the number of vertices of degree i in Rn. More precisely, Definition 2.1. The degree enumerator polynomials gn(x) of Rn is defined for n ≥1 by gn(x) = ∑ v∈Rn xdeg(v) . (2) Similar polynomials are defined to keep track of the up- and down-degree sequences as well. In particular the down-degree enumerator polynomial and the up-degree enumerator polynomial of Rn are defined as ∑ v∈Rn ddegdown(v) , and ∑ v∈Rn udegup(v) , respectively. The generating function of the sequence of down-degree enumerator polynomials of Rn is ∑ n≥1 tn ∑ v∈Rn ddegdown(v) . The generating functions of the sequence of up-degree enumerator polynomials and the degree enumerator polynomials are defined similarly. The nature of the distribution of the up-degrees and the down-degrees are most easily seen from the Hasse diagram of Rn for which degup(v) and degdown(v) are simply the number of edges emanating up and down from v ∈Rn, respectively. Inspecting Fig. 2, we see that the down-degree, up-degree and the degree enumerator polynomials of R4 are given respectively by 1 + 4d + 3d2, 3 + 2u + 2u2 + u4, 5x2 + 2x3 + x4 . (3) 58 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Fig. 3. The calculation of the bivariate up–down degree enumerator polynomial 3d2 + 2du + 2du2 + u4 of R4. It was determined in [5, Proposition 7.1] and [5, Proposition 7.2] that generating function for down-degree enumerator polynomials of Rn is t(1 + d + dt + (d2 −1)t2 + d(d −1)t3 + d(d −1)t4) 1 −t −t2 −(d −1)t3 −d(d −1)t5 , (4) and the generating function for up-degree enumerator polynomials of Rn is t(1 + u −(u −2)t −2ut2 + t3 −(u −1)t5 −(u −1)t6) 1 −ut −2t2 + (2u −1)t3 + t4 −(u −1)t5 + (u −1)t7 . (5) Our general aim in this paper is to study the generating function of the bivariate polynomials ∑ v∈Rn udegup(v)ddegdown(v) (6) that we refer to as the up–down degree enumerator of Rn. For example, for n = 4, this polynomial is given by 3d2 + 2du + 2du2 + u4 , (7) as can be verified by inspecting R4 in Fig. 3, where the term contributed by each vertex is indicated in a box. The up–down degree enumerator polynomial is a refinement of the both up- and down-degree enumerator polynomi-als and of the degree enumerator polynomial gn(x). For instance the polynomial in (7) specializes to the first polynomial in (3) for u = 1, to the second one for d = 1, and to the last for u = d = x. More generally, the generating function of the up–down degree enumerator polynomials for Rn specializes to the generating functions (4) and (5) as corollaries, and provide the generating function for the degree enumerator polynomials {gn(x)}n≥1 itself for u = d = x. Example 2.2. For the graphs R1 through R8, the up–down degree enumerator polynomials are as follows: d + u 2d + u2 d + d2 + 2du + u3 3d2 + 2du + 2du2 + u4 5d2 + 2d2u + 3du2 + 2du3 + u5 4d2 + 2d3 + 6d2u + 2d2u2 + 4du3 + 2du4 + u6 3d2 + 7d3 + 5d2u + 9d2u2 + 2d2u3 + 5du4 + 2du5 + u7 2d2 + 10d3 + d4 + 4d2u + 8d3u + 7d2u2 + 12d2u3 + 2d2u4 + 6du5 + 2du6 + u8 We define the generating function GF of the up–down degree enumerator polynomials of Rn formally as follows: Definition 2.3. The generating function of the sequence of the up–down degree enumerator polynomials is defined as GF = GF(u, d; t) = ∑ n≥1 tn ∑ v∈Rn udegup(v)ddegdown(v) . (8) 59 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 3. Decomposition of run-constrained strings and a preparatory result In this section, we present some preliminary results about formal power series, which are needed for the computation of our up–down degree enumerators and their generating function GF. We also need another description of run-constrained binary strings. Define the set S = {100, 11000, 1110000, . . .} , which is used in the rest of the paper. Every run-constrained binary string consists of words from S interspersed with runs of 0s, including a prefix and a suffix which may also be runs of 0s. Let s∗denote an arbitrary string of zero or more words from S, and s+ denote an arbitrary string of one or more words from S. So we have s+ = ss∗, and as another example, s2s∗denotes all strings obtained by the concatenation of two or more words from S. Note that this notation is consistent with its usage in formal languages. Every non-trivial (i.e. not consisting only of 0s) run-constrained binary string can be written in the form 0i0s+0i1s+0i2 · · · 0iks+0ik+1, where k ≥0, i0, ik+1 ≥0, i1, i2, . . . , ik ≥1. The runs 0i1, 0i2, . . . , 0ik are called internal runs, the initial string 0i0 is the pre-run, and the final string 0ik+1 is the post-run of the word. Note that for the latter two, we do not rule out the possibility that they have length zero (i.e. i0 = 0 or ik+1 = 0.) So in that sense they are not ‘‘real’’ runs like the interior runs of the string, which are the portions in between the letters of S that appear in the word, and must have positive length. Note that the up–down generating function G of the words in S itself is G = dt3 + d2t5 + d2t7 + · · · = dt3 + d2t5 1 −t2 . (9) Here we are keeping the trailing pair of zeros in a run-constrained string into account as the exponent of t so that the exponent of t starts at 3. This in contrast with definition (8) in which the trailing zeros are not considered and the exponent of t starts at 1. The reason for this is that it is somewhat easier to explain the steps of the proof of our Theorem 5.1 if we keep the trailing pair of zeros in the representation of run-constrained strings. The generating function GFX we obtain this way differs from our target generating function GF in (8) in that it accounts for two extra words 0 and 00, and the exponent of t for all other words in GFX is 2 more than those in GF. Algebraically this is expressed in (20), and once we have GFX, we immediately obtain GF from it. We make us of the following two preparatory results. Consider the alphabet Σ = {a, b} and let Σ∗denote all words over Σ. The length of u ∈Σ∗is denoted by \|u\|. Let \|u\|a and \|u\|b denote the number of occurrences of a and b in u, respectively. Let also \|u\|aa, \|u\|ab, \|u\|ba, \|u\|bb denote the number of appearances of the words aa, ab, ba, bb in u, respectively. For n ≥0, let aΣna = {awa \| w ∈Σ∗, \|w\| = n}. Similarly, we define the sets of strings aΣnb, bΣna, and bΣnb. For a word u with \|u\| ≥2, define m(u) = x\|u\|ay\|u\|bα\|u\|aa+\|u\|baβ\|u\|ab+\|u\|bb . (10) Example 3.1. For the word u = aababbaaa ∈aΣ7a, we have \|u\| = 9, \|u\|a = 6, \|u\|b = 3, \|u\|aa +\|u\|ba = 5, \|u\|ab +\|u\|bb = 3, and consequently m(u) = x6y3α5β3 . Next, we prove the following proposition, to be used for the calculation of the generating function GF of the up–down degree enumerator polynomials. Proposition 3.2. Let n ≥0 be an integer and let w ∈Σ∗, where Σ = {a, b}. Then ∑ \|w\|=n m(awa) = αx2(αx + βy)n, (11) ∑ \|w\|=n m(awb) = βxy(αx + βy)n, (12) ∑ \|w\|=n m(bwa) = αxy(αx + βy)n, (13) ∑ \|w\|=n m(bwb) = βy2(αx + βy)n. (14) Proof. Consider the first identity. For n = 0, there is only one word aa, and both sides are αx2 in this case. If n > 0 and u = awa, then note that the number \|u\|aa + \|u\|ba is the number of a’s in wa and \|u\|ab + \|u\|bb is the number of b’s in w. Given a word w with \|w\|a = k and \|w\|b = n −k, we calculate m(u) as m(u) = xk+2yn−kαk+1βn−k = αx2xkyn−kαkβn−k . 60 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Since there are (n k ) such strings w, we obtain ∑ \|w\|=n m(aua) = αx2 n ∑ k=0 ( n k ) xkyn−kαkβn−k = αx2(αx + βy)n . The proofs of the other identities are similar. □ By summing each identity in Proposition 3.2 over all nonnegative integers n, we obtain the following formulas. Corollary 3.3. Let Σ = {a, b} and m be as defined in (10). Then ∑ w∈Σ∗ m(awa) = αx2 1 −(αx + βy), (15) ∑ w∈Σ∗ m(awb) = βxy 1 −(αx + βy), (16) ∑ w∈Σ∗ m(bwa) = αxy 1 −(αx + βy), (17) ∑ w∈Σ∗ m(bwb) = βy2 1 −(αx + βy). (18) 4. Up–down degree polynomials In the general case we aim to keep track of the terms udegup(v)ddegdown(v)t\|v\| (19) for every run-constrained binary string v. If we are to only keep track of the down-degree of a run-constrained binary string, then the problem is considerably simpler. In this case we are flipping 1s in the string to 0s, and the contribution of every word s ∈S in the observed run-constrained binary string v, independently of where it is located in v, is either 1 (s = 100) or 2 (s ∈S \ {100}). The difficulty with keeping track of the up-degree arises in the following situations. As an example consider the subword s10s1 that appears somewhere in the string, where s1 = 100, and use parentheses to highlight the words from S: · · · (100) 0 (100) · · · The 0 in the middle can be flipped to 1 in the case that the 100 on the right is followed by a 0: · · · (100) 0 (100) 0 · · · but not if the 100 on the right is followed by another word from S: · · · (100) 0 (100) (11000) · · · As another example, consider the last subword s ∈S in the word. It may be followed by zero or more 0s. For example, if s = 100, · · · (100), · · · (100) 0, · · · (100) 00, · · · (100) 000, · · · (100) 0000, . . . In the first two cases, the 0 immediately to the right of 1 cannot be flipped to a 1 but in all other cases it can be. Additionally, whenever we have r ≥3 trailing 0s in a case like this, r −2 of those can be flipped to a 1. 5. Calculation of GF Theorem 5.1. The generating function for the up–down degree enumerator polynomials of the graphs Rn is given by GF = Nu,d/Du,d where Nu,d = (d + u)t −d(u −2)t2 + (d2 −d −2u)t3 −(d −2)d(u −2)t4 −(d −1)(u −d + du)t5 −d(d + u −2)t6 + d(1 −2d + 2d2 + du −2d2u)t7 −2(d −1)d2(u −1)t8 −(d −1)d2(d + 1)(u −1)t9 −(d −1)2d2(u −1)t10 −(d −1)2d2(u −1)t11 , and Du,d = 1 −ut −2t2 + (2u −d)t3 + t4 + (2d −d2 −u)t5 + d(du −1)t7 + 2(d −1)d2(u −1)t9 + (d −1)2d2(u −1)t11 . 61 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Proof. We consider the cases according to the type of word from S that precedes the leftmost internal run of 0s, and the word from S that follows the rightmost internal run of 0s. There is also the case, where the run-constrained binary string does not have internal runs of 0s. Thus there are altogether five cases to consider. In the cases (A) through (D), k ≥1, i0, ik+1 ≥0, and i1, i2, . . . , ik > 0. Note that s simply denotes a word from S, and this word may be different at different places in the below schematic description of the cases. Similarly, s2s∗denotes a string of at least two words from S. (A) The string is of the form 0i0 s 0i1 · · · 0ik s 0ik+1 (B) The string is of the form 0i0 s2s∗0i1 · · · 0ik s 0ik+1 (C) The string is of the form 0i0 s 0i1 · · · 0ik s2s∗0ik+1 (D) The string is of the form 0i0 s2s∗0i1 · · · 0ik s2s∗0ik+1 (E) The string has no internal runs of 0s. Let GFA, GFB, GFC, GFD, GFE denote the generating functions of the classes of strings in the cases (A) through (E), respectively. Then the generating function for the run-constrained binary strings with the statistic (19) is GFX = GFA + GFB + GFC + GFD + GFE, and the generating function GF for the same statistic for V(Rn) is (after getting rid of the strings 0 and 00, and removing the trailing 00 in all strings) GF = (GFX −t −t2)/t2 . (20) To simplify the notation in the remaining part of the proof, we define the following quantities: G = dt3 + d2t5 1 −t2 , α = ut 1 −ut , β = t 1 −ut , x = G, y = G2 1 −G . Calculation of GFE We first consider the calculation of GFE, which is the most straightforward. Since there are no internal runs of 0s in case (E), we can partition the strings in (E) into three classes and calculate the generating function for each. (1) The extended Fibonacci string is all 0s: The generating function for these is t + t2 1 −ut . (21) (2) The extended Fibonacci string has a single word from S: The generating function is ( 1 + 2t 1 −ut + t2 (1 −ut)2 ) G . (22) To see this, note that words in this set are of the form 0is0j with i, j ≥0. For j = 0 the generating function is the product of G and 1 + t + t2 + ut3 + u2t4 + · · · = 1 + t + t2 1 −ut . For j = 1 it is the product of G and t + ut2 + ut3 + u2t4 + · · · = t + ut2 + ut3 1 −ut . For j = 2 it is the product of G and ut2 + u2t3 + u2t4 + u3t5 + · · · = ut2 + u2t3 + u3t5 1 −ut , and so on. Adding the fractional expressions over j yields ∑ j≥0 ujtj+2 1 −ut = t2 1 −ut ∑ j≥0 ujtj = t2 (1 −ut)2 . (23) 62 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 The sum of the pairs of terms 1 + t, t + ut2, ut2 + u2t3, etc. over j ≥0 is calculated to be 1 + 2t 1 −ut . (24) Adding the contributions of (23) and (24) proves (22). (3) The extended Fibonacci string has two or more words from S: We prove that in this case the generating function is ( 1 + t + t + t2 1 −ut + t2 1 −ut + t3 (1 −ut)2 ) G2 1 −G . (25) The strings here are of the form 0is2s∗0j with i, j ≥0. The generating function of the strings of at least two words from S is G2/(1 −G). We again calculate the contribution to the generating function for j = 0, 1, 2, . . .. For j = 0, we have 1 + t + t2 + ut3 + u2t4 + · · · = 1 + t + t2 1 −ut , for j = 1 t + t2 + t3 + ut4 + u2t5 + · · · = t + t2 + t3 1 −ut , for j = 2 ut2 + ut3 + ut4 + u2t5 + · · · = ut2 + ut3 + ut4 1 −ut , etc. Adding the fractional terms gives t2 1 −ut + ∑ j≥1 uj−1tj+2 1 −ut = t2 1 −ut + t3 (1 −ut)2 . The sum of the pairs of remaining terms that appear for each j is 1 + t 1 −ut + t + t2 1 −ut and adding these up gives (25). Finally, adding up and simplifying the contributions of (21), (22) and (25), we obtain GFE = t(1 + (1 −u)t) 1 −ut + (1 + (1 −u)t)2 (1 −ut)2 G (26) + (1 + (1 −u)t)(1 + (1 −u)t + (1 −u)t2) (1 −ut)2 G2 1 −G . Calculation of GFA We first consider the generating function GFA on an example. We will see that the observations made on it can be used on a general string. Take a word of the type 0i0s0i1s0i2s2s∗0i3s2s∗0i4s0i5s0i6 . (27) In this example, k = 5, i0, i6 ≥0, i1, i2, i3, i4, i5 > 0. The contribution of the pre-run in such a word is the factor 1 + ut + ut2 1 −ut . Note that we are using the fact that the pre-run is followed by s0. The contribution of the first interior run 0i1 is α = ut 1 −ut as it is located in the context s0i1s0. The contribution of the second interior run 0i2 is β = t 1 −ut as it appears in the context s0i2s2s∗. Continuing, the contribution of the third interior run 0i3 is β = t 1 −ut 63 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 because it appears in the context s2s∗0i3s2s∗. The contribution of the fourth interior run 0i4 is α = ut 1 −ut as it appears in the context s2s∗0i4s. We can summarize the situation with the contribution of the internal runs as follows: the internal runs located in the context of s0is0 and s2s∗0is0 contribute α, the internal runs located in the context of s0is2s∗and s2s∗0is2s∗contribute β. Of course each occurrence of s contributes x = G and each occurrence of s2s∗contributes y = G2/(1−G). For our example, this leaves the contribution of the last interior run and the post-run. Note that the last interior run contributes β if the post-run has length zero, and α if the post-run has positive length. In the first case the contribution of the post-run is 1, and in the second it is t 1 −ut . Adding the two contributions, the last interior run and the post-run together contribute β + α t 1 −ut = α u−1 1 −ut . Therefore we can ‘‘charge" α as the contribution of the last interior run if we make the contribution of the post-run equal to u−1 1 −ut . We can simplify the situation further. Encode the strings of type (27) by the word aabbaa over the two letter alphabet {a, b}, ignoring the runs of 0s altogether, and encoding s by a and s2s∗by b. Since we are in case (A), these words start and end with the letter a. Then we have the following: (1) The contribution of the pre-run is 1 + ut + ut2 1 −ut . (2) The contribution of the post-run is u−1 1 −ut . (28) (3) The contribution of each letter a is x = G. (4) The contribution of each letter b is y = G2/(1 −G). (5) The contribution of each letter pair aa or ba is α. (6) The contribution of each letter pair bb or ab is β. In our example aabbaa, there are four a’s, two b’s, two aa’s, one ba, one bb and one ab. So the generating function of the words encoded by aabbaa is ( 1 + ut + ut2 1 −ut ) ( u−1 1 −ut ) x4y2α3β2 . Note that x4y2α3β2 = m(aabbaa) as defined in (10). By (11), ∑ \|w\|=4 m(awa) = αx2(αx + βy)4 . Generalizing these results and using (15), the generating function GFA is given by GFA = ( 1 + ut + ut2 1 −ut ) ( u−1 1 −ut ) αx2 1 −(αx + βy). (29) Calculation of GFB Let us consider how this case differs from the computation of GFA. This time the contribution of the pre-run is 1 + t + t2 + ut3 + u2t4 + · · · = 1 + t + t2 1 −ut . 64 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Again the contribution of the post-run is taken to be (28). Each adjacent pair aa or ba contributes α, and each pair bb or ab contributes β. In this case the encoding words start with b and end with a. Using (12) and (16), we find GFB = ( 1 + t + t2 1 −ut ) ( u−1 1 −ut ) ∑ n≥0 αxy(αx + βy)n (30) = ( 1 + t + t2 1 −ut ) ( u−1 1 −ut ) αxy 1 −(αx + βy) . Calculation of GFC Here the contribution of the pre-run is as in case (A), but the contribution of the post-run is 1 + t 1 −ut . The encoding words over {a, b} start with a and end with b. Therefore by (15) GFC = ( 1 + ut + ut2 1 −ut ) ( 1 + t 1 −ut ) βxy 1 −(αx + βy) . (31) Calculation of GFD In this case the contribution of the pre-run is such as in case (B), and the contribution of the post-run is as in case (C). The encoding words over {a, b} start and end with b. Therefore by (17), we have ( 1 + t + t2 1 −ut ) ( 1 + t 1 −ut ) βy2 1 −(αx + βy) . (32) Finally, adding up the contributions from (29), (30), (31), (32), (26) (by Mathematica) gives the generating function GFX and via (20), the generating function GF given in the theorem. □ 6. Consequences of the up–down degree enumerator The degree enumerator polynomials gn(x) for R1 through R10 (computed by Mathematica) are as follows: g1(x) = 2x g2(x) = x2 + 2x g3(x) = x3 + 3x2 + x g4(x) = x4 + 2x3 + 5x2 g5(x) = x5 + 2x4 + 5x3 + 5x2 g6(x) = x6 + 2x5 + 6x4 + 8x3 + 4x2 g7(x) = x7 + 2x6 + 7x5 + 9x4 + 12x3 + 3x2 (33) g8(x) = x8 + 2x7 + 8x6 + 12x5 + 16x4 + 14x3 + 2x2 g9(x) = x9 + 2x8 + 9x7 + 15x6 + 22x5 + 24x4 + 14x3 + 2x2 g10(x) = x10 + 2x9 + 10x8 + 18x7 + 30x6 + 32x5 + 39x4 + 10x3 + 2x2 g11(x) = x11 + 2x10 + 11x9 + 21x8 + 39x7 + 48x6 + 57x5 + 42x4 + 10x3 + 2x2 g12(x) = x12 + 2x11 + 12x10 + 24x9 + 49x8 + 68x7 + 81x6 + 84x5 + 46x4 + 8x3 + 2x2 Making the substitutions u →x and d →x in the generating function of the up–down degree enumerator polynomials, we find the generating function of the degree enumerator polynomials of the Rn to be as follows. Theorem 6.1. The generating function for the degree enumerator polynomials of Fibonacci-run graphs f (t, x) = ∑ n≥1 gn(x)tn is given in closed form by Nx/Dx, where Nx = xt ( 2 −(x −2)t + (x −3)t2 −(x −2)2t3 −x(x −1)t4 −2(x −1)t5 −(x −1)(2x2 −x + 1)t6 −2x(x −1)2t7 −x(x −1)2(x + 1)t8 −x(x −1)3t9 −x(x −1)3t10) 65 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 and Dx = 1 −xt −2t2 + xt3 + t4 −x(x −1)t5 + x(x −1)(x + 1)t7 + 2x2(x −1)2t9 + x2(x −1)3t11. Remark 6.2. To find the generating function of the number of vertices with degree k, we can take dk dxk f (t, x), and then set x = 0. The resulting series divided by k! is then the generating function of the number of vertices of degree k in Rn. We can denote this series by 1 k! Dkf (t, x) \|x=0 . (34) In other words, the coefficient of tn in (34) is the number of vertices of degree k in Rn. In the following examples we calculate the number of vertices of small degree in Fibonacci-run graphs. Example 6.3. We differentiate the generating function f (t, x) of Theorem 6.1 with respect to x twice using Mathematica, and then set x = 0. Dividing the resulting expression by 2 gives 1 2 D2f (t, x) \|x=0= t2 + 3t3 + 5t4 + 5t5 + 4t6 + 3t7 + 2t8 1 −t , confirming that for n ≥8, Rn has exactly two vertices of degree 2. If n is even, then these two vertices are 01n/20n/2+1 and 1n/20n/2+2. If n is odd, then they are 001⌊n/2⌋0⌈n/2⌉and 1⌈n/2⌉0⌊n/2⌋+2. Example 6.4. Continuing computing with Mathematica, we find 1 6 D3f (t, x) \|x=0= t3 + 2t4 + 5t5 + 8t6 + 12t7 + 14t8 + 14t9 + 10t10 + 10t11 1 −t2 + 8t12 1 −t2 which means that for n ≥11, the number of vertices of degree 3 in Rn is 10 if n is odd, and 8 if n is even. Example 6.5. For k = 4, we get 1 24 D4f (t, x) \|x=0 = t4 + 2t5 + 6t6 + 9t7 + 16t8 + 24t9 + 39t10 + 42t11 + 46t12 + 39t13 + 43t14 + t15(39 + 45t −35t2 −42t3) (1 −t2)2 from which we compute that for n ≥15, the number of vertices of degree 4 in Rn is 2n + 9 if n is odd and 3n/2 + 21 if n is even. Example 6.6. Similar calculations give that the number of vertices of degree 5 in Rn for n ≥18 is 9n −1 if n is odd, and 12n −48 if n is even. The presented examples lead to the following conjecture. Conjecture 6.7. For a given k, the generating function of the number of vertices of degree k in Rn is of the form pk(t) (1 −t2)k+1 where pk(t) is a polynomial of degree 1 2(15k + 8) if k is even, and of degree 1 2(15k + 7) if k is odd. We already have the first few degree enumerator polynomials as given in (33). From the denominator of their generating function in Theorem 6.1, we get the following result. Theorem 6.8. If gn = gn(x) is the degree enumerator polynomial for Rn as defined in Definition 2.1 with the initial values given by (33), then for n ≥12 gn = xgn−1 + 2gn−2 −xgn−3 −gn−4 + x(x −1)gn−5 −x(x2 −1)gn−7 −2x2(x −1)2gn−9 −x2(x −1)3gn−11 . Two other specializations of the generating function of the up–down degree enumerator polynomials are obtained in the following way. (1) Setting u = 1, we obtain the generating function of the down-degree enumerator polynomials in (4), which we had already computed [5, Proposition 7.1]. (2) Setting d = 1, we obtain the generating function of the up-degree enumerator polynomials shown in (5), also already computed in [5, Proposition 7.2]. 66 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Fig. 4. The Hasse diagram of the graph R6 as a poset. The trailing 00 of the vertex labels have been truncated. The dark lines show the interval [000000, 110001] which is isomorphic to the 3-dimensional hypercube Q3. The six maximal elements of the poset are underlined. Note that Theorem 5.1 can be used to recalculate another result. The down-degree enumerator generating function differentiated with respect to d is t(1 −t2)(1 + (2d −1)t2) (1 −t −t2 −(d −1)t3 −d(d −1)t5)2 which for d = 1 gives the generating function of the number of edges of Rn, already determined in as t(1 −t4) (1 −t −t2)2 . (35) Clearly, this result could also be obtained by differentiating the up-degree enumerator generating function with respect to u, and then evaluating it at u = 1. 7. Fibonacci-run graphs as partially ordered sets Naturally, one might view the vertices of a Fibonacci-run graph as a partially ordered set (poset for short). This poset is defined as (Rn, ≤), where the covering relation is given as follows: for u, v ∈Rn, v covers u if and only if v is obtained from u by flipping a 0 in u to a 1. The binary relation ‘‘≤" is the transitive closure of this covering relation. We have shown the Hasse diagram of R4 in Fig. 2. Fig. 4 depicts the Hasse diagram of R6. Note that the number of vertices v that cover a vertex u ∈Rn is precisely the up-degree degup(u). The number of vertices u ∈Rn that are covered by v is the down-degree degdown(v) of the vertex v, as indicated in Fig. 3. Notice that (Rn, ≤) is a ranked poset, where the rank of u ∈Rn is its Hamming weight \|u\|1. Since the number of vertices in Rn of weight w is given by ( n −w + 1 w ) (36) for 0 ≤w ≤⌈n/2⌉[5, Corollary 3.2], the rank generating polynomial of the poset Rn is F(Rn, x) = ⌈n/2⌉ ∑ k=0 ( n −k + 1 k ) xk . (37) Next we consider the maximal elements of Rn and determine its Möbius function. 67 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 7.1. Maximal elements The maximal elements of a poset are those elements which are not smaller than any other element of the set. The following table lists the run-constrained binary strings of lengths 3, 4, . . . , 8 which correspond to maximal elements of R1 through R6, viewed as a poset. R1 R2 R3 R4 R5 R6 100010000100011000001100000100100 100011000100100010010001001000 110000100010001110000 100100010011000 111000011000100 11100000 The sequence {Mn}n≥1 of the number of maximal elements of Rn starts as 1, 2, 2, 3, 5, 6, 10, 13, 20, 27, 40, 56, 80, . . . We can obtain the generating function of this sequence by setting d = 1 and u = 0 in the up–down degree enumerator generating function in Theorem 5.1. This specialization gives the following result. Corollary 7.1. If Mn denotes the number of maximal elements of the poset of Rn, then the generating function of the sequence {Mn}n≥1 is given by ∑ n≥1 Mntn = t(1 + 2t −2t3 + t5 + t6) 1 −2t2 −t3 + t4 + t5 −t7 . 7.2. The Möbius function We can easily compute the Möbius function µ of the poset (Rn, ≤), since every interval [u, v] is isomorphic to a cube (Boolean algebra) and therefore has the same Möbius function. Denoting the weights of u and v by \|u\|1 and \|v\|1, we have µ(u, v) = { (−1)\|v\|1−\|u\|1 if u ≤v, 0 if u ≰v. 8. Inversion generating function In this section, we present an observation about a combinatorial property of run-constrained binary strings that we use to define Fibonacci-run graphs. For n ≥1, consider the inversion enumerator polynomial Qn(x, q) defined by Qn(x, q) = ∑ w∈V(Rn) x\|w\|1qinv(w), (38) where inv(w) denotes the number of inversions of w and \|w\|1 is the Hamming weight, or the rank of w. Recall that for w = w1w2 · · · wn, inv(w) is the number of pairs 1 ≤i < j ≤n with wi > wj. Note that in (38), the trailing pair of zeros are not taken in the representation of the vertices V(Rn), as indicated in Fig. 1. A few of these polynomials are as shown below: Q1(x, q) = 1 + x, Q2(x, q) = 1 + (q + 1)x, Q3(x, q) = 1 + ( q2 + q + 1) x + x2, Q4(x, q) = 1 + ( q3 + q2 + q + 1) x + ( 2q2 + 1) x2, Q5(x, q) = 1 + ( q4 + q3 + q2 + q + 1) x + ( 2q4 + q3 + 2q2 + 1) x2 + x3 . Proposition 8.1. Set Q−2(x, q) = Q−1(x, q) = Q0(x, q) = 1 with Q−n(x, q) = 0 for n ≥3. Then for n ≥1 Qn(x, q) = ∑ k≥0 xkQn−1−2k(xqk+1, q) . (39) Proof. The proof is a consequence of the fundamental decomposition of Fibonacci-run graphs given in Theorem [5, Lemma 4.1]. Note that the terms in the sum vanish for k ≥⌈n+1 2 ⌉. □ 68 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Define H(x, q; t) = ∑ n≥1 Qn(x, q)tn . Then a consequence of (39) is the functional identity H(x, q; t) = t(1 + x + xt) 1 −xt2 + t ∑ k≥0 xkt2kH(xqk+1, q; t) . (40) This identity can be proved by multiplying both sides of (39) by tn for n ≥1, summing over n, and changing the order of summation on the right hand side. We omit the details. We also note that substituting q = 1 in Qn(x, q) gives the rank generating polynomial F(Rn, x) of Rn defined in (37). In other words, the coefficient of xk becomes the number of words in V(Rn) with weight k as given by (36). For instance, for n = 7 and q = 1, we have 1 + 7x + 15x2 + 10x3 + x4, for which the coefficients are ( 8 0 ) = 1, ( 7 1 ) = 7, ( 6 2 ) = 15, ( 5 3 ) = 10, ( 4 4 ) = 1. Indeed, taking q = 1 in (40), we see that H satisfies H(x, 1; t) = t(1 + x + xt) 1 −xt2 + tH(x, 1; t) 1 −xt2 , so that H(x, 1; t) = t(1 + x + xt) 1 −t −xt2 . We thus get the generating function of the rank generating polynomials F(Rn, x) (see (37)) of Fibonacci-run graphs as a poset as t(1 + x + xt) 1 −t −xt2 = ∑ n≥1 F(Rn, x)tn = (1 + x)t + (1 + 2x)t2 + (1 + 3x + x2)t3 + (1 + 4x + 3x2)t4 + · · · Inversions and the major index statistics of a similar flavor for a class of related Fibonacci strings can be found in . 9. Embedding related results We can encode a binary string of length n as a run-constrained binary string of length 3n + 1. Let si = 1i0i+1 for i ≥1. Given a binary string with k runs of 1s, w = 0j01i10j11i2 · · · 1ik0jk, we first encode the runs as 0sj0si1sj1 · · · siksjk if j0 > 0 (i.e. the binary string starts with 0), and by si1sj1 · · · siksjk if j0 = 0 (i.e. the binary string starts with 1). Then, if necessary, we append 0s at the end to make the length of the encoding 3n + 1. For n = 3, this works as follows 000 → 0s3 → 01110000 00 001 → 0s2s1 → 011000100 0 010 → 0s1s1s1 → 0100100100 100 → s1s2 → 10011000 00 011 → 0s1s2 → 010011000 0 101 → s1s1s1 → 100100100 0 110 → s2s1 → 11000100 00 111 → s3 → 1110000 000 This encoding can be carried out for arbitrary n, resulting in an embedding of the hypercube Qn into R3n−1 (the trailing 00 is not included in the vertex labels here). 69 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 Question 9.1. What is the dilation of this embedding, i.e. the value of max uv∈E(Qn) dR3n−1(u′, v′) , where the prime denotes the image under the encoding described above? We note that although it is intuitive, the embedding described above does not seem to give the smallest dimensional Fibonacci-run graph in which it is possible to embed Qn. The study of hypercubes of various dimensions which are subgraphs of a Fibonacci-run graphs are interesting in its own right. An analogous question has been studied for Fibonacci cubes, cf. results about cube polynomial listed in , and generalizations in . Let hn,k denote the number of k-dimensional hypercubes Qk in Rn. A corollary of [5, Proposition 8.2], obtained by taking q = 1 in that proposition is the following. The generating function of the cube polynomials of Rn is given by ∑ n≥1 tn ∑ k≥0 hn,kxk = t(2 + x + (x + 1)t + x(x + 2)t2 + x(x + 1)t3 + x(x + 1)t4) 1 −t −t2 −xt3 −x(x + 1)t5 . (41) In the series expansion of this generating function in powers of t, the largest m for which the term xm appears as a coefficient of tn, gives the dimension of the largest hypercube Qm that can be embedded in Rn. Calculations on (41), using high order derivatives with respect to x (with Mathematica) suggest that for m ≥0, the hypercube Q2m+1 embeds in R5m+1 and the hypercube Q2m+2 embeds in R5m+3, and these are the smallest possible Fibonacci-run graphs with this property. So it appears that the hypercube graph Qn can be embedded into R⌈(5n−4)/2⌉, and this is the smallest possible run graph with this property. Conjecture 9.2. The smallest m for which the hypercube graph Qn can be embedded into Rm is m = ⌈5n−4 2 ⌉ . 10. Further directions In the final section, we list various questions and conjectures, which are of interest in the further study of Fibonacci-run graphs. These are in addition to Conjecture 9.2 on the determination of the smallest dimensional Fibonacci-run graph that contains Qn, Question 9.1 on the dilation of the mapping described at the start of Section 9, and Conjecture 6.7 on the form of the generating function of the number of vertices of a given degree in Rn. For Conjecture 6.7, the extraction of the coefficients in the generating function of the degree enumerator polynomials f (t, x) of Theorem 6.1, one may use the mechanism described in Remark 6.2, and the examples presented afterwards, along with the Leibniz formula for higher derivatives and the reciprocal differentiation result in . Considering the results obtained in Examples 6.3–6.6 in Section 6, the following general question arises. Question 10.1. What is the number of vertices of degree k in Rn? The analysis and the conjecture on the diameter of Rn can be found in . In relation to this, a natural question that arises is the following: Question 10.2. What is the radius of Rn? It may be possible to consider Question 10.2 in conjunction with the analysis for the exact diameter of Rn in [5, Section 5]. An irregularity measure of graphs was defined by Albertson , and recently studied for various families of graphs in , and generalized to a polynomial enumerator in . Question 10.3. What is the irregularity, or more generally the irregularity polynomial, of Rn, as defined in ? Acknowledgments We would like to thank Prof. Sandi Klavžar for enabling the cooperation of the authors of this article, and his early interest in the topic. The first author would like to acknowledge the hospitality of Reykjavik University during his sabbatical stay there, during which a portion of this research was carried out. The second author acknowledges the financial support from the Slovenian Research Agency (project N1-0095). References M.O. Albertson, The irregularity of a graph, Ars Combin. 46 (1997) 219–225. Y. Alizadeh, E. Deutsch, S. Klavžar, On the irregularity of π-permutation graphs, Fibonacci cubes, and trees, Bull. Malays. Math. Sci. Soc. 43 (6) (2020) 4443–4456. 70 Ö. Eğecioğlu and V. Iršič Discrete Applied Mathematics 300 (2021) 56–71 J. Azarija, S. Klavžar, Y. Rho, S. Sim, On domination-type invariants of Fibonacci cubes and hypercubes, Ars Math. Contemp. 14 (2) (2018) 387–395. O. Eğecioğlu, Statistics on restricted Fibonacci words, Trans. Comb. 10 (1) (2021) 31–42. O. Eğecioğlu, V. Iršič, Fibonacci-run graphs i: basic properties, Discrete Appl. Math. 295 (2021) 70–84. O. Eğecioğlu, E. Saygı, Z. Saygi, The irregularity polynomials of Fibonacci and Lucas cubes, Bull. Malays. Math. Sci. Soc. 44 (2) (2021) 753–765. W.J. Hsu, Fibonacci cubes - new interconnection topology, IEEE Trans. Parallel Distrib. Syst. 4 (1993) 3–12. S. Klavžar, Structure of fibonacci cubes: A survey, J. Comb. Optim. 25 (2013) 4. S. Klavžar, M. Mollard, M. Petkovšek, The degree sequence of Fibonacci and Lucas cubes, Discrete Math. 311 (14) (2011) 1310–1322. R.A. Leslie, How not to repeatedly differentiate a reciprocal, Amer. Math. Monthly 98 (8) (1991) 732–735. M. Mollard, Non covered vertices in Fibonacci cubes by a maximum set of disjoint hypercubes, Discrete Appl. Math. 219 (2017) 219–221. K.S. Savitha, A. Vijayakumar, Some diameter notions of Fibonacci cubes, Asian-Eur. J. Math. 13 (3) (2020) 2050057, 9. E. Saygı, O. Eğecioğlu, q-cube enumerator polynomial of Fibonacci cubes, Discrete Appl. Math. 226 (2017) 127–137. E. Saygı, O. Eğecioğlu, Boundary enumerator polynomial of hypercubes in Fibonacci cubes, Discrete Appl. Math. 266 (2019) 191–199. 71
2214	https://worldjusticeproject.org/rule-of-law-index/downloads/WJPIndex2023.pdf	2023 RULE OF LAW INDEX® The World Justice Project (WJP) Rule of Law Index® 2023 report was prepared by the World Justice Project. The Index’s conceptual framework and methodology were developed by Juan Carlos Botero, Mark David Agrast, and Alejandro Ponce. Data collection and analysis for the 2023 report was performed by Erin Campbell, James Davis, Alicia Evangelides, Joshua Fuller, Natalia Jardon, Lauren Littlejohn, Gustavo Núñez Peralta, Alejandro Ponce, Hannah Rigazzi, Natalia Rodríguez Cajamarca, Victoria Thomaides, Carlos Toruño Paniagua, Santiago Pardo, and Moss Woodbury, with the assistance of Said Aarji, Lloyd Cleary, John Cullen, Allyse Feitzinger, Skye Jacobs, and Helen Souki Reyes. The graphic design team for this report included Enrique Paulin, Irene Heras, Mariana Lopez, and Raquel Medina. Pitch Interactive served as lead website designer, with assistance from Irene Heras, Natalia Jardon, Mariana Lopez, Raquel Medina, and Enrique Paulin, ISBN (print version): 978-0-615-40781-4 ISBN (digital version): 978-0-615-51219-8 Board of Directors: Sheikha Abdulla Al-Misnad; Kamel Ayadi; Adam Bodnar; Michael Chu; William C. Hubbard; Hassan Bubacar Jallow; Suet-Fern Lee; Mondli Makhanya; M. Margaret McKeown; John Nery; William H. Neukom; Ellen Gracie Northfleet; and James R. Silkenat. Directors Emeritus: Ashraf Ghani Ahmadzai, Emil Constantinescu, and Petar Stoyanov Officers: William C. Hubbard, Co-Founder and Chairman of the Board; William H. Neukom, Co-Founder and CEO; Mark D. Agrast, Vice President; Deborah Enix-Ross, Vice President; Judy Perry Martinez, Vice President; Nancy Ward, Vice President; James R. Silkenat, Director and Treasurer; and Gerold W. Libby, General Counsel and Secretary. Executive Director: Elizabeth Andersen Chief Research Officer: Alejandro Ponce The WJP Rule of Law Index 2023 report was made possible by the generous supporters of the work of the World Justice Project listed in this report on page 217. © Copyright 2023 by the World Justice Project. The WJP Rule of Law Index and the World Justice Project Rule of Law Index are trademarks of the World Justice Project. All rights reserved. Requests to reproduce this document should be sent to: WJP Rule of Law Index Permissions World Justice Project 1025 Vermont Avenue, N.W., Suite 1200 Washington, D.C. 20005 U.S.A. E-mail: wjp@worldjusticeproject.org Subject line: WJP Rule of Law Index Permissions The World Justice Project The World Justice Project Rule of Law Index ® 2023 2023 RULE OF LAW INDEX® About the WJP Rule of Law Index Scores and Rankings 01 02 8 Foreword 10 Overview of Scores and Rankings 12 Features of the WJP Rule of Law Index 13 Defining the Rule of Law 14 The Four Universal Principles of the Rule of Law 15 Conceptual Framework of the WJP Rule of Law Index 16 Indicators of the WJP Rule of Law Index 22 Rule of Law Around the World 24 Rule of Law by Region 26 Rule of Law by Income 28 Rule of Law by Factor Table of Contents Country Profiles Behind the Numbers 03 04 38 How to Read the Country Profiles 39 Country Profiles 182 Methodology Snapshot 183 Methodology 190 Contributing Experts 215 Acknowledgements 216 About the WJP 218 More From WJP 6 7 WJP Rule of Law Index 2023 8 Foreword 10 Overview of Scores and Rankings 12 Features of the WJP Rule of Law Index 13 Defining the Rule of Law 14 The Four Universal Principles of the Rule of Law 15 Conceptual Framework of the WJP Rule of Law Index 16 Indicators of the WJP Rule of Law Index 8 The rule of law is internationally recognized as a foundational element in guaranteeing peace, justice, human rights, effective democracy, and sustainable development. Around the world, however, the rule of law continues to weaken. This 2023 edition of the World Justice Project (WJP) Rule of Law Index shows that over 6 billion people now live in countries where the rule of law is declining. Amid widespread institutional stagnation, a majority of countries continue to experience rule of law backsliding characterized by executive overreach, diminished human rights, and justice systems that are failing to meet people’s needs. In the face of these challenges, there is a need for a shared understanding of the landscape we face and a systematic, robust, and actionable assessment of adherence to the rule of law around the world. To this end, the World Justice Project is pleased to present the latest edition of the WJP Rule of Law Index so that a wide variety of stakeholders can identify rule of law strengths, weaknesses, progress, and setbacks across 142 countries and jurisdictions. The Index offers original, independent data organized into eight factors that encompass the concept of the rule of law: Constraints on Government Powers, Absence of Corruption, Open Government, Fundamental Rights, Order and Security, Regulatory Enforcement, Civil Justice, and Criminal Justice. Index factor scores reflect the perspectives and experiences of more than 149,000 everyday people and 3,400 legal experts around the world, and they are backed by a rigorous process of validation and analysis. In 2023, overall rule of law has declined in a majority of countries yet again. The authoritarian trends that first spurred the global rule of law recession in 2016 persist in every region of the world today. However, these declines are less widespread and extreme for the second year in a row. Some countries have managed to reverse authoritarian trends, while others have made sustained progress in justice, anti-corruption, and human rights. The Index data proves that progress is possible. There is still much work to be done to advance the rule of law worldwide and to prevent the arbitrary exercise of power. As a global leader, the World Justice Project reaffirms its commitment to work hand in hand with governments, policy makers, political actors, the private sector, civil society organizations, the media, donors, and academia to support reform efforts aimed at advancing the rule of law. This edition of the WJP Rule of Law Index is a small but important step in this direction. Dr. Alejandro Ponce Chief Research Officer World Justice Project Foreword 9 WJP Rule of Law Index 2023 .... 10 Afghanistan 0.32 -4.0% 140 0 Albania 0.48 -0.7% 91 2 Algeria 0.49 0.6% 84 7 Angola 0.43 -0.3% 115 0 Antigua and Barbuda 0.63 0.6% 38 2 Argentina 0.55 -0.5% 63 2 Australia 0.80 0.5% 13 0 Austria 0.80 -0.3% 11 0 The Bahamas 0.59 -2.0% 50 2 Bangladesh 0.38 -1.5% 127 2 Barbados 0.66 0.0% 35 1 Belarus 0.45 -1.9% 104 3 Belgium 0.78 -1.0% 16 2 Belize 0.49 0.9% 80 7 Benin 0.48 -0.6% 90 0 Bolivia 0.37 -0.5% 131 1 Bosnia and Herzegovina 0.51 -1.2% 75 3 Botswana 0.59 0.0% 51 0 Brazil 0.49 -0.9% 83 0 Bulgaria 0.56 1.7% 59 3 Burkina Faso 0.47 -3.7% 95 11 Cambodia 0.31 -0.1% 141 0 Cameroon 0.35 -0.3% 134 2 Canada 0.80 0.0% 12 0 Chile 0.66 -0.1% 33 0 China 0.47 -0.9% 97 0 Colombia 0.48 -0.4% 94 1 Congo, Dem. Rep. 0.34 0.8% 138 1 Congo, Rep. 0.40 -1.2% 122 0 Costa Rica 0.68 0.0% 29 0 Côte d'Ivoire 0.45 0.7% 106 4 Croatia 0.61 0.3% 45 2 Cyprus 0.68 -0.9% 31 3 Czechia 0.73 0.1% 20 0 Denmark 0.90 -0.3% 1 0 Overview of Overall Scores and Rankings The table below shows the overall scores and rankings of the WJP Rule of Law Index 2023 in alphabetical order. Scores range from 0 to 1, with 1 indicating the strongest adherence to the rule of law. Dominica 0.58 0.1% 53 1 Dominican Republic 0.49 1.0% 86 8 Ecuador 0.47 -2.1% 96 1 Egypt, Arab Rep. 0.35 -1.2% 136 1 El Salvador 0.45 -2.5% 108 4 Estonia 0.82 0.0% 9 0 Ethiopia 0.38 -3.1% 129 4 Finland 0.87 0.4% 3 0 France 0.73 -0.4% 21 0 Gabon 0.39 0.3% 124 4 The Gambia 0.49 0.1% 85 3 Georgia 0.60 0.3% 48 1 Germany 0.83 0.0% 5 1 Ghana 0.55 -0.5% 61 1 Greece 0.61 -1.4% 47 3 Grenada 0.60 1.4% 49 1 Guatemala 0.44 -0.3% 111 1 Guinea 0.41 0.8% 118 1 Guyana 0.50 0.2% 76 1 Haiti 0.34 -3.5% 139 1 Honduras 0.41 1.6% 119 4 Hong Kong SAR, China 0.73 -0.2% 23 1 Hungary 0.51 -0.2% 73 2 India 0.49 -0.7% 79 0 Indonesia 0.53 0.2% 66 0 Iran, Islamic Rep. 0.39 -5.0% 126 5 Ireland 0.81 0.3% 10 0 Italy 0.67 0.0% 32 0 Jamaica 0.57 -1.1% 54 1 Japan 0.79 0.0% 14 2 Jordan 0.55 1.4% 62 1 Kazakhstan 0.53 1.0% 65 2 Kenya 0.46 1.6% 101 5 Korea, Rep. 0.74 0.5% 19 0 Kosovo 0.56 0.4% 58 1 Kuwait 0.58 -52 -% Change in Overall Score Overall Score Country/Jurisdiction Global Rank Change in Global Rank† Scores are rounded to two decimal places and percentage changes in score are rounded to one decimal place. † The change in rankings was calculated by comparing the positions of the 140 countries and jurisdictions measured in the 2022 Index with the rankings of the same 140 countries and jurisdictions in 2023, exclusive of the two new additions to the 2023 Index, Kuwait and Montenegro. % Change in Overall Score Overall Score Country/Jurisdiction Global Rank Change in Global Rank† 11 WJP Rule of Law Index 2023 Kyrgyz Republic 0.45 -1.4% 103 1 Latvia 0.73 0.7% 22 2 Lebanon 0.45 0.0% 107 1 Liberia 0.44 0.8% 112 2 Lithuania 0.77 0.4% 18 0 Luxembourg 0.83 0.8% 6 2 Madagascar 0.43 -1.4% 114 1 Malawi 0.52 -0.5% 69 1 Malaysia 0.57 0.8% 55 1 Mali 0.40 -5.3% 121 5 Malta 0.68 0.1% 30 0 Mauritania 0.36 -1.4% 133 0 Mauritius 0.61 -0.4% 46 1 Mexico 0.42 -1.3% 116 1 Moldova 0.53 1.3% 68 2 Mongolia 0.53 -0.6% 64 0 Montenegro 0.56 -57 -Morocco 0.48 0.9% 92 4 Mozambique 0.38 -3.0% 128 4 Myanmar 0.35 -3.7% 135 1 Namibia 0.61 0.5% 44 2 Nepal 0.52 -0.9% 71 0 Netherlands 0.83 -0.3% 7 2 New Zealand 0.83 -0.2% 8 1 Nicaragua 0.35 -4.4% 137 2 Niger 0.44 -0.6% 109 2 Nigeria 0.41 0.8% 120 0 North Macedonia 0.53 -0.9% 67 2 Norway 0.89 0.3% 2 0 Pakistan 0.38 -2.3% 130 1 Panama 0.51 -0.5% 74 0 Paraguay 0.46 -1.2% 99 1 Peru 0.49 0.2% 88 4 Philippines 0.46 -1.5% 100 1 Poland 0.64 -0.6% 36 0 Portugal 0.68 -0.9% 28 1 Romania 0.63 -0.4% 40 2 Russian Federation 0.44 -2.2% 113 4 Rwanda 0.63 0.1% 41 1 Senegal 0.55 -1.3% 60 2 Serbia 0.48 -1.6% 93 8 Sierra Leone 0.44 -2.1% 110 3 Singapore 0.78 -0.1% 17 0 Slovak Republic 0.66 0.3% 34 1 Slovenia 0.69 1.6% 27 4 South Africa 0.57 -1.2% 56 1 Spain 0.72 -0.8% 24 1 Sri Lanka 0.50 -0.5% 77 1 St. Kitts and Nevis 0.63 0.1% 39 0 St. Lucia 0.62 0.4% 43 0 St. Vincent and the Grenadines 0.63 -0.8% 42 1 Sudan 0.36 -7.4% 132 5 Suriname 0.49 -0.9% 81 0 Sweden 0.85 -0.4% 4 0 Tanzania 0.47 0.9% 98 2 Thailand 0.49 -1.0% 82 0 Togo 0.45 -1.1% 102 1 Trinidad and Tobago 0.52 -0.8% 70 1 Tunisia 0.52 -0.6% 72 1 Türkiye 0.41 -0.7% 117 1 Uganda 0.39 0.0% 125 5 Ukraine 0.49 -2.9% 89 11 United Arab Emirates 0.64 0.2% 37 0 United Kingdom 0.78 -0.4% 15 0 United States 0.70 -0.6% 26 0 Uruguay 0.72 0.4% 25 0 Uzbekistan 0.50 0.1% 78 2 Venezuela, RB 0.26 0.9% 142 0 Vietnam 0.49 -0.6% 87 1 Zambia 0.45 0.0% 105 0 Zimbabwe 0.40 1.3% 123 3 % Change in Overall Score Overall Score Country/Jurisdiction Global Rank Change in Global Rank† % Change in Overall Score Overall Score Country/Jurisdiction Global Rank Change in Global Rank† Scores are rounded to two decimal places and percentage changes in score are rounded to one decimal place. † The change in rankings was calculated by comparing the positions of the 140 countries and jurisdictions measured in the 2022 Index with the rankings of the same 140 countries and jurisdictions in 2023, exclusive of the two new additions to the 2023 Index, Kuwait and Montenegro. 12 The scores and rankings of the eight factors and 44 sub- factors of the Index draw from two sources of data collected by the WJP: 1. A General Population Poll (GPP) conducted by leading local polling companies, using a representative sample of 1,0001 respondents in each country and jurisdiction. 2. Qualified Respondents’ Questionnaires (QRQs) consisting of closed-ended questions completed by in-country legal practitioners, experts, and academics with expertise in civil and commercial law; constitutional law, civil liberties, and criminal law; labor law; and public health.2 Taken together, these two data sources provide current, original information reflecting the experiences and perceptions of the general public and in-country legal practitioners and experts in 142 countries and jurisdictions worldwide. Rule of Law in Practice The Index measures adherence to the rule of law by looking at policy outcomes, such as whether people have access to courts or whether crime is effectively controlled. This stands in contrast to efforts that focus on the written legal code, or the institutional means by which a society may seek to achieve these policy outcomes. Comprehensive and Multi-Dimensional While other indices cover particular aspects of the rule of law, such as absence of corruption or human rights, they do not yield a full picture of the state of the rule of law. The WJP Rule of Law Index is the only global instrument that looks at the rule of law comprehensively. Perspective of Ordinary People The WJP Rule of Law Index puts people at its core. It looks at a country’s adherence to the rule of law from the perspec-tive of ordinary individuals and their experiences with the rule of law in their societies. The Index examines practical, everyday situations, such as whether people can access public services and whether a dispute among neighbors can be resolved peacefully and cost-effectively by an independent adjudicator. New Data Anchored in Actual Experiences The Index is the only comprehensive set of indicators on the rule of law that is based on primary data. Index scores are built from the assessments of residents (generally 1,000 respondents per country or jurisdiction) and local legal practitioners and experts, which ensure that the findings reflect the conditions experienced by actual people, includ-ing residents from marginalized sectors of society. Culturally Competent The Index has been designed to be applied in countries and jurisdictions with vastly different social, cultural, economic, and political systems. No society has ever attained—let alone sustained—a perfect realization of the rule of law. Every country faces the perpetual challenge of building and renewing the structures, institutions, and norms that can support and sustain a rule of law culture. Country-Specific Data and Online Tools In addition to this written report, an interactive online platform for country-specific WJP Rule of Law Index data is available at: worldjusticeproject.org. The interactive data site invites viewers to browse each of the 142 country and jurisdiction profiles and explore overall and factor scores. Features of the WJP Rule of Law Index The World Justice Project (WJP) developed the WJP Rule of Law Index to serve as a quantitative tool that measures the rule of law in practice. The Index’s methodology and comprehensive definition of the rule of law are the products of intensive consultation and vetting with academics, practitioners, and community leaders from more than 100 countries and jurisdictions and 17 professional disciplines. 1. Due to small populations or obstacles to data collection in certain countries and jurisdictions, the sampling plan was adjusted in some cases. For more information on specific countries and jurisdictions and sample sizes, see pages 186-189. 2. Please see the “Methodology” section on page 180 of this report for more detailed information regarding data collection and score computation. Scan the QR code or visit worldjusticeproject.org/index to view our interactive data portal. 13 WJP Rule of Law Index 2023 Defining the Rule of Law Effective rule of law reduces corruption, combats poverty and disease, and protects people from injustices large and small. It is the foundation for communities of justice, opportunity, and peace—underpinning development, accountable government, and respect for fundamental rights. Traditionally, the rule of law has been viewed as the domain of lawyers and judges. However, everyday issues of safety, rights, justice, and governance affect us all; everyone is a stakeholder in the rule of law. Despite its profound importance for fair and functioning societies, the rule of law is notoriously difficult to define and measure. A simple way of approaching it is to examine a set of outcomes that the rule of law brings to societies, each of which reflects one aspect of the complex concept of the rule of law. The WJP Rule of Law Index seeks to embody these outcomes within a simple and coherent framework. The WJP Rule of Law Index captures adherence to the rule of law as defined by the WJP’s universal principles (see following page) through a comprehensive and multi-dimen-sional set of outcome indicators, each of which reflects a particular aspect of this complex concept. The theoretical framework linking these outcome indicators draws upon two main principles pertaining to the relationship between the state and the governed. The first principle measures whether the law imposes limits on the exercise of power by the state and its agents, as well as individuals and private entities. This is measured in Factors One, Two, Three, and Four of the Index. The second princi-ple measures whether the state limits the actions of members of society and fulfills its basic duties towards its population so that the public interest is served, people are protected from violence, and all members of society have access to dispute settlement and grievance mechanisms. This is measured in Factors Five, Six, Seven, and Eight of the Index. Although broad in scope, this framework assumes very little about the functions of the state, and when it does, it incorporates functions that are recognized by practically all societies, such as the provision of justice or the guaran-tee of order and security. The resulting set of indicators is also an effort to strike a balance between what scholars call a “thin” or minimalist conception of the rule of law that focuses on formal, procedural rules, and a “thick” conception that includes substantive characteristics, such as self-governance and various fundamental rights and freedoms. Striking this balance between “thin” and “thick” conceptions of the rule of law enables the Index to apply to different types of social and political systems, including those that lack many of the features that characterize democratic nations, while includ-ing sufficient substantive characteristics to render the rule of law as more than a system of rules. The Index recognizes that a system of law that fails to respect core human rights guaranteed under international law is at best “rule by law” and does not deserve to be called a rule of law system. The rule of law affects all of us in our everyday lives. Although we may not be aware of it, the rule of law is profoundly important—and not just for lawyers or judges. Every sector of society is a stakeholder in the rule of law. Below are a few examples: Business Environment Imagine an investor seeking to commit resources abroad. She would probably think twice before investing in a country where corruption is rampant, property rights are ill-defined, and contracts are difficult to enforce. Uneven enforcement of regulations, corruption, insecure property rights, and ineffective means to settle disputes undermine legitimate business and deter both domestic and foreign investment. Public Works Consider the bridges, roads, or runways we traverse daily— or the offices and buildings in which we live, work, and play. What would happen if building codes governing design and safety were not enforced? What if government officials and contractors used low-quality materials to pocket the surplus? Weak regulatory enforcement and corruption decrease the security of physical infrastructure and waste scarce resources, which are essential to a thriving economy. Public Health and Environment Consider the implications of pollution, wildlife poaching, and deforestation for public health and the environment. What would happen if a company were pouring harmful chemicals into a river in a highly populated area and the environmen-tal inspector ignored these actions in exchange for a bribe? Adherence to the rule of law is essential to holding govern-ments, businesses, civil society organizations, and communi-ties accountable for protecting public health and the environment. 14 The Four Universal Principles of the Rule of Law The World Justice Project defines the rule of law as a durable system of laws, institutions, norms, and community commitment that delivers: Accountability Universal Principle One Universal Principle Two Just Law Universal Principle Three Open Government Universal Principle Four Accessible and Impartial Justice The government as well as private actors are accountable under the law. The law is clear, publicized, and stable and is applied evenly. It ensures human rights as well as property, contract, and procedural rights. The processes by which the law is adopted, administered, adjudicated, and enforced are accessible, fair, and efficient. Justice is delivered timely by competent, ethical, and independent representatives and neutrals who are accessible, have adequate resources, and reflect the makeup of the communities they serve. The four universal principles are further developed in the following eight factors of the annual WJP Rule of Law Index: Constraints on Government Powers, Absence of Corruption, Open Government, Fundamental Rights, Order and Security, Regulatory Enforcement, Civil Justice, and Criminal Justice. 15 WJP Rule of Law Index 2023 Conceptual Framework of the WJP Rule of Law Index The conceptual framework of the WJP Rule of Law Index is comprised of eight factors further disaggregated into 44 sub-factors. These factors and sub-factors are presented below and described in detail in the section that follows. 1.1 Government powers are effectively limited by the legislature 1.2 Government powers are effectively limited by the judiciary 1.3 Government powers are effectively limited by independent auditing and review 1.4 Government officials are sanctioned for misconduct 1.5 Government powers are subject to non-governmental checks 1.6 Transition of power is subject to the law Constraints on Government Powers Factor 1 2.1 Government officials in the executive branch do not use public office for private gain 2.2 Government officials in the judicial branch do not use public office for private gain 2.3 Government officials in the police and the military do not use public office for private gain 2.4 Government officials in the legislative branch do not use public office for private gain Absence of Corruption Factor 2 3.1 Publicized laws and government data 3.2 Right to information 3.3 Civic participation 3.4 Complaint mechanisms Open Government Factor 3 4.1 Equal treatment and absence of discrimination 4.2 The right to life and security of the person is effectively guaranteed 4.3 Due process of the law and rights of the accused 4.4 Freedom of opinion and expression is effectively guaranteed 4.5 Freedom of belief and religion is effectively guaranteed 4.6 Freedom from arbitrary interference with privacy is effectively guaranteed 4.7 Freedom of assembly and association is effectively guaranteed 4.8 Fundamental labor rights are effectively guaranteed Fundamental Rights Factor 4 5.1 Crime is effectively controlled 5.2 Civil conflict is effectively limited 5.3 People do not resort to violence to redress personal grievances Order and Security Factor 5 6.1 Government regulations are effectively enforced 6.2 Government regulations are applied and enforced without improper influence 6.3 Administrative proceedings are conducted without unreasonable delay 6.4 Due process is respected in administrative proceedings 6.5 The government does not expropriate without lawful process and adequate compensation Regulatory Enforcement Factor 6 7.1 People can access and afford civil justice 7.2 Civil justice is free of discrimination 7.3 Civil justice is free of corruption 7.4 Civil justice is free of improper government influence 7.5 Civil justice is not subject to unreasonable delay 7.6 Civil justice is effectively enforced 7.7 Alternative dispute resolution mechanisms are accessible, impartial, and effective Civil Justice Factor 7 8.1 Criminal investigative system is effective 8.2 Criminal adjudication system is timely and effective 8.3 Correctional system is effective in reducing criminal behavior 8.4 Criminal justice is impartial 8.5 Criminal justice is free of corruption 8.6 Criminal justice is free of improper government influence 8.7 Due process of the law and rights of the accused Criminal Justice Factor 8 16 Informal Justice and the Rule of Law The conceptual framework of the Index includes a ninth factor on informal justice that is not included in the Index’s aggregate scores and rankings. Informal justice systems often play a large role in countries and jurisdictions where formal legal institutions are weak, remote, or perceived as ineffective. As such, the WJP has devoted effort to collecting data on informal justice through our surveys. Nonetheless, the complexities of these systems and the difficulties of systematically measuring their fairness and effectiveness make cross-country assessments extraordinarily challenging. For this reason, the informal justice factor is not included in the Index scores and rankings. Factor 9: Informal Justice 9.1 Informal justice is timely and effective 9.2 Informal justice is impartial and free of improper influence 9.3 Informal justice respects and protects fundamental rights 1.1 Government powers are effectively limited by the legislature Measures whether legislative bodies have the independence and the ability in practice to exercise effective checks on and oversight of the government. 1.2 Government powers are effectively limited by the judiciary Measures whether the judiciary has the independence and the ability in practice to exercise effective checks on and oversight of the government. 1.3 Government powers are effectively limited by independent auditing and review Measures whether comptrollers or auditors, as well as national human rights ombudsman agencies, have sufficient independence and the ability to exercise effective checks on and oversight of the government. 1.4 Government officials are sanctioned for misconduct Measures whether government officials in the executive, legislature, judiciary, and the police are investigated, prosecuted, and punished for official misconduct and other violations. 1.5 Government powers are subject to non-governmental checks Measures whether an independent media, civil society organizations, political parties, and individuals are free to report and comment on government policies without fear of retaliation. 1.6 Transition of power is subject to the law Measures whether government officials are elected or appointed in accordance with the rules and procedures set forth in the constitution. Where elections take place, it also measures the integrity of the electoral process, including access to the ballot, the absence of intimidation, and public scrutiny of election results. 2.1 Government officials in the executive branch do not use public office for private gain Measures the prevalence of bribery, informal payments, and other inducements in the delivery of public services and the enforcement of regulations. It also measures whether government procurement and public works contracts are awarded through an open and competitive bidding process, and whether government officials at various levels of the executive branch refrain from embezzling public funds. 2.2 Government officials in the judicial branch do not use public office for private gain Measures whether judges and judicial officials refrain from soliciting and accepting bribes to perform duties or expedite processes, and whether the judiciary and judicial rulings are free of improper influence by the government, private interests, or criminal organizations. 2.3 Government officials in the police and the military do not use public office for private gain Measures whether police officers and criminal investigators refrain from soliciting and accepting bribes to perform basic police services or to investigate crimes, and whether government officials in the police and the military are free of improper influence by private interests or criminal organizations. 2.4 Government officials in the legislative branch do not use public office for private gain Measures whether members of the legislature refrain from soliciting or accepting bribes or other inducements in exchange for political favors or favorable votes on legislation. Indicators of the WJP Rule of Law Index FACTOR ONE: Constraints on Government Powers FACTOR TWO: Absence of Corruption 17 WJP Rule of Law Index 2023 3.1 Publicized laws and government data Measures whether basic laws and information on legal rights are publicly available, presented in plain language, and made accessible in all languages used in the country or jurisdiction. It also measures the quality and accessibility of information published by the government in print or online, and whether administrative regulations, drafts of legislation, and high court decisions are made accessible to the public in a timely manner. 3.2 Right to information Measures whether requests for information held by a government agency are granted, whether these requests are granted within a reasonable time period, if the information provided is pertinent and complete, and if requests for information are granted at a reasonable cost and without having to pay a bribe. It also measures whether people are aware of their right to information, and whether relevant records are accessible to the public upon request. 3.3 Civic participation Measures the effectiveness of civic participation mechanisms, including the protection of the freedoms of opinion and expression, assembly and association, and the right to petition the government. It also measures whether people can voice concerns to various government officers and whether government officials provide sufficient information and notice about decisions affecting the community. 3.4 Complaint mechanisms Measures whether people are able to bring specific complaints to the government about the provision of public services or the performance of government officers in carrying out their legal duties in practice, and how government officials respond to such complaints. 4.1 Equal treatment and absence of discrimination Measures whether individuals are free from discrimination— based on socio-economic status, gender, ethnicity, religion, national origin, sexual orientation, or gender identity—with respect to public services, employment, court proceedings, and the justice system. 4.2 The right to life and security of the person is effectively guaranteed Measures whether the police inflict physical harm upon criminal suspects during arrest and interrogation, and whether political dissidents or members of the media are subjected to unreasonable searches, arrest, detention, imprisonment, threats, abusive treatment, or violence. 4.3 Due process of the law and rights of the accused Measures whether the basic rights of criminal suspects are respected, including the presumption of innocence and the freedom from arbitrary arrest and unreasonable pre-trial detention. It also measures whether criminal suspects are able to access and challenge evidence used against them, whether they are subject to abusive treatment, and whether they are provided with adequate legal assistance. In addition, it measures whether the basic rights of prisoners are respected once they have been convicted of a crime. 4.4 Freedom of opinion and expression is effectively guaranteed Measures whether an independent media, civil society organizations, political parties, and individuals are free to report and comment on government policies without fear of retaliation. 4.5 Freedom of belief and religion is effectively guaranteed Measures whether members of religious minorities can worship and conduct religious practices freely and publicly, and whether non-adherents are protected from having to submit to religious laws. 4.6 Freedom from arbitrary interference with privacy is effectively guaranteed Measures whether the police or other government officials conduct physical searches without warrants, or intercept electronic communications of private individuals without judicial authorization. 4.7 Freedom of assembly and association is effectively guaranteed Measures whether people can freely attend community meetings, join political organizations, hold peaceful public demonstrations, sign petitions, and express opinions against government policies and actions without fear of retaliation. 4.8 Fundamental labor rights are effectively guaranteed Measures the effective enforcement of fundamental labor rights, including freedom of association and the right to collective bargaining, the absence of discrimination with respect to employment, and freedom from forced labor and child labor. FACTOR THREE: Open Government FACTOR FOUR: Fundamental Rights 18 18 5.1 Crime is effectively controlled Measures the prevalence of common crimes, including homicide, kidnapping, burglary and theft, armed robbery, and extortion, as well as people’s general perceptions of safety in their communities. 5.2 Civil conflict is effectively limited Measures whether people are effectively protected from armed conflict and terrorism. 5.3 People do not resort to violence to redress personal grievances Measures whether people resort to intimidation or violence to resolve civil disputes amongst themselves or to seek redress from the government, and whether people are free from mob violence. 6.1 Government regulations are effectively enforced Measures whether government regulations, such as labor, environmental, public health, commercial, and consumer protection regulations are effectively enforced. 6.2 Government regulations are applied and enforced without improper influence Measures whether the enforcement of regulations is subject to bribery or improper influence by private interests, and whether public services, such as the issuance of permits and licenses and the administration of public health services, are provided without bribery or other illegal inducements. 6.3 Administrative proceedings are conducted without unreasonable delay Measures whether administrative proceedings at the national and local levels are conducted without unreasonable delay. 6.4 Due process is respected in administrative proceedings Measures whether due process of the law is respected in administrative proceedings conducted by national and local authorities in issue areas such as the environment, taxes, and labor. 6.5 The government does not expropriate without lawful process and adequate compensation Measures whether the government respects the property rights of people and corporations, refrains from the illegal seizure of private property, and provides adequate compensation when property is legally expropriated. 7.1 People can access and afford civil justice Measures the accessibility and affordability of civil courts, including whether people are aware of available remedies; can access and afford legal advice and representation; and can access the court system without incurring unreasonable fees, encountering unreasonable procedural hurdles, or experiencing physical or linguistic barriers. 7.2 Civil justice is free of discrimination Measures whether the civil justice system discriminates in practice based on socio-economic status, gender, ethnicity, religion, national origin, sexual orientation, or gender identity. 7.3 Civil justice is free of corruption Measures whether the civil justice system is free of bribery and improper influence by private interests. 7.4 Civil justice is free of improper government influence Measures whether the civil justice system is free of improper government or political influence. 7.5 Civil justice is not subject to unreasonable delay Measures whether civil justice proceedings are conducted and judgments are produced in a timely manner without unreasonable delay. 7.6 Civil justice is effectively enforced Measures the effectiveness and timeliness of the enforcement of civil justice decisions and judgments in practice. 7.7 Alternative dispute resolution mechanisms are accessible, impartial, and effective Measures whether alternative dispute resolution mechanisms (ADRs) are affordable, efficient, enforceable, and free of corruption. FACTOR SEVEN: Civil Justice FACTOR SIX: Regulatory Enforcement FACTOR FIVE: Order and Security 19 WJP Rule of Law Index 2023 8.1 Criminal investigative system is effective Measures whether perpetrators of crimes are effectively apprehended and charged. It also measures whether police, investigators, and prosecutors have adequate resources, are free of corruption, and perform their duties competently. 8.2 Criminal adjudication system is timely and effective Measures whether perpetrators of crimes are effectively prosecuted and punished. It also measures whether criminal judges and other judicial officers are competent and produce speedy decisions. 8.3 Correctional system is effective in reducing criminal behavior Measures whether correctional institutions are secure, respect prisoners’ rights, and are effective in preventing recidivism. 8.4 Criminal justice is impartial Measures whether the police and criminal judges are impartial and whether they discriminate in practice based on socio-economic status, gender, ethnicity, religion, national origin, sexual orientation, or gender identity. 8.5 Criminal justice is free of corruption Measures whether the police, prosecutors, and judges are free of bribery and improper influence from criminal organizations. 8.6 Criminal justice is free of improper government influence Measures whether the criminal justice system is independent from government or political influence. 8.7 Due process of the law and rights of the accused Measures whether the basic rights of criminal suspects are respected, including the presumption of innocence and the freedom from arbitrary arrest and unreasonable pre-trial detention. It also measures whether criminal suspects can access and challenge evidence used against them, whether they are subject to abusive treatment, and whether they are provided with adequate legal assistance. In addition, it measures whether the basic rights of prisoners are respected once they have been convicted of a crime. FACTOR EIGHT: Criminal Justice 20 21 WJP Rule of Law Index 2023 22 Rule of Law Around the World 24 Rule of Law by Region 26 Rule of Law by Income 28 Rule of Law by Factor 22 Scores are rounded to two decimal places. Rule of Law Around the World Denmark 0.90 1 Norway 0.89 2 Finland 0.87 3 Sweden 0.85 4 Germany 0.83 5 Luxembourg 0.83 6 Netherlands 0.83 7 New Zealand 0.83 8 Estonia 0.82 9 Ireland 0.81 10 Austria 0.80 11 Canada 0.80 12 Australia 0.80 13 Japan 0.79 14 United Kingdom 0.78 15 Belgium 0.78 16 Singapore 0.78 17 Lithuania 0.77 18 Korea, Rep. 0.74 19 Czechia 0.73 20 France 0.73 21 Latvia 0.73 22 Hong Kong SAR, China 0.73 23 Spain 0.72 24 Uruguay 0.72 25 United States 0.70 26 Slovenia 0.69 27 Portugal 0.68 28 Costa Rica 0.68 29 Malta 0.68 30 Cyprus 0.68 31 Italy 0.67 32 Chile 0.66 33 Slovak Republic 0.66 34 Barbados 0.66 35 Poland 0.64 36 United Arab Emirates 0.64 37 Antigua and Barbuda 0.63 38 St. Kitts and Nevis 0.63 39 Romania 0.63 40 Rwanda 0.63 41 St. Vincent and the Grenadines 0.63 42 St. Lucia 0.62 43 Namibia 0.61 44 Croatia 0.61 45 Mauritius 0.61 46 Greece 0.61 47 Georgia 0.60 48 Grenada 0.60 49 The Bahamas 0.59 50 Botswana 0.59 51 Kuwait 0.58 52 Dominica 0.58 53 Jamaica 0.57 54 Malaysia 0.57 55 South Africa 0.57 56 Montenegro 0.56 57 Kosovo 0.56 58 Bulgaria 0.56 59 Senegal 0.55 60 Ghana 0.55 61 Jordan 0.55 62 Argentina 0.55 63 Mongolia 0.53 64 Kazakhstan 0.53 65 Indonesia 0.53 66 North Macedonia 0.53 67 Moldova 0.53 68 Malawi 0.52 69 Trinidad and Tobago 0.52 70 Nepal 0.52 71 Tunisia 0.52 72 Hungary 0.51 73 Panama 0.51 74 Bosnia and Herzegovina 0.51 75 Guyana 0.50 76 Sri Lanka 0.50 77 Uzbekistan 0.50 78 India 0.49 79 Belize 0.49 80 Suriname 0.49 81 Thailand 0.49 82 Brazil 0.49 83 Algeria 0.49 84 The Gambia 0.49 85 Overall Score Global Rank Country/Jurisdiction Overall Score Global Rank Country/Jurisdiction Overall Score Global Rank Country/Jurisdiction The table below shows the overall scores and rankings of the WJP Rule of Law Index ® 2023 by country rank. Scores range from 0 to 1, with 1 indicating the strongest adherence to the rule of law. 23 WJP Rule of Law Index 2023 Scores are rounded to two decimal places. Dominican Republic 0.49 86 Vietnam 0.49 87 Peru 0.49 88 Ukraine 0.49 89 Benin 0.48 90 Albania 0.48 91 Morocco 0.48 92 Serbia 0.48 93 Colombia 0.48 94 Burkina Faso 0.47 95 Ecuador 0.47 96 China 0.47 97 Tanzania 0.47 98 Paraguay 0.46 99 Philippines 0.46 100 Kenya 0.46 101 Togo 0.45 102 Kyrgyz Republic 0.45 103 Belarus 0.45 104 Zambia 0.45 105 Côte d'Ivoire 0.45 106 Lebanon 0.45 107 El Salvador 0.45 108 Niger 0.44 109 Sierra Leone 0.44 110 Guatemala 0.44 111 Liberia 0.44 112 Russian Federation 0.44 113 Madagascar 0.43 114 Angola 0.43 115 Mexico 0.42 116 Türkiye 0.41 117 Guinea 0.41 118 Honduras 0.41 119 Nigeria 0.41 120 Mali 0.40 121 Congo, Rep. 0.40 122 Zimbabwe 0.40 123 Gabon 0.39 124 Uganda 0.39 125 Iran, Islamic Rep. 0.39 126 Bangladesh 0.38 127 Mozambique 0.38 128 Ethiopia 0.38 129 Pakistan 0.38 130 Bolivia 0.37 131 Sudan 0.36 132 Mauritania 0.36 133 Cameroon 0.35 134 Myanmar 0.35 135 Egypt, Arab Rep. 0.35 136 Nicaragua 0.35 137 Congo, Dem. Rep. 0.34 138 Haiti 0.34 139 Afghanistan 0.32 140 Cambodia 0.31 141 Venezuela, RB 0.26 142 Overall Score Global Rank Country/Jurisdiction Overall Score Global Rank Country/Jurisdiction Overall Score Global Rank Country/Jurisdiction 24 Rule of Law Around the World by Region East Asia and Pacific Eastern Europe and Central Asia EU, EFTA, and North America 1/15 New Zealand 0.83 -0.2% -0.4% 8 1 2/15 Australia 0.80 0.5% -1.3% 13 0 3/15 Japan 0.79 0.0% -0.2% 14 2 4/15 Singapore 0.78 -0.1% -2.6% 17 0 5/15 Korea, Rep. 0.74 0.5% 2.0% 19 0 6/15 Hong Kong SAR, China 0.73 -0.2% -6.0% 23 1 7/15 Malaysia 0.57 0.8% 6.1% 55 1 8/15 Mongolia 0.53 -0.6% -2.0% 64 0 9/15 Indonesia 0.53 0.2% 2.6% 66 0 10/15 Thailand 0.49 -1.0% -3.0% 82 0 11/15 Vietnam 0.49 -0.6% -2.9% 87 1 12/15 China 0.47 -0.9% -6.7% 97 0 13/15 Philippines 0.46 -1.5% -2.0% 100 1 14/15 Myanmar 0.35 -3.7% -17.2% 135 1 15/15 Cambodia 0.31 -0.1% -3.5% 141 0 1/15 Georgia 0.60 0.3% -1.2% 48 1 2/15 Montenegro 0.56 --57 -3/15 Kosovo 0.56 0.4% -58 1 4/15 Kazakhstan 0.53 1.0% 3.1% 65 2 5/15 North Macedonia 0.53 -0.9% -0.4% 67 2 6/15 Moldova 0.53 1.3% 7.7% 68 2 7/15 Bosnia and Herzegovina 0.51 -1.2% -3.5% 75 3 8/15 Uzbekistan 0.50 0.1% 7.7% 78 2 9/15 Ukraine 0.49 -2.9% -2.7% 89 11 10/15 Albania 0.48 -0.7% -5.2% 91 2 11/15 Serbia 0.48 -1.6% -4.0% 93 8 12/15 Kyrgyz Republic 0.45 -1.4% -4.5% 103 1 13/15 Belarus 0.45 -1.9% -12.0% 104 3 14/15 Russian Federation 0.44 -2.2% -7.0% 113 4 15/15 Türkiye 0.41 -0.7% -1.3% 117 1 1/31 Denmark 0.90 -0.3% 0.7% 1 0 2/31 Norway 0.89 0.3% 0.3% 2 0 3/31 Finland 0.87 0.4% 0.3% 3 0 4/31 Sweden 0.85 -0.4% -1.1% 4 0 5/31 Germany 0.83 0.0% -0.2% 5 1 6/31 Luxembourg 0.83 0.8% -6 2 7/31 Netherlands 0.83 -0.3% -2.5% 7 2 8/31 Estonia 0.82 0.0% 2.3% 9 0 9/31 Ireland 0.81 0.3% -10 0 10/31 Austria 0.80 -0.3% -2.0% 11 0 11/31 Canada 0.80 0.0% -1.6% 12 0 12/31 United Kingdom 0.78 -0.4% -3.1% 15 0 13/31 Belgium 0.78 -1.0% 1.2% 16 2 14/31 Lithuania 0.77 0.4% -18 0 15/31 Czechia 0.73 0.1% -1.3% 20 0 16/31 France 0.73 -0.4% -1.4% 21 0 17/31 Latvia 0.73 0.7% -22 2 18/31 Spain 0.72 -0.8% 2.4% 24 1 19/31 United States 0.70 -0.6% -4.0% 26 0 20/31 Slovenia 0.69 1.6% 2.3% 27 4 21/31 Portugal 0.68 -0.9% -5.2% 28 1 22/31 Malta 0.68 0.1% -30 0 23/31 Cyprus 0.68 -0.9% -31 3 24/31 Italy 0.67 0.0% 2.0% 32 0 25/31 Slovak Republic 0.66 0.3% -34 1 26/31 Poland 0.64 -0.6% -5.3% 36 0 27/31 Romania 0.63 -0.4% -4.1% 40 2 28/31 Croatia 0.61 0.3% 0.2% 45 2 29/31 Greece 0.61 -1.4% -0.3% 47 3 30/31 Bulgaria 0.56 1.7% 3.9% 59 3 31/31 Hungary 0.51 -0.2% -6.0% 73 2 (European Union, European Free Trade Association, and North America) Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Scores are rounded to two decimal places. Percentage changes in score are rounded to one decimal place. † The change in rankings was calculated by comparing the positions of the 140 countries and jurisdictions measured in the 2022 Index with the rankings of the same 140 countries and jurisdictions in 2023, exclusive of the two new additions to the 2023 Index, Kuwait and Montenegro. 25 WJP Rule of Law Index 2023 Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Country/Jurisdiction Change in Global Rank† Annual % Change in Overall Score 5-year % Change in Overall Score Global Rank Overall Score Regional Rank Middle East and North Africa South Asia Sub-Saharan Africa Latin America and Caribbean 1/32 Uruguay 0.72 0.4% 0.7% 25 0 2/32 Costa Rica 0.68 0.0% -0.6% 29 0 3/32 Chile 0.66 -0.1% -0.7% 33 0 4/32 Barbados 0.66 0.0% 1.1% 35 1 5/32 Antigua and Barbuda 0.63 0.6% 0.7% 38 2 6/32 St. Kitts and Nevis 0.63 0.1% -4.3% 39 0 7/32 St. Vincent and the Grenadines 0.63 -0.8% 2.3% 42 1 8/32 St. Lucia 0.62 0.4% -2.0% 43 0 9/32 Grenada 0.60 1.4% -1.6% 49 1 10/32 The Bahamas 0.59 -2.0% -1.4% 50 2 11/32 Dominica 0.58 0.1% -3.6% 53 1 12/32 Jamaica 0.57 -1.1% -0.4% 54 1 13/32 Argentina 0.55 -0.5% -6.4% 63 2 14/32 Trinidad and Tobago 0.52 -0.8% -7.0% 70 1 15/32 Panama 0.51 -0.5% -1.2% 74 0 16/32 Guyana 0.50 0.2% -0.5% 76 1 17/32 Belize 0.49 0.9% 3.3% 80 7 18/32 Suriname 0.49 -0.9% -3.6% 81 0 19/32 Brazil 0.49 -0.9% -8.9% 83 0 20/32 Dominican Republic 0.49 1.0% 4.4% 86 8 21/32 Peru 0.49 0.2% -7.1% 88 4 22/32 Colombia 0.48 -0.4% -4.7% 94 1 23/32 Ecuador 0.47 -2.1% -0.2% 96 1 24/32 Paraguay 0.46 -1.2% -99 1 25/32 El Salvador 0.45 -2.5% -7.4% 108 4 26/32 Guatemala 0.44 -0.3% -1.0% 111 1 27/32 Mexico 0.42 -1.3% -8.7% 116 1 28/32 Honduras 0.41 1.6% 1.6% 119 4 29/32 Bolivia 0.37 -0.5% -2.9% 131 1 30/32 Nicaragua 0.35 -4.4% -20.8% 137 2 31/32 Haiti 0.34 -3.5% -139 1 32/32 Venezuela, RB 0.26 0.9% -7.8% 142 0 1/9 United Arab Emirates 0.64 0.2% -1.9% 37 0 2/9 Kuwait 0.58 --52 -3/9 Jordan 0.55 1.4% -8.1% 62 1 4/9 Tunisia 0.52 -0.6% -3.6% 72 1 5/9 Algeria 0.49 0.6% -84 7 6/9 Morocco 0.48 0.9% -5.0% 92 4 7/9 Lebanon 0.45 0.0% -5.0% 107 1 8/9 Iran, Islamic Rep. 0.39 -5.0% -18.9% 126 5 9/9 Egypt, Arab Rep. 0.35 -1.2% -4.4% 136 1 1/6 Nepal 0.52 -0.9% -2.2% 71 0 2/6 Sri Lanka 0.50 -0.5% -5.2% 77 1 3/6 India 0.49 -0.7% -4.8% 79 0 4/6 Bangladesh 0.38 -1.5% -6.2% 127 2 5/6 Pakistan 0.38 -2.3% -4.2% 130 1 6/6 Afghanistan 0.32 -4.0% -8.3% 140 0 1/34 Rwanda 0.63 0.1% -41 1 2/34 Namibia 0.61 0.5% -44 2 3/34 Mauritius 0.61 -0.4% -46 1 4/34 Botswana 0.59 0.0% 1.3% 51 0 5/34 South Africa 0.57 -1.2% -3.3% 56 1 6/34 Senegal 0.55 -1.3% 0.2% 60 2 7/34 Ghana 0.55 -0.5% -7.4% 61 1 8/34 Malawi 0.52 -0.5% 2.1% 69 1 9/34 The Gambia 0.49 0.1% -85 3 10/34 Benin 0.48 -0.6% -90 0 11/34 Burkina Faso 0.47 -3.7% -6.7% 95 11 12/34 Tanzania 0.47 0.9% -0.9% 98 2 13/34 Kenya 0.46 1.6% 1.7% 101 5 14/34 Togo 0.45 -1.1% -102 1 15/34 Zambia 0.45 0.0% -4.9% 105 0 16/34 Côte d'Ivoire 0.45 0.7% -5.2% 106 4 17/34 Niger 0.44 -0.6% -109 2 18/34 Sierra Leone 0.44 -2.1% -3.1% 110 3 19/34 Liberia 0.44 0.8% -3.6% 112 2 20/34 Madagascar 0.43 -1.4% -1.7% 114 1 21/34 Angola 0.43 -0.3% -115 0 22/34 Guinea 0.41 0.8% -118 1 23/34 Nigeria 0.41 0.8% -6.3% 120 0 24/34 Mali 0.40 -5.3% -121 5 25/34 Congo, Rep. 0.40 -1.2% -122 0 26/34 Zimbabwe 0.40 1.3% 5.9% 123 3 27/34 Gabon 0.39 0.3% -124 4 28/34 Uganda 0.39 0.0% -3.1% 125 5 29/34 Mozambique 0.38 -3.0% -128 4 30/34 Ethiopia 0.38 -3.1% 0.2% 129 4 31/34 Sudan 0.36 -7.4% -132 5 32/34 Mauritania 0.36 -1.4% -133 0 33/34 Cameroon 0.35 -0.3% -4.0% 134 2 34/34 Congo, Dem. Rep. 0.34 0.8% -138 1 Scores are rounded to two decimal places. Percentage changes in score are rounded to one decimal place. † The change in rankings was calculated by comparing the positions of the 140 countries and jurisdictions measured in the 2022 Index with the rankings of the same 140 countries and jurisdictions in 2023, exclusive of the two new additions to the 2023 Index, Kuwait and Montenegro. 26 1/18 Rwanda 0.63 0.1% -41 1 2/18 Malawi 0.52 -0.5% 2.1% 69 1 3/18 The Gambia 0.49 0.1% -85 3 4/18 Burkina Faso 0.47 -3.7% -6.7% 95 11 5/18 Togo 0.45 -1.1% -102 1 6/18 Zambia 0.45 0.0% -4.9% 105 0 7/18 Niger 0.44 -0.6% -109 2 8/18 Sierra Leone 0.44 -2.1% -3.1% 110 3 9/18 Liberia 0.44 0.8% -3.6% 112 2 10/18 Madagascar 0.43 -1.4% -1.7% 114 1 11/18 Guinea 0.41 0.8% -118 1 12/18 Mali 0.40 -5.3% -121 5 13/18 Uganda 0.39 0.0% -3.1% 125 5 14/18 Mozambique 0.38 -3.0% -128 4 15/18 Ethiopia 0.38 -3.1% 0.2% 129 4 16/18 Sudan 0.36 -7.4% -132 5 17/18 Congo, Dem. Rep. 0.34 0.8% -138 1 18/18 Afghanistan 0.32 -4.0% -8.3% 140 0 1/37 Senegal 0.55 -1.3% 0.2% 60 2 2/37 Ghana 0.55 -0.5% -7.4% 61 1 3/37 Mongolia 0.53 -0.6% -2.0% 64 0 4/37 Indonesia 0.53 0.2% 2.6% 66 0 5/37 Nepal 0.52 -0.9% -2.2% 71 0 6/37 Tunisia 0.52 -0.6% -3.6% 72 1 7/37 Sri Lanka 0.50 -0.5% -5.2% 77 1 8/37 Uzbekistan 0.50 0.1% 7.7% 78 2 9/37 India 0.49 -0.7% -4.8% 79 0 10/37 Algeria 0.49 0.6% -84 7 11/37 Vietnam 0.49 -0.6% -2.9% 87 1 12/37 Ukraine 0.49 -2.9% -2.7% 89 11 13/37 Benin 0.48 -0.6% -90 0 14/37 Morocco 0.48 0.9% -5.0% 92 4 15/37 Tanzania 0.47 0.9% -0.9% 98 2 16/37 Philippines 0.46 -1.5% -2.0% 100 1 17/37 Kenya 0.46 1.6% 1.7% 101 5 18/37 Kyrgyz Republic 0.45 -1.4% -4.5% 103 1 19/37 Côte d'Ivoire 0.45 0.7% -5.2% 106 4 20/37 Lebanon 0.45 0.0% -5.0% 107 1 21/37 El Salvador 0.45 -2.5% -7.4% 108 4 22/37 Angola 0.43 -0.3% -115 0 23/37 Honduras 0.41 1.6% 1.6% 119 4 24/37 Nigeria 0.41 0.8% -6.3% 120 0 25/37 Congo, Rep. 0.40 -1.2% -122 0 26/37 Zimbabwe 0.40 1.3% 5.9% 123 3 27/37 Iran, Islamic Rep. 0.39 -5.0% -18.9% 126 5 28/37 Bangladesh 0.38 -1.5% -6.2% 127 2 29/37 Pakistan 0.38 -2.3% -4.2% 130 1 30/37 Bolivia 0.37 -0.5% -2.9% 131 1 31/37 Mauritania 0.36 -1.4% -133 0 32/37 Cameroon 0.35 -0.3% -4.0% 134 2 33/37 Myanmar 0.35 -3.7% -17.2% 135 1 34/37 Egypt, Arab Rep. 0.35 -1.2% -4.4% 136 1 35/37 Nicaragua 0.35 -4.4% -20.8% 137 2 36/37 Haiti 0.34 -3.5% -139 1 37/37 Cambodia 0.31 -0.1% -3.5% 141 0 Low Income Lower-Middle Income Income groups used in this year’s report are based on the World Bank’s 2023 fiscal year income group classifications. For the 2023 fiscal year, low-income economies are defined as those with a GNI per capita, calculated using the World Bank Atlas method, of $1,085 or less in 2021; lower middle-income economies are those with a GNI per capita between $1,086 and $4,255; upper middle-income economies are those with a GNI per capita between $4,256 and $13,205; high-income economies are those with a GNI per capita of more than $13,205. https:/ /datahelpdesk.worldbank.org/knowledgebase/articles/906519-world-bank-country-and-lending-groups † Scores are rounded to two decimal places. Percentage changes in score are rounded to one decimal place. ‡ The change in rankings was calculated by comparing the positions of the 140 countries and jurisdictions measured in the 2022 Index with the rankings of the same 140 countries and jurisdictions in 2023, exclusive of the two new additions to the 2023 Index, Kuwait and Montenegro. Rule of Law Around the World by Income Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank 27 WJP Rule of Law Index 2023 1/41 Costa Rica 0.68 0.0% -0.6% 29 0 2/41 St. Vincent and the Grenadines 0.63 -0.8% 2.3% 42 1 3/41 St. Lucia 0.62 0.4% -2.0% 43 0 4/41 Namibia 0.61 0.5% -44 2 5/41 Mauritius 0.61 -0.4% -46 1 6/41 Georgia 0.60 0.3% -1.2% 48 1 7/41 Grenada 0.60 1.4% -1.6% 49 1 8/41 Botswana 0.59 0.0% 1.3% 51 0 9/41 Dominica 0.58 0.1% -3.6% 53 1 10/41 Jamaica 0.57 -1.1% -0.4% 54 1 11/41 Malaysia 0.57 0.8% 6.1% 55 1 12/41 South Africa 0.57 -1.2% -3.3% 56 1 13/41 Montenegro 0.56 --57 -14/41 Kosovo 0.56 0.4% -58 1 15/41 Bulgaria 0.56 1.7% 3.9% 59 3 16/41 Jordan 0.55 1.4% -8.1% 62 1 17/41 Argentina 0.55 -0.5% -6.4% 63 2 18/41 Kazakhstan 0.53 1.0% 3.1% 65 2 19/41 North Macedonia 0.53 -0.9% -0.4% 67 2 20/41 Moldova 0.53 1.3% 7.7% 68 2 21/41 Bosnia and Herzegovina 0.51 -1.2% -3.5% 75 3 22/41 Guyana 0.50 0.2% -0.5% 76 1 23/41 Belize 0.49 0.9% 3.3% 80 7 24/41 Suriname 0.49 -0.9% -3.6% 81 0 25/41 Thailand 0.49 -1.0% -3.0% 82 0 26/41 Brazil 0.49 -0.9% -8.9% 83 0 27/41 Dominican Republic 0.49 1.0% 4.4% 86 8 28/41 Peru 0.49 0.2% -7.1% 88 4 29/41 Albania 0.48 -0.7% -5.2% 91 2 30/41 Serbia 0.48 -1.6% -4.0% 93 8 31/41 Colombia 0.48 -0.4% -4.7% 94 1 32/41 Ecuador 0.47 -2.1% -0.2% 96 1 33/41 China 0.47 -0.9% -6.7% 97 0 34/41 Paraguay 0.46 -1.2% -99 1 35/41 Belarus 0.45 -1.9% -12.0% 104 3 36/41 Guatemala 0.44 -0.3% -1.0% 111 1 37/41 Russian Federation 0.44 -2.2% -7.0% 113 4 38/41 Mexico 0.42 -1.3% -8.7% 116 1 39/41 Türkiye 0.41 -0.7% -1.3% 117 1 40/41 Gabon 0.39 0.3% -124 4 41/41 Venezuela, RB 0.26 0.9% -7.8% 142 0 1/46 Denmark 0.90 -0.3% 0.7% 1 0 2/46 Norway 0.89 0.3% 0.3% 2 0 3/46 Finland 0.87 0.4% 0.3% 3 0 4/46 Sweden 0.85 -0.4% -1.1% 4 0 5/46 Germany 0.83 0.0% -0.2% 5 1 6/46 Luxembourg 0.83 0.8% -6 2 7/46 Netherlands 0.83 -0.3% -2.5% 7 2 8/46 New Zealand 0.83 -0.2% -0.4% 8 1 9/46 Estonia 0.82 0.0% 2.3% 9 0 10/46 Ireland 0.81 0.3% -10 0 11/46 Austria 0.80 -0.3% -2.0% 11 0 12/46 Canada 0.80 0.0% -1.6% 12 0 13/46 Australia 0.80 0.5% -1.3% 13 0 14/46 Japan 0.79 0.0% -0.2% 14 2 15/46 United Kingdom 0.78 -0.4% -3.1% 15 0 16/46 Belgium 0.78 -1.0% 1.2% 16 2 17/46 Singapore 0.78 -0.1% -2.6% 17 0 18/46 Lithuania 0.77 0.4% -18 0 19/46 Korea, Rep. 0.74 0.5% 2.0% 19 0 20/46 Czechia 0.73 0.1% -1.3% 20 0 21/46 France 0.73 -0.4% -1.4% 21 0 22/46 Latvia 0.73 0.7% -22 2 23/46 Hong Kong SAR, China 0.73 -0.2% -6.0% 23 1 24/46 Spain 0.72 -0.8% 2.4% 24 1 25/46 Uruguay 0.72 0.4% 0.7% 25 0 26/46 United States 0.70 -0.6% -4.0% 26 0 27/46 Slovenia 0.69 1.6% 2.3% 27 4 28/46 Portugal 0.68 -0.9% -5.2% 28 1 29/46 Malta 0.68 0.1% -30 0 30/46 Cyprus 0.68 -0.9% -31 3 31/46 Italy 0.67 0.0% 2.0% 32 0 32/46 Chile 0.66 -0.1% -0.7% 33 0 33/46 Slovak Republic 0.66 0.3% -34 1 34/46 Barbados 0.66 0.0% 1.1% 35 1 35/46 Poland 0.64 -0.6% -5.3% 36 0 36/46 United Arab Emirates 0.64 0.2% -1.9% 37 0 37/46 Antigua and Barbuda 0.63 0.6% 0.7% 38 2 38/46 St. Kitts and Nevis 0.63 0.1% -4.3% 39 0 39/46 Romania 0.63 -0.4% -4.1% 40 2 40/46 Croatia 0.61 0.3% 0.2% 45 2 41/46 Greece 0.61 -1.4% -0.3% 47 3 42/46 The Bahamas 0.59 -2.0% -1.4% 50 2 43/46 Kuwait 0.58 --52 -44/46 Trinidad and Tobago 0.52 -0.8% -7.0% 70 1 45/46 Hungary 0.51 -0.2% -6.0% 73 2 46/46 Panama 0.51 -0.5% -1.2% 74 0 Upper-Middle Income High Income † Scores are rounded to two decimal places. Percentage changes in score are rounded to one decimal place. ‡ The change in rankings was calculated by comparing the positions of the 140 countries and jurisdictions measured in the 2022 Index with the rankings of the same 140 countries and jurisdictions in 2023, exclusive of the two new additions to the 2023 Index, Kuwait and Montenegro. Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank Country/Jurisdiction Change in Global Rank ‡ Annual % Change in Overall Score† 5-year % Change in Overall Score† Global Rank Overall Score† Income Rank 28 Denmark 0.95 1 Norway 0.92 2 Finland 0.92 3 Germany 0.86 4 Sweden 0.86 5 New Zealand 0.85 6 Netherlands 0.85 7 Austria 0.84 8 Ireland 0.83 9 Estonia 0.83 10 Canada 0.82 11 Belgium 0.82 12 Australia 0.82 13 Luxembourg 0.82 14 United Kingdom 0.81 15 Costa Rica 0.77 16 Lithuania 0.76 17 Uruguay 0.76 18 Portugal 0.76 19 Czechia 0.74 20 Japan 0.73 21 France 0.72 22 Spain 0.72 23 Italy 0.71 24 Chile 0.71 25 Latvia 0.71 26 Korea, Rep. 0.71 27 United States 0.69 28 Slovak Republic 0.67 29 Greece 0.67 30 Singapore 0.67 31 Cyprus 0.66 32 Namibia 0.66 33 Indonesia 0.66 34 Ghana 0.66 35 Slovenia 0.65 36 Barbados 0.64 37 Malta 0.64 38 Jamaica 0.64 39 South Africa 0.62 40 St. Kitts and Nevis 0.61 41 Antigua and Barbuda 0.61 42 Romania 0.61 43 Botswana 0.61 44 Rwanda 0.60 45 St. Lucia 0.60 46 The Bahamas 0.59 47 Nepal 0.59 48 Peru 0.59 49 St. Vincent and the Grenadines 0.58 50 Mauritius 0.58 51 Grenada 0.58 52 Croatia 0.58 53 United Arab Emirates 0.57 54 Malawi 0.57 55 Malaysia 0.57 56 The Gambia 0.57 57 India 0.57 58 Hong Kong SAR, China 0.57 59 Argentina 0.56 60 Kosovo 0.56 61 Kuwait 0.55 62 Senegal 0.55 63 Tunisia 0.54 64 Guyana 0.54 65 Poland 0.53 66 Panama 0.53 67 Trinidad and Tobago 0.53 68 Georgia 0.53 69 Burkina Faso 0.53 70 Mongolia 0.53 71 Montenegro 0.52 72 Dominica 0.52 73 Moldova 0.51 74 Bulgaria 0.51 75 Morocco 0.51 76 Dominican Republic 0.51 77 Brazil 0.51 78 Tanzania 0.50 79 Paraguay 0.50 80 Sri Lanka 0.50 81 Benin 0.50 82 Kenya 0.50 83 Colombia 0.50 84 Nigeria 0.50 85 Guatemala 0.49 86 Liberia 0.49 87 Ecuador 0.49 88 Lebanon 0.49 89 Suriname 0.48 90 Belize 0.48 91 Sierra Leone 0.48 92 Philippines 0.47 93 Ukraine 0.47 94 Pakistan 0.47 95 Algeria 0.47 96 North Macedonia 0.46 97 Zambia 0.46 98 Jordan 0.46 99 Thailand 0.45 100 Bosnia and Herzegovina 0.45 101 Vietnam 0.45 102 Kazakhstan 0.45 103 Mexico 0.44 104 Kyrgyz Republic 0.43 105 Madagascar 0.43 106 Albania 0.43 107 Constraints on Government Powers Factor 1 measures the extent to which those who govern are bound by law. It comprises the means, both constitutional and institutional, by which the powers of the government and its officials and agents are limited and held accountable under the law. It also includes non-governmental checks on the government’s power, such as a free and independent press. For a further breakdown of Constraints on Government Powers by sub-factor, please refer to page 16. Factor Score Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction Scores are rounded to two decimal places. Mali 0.41 108 Mozambique 0.41 109 El Salvador 0.41 110 Angola 0.41 111 Congo, Dem. Rep. 0.40 112 Niger 0.39 113 Uganda 0.39 114 Guinea 0.38 115 Honduras 0.38 116 Côte d’Ivoire 0.38 117 Togo 0.38 118 Uzbekistan 0.38 119 Bangladesh 0.37 120 Cameroon 0.37 121 Afghanistan 0.37 122 Hungary 0.37 123 Congo, Rep. 0.36 124 Haiti 0.36 125 Bolivia 0.36 126 Serbia 0.35 127 Ethiopia 0.35 128 Mauritania 0.33 129 Zimbabwe 0.33 130 Iran, Islamic Rep. 0.32 131 Russian Federation 0.32 132 Gabon 0.32 133 China 0.31 134 Myanmar 0.29 135 Sudan 0.29 136 Türkiye 0.28 137 Cambodia 0.26 138 Belarus 0.26 139 Egypt, Arab Rep. 0.24 140 Nicaragua 0.23 141 Venezuela, RB 0.18 142 29 WJP Rule of Law Index 2023 Factor Score Factor Rank Denmark 0.96 1 Norway 0.94 2 Singapore 0.91 3 Sweden 0.90 4 Finland 0.89 5 New Zealand 0.87 6 Netherlands 0.87 7 Luxembourg 0.85 8 Hong Kong SAR, China 0.83 9 Canada 0.83 10 United Kingdom 0.83 11 Germany 0.82 12 Japan 0.82 13 Ireland 0.82 14 Australia 0.81 15 Estonia 0.81 16 Austria 0.80 17 Belgium 0.79 18 United Arab Emirates 0.78 19 France 0.75 20 Uruguay 0.73 21 United States 0.73 22 Spain 0.73 23 Lithuania 0.72 24 Poland 0.72 25 Portugal 0.71 26 Barbados 0.70 27 Chile 0.69 28 St. Vincent and the Grenadines 0.68 29 Latvia 0.68 30 Malta 0.68 31 Rwanda 0.68 32 Georgia 0.68 33 Korea, Rep. 0.67 34 Slovenia 0.67 35 Kuwait 0.66 36 Czechia 0.66 37 Grenada 0.66 38 St. Kitts and Nevis 0.65 39 Cyprus 0.65 40 Italy 0.65 41 Costa Rica 0.64 42 The Bahamas 0.64 43 St. Lucia 0.64 44 Antigua and Barbuda 0.62 45 Dominica 0.60 46 Jordan 0.58 47 Mauritius 0.58 48 Botswana 0.57 49 Croatia 0.57 50 Malaysia 0.57 51 Romania 0.56 52 Greece 0.56 53 Senegal 0.55 54 Jamaica 0.53 55 Slovak Republic 0.53 56 China 0.53 57 Namibia 0.52 58 Hungary 0.50 59 Belarus 0.49 60 Trinidad and Tobago 0.49 61 Montenegro 0.48 62 Kazakhstan 0.48 63 Kosovo 0.48 64 Sri Lanka 0.47 65 Uzbekistan 0.47 66 The Gambia 0.47 67 Tunisia 0.47 68 South Africa 0.47 69 Argentina 0.46 70 Bulgaria 0.45 71 Thailand 0.45 72 Belize 0.45 73 Algeria 0.45 74 North Macedonia 0.45 75 Guyana 0.45 76 Türkiye 0.44 77 Ethiopia 0.44 78 Angola 0.44 79 Suriname 0.43 80 Brazil 0.43 81 Togo 0.43 82 Philippines 0.43 83 Burkina Faso 0.43 84 Myanmar 0.42 85 Mongolia 0.42 86 Bosnia and Herzegovina 0.42 87 Morocco 0.42 88 Vietnam 0.42 89 Malawi 0.42 90 Serbia 0.42 91 Russian Federation 0.41 92 Tanzania 0.41 93 Nepal 0.40 94 Indonesia 0.40 95 India 0.40 96 Benin 0.40 97 Panama 0.40 98 Dominican Republic 0.39 99 Niger 0.39 100 Ghana 0.39 101 Egypt, Arab Rep. 0.38 102 Colombia 0.38 103 Zambia 0.38 104 Ecuador 0.38 105 Moldova 0.38 106 Country/ Jurisdiction Factor 2 measures the absence of corruption in government. The factor considers three forms of corruption: bribery, improper influence by public or private interests, and misappropriation of public funds or other resources. These three forms of corruption are examined with respect to government officers in the executive branch, the judiciary, the military, police, and the legislature. For a further breakdown of Absence of Corruption by sub-factor, please refer to page 16. Absence of Corruption Iran, Islamic Rep. 0.37 107 Albania 0.36 108 Lebanon 0.36 109 Sierra Leone 0.36 110 Côte d’Ivoire 0.35 111 Sudan 0.35 112 Mozambique 0.35 113 El Salvador 0.35 114 Guatemala 0.34 115 Peru 0.33 116 Bangladesh 0.33 117 Ukraine 0.33 118 Honduras 0.32 119 Zimbabwe 0.32 120 Nigeria 0.32 121 Mali 0.31 122 Pakistan 0.31 123 Liberia 0.30 124 Afghanistan 0.30 125 Congo, Rep. 0.30 126 Nicaragua 0.30 127 Mauritania 0.29 128 Guinea 0.29 129 Kyrgyz Republic 0.29 130 Paraguay 0.29 131 Venezuela, RB 0.28 132 Kenya 0.27 133 Madagascar 0.27 134 Uganda 0.26 135 Mexico 0.26 136 Bolivia 0.25 137 Cameroon 0.24 138 Gabon 0.24 139 Haiti 0.24 140 Cambodia 0.23 141 Congo, Dem. Rep. 0.17 142 Scores are rounded to two decimal places. Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction 30 Factor 3 measures the openness of government defined by the extent to which a government shares information, empowers people with tools to hold the government accountable, and fosters citizen participation in public policy deliberations. This factor measures whether basic laws and information on legal rights are publicized and evaluates the quality of information published by the government. For a further breakdown of Open Government by sub-factor, please refer to page 17. Norway 0.88 1 Denmark 0.86 2 Finland 0.86 3 Sweden 0.84 4 Netherlands 0.83 5 New Zealand 0.82 6 Luxembourg 0.82 7 Australia 0.81 8 Estonia 0.81 9 Canada 0.80 10 Ireland 0.79 11 United Kingdom 0.79 12 Germany 0.79 13 Belgium 0.76 14 United States 0.76 15 France 0.75 16 Lithuania 0.75 17 Uruguay 0.73 18 Latvia 0.72 19 Korea, Rep. 0.72 20 Japan 0.70 21 Spain 0.70 22 Chile 0.70 23 Costa Rica 0.70 24 Austria 0.70 25 Hong Kong SAR, China 0.69 26 Slovak Republic 0.69 27 Czechia 0.69 28 Slovenia 0.66 29 Portugal 0.64 30 Malta 0.64 31 Italy 0.63 32 Romania 0.63 33 South Africa 0.62 34 Singapore 0.61 35 Colombia 0.61 36 Croatia 0.61 37 Argentina 0.61 38 Greece 0.61 39 Cyprus 0.60 40 Brazil 0.59 41 India 0.59 42 Georgia 0.59 43 Poland 0.58 44 Mexico 0.58 45 Namibia 0.58 46 Kosovo 0.58 47 Bulgaria 0.57 48 Moldova 0.57 49 Jamaica 0.57 50 Rwanda 0.57 51 Paraguay 0.56 52 Panama 0.56 53 Indonesia 0.55 54 Ukraine 0.55 55 Dominican Republic 0.55 56 Barbados 0.55 57 Peru 0.54 58 Montenegro 0.53 59 Trinidad and Tobago 0.53 60 Mauritius 0.53 61 Antigua and Barbuda 0.52 62 Ecuador 0.52 63 Guatemala 0.52 64 Nepal 0.52 65 St. Vincent and the Grenadines 0.51 66 Sri Lanka 0.51 67 Ghana 0.51 68 Kyrgyz Republic 0.50 69 North Macedonia 0.50 70 Kenya 0.50 71 St. Lucia 0.50 72 Dominica 0.49 73 Mongolia 0.49 74 Tunisia 0.48 75 The Bahamas 0.48 76 Thailand 0.48 77 Burkina Faso 0.48 78 Philippines 0.47 79 Albania 0.47 80 Botswana 0.47 81 El Salvador 0.47 82 Bosnia and Herzegovina 0.47 83 Kazakhstan 0.47 84 Liberia 0.46 85 Malawi 0.46 86 Grenada 0.46 87 Mali 0.45 88 Russian Federation 0.45 89 Serbia 0.45 90 Madagascar 0.45 91 Honduras 0.45 92 Guyana 0.45 93 St. Kitts and Nevis 0.45 94 Hungary 0.45 95 Vietnam 0.45 96 Belize 0.44 97 Senegal 0.44 98 Lebanon 0.44 99 Bolivia 0.44 100 Morocco 0.43 101 Kuwait 0.43 102 Malaysia 0.42 103 Nigeria 0.42 104 Pakistan 0.41 105 Bangladesh 0.40 106 Country/ Jurisdiction Türkiye 0.40 107 China 0.40 108 Jordan 0.39 109 Zambia 0.39 110 Angola 0.39 111 Uganda 0.39 112 Tanzania 0.39 113 Sierra Leone 0.39 114 Suriname 0.38 115 The Gambia 0.38 116 Benin 0.38 117 Gabon 0.37 118 Uzbekistan 0.37 119 Niger 0.37 120 Guinea 0.37 121 Mozambique 0.36 122 Algeria 0.36 123 Côte d’Ivoire 0.36 124 Afghanistan 0.35 125 Haiti 0.35 126 Sudan 0.35 127 Congo, Dem. Rep. 0.34 128 United Arab Emirates 0.34 129 Zimbabwe 0.33 130 Congo, Rep. 0.33 131 Cameroon 0.33 132 Nicaragua 0.32 133 Togo 0.32 134 Belarus 0.31 135 Ethiopia 0.31 136 Myanmar 0.30 137 Mauritania 0.28 138 Venezuela, RB 0.28 139 Iran, Islamic Rep. 0.27 140 Cambodia 0.24 141 Egypt, Arab Rep. 0.23 142 Factor Score Factor Rank Scores are rounded to two decimal places. Open Government Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction 31 WJP Rule of Law Index 2023 Factor 4 recognizes that a system of positive law that fails to respect core human rights established under international law is at best “rule by law,” and does not deserve to be called a rule of law system. Since there are many other indices that address human rights, and because it would be impossible for the Index to assess adherence to the full range of rights, this factor focuses on a relatively modest menu of rights that are firmly established under the United Nations Universal Declaration of Human Rights and are most closely related to rule of law concerns. For a further breakdown of Fundamental Rights by sub-factor, please refer to page 17. Denmark 0.92 1 Norway 0.91 2 Finland 0.90 3 Sweden 0.87 4 Germany 0.86 5 Luxembourg 0.85 6 Belgium 0.84 7 Austria 0.84 8 Netherlands 0.84 9 Estonia 0.83 10 Ireland 0.82 11 New Zealand 0.82 12 Canada 0.81 13 Uruguay 0.80 14 United Kingdom 0.80 15 Costa Rica 0.79 16 Spain 0.79 17 Lithuania 0.78 18 Japan 0.78 19 Czechia 0.78 20 Australia 0.78 21 Latvia 0.77 22 Portugal 0.76 23 Slovenia 0.75 24 Korea, Rep. 0.75 25 Malta 0.74 26 France 0.74 27 Barbados 0.73 28 Slovak Republic 0.73 29 Italy 0.73 30 Chile 0.72 31 Cyprus 0.72 32 Antigua and Barbuda 0.71 33 St. Vincent and the Grenadines 0.70 34 Argentina 0.69 35 St. Kitts and Nevis 0.69 36 Croatia 0.68 37 United States 0.68 38 Romania 0.67 39 Singapore 0.67 40 Montenegro 0.67 41 St. Lucia 0.66 42 Namibia 0.66 43 Greece 0.65 44 Mauritius 0.64 45 Jamaica 0.64 46 The Bahamas 0.64 47 South Africa 0.63 48 Panama 0.62 49 Grenada 0.62 50 Georgia 0.62 51 Dominica 0.62 52 Bulgaria 0.61 53 Poland 0.61 54 Kosovo 0.61 55 North Macedonia 0.60 56 Peru 0.60 57 Hong Kong SAR, China 0.60 58 Ukraine 0.59 59 Senegal 0.59 60 Trinidad and Tobago 0.59 61 Botswana 0.59 62 Moldova 0.59 63 Bosnia and Herzegovina 0.58 64 Malawi 0.58 65 Ghana 0.58 66 Dominican Republic 0.58 67 Albania 0.57 68 Mongolia 0.56 69 Guyana 0.56 70 Hungary 0.55 71 Serbia 0.55 72 Burkina Faso 0.53 73 Guatemala 0.53 74 Ecuador 0.53 75 Rwanda 0.52 76 Belize 0.52 77 Benin 0.52 78 The Gambia 0.52 79 Nepal 0.52 80 Suriname 0.51 81 Liberia 0.51 82 Paraguay 0.51 83 Tunisia 0.51 84 Indonesia 0.50 85 Sri Lanka 0.50 86 Mali 0.50 87 Guinea 0.50 88 Colombia 0.49 89 Malaysia 0.49 90 Brazil 0.49 91 Sierra Leone 0.48 92 Mexico 0.48 93 Kenya 0.47 94 Niger 0.47 95 Kyrgyz Republic 0.46 96 Lebanon 0.46 97 Madagascar 0.46 98 India 0.46 99 Kazakhstan 0.46 100 Thailand 0.46 101 El Salvador 0.46 102 Jordan 0.46 103 Togo 0.46 104 Côte d’Ivoire 0.45 105 Honduras 0.45 106 Kuwait 0.45 107 United Arab Emirates 0.45 108 Country/ Jurisdiction Uzbekistan 0.45 109 Vietnam 0.45 110 Bolivia 0.44 111 Tanzania 0.44 112 Algeria 0.43 113 Morocco 0.43 114 Gabon 0.43 115 Nigeria 0.42 116 Haiti 0.42 117 Congo, Dem. Rep. 0.41 118 Zambia 0.40 119 Philippines 0.40 120 Congo, Rep. 0.40 121 Mozambique 0.39 122 Mauritania 0.38 123 Russian Federation 0.38 124 Pakistan 0.38 125 Angola 0.37 126 Cameroon 0.36 127 Belarus 0.36 128 Uganda 0.35 129 Zimbabwe 0.34 130 Cambodia 0.33 131 Sudan 0.33 132 Türkiye 0.30 133 Venezuela, RB 0.30 134 Ethiopia 0.30 135 Nicaragua 0.30 136 Bangladesh 0.29 137 Afghanistan 0.28 138 China 0.25 139 Egypt, Arab Rep. 0.24 140 Myanmar 0.20 141 Iran, Islamic Rep. 0.20 142 Factor Score Factor Rank Scores are rounded to two decimal places. Fundamental Rights Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction 32 Factor 5 measures how well a society ensures the security of persons and property. Security is one of the defining aspects of any rule of law society and is a fundamental function of the state. It is also a precondition for the realization of the rights and freedoms that the rule of law seeks to advance. For a further breakdown of Order and Security by sub-factor, please refer to page 18. Luxembourg 0.95 1 Ireland 0.95 2 Singapore 0.93 3 Denmark 0.93 4 Norway 0.93 5 Hong Kong SAR, China 0.93 6 Sweden 0.92 7 Japan 0.92 8 Finland 0.92 9 United Arab Emirates 0.91 10 Malta 0.91 11 Austria 0.91 12 Slovak Republic 0.90 13 Hungary 0.90 14 Canada 0.90 15 Estonia 0.90 16 Uzbekistan 0.90 17 Slovenia 0.89 18 Czechia 0.89 19 Lithuania 0.89 20 Germany 0.89 21 New Zealand 0.88 22 Australia 0.87 23 Kuwait 0.87 24 Latvia 0.86 25 Poland 0.86 26 Rwanda 0.85 27 Netherlands 0.85 28 Korea, Rep. 0.84 29 United Kingdom 0.84 30 Croatia 0.84 31 Kosovo 0.83 32 United States 0.83 33 Romania 0.83 34 Spain 0.83 35 Montenegro 0.82 36 Belgium 0.82 37 Cyprus 0.81 38 China 0.81 39 Moldova 0.81 40 Belarus 0.81 41 Barbados 0.80 42 North Macedonia 0.80 43 Kazakhstan 0.80 44 Grenada 0.80 45 France 0.79 46 Georgia 0.79 47 Antigua and Barbuda 0.79 48 Portugal 0.78 49 Malaysia 0.78 50 Vietnam 0.78 51 Bulgaria 0.78 52 Albania 0.78 53 St. Kitts and Nevis 0.78 54 Bosnia and Herzegovina 0.77 55 Jordan 0.76 56 Serbia 0.76 57 Mongolia 0.76 58 St. Vincent and the Grenadines 0.76 59 Mauritius 0.75 60 Algeria 0.75 61 Italy 0.75 62 Namibia 0.75 63 Benin 0.74 64 Thailand 0.74 65 The Bahamas 0.73 66 St. Lucia 0.73 67 Kyrgyz Republic 0.73 68 Dominica 0.73 69 Nepal 0.73 70 Tunisia 0.72 71 Botswana 0.72 72 Greece 0.72 73 Guinea 0.72 74 Türkiye 0.72 75 Ghana 0.71 76 Belize 0.71 77 Uruguay 0.71 78 Indonesia 0.71 79 Madagascar 0.71 80 Paraguay 0.70 81 Senegal 0.70 82 Tanzania 0.70 83 Panama 0.70 84 Costa Rica 0.70 85 Togo 0.70 86 The Gambia 0.70 87 Nicaragua 0.70 88 Morocco 0.69 89 Zambia 0.69 90 Sri Lanka 0.68 91 Malawi 0.68 92 Lebanon 0.68 93 Philippines 0.67 94 Zimbabwe 0.67 95 Chile 0.67 96 Côte d’Ivoire 0.67 97 Sierra Leone 0.67 98 Cambodia 0.66 99 El Salvador 0.66 100 Russian Federation 0.66 101 Trinidad and Tobago 0.65 102 Suriname 0.65 103 Mauritania 0.65 104 India 0.64 105 Dominican Republic 0.64 106 Honduras 0.64 107 Country/ Jurisdiction Bangladesh 0.63 108 Iran, Islamic Rep. 0.63 109 Gabon 0.63 110 Guyana 0.63 111 Egypt, Arab Rep. 0.62 112 Niger 0.62 113 Jamaica 0.62 114 Peru 0.62 115 Argentina 0.61 116 Congo, Rep. 0.61 117 Ukraine 0.61 118 South Africa 0.60 119 Brazil 0.60 120 Liberia 0.60 121 Bolivia 0.59 122 Ecuador 0.59 123 Guatemala 0.59 124 Myanmar 0.58 125 Kenya 0.58 126 Angola 0.57 127 Uganda 0.57 128 Colombia 0.56 129 Sudan 0.56 130 Ethiopia 0.53 131 Venezuela, RB 0.51 132 Mexico 0.50 133 Cameroon 0.48 134 Haiti 0.48 135 Congo, Dem. Rep. 0.45 136 Burkina Faso 0.43 137 Mozambique 0.43 138 Nigeria 0.37 139 Mali 0.37 140 Pakistan 0.33 141 Afghanistan 0.30 142 Factor Score Factor Rank Scores are rounded to two decimal places. Order and Security Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction 33 WJP Rule of Law Index 2023 Factor 6 measures the extent to which regulations are fairly and effectively implemented and enforced. Regulations, both legal and administrative, structure behaviors within and outside of the government. This factor does not assess which activities a government chooses to regulate, nor does it consider how much regulation of a particular activity is appropriate. Rather, it examines how regulations are implemented and enforced. For a further breakdown of Regulatory Enforcement by sub-factor, please refer to page 18. Denmark 0.88 1 Norway 0.88 2 Finland 0.87 3 Luxembourg 0.87 4 Singapore 0.86 5 Netherlands 0.85 6 New Zealand 0.84 7 Germany 0.84 8 Sweden 0.83 9 Ireland 0.82 10 Australia 0.82 11 Canada 0.81 12 Estonia 0.81 13 Austria 0.80 14 Japan 0.80 15 Belgium 0.79 16 Hong Kong SAR, China 0.79 17 United Kingdom 0.79 18 Lithuania 0.76 19 France 0.75 20 Korea, Rep. 0.74 21 United States 0.72 22 Czechia 0.71 23 Latvia 0.71 24 Uruguay 0.71 25 United Arab Emirates 0.70 26 Spain 0.69 27 Costa Rica 0.68 28 Cyprus 0.66 29 Kuwait 0.66 30 Slovenia 0.65 31 Chile 0.64 32 Italy 0.64 33 Poland 0.63 34 Mauritius 0.62 35 Barbados 0.62 36 Slovak Republic 0.62 37 Antigua and Barbuda 0.62 38 St. Kitts and Nevis 0.62 39 St. Lucia 0.61 40 Rwanda 0.60 41 Portugal 0.60 42 Malta 0.59 43 Romania 0.59 44 Botswana 0.58 45 Namibia 0.58 46 Georgia 0.58 47 Indonesia 0.57 48 Senegal 0.56 49 Croatia 0.56 50 Greece 0.55 51 Jamaica 0.55 52 Grenada 0.55 53 Jordan 0.55 54 Malaysia 0.55 55 St. Vincent and the Grenadines 0.53 56 Ghana 0.53 57 Dominica 0.53 58 Bulgaria 0.53 59 South Africa 0.52 60 Kazakhstan 0.52 61 Côte d’Ivoire 0.52 62 Colombia 0.52 63 The Bahamas 0.51 64 Togo 0.51 65 Morocco 0.50 66 Tunisia 0.50 67 Panama 0.49 68 Argentina 0.49 69 Algeria 0.49 70 Bosnia and Herzegovina 0.49 71 Trinidad and Tobago 0.49 72 Nepal 0.49 73 China 0.49 74 Sri Lanka 0.48 75 Brazil 0.48 76 El Salvador 0.48 77 Benin 0.48 78 Mongolia 0.48 79 Kosovo 0.48 80 Peru 0.48 81 Montenegro 0.48 82 India 0.48 83 Philippines 0.47 84 Niger 0.47 85 Guyana 0.47 86 Mali 0.47 87 Ecuador 0.47 88 Burkina Faso 0.46 89 Malawi 0.46 90 Paraguay 0.46 91 Russian Federation 0.46 92 Gabon 0.46 93 North Macedonia 0.46 94 Suriname 0.46 95 Moldova 0.46 96 Serbia 0.46 97 Hungary 0.45 98 Uzbekistan 0.45 99 Kenya 0.45 100 Tanzania 0.44 101 Mexico 0.44 102 Thailand 0.44 103 Vietnam 0.44 104 Belarus 0.44 105 Iran, Islamic Rep. 0.44 106 Country/ Jurisdiction Lebanon 0.44 107 Ukraine 0.43 108 Angola 0.43 109 Congo, Rep. 0.43 110 Uganda 0.43 111 Albania 0.43 112 Zambia 0.43 113 Dominican Republic 0.42 114 Myanmar 0.42 115 Türkiye 0.42 116 Kyrgyz Republic 0.41 117 Belize 0.41 118 Nigeria 0.41 119 Mozambique 0.41 120 Cameroon 0.40 121 Guatemala 0.40 122 Liberia 0.40 123 Honduras 0.39 124 Bolivia 0.39 125 Bangladesh 0.38 126 Madagascar 0.38 127 Pakistan 0.38 128 The Gambia 0.37 129 Sierra Leone 0.37 130 Guinea 0.37 131 Ethiopia 0.36 132 Egypt, Arab Rep. 0.36 133 Nicaragua 0.36 134 Zimbabwe 0.35 135 Afghanistan 0.35 136 Congo, Dem. Rep. 0.34 137 Sudan 0.33 138 Mauritania 0.28 139 Haiti 0.28 140 Cambodia 0.26 141 Venezuela, RB 0.19 142 Factor Score Factor Rank Scores are rounded to two decimal places. Regulatory Enforcement Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction 34 Factor 7 measures whether ordinary people can resolve their grievances peacefully and effectively through the civil justice system. It measures whether civil justice systems are accessible and affordable as well as free of discrimination, corruption, and improper influence by public officials. It examines whether court proceedings are conducted without unreasonable delays and whether decisions are enforced effectively. It also measures the accessibility, impartiality, and effectiveness of alternative dispute resolution mechanisms. For a further breakdown of Civil Justice by sub-factor, please refer to page 18. Norway 0.86 1 Denmark 0.86 2 Netherlands 0.84 3 Germany 0.83 4 Sweden 0.82 5 Finland 0.81 6 Estonia 0.81 7 Lithuania 0.79 8 Singapore 0.79 9 Luxembourg 0.78 10 New Zealand 0.78 11 Japan 0.76 12 Korea, Rep. 0.75 13 Australia 0.75 14 Belgium 0.74 15 Austria 0.74 16 Ireland 0.73 17 Uruguay 0.72 18 United Kingdom 0.71 19 Hong Kong SAR, China 0.71 20 Czechia 0.69 21 France 0.69 22 Latvia 0.69 23 Canada 0.69 24 St. Kitts and Nevis 0.68 25 Antigua and Barbuda 0.68 26 Slovenia 0.67 27 United Arab Emirates 0.66 28 St. Lucia 0.65 29 Spain 0.65 30 Rwanda 0.65 31 Portugal 0.65 32 St. Vincent and the Grenadines 0.64 33 Mauritius 0.63 34 Romania 0.63 35 Kazakhstan 0.63 36 Barbados 0.63 37 United States 0.62 38 Namibia 0.62 39 Cyprus 0.62 40 Chile 0.62 41 Malaysia 0.62 42 Grenada 0.61 43 Jordan 0.61 44 Poland 0.61 45 Botswana 0.61 46 Costa Rica 0.61 47 Malta 0.60 48 South Africa 0.59 49 Greece 0.58 50 Italy 0.58 51 Kuwait 0.58 52 The Bahamas 0.57 53 Ghana 0.57 54 Dominica 0.57 55 Malawi 0.56 56 Senegal 0.56 57 Croatia 0.56 58 Trinidad and Tobago 0.56 59 Belarus 0.56 60 Jamaica 0.56 61 Algeria 0.55 62 Argentina 0.55 63 Slovak Republic 0.55 64 Bulgaria 0.54 65 Mongolia 0.54 66 Montenegro 0.54 67 Georgia 0.53 68 Ukraine 0.53 69 Belize 0.52 70 Guyana 0.52 71 Morocco 0.52 72 China 0.52 73 Uzbekistan 0.52 74 North Macedonia 0.52 75 Iran, Islamic Rep. 0.51 76 Côte d’Ivoire 0.51 77 Russian Federation 0.51 78 Moldova 0.50 79 Kosovo 0.50 80 Tunisia 0.49 81 El Salvador 0.49 82 The Gambia 0.49 83 Brazil 0.49 84 Kenya 0.49 85 Suriname 0.49 86 Thailand 0.49 87 Togo 0.48 88 Kyrgyz Republic 0.48 89 Tanzania 0.47 90 Colombia 0.47 91 Serbia 0.47 92 Indonesia 0.47 93 Zimbabwe 0.46 94 Panama 0.46 95 Ecuador 0.46 96 Bosnia and Herzegovina 0.46 97 Zambia 0.46 98 Niger 0.46 99 Nigeria 0.46 100 Burkina Faso 0.45 101 Albania 0.45 102 Philippines 0.45 103 Vietnam 0.45 104 Hungary 0.45 105 Congo, Rep. 0.44 106 Nepal 0.44 107 Angola 0.44 108 Sri Lanka 0.43 109 Dominican Republic 0.43 110 India 0.43 111 Benin 0.43 112 Madagascar 0.43 113 Sierra Leone 0.43 114 Uganda 0.42 115 Mozambique 0.42 116 Paraguay 0.42 117 Ethiopia 0.42 118 Türkiye 0.41 119 Mali 0.41 120 Peru 0.41 121 Cameroon 0.40 122 Honduras 0.40 123 Lebanon 0.40 124 Liberia 0.40 125 Mauritania 0.40 126 Gabon 0.40 127 Guinea 0.39 128 Pakistan 0.38 129 Egypt, Arab Rep. 0.38 130 Mexico 0.37 131 Congo, Dem. Rep. 0.36 132 Bangladesh 0.36 133 Sudan 0.36 134 Haiti 0.35 135 Guatemala 0.33 136 Afghanistan 0.33 137 Myanmar 0.32 138 Nicaragua 0.32 139 Bolivia 0.31 140 Venezuela, RB 0.26 141 Cambodia 0.25 142 Country/ Jurisdiction Factor Score Factor Rank Scores are rounded to two decimal places. Civil Justice Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction 35 WJP Rule of Law Index 2023 Factor 8 evaluates a country’s criminal justice system. An effective criminal justice system is a key aspect of the rule of law, as it constitutes the conventional mechanism to redress grievances and bring action against individuals for offenses against society. An assessment of the delivery of criminal justice should take into consideration the entire system, including the police, lawyers, prosecutors, judges, and prison officers. For a further breakdown of Criminal Justice by sub-factor, please refer to page 19. Scores are rounded to two decimal places. Finland 0.84 1 Denmark 0.83 2 Norway 0.83 3 Sweden 0.79 4 Austria 0.79 5 Germany 0.78 6 Singapore 0.77 7 Japan 0.76 8 Estonia 0.75 9 Netherlands 0.74 10 New Zealand 0.73 11 Canada 0.73 12 Luxembourg 0.73 13 Australia 0.73 14 Ireland 0.72 15 Korea, Rep. 0.71 16 Belgium 0.71 17 United Kingdom 0.70 18 Czechia 0.70 19 Hong Kong SAR, China 0.69 20 Lithuania 0.69 21 Latvia 0.68 22 Cyprus 0.68 23 United Arab Emirates 0.68 24 Spain 0.66 25 Italy 0.64 26 Malta 0.63 27 France 0.63 28 United States 0.60 29 St. Vincent and the Grenadines 0.60 30 St. Kitts and Nevis 0.59 31 Botswana 0.59 32 The Bahamas 0.58 33 Barbados 0.58 34 Slovak Republic 0.58 35 Costa Rica 0.58 36 Poland 0.58 37 Uruguay 0.58 38 Jordan 0.57 39 Slovenia 0.56 40 Portugal 0.56 41 Rwanda 0.56 42 Malaysia 0.56 43 Dominica 0.56 44 Namibia 0.55 45 St. Lucia 0.54 46 Mauritius 0.54 47 Chile 0.54 48 Grenada 0.54 49 Antigua and Barbuda 0.53 50 Suriname 0.53 51 Romania 0.52 52 Georgia 0.52 53 Croatia 0.51 54 South Africa 0.50 55 Greece 0.50 56 Jamaica 0.49 57 Mongolia 0.48 58 Burkina Faso 0.48 59 Montenegro 0.47 60 Kazakhstan 0.47 61 Kuwait 0.47 62 Bosnia and Herzegovina 0.47 63 Senegal 0.47 64 Kosovo 0.46 65 Vietnam 0.46 66 Hungary 0.45 67 Nepal 0.45 68 Ghana 0.45 69 Bulgaria 0.44 70 Uzbekistan 0.44 71 North Macedonia 0.44 72 Malawi 0.44 73 China 0.43 74 Benin 0.43 75 Thailand 0.41 76 Tunisia 0.41 77 Moldova 0.41 78 Algeria 0.41 79 Sri Lanka 0.41 80 Guyana 0.40 81 The Gambia 0.40 82 Indonesia 0.40 83 Zambia 0.40 84 Belize 0.39 85 Nigeria 0.39 86 Argentina 0.39 87 Kenya 0.39 88 Belarus 0.39 89 Serbia 0.39 90 Albania 0.38 91 Dominican Republic 0.38 92 India 0.37 93 Tanzania 0.37 94 Morocco 0.37 95 Togo 0.36 96 Ukraine 0.36 97 Zimbabwe 0.36 98 Pakistan 0.36 99 Côte d’Ivoire 0.36 100 Angola 0.35 101 Niger 0.35 102 Sierra Leone 0.35 103 Ethiopia 0.35 104 Sudan 0.34 105 Panama 0.34 106 Türkiye 0.34 107 Congo, Rep. 0.34 108 Ecuador 0.33 109 Iran, Islamic Rep. 0.33 110 Egypt, Arab Rep. 0.33 111 Peru 0.33 112 Liberia 0.33 113 Brazil 0.32 114 Colombia 0.32 115 Kyrgyz Republic 0.32 116 Madagascar 0.32 117 Lebanon 0.32 118 Uganda 0.32 119 Philippines 0.31 120 Trinidad and Tobago 0.31 121 Mozambique 0.31 122 Bangladesh 0.30 123 Guatemala 0.30 124 Guinea 0.29 125 Gabon 0.29 126 Russian Federation 0.29 127 Congo, Dem. Rep. 0.28 128 Mali 0.28 129 Mauritania 0.27 130 Paraguay 0.26 131 Mexico 0.26 132 Honduras 0.26 133 Cambodia 0.26 134 Afghanistan 0.25 135 Myanmar 0.25 136 Nicaragua 0.25 137 Haiti 0.25 138 El Salvador 0.25 139 Cameroon 0.24 140 Bolivia 0.21 141 Venezuela, RB 0.12 142 Country/ Jurisdiction Factor Score Factor Rank Criminal Justice Factor Score Factor Score Factor Score Factor Rank Factor Rank Factor Rank Country/ Jurisdiction Country/ Jurisdiction Country/ Jurisdiction 36 37 WJP Rule of Law Index 2023 38 How to Read the Country Profiles 39 Country Profiles 38 38 How to Read the Country Profiles This section presents profiles for the 142 countries and jurisdictions included in the WJP Rule of Law Index® 2023 report. Each profile presents the featured country’s scores for each of the WJP Rule of Law Index’s factors and sub-factors and draws comparisons between the scores of the featured country and the scores of other indexed countries in the same regional and income groups. The scores range from 0 to 1, where 1 signifies the highest possible score (strong adherence to rule of law) and 0 signifies the lowest possible score (weak adherence to rule of law). The country profiles consist of four sections, outlined below. Indonesia Region: East Asia and PaciØc Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.53 9/15 4/37 66/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.66 0.00 6/15 1/37 34/142 Absence of Corruption 0.40 0.00 14/15 14/37 95/142 Open Government 0.55 0.00 7/15 2/37 54/142 Fundamental Rights 0.50 0.00 8/15 8/37 85/142 Order and Security 0.71 0.00 12/15 10/37 79/142 Regulatory Enforcement 0.57 0.00 7/15 1/37 48/142 Civil Justice 0.47 0.00 11/15 15/37 93/142 Criminal Justice 0.40 0.01 12/15 11/37 83/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Indonesia East Asia and PaciØc Lower-Middle Constraints on Government Powers 1.1 0.78 Limits by legislature 1.2 0.67 Limits by judiciary 1.3 0.58 Independent auditing 1.4 0.60 Sanctions for ofØcial misconduct 1.5 0.64 Non-governmental checks 1.6 0.70 Lawful transition of power Absence of Corruption 2.1 0.49 In the executive branch 2.2 0.33 In the judiciary 2.3 0.49 In the police/military 2.4 0.30 In the legislature Open Government 3.1 0.39 Publicized laws and gov't data 3.2 0.57 Right to information 3.3 0.61 Civic participation 3.4 0.64 Complaint mechanisms Fundamental Rights 4.1 0.46 No discrimination 4.2 0.48 Right to life and security 4.3 0.40 Due process of law 4.4 0.64 Freedom of expression 4.5 0.42 Freedom of religion 4.6 0.36 Right to privacy 4.7 0.63 Freedom of association 4.8 0.63 Labor rights Order and Security 5.1 0.85 Absence of crime 5.2 0.85 Absence of civil conÙict 5.3 0.41 Absence of violent redress Regulatory Enforcement 6.1 0.58 Effective regulatory enforcement 6.2 0.68 No improper inÙuence 6.3 0.50 No unreasonable delay 6.4 0.48 Respect for due process 6.5 0.60 No expropriation w/out adequate compensation Civil Justice 7.1 0.51 Accessibility and affordability 7.2 0.33 No discrimination 7.3 0.43 No corruption 7.4 0.50 No improper gov't inÙuence 7.5 0.52 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.56 Impartial and effective ADRs Criminal Justice 8.1 0.35 Effective investigations 8.2 0.54 Timely and effective adjudication 8.3 0.33 Effective correctional system 8.4 0.26 No discrimination 8.5 0.46 No corruption 8.6 0.45 No improper gov't inÙuence 8.7 0.40 Due process of law Displays the country’s overall rule of law score; its overall global, income, and regional ranks; and its change in score and rank from the 2022 edition of the Index. Section 1 Section 2 Section 3 Section 4 Displays the featured country’s individual factor scores, along with its global, regional, and income group rankings. The global, regional, and income rankings are distributed across three tiers — high, medium, and low — as indicated by the color of the box where the score is found. Displays the country’s disaggregated scores for each of the sub-factors that compose the WJP Rule of Law Index. The featured country’s score is represented by the purple bar and labeled at the end of the bar. The average score of the country’s region is represented by the orange line. The average score of the country’s income group is represent-ed by the green line. Presents the individual sub-factor scores underlying each of the factors listed in Section 3 of the country profile. Each of the 44 sub-factors is repre-sented by a gray line drawn from the center to the periphery of the circle. The center of the circle corresponds to the worst possible score for each sub-factor (0), and the outer edge of the circle marks the best possible score for each sub-factor (1). The featured country’s scores for 2023 are represented by the purple line. The featured country’s scores for 2022 are represented by the gray line. Afghanistan Region: South Asia Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.32 6/6 18/18 140/142 Score Change Rank Change -0.01 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.37 -0.03 6/6 16/18 122/142 Absence of Corruption 0.30 0.00 6/6 14/18 125/142 Open Government 0.35 -0.02 6/6 14/18 125/142 Fundamental Rights 0.28 -0.04 6/6 18/18 138/142 Order and Security 0.30 0.00 6/6 18/18 142/142 Regulatory Enforcement 0.35 -0.01 6/6 16/18 136/142 Civil Justice 0.33 -0.01 6/6 18/18 137/142 Criminal Justice 0.25 -0.01 6/6 18/18 135/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Afghanistan South Asia Low Constraints on Government Powers 1.1 0.50 Limits by legislature 1.2 0.31 Limits by judiciary 1.3 0.37 Independent auditing 1.4 0.31 Sanctions for ofØcial misconduct 1.5 0.46 Non-governmental checks 1.6 0.25 Lawful transition of power Absence of Corruption 2.1 0.36 In the executive branch 2.2 0.20 In the judiciary 2.3 0.42 In the police/military 2.4 0.24 In the legislature Open Government 3.1 0.21 Publicized laws and gov't data 3.2 0.38 Right to information 3.3 0.46 Civic participation 3.4 0.36 Complaint mechanisms Fundamental Rights 4.1 0.25 No discrimination 4.2 0.13 Right to life and security 4.3 0.25 Due process of law 4.4 0.46 Freedom of expression 4.5 0.19 Freedom of religion 4.6 0.17 Right to privacy 4.7 0.45 Freedom of association 4.8 0.37 Labor rights Order and Security 5.1 0.47 Absence of crime 5.2 0.06 Absence of civil conÙict 5.3 0.37 Absence of violent redress Regulatory Enforcement 6.1 0.32 Effective regulatory enforcement 6.2 0.42 No improper inÙuence 6.3 0.41 No unreasonable delay 6.4 0.21 Respect for due process 6.5 0.39 No expropriation w/out adequate compensation Civil Justice 7.1 0.42 Accessibility and affordability 7.2 0.11 No discrimination 7.3 0.20 No corruption 7.4 0.26 No improper gov't inÙuence 7.5 0.38 No unreasonable delay 7.6 0.52 Effective enforcement 7.7 0.44 Impartial and effective ADRs Criminal Justice 8.1 0.30 Effective investigations 8.2 0.40 Timely and effective adjudication 8.3 0.23 Effective correctional system 8.4 0.20 No discrimination 8.5 0.27 No corruption 8.6 0.12 No improper gov't inÙuence 8.7 0.25 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 39 WJP Rule of Law Index 2023 Albania Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.48 10/15 29/41 91/142 Score Change Rank Change 0.00 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.43 0.00 10/15 34/41 107/142 Absence of Corruption 0.36 0.00 13/15 35/41 108/142 Open Government 0.47 0.01 8/15 25/41 80/142 Fundamental Rights 0.57 -0.01 8/15 21/41 68/142 Order and Security 0.78 0.00 9/15 12/41 53/142 Regulatory Enforcement 0.43 0.00 13/15 36/41 112/142 Civil Justice 0.45 -0.01 14/15 33/41 102/142 Criminal Justice 0.38 -0.02 11/15 29/41 91/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Albania Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.45 Limits by legislature 1.2 0.29 Limits by judiciary 1.3 0.50 Independent auditing 1.4 0.37 Sanctions for ofØcial misconduct 1.5 0.50 Non-governmental checks 1.6 0.45 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.35 In the judiciary 2.3 0.52 In the police/military 2.4 0.21 In the legislature Open Government 3.1 0.46 Publicized laws and gov't data 3.2 0.46 Right to information 3.3 0.43 Civic participation 3.4 0.53 Complaint mechanisms Fundamental Rights 4.1 0.53 No discrimination 4.2 0.73 Right to life and security 4.3 0.50 Due process of law 4.4 0.50 Freedom of expression 4.5 0.76 Freedom of religion 4.6 0.58 Right to privacy 4.7 0.51 Freedom of association 4.8 0.48 Labor rights Order and Security 5.1 0.88 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.47 No improper inÙuence 6.3 0.50 No unreasonable delay 6.4 0.28 Respect for due process 6.5 0.42 No expropriation w/out adequate compensation Civil Justice 7.1 0.56 Accessibility and affordability 7.2 0.47 No discrimination 7.3 0.31 No corruption 7.4 0.32 No improper gov't inÙuence 7.5 0.36 No unreasonable delay 7.6 0.51 Effective enforcement 7.7 0.63 Impartial and effective ADRs Criminal Justice 8.1 0.43 Effective investigations 8.2 0.35 Timely and effective adjudication 8.3 0.36 Effective correctional system 8.4 0.45 No discrimination 8.5 0.33 No corruption 8.6 0.28 No improper gov't inÙuence 8.7 0.50 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 40 Algeria Region: Middle East and North Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 5/9 10/37 84/142 Score Change Rank Change 0.00 7 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.47 0.00 6/9 18/37 96/142 Absence of Corruption 0.45 0.01 5/9 5/37 74/142 Open Government 0.36 0.01 6/9 26/37 123/142 Fundamental Rights 0.43 0.00 6/9 21/37 113/142 Order and Security 0.75 0.00 4/9 4/37 61/142 Regulatory Enforcement 0.49 0.01 6/9 7/37 70/142 Civil Justice 0.55 0.00 4/9 3/37 62/142 Criminal Justice 0.41 0.01 5/9 9/37 79/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Algeria Middle East and North Africa Lower-Middle Constraints on Government Powers 1.1 0.49 Limits by legislature 1.2 0.42 Limits by judiciary 1.3 0.49 Independent auditing 1.4 0.53 Sanctions for ofØcial misconduct 1.5 0.39 Non-governmental checks 1.6 0.47 Lawful transition of power Absence of Corruption 2.1 0.40 In the executive branch 2.2 0.56 In the judiciary 2.3 0.46 In the police/military 2.4 0.39 In the legislature Open Government 3.1 0.24 Publicized laws and gov't data 3.2 0.34 Right to information 3.3 0.37 Civic participation 3.4 0.50 Complaint mechanisms Fundamental Rights 4.1 0.55 No discrimination 4.2 0.47 Right to life and security 4.3 0.47 Due process of law 4.4 0.39 Freedom of expression 4.5 0.21 Freedom of religion 4.6 0.47 Right to privacy 4.7 0.40 Freedom of association 4.8 0.49 Labor rights Order and Security 5.1 0.67 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.59 Absence of violent redress Regulatory Enforcement 6.1 0.44 Effective regulatory enforcement 6.2 0.53 No improper inÙuence 6.3 0.52 No unreasonable delay 6.4 0.42 Respect for due process 6.5 0.54 No expropriation w/out adequate compensation Civil Justice 7.1 0.60 Accessibility and affordability 7.2 0.44 No discrimination 7.3 0.61 No corruption 7.4 0.42 No improper gov't inÙuence 7.5 0.73 No unreasonable delay 7.6 0.50 Effective enforcement 7.7 0.56 Impartial and effective ADRs Criminal Justice 8.1 0.27 Effective investigations 8.2 0.45 Timely and effective adjudication 8.3 0.46 Effective correctional system 8.4 0.60 No discrimination 8.5 0.42 No corruption 8.6 0.20 No improper gov't inÙuence 8.7 0.47 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 41 WJP Rule of Law Index 2023 Angola Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.43 21/34 22/37 115/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.41 0.00 21/34 22/37 111/142 Absence of Corruption 0.44 0.00 9/34 6/37 79/142 Open Government 0.39 0.01 16/34 22/37 111/142 Fundamental Rights 0.37 -0.01 29/34 29/37 126/142 Order and Security 0.57 -0.02 25/34 33/37 127/142 Regulatory Enforcement 0.43 0.01 18/34 22/37 109/142 Civil Justice 0.44 -0.01 20/34 22/37 108/142 Criminal Justice 0.35 0.01 19/34 21/37 101/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Angola Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.50 Limits by legislature 1.2 0.40 Limits by judiciary 1.3 0.39 Independent auditing 1.4 0.35 Sanctions for of cial misconduct 1.5 0.39 Non-governmental checks 1.6 0.41 Lawful transition of power Absence of Corruption 2.1 0.36 In the executive branch 2.2 0.51 In the judiciary 2.3 0.48 In the police/military 2.4 0.40 In the legislature Open Government 3.1 0.17 Publicized laws and gov't data 3.2 0.43 Right to information 3.3 0.38 Civic participation 3.4 0.59 Complaint mechanisms Fundamental Rights 4.1 0.44 No discrimination 4.2 0.24 Right to life and security 4.3 0.33 Due process of law 4.4 0.39 Freedom of expression 4.5 0.49 Freedom of religion 4.6 0.18 Right to privacy 4.7 0.41 Freedom of association 4.8 0.47 Labor rights Order and Security 5.1 0.48 Absence of crime 5.2 0.94 Absence of civil con ct 5.3 0.31 Absence of violent redress Regulatory Enforcement 6.1 0.42 Effective regulatory enforcement 6.2 0.50 No improper in ence 6.3 0.33 No unreasonable delay 6.4 0.48 Respect for due process 6.5 0.41 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.53 No discrimination 7.3 0.46 No corruption 7.4 0.31 No improper gov't in ence 7.5 0.28 No unreasonable delay 7.6 0.37 Effective enforcement 7.7 0.60 Impartial and effective ADRs Criminal Justice 8.1 0.37 Effective investigations 8.2 0.36 Timely and effective adjudication 8.3 0.25 Effective correctional system 8.4 0.38 No discrimination 8.5 0.50 No corruption 8.6 0.28 No improper gov't in ence 8.7 0.33 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 42 Antigua and Barbuda Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.63 5/32 37/46 38/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.61 0.00 7/32 36/46 42/142 Absence of Corruption 0.62 0.01 10/32 39/46 45/142 Open Government 0.52 0.00 15/32 41/46 62/142 Fundamental Rights 0.71 0.01 5/32 32/46 33/142 Order and Security 0.79 -0.01 3/32 37/46 48/142 Regulatory Enforcement 0.62 0.01 5/32 36/46 38/142 Civil Justice 0.68 -0.01 3/32 26/46 26/142 Criminal Justice 0.53 0.01 11/32 39/46 50/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Antigua and Barbuda Latin America and Caribbean High Constraints on Government Powers 1.1 0.56 Limits by legislature 1.2 0.73 Limits by judiciary 1.3 0.55 Independent auditing 1.4 0.42 Sanctions for of cial misconduct 1.5 0.63 Non-governmental checks 1.6 0.77 Lawful transition of power Absence of Corruption 2.1 0.58 In the executive branch 2.2 0.80 In the judiciary 2.3 0.74 In the police/military 2.4 0.38 In the legislature Open Government 3.1 0.30 Publicized laws and gov't data 3.2 0.56 Right to information 3.3 0.66 Civic participation 3.4 0.56 Complaint mechanisms Fundamental Rights 4.1 0.71 No discrimination 4.2 0.82 Right to life and security 4.3 0.58 Due process of law 4.4 0.63 Freedom of expression 4.5 0.65 Freedom of religion 4.6 0.73 Right to privacy 4.7 0.72 Freedom of association 4.8 0.85 Labor rights Order and Security 5.1 0.80 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.55 Absence of violent redress Regulatory Enforcement 6.1 0.54 Effective regulatory enforcement 6.2 0.78 No improper in ence 6.3 0.50 No unreasonable delay 6.4 0.58 Respect for due process 6.5 0.68 No expropriation w/out adequate compensation Civil Justice 7.1 0.73 Accessibility and affordability 7.2 0.66 No discrimination 7.3 0.82 No corruption 7.4 0.76 No improper gov't in ence 7.5 0.47 No unreasonable delay 7.6 0.52 Effective enforcement 7.7 0.77 Impartial and effective ADRs Criminal Justice 8.1 0.48 Effective investigations 8.2 0.53 Timely and effective adjudication 8.3 0.38 Effective correctional system 8.4 0.48 No discrimination 8.5 0.71 No corruption 8.6 0.56 No improper gov't in ence 8.7 0.58 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 43 WJP Rule of Law Index 2023 Argentina Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.55 13/32 17/41 63/142 Score Change Rank Change 0.00 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.56 0.00 13/32 12/41 60/142 Absence of Corruption 0.46 0.00 14/32 19/41 70/142 Open Government 0.61 -0.01 5/32 4/41 38/142 Fundamental Rights 0.69 0.01 7/32 3/41 35/142 Order and Security 0.61 0.00 24/32 34/41 116/142 Regulatory Enforcement 0.49 0.00 15/32 17/41 69/142 Civil Justice 0.55 0.00 14/32 15/41 63/142 Criminal Justice 0.39 -0.02 16/32 26/41 87/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Argentina Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.61 Limits by legislature 1.2 0.43 Limits by judiciary 1.3 0.54 Independent auditing 1.4 0.36 Sanctions for ofØcial misconduct 1.5 0.72 Non-governmental checks 1.6 0.72 Lawful transition of power Absence of Corruption 2.1 0.40 In the executive branch 2.2 0.63 In the judiciary 2.3 0.59 In the police/military 2.4 0.21 In the legislature Open Government 3.1 0.67 Publicized laws and gov't data 3.2 0.49 Right to information 3.3 0.69 Civic participation 3.4 0.58 Complaint mechanisms Fundamental Rights 4.1 0.62 No discrimination 4.2 0.79 Right to life and security 4.3 0.55 Due process of law 4.4 0.72 Freedom of expression 4.5 0.79 Freedom of religion 4.6 0.64 Right to privacy 4.7 0.78 Freedom of association 4.8 0.65 Labor rights Order and Security 5.1 0.57 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.28 Absence of violent redress Regulatory Enforcement 6.1 0.50 Effective regulatory enforcement 6.2 0.60 No improper inÙuence 6.3 0.42 No unreasonable delay 6.4 0.47 Respect for due process 6.5 0.48 No expropriation w/out adequate compensation Civil Justice 7.1 0.72 Accessibility and affordability 7.2 0.64 No discrimination 7.3 0.53 No corruption 7.4 0.44 No improper gov't inÙuence 7.5 0.28 No unreasonable delay 7.6 0.51 Effective enforcement 7.7 0.72 Impartial and effective ADRs Criminal Justice 8.1 0.29 Effective investigations 8.2 0.33 Timely and effective adjudication 8.3 0.28 Effective correctional system 8.4 0.50 No discrimination 8.5 0.45 No corruption 8.6 0.34 No improper gov't inÙuence 8.7 0.55 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 44 Australia Region: East Asia and Paci c Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.80 2/15 13/46 13/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.82 0.00 2/15 13/46 13/142 Absence of Corruption 0.81 0.00 5/15 15/46 15/142 Open Government 0.81 0.01 2/15 8/46 8/142 Fundamental Rights 0.78 0.00 3/15 20/46 21/142 Order and Security 0.87 0.00 5/15 22/46 23/142 Regulatory Enforcement 0.82 0.01 3/15 11/46 11/142 Civil Justice 0.75 0.01 5/15 14/46 14/142 Criminal Justice 0.73 0.00 4/15 14/46 14/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Australia East Asia and Paci c High Constraints on Government Powers 1.1 0.85 Limits by legislature 1.2 0.83 Limits by judiciary 1.3 0.76 Independent auditing 1.4 0.75 Sanctions for of cial misconduct 1.5 0.79 Non-governmental checks 1.6 0.91 Lawful transition of power Absence of Corruption 2.1 0.78 In the executive branch 2.2 0.95 In the judiciary 2.3 0.90 In the police/military 2.4 0.61 In the legislature Open Government 3.1 0.90 Publicized laws and gov't data 3.2 0.68 Right to information 3.3 0.78 Civic participation 3.4 0.87 Complaint mechanisms Fundamental Rights 4.1 0.66 No discrimination 4.2 0.89 Right to life and security 4.3 0.76 Due process of law 4.4 0.79 Freedom of expression 4.5 0.83 Freedom of religion 4.6 0.81 Right to privacy 4.7 0.81 Freedom of association 4.8 0.70 Labor rights Order and Security 5.1 0.90 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.72 Absence of violent redress Regulatory Enforcement 6.1 0.71 Effective regulatory enforcement 6.2 0.91 No improper in ence 6.3 0.75 No unreasonable delay 6.4 0.86 Respect for due process 6.5 0.86 No expropriation w/out adequate compensation Civil Justice 7.1 0.59 Accessibility and affordability 7.2 0.65 No discrimination 7.3 0.87 No corruption 7.4 0.89 No improper gov't in ence 7.5 0.67 No unreasonable delay 7.6 0.78 Effective enforcement 7.7 0.81 Impartial and effective ADRs Criminal Justice 8.1 0.66 Effective investigations 8.2 0.70 Timely and effective adjudication 8.3 0.63 Effective correctional system 8.4 0.58 No discrimination 8.5 0.85 No corruption 8.6 0.91 No improper gov't in ence 8.7 0.76 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 45 WJP Rule of Law Index 2023 Austria Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.80 10/31 11/46 11/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.84 0.00 7/31 8/46 8/142 Absence of Corruption 0.80 0.00 12/31 17/46 17/142 Open Government 0.70 -0.01 18/31 24/46 25/142 Fundamental Rights 0.84 0.00 8/31 8/46 8/142 Order and Security 0.91 0.00 8/31 12/46 12/142 Regulatory Enforcement 0.80 -0.01 11/31 14/46 14/142 Civil Justice 0.74 -0.01 11/31 16/46 16/142 Criminal Justice 0.79 0.00 5/31 5/46 5/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Austria EU, EFTA, and North America High Constraints on Government Powers 1.1 0.83 Limits by legislature 1.2 0.82 Limits by judiciary 1.3 0.81 Independent auditing 1.4 0.79 Sanctions for of cial misconduct 1.5 0.83 Non-governmental checks 1.6 0.94 Lawful transition of power Absence of Corruption 2.1 0.74 In the executive branch 2.2 0.94 In the judiciary 2.3 0.91 In the police/military 2.4 0.61 In the legislature Open Government 3.1 0.71 Publicized laws and gov't data 3.2 0.61 Right to information 3.3 0.79 Civic participation 3.4 0.68 Complaint mechanisms Fundamental Rights 4.1 0.70 No discrimination 4.2 0.95 Right to life and security 4.3 0.84 Due process of law 4.4 0.83 Freedom of expression 4.5 0.83 Freedom of religion 4.6 0.88 Right to privacy 4.7 0.87 Freedom of association 4.8 0.83 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.81 Absence of violent redress Regulatory Enforcement 6.1 0.84 Effective regulatory enforcement 6.2 0.91 No improper in ence 6.3 0.66 No unreasonable delay 6.4 0.78 Respect for due process 6.5 0.80 No expropriation w/out adequate compensation Civil Justice 7.1 0.68 Accessibility and affordability 7.2 0.67 No discrimination 7.3 0.87 No corruption 7.4 0.84 No improper gov't in ence 7.5 0.62 No unreasonable delay 7.6 0.78 Effective enforcement 7.7 0.70 Impartial and effective ADRs Criminal Justice 8.1 0.64 Effective investigations 8.2 0.78 Timely and effective adjudication 8.3 0.82 Effective correctional system 8.4 0.69 No discrimination 8.5 0.87 No corruption 8.6 0.86 No improper gov't in ence 8.7 0.84 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 46 The Bahamas Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.59 10/32 42/46 50/142 Score Change Rank Change -0.01 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.59 0.00 9/32 38/46 47/142 Absence of Corruption 0.64 -0.01 8/32 38/46 43/142 Open Government 0.48 -0.01 21/32 42/46 76/142 Fundamental Rights 0.64 -0.02 11/32 39/46 47/142 Order and Security 0.73 -0.01 6/32 41/46 66/142 Regulatory Enforcement 0.51 -0.01 13/32 43/46 64/142 Civil Justice 0.57 -0.01 10/32 41/46 53/142 Criminal Justice 0.58 -0.02 3/32 31/46 33/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 The Bahamas Latin America and Caribbean High Constraints on Government Powers 1.1 0.59 Limits by legislature 1.2 0.60 Limits by judiciary 1.3 0.52 Independent auditing 1.4 0.45 Sanctions for ofØcial misconduct 1.5 0.64 Non-governmental checks 1.6 0.77 Lawful transition of power Absence of Corruption 2.1 0.50 In the executive branch 2.2 0.84 In the judiciary 2.3 0.81 In the police/military 2.4 0.39 In the legislature Open Government 3.1 0.30 Publicized laws and gov't data 3.2 0.44 Right to information 3.3 0.64 Civic participation 3.4 0.54 Complaint mechanisms Fundamental Rights 4.1 0.57 No discrimination 4.2 0.78 Right to life and security 4.3 0.52 Due process of law 4.4 0.64 Freedom of expression 4.5 0.66 Freedom of religion 4.6 0.57 Right to privacy 4.7 0.71 Freedom of association 4.8 0.65 Labor rights Order and Security 5.1 0.77 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.42 Absence of violent redress Regulatory Enforcement 6.1 0.40 Effective regulatory enforcement 6.2 0.70 No improper inÙuence 6.3 0.41 No unreasonable delay 6.4 0.51 Respect for due process 6.5 0.53 No expropriation w/out adequate compensation Civil Justice 7.1 0.60 Accessibility and affordability 7.2 0.50 No discrimination 7.3 0.70 No corruption 7.4 0.64 No improper gov't inÙuence 7.5 0.40 No unreasonable delay 7.6 0.43 Effective enforcement 7.7 0.71 Impartial and effective ADRs Criminal Justice 8.1 0.49 Effective investigations 8.2 0.51 Timely and effective adjudication 8.3 0.45 Effective correctional system 8.4 0.55 No discrimination 8.5 0.78 No corruption 8.6 0.80 No improper gov't inÙuence 8.7 0.52 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 47 WJP Rule of Law Index 2023 Bangladesh Region: South Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.38 4/6 28/37 127/142 Score Change Rank Change -0.01 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.37 0.00 5/6 26/37 120/142 Absence of Corruption 0.33 -0.01 4/6 23/37 117/142 Open Government 0.40 0.00 5/6 21/37 106/142 Fundamental Rights 0.29 -0.01 5/6 34/37 137/142 Order and Security 0.63 0.00 4/6 25/37 108/142 Regulatory Enforcement 0.38 -0.01 4/6 30/37 126/142 Civil Justice 0.36 -0.01 5/6 32/37 133/142 Criminal Justice 0.30 -0.01 5/6 28/37 123/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Bangladesh South Asia Lower-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.47 Limits by judiciary 1.3 0.34 Independent auditing 1.4 0.34 Sanctions for of cial misconduct 1.5 0.25 Non-governmental checks 1.6 0.31 Lawful transition of power Absence of Corruption 2.1 0.38 In the executive branch 2.2 0.31 In the judiciary 2.3 0.27 In the police/military 2.4 0.35 In the legislature Open Government 3.1 0.34 Publicized laws and gov't data 3.2 0.52 Right to information 3.3 0.29 Civic participation 3.4 0.45 Complaint mechanisms Fundamental Rights 4.1 0.41 No discrimination 4.2 0.13 Right to life and security 4.3 0.27 Due process of law 4.4 0.25 Freedom of expression 4.5 0.48 Freedom of religion 4.6 0.06 Right to privacy 4.7 0.31 Freedom of association 4.8 0.44 Labor rights Order and Security 5.1 0.77 Absence of crime 5.2 0.92 Absence of civil con ct 5.3 0.21 Absence of violent redress Regulatory Enforcement 6.1 0.44 Effective regulatory enforcement 6.2 0.39 No improper in ence 6.3 0.25 No unreasonable delay 6.4 0.28 Respect for due process 6.5 0.57 No expropriation w/out adequate compensation Civil Justice 7.1 0.43 Accessibility and affordability 7.2 0.29 No discrimination 7.3 0.34 No corruption 7.4 0.40 No improper gov't in ence 7.5 0.19 No unreasonable delay 7.6 0.31 Effective enforcement 7.7 0.57 Impartial and effective ADRs Criminal Justice 8.1 0.40 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.31 Effective correctional system 8.4 0.20 No discrimination 8.5 0.32 No corruption 8.6 0.22 No improper gov't in ence 8.7 0.27 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 48 Barbados Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.66 4/32 34/46 35/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.64 0.00 4/32 33/46 37/142 Absence of Corruption 0.70 0.00 2/32 27/46 27/142 Open Government 0.55 0.00 12/32 39/46 57/142 Fundamental Rights 0.73 0.00 3/32 27/46 28/142 Order and Security 0.80 0.01 1/32 35/46 42/142 Regulatory Enforcement 0.62 0.00 4/32 34/46 36/142 Civil Justice 0.63 -0.01 6/32 32/46 37/142 Criminal Justice 0.58 0.00 4/32 32/46 34/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Barbados Latin America and Caribbean High Constraints on Government Powers 1.1 0.76 Limits by legislature 1.2 0.77 Limits by judiciary 1.3 0.40 Independent auditing 1.4 0.41 Sanctions for of cial misconduct 1.5 0.73 Non-governmental checks 1.6 0.80 Lawful transition of power Absence of Corruption 2.1 0.64 In the executive branch 2.2 0.90 In the judiciary 2.3 0.81 In the police/military 2.4 0.45 In the legislature Open Government 3.1 0.26 Publicized laws and gov't data 3.2 0.52 Right to information 3.3 0.74 Civic participation 3.4 0.66 Complaint mechanisms Fundamental Rights 4.1 0.66 No discrimination 4.2 0.82 Right to life and security 4.3 0.54 Due process of law 4.4 0.73 Freedom of expression 4.5 0.79 Freedom of religion 4.6 0.77 Right to privacy 4.7 0.81 Freedom of association 4.8 0.74 Labor rights Order and Security 5.1 0.90 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.51 Absence of violent redress Regulatory Enforcement 6.1 0.60 Effective regulatory enforcement 6.2 0.84 No improper in ence 6.3 0.39 No unreasonable delay 6.4 0.65 Respect for due process 6.5 0.64 No expropriation w/out adequate compensation Civil Justice 7.1 0.73 Accessibility and affordability 7.2 0.64 No discrimination 7.3 0.87 No corruption 7.4 0.70 No improper gov't in ence 7.5 0.28 No unreasonable delay 7.6 0.47 Effective enforcement 7.7 0.72 Impartial and effective ADRs Criminal Justice 8.1 0.52 Effective investigations 8.2 0.44 Timely and effective adjudication 8.3 0.58 Effective correctional system 8.4 0.50 No discrimination 8.5 0.79 No corruption 8.6 0.71 No improper gov't in ence 8.7 0.54 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 49 WJP Rule of Law Index 2023 Belarus Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.45 13/15 35/41 104/142 Score Change Rank Change -0.01 -3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.26 -0.01 15/15 40/41 139/142 Absence of Corruption 0.49 -0.01 2/15 14/41 60/142 Open Government 0.31 0.00 15/15 40/41 135/142 Fundamental Rights 0.36 -0.02 14/15 38/41 128/142 Order and Security 0.81 0.00 5/15 5/41 41/142 Regulatory Enforcement 0.44 0.00 11/15 35/41 105/142 Civil Justice 0.56 0.00 2/15 13/41 60/142 Criminal Justice 0.39 -0.03 9/15 27/41 89/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Belarus Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.26 Limits by legislature 1.2 0.25 Limits by judiciary 1.3 0.27 Independent auditing 1.4 0.40 Sanctions for of cial misconduct 1.5 0.19 Non-governmental checks 1.6 0.19 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.70 In the judiciary 2.3 0.61 In the police/military 2.4 0.27 In the legislature Open Government 3.1 0.21 Publicized laws and gov't data 3.2 0.31 Right to information 3.3 0.28 Civic participation 3.4 0.45 Complaint mechanisms Fundamental Rights 4.1 0.62 No discrimination 4.2 0.33 Right to life and security 4.3 0.34 Due process of law 4.4 0.19 Freedom of expression 4.5 0.48 Freedom of religion 4.6 0.12 Right to privacy 4.7 0.33 Freedom of association 4.8 0.44 Labor rights Order and Security 5.1 0.89 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.53 Absence of violent redress Regulatory Enforcement 6.1 0.59 Effective regulatory enforcement 6.2 0.54 No improper in ence 6.3 0.56 No unreasonable delay 6.4 0.24 Respect for due process 6.5 0.27 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.68 No discrimination 7.3 0.58 No corruption 7.4 0.20 No improper gov't in ence 7.5 0.67 No unreasonable delay 7.6 0.57 Effective enforcement 7.7 0.64 Impartial and effective ADRs Criminal Justice 8.1 0.47 Effective investigations 8.2 0.47 Timely and effective adjudication 8.3 0.41 Effective correctional system 8.4 0.49 No discrimination 8.5 0.51 No corruption 8.6 0.03 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 50 Belgium Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.78 13/31 16/46 16/142 Score Change Rank Change -0.01 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.82 -0.01 11/31 12/46 12/142 Absence of Corruption 0.79 -0.01 13/31 18/46 18/142 Open Government 0.76 -0.02 12/31 14/46 14/142 Fundamental Rights 0.84 0.00 7/31 7/46 7/142 Order and Security 0.82 0.00 25/31 33/46 37/142 Regulatory Enforcement 0.79 -0.01 12/31 16/46 16/142 Civil Justice 0.74 -0.01 10/31 15/46 15/142 Criminal Justice 0.71 0.00 12/31 17/46 17/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Belgium EU, EFTA, and North America High Constraints on Government Powers 1.1 0.84 Limits by legislature 1.2 0.78 Limits by judiciary 1.3 0.85 Independent auditing 1.4 0.75 Sanctions for ofØcial misconduct 1.5 0.81 Non-governmental checks 1.6 0.89 Lawful transition of power Absence of Corruption 2.1 0.78 In the executive branch 2.2 0.93 In the judiciary 2.3 0.88 In the police/military 2.4 0.58 In the legislature Open Government 3.1 0.70 Publicized laws and gov't data 3.2 0.68 Right to information 3.3 0.79 Civic participation 3.4 0.87 Complaint mechanisms Fundamental Rights 4.1 0.77 No discrimination 4.2 0.95 Right to life and security 4.3 0.80 Due process of law 4.4 0.81 Freedom of expression 4.5 0.85 Freedom of religion 4.6 0.89 Right to privacy 4.7 0.86 Freedom of association 4.8 0.81 Labor rights Order and Security 5.1 0.86 Absence of crime 5.2 0.92 Absence of civil conÙict 5.3 0.68 Absence of violent redress Regulatory Enforcement 6.1 0.75 Effective regulatory enforcement 6.2 0.89 No improper inÙuence 6.3 0.68 No unreasonable delay 6.4 0.76 Respect for due process 6.5 0.89 No expropriation w/out adequate compensation Civil Justice 7.1 0.73 Accessibility and affordability 7.2 0.78 No discrimination 7.3 0.83 No corruption 7.4 0.86 No improper gov't inÙuence 7.5 0.43 No unreasonable delay 7.6 0.75 Effective enforcement 7.7 0.79 Impartial and effective ADRs Criminal Justice 8.1 0.58 Effective investigations 8.2 0.67 Timely and effective adjudication 8.3 0.54 Effective correctional system 8.4 0.67 No discrimination 8.5 0.82 No corruption 8.6 0.85 No improper gov't inÙuence 8.7 0.80 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 51 WJP Rule of Law Index 2023 Belize Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 17/32 23/41 80/142 Score Change Rank Change 0.00 7 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.48 0.01 25/32 27/41 91/142 Absence of Corruption 0.45 0.00 15/32 22/41 73/142 Open Government 0.44 0.00 27/32 33/41 97/142 Fundamental Rights 0.52 0.01 21/32 26/41 77/142 Order and Security 0.71 0.00 9/32 24/41 77/142 Regulatory Enforcement 0.41 0.00 26/32 39/41 118/142 Civil Justice 0.52 0.00 15/32 19/41 70/142 Criminal Justice 0.39 0.00 15/32 25/41 85/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Belize Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.53 Limits by legislature 1.2 0.53 Limits by judiciary 1.3 0.33 Independent auditing 1.4 0.28 Sanctions for ofØcial misconduct 1.5 0.56 Non-governmental checks 1.6 0.64 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.63 In the judiciary 2.3 0.56 In the police/military 2.4 0.22 In the legislature Open Government 3.1 0.35 Publicized laws and gov't data 3.2 0.40 Right to information 3.3 0.52 Civic participation 3.4 0.50 Complaint mechanisms Fundamental Rights 4.1 0.52 No discrimination 4.2 0.66 Right to life and security 4.3 0.40 Due process of law 4.4 0.56 Freedom of expression 4.5 0.52 Freedom of religion 4.6 0.47 Right to privacy 4.7 0.55 Freedom of association 4.8 0.50 Labor rights Order and Security 5.1 0.70 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.44 Absence of violent redress Regulatory Enforcement 6.1 0.40 Effective regulatory enforcement 6.2 0.51 No improper inÙuence 6.3 0.39 No unreasonable delay 6.4 0.25 Respect for due process 6.5 0.50 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.51 No discrimination 7.3 0.61 No corruption 7.4 0.53 No improper gov't inÙuence 7.5 0.45 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.33 Effective investigations 8.2 0.34 Timely and effective adjudication 8.3 0.20 Effective correctional system 8.4 0.44 No discrimination 8.5 0.56 No corruption 8.6 0.50 No improper gov't inÙuence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 52 Benin Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.48 10/34 13/37 90/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.50 -0.01 12/34 11/37 82/142 Absence of Corruption 0.40 0.01 14/34 16/37 97/142 Open Government 0.38 0.00 21/34 24/37 117/142 Fundamental Rights 0.52 0.00 10/34 5/37 78/142 Order and Security 0.74 -0.03 4/34 5/37 64/142 Regulatory Enforcement 0.48 0.00 10/34 11/37 78/142 Civil Justice 0.43 0.00 21/34 25/37 112/142 Criminal Justice 0.43 0.01 10/34 7/37 75/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Benin Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.49 Limits by legislature 1.2 0.43 Limits by judiciary 1.3 0.41 Independent auditing 1.4 0.49 Sanctions for ofØcial misconduct 1.5 0.49 Non-governmental checks 1.6 0.68 Lawful transition of power Absence of Corruption 2.1 0.46 In the executive branch 2.2 0.46 In the judiciary 2.3 0.46 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.31 Publicized laws and gov't data 3.2 0.32 Right to information 3.3 0.53 Civic participation 3.4 0.37 Complaint mechanisms Fundamental Rights 4.1 0.64 No discrimination 4.2 0.37 Right to life and security 4.3 0.48 Due process of law 4.4 0.49 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.24 Right to privacy 4.7 0.61 Freedom of association 4.8 0.61 Labor rights Order and Security 5.1 0.77 Absence of crime 5.2 0.94 Absence of civil conÙict 5.3 0.51 Absence of violent redress Regulatory Enforcement 6.1 0.52 Effective regulatory enforcement 6.2 0.61 No improper inÙuence 6.3 0.41 No unreasonable delay 6.4 0.38 Respect for due process 6.5 0.48 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.59 No discrimination 7.3 0.40 No corruption 7.4 0.25 No improper gov't inÙuence 7.5 0.38 No unreasonable delay 7.6 0.35 Effective enforcement 7.7 0.54 Impartial and effective ADRs Criminal Justice 8.1 0.38 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.32 Effective correctional system 8.4 0.68 No discrimination 8.5 0.42 No corruption 8.6 0.30 No improper gov't inÙuence 8.7 0.48 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 53 WJP Rule of Law Index 2023 Bolivia Region: Latin America and Caribbean Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.37 29/32 30/37 131/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.36 0.00 30/32 30/37 126/142 Absence of Corruption 0.25 0.00 31/32 34/37 137/142 Open Government 0.44 0.01 28/32 17/37 100/142 Fundamental Rights 0.44 -0.02 29/32 19/37 111/142 Order and Security 0.59 0.00 26/32 30/37 122/142 Regulatory Enforcement 0.39 -0.01 29/32 29/37 125/142 Civil Justice 0.31 0.00 31/32 36/37 140/142 Criminal Justice 0.21 0.00 31/32 37/37 141/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Bolivia Latin America and Caribbean Lower-Middle Constraints on Government Powers 1.1 0.38 Limits by legislature 1.2 0.27 Limits by judiciary 1.3 0.30 Independent auditing 1.4 0.27 Sanctions for of cial misconduct 1.5 0.48 Non-governmental checks 1.6 0.45 Lawful transition of power Absence of Corruption 2.1 0.32 In the executive branch 2.2 0.18 In the judiciary 2.3 0.28 In the police/military 2.4 0.21 In the legislature Open Government 3.1 0.30 Publicized laws and gov't data 3.2 0.45 Right to information 3.3 0.49 Civic participation 3.4 0.51 Complaint mechanisms Fundamental Rights 4.1 0.37 No discrimination 4.2 0.37 Right to life and security 4.3 0.29 Due process of law 4.4 0.48 Freedom of expression 4.5 0.64 Freedom of religion 4.6 0.34 Right to privacy 4.7 0.57 Freedom of association 4.8 0.50 Labor rights Order and Security 5.1 0.58 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.19 Absence of violent redress Regulatory Enforcement 6.1 0.46 Effective regulatory enforcement 6.2 0.45 No improper in ence 6.3 0.41 No unreasonable delay 6.4 0.19 Respect for due process 6.5 0.43 No expropriation w/out adequate compensation Civil Justice 7.1 0.46 Accessibility and affordability 7.2 0.29 No discrimination 7.3 0.18 No corruption 7.4 0.20 No improper gov't in ence 7.5 0.21 No unreasonable delay 7.6 0.29 Effective enforcement 7.7 0.55 Impartial and effective ADRs Criminal Justice 8.1 0.21 Effective investigations 8.2 0.17 Timely and effective adjudication 8.3 0.20 Effective correctional system 8.4 0.28 No discrimination 8.5 0.20 No corruption 8.6 0.11 No improper gov't in ence 8.7 0.29 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 54 Bosnia and Herzegovina Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.51 7/15 21/41 75/142 Score Change Rank Change -0.01 -3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.45 -0.01 7/15 31/41 101/142 Absence of Corruption 0.42 0.00 9/15 28/41 87/142 Open Government 0.47 0.00 9/15 27/41 83/142 Fundamental Rights 0.58 -0.02 7/15 19/41 64/142 Order and Security 0.77 0.00 10/15 13/41 55/142 Regulatory Enforcement 0.49 0.00 3/15 18/41 71/142 Civil Justice 0.46 -0.01 13/15 32/41 97/142 Criminal Justice 0.47 -0.01 4/15 17/41 63/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Bosnia and Herzegovina Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.38 Limits by judiciary 1.3 0.50 Independent auditing 1.4 0.35 Sanctions for ofØcial misconduct 1.5 0.44 Non-governmental checks 1.6 0.50 Lawful transition of power Absence of Corruption 2.1 0.36 In the executive branch 2.2 0.52 In the judiciary 2.3 0.57 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.42 Publicized laws and gov't data 3.2 0.50 Right to information 3.3 0.44 Civic participation 3.4 0.51 Complaint mechanisms Fundamental Rights 4.1 0.56 No discrimination 4.2 0.68 Right to life and security 4.3 0.62 Due process of law 4.4 0.44 Freedom of expression 4.5 0.63 Freedom of religion 4.6 0.54 Right to privacy 4.7 0.57 Freedom of association 4.8 0.61 Labor rights Order and Security 5.1 0.86 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.47 Absence of violent redress Regulatory Enforcement 6.1 0.41 Effective regulatory enforcement 6.2 0.52 No improper inÙuence 6.3 0.42 No unreasonable delay 6.4 0.46 Respect for due process 6.5 0.63 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.55 No discrimination 7.3 0.45 No corruption 7.4 0.41 No improper gov't inÙuence 7.5 0.27 No unreasonable delay 7.6 0.31 Effective enforcement 7.7 0.67 Impartial and effective ADRs Criminal Justice 8.1 0.42 Effective investigations 8.2 0.55 Timely and effective adjudication 8.3 0.42 Effective correctional system 8.4 0.47 No discrimination 8.5 0.46 No corruption 8.6 0.33 No improper gov't inÙuence 8.7 0.62 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 55 WJP Rule of Law Index 2023 Botswana Region: Sub-Saharan Africa Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.59 4/34 8/41 51/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.61 0.00 4/34 5/41 44/142 Absence of Corruption 0.57 0.00 3/34 9/41 49/142 Open Government 0.47 0.00 8/34 26/41 81/142 Fundamental Rights 0.59 0.00 5/34 17/41 62/142 Order and Security 0.72 0.00 5/34 22/41 72/142 Regulatory Enforcement 0.58 -0.01 3/34 4/41 45/142 Civil Justice 0.61 0.01 4/34 9/41 46/142 Criminal Justice 0.59 0.00 1/34 2/41 32/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Botswana Sub-Saharan Africa Upper-Middle Constraints on Government Powers 1.1 0.62 Limits by legislature 1.2 0.63 Limits by judiciary 1.3 0.55 Independent auditing 1.4 0.53 Sanctions for ofØcial misconduct 1.5 0.59 Non-governmental checks 1.6 0.72 Lawful transition of power Absence of Corruption 2.1 0.54 In the executive branch 2.2 0.76 In the judiciary 2.3 0.71 In the police/military 2.4 0.30 In the legislature Open Government 3.1 0.22 Publicized laws and gov't data 3.2 0.51 Right to information 3.3 0.61 Civic participation 3.4 0.54 Complaint mechanisms Fundamental Rights 4.1 0.61 No discrimination 4.2 0.68 Right to life and security 4.3 0.58 Due process of law 4.4 0.59 Freedom of expression 4.5 0.62 Freedom of religion 4.6 0.40 Right to privacy 4.7 0.66 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.68 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.48 Absence of violent redress Regulatory Enforcement 6.1 0.52 Effective regulatory enforcement 6.2 0.71 No improper inÙuence 6.3 0.43 No unreasonable delay 6.4 0.57 Respect for due process 6.5 0.69 No expropriation w/out adequate compensation Civil Justice 7.1 0.54 Accessibility and affordability 7.2 0.55 No discrimination 7.3 0.67 No corruption 7.4 0.64 No improper gov't inÙuence 7.5 0.52 No unreasonable delay 7.6 0.65 Effective enforcement 7.7 0.69 Impartial and effective ADRs Criminal Justice 8.1 0.42 Effective investigations 8.2 0.57 Timely and effective adjudication 8.3 0.55 Effective correctional system 8.4 0.63 No discrimination 8.5 0.69 No corruption 8.6 0.68 No improper gov't inÙuence 8.7 0.58 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 56 Brazil Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 19/32 26/41 83/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.51 0.00 19/32 21/41 78/142 Absence of Corruption 0.43 0.00 18/32 27/41 81/142 Open Government 0.59 0.00 6/32 5/41 41/142 Fundamental Rights 0.49 0.01 25/32 31/41 91/142 Order and Security 0.60 -0.04 25/32 36/41 120/142 Regulatory Enforcement 0.48 0.00 17/32 20/41 76/142 Civil Justice 0.49 0.00 18/32 26/41 84/142 Criminal Justice 0.32 0.00 21/32 34/41 114/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Brazil Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.64 Limits by legislature 1.2 0.57 Limits by judiciary 1.3 0.40 Independent auditing 1.4 0.27 Sanctions for ofØcial misconduct 1.5 0.54 Non-governmental checks 1.6 0.63 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.68 In the judiciary 2.3 0.59 In the police/military 2.4 0.07 In the legislature Open Government 3.1 0.72 Publicized laws and gov't data 3.2 0.58 Right to information 3.3 0.53 Civic participation 3.4 0.53 Complaint mechanisms Fundamental Rights 4.1 0.46 No discrimination 4.2 0.44 Right to life and security 4.3 0.35 Due process of law 4.4 0.54 Freedom of expression 4.5 0.54 Freedom of religion 4.6 0.51 Right to privacy 4.7 0.60 Freedom of association 4.8 0.47 Labor rights Order and Security 5.1 0.56 Absence of crime 5.2 0.88 Absence of civil conÙict 5.3 0.36 Absence of violent redress Regulatory Enforcement 6.1 0.51 Effective regulatory enforcement 6.2 0.62 No improper inÙuence 6.3 0.25 No unreasonable delay 6.4 0.43 Respect for due process 6.5 0.60 No expropriation w/out adequate compensation Civil Justice 7.1 0.64 Accessibility and affordability 7.2 0.47 No discrimination 7.3 0.64 No corruption 7.4 0.50 No improper gov't inÙuence 7.5 0.25 No unreasonable delay 7.6 0.33 Effective enforcement 7.7 0.63 Impartial and effective ADRs Criminal Justice 8.1 0.31 Effective investigations 8.2 0.27 Timely and effective adjudication 8.3 0.18 Effective correctional system 8.4 0.11 No discrimination 8.5 0.53 No corruption 8.6 0.51 No improper gov't inÙuence 8.7 0.35 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 57 WJP Rule of Law Index 2023 Bulgaria Region: EU, EFTA, and North America Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.56 30/31 15/41 59/142 Score Change Rank Change 0.01 3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.51 0.02 30/31 19/41 75/142 Absence of Corruption 0.45 0.01 31/31 20/41 71/142 Open Government 0.57 0.02 30/31 10/41 48/142 Fundamental Rights 0.61 0.02 29/31 13/41 53/142 Order and Security 0.78 0.00 29/31 11/41 52/142 Regulatory Enforcement 0.53 0.00 30/31 13/41 59/142 Civil Justice 0.54 0.00 30/31 16/41 65/142 Criminal Justice 0.44 0.01 31/31 19/41 70/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Bulgaria EU, EFTA, and North America Upper-Middle Constraints on Government Powers 1.1 0.58 Limits by legislature 1.2 0.44 Limits by judiciary 1.3 0.48 Independent auditing 1.4 0.36 Sanctions for ofØcial misconduct 1.5 0.63 Non-governmental checks 1.6 0.59 Lawful transition of power Absence of Corruption 2.1 0.43 In the executive branch 2.2 0.58 In the judiciary 2.3 0.62 In the police/military 2.4 0.18 In the legislature Open Government 3.1 0.56 Publicized laws and gov't data 3.2 0.50 Right to information 3.3 0.61 Civic participation 3.4 0.62 Complaint mechanisms Fundamental Rights 4.1 0.59 No discrimination 4.2 0.64 Right to life and security 4.3 0.54 Due process of law 4.4 0.63 Freedom of expression 4.5 0.76 Freedom of religion 4.6 0.44 Right to privacy 4.7 0.68 Freedom of association 4.8 0.63 Labor rights Order and Security 5.1 0.85 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.49 Absence of violent redress Regulatory Enforcement 6.1 0.65 Effective regulatory enforcement 6.2 0.63 No improper inÙuence 6.3 0.54 No unreasonable delay 6.4 0.36 Respect for due process 6.5 0.45 No expropriation w/out adequate compensation Civil Justice 7.1 0.64 Accessibility and affordability 7.2 0.58 No discrimination 7.3 0.52 No corruption 7.4 0.47 No improper gov't inÙuence 7.5 0.38 No unreasonable delay 7.6 0.54 Effective enforcement 7.7 0.68 Impartial and effective ADRs Criminal Justice 8.1 0.32 Effective investigations 8.2 0.53 Timely and effective adjudication 8.3 0.30 Effective correctional system 8.4 0.48 No discrimination 8.5 0.49 No corruption 8.6 0.45 No improper gov't inÙuence 8.7 0.54 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 58 Burkina Faso Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.47 11/34 4/18 95/142 Score Change Rank Change -0.02 -11 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.53 -0.03 10/34 4/18 70/142 Absence of Corruption 0.43 0.00 11/34 5/18 84/142 Open Government 0.48 -0.01 7/34 2/18 78/142 Fundamental Rights 0.53 -0.02 8/34 2/18 73/142 Order and Security 0.43 -0.09 31/34 15/18 137/142 Regulatory Enforcement 0.46 -0.01 13/34 5/18 89/142 Civil Justice 0.45 0.00 18/34 7/18 101/142 Criminal Justice 0.48 0.00 6/34 2/18 59/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Burkina Faso Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.62 Limits by legislature 1.2 0.52 Limits by judiciary 1.3 0.49 Independent auditing 1.4 0.43 Sanctions for ofØcial misconduct 1.5 0.58 Non-governmental checks 1.6 0.53 Lawful transition of power Absence of Corruption 2.1 0.44 In the executive branch 2.2 0.50 In the judiciary 2.3 0.56 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.32 Publicized laws and gov't data 3.2 0.49 Right to information 3.3 0.57 Civic participation 3.4 0.51 Complaint mechanisms Fundamental Rights 4.1 0.64 No discrimination 4.2 0.46 Right to life and security 4.3 0.47 Due process of law 4.4 0.58 Freedom of expression 4.5 0.63 Freedom of religion 4.6 0.32 Right to privacy 4.7 0.61 Freedom of association 4.8 0.53 Labor rights Order and Security 5.1 0.69 Absence of crime 5.2 0.17 Absence of civil conÙict 5.3 0.43 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.55 No improper inÙuence 6.3 0.37 No unreasonable delay 6.4 0.37 Respect for due process 6.5 0.58 No expropriation w/out adequate compensation Civil Justice 7.1 0.36 Accessibility and affordability 7.2 0.47 No discrimination 7.3 0.40 No corruption 7.4 0.43 No improper gov't inÙuence 7.5 0.48 No unreasonable delay 7.6 0.38 Effective enforcement 7.7 0.63 Impartial and effective ADRs Criminal Justice 8.1 0.41 Effective investigations 8.2 0.48 Timely and effective adjudication 8.3 0.41 Effective correctional system 8.4 0.56 No discrimination 8.5 0.48 No corruption 8.6 0.56 No improper gov't inÙuence 8.7 0.47 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 59 WJP Rule of Law Index 2023 Cambodia Region: East Asia and PaciØc Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.31 15/15 37/37 141/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.26 -0.01 15/15 35/37 138/142 Absence of Corruption 0.23 0.00 15/15 37/37 141/142 Open Government 0.24 0.01 15/15 36/37 141/142 Fundamental Rights 0.33 0.00 13/15 32/37 131/142 Order and Security 0.66 -0.01 14/15 20/37 99/142 Regulatory Enforcement 0.26 0.00 15/15 37/37 141/142 Civil Justice 0.25 -0.01 15/15 37/37 142/142 Criminal Justice 0.26 0.00 14/15 31/37 134/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Cambodia East Asia and PaciØc Lower-Middle Constraints on Government Powers 1.1 0.38 Limits by legislature 1.2 0.24 Limits by judiciary 1.3 0.18 Independent auditing 1.4 0.24 Sanctions for ofØcial misconduct 1.5 0.28 Non-governmental checks 1.6 0.26 Lawful transition of power Absence of Corruption 2.1 0.26 In the executive branch 2.2 0.13 In the judiciary 2.3 0.21 In the police/military 2.4 0.30 In the legislature Open Government 3.1 0.20 Publicized laws and gov't data 3.2 0.22 Right to information 3.3 0.31 Civic participation 3.4 0.25 Complaint mechanisms Fundamental Rights 4.1 0.43 No discrimination 4.2 0.28 Right to life and security 4.3 0.29 Due process of law 4.4 0.28 Freedom of expression 4.5 0.42 Freedom of religion 4.6 0.15 Right to privacy 4.7 0.37 Freedom of association 4.8 0.45 Labor rights Order and Security 5.1 0.82 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.17 Absence of violent redress Regulatory Enforcement 6.1 0.24 Effective regulatory enforcement 6.2 0.22 No improper inÙuence 6.3 0.41 No unreasonable delay 6.4 0.14 Respect for due process 6.5 0.27 No expropriation w/out adequate compensation Civil Justice 7.1 0.33 Accessibility and affordability 7.2 0.22 No discrimination 7.3 0.11 No corruption 7.4 0.16 No improper gov't inÙuence 7.5 0.24 No unreasonable delay 7.6 0.23 Effective enforcement 7.7 0.45 Impartial and effective ADRs Criminal Justice 8.1 0.35 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.27 Effective correctional system 8.4 0.24 No discrimination 8.5 0.16 No corruption 8.6 0.11 No improper gov't inÙuence 8.7 0.29 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 60 Cameroon Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.35 33/34 32/37 134/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.37 0.00 28/34 27/37 121/142 Absence of Corruption 0.24 0.00 32/34 35/37 138/142 Open Government 0.33 0.00 31/34 31/37 132/142 Fundamental Rights 0.36 0.00 30/34 30/37 127/142 Order and Security 0.48 -0.01 29/34 34/37 134/142 Regulatory Enforcement 0.40 0.00 24/34 27/37 121/142 Civil Justice 0.40 0.00 28/34 26/37 122/142 Criminal Justice 0.24 -0.02 34/34 36/37 140/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Cameroon Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.40 Limits by legislature 1.2 0.30 Limits by judiciary 1.3 0.43 Independent auditing 1.4 0.37 Sanctions for of cial misconduct 1.5 0.35 Non-governmental checks 1.6 0.37 Lawful transition of power Absence of Corruption 2.1 0.27 In the executive branch 2.2 0.29 In the judiciary 2.3 0.24 In the police/military 2.4 0.19 In the legislature Open Government 3.1 0.21 Publicized laws and gov't data 3.2 0.34 Right to information 3.3 0.38 Civic participation 3.4 0.38 Complaint mechanisms Fundamental Rights 4.1 0.45 No discrimination 4.2 0.20 Right to life and security 4.3 0.30 Due process of law 4.4 0.35 Freedom of expression 4.5 0.56 Freedom of religion 4.6 0.08 Right to privacy 4.7 0.44 Freedom of association 4.8 0.49 Labor rights Order and Security 5.1 0.59 Absence of crime 5.2 0.54 Absence of civil con ct 5.3 0.31 Absence of violent redress Regulatory Enforcement 6.1 0.44 Effective regulatory enforcement 6.2 0.45 No improper in ence 6.3 0.26 No unreasonable delay 6.4 0.39 Respect for due process 6.5 0.47 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.45 No discrimination 7.3 0.31 No corruption 7.4 0.24 No improper gov't in ence 7.5 0.40 No unreasonable delay 7.6 0.39 Effective enforcement 7.7 0.55 Impartial and effective ADRs Criminal Justice 8.1 0.29 Effective investigations 8.2 0.26 Timely and effective adjudication 8.3 0.15 Effective correctional system 8.4 0.36 No discrimination 8.5 0.22 No corruption 8.6 0.11 No improper gov't in ence 8.7 0.30 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 61 WJP Rule of Law Index 2023 Canada Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.80 11/31 12/46 12/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.82 0.00 10/31 11/46 11/142 Absence of Corruption 0.83 0.00 7/31 10/46 10/142 Open Government 0.80 0.01 8/31 10/46 10/142 Fundamental Rights 0.81 0.00 12/31 13/46 13/142 Order and Security 0.90 0.00 11/31 15/46 15/142 Regulatory Enforcement 0.81 0.00 9/31 12/46 12/142 Civil Justice 0.69 -0.01 17/31 24/46 24/142 Criminal Justice 0.73 0.00 9/31 12/46 12/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Canada EU, EFTA, and North America High Constraints on Government Powers 1.1 0.79 Limits by legislature 1.2 0.84 Limits by judiciary 1.3 0.76 Independent auditing 1.4 0.76 Sanctions for of cial misconduct 1.5 0.85 Non-governmental checks 1.6 0.95 Lawful transition of power Absence of Corruption 2.1 0.80 In the executive branch 2.2 0.95 In the judiciary 2.3 0.88 In the police/military 2.4 0.68 In the legislature Open Government 3.1 0.87 Publicized laws and gov't data 3.2 0.65 Right to information 3.3 0.84 Civic participation 3.4 0.83 Complaint mechanisms Fundamental Rights 4.1 0.62 No discrimination 4.2 0.95 Right to life and security 4.3 0.76 Due process of law 4.4 0.85 Freedom of expression 4.5 0.83 Freedom of religion 4.6 0.91 Right to privacy 4.7 0.88 Freedom of association 4.8 0.72 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.80 Absence of violent redress Regulatory Enforcement 6.1 0.78 Effective regulatory enforcement 6.2 0.95 No improper in ence 6.3 0.66 No unreasonable delay 6.4 0.91 Respect for due process 6.5 0.77 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.53 No discrimination 7.3 0.90 No corruption 7.4 0.88 No improper gov't in ence 7.5 0.46 No unreasonable delay 7.6 0.71 Effective enforcement 7.7 0.76 Impartial and effective ADRs Criminal Justice 8.1 0.68 Effective investigations 8.2 0.68 Timely and effective adjudication 8.3 0.68 Effective correctional system 8.4 0.54 No discrimination 8.5 0.85 No corruption 8.6 0.94 No improper gov't in ence 8.7 0.76 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 62 Chile Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.66 3/32 32/46 33/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.71 0.00 3/32 24/46 25/142 Absence of Corruption 0.69 0.00 3/32 28/46 28/142 Open Government 0.70 0.00 2/32 23/46 23/142 Fundamental Rights 0.72 0.00 4/32 30/46 31/142 Order and Security 0.67 0.00 15/32 45/46 96/142 Regulatory Enforcement 0.64 0.01 3/32 31/46 32/142 Civil Justice 0.62 0.00 7/32 35/46 41/142 Criminal Justice 0.54 -0.01 9/32 38/46 48/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Chile Latin America and Caribbean High Constraints on Government Powers 1.1 0.70 Limits by legislature 1.2 0.59 Limits by judiciary 1.3 0.81 Independent auditing 1.4 0.57 Sanctions for of cial misconduct 1.5 0.72 Non-governmental checks 1.6 0.89 Lawful transition of power Absence of Corruption 2.1 0.62 In the executive branch 2.2 0.83 In the judiciary 2.3 0.81 In the police/military 2.4 0.50 In the legislature Open Government 3.1 0.66 Publicized laws and gov't data 3.2 0.73 Right to information 3.3 0.68 Civic participation 3.4 0.74 Complaint mechanisms Fundamental Rights 4.1 0.57 No discrimination 4.2 0.80 Right to life and security 4.3 0.62 Due process of law 4.4 0.72 Freedom of expression 4.5 0.79 Freedom of religion 4.6 0.82 Right to privacy 4.7 0.74 Freedom of association 4.8 0.73 Labor rights Order and Security 5.1 0.74 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.26 Absence of violent redress Regulatory Enforcement 6.1 0.60 Effective regulatory enforcement 6.2 0.77 No improper in ence 6.3 0.53 No unreasonable delay 6.4 0.54 Respect for due process 6.5 0.76 No expropriation w/out adequate compensation Civil Justice 7.1 0.67 Accessibility and affordability 7.2 0.65 No discrimination 7.3 0.64 No corruption 7.4 0.69 No improper gov't in ence 7.5 0.39 No unreasonable delay 7.6 0.58 Effective enforcement 7.7 0.71 Impartial and effective ADRs Criminal Justice 8.1 0.34 Effective investigations 8.2 0.55 Timely and effective adjudication 8.3 0.35 Effective correctional system 8.4 0.51 No discrimination 8.5 0.67 No corruption 8.6 0.74 No improper gov't in ence 8.7 0.62 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 63 WJP Rule of Law Index 2023 China Region: East Asia and Paci c Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.47 12/15 33/41 97/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.31 -0.01 13/15 38/41 134/142 Absence of Corruption 0.53 -0.01 8/15 12/41 57/142 Open Government 0.40 0.00 13/15 36/41 108/142 Fundamental Rights 0.25 -0.01 14/15 41/41 139/142 Order and Security 0.81 0.00 7/15 3/41 39/142 Regulatory Enforcement 0.49 0.00 9/15 19/41 74/142 Civil Justice 0.52 0.00 9/15 21/41 73/142 Criminal Justice 0.43 -0.02 10/15 21/41 74/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 China East Asia and Paci c Upper-Middle Constraints on Government Powers 1.1 0.35 Limits by legislature 1.2 0.26 Limits by judiciary 1.3 0.43 Independent auditing 1.4 0.47 Sanctions for of cial misconduct 1.5 0.10 Non-governmental checks 1.6 0.25 Lawful transition of power Absence of Corruption 2.1 0.53 In the executive branch 2.2 0.51 In the judiciary 2.3 0.64 In the police/military 2.4 0.43 In the legislature Open Government 3.1 0.44 Publicized laws and gov't data 3.2 0.49 Right to information 3.3 0.17 Civic participation 3.4 0.48 Complaint mechanisms Fundamental Rights 4.1 0.46 No discrimination 4.2 0.29 Right to life and security 4.3 0.44 Due process of law 4.4 0.10 Freedom of expression 4.5 0.20 Freedom of religion 4.6 0.12 Right to privacy 4.7 0.10 Freedom of association 4.8 0.32 Labor rights Order and Security 5.1 0.81 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.62 Absence of violent redress Regulatory Enforcement 6.1 0.53 Effective regulatory enforcement 6.2 0.60 No improper in ence 6.3 0.60 No unreasonable delay 6.4 0.33 Respect for due process 6.5 0.36 No expropriation w/out adequate compensation Civil Justice 7.1 0.66 Accessibility and affordability 7.2 0.40 No discrimination 7.3 0.43 No corruption 7.4 0.19 No improper gov't in ence 7.5 0.72 No unreasonable delay 7.6 0.55 Effective enforcement 7.7 0.66 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.48 Timely and effective adjudication 8.3 0.56 Effective correctional system 8.4 0.37 No discrimination 8.5 0.52 No corruption 8.6 0.16 No improper gov't in ence 8.7 0.44 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 64 Colombia Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.48 22/32 31/41 94/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.50 -0.01 21/32 23/41 84/142 Absence of Corruption 0.38 0.00 21/32 32/41 103/142 Open Government 0.61 -0.01 4/32 3/41 36/142 Fundamental Rights 0.49 0.00 24/32 29/41 89/142 Order and Security 0.56 0.02 29/32 39/41 129/142 Regulatory Enforcement 0.52 -0.01 12/32 16/41 63/142 Civil Justice 0.47 0.00 20/32 29/41 91/142 Criminal Justice 0.32 0.00 22/32 35/41 115/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Colombia Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.57 Limits by legislature 1.2 0.48 Limits by judiciary 1.3 0.49 Independent auditing 1.4 0.35 Sanctions for ofØcial misconduct 1.5 0.51 Non-governmental checks 1.6 0.58 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.55 In the judiciary 2.3 0.49 In the police/military 2.4 0.09 In the legislature Open Government 3.1 0.64 Publicized laws and gov't data 3.2 0.61 Right to information 3.3 0.52 Civic participation 3.4 0.67 Complaint mechanisms Fundamental Rights 4.1 0.46 No discrimination 4.2 0.50 Right to life and security 4.3 0.38 Due process of law 4.4 0.51 Freedom of expression 4.5 0.60 Freedom of religion 4.6 0.49 Right to privacy 4.7 0.55 Freedom of association 4.8 0.46 Labor rights Order and Security 5.1 0.47 Absence of crime 5.2 0.94 Absence of civil conÙict 5.3 0.28 Absence of violent redress Regulatory Enforcement 6.1 0.51 Effective regulatory enforcement 6.2 0.64 No improper inÙuence 6.3 0.35 No unreasonable delay 6.4 0.46 Respect for due process 6.5 0.62 No expropriation w/out adequate compensation Civil Justice 7.1 0.56 Accessibility and affordability 7.2 0.45 No discrimination 7.3 0.49 No corruption 7.4 0.50 No improper gov't inÙuence 7.5 0.21 No unreasonable delay 7.6 0.40 Effective enforcement 7.7 0.70 Impartial and effective ADRs Criminal Justice 8.1 0.20 Effective investigations 8.2 0.31 Timely and effective adjudication 8.3 0.22 Effective correctional system 8.4 0.29 No discrimination 8.5 0.40 No corruption 8.6 0.46 No improper gov't inÙuence 8.7 0.38 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 65 WJP Rule of Law Index 2023 Congo, Dem. Rep. Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.34 34/34 17/18 138/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.40 0.01 22/34 11/18 112/142 Absence of Corruption 0.17 0.00 34/34 18/18 142/142 Open Government 0.34 0.01 28/34 16/18 128/142 Fundamental Rights 0.41 0.01 24/34 12/18 118/142 Order and Security 0.45 -0.01 30/34 14/18 136/142 Regulatory Enforcement 0.34 -0.01 32/34 17/18 137/142 Civil Justice 0.36 0.00 33/34 16/18 132/142 Criminal Justice 0.28 0.01 31/34 16/18 128/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Congo, Dem. Rep. Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.51 Limits by legislature 1.2 0.38 Limits by judiciary 1.3 0.35 Independent auditing 1.4 0.37 Sanctions for of cial misconduct 1.5 0.39 Non-governmental checks 1.6 0.40 Lawful transition of power Absence of Corruption 2.1 0.23 In the executive branch 2.2 0.17 In the judiciary 2.3 0.19 In the police/military 2.4 0.09 In the legislature Open Government 3.1 0.16 Publicized laws and gov't data 3.2 0.35 Right to information 3.3 0.41 Civic participation 3.4 0.45 Complaint mechanisms Fundamental Rights 4.1 0.53 No discrimination 4.2 0.20 Right to life and security 4.3 0.32 Due process of law 4.4 0.39 Freedom of expression 4.5 0.63 Freedom of religion 4.6 0.21 Right to privacy 4.7 0.49 Freedom of association 4.8 0.51 Labor rights Order and Security 5.1 0.52 Absence of crime 5.2 0.50 Absence of civil con ct 5.3 0.33 Absence of violent redress Regulatory Enforcement 6.1 0.33 Effective regulatory enforcement 6.2 0.35 No improper in ence 6.3 0.35 No unreasonable delay 6.4 0.38 Respect for due process 6.5 0.31 No expropriation w/out adequate compensation Civil Justice 7.1 0.39 Accessibility and affordability 7.2 0.45 No discrimination 7.3 0.17 No corruption 7.4 0.26 No improper gov't in ence 7.5 0.42 No unreasonable delay 7.6 0.34 Effective enforcement 7.7 0.50 Impartial and effective ADRs Criminal Justice 8.1 0.23 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.17 Effective correctional system 8.4 0.45 No discrimination 8.5 0.19 No corruption 8.6 0.24 No improper gov't in ence 8.7 0.32 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 66 Congo, Rep. Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.40 25/34 25/37 122/142 Score Change Rank Change -0.01 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.36 0.00 29/34 28/37 124/142 Absence of Corruption 0.30 0.00 26/34 29/37 126/142 Open Government 0.33 -0.01 30/34 30/37 131/142 Fundamental Rights 0.40 0.00 26/34 26/37 121/142 Order and Security 0.61 0.00 21/34 28/37 117/142 Regulatory Enforcement 0.43 -0.03 19/34 23/37 110/142 Civil Justice 0.44 0.00 19/34 20/37 106/142 Criminal Justice 0.34 0.00 24/34 22/37 108/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Congo, Rep. Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.38 Limits by legislature 1.2 0.33 Limits by judiciary 1.3 0.34 Independent auditing 1.4 0.35 Sanctions for of cial misconduct 1.5 0.28 Non-governmental checks 1.6 0.48 Lawful transition of power Absence of Corruption 2.1 0.31 In the executive branch 2.2 0.43 In the judiciary 2.3 0.30 In the police/military 2.4 0.16 In the legislature Open Government 3.1 0.16 Publicized laws and gov't data 3.2 0.38 Right to information 3.3 0.35 Civic participation 3.4 0.42 Complaint mechanisms Fundamental Rights 4.1 0.49 No discrimination 4.2 0.29 Right to life and security 4.3 0.39 Due process of law 4.4 0.28 Freedom of expression 4.5 0.66 Freedom of religion 4.6 0.17 Right to privacy 4.7 0.43 Freedom of association 4.8 0.46 Labor rights Order and Security 5.1 0.51 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.33 Absence of violent redress Regulatory Enforcement 6.1 0.39 Effective regulatory enforcement 6.2 0.47 No improper in ence 6.3 0.37 No unreasonable delay 6.4 0.45 Respect for due process 6.5 0.45 No expropriation w/out adequate compensation Civil Justice 7.1 0.43 Accessibility and affordability 7.2 0.46 No discrimination 7.3 0.30 No corruption 7.4 0.42 No improper gov't in ence 7.5 0.53 No unreasonable delay 7.6 0.43 Effective enforcement 7.7 0.52 Impartial and effective ADRs Criminal Justice 8.1 0.22 Effective investigations 8.2 0.38 Timely and effective adjudication 8.3 0.29 Effective correctional system 8.4 0.43 No discrimination 8.5 0.32 No corruption 8.6 0.31 No improper gov't in ence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 67 WJP Rule of Law Index 2023 Costa Rica Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.68 2/32 1/41 29/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.77 -0.01 1/32 1/41 16/142 Absence of Corruption 0.64 0.01 7/32 4/41 42/142 Open Government 0.70 0.00 3/32 1/41 24/142 Fundamental Rights 0.79 0.00 2/32 1/41 16/142 Order and Security 0.70 0.00 13/32 26/41 85/142 Regulatory Enforcement 0.68 0.00 2/32 1/41 28/142 Civil Justice 0.61 0.00 9/32 10/41 47/142 Criminal Justice 0.58 0.00 5/32 3/41 36/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Costa Rica Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.77 Limits by legislature 1.2 0.72 Limits by judiciary 1.3 0.80 Independent auditing 1.4 0.60 Sanctions for ofØcial misconduct 1.5 0.83 Non-governmental checks 1.6 0.92 Lawful transition of power Absence of Corruption 2.1 0.62 In the executive branch 2.2 0.79 In the judiciary 2.3 0.74 In the police/military 2.4 0.40 In the legislature Open Government 3.1 0.48 Publicized laws and gov't data 3.2 0.72 Right to information 3.3 0.78 Civic participation 3.4 0.80 Complaint mechanisms Fundamental Rights 4.1 0.70 No discrimination 4.2 0.92 Right to life and security 4.3 0.73 Due process of law 4.4 0.83 Freedom of expression 4.5 0.83 Freedom of religion 4.6 0.83 Right to privacy 4.7 0.84 Freedom of association 4.8 0.63 Labor rights Order and Security 5.1 0.72 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.37 Absence of violent redress Regulatory Enforcement 6.1 0.70 Effective regulatory enforcement 6.2 0.75 No improper inÙuence 6.3 0.47 No unreasonable delay 6.4 0.70 Respect for due process 6.5 0.78 No expropriation w/out adequate compensation Civil Justice 7.1 0.66 Accessibility and affordability 7.2 0.71 No discrimination 7.3 0.72 No corruption 7.4 0.72 No improper gov't inÙuence 7.5 0.22 No unreasonable delay 7.6 0.46 Effective enforcement 7.7 0.76 Impartial and effective ADRs Criminal Justice 8.1 0.43 Effective investigations 8.2 0.44 Timely and effective adjudication 8.3 0.41 Effective correctional system 8.4 0.71 No discrimination 8.5 0.66 No corruption 8.6 0.67 No improper gov't inÙuence 8.7 0.73 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 68 Côte d'Ivoire Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.45 16/34 19/37 106/142 Score Change Rank Change 0.00 4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.38 0.00 26/34 24/37 117/142 Absence of Corruption 0.35 0.01 19/34 21/37 111/142 Open Government 0.36 0.00 26/34 27/37 124/142 Fundamental Rights 0.45 0.00 20/34 15/37 105/142 Order and Security 0.67 0.00 16/34 19/37 97/142 Regulatory Enforcement 0.52 0.01 8/34 4/37 62/142 Civil Justice 0.51 0.01 9/34 9/37 77/142 Criminal Justice 0.36 0.00 18/34 20/37 100/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Côte d'Ivoire Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.40 Limits by legislature 1.2 0.36 Limits by judiciary 1.3 0.44 Independent auditing 1.4 0.36 Sanctions for of cial misconduct 1.5 0.32 Non-governmental checks 1.6 0.39 Lawful transition of power Absence of Corruption 2.1 0.35 In the executive branch 2.2 0.38 In the judiciary 2.3 0.43 In the police/military 2.4 0.26 In the legislature Open Government 3.1 0.20 Publicized laws and gov't data 3.2 0.45 Right to information 3.3 0.42 Civic participation 3.4 0.36 Complaint mechanisms Fundamental Rights 4.1 0.60 No discrimination 4.2 0.33 Right to life and security 4.3 0.39 Due process of law 4.4 0.32 Freedom of expression 4.5 0.71 Freedom of religion 4.6 0.13 Right to privacy 4.7 0.52 Freedom of association 4.8 0.61 Labor rights Order and Security 5.1 0.66 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.34 Absence of violent redress Regulatory Enforcement 6.1 0.51 Effective regulatory enforcement 6.2 0.53 No improper in ence 6.3 0.44 No unreasonable delay 6.4 0.55 Respect for due process 6.5 0.55 No expropriation w/out adequate compensation Civil Justice 7.1 0.52 Accessibility and affordability 7.2 0.58 No discrimination 7.3 0.37 No corruption 7.4 0.28 No improper gov't in ence 7.5 0.55 No unreasonable delay 7.6 0.61 Effective enforcement 7.7 0.66 Impartial and effective ADRs Criminal Justice 8.1 0.35 Effective investigations 8.2 0.49 Timely and effective adjudication 8.3 0.33 Effective correctional system 8.4 0.46 No discrimination 8.5 0.38 No corruption 8.6 0.13 No improper gov't in ence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 69 WJP Rule of Law Index 2023 Croatia Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.61 28/31 40/46 45/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.58 0.02 28/31 39/46 53/142 Absence of Corruption 0.57 0.00 26/31 40/46 50/142 Open Government 0.61 0.00 26/31 34/46 37/142 Fundamental Rights 0.68 0.01 25/31 34/46 37/142 Order and Security 0.84 -0.01 21/31 29/46 31/142 Regulatory Enforcement 0.56 -0.01 28/31 41/46 50/142 Civil Justice 0.56 0.00 28/31 42/46 58/142 Criminal Justice 0.51 0.00 28/31 41/46 54/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Croatia EU, EFTA, and North America High Constraints on Government Powers 1.1 0.57 Limits by legislature 1.2 0.47 Limits by judiciary 1.3 0.59 Independent auditing 1.4 0.49 Sanctions for of cial misconduct 1.5 0.65 Non-governmental checks 1.6 0.71 Lawful transition of power Absence of Corruption 2.1 0.49 In the executive branch 2.2 0.71 In the judiciary 2.3 0.76 In the police/military 2.4 0.33 In the legislature Open Government 3.1 0.55 Publicized laws and gov't data 3.2 0.59 Right to information 3.3 0.64 Civic participation 3.4 0.66 Complaint mechanisms Fundamental Rights 4.1 0.61 No discrimination 4.2 0.74 Right to life and security 4.3 0.66 Due process of law 4.4 0.65 Freedom of expression 4.5 0.68 Freedom of religion 4.6 0.60 Right to privacy 4.7 0.75 Freedom of association 4.8 0.75 Labor rights Order and Security 5.1 0.95 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.56 Absence of violent redress Regulatory Enforcement 6.1 0.60 Effective regulatory enforcement 6.2 0.67 No improper in ence 6.3 0.44 No unreasonable delay 6.4 0.50 Respect for due process 6.5 0.59 No expropriation w/out adequate compensation Civil Justice 7.1 0.69 Accessibility and affordability 7.2 0.70 No discrimination 7.3 0.57 No corruption 7.4 0.54 No improper gov't in ence 7.5 0.27 No unreasonable delay 7.6 0.46 Effective enforcement 7.7 0.71 Impartial and effective ADRs Criminal Justice 8.1 0.49 Effective investigations 8.2 0.45 Timely and effective adjudication 8.3 0.44 Effective correctional system 8.4 0.47 No discrimination 8.5 0.61 No corruption 8.6 0.47 No improper gov't in ence 8.7 0.66 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 70 Cyprus Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.68 23/31 30/46 31/142 Score Change Rank Change -0.01 -3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.66 0.00 24/31 31/46 32/142 Absence of Corruption 0.65 -0.03 24/31 36/46 40/142 Open Government 0.60 0.00 28/31 36/46 40/142 Fundamental Rights 0.72 0.00 24/31 31/46 32/142 Order and Security 0.81 0.00 26/31 34/46 38/142 Regulatory Enforcement 0.66 0.00 20/31 28/46 29/142 Civil Justice 0.62 -0.01 23/31 34/46 40/142 Criminal Justice 0.68 -0.02 17/31 23/46 23/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Cyprus EU, EFTA, and North America High Constraints on Government Powers 1.1 0.73 Limits by legislature 1.2 0.62 Limits by judiciary 1.3 0.68 Independent auditing 1.4 0.53 Sanctions for ofØcial misconduct 1.5 0.63 Non-governmental checks 1.6 0.80 Lawful transition of power Absence of Corruption 2.1 0.63 In the executive branch 2.2 0.86 In the judiciary 2.3 0.71 In the police/military 2.4 0.40 In the legislature Open Government 3.1 0.61 Publicized laws and gov't data 3.2 0.56 Right to information 3.3 0.61 Civic participation 3.4 0.63 Complaint mechanisms Fundamental Rights 4.1 0.61 No discrimination 4.2 0.81 Right to life and security 4.3 0.68 Due process of law 4.4 0.63 Freedom of expression 4.5 0.81 Freedom of religion 4.6 0.82 Right to privacy 4.7 0.75 Freedom of association 4.8 0.63 Labor rights Order and Security 5.1 0.83 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.61 Absence of violent redress Regulatory Enforcement 6.1 0.65 Effective regulatory enforcement 6.2 0.69 No improper inÙuence 6.3 0.58 No unreasonable delay 6.4 0.65 Respect for due process 6.5 0.75 No expropriation w/out adequate compensation Civil Justice 7.1 0.67 Accessibility and affordability 7.2 0.64 No discrimination 7.3 0.71 No corruption 7.4 0.76 No improper gov't inÙuence 7.5 0.29 No unreasonable delay 7.6 0.57 Effective enforcement 7.7 0.71 Impartial and effective ADRs Criminal Justice 8.1 0.71 Effective investigations 8.2 0.64 Timely and effective adjudication 8.3 0.71 Effective correctional system 8.4 0.60 No discrimination 8.5 0.68 No corruption 8.6 0.76 No improper gov't inÙuence 8.7 0.68 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 71 WJP Rule of Law Index 2023 Czechia Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.73 15/31 20/46 20/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.74 0.00 16/31 19/46 20/142 Absence of Corruption 0.66 0.00 23/31 34/46 37/142 Open Government 0.69 0.00 20/31 27/46 28/142 Fundamental Rights 0.78 0.00 16/31 19/46 20/142 Order and Security 0.89 0.00 14/31 18/46 19/142 Regulatory Enforcement 0.71 0.00 17/31 23/46 23/142 Civil Justice 0.69 0.00 14/31 21/46 21/142 Criminal Justice 0.70 0.00 14/31 19/46 19/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Czechia EU, EFTA, and North America High Constraints on Government Powers 1.1 0.72 Limits by legislature 1.2 0.72 Limits by judiciary 1.3 0.75 Independent auditing 1.4 0.62 Sanctions for ofØcial misconduct 1.5 0.74 Non-governmental checks 1.6 0.87 Lawful transition of power Absence of Corruption 2.1 0.62 In the executive branch 2.2 0.87 In the judiciary 2.3 0.79 In the police/military 2.4 0.36 In the legislature Open Government 3.1 0.67 Publicized laws and gov't data 3.2 0.63 Right to information 3.3 0.71 Civic participation 3.4 0.73 Complaint mechanisms Fundamental Rights 4.1 0.72 No discrimination 4.2 0.94 Right to life and security 4.3 0.80 Due process of law 4.4 0.74 Freedom of expression 4.5 0.78 Freedom of religion 4.6 0.74 Right to privacy 4.7 0.79 Freedom of association 4.8 0.73 Labor rights Order and Security 5.1 0.89 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.78 Absence of violent redress Regulatory Enforcement 6.1 0.73 Effective regulatory enforcement 6.2 0.87 No improper inÙuence 6.3 0.62 No unreasonable delay 6.4 0.60 Respect for due process 6.5 0.74 No expropriation w/out adequate compensation Civil Justice 7.1 0.65 Accessibility and affordability 7.2 0.71 No discrimination 7.3 0.80 No corruption 7.4 0.74 No improper gov't inÙuence 7.5 0.52 No unreasonable delay 7.6 0.63 Effective enforcement 7.7 0.81 Impartial and effective ADRs Criminal Justice 8.1 0.61 Effective investigations 8.2 0.66 Timely and effective adjudication 8.3 0.60 Effective correctional system 8.4 0.66 No discrimination 8.5 0.75 No corruption 8.6 0.82 No improper gov't inÙuence 8.7 0.80 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 72 Denmark Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.90 1/31 1/46 1/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.95 0.00 1/31 1/46 1/142 Absence of Corruption 0.96 0.00 1/31 1/46 1/142 Open Government 0.86 0.00 2/31 2/46 2/142 Fundamental Rights 0.92 0.00 1/31 1/46 1/142 Order and Security 0.93 0.00 3/31 4/46 4/142 Regulatory Enforcement 0.88 -0.01 1/31 1/46 1/142 Civil Justice 0.86 -0.02 2/31 2/46 2/142 Criminal Justice 0.83 0.00 2/31 2/46 2/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Denmark EU, EFTA, and North America High Constraints on Government Powers 1.1 0.91 Limits by legislature 1.2 0.95 Limits by judiciary 1.3 0.94 Independent auditing 1.4 0.94 Sanctions for of cial misconduct 1.5 0.97 Non-governmental checks 1.6 0.99 Lawful transition of power Absence of Corruption 2.1 0.94 In the executive branch 2.2 0.99 In the judiciary 2.3 0.98 In the police/military 2.4 0.92 In the legislature Open Government 3.1 0.85 Publicized laws and gov't data 3.2 0.80 Right to information 3.3 0.94 Civic participation 3.4 0.87 Complaint mechanisms Fundamental Rights 4.1 0.80 No discrimination 4.2 0.99 Right to life and security 4.3 0.91 Due process of law 4.4 0.97 Freedom of expression 4.5 0.82 Freedom of religion 4.6 0.97 Right to privacy 4.7 0.98 Freedom of association 4.8 0.95 Labor rights Order and Security 5.1 0.95 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.84 Absence of violent redress Regulatory Enforcement 6.1 0.86 Effective regulatory enforcement 6.2 0.96 No improper in ence 6.3 0.86 No unreasonable delay 6.4 0.88 Respect for due process 6.5 0.85 No expropriation w/out adequate compensation Civil Justice 7.1 0.78 Accessibility and affordability 7.2 0.87 No discrimination 7.3 0.99 No corruption 7.4 0.91 No improper gov't in ence 7.5 0.70 No unreasonable delay 7.6 0.90 Effective enforcement 7.7 0.85 Impartial and effective ADRs Criminal Justice 8.1 0.68 Effective investigations 8.2 0.75 Timely and effective adjudication 8.3 0.77 Effective correctional system 8.4 0.79 No discrimination 8.5 0.99 No corruption 8.6 0.94 No improper gov't in ence 8.7 0.91 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 73 WJP Rule of Law Index 2023 Dominica Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.58 11/32 9/41 53/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.52 0.00 17/32 17/41 73/142 Absence of Corruption 0.60 0.01 11/32 6/41 46/142 Open Government 0.49 -0.01 20/32 23/41 73/142 Fundamental Rights 0.62 0.00 14/32 12/41 52/142 Order and Security 0.73 0.01 8/32 21/41 69/142 Regulatory Enforcement 0.53 0.00 11/32 12/41 58/142 Civil Justice 0.57 0.00 11/32 12/41 55/142 Criminal Justice 0.56 0.00 7/32 6/41 44/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Dominica Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.42 Limits by legislature 1.2 0.57 Limits by judiciary 1.3 0.67 Independent auditing 1.4 0.42 Sanctions for ofØcial misconduct 1.5 0.51 Non-governmental checks 1.6 0.54 Lawful transition of power Absence of Corruption 2.1 0.51 In the executive branch 2.2 0.86 In the judiciary 2.3 0.76 In the police/military 2.4 0.29 In the legislature Open Government 3.1 0.38 Publicized laws and gov't data 3.2 0.51 Right to information 3.3 0.57 Civic participation 3.4 0.52 Complaint mechanisms Fundamental Rights 4.1 0.55 No discrimination 4.2 0.79 Right to life and security 4.3 0.61 Due process of law 4.4 0.51 Freedom of expression 4.5 0.74 Freedom of religion 4.6 0.57 Right to privacy 4.7 0.61 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.84 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.34 Absence of violent redress Regulatory Enforcement 6.1 0.40 Effective regulatory enforcement 6.2 0.71 No improper inÙuence 6.3 0.45 No unreasonable delay 6.4 0.54 Respect for due process 6.5 0.54 No expropriation w/out adequate compensation Civil Justice 7.1 0.76 Accessibility and affordability 7.2 0.58 No discrimination 7.3 0.74 No corruption 7.4 0.73 No improper gov't inÙuence 7.5 0.26 No unreasonable delay 7.6 0.21 Effective enforcement 7.7 0.67 Impartial and effective ADRs Criminal Justice 8.1 0.50 Effective investigations 8.2 0.48 Timely and effective adjudication 8.3 0.34 Effective correctional system 8.4 0.64 No discrimination 8.5 0.71 No corruption 8.6 0.63 No improper gov't inÙuence 8.7 0.61 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 74 Dominican Republic Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 20/32 27/41 86/142 Score Change Rank Change 0.00 8 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.51 0.01 18/32 20/41 77/142 Absence of Corruption 0.39 0.00 20/32 31/41 99/142 Open Government 0.55 0.00 11/32 14/41 56/142 Fundamental Rights 0.58 0.01 17/32 20/41 67/142 Order and Security 0.64 0.00 19/32 29/41 106/142 Regulatory Enforcement 0.42 0.00 25/32 37/41 114/142 Civil Justice 0.43 0.00 23/32 34/41 110/142 Criminal Justice 0.38 0.02 17/32 30/41 92/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Dominican Republic Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.58 Limits by legislature 1.2 0.46 Limits by judiciary 1.3 0.36 Independent auditing 1.4 0.36 Sanctions for ofØcial misconduct 1.5 0.66 Non-governmental checks 1.6 0.63 Lawful transition of power Absence of Corruption 2.1 0.41 In the executive branch 2.2 0.49 In the judiciary 2.3 0.47 In the police/military 2.4 0.20 In the legislature Open Government 3.1 0.36 Publicized laws and gov't data 3.2 0.59 Right to information 3.3 0.61 Civic participation 3.4 0.65 Complaint mechanisms Fundamental Rights 4.1 0.50 No discrimination 4.2 0.65 Right to life and security 4.3 0.40 Due process of law 4.4 0.66 Freedom of expression 4.5 0.64 Freedom of religion 4.6 0.47 Right to privacy 4.7 0.70 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.63 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.28 Absence of violent redress Regulatory Enforcement 6.1 0.43 Effective regulatory enforcement 6.2 0.62 No improper inÙuence 6.3 0.38 No unreasonable delay 6.4 0.24 Respect for due process 6.5 0.45 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.56 No discrimination 7.3 0.46 No corruption 7.4 0.38 No improper gov't inÙuence 7.5 0.24 No unreasonable delay 7.6 0.36 Effective enforcement 7.7 0.56 Impartial and effective ADRs Criminal Justice 8.1 0.35 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.25 Effective correctional system 8.4 0.42 No discrimination 8.5 0.46 No corruption 8.6 0.37 No improper gov't inÙuence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 75 WJP Rule of Law Index 2023 Ecuador Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.47 23/32 32/41 96/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.49 0.00 23/32 25/41 88/142 Absence of Corruption 0.38 -0.03 22/32 33/41 105/142 Open Government 0.52 0.00 16/32 18/41 63/142 Fundamental Rights 0.53 0.00 20/32 25/41 75/142 Order and Security 0.59 -0.01 27/32 37/41 123/142 Regulatory Enforcement 0.47 -0.02 21/32 25/41 88/142 Civil Justice 0.46 -0.02 22/32 31/41 96/142 Criminal Justice 0.33 -0.02 19/32 32/41 109/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Ecuador Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.56 Limits by legislature 1.2 0.44 Limits by judiciary 1.3 0.51 Independent auditing 1.4 0.35 Sanctions for of cial misconduct 1.5 0.53 Non-governmental checks 1.6 0.55 Lawful transition of power Absence of Corruption 2.1 0.38 In the executive branch 2.2 0.38 In the judiciary 2.3 0.52 In the police/military 2.4 0.24 In the legislature Open Government 3.1 0.37 Publicized laws and gov't data 3.2 0.53 Right to information 3.3 0.53 Civic participation 3.4 0.64 Complaint mechanisms Fundamental Rights 4.1 0.40 No discrimination 4.2 0.61 Right to life and security 4.3 0.42 Due process of law 4.4 0.53 Freedom of expression 4.5 0.65 Freedom of religion 4.6 0.44 Right to privacy 4.7 0.62 Freedom of association 4.8 0.55 Labor rights Order and Security 5.1 0.51 Absence of crime 5.2 0.98 Absence of civil con ct 5.3 0.29 Absence of violent redress Regulatory Enforcement 6.1 0.53 Effective regulatory enforcement 6.2 0.52 No improper in ence 6.3 0.42 No unreasonable delay 6.4 0.33 Respect for due process 6.5 0.53 No expropriation w/out adequate compensation Civil Justice 7.1 0.56 Accessibility and affordability 7.2 0.39 No discrimination 7.3 0.38 No corruption 7.4 0.40 No improper gov't in ence 7.5 0.40 No unreasonable delay 7.6 0.44 Effective enforcement 7.7 0.66 Impartial and effective ADRs Criminal Justice 8.1 0.22 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.21 Effective correctional system 8.4 0.33 No discrimination 8.5 0.38 No corruption 8.6 0.39 No improper gov't in ence 8.7 0.42 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 76 Egypt, Arab Rep. Region: Middle East and North Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.35 9/9 34/37 136/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.24 -0.01 9/9 36/37 140/142 Absence of Corruption 0.38 0.00 7/9 18/37 102/142 Open Government 0.23 0.00 9/9 37/37 142/142 Fundamental Rights 0.24 -0.01 8/9 35/37 140/142 Order and Security 0.62 0.00 9/9 27/37 112/142 Regulatory Enforcement 0.36 0.00 9/9 32/37 133/142 Civil Justice 0.38 0.00 9/9 31/37 130/142 Criminal Justice 0.33 -0.01 8/9 24/37 111/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Egypt, Arab Rep. Middle East and North Africa Lower-Middle Constraints on Government Powers 1.1 0.07 Limits by legislature 1.2 0.32 Limits by judiciary 1.3 0.27 Independent auditing 1.4 0.38 Sanctions for of cial misconduct 1.5 0.08 Non-governmental checks 1.6 0.33 Lawful transition of power Absence of Corruption 2.1 0.38 In the executive branch 2.2 0.46 In the judiciary 2.3 0.43 In the police/military 2.4 0.26 In the legislature Open Government 3.1 0.31 Publicized laws and gov't data 3.2 0.12 Right to information 3.3 0.17 Civic participation 3.4 0.32 Complaint mechanisms Fundamental Rights 4.1 0.44 No discrimination 4.2 0.16 Right to life and security 4.3 0.28 Due process of law 4.4 0.08 Freedom of expression 4.5 0.22 Freedom of religion 4.6 0.14 Right to privacy 4.7 0.22 Freedom of association 4.8 0.37 Labor rights Order and Security 5.1 0.77 Absence of crime 5.2 0.83 Absence of civil con ct 5.3 0.28 Absence of violent redress Regulatory Enforcement 6.1 0.44 Effective regulatory enforcement 6.2 0.47 No improper in ence 6.3 0.20 No unreasonable delay 6.4 0.28 Respect for due process 6.5 0.41 No expropriation w/out adequate compensation Civil Justice 7.1 0.46 Accessibility and affordability 7.2 0.30 No discrimination 7.3 0.47 No corruption 7.4 0.35 No improper gov't in ence 7.5 0.27 No unreasonable delay 7.6 0.28 Effective enforcement 7.7 0.52 Impartial and effective ADRs Criminal Justice 8.1 0.40 Effective investigations 8.2 0.34 Timely and effective adjudication 8.3 0.24 Effective correctional system 8.4 0.34 No discrimination 8.5 0.44 No corruption 8.6 0.27 No improper gov't in ence 8.7 0.28 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 77 WJP Rule of Law Index 2023 El Salvador Region: Latin America and Caribbean Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.45 25/32 21/37 108/142 Score Change Rank Change -0.01 -4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.41 -0.01 27/32 21/37 110/142 Absence of Corruption 0.35 -0.01 23/32 22/37 114/142 Open Government 0.47 -0.01 22/32 12/37 82/142 Fundamental Rights 0.46 -0.01 27/32 14/37 102/142 Order and Security 0.66 -0.02 16/32 21/37 100/142 Regulatory Enforcement 0.48 -0.01 18/32 10/37 77/142 Civil Justice 0.49 0.00 17/32 11/37 82/142 Criminal Justice 0.25 -0.02 30/32 35/37 139/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 El Salvador Latin America and Caribbean Lower-Middle Constraints on Government Powers 1.1 0.47 Limits by legislature 1.2 0.39 Limits by judiciary 1.3 0.27 Independent auditing 1.4 0.30 Sanctions for of cial misconduct 1.5 0.47 Non-governmental checks 1.6 0.54 Lawful transition of power Absence of Corruption 2.1 0.32 In the executive branch 2.2 0.48 In the judiciary 2.3 0.52 In the police/military 2.4 0.08 In the legislature Open Government 3.1 0.41 Publicized laws and gov't data 3.2 0.50 Right to information 3.3 0.44 Civic participation 3.4 0.53 Complaint mechanisms Fundamental Rights 4.1 0.52 No discrimination 4.2 0.49 Right to life and security 4.3 0.27 Due process of law 4.4 0.47 Freedom of expression 4.5 0.57 Freedom of religion 4.6 0.36 Right to privacy 4.7 0.47 Freedom of association 4.8 0.52 Labor rights Order and Security 5.1 0.64 Absence of crime 5.2 0.90 Absence of civil con ct 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.40 Effective regulatory enforcement 6.2 0.63 No improper in ence 6.3 0.48 No unreasonable delay 6.4 0.36 Respect for due process 6.5 0.53 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.62 No discrimination 7.3 0.47 No corruption 7.4 0.34 No improper gov't in ence 7.5 0.36 No unreasonable delay 7.6 0.49 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.13 Effective investigations 8.2 0.28 Timely and effective adjudication 8.3 0.23 Effective correctional system 8.4 0.16 No discrimination 8.5 0.39 No corruption 8.6 0.27 No improper gov't in ence 8.7 0.27 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 78 Estonia Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.82 8/31 9/46 9/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.83 0.00 9/31 10/46 10/142 Absence of Corruption 0.81 0.00 11/31 16/46 16/142 Open Government 0.81 0.00 7/31 9/46 9/142 Fundamental Rights 0.83 0.00 10/31 10/46 10/142 Order and Security 0.90 0.01 12/31 16/46 16/142 Regulatory Enforcement 0.81 0.00 10/31 13/46 13/142 Civil Justice 0.81 0.00 7/31 7/46 7/142 Criminal Justice 0.75 0.01 7/31 9/46 9/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Estonia EU, EFTA, and North America High Constraints on Government Powers 1.1 0.81 Limits by legislature 1.2 0.83 Limits by judiciary 1.3 0.88 Independent auditing 1.4 0.79 Sanctions for ofØcial misconduct 1.5 0.79 Non-governmental checks 1.6 0.89 Lawful transition of power Absence of Corruption 2.1 0.75 In the executive branch 2.2 0.96 In the judiciary 2.3 0.92 In the police/military 2.4 0.60 In the legislature Open Government 3.1 0.88 Publicized laws and gov't data 3.2 0.78 Right to information 3.3 0.78 Civic participation 3.4 0.79 Complaint mechanisms Fundamental Rights 4.1 0.83 No discrimination 4.2 0.93 Right to life and security 4.3 0.81 Due process of law 4.4 0.79 Freedom of expression 4.5 0.82 Freedom of religion 4.6 0.88 Right to privacy 4.7 0.83 Freedom of association 4.8 0.72 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.78 Absence of violent redress Regulatory Enforcement 6.1 0.85 Effective regulatory enforcement 6.2 0.90 No improper inÙuence 6.3 0.81 No unreasonable delay 6.4 0.69 Respect for due process 6.5 0.78 No expropriation w/out adequate compensation Civil Justice 7.1 0.70 Accessibility and affordability 7.2 0.86 No discrimination 7.3 0.90 No corruption 7.4 0.87 No improper gov't inÙuence 7.5 0.76 No unreasonable delay 7.6 0.72 Effective enforcement 7.7 0.85 Impartial and effective ADRs Criminal Justice 8.1 0.62 Effective investigations 8.2 0.56 Timely and effective adjudication 8.3 0.73 Effective correctional system 8.4 0.78 No discrimination 8.5 0.88 No corruption 8.6 0.86 No improper gov't inÙuence 8.7 0.81 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 79 WJP Rule of Law Index 2023 Ethiopia Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.38 30/34 15/18 129/142 Score Change Rank Change -0.01 -4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.35 -0.03 30/34 17/18 128/142 Absence of Corruption 0.44 -0.01 8/34 3/18 78/142 Open Government 0.31 -0.02 33/34 18/18 136/142 Fundamental Rights 0.30 -0.02 34/34 17/18 135/142 Order and Security 0.53 0.00 28/34 13/18 131/142 Regulatory Enforcement 0.36 -0.01 30/34 15/18 132/142 Civil Justice 0.42 -0.01 26/34 12/18 118/142 Criminal Justice 0.35 0.00 22/34 9/18 104/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Ethiopia Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.43 Limits by legislature 1.2 0.32 Limits by judiciary 1.3 0.39 Independent auditing 1.4 0.37 Sanctions for of cial misconduct 1.5 0.26 Non-governmental checks 1.6 0.32 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.41 In the judiciary 2.3 0.46 In the police/military 2.4 0.51 In the legislature Open Government 3.1 0.21 Publicized laws and gov't data 3.2 0.37 Right to information 3.3 0.26 Civic participation 3.4 0.38 Complaint mechanisms Fundamental Rights 4.1 0.42 No discrimination 4.2 0.17 Right to life and security 4.3 0.32 Due process of law 4.4 0.26 Freedom of expression 4.5 0.42 Freedom of religion 4.6 0.11 Right to privacy 4.7 0.30 Freedom of association 4.8 0.37 Labor rights Order and Security 5.1 0.62 Absence of crime 5.2 0.50 Absence of civil con ct 5.3 0.47 Absence of violent redress Regulatory Enforcement 6.1 0.33 Effective regulatory enforcement 6.2 0.49 No improper in ence 6.3 0.37 No unreasonable delay 6.4 0.21 Respect for due process 6.5 0.40 No expropriation w/out adequate compensation Civil Justice 7.1 0.42 Accessibility and affordability 7.2 0.36 No discrimination 7.3 0.34 No corruption 7.4 0.27 No improper gov't in ence 7.5 0.41 No unreasonable delay 7.6 0.49 Effective enforcement 7.7 0.61 Impartial and effective ADRs Criminal Justice 8.1 0.34 Effective investigations 8.2 0.37 Timely and effective adjudication 8.3 0.39 Effective correctional system 8.4 0.36 No discrimination 8.5 0.42 No corruption 8.6 0.23 No improper gov't in ence 8.7 0.32 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 80 Finland Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.87 3/31 3/46 3/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.92 0.01 3/31 3/46 3/142 Absence of Corruption 0.89 0.00 4/31 5/46 5/142 Open Government 0.86 0.00 3/31 3/46 3/142 Fundamental Rights 0.90 0.01 3/31 3/46 3/142 Order and Security 0.92 0.00 6/31 9/46 9/142 Regulatory Enforcement 0.87 0.01 3/31 3/46 3/142 Civil Justice 0.81 0.00 6/31 6/46 6/142 Criminal Justice 0.84 0.00 1/31 1/46 1/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Finland EU, EFTA, and North America High Constraints on Government Powers 1.1 0.87 Limits by legislature 1.2 0.90 Limits by judiciary 1.3 0.93 Independent auditing 1.4 0.92 Sanctions for of cial misconduct 1.5 0.89 Non-governmental checks 1.6 0.99 Lawful transition of power Absence of Corruption 2.1 0.89 In the executive branch 2.2 0.98 In the judiciary 2.3 0.97 In the police/military 2.4 0.71 In the legislature Open Government 3.1 0.91 Publicized laws and gov't data 3.2 0.78 Right to information 3.3 0.89 Civic participation 3.4 0.86 Complaint mechanisms Fundamental Rights 4.1 0.84 No discrimination 4.2 0.97 Right to life and security 4.3 0.92 Due process of law 4.4 0.89 Freedom of expression 4.5 0.87 Freedom of religion 4.6 0.96 Right to privacy 4.7 0.92 Freedom of association 4.8 0.85 Labor rights Order and Security 5.1 0.94 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.81 Absence of violent redress Regulatory Enforcement 6.1 0.82 Effective regulatory enforcement 6.2 0.97 No improper in ence 6.3 0.81 No unreasonable delay 6.4 0.94 Respect for due process 6.5 0.80 No expropriation w/out adequate compensation Civil Justice 7.1 0.71 Accessibility and affordability 7.2 0.86 No discrimination 7.3 0.94 No corruption 7.4 0.90 No improper gov't in ence 7.5 0.59 No unreasonable delay 7.6 0.89 Effective enforcement 7.7 0.78 Impartial and effective ADRs Criminal Justice 8.1 0.60 Effective investigations 8.2 0.78 Timely and effective adjudication 8.3 0.84 Effective correctional system 8.4 0.80 No discrimination 8.5 0.93 No corruption 8.6 0.99 No improper gov't in ence 8.7 0.92 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 81 WJP Rule of Law Index 2023 France Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.73 16/31 21/46 21/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.72 -0.01 17/31 21/46 22/142 Absence of Corruption 0.75 0.00 14/31 20/46 20/142 Open Government 0.75 -0.01 14/31 16/46 16/142 Fundamental Rights 0.74 0.00 21/31 26/46 27/142 Order and Security 0.79 0.00 27/31 36/46 46/142 Regulatory Enforcement 0.75 -0.01 15/31 20/46 20/142 Civil Justice 0.69 -0.01 15/31 22/46 22/142 Criminal Justice 0.63 0.00 21/31 28/46 28/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 France EU, EFTA, and North America High Constraints on Government Powers 1.1 0.71 Limits by legislature 1.2 0.67 Limits by judiciary 1.3 0.72 Independent auditing 1.4 0.64 Sanctions for ofØcial misconduct 1.5 0.72 Non-governmental checks 1.6 0.89 Lawful transition of power Absence of Corruption 2.1 0.74 In the executive branch 2.2 0.90 In the judiciary 2.3 0.85 In the police/military 2.4 0.50 In the legislature Open Government 3.1 0.84 Publicized laws and gov't data 3.2 0.65 Right to information 3.3 0.73 Civic participation 3.4 0.78 Complaint mechanisms Fundamental Rights 4.1 0.66 No discrimination 4.2 0.84 Right to life and security 4.3 0.68 Due process of law 4.4 0.72 Freedom of expression 4.5 0.74 Freedom of religion 4.6 0.70 Right to privacy 4.7 0.82 Freedom of association 4.8 0.78 Labor rights Order and Security 5.1 0.82 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.56 Absence of violent redress Regulatory Enforcement 6.1 0.70 Effective regulatory enforcement 6.2 0.85 No improper inÙuence 6.3 0.67 No unreasonable delay 6.4 0.74 Respect for due process 6.5 0.78 No expropriation w/out adequate compensation Civil Justice 7.1 0.63 Accessibility and affordability 7.2 0.67 No discrimination 7.3 0.76 No corruption 7.4 0.76 No improper gov't inÙuence 7.5 0.53 No unreasonable delay 7.6 0.69 Effective enforcement 7.7 0.78 Impartial and effective ADRs Criminal Justice 8.1 0.60 Effective investigations 8.2 0.60 Timely and effective adjudication 8.3 0.54 Effective correctional system 8.4 0.54 No discrimination 8.5 0.77 No corruption 8.6 0.67 No improper gov't inÙuence 8.7 0.68 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 82 Gabon Region: Sub-Saharan Africa Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.39 27/34 40/41 124/142 Score Change Rank Change 0.00 4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.32 0.01 33/34 37/41 133/142 Absence of Corruption 0.24 0.00 33/34 41/41 139/142 Open Government 0.37 0.01 22/34 39/41 118/142 Fundamental Rights 0.43 -0.01 22/34 36/41 115/142 Order and Security 0.63 0.00 19/34 30/41 110/142 Regulatory Enforcement 0.46 -0.01 15/34 28/41 93/142 Civil Justice 0.40 0.03 31/34 38/41 127/142 Criminal Justice 0.29 -0.01 30/34 37/41 126/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Gabon Sub-Saharan Africa Upper-Middle Constraints on Government Powers 1.1 0.38 Limits by legislature 1.2 0.29 Limits by judiciary 1.3 0.37 Independent auditing 1.4 0.25 Sanctions for of cial misconduct 1.5 0.30 Non-governmental checks 1.6 0.30 Lawful transition of power Absence of Corruption 2.1 0.25 In the executive branch 2.2 0.36 In the judiciary 2.3 0.26 In the police/military 2.4 0.09 In the legislature Open Government 3.1 0.37 Publicized laws and gov't data 3.2 0.25 Right to information 3.3 0.38 Civic participation 3.4 0.50 Complaint mechanisms Fundamental Rights 4.1 0.48 No discrimination 4.2 0.45 Right to life and security 4.3 0.39 Due process of law 4.4 0.30 Freedom of expression 4.5 0.60 Freedom of religion 4.6 0.13 Right to privacy 4.7 0.47 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.53 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.36 Absence of violent redress Regulatory Enforcement 6.1 0.50 Effective regulatory enforcement 6.2 0.50 No improper in ence 6.3 0.35 No unreasonable delay 6.4 0.48 Respect for due process 6.5 0.45 No expropriation w/out adequate compensation Civil Justice 7.1 0.46 Accessibility and affordability 7.2 0.52 No discrimination 7.3 0.27 No corruption 7.4 0.25 No improper gov't in ence 7.5 0.39 No unreasonable delay 7.6 0.38 Effective enforcement 7.7 0.53 Impartial and effective ADRs Criminal Justice 8.1 0.28 Effective investigations 8.2 0.22 Timely and effective adjudication 8.3 0.27 Effective correctional system 8.4 0.49 No discrimination 8.5 0.28 No corruption 8.6 0.13 No improper gov't in ence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 83 WJP Rule of Law Index 2023 The Gambia Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.49 9/34 3/18 85/142 Score Change Rank Change 0.00 3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.57 0.00 8/34 3/18 57/142 Absence of Corruption 0.47 0.00 6/34 2/18 67/142 Open Government 0.38 0.01 20/34 10/18 116/142 Fundamental Rights 0.52 0.01 11/34 4/18 79/142 Order and Security 0.70 -0.01 12/34 5/18 87/142 Regulatory Enforcement 0.37 0.01 27/34 12/18 129/142 Civil Justice 0.49 0.00 10/34 3/18 83/142 Criminal Justice 0.40 0.00 11/34 4/18 82/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 The Gambia Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.71 Limits by legislature 1.2 0.66 Limits by judiciary 1.3 0.31 Independent auditing 1.4 0.40 Sanctions for ofØcial misconduct 1.5 0.63 Non-governmental checks 1.6 0.72 Lawful transition of power Absence of Corruption 2.1 0.40 In the executive branch 2.2 0.57 In the judiciary 2.3 0.49 In the police/military 2.4 0.42 In the legislature Open Government 3.1 0.14 Publicized laws and gov't data 3.2 0.40 Right to information 3.3 0.59 Civic participation 3.4 0.40 Complaint mechanisms Fundamental Rights 4.1 0.49 No discrimination 4.2 0.40 Right to life and security 4.3 0.39 Due process of law 4.4 0.63 Freedom of expression 4.5 0.71 Freedom of religion 4.6 0.41 Right to privacy 4.7 0.66 Freedom of association 4.8 0.46 Labor rights Order and Security 5.1 0.56 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.53 Absence of violent redress Regulatory Enforcement 6.1 0.36 Effective regulatory enforcement 6.2 0.50 No improper inÙuence 6.3 0.30 No unreasonable delay 6.4 0.27 Respect for due process 6.5 0.46 No expropriation w/out adequate compensation Civil Justice 7.1 0.44 Accessibility and affordability 7.2 0.67 No discrimination 7.3 0.59 No corruption 7.4 0.55 No improper gov't inÙuence 7.5 0.26 No unreasonable delay 7.6 0.35 Effective enforcement 7.7 0.60 Impartial and effective ADRs Criminal Justice 8.1 0.41 Effective investigations 8.2 0.40 Timely and effective adjudication 8.3 0.12 Effective correctional system 8.4 0.35 No discrimination 8.5 0.55 No corruption 8.6 0.57 No improper gov't inÙuence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 84 Georgia Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.60 1/15 6/41 48/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.53 0.00 2/15 15/41 69/142 Absence of Corruption 0.68 0.00 1/15 2/41 33/142 Open Government 0.59 0.00 1/15 6/41 43/142 Fundamental Rights 0.62 0.01 2/15 11/41 51/142 Order and Security 0.79 0.00 8/15 9/41 47/142 Regulatory Enforcement 0.58 0.01 1/15 6/41 47/142 Civil Justice 0.53 0.00 4/15 18/41 68/142 Criminal Justice 0.52 0.00 1/15 12/41 53/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Georgia Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.57 Limits by legislature 1.2 0.44 Limits by judiciary 1.3 0.58 Independent auditing 1.4 0.39 Sanctions for of cial misconduct 1.5 0.66 Non-governmental checks 1.6 0.55 Lawful transition of power Absence of Corruption 2.1 0.59 In the executive branch 2.2 0.73 In the judiciary 2.3 0.88 In the police/military 2.4 0.50 In the legislature Open Government 3.1 0.52 Publicized laws and gov't data 3.2 0.66 Right to information 3.3 0.61 Civic participation 3.4 0.56 Complaint mechanisms Fundamental Rights 4.1 0.57 No discrimination 4.2 0.76 Right to life and security 4.3 0.59 Due process of law 4.4 0.66 Freedom of expression 4.5 0.66 Freedom of religion 4.6 0.47 Right to privacy 4.7 0.69 Freedom of association 4.8 0.58 Labor rights Order and Security 5.1 0.92 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.57 Effective regulatory enforcement 6.2 0.80 No improper in ence 6.3 0.51 No unreasonable delay 6.4 0.41 Respect for due process 6.5 0.60 No expropriation w/out adequate compensation Civil Justice 7.1 0.62 Accessibility and affordability 7.2 0.54 No discrimination 7.3 0.60 No corruption 7.4 0.39 No improper gov't in ence 7.5 0.34 No unreasonable delay 7.6 0.50 Effective enforcement 7.7 0.75 Impartial and effective ADRs Criminal Justice 8.1 0.36 Effective investigations 8.2 0.56 Timely and effective adjudication 8.3 0.63 Effective correctional system 8.4 0.49 No discrimination 8.5 0.72 No corruption 8.6 0.26 No improper gov't in ence 8.7 0.59 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 85 WJP Rule of Law Index 2023 Germany Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.83 5/31 5/46 5/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.86 0.01 4/31 4/46 4/142 Absence of Corruption 0.82 0.00 9/31 12/46 12/142 Open Government 0.79 0.00 11/31 13/46 13/142 Fundamental Rights 0.86 0.01 5/31 5/46 5/142 Order and Security 0.89 0.00 16/31 20/46 21/142 Regulatory Enforcement 0.84 -0.01 6/31 8/46 8/142 Civil Justice 0.83 0.00 4/31 4/46 4/142 Criminal Justice 0.78 0.00 6/31 6/46 6/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Germany EU, EFTA, and North America High Constraints on Government Powers 1.1 0.84 Limits by legislature 1.2 0.82 Limits by judiciary 1.3 0.89 Independent auditing 1.4 0.82 Sanctions for ofØcial misconduct 1.5 0.84 Non-governmental checks 1.6 0.96 Lawful transition of power Absence of Corruption 2.1 0.81 In the executive branch 2.2 0.95 In the judiciary 2.3 0.91 In the police/military 2.4 0.62 In the legislature Open Government 3.1 0.74 Publicized laws and gov't data 3.2 0.74 Right to information 3.3 0.86 Civic participation 3.4 0.84 Complaint mechanisms Fundamental Rights 4.1 0.77 No discrimination 4.2 0.96 Right to life and security 4.3 0.84 Due process of law 4.4 0.84 Freedom of expression 4.5 0.86 Freedom of religion 4.6 0.84 Right to privacy 4.7 0.90 Freedom of association 4.8 0.85 Labor rights Order and Security 5.1 0.90 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.77 Absence of violent redress Regulatory Enforcement 6.1 0.77 Effective regulatory enforcement 6.2 0.86 No improper inÙuence 6.3 0.80 No unreasonable delay 6.4 0.88 Respect for due process 6.5 0.90 No expropriation w/out adequate compensation Civil Justice 7.1 0.76 Accessibility and affordability 7.2 0.83 No discrimination 7.3 0.89 No corruption 7.4 0.90 No improper gov't inÙuence 7.5 0.74 No unreasonable delay 7.6 0.85 Effective enforcement 7.7 0.83 Impartial and effective ADRs Criminal Justice 8.1 0.59 Effective investigations 8.2 0.77 Timely and effective adjudication 8.3 0.79 Effective correctional system 8.4 0.72 No discrimination 8.5 0.85 No corruption 8.6 0.90 No improper gov't inÙuence 8.7 0.84 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 86 Ghana Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.55 7/34 2/37 61/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.66 0.00 2/34 2/37 35/142 Absence of Corruption 0.39 0.00 16/34 17/37 101/142 Open Government 0.51 0.00 5/34 6/37 68/142 Fundamental Rights 0.58 0.00 7/34 3/37 66/142 Order and Security 0.71 -0.01 7/34 9/37 76/142 Regulatory Enforcement 0.53 0.00 6/34 3/37 57/142 Civil Justice 0.57 -0.01 6/34 1/37 54/142 Criminal Justice 0.45 0.00 8/34 5/37 69/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Ghana Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.76 Limits by legislature 1.2 0.64 Limits by judiciary 1.3 0.63 Independent auditing 1.4 0.50 Sanctions for ofØcial misconduct 1.5 0.70 Non-governmental checks 1.6 0.72 Lawful transition of power Absence of Corruption 2.1 0.38 In the executive branch 2.2 0.54 In the judiciary 2.3 0.37 In the police/military 2.4 0.27 In the legislature Open Government 3.1 0.25 Publicized laws and gov't data 3.2 0.43 Right to information 3.3 0.68 Civic participation 3.4 0.66 Complaint mechanisms Fundamental Rights 4.1 0.60 No discrimination 4.2 0.52 Right to life and security 4.3 0.40 Due process of law 4.4 0.70 Freedom of expression 4.5 0.69 Freedom of religion 4.6 0.44 Right to privacy 4.7 0.72 Freedom of association 4.8 0.55 Labor rights Order and Security 5.1 0.76 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.38 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.53 No improper inÙuence 6.3 0.44 No unreasonable delay 6.4 0.60 Respect for due process 6.5 0.63 No expropriation w/out adequate compensation Civil Justice 7.1 0.58 Accessibility and affordability 7.2 0.58 No discrimination 7.3 0.55 No corruption 7.4 0.65 No improper gov't inÙuence 7.5 0.40 No unreasonable delay 7.6 0.51 Effective enforcement 7.7 0.71 Impartial and effective ADRs Criminal Justice 8.1 0.39 Effective investigations 8.2 0.43 Timely and effective adjudication 8.3 0.37 Effective correctional system 8.4 0.54 No discrimination 8.5 0.40 No corruption 8.6 0.59 No improper gov't inÙuence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 87 WJP Rule of Law Index 2023 Greece Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.61 29/31 41/46 47/142 Score Change Rank Change -0.01 -3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.67 0.00 23/31 29/46 30/142 Absence of Corruption 0.56 -0.02 28/31 42/46 53/142 Open Government 0.61 0.00 27/31 35/46 39/142 Fundamental Rights 0.65 -0.01 28/31 38/46 44/142 Order and Security 0.72 0.00 31/31 42/46 73/142 Regulatory Enforcement 0.55 -0.01 29/31 42/46 51/142 Civil Justice 0.58 -0.01 26/31 38/46 50/142 Criminal Justice 0.50 -0.02 29/31 42/46 56/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Greece EU, EFTA, and North America High Constraints on Government Powers 1.1 0.63 Limits by legislature 1.2 0.64 Limits by judiciary 1.3 0.69 Independent auditing 1.4 0.53 Sanctions for of cial misconduct 1.5 0.68 Non-governmental checks 1.6 0.86 Lawful transition of power Absence of Corruption 2.1 0.51 In the executive branch 2.2 0.78 In the judiciary 2.3 0.74 In the police/military 2.4 0.20 In the legislature Open Government 3.1 0.55 Publicized laws and gov't data 3.2 0.62 Right to information 3.3 0.64 Civic participation 3.4 0.62 Complaint mechanisms Fundamental Rights 4.1 0.58 No discrimination 4.2 0.75 Right to life and security 4.3 0.55 Due process of law 4.4 0.68 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.61 Right to privacy 4.7 0.76 Freedom of association 4.8 0.57 Labor rights Order and Security 5.1 0.82 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.34 Absence of violent redress Regulatory Enforcement 6.1 0.61 Effective regulatory enforcement 6.2 0.62 No improper in ence 6.3 0.42 No unreasonable delay 6.4 0.45 Respect for due process 6.5 0.68 No expropriation w/out adequate compensation Civil Justice 7.1 0.61 Accessibility and affordability 7.2 0.61 No discrimination 7.3 0.71 No corruption 7.4 0.60 No improper gov't in ence 7.5 0.25 No unreasonable delay 7.6 0.52 Effective enforcement 7.7 0.78 Impartial and effective ADRs Criminal Justice 8.1 0.48 Effective investigations 8.2 0.48 Timely and effective adjudication 8.3 0.33 Effective correctional system 8.4 0.43 No discrimination 8.5 0.66 No corruption 8.6 0.55 No improper gov't in ence 8.7 0.55 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 88 Grenada Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.60 9/32 7/41 49/142 Score Change Rank Change 0.01 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.58 0.02 12/32 10/41 52/142 Absence of Corruption 0.66 0.01 5/32 3/41 38/142 Open Government 0.46 0.01 23/32 29/41 87/142 Fundamental Rights 0.62 0.01 13/32 10/41 50/142 Order and Security 0.80 0.01 2/32 8/41 45/142 Regulatory Enforcement 0.55 -0.01 9/32 8/41 53/142 Civil Justice 0.61 0.00 8/32 7/41 43/142 Criminal Justice 0.54 0.02 10/32 10/41 49/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Grenada Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.52 Limits by legislature 1.2 0.70 Limits by judiciary 1.3 0.41 Independent auditing 1.4 0.50 Sanctions for of cial misconduct 1.5 0.64 Non-governmental checks 1.6 0.71 Lawful transition of power Absence of Corruption 2.1 0.64 In the executive branch 2.2 0.84 In the judiciary 2.3 0.75 In the police/military 2.4 0.42 In the legislature Open Government 3.1 0.29 Publicized laws and gov't data 3.2 0.41 Right to information 3.3 0.63 Civic participation 3.4 0.49 Complaint mechanisms Fundamental Rights 4.1 0.69 No discrimination 4.2 0.71 Right to life and security 4.3 0.48 Due process of law 4.4 0.64 Freedom of expression 4.5 0.64 Freedom of religion 4.6 0.40 Right to privacy 4.7 0.71 Freedom of association 4.8 0.72 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.47 Absence of violent redress Regulatory Enforcement 6.1 0.51 Effective regulatory enforcement 6.2 0.79 No improper in ence 6.3 0.52 No unreasonable delay 6.4 0.43 Respect for due process 6.5 0.51 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.69 No discrimination 7.3 0.84 No corruption 7.4 0.72 No improper gov't in ence 7.5 0.37 No unreasonable delay 7.6 0.35 Effective enforcement 7.7 0.75 Impartial and effective ADRs Criminal Justice 8.1 0.50 Effective investigations 8.2 0.56 Timely and effective adjudication 8.3 0.41 Effective correctional system 8.4 0.49 No discrimination 8.5 0.70 No corruption 8.6 0.60 No improper gov't in ence 8.7 0.48 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 89 WJP Rule of Law Index 2023 Guatemala Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.44 26/32 36/41 111/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.49 -0.01 22/32 24/41 86/142 Absence of Corruption 0.34 0.01 24/32 36/41 115/142 Open Government 0.52 0.00 17/32 19/41 64/142 Fundamental Rights 0.53 0.00 19/32 24/41 74/142 Order and Security 0.59 0.00 28/32 38/41 124/142 Regulatory Enforcement 0.40 -0.01 27/32 40/41 122/142 Civil Justice 0.33 -0.01 29/32 40/41 136/142 Criminal Justice 0.30 0.01 24/32 36/41 124/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Guatemala Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.61 Limits by legislature 1.2 0.47 Limits by judiciary 1.3 0.37 Independent auditing 1.4 0.31 Sanctions for of cial misconduct 1.5 0.62 Non-governmental checks 1.6 0.60 Lawful transition of power Absence of Corruption 2.1 0.35 In the executive branch 2.2 0.42 In the judiciary 2.3 0.52 In the police/military 2.4 0.07 In the legislature Open Government 3.1 0.31 Publicized laws and gov't data 3.2 0.55 Right to information 3.3 0.57 Civic participation 3.4 0.63 Complaint mechanisms Fundamental Rights 4.1 0.38 No discrimination 4.2 0.59 Right to life and security 4.3 0.40 Due process of law 4.4 0.62 Freedom of expression 4.5 0.65 Freedom of religion 4.6 0.56 Right to privacy 4.7 0.64 Freedom of association 4.8 0.39 Labor rights Order and Security 5.1 0.57 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.20 Absence of violent redress Regulatory Enforcement 6.1 0.41 Effective regulatory enforcement 6.2 0.56 No improper in ence 6.3 0.29 No unreasonable delay 6.4 0.27 Respect for due process 6.5 0.47 No expropriation w/out adequate compensation Civil Justice 7.1 0.33 Accessibility and affordability 7.2 0.29 No discrimination 7.3 0.42 No corruption 7.4 0.32 No improper gov't in ence 7.5 0.13 No unreasonable delay 7.6 0.24 Effective enforcement 7.7 0.60 Impartial and effective ADRs Criminal Justice 8.1 0.20 Effective investigations 8.2 0.25 Timely and effective adjudication 8.3 0.12 Effective correctional system 8.4 0.31 No discrimination 8.5 0.45 No corruption 8.6 0.34 No improper gov't in ence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 90 Guinea Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.41 22/34 11/18 118/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.38 0.00 25/34 14/18 115/142 Absence of Corruption 0.29 0.01 28/34 15/18 129/142 Open Government 0.37 0.00 24/34 12/18 121/142 Fundamental Rights 0.50 0.00 14/34 7/18 88/142 Order and Security 0.72 0.00 6/34 2/18 74/142 Regulatory Enforcement 0.37 0.01 29/34 14/18 131/142 Civil Justice 0.39 0.00 32/34 15/18 128/142 Criminal Justice 0.29 0.00 29/34 15/18 125/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Guinea Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.45 Limits by legislature 1.2 0.31 Limits by judiciary 1.3 0.36 Independent auditing 1.4 0.31 Sanctions for of cial misconduct 1.5 0.50 Non-governmental checks 1.6 0.37 Lawful transition of power Absence of Corruption 2.1 0.30 In the executive branch 2.2 0.27 In the judiciary 2.3 0.34 In the police/military 2.4 0.26 In the legislature Open Government 3.1 0.19 Publicized laws and gov't data 3.2 0.36 Right to information 3.3 0.49 Civic participation 3.4 0.43 Complaint mechanisms Fundamental Rights 4.1 0.57 No discrimination 4.2 0.44 Right to life and security 4.3 0.35 Due process of law 4.4 0.50 Freedom of expression 4.5 0.70 Freedom of religion 4.6 0.25 Right to privacy 4.7 0.58 Freedom of association 4.8 0.58 Labor rights Order and Security 5.1 0.75 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.40 Absence of violent redress Regulatory Enforcement 6.1 0.38 Effective regulatory enforcement 6.2 0.41 No improper in ence 6.3 0.36 No unreasonable delay 6.4 0.34 Respect for due process 6.5 0.35 No expropriation w/out adequate compensation Civil Justice 7.1 0.40 Accessibility and affordability 7.2 0.49 No discrimination 7.3 0.15 No corruption 7.4 0.28 No improper gov't in ence 7.5 0.46 No unreasonable delay 7.6 0.44 Effective enforcement 7.7 0.49 Impartial and effective ADRs Criminal Justice 8.1 0.24 Effective investigations 8.2 0.34 Timely and effective adjudication 8.3 0.15 Effective correctional system 8.4 0.44 No discrimination 8.5 0.28 No corruption 8.6 0.25 No improper gov't in ence 8.7 0.35 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 91 WJP Rule of Law Index 2023 Guyana Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.50 16/32 22/41 76/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.54 0.00 14/32 14/41 65/142 Absence of Corruption 0.45 0.00 16/32 24/41 76/142 Open Government 0.45 0.01 25/32 32/41 93/142 Fundamental Rights 0.56 -0.01 18/32 22/41 70/142 Order and Security 0.63 0.00 21/32 31/41 111/142 Regulatory Enforcement 0.47 0.01 20/32 24/41 86/142 Civil Justice 0.52 0.00 16/32 20/41 71/142 Criminal Justice 0.40 0.00 14/32 24/41 81/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Guyana Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.61 Limits by legislature 1.2 0.60 Limits by judiciary 1.3 0.46 Independent auditing 1.4 0.39 Sanctions for of cial misconduct 1.5 0.58 Non-governmental checks 1.6 0.59 Lawful transition of power Absence of Corruption 2.1 0.41 In the executive branch 2.2 0.62 In the judiciary 2.3 0.41 In the police/military 2.4 0.34 In the legislature Open Government 3.1 0.30 Publicized laws and gov't data 3.2 0.42 Right to information 3.3 0.54 Civic participation 3.4 0.54 Complaint mechanisms Fundamental Rights 4.1 0.54 No discrimination 4.2 0.57 Right to life and security 4.3 0.41 Due process of law 4.4 0.58 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.43 Right to privacy 4.7 0.59 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.64 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.24 Absence of violent redress Regulatory Enforcement 6.1 0.49 Effective regulatory enforcement 6.2 0.61 No improper in ence 6.3 0.40 No unreasonable delay 6.4 0.38 Respect for due process 6.5 0.48 No expropriation w/out adequate compensation Civil Justice 7.1 0.56 Accessibility and affordability 7.2 0.39 No discrimination 7.3 0.59 No corruption 7.4 0.60 No improper gov't in ence 7.5 0.41 No unreasonable delay 7.6 0.50 Effective enforcement 7.7 0.60 Impartial and effective ADRs Criminal Justice 8.1 0.35 Effective investigations 8.2 0.43 Timely and effective adjudication 8.3 0.17 Effective correctional system 8.4 0.38 No discrimination 8.5 0.48 No corruption 8.6 0.60 No improper gov't in ence 8.7 0.41 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 92 Haiti Region: Latin America and Caribbean Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.34 31/32 36/37 139/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.36 0.00 29/32 29/37 125/142 Absence of Corruption 0.24 -0.01 32/32 36/37 140/142 Open Government 0.35 -0.02 30/32 28/37 126/142 Fundamental Rights 0.42 0.00 30/32 24/37 117/142 Order and Security 0.48 -0.01 32/32 35/37 135/142 Regulatory Enforcement 0.28 -0.03 31/32 36/37 140/142 Civil Justice 0.35 -0.03 28/32 33/37 135/142 Criminal Justice 0.25 0.00 29/32 34/37 138/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Haiti Latin America and Caribbean Lower-Middle Constraints on Government Powers 1.1 0.38 Limits by legislature 1.2 0.36 Limits by judiciary 1.3 0.34 Independent auditing 1.4 0.23 Sanctions for of cial misconduct 1.5 0.51 Non-governmental checks 1.6 0.34 Lawful transition of power Absence of Corruption 2.1 0.23 In the executive branch 2.2 0.35 In the judiciary 2.3 0.32 In the police/military 2.4 0.05 In the legislature Open Government 3.1 0.25 Publicized laws and gov't data 3.2 0.38 Right to information 3.3 0.45 Civic participation 3.4 0.32 Complaint mechanisms Fundamental Rights 4.1 0.48 No discrimination 4.2 0.31 Right to life and security 4.3 0.29 Due process of law 4.4 0.51 Freedom of expression 4.5 0.50 Freedom of religion 4.6 0.27 Right to privacy 4.7 0.50 Freedom of association 4.8 0.48 Labor rights Order and Security 5.1 0.42 Absence of crime 5.2 0.88 Absence of civil con ct 5.3 0.14 Absence of violent redress Regulatory Enforcement 6.1 0.31 Effective regulatory enforcement 6.2 0.40 No improper in ence 6.3 0.26 No unreasonable delay 6.4 0.09 Respect for due process 6.5 0.31 No expropriation w/out adequate compensation Civil Justice 7.1 0.37 Accessibility and affordability 7.2 0.35 No discrimination 7.3 0.33 No corruption 7.4 0.20 No improper gov't in ence 7.5 0.35 No unreasonable delay 7.6 0.37 Effective enforcement 7.7 0.50 Impartial and effective ADRs Criminal Justice 8.1 0.30 Effective investigations 8.2 0.22 Timely and effective adjudication 8.3 0.15 Effective correctional system 8.4 0.30 No discrimination 8.5 0.26 No corruption 8.6 0.22 No improper gov't in ence 8.7 0.29 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 93 WJP Rule of Law Index 2023 Honduras Region: Latin America and Caribbean Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.41 28/32 23/37 119/142 Score Change Rank Change 0.01 4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.38 0.02 28/32 23/37 116/142 Absence of Corruption 0.32 0.00 26/32 25/37 119/142 Open Government 0.45 0.01 24/32 13/37 92/142 Fundamental Rights 0.45 0.02 28/32 16/37 106/142 Order and Security 0.64 -0.01 20/32 24/37 107/142 Regulatory Enforcement 0.39 0.00 28/32 28/37 124/142 Civil Justice 0.40 0.00 26/32 27/37 123/142 Criminal Justice 0.26 0.00 27/32 30/37 133/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Honduras Latin America and Caribbean Lower-Middle Constraints on Government Powers 1.1 0.44 Limits by legislature 1.2 0.33 Limits by judiciary 1.3 0.33 Independent auditing 1.4 0.24 Sanctions for of cial misconduct 1.5 0.50 Non-governmental checks 1.6 0.43 Lawful transition of power Absence of Corruption 2.1 0.35 In the executive branch 2.2 0.38 In the judiciary 2.3 0.47 In the police/military 2.4 0.10 In the legislature Open Government 3.1 0.29 Publicized laws and gov't data 3.2 0.48 Right to information 3.3 0.50 Civic participation 3.4 0.53 Complaint mechanisms Fundamental Rights 4.1 0.42 No discrimination 4.2 0.37 Right to life and security 4.3 0.33 Due process of law 4.4 0.50 Freedom of expression 4.5 0.60 Freedom of religion 4.6 0.31 Right to privacy 4.7 0.58 Freedom of association 4.8 0.49 Labor rights Order and Security 5.1 0.56 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.35 Absence of violent redress Regulatory Enforcement 6.1 0.36 Effective regulatory enforcement 6.2 0.51 No improper in ence 6.3 0.35 No unreasonable delay 6.4 0.23 Respect for due process 6.5 0.51 No expropriation w/out adequate compensation Civil Justice 7.1 0.47 Accessibility and affordability 7.2 0.37 No discrimination 7.3 0.37 No corruption 7.4 0.29 No improper gov't in ence 7.5 0.26 No unreasonable delay 7.6 0.42 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.21 Effective investigations 8.2 0.30 Timely and effective adjudication 8.3 0.14 Effective correctional system 8.4 0.26 No discrimination 8.5 0.40 No corruption 8.6 0.18 No improper gov't in ence 8.7 0.33 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 94 Hong Kong SAR, China Region: East Asia and PaciØc Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.73 6/15 23/46 23/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.57 -0.01 8/15 41/46 59/142 Absence of Corruption 0.83 0.00 3/15 9/46 9/142 Open Government 0.69 -0.01 5/15 25/46 26/142 Fundamental Rights 0.60 0.01 6/15 42/46 58/142 Order and Security 0.93 0.00 2/15 6/46 6/142 Regulatory Enforcement 0.79 0.00 5/15 17/46 17/142 Civil Justice 0.71 -0.01 6/15 20/46 20/142 Criminal Justice 0.69 0.00 6/15 20/46 20/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Hong Kong SAR, China East Asia and PaciØc High Constraints on Government Powers 1.1 0.57 Limits by legislature 1.2 0.64 Limits by judiciary 1.3 0.52 Independent auditing 1.4 0.68 Sanctions for ofØcial misconduct 1.5 0.47 Non-governmental checks 1.6 0.54 Lawful transition of power Absence of Corruption 2.1 0.79 In the executive branch 2.2 0.93 In the judiciary 2.3 0.89 In the police/military 2.4 0.73 In the legislature Open Government 3.1 0.72 Publicized laws and gov't data 3.2 0.77 Right to information 3.3 0.53 Civic participation 3.4 0.74 Complaint mechanisms Fundamental Rights 4.1 0.69 No discrimination 4.2 0.63 Right to life and security 4.3 0.69 Due process of law 4.4 0.47 Freedom of expression 4.5 0.78 Freedom of religion 4.6 0.42 Right to privacy 4.7 0.48 Freedom of association 4.8 0.64 Labor rights Order and Security 5.1 0.94 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.84 Absence of violent redress Regulatory Enforcement 6.1 0.73 Effective regulatory enforcement 6.2 0.98 No improper inÙuence 6.3 0.74 No unreasonable delay 6.4 0.81 Respect for due process 6.5 0.71 No expropriation w/out adequate compensation Civil Justice 7.1 0.63 Accessibility and affordability 7.2 0.63 No discrimination 7.3 0.93 No corruption 7.4 0.63 No improper gov't inÙuence 7.5 0.63 No unreasonable delay 7.6 0.72 Effective enforcement 7.7 0.77 Impartial and effective ADRs Criminal Justice 8.1 0.67 Effective investigations 8.2 0.63 Timely and effective adjudication 8.3 0.76 Effective correctional system 8.4 0.63 No discrimination 8.5 0.88 No corruption 8.6 0.55 No improper gov't inÙuence 8.7 0.69 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 95 WJP Rule of Law Index 2023 Hungary Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.51 31/31 45/46 73/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.37 0.00 31/31 46/46 123/142 Absence of Corruption 0.50 0.01 30/31 44/46 59/142 Open Government 0.45 0.00 31/31 44/46 95/142 Fundamental Rights 0.55 -0.01 31/31 44/46 71/142 Order and Security 0.90 0.00 10/31 14/46 14/142 Regulatory Enforcement 0.45 0.00 31/31 46/46 98/142 Civil Justice 0.45 -0.01 31/31 46/46 105/142 Criminal Justice 0.45 0.00 30/31 44/46 67/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Hungary EU, EFTA, and North America High Constraints on Government Powers 1.1 0.35 Limits by legislature 1.2 0.39 Limits by judiciary 1.3 0.22 Independent auditing 1.4 0.30 Sanctions for of cial misconduct 1.5 0.48 Non-governmental checks 1.6 0.46 Lawful transition of power Absence of Corruption 2.1 0.40 In the executive branch 2.2 0.68 In the judiciary 2.3 0.67 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.54 Publicized laws and gov't data 3.2 0.33 Right to information 3.3 0.43 Civic participation 3.4 0.49 Complaint mechanisms Fundamental Rights 4.1 0.40 No discrimination 4.2 0.74 Right to life and security 4.3 0.55 Due process of law 4.4 0.48 Freedom of expression 4.5 0.57 Freedom of religion 4.6 0.56 Right to privacy 4.7 0.48 Freedom of association 4.8 0.67 Labor rights Order and Security 5.1 0.93 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.78 Absence of violent redress Regulatory Enforcement 6.1 0.51 Effective regulatory enforcement 6.2 0.62 No improper in ence 6.3 0.49 No unreasonable delay 6.4 0.25 Respect for due process 6.5 0.41 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.25 No discrimination 7.3 0.66 No corruption 7.4 0.33 No improper gov't in ence 7.5 0.32 No unreasonable delay 7.6 0.36 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.49 Timely and effective adjudication 8.3 0.43 Effective correctional system 8.4 0.25 No discrimination 8.5 0.64 No corruption 8.6 0.31 No improper gov't in ence 8.7 0.55 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 96 India Region: South Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 3/6 9/37 79/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.57 -0.01 2/6 4/37 58/142 Absence of Corruption 0.40 0.00 3/6 15/37 96/142 Open Government 0.59 0.00 1/6 1/37 42/142 Fundamental Rights 0.46 -0.01 3/6 13/37 99/142 Order and Security 0.64 0.00 3/6 23/37 105/142 Regulatory Enforcement 0.48 0.01 3/6 13/37 83/142 Civil Justice 0.43 0.00 3/6 24/37 111/142 Criminal Justice 0.37 -0.01 3/6 14/37 93/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 India South Asia Lower-Middle Constraints on Government Powers 1.1 0.66 Limits by legislature 1.2 0.61 Limits by judiciary 1.3 0.54 Independent auditing 1.4 0.37 Sanctions for ofØcial misconduct 1.5 0.52 Non-governmental checks 1.6 0.71 Lawful transition of power Absence of Corruption 2.1 0.41 In the executive branch 2.2 0.47 In the judiciary 2.3 0.42 In the police/military 2.4 0.31 In the legislature Open Government 3.1 0.55 Publicized laws and gov't data 3.2 0.58 Right to information 3.3 0.53 Civic participation 3.4 0.70 Complaint mechanisms Fundamental Rights 4.1 0.45 No discrimination 4.2 0.36 Right to life and security 4.3 0.37 Due process of law 4.4 0.52 Freedom of expression 4.5 0.54 Freedom of religion 4.6 0.39 Right to privacy 4.7 0.54 Freedom of association 4.8 0.50 Labor rights Order and Security 5.1 0.78 Absence of crime 5.2 0.83 Absence of civil conÙict 5.3 0.32 Absence of violent redress Regulatory Enforcement 6.1 0.40 Effective regulatory enforcement 6.2 0.44 No improper inÙuence 6.3 0.42 No unreasonable delay 6.4 0.50 Respect for due process 6.5 0.61 No expropriation w/out adequate compensation Civil Justice 7.1 0.40 Accessibility and affordability 7.2 0.37 No discrimination 7.3 0.48 No corruption 7.4 0.59 No improper gov't inÙuence 7.5 0.20 No unreasonable delay 7.6 0.39 Effective enforcement 7.7 0.61 Impartial and effective ADRs Criminal Justice 8.1 0.25 Effective investigations 8.2 0.35 Timely and effective adjudication 8.3 0.36 Effective correctional system 8.4 0.35 No discrimination 8.5 0.45 No corruption 8.6 0.48 No improper gov't inÙuence 8.7 0.37 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 97 WJP Rule of Law Index 2023 Indonesia Region: East Asia and PaciØc Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.53 9/15 4/37 66/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.66 0.00 6/15 1/37 34/142 Absence of Corruption 0.40 0.00 14/15 14/37 95/142 Open Government 0.55 0.00 7/15 2/37 54/142 Fundamental Rights 0.50 0.00 8/15 8/37 85/142 Order and Security 0.71 0.00 12/15 10/37 79/142 Regulatory Enforcement 0.57 0.00 7/15 1/37 48/142 Civil Justice 0.47 0.00 11/15 15/37 93/142 Criminal Justice 0.40 0.01 12/15 11/37 83/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Indonesia East Asia and PaciØc Lower-Middle Constraints on Government Powers 1.1 0.78 Limits by legislature 1.2 0.67 Limits by judiciary 1.3 0.58 Independent auditing 1.4 0.60 Sanctions for ofØcial misconduct 1.5 0.64 Non-governmental checks 1.6 0.70 Lawful transition of power Absence of Corruption 2.1 0.49 In the executive branch 2.2 0.33 In the judiciary 2.3 0.49 In the police/military 2.4 0.30 In the legislature Open Government 3.1 0.39 Publicized laws and gov't data 3.2 0.57 Right to information 3.3 0.61 Civic participation 3.4 0.64 Complaint mechanisms Fundamental Rights 4.1 0.46 No discrimination 4.2 0.48 Right to life and security 4.3 0.40 Due process of law 4.4 0.64 Freedom of expression 4.5 0.42 Freedom of religion 4.6 0.36 Right to privacy 4.7 0.63 Freedom of association 4.8 0.63 Labor rights Order and Security 5.1 0.85 Absence of crime 5.2 0.85 Absence of civil conÙict 5.3 0.41 Absence of violent redress Regulatory Enforcement 6.1 0.58 Effective regulatory enforcement 6.2 0.68 No improper inÙuence 6.3 0.50 No unreasonable delay 6.4 0.48 Respect for due process 6.5 0.60 No expropriation w/out adequate compensation Civil Justice 7.1 0.51 Accessibility and affordability 7.2 0.33 No discrimination 7.3 0.43 No corruption 7.4 0.50 No improper gov't inÙuence 7.5 0.52 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.56 Impartial and effective ADRs Criminal Justice 8.1 0.35 Effective investigations 8.2 0.54 Timely and effective adjudication 8.3 0.33 Effective correctional system 8.4 0.26 No discrimination 8.5 0.46 No corruption 8.6 0.45 No improper gov't inÙuence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 98 Iran, Islamic Rep. Region: Middle East and North Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.39 8/9 27/37 126/142 Score Change Rank Change -0.02 -5 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.32 -0.01 8/9 33/37 131/142 Absence of Corruption 0.37 -0.02 8/9 19/37 107/142 Open Government 0.27 0.00 8/9 35/37 140/142 Fundamental Rights 0.20 0.00 9/9 37/37 142/142 Order and Security 0.63 -0.08 8/9 26/37 109/142 Regulatory Enforcement 0.44 -0.01 7/9 19/37 106/142 Civil Justice 0.51 -0.03 6/9 8/37 76/142 Criminal Justice 0.33 -0.02 7/9 23/37 110/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Iran, Islamic Rep. Middle East and North Africa Lower-Middle Constraints on Government Powers 1.1 0.36 Limits by legislature 1.2 0.34 Limits by judiciary 1.3 0.36 Independent auditing 1.4 0.34 Sanctions for of cial misconduct 1.5 0.17 Non-governmental checks 1.6 0.37 Lawful transition of power Absence of Corruption 2.1 0.34 In the executive branch 2.2 0.46 In the judiciary 2.3 0.47 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.31 Publicized laws and gov't data 3.2 0.30 Right to information 3.3 0.16 Civic participation 3.4 0.30 Complaint mechanisms Fundamental Rights 4.1 0.38 No discrimination 4.2 0.21 Right to life and security 4.3 0.36 Due process of law 4.4 0.17 Freedom of expression 4.5 0.03 Freedom of religion 4.6 0.13 Right to privacy 4.7 0.10 Freedom of association 4.8 0.23 Labor rights Order and Security 5.1 0.76 Absence of crime 5.2 0.58 Absence of civil con ct 5.3 0.56 Absence of violent redress Regulatory Enforcement 6.1 0.35 Effective regulatory enforcement 6.2 0.42 No improper in ence 6.3 0.50 No unreasonable delay 6.4 0.41 Respect for due process 6.5 0.50 No expropriation w/out adequate compensation Civil Justice 7.1 0.63 Accessibility and affordability 7.2 0.35 No discrimination 7.3 0.41 No corruption 7.4 0.42 No improper gov't in ence 7.5 0.56 No unreasonable delay 7.6 0.50 Effective enforcement 7.7 0.72 Impartial and effective ADRs Criminal Justice 8.1 0.26 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.46 Effective correctional system 8.4 0.26 No discrimination 8.5 0.36 No corruption 8.6 0.23 No improper gov't in ence 8.7 0.36 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 99 WJP Rule of Law Index 2023 Ireland Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.81 9/31 10/46 10/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.83 0.00 8/31 9/46 9/142 Absence of Corruption 0.82 0.01 10/31 14/46 14/142 Open Government 0.79 0.00 9/31 11/46 11/142 Fundamental Rights 0.82 0.00 11/31 11/46 11/142 Order and Security 0.95 0.01 2/31 2/46 2/142 Regulatory Enforcement 0.82 0.00 8/31 10/46 10/142 Civil Justice 0.73 0.00 12/31 17/46 17/142 Criminal Justice 0.72 0.00 11/31 15/46 15/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Ireland EU, EFTA, and North America High Constraints on Government Powers 1.1 0.87 Limits by legislature 1.2 0.83 Limits by judiciary 1.3 0.85 Independent auditing 1.4 0.71 Sanctions for ofØcial misconduct 1.5 0.81 Non-governmental checks 1.6 0.95 Lawful transition of power Absence of Corruption 2.1 0.81 In the executive branch 2.2 0.97 In the judiciary 2.3 0.84 In the police/military 2.4 0.65 In the legislature Open Government 3.1 0.86 Publicized laws and gov't data 3.2 0.68 Right to information 3.3 0.81 Civic participation 3.4 0.83 Complaint mechanisms Fundamental Rights 4.1 0.71 No discrimination 4.2 0.88 Right to life and security 4.3 0.81 Due process of law 4.4 0.81 Freedom of expression 4.5 0.78 Freedom of religion 4.6 0.90 Right to privacy 4.7 0.84 Freedom of association 4.8 0.86 Labor rights Order and Security 5.1 0.92 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.92 Absence of violent redress Regulatory Enforcement 6.1 0.72 Effective regulatory enforcement 6.2 0.92 No improper inÙuence 6.3 0.63 No unreasonable delay 6.4 0.92 Respect for due process 6.5 0.92 No expropriation w/out adequate compensation Civil Justice 7.1 0.61 Accessibility and affordability 7.2 0.70 No discrimination 7.3 0.88 No corruption 7.4 0.93 No improper gov't inÙuence 7.5 0.65 No unreasonable delay 7.6 0.52 Effective enforcement 7.7 0.81 Impartial and effective ADRs Criminal Justice 8.1 0.69 Effective investigations 8.2 0.65 Timely and effective adjudication 8.3 0.69 Effective correctional system 8.4 0.56 No discrimination 8.5 0.82 No corruption 8.6 0.85 No improper gov't inÙuence 8.7 0.81 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 100 Italy Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.67 24/31 31/46 32/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.71 -0.01 19/31 23/46 24/142 Absence of Corruption 0.65 0.00 25/31 37/46 41/142 Open Government 0.63 0.00 24/31 31/46 32/142 Fundamental Rights 0.73 0.00 23/31 29/46 30/142 Order and Security 0.75 0.00 30/31 40/46 62/142 Regulatory Enforcement 0.64 0.00 22/31 32/46 33/142 Civil Justice 0.58 0.00 27/31 39/46 51/142 Criminal Justice 0.64 0.00 19/31 26/46 26/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Italy EU, EFTA, and North America High Constraints on Government Powers 1.1 0.72 Limits by legislature 1.2 0.70 Limits by judiciary 1.3 0.76 Independent auditing 1.4 0.62 Sanctions for ofØcial misconduct 1.5 0.68 Non-governmental checks 1.6 0.80 Lawful transition of power Absence of Corruption 2.1 0.59 In the executive branch 2.2 0.83 In the judiciary 2.3 0.86 In the police/military 2.4 0.31 In the legislature Open Government 3.1 0.60 Publicized laws and gov't data 3.2 0.66 Right to information 3.3 0.66 Civic participation 3.4 0.61 Complaint mechanisms Fundamental Rights 4.1 0.68 No discrimination 4.2 0.84 Right to life and security 4.3 0.70 Due process of law 4.4 0.68 Freedom of expression 4.5 0.74 Freedom of religion 4.6 0.81 Right to privacy 4.7 0.78 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.81 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.64 Effective regulatory enforcement 6.2 0.82 No improper inÙuence 6.3 0.45 No unreasonable delay 6.4 0.62 Respect for due process 6.5 0.66 No expropriation w/out adequate compensation Civil Justice 7.1 0.61 Accessibility and affordability 7.2 0.66 No discrimination 7.3 0.69 No corruption 7.4 0.70 No improper gov't inÙuence 7.5 0.32 No unreasonable delay 7.6 0.35 Effective enforcement 7.7 0.72 Impartial and effective ADRs Criminal Justice 8.1 0.48 Effective investigations 8.2 0.56 Timely and effective adjudication 8.3 0.53 Effective correctional system 8.4 0.62 No discrimination 8.5 0.75 No corruption 8.6 0.83 No improper gov't inÙuence 8.7 0.70 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 101 WJP Rule of Law Index 2023 Jamaica Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.57 12/32 10/41 54/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.64 -0.01 5/32 3/41 39/142 Absence of Corruption 0.53 -0.02 12/32 11/41 55/142 Open Government 0.57 0.00 8/32 12/41 50/142 Fundamental Rights 0.64 -0.01 10/32 8/41 46/142 Order and Security 0.62 -0.01 22/32 32/41 114/142 Regulatory Enforcement 0.55 0.00 8/32 7/41 52/142 Civil Justice 0.56 0.01 13/32 14/41 61/142 Criminal Justice 0.49 -0.01 13/32 14/41 57/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Jamaica Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.62 Limits by legislature 1.2 0.69 Limits by judiciary 1.3 0.68 Independent auditing 1.4 0.44 Sanctions for ofØcial misconduct 1.5 0.64 Non-governmental checks 1.6 0.74 Lawful transition of power Absence of Corruption 2.1 0.46 In the executive branch 2.2 0.84 In the judiciary 2.3 0.59 In the police/military 2.4 0.24 In the legislature Open Government 3.1 0.38 Publicized laws and gov't data 3.2 0.62 Right to information 3.3 0.64 Civic participation 3.4 0.64 Complaint mechanisms Fundamental Rights 4.1 0.63 No discrimination 4.2 0.61 Right to life and security 4.3 0.50 Due process of law 4.4 0.64 Freedom of expression 4.5 0.73 Freedom of religion 4.6 0.68 Right to privacy 4.7 0.70 Freedom of association 4.8 0.64 Labor rights Order and Security 5.1 0.63 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.23 Absence of violent redress Regulatory Enforcement 6.1 0.53 Effective regulatory enforcement 6.2 0.70 No improper inÙuence 6.3 0.37 No unreasonable delay 6.4 0.53 Respect for due process 6.5 0.63 No expropriation w/out adequate compensation Civil Justice 7.1 0.52 Accessibility and affordability 7.2 0.59 No discrimination 7.3 0.76 No corruption 7.4 0.74 No improper gov't inÙuence 7.5 0.26 No unreasonable delay 7.6 0.36 Effective enforcement 7.7 0.67 Impartial and effective ADRs Criminal Justice 8.1 0.37 Effective investigations 8.2 0.35 Timely and effective adjudication 8.3 0.32 Effective correctional system 8.4 0.50 No discrimination 8.5 0.65 No corruption 8.6 0.72 No improper gov't inÙuence 8.7 0.50 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 102 Japan Region: East Asia and PaciØc Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.79 3/15 14/46 14/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.73 0.00 3/15 20/46 21/142 Absence of Corruption 0.82 0.00 4/15 13/46 13/142 Open Government 0.70 0.00 4/15 21/46 21/142 Fundamental Rights 0.78 0.00 2/15 18/46 19/142 Order and Security 0.92 0.00 3/15 8/46 8/142 Regulatory Enforcement 0.80 0.00 4/15 15/46 15/142 Civil Justice 0.76 -0.01 3/15 12/46 12/142 Criminal Justice 0.76 0.00 2/15 8/46 8/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Japan East Asia and PaciØc High Constraints on Government Powers 1.1 0.63 Limits by legislature 1.2 0.71 Limits by judiciary 1.3 0.77 Independent auditing 1.4 0.71 Sanctions for ofØcial misconduct 1.5 0.72 Non-governmental checks 1.6 0.84 Lawful transition of power Absence of Corruption 2.1 0.79 In the executive branch 2.2 0.98 In the judiciary 2.3 0.93 In the police/military 2.4 0.59 In the legislature Open Government 3.1 0.76 Publicized laws and gov't data 3.2 0.63 Right to information 3.3 0.69 Civic participation 3.4 0.74 Complaint mechanisms Fundamental Rights 4.1 0.81 No discrimination 4.2 0.91 Right to life and security 4.3 0.76 Due process of law 4.4 0.72 Freedom of expression 4.5 0.69 Freedom of religion 4.6 0.89 Right to privacy 4.7 0.73 Freedom of association 4.8 0.75 Labor rights Order and Security 5.1 0.93 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.84 Absence of violent redress Regulatory Enforcement 6.1 0.71 Effective regulatory enforcement 6.2 0.95 No improper inÙuence 6.3 0.78 No unreasonable delay 6.4 0.79 Respect for due process 6.5 0.76 No expropriation w/out adequate compensation Civil Justice 7.1 0.65 Accessibility and affordability 7.2 0.76 No discrimination 7.3 0.95 No corruption 7.4 0.77 No improper gov't inÙuence 7.5 0.70 No unreasonable delay 7.6 0.70 Effective enforcement 7.7 0.81 Impartial and effective ADRs Criminal Justice 8.1 0.67 Effective investigations 8.2 0.66 Timely and effective adjudication 8.3 0.81 Effective correctional system 8.4 0.78 No discrimination 8.5 0.91 No corruption 8.6 0.77 No improper gov't inÙuence 8.7 0.76 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 103 WJP Rule of Law Index 2023 Jordan Region: Middle East and North Africa Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.55 3/9 16/41 62/142 Score Change Rank Change 0.01 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.46 0.01 7/9 29/41 99/142 Absence of Corruption 0.58 0.00 3/9 7/41 47/142 Open Government 0.39 0.01 5/9 37/41 109/142 Fundamental Rights 0.46 0.00 3/9 35/41 103/142 Order and Security 0.76 0.00 3/9 14/41 56/142 Regulatory Enforcement 0.55 0.01 3/9 9/41 54/142 Civil Justice 0.61 0.01 2/9 8/41 44/142 Criminal Justice 0.57 0.02 2/9 4/41 39/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Jordan Middle East and North Africa Upper-Middle Constraints on Government Powers 1.1 0.43 Limits by legislature 1.2 0.47 Limits by judiciary 1.3 0.53 Independent auditing 1.4 0.52 Sanctions for ofØcial misconduct 1.5 0.39 Non-governmental checks 1.6 0.42 Lawful transition of power Absence of Corruption 2.1 0.54 In the executive branch 2.2 0.71 In the judiciary 2.3 0.76 In the police/military 2.4 0.31 In the legislature Open Government 3.1 0.35 Publicized laws and gov't data 3.2 0.54 Right to information 3.3 0.34 Civic participation 3.4 0.35 Complaint mechanisms Fundamental Rights 4.1 0.61 No discrimination 4.2 0.49 Right to life and security 4.3 0.50 Due process of law 4.4 0.39 Freedom of expression 4.5 0.41 Freedom of religion 4.6 0.34 Right to privacy 4.7 0.37 Freedom of association 4.8 0.53 Labor rights Order and Security 5.1 0.85 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.44 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.71 No improper inÙuence 6.3 0.58 No unreasonable delay 6.4 0.47 Respect for due process 6.5 0.53 No expropriation w/out adequate compensation Civil Justice 7.1 0.58 Accessibility and affordability 7.2 0.72 No discrimination 7.3 0.74 No corruption 7.4 0.59 No improper gov't inÙuence 7.5 0.37 No unreasonable delay 7.6 0.58 Effective enforcement 7.7 0.70 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.64 Timely and effective adjudication 8.3 0.51 Effective correctional system 8.4 0.57 No discrimination 8.5 0.68 No corruption 8.6 0.56 No improper gov't inÙuence 8.7 0.50 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 104 Kazakhstan Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.53 4/15 18/41 65/142 Score Change Rank Change 0.01 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.45 0.02 8/15 32/41 103/142 Absence of Corruption 0.48 0.00 4/15 16/41 63/142 Open Government 0.47 0.01 10/15 28/41 84/142 Fundamental Rights 0.46 0.00 11/15 33/41 100/142 Order and Security 0.80 0.00 7/15 7/41 44/142 Regulatory Enforcement 0.52 0.00 2/15 15/41 61/142 Civil Justice 0.63 0.02 1/15 4/41 36/142 Criminal Justice 0.47 0.00 3/15 16/41 61/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Kazakhstan Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.48 Limits by legislature 1.2 0.44 Limits by judiciary 1.3 0.48 Independent auditing 1.4 0.47 Sanctions for ofØcial misconduct 1.5 0.38 Non-governmental checks 1.6 0.42 Lawful transition of power Absence of Corruption 2.1 0.48 In the executive branch 2.2 0.56 In the judiciary 2.3 0.53 In the police/military 2.4 0.36 In the legislature Open Government 3.1 0.53 Publicized laws and gov't data 3.2 0.43 Right to information 3.3 0.38 Civic participation 3.4 0.53 Complaint mechanisms Fundamental Rights 4.1 0.59 No discrimination 4.2 0.51 Right to life and security 4.3 0.46 Due process of law 4.4 0.38 Freedom of expression 4.5 0.54 Freedom of religion 4.6 0.28 Right to privacy 4.7 0.37 Freedom of association 4.8 0.54 Labor rights Order and Security 5.1 0.79 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.60 Absence of violent redress Regulatory Enforcement 6.1 0.58 Effective regulatory enforcement 6.2 0.65 No improper inÙuence 6.3 0.60 No unreasonable delay 6.4 0.30 Respect for due process 6.5 0.46 No expropriation w/out adequate compensation Civil Justice 7.1 0.61 Accessibility and affordability 7.2 0.56 No discrimination 7.3 0.56 No corruption 7.4 0.39 No improper gov't inÙuence 7.5 0.84 No unreasonable delay 7.6 0.69 Effective enforcement 7.7 0.75 Impartial and effective ADRs Criminal Justice 8.1 0.44 Effective investigations 8.2 0.64 Timely and effective adjudication 8.3 0.46 Effective correctional system 8.4 0.45 No discrimination 8.5 0.53 No corruption 8.6 0.32 No improper gov't inÙuence 8.7 0.46 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 105 WJP Rule of Law Index 2023 Kenya Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.46 13/34 17/37 101/142 Score Change Rank Change 0.01 5 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.50 0.00 13/34 12/37 83/142 Absence of Corruption 0.27 -0.01 29/34 33/37 133/142 Open Government 0.50 0.01 6/34 8/37 71/142 Fundamental Rights 0.47 0.00 16/34 10/37 94/142 Order and Security 0.58 0.03 24/34 32/37 126/142 Regulatory Enforcement 0.45 0.00 16/34 16/37 100/142 Civil Justice 0.49 0.02 11/34 12/37 85/142 Criminal Justice 0.39 0.00 14/34 13/37 88/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Kenya Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.63 Limits by legislature 1.2 0.52 Limits by judiciary 1.3 0.45 Independent auditing 1.4 0.33 Sanctions for ofØcial misconduct 1.5 0.54 Non-governmental checks 1.6 0.51 Lawful transition of power Absence of Corruption 2.1 0.28 In the executive branch 2.2 0.45 In the judiciary 2.3 0.26 In the police/military 2.4 0.08 In the legislature Open Government 3.1 0.31 Publicized laws and gov't data 3.2 0.41 Right to information 3.3 0.58 Civic participation 3.4 0.70 Complaint mechanisms Fundamental Rights 4.1 0.49 No discrimination 4.2 0.30 Right to life and security 4.3 0.39 Due process of law 4.4 0.54 Freedom of expression 4.5 0.66 Freedom of religion 4.6 0.23 Right to privacy 4.7 0.62 Freedom of association 4.8 0.54 Labor rights Order and Security 5.1 0.58 Absence of crime 5.2 0.83 Absence of civil conÙict 5.3 0.33 Absence of violent redress Regulatory Enforcement 6.1 0.44 Effective regulatory enforcement 6.2 0.46 No improper inÙuence 6.3 0.42 No unreasonable delay 6.4 0.37 Respect for due process 6.5 0.57 No expropriation w/out adequate compensation Civil Justice 7.1 0.47 Accessibility and affordability 7.2 0.54 No discrimination 7.3 0.43 No corruption 7.4 0.51 No improper gov't inÙuence 7.5 0.33 No unreasonable delay 7.6 0.48 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.35 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.45 Effective correctional system 8.4 0.40 No discrimination 8.5 0.31 No corruption 8.6 0.44 No improper gov't inÙuence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 106 Korea, Rep. Region: East Asia and PaciØc Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.74 5/15 19/46 19/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.71 0.01 4/15 26/46 27/142 Absence of Corruption 0.67 0.00 6/15 31/46 34/142 Open Government 0.72 0.01 3/15 20/46 20/142 Fundamental Rights 0.75 0.00 4/15 24/46 25/142 Order and Security 0.84 0.00 6/15 27/46 29/142 Regulatory Enforcement 0.74 0.00 6/15 21/46 21/142 Civil Justice 0.75 0.00 4/15 13/46 13/142 Criminal Justice 0.71 0.01 5/15 16/46 16/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Korea, Rep. East Asia and PaciØc High Constraints on Government Powers 1.1 0.67 Limits by legislature 1.2 0.60 Limits by judiciary 1.3 0.82 Independent auditing 1.4 0.66 Sanctions for ofØcial misconduct 1.5 0.66 Non-governmental checks 1.6 0.83 Lawful transition of power Absence of Corruption 2.1 0.70 In the executive branch 2.2 0.88 In the judiciary 2.3 0.78 In the police/military 2.4 0.31 In the legislature Open Government 3.1 0.71 Publicized laws and gov't data 3.2 0.75 Right to information 3.3 0.67 Civic participation 3.4 0.74 Complaint mechanisms Fundamental Rights 4.1 0.73 No discrimination 4.2 0.92 Right to life and security 4.3 0.80 Due process of law 4.4 0.66 Freedom of expression 4.5 0.73 Freedom of religion 4.6 0.86 Right to privacy 4.7 0.72 Freedom of association 4.8 0.61 Labor rights Order and Security 5.1 0.92 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.61 Absence of violent redress Regulatory Enforcement 6.1 0.60 Effective regulatory enforcement 6.2 0.86 No improper inÙuence 6.3 0.80 No unreasonable delay 6.4 0.67 Respect for due process 6.5 0.76 No expropriation w/out adequate compensation Civil Justice 7.1 0.70 Accessibility and affordability 7.2 0.74 No discrimination 7.3 0.75 No corruption 7.4 0.73 No improper gov't inÙuence 7.5 0.74 No unreasonable delay 7.6 0.76 Effective enforcement 7.7 0.86 Impartial and effective ADRs Criminal Justice 8.1 0.58 Effective investigations 8.2 0.75 Timely and effective adjudication 8.3 0.75 Effective correctional system 8.4 0.73 No discrimination 8.5 0.67 No corruption 8.6 0.70 No improper gov't inÙuence 8.7 0.80 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 107 WJP Rule of Law Index 2023 Kosovo Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.56 3/15 14/41 58/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.56 0.00 1/15 13/41 61/142 Absence of Corruption 0.48 -0.01 5/15 17/41 64/142 Open Government 0.58 0.00 2/15 9/41 47/142 Fundamental Rights 0.61 0.01 3/15 14/41 55/142 Order and Security 0.83 0.01 2/15 1/41 32/142 Regulatory Enforcement 0.48 0.00 4/15 21/41 80/142 Civil Justice 0.50 0.01 10/15 25/41 80/142 Criminal Justice 0.46 0.00 5/15 18/41 65/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Kosovo Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.62 Limits by legislature 1.2 0.46 Limits by judiciary 1.3 0.57 Independent auditing 1.4 0.46 Sanctions for ofØcial misconduct 1.5 0.59 Non-governmental checks 1.6 0.67 Lawful transition of power Absence of Corruption 2.1 0.48 In the executive branch 2.2 0.48 In the judiciary 2.3 0.61 In the police/military 2.4 0.33 In the legislature Open Government 3.1 0.49 Publicized laws and gov't data 3.2 0.54 Right to information 3.3 0.58 Civic participation 3.4 0.70 Complaint mechanisms Fundamental Rights 4.1 0.61 No discrimination 4.2 0.70 Right to life and security 4.3 0.60 Due process of law 4.4 0.59 Freedom of expression 4.5 0.70 Freedom of religion 4.6 0.53 Right to privacy 4.7 0.63 Freedom of association 4.8 0.50 Labor rights Order and Security 5.1 0.89 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.61 Absence of violent redress Regulatory Enforcement 6.1 0.38 Effective regulatory enforcement 6.2 0.58 No improper inÙuence 6.3 0.47 No unreasonable delay 6.4 0.37 Respect for due process 6.5 0.58 No expropriation w/out adequate compensation Civil Justice 7.1 0.62 Accessibility and affordability 7.2 0.53 No discrimination 7.3 0.44 No corruption 7.4 0.51 No improper gov't inÙuence 7.5 0.26 No unreasonable delay 7.6 0.49 Effective enforcement 7.7 0.64 Impartial and effective ADRs Criminal Justice 8.1 0.33 Effective investigations 8.2 0.33 Timely and effective adjudication 8.3 0.56 Effective correctional system 8.4 0.56 No discrimination 8.5 0.48 No corruption 8.6 0.39 No improper gov't inÙuence 8.7 0.60 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 108 Kuwait Region: Middle East and North Africa Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.58 2/9 43/46 52/142 Score Change Rank Change Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.55 2/9 42/46 62/142 Absence of Corruption 0.66 — — 2/9 33/46 36/142 Open Government 0.43 — 4/9 45/46 102/142 Fundamental Rights 0.45 — 4/9 45/46 107/142 Order and Security 0.87 — 2/9 23/46 24/142 Regulatory Enforcement 0.66 — 2/9 29/46 30/142 Civil Justice 0.58 — 3/9 40/46 52/142 Criminal Justice 0.47 — 3/9 43/46 62/142 Indicates statistically signi cant change at the 10 percent level Low Medium High Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 2023 Score 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Kuwait Middle East and North Africa High Constraints on Government Powers 1.1 0.61 Limits by legislature 1.2 0.58 Limits by judiciary 1.3 0.37 Independent auditing 1.4 0.63 Sanctions for of cial misconduct 1.5 0.54 Non-governmental checks 1.6 0.57 Lawful transition of power Absence of Corruption 2.1 0.60 In the executive branch 2.2 0.71 In the judiciary 2.3 0.78 In the police/military 2.4 0.57 In the legislature Open Government 3.1 0.30 Publicized laws and gov't data 3.2 0.38 Right to information 3.3 0.50 Civic participation 3.4 0.52 Complaint mechanisms Fundamental Rights 4.1 0.36 No discrimination 4.2 0.48 Right to life and security 4.3 0.50 Due process of law 4.4 0.54 Freedom of expression 4.5 0.31 Freedom of religion 4.6 0.35 Right to privacy 4.7 0.51 Freedom of association 4.8 0.55 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.71 Absence of violent redress Regulatory Enforcement 6.1 0.79 Effective regulatory enforcement 6.2 0.74 No improper in ence 6.3 0.62 No unreasonable delay 6.4 0.46 Respect for due process 6.5 0.69 No expropriation w/out adequate compensation Civil Justice 7.1 0.61 Accessibility and affordability 7.2 0.49 No discrimination 7.3 0.70 No corruption 7.4 0.63 No improper gov't in ence 7.5 0.30 No unreasonable delay 7.6 0.54 Effective enforcement 7.7 0.75 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.38 Timely and effective adjudication 8.3 0.60 Effective correctional system 8.4 0.24 No discrimination 8.5 0.78 No corruption 8.6 0.27 No improper gov't in ence 8.7 0.50 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 109 WJP Rule of Law Index 2023 Kyrgyz Republic Region: Eastern Europe and Central Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.45 12/15 18/37 103/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.43 -0.01 9/15 20/37 105/142 Absence of Corruption 0.29 -0.01 15/15 32/37 130/142 Open Government 0.50 -0.01 6/15 7/37 69/142 Fundamental Rights 0.46 -0.01 10/15 11/37 96/142 Order and Security 0.73 -0.01 12/15 6/37 68/142 Regulatory Enforcement 0.41 -0.01 15/15 25/37 117/142 Civil Justice 0.48 0.01 11/15 13/37 89/142 Criminal Justice 0.32 0.00 14/15 25/37 116/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Kyrgyz Republic Eastern Europe and Central Asia Lower-Middle Constraints on Government Powers 1.1 0.50 Limits by legislature 1.2 0.28 Limits by judiciary 1.3 0.40 Independent auditing 1.4 0.42 Sanctions for of cial misconduct 1.5 0.54 Non-governmental checks 1.6 0.43 Lawful transition of power Absence of Corruption 2.1 0.34 In the executive branch 2.2 0.34 In the judiciary 2.3 0.35 In the police/military 2.4 0.13 In the legislature Open Government 3.1 0.51 Publicized laws and gov't data 3.2 0.50 Right to information 3.3 0.54 Civic participation 3.4 0.47 Complaint mechanisms Fundamental Rights 4.1 0.51 No discrimination 4.2 0.45 Right to life and security 4.3 0.34 Due process of law 4.4 0.54 Freedom of expression 4.5 0.56 Freedom of religion 4.6 0.19 Right to privacy 4.7 0.59 Freedom of association 4.8 0.52 Labor rights Order and Security 5.1 0.80 Absence of crime 5.2 0.94 Absence of civil con ct 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.48 Effective regulatory enforcement 6.2 0.39 No improper in ence 6.3 0.49 No unreasonable delay 6.4 0.29 Respect for due process 6.5 0.41 No expropriation w/out adequate compensation Civil Justice 7.1 0.60 Accessibility and affordability 7.2 0.52 No discrimination 7.3 0.34 No corruption 7.4 0.30 No improper gov't in ence 7.5 0.51 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.67 Impartial and effective ADRs Criminal Justice 8.1 0.37 Effective investigations 8.2 0.52 Timely and effective adjudication 8.3 0.27 Effective correctional system 8.4 0.29 No discrimination 8.5 0.27 No corruption 8.6 0.19 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 110 Latvia Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.73 17/31 22/46 22/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.71 0.00 20/31 25/46 26/142 Absence of Corruption 0.68 0.01 20/31 29/46 30/142 Open Government 0.72 0.00 16/31 19/46 19/142 Fundamental Rights 0.77 0.00 17/31 21/46 22/142 Order and Security 0.86 0.01 17/31 24/46 25/142 Regulatory Enforcement 0.71 0.00 18/31 24/46 24/142 Civil Justice 0.69 0.01 16/31 23/46 23/142 Criminal Justice 0.68 0.01 16/31 22/46 22/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Latvia EU, EFTA, and North America High Constraints on Government Powers 1.1 0.75 Limits by legislature 1.2 0.65 Limits by judiciary 1.3 0.73 Independent auditing 1.4 0.60 Sanctions for ofØcial misconduct 1.5 0.70 Non-governmental checks 1.6 0.81 Lawful transition of power Absence of Corruption 2.1 0.64 In the executive branch 2.2 0.84 In the judiciary 2.3 0.73 In the police/military 2.4 0.50 In the legislature Open Government 3.1 0.79 Publicized laws and gov't data 3.2 0.66 Right to information 3.3 0.68 Civic participation 3.4 0.75 Complaint mechanisms Fundamental Rights 4.1 0.74 No discrimination 4.2 0.91 Right to life and security 4.3 0.72 Due process of law 4.4 0.70 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.84 Right to privacy 4.7 0.75 Freedom of association 4.8 0.78 Labor rights Order and Security 5.1 0.84 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.75 Absence of violent redress Regulatory Enforcement 6.1 0.78 Effective regulatory enforcement 6.2 0.82 No improper inÙuence 6.3 0.68 No unreasonable delay 6.4 0.56 Respect for due process 6.5 0.71 No expropriation w/out adequate compensation Civil Justice 7.1 0.62 Accessibility and affordability 7.2 0.73 No discrimination 7.3 0.72 No corruption 7.4 0.73 No improper gov't inÙuence 7.5 0.48 No unreasonable delay 7.6 0.82 Effective enforcement 7.7 0.72 Impartial and effective ADRs Criminal Justice 8.1 0.56 Effective investigations 8.2 0.56 Timely and effective adjudication 8.3 0.69 Effective correctional system 8.4 0.76 No discrimination 8.5 0.69 No corruption 8.6 0.80 No improper gov't inÙuence 8.7 0.72 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 111 WJP Rule of Law Index 2023 Lebanon Region: Middle East and North Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.45 7/9 20/37 107/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.49 0.01 5/9 14/37 89/142 Absence of Corruption 0.36 0.00 9/9 20/37 109/142 Open Government 0.44 0.01 2/9 16/37 99/142 Fundamental Rights 0.46 0.00 2/9 12/37 97/142 Order and Security 0.68 0.00 7/9 16/37 93/142 Regulatory Enforcement 0.44 0.00 8/9 20/37 107/142 Civil Justice 0.40 -0.01 8/9 28/37 124/142 Criminal Justice 0.32 -0.01 9/9 26/37 118/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Lebanon Middle East and North Africa Lower-Middle Constraints on Government Powers 1.1 0.67 Limits by legislature 1.2 0.44 Limits by judiciary 1.3 0.48 Independent auditing 1.4 0.32 Sanctions for of cial misconduct 1.5 0.55 Non-governmental checks 1.6 0.46 Lawful transition of power Absence of Corruption 2.1 0.34 In the executive branch 2.2 0.41 In the judiciary 2.3 0.52 In the police/military 2.4 0.16 In the legislature Open Government 3.1 0.29 Publicized laws and gov't data 3.2 0.47 Right to information 3.3 0.52 Civic participation 3.4 0.48 Complaint mechanisms Fundamental Rights 4.1 0.37 No discrimination 4.2 0.47 Right to life and security 4.3 0.38 Due process of law 4.4 0.55 Freedom of expression 4.5 0.45 Freedom of religion 4.6 0.41 Right to privacy 4.7 0.61 Freedom of association 4.8 0.45 Labor rights Order and Security 5.1 0.70 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.33 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.40 No improper in ence 6.3 0.38 No unreasonable delay 6.4 0.50 Respect for due process 6.5 0.44 No expropriation w/out adequate compensation Civil Justice 7.1 0.55 Accessibility and affordability 7.2 0.43 No discrimination 7.3 0.35 No corruption 7.4 0.32 No improper gov't in ence 7.5 0.29 No unreasonable delay 7.6 0.40 Effective enforcement 7.7 0.49 Impartial and effective ADRs Criminal Justice 8.1 0.43 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.24 Effective correctional system 8.4 0.18 No discrimination 8.5 0.41 No corruption 8.6 0.19 No improper gov't in ence 8.7 0.38 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 112 Liberia Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.44 19/34 9/18 112/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.49 0.00 15/34 5/18 87/142 Absence of Corruption 0.30 0.01 25/34 13/18 124/142 Open Government 0.46 0.00 9/34 3/18 85/142 Fundamental Rights 0.51 0.01 12/34 5/18 82/142 Order and Security 0.60 0.00 23/34 10/18 121/142 Regulatory Enforcement 0.40 0.00 25/34 10/18 123/142 Civil Justice 0.40 0.00 29/34 14/18 125/142 Criminal Justice 0.33 0.00 25/34 11/18 113/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Liberia Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.51 Limits by legislature 1.2 0.49 Limits by judiciary 1.3 0.44 Independent auditing 1.4 0.35 Sanctions for of cial misconduct 1.5 0.57 Non-governmental checks 1.6 0.60 Lawful transition of power Absence of Corruption 2.1 0.31 In the executive branch 2.2 0.29 In the judiciary 2.3 0.39 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.17 Publicized laws and gov't data 3.2 0.45 Right to information 3.3 0.57 Civic participation 3.4 0.64 Complaint mechanisms Fundamental Rights 4.1 0.46 No discrimination 4.2 0.53 Right to life and security 4.3 0.40 Due process of law 4.4 0.57 Freedom of expression 4.5 0.62 Freedom of religion 4.6 0.45 Right to privacy 4.7 0.62 Freedom of association 4.8 0.46 Labor rights Order and Security 5.1 0.47 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.33 Absence of violent redress Regulatory Enforcement 6.1 0.40 Effective regulatory enforcement 6.2 0.40 No improper in ence 6.3 0.37 No unreasonable delay 6.4 0.46 Respect for due process 6.5 0.37 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.43 No discrimination 7.3 0.31 No corruption 7.4 0.32 No improper gov't in ence 7.5 0.36 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.49 Impartial and effective ADRs Criminal Justice 8.1 0.30 Effective investigations 8.2 0.34 Timely and effective adjudication 8.3 0.25 Effective correctional system 8.4 0.35 No discrimination 8.5 0.38 No corruption 8.6 0.29 No improper gov't in ence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 113 WJP Rule of Law Index 2023 Lithuania Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.77 14/31 18/46 18/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.76 0.00 14/31 16/46 17/142 Absence of Corruption 0.72 0.00 17/31 24/46 24/142 Open Government 0.75 0.00 15/31 17/46 17/142 Fundamental Rights 0.78 0.01 15/31 17/46 18/142 Order and Security 0.89 0.00 15/31 19/46 20/142 Regulatory Enforcement 0.76 0.02 14/31 19/46 19/142 Civil Justice 0.79 0.00 8/31 8/46 8/142 Criminal Justice 0.69 0.00 15/31 21/46 21/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Lithuania EU, EFTA, and North America High Constraints on Government Powers 1.1 0.79 Limits by legislature 1.2 0.69 Limits by judiciary 1.3 0.78 Independent auditing 1.4 0.71 Sanctions for ofØcial misconduct 1.5 0.75 Non-governmental checks 1.6 0.87 Lawful transition of power Absence of Corruption 2.1 0.69 In the executive branch 2.2 0.84 In the judiciary 2.3 0.75 In the police/military 2.4 0.60 In the legislature Open Government 3.1 0.81 Publicized laws and gov't data 3.2 0.68 Right to information 3.3 0.74 Civic participation 3.4 0.77 Complaint mechanisms Fundamental Rights 4.1 0.82 No discrimination 4.2 0.92 Right to life and security 4.3 0.71 Due process of law 4.4 0.75 Freedom of expression 4.5 0.81 Freedom of religion 4.6 0.75 Right to privacy 4.7 0.78 Freedom of association 4.8 0.74 Labor rights Order and Security 5.1 0.88 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.79 Absence of violent redress Regulatory Enforcement 6.1 0.81 Effective regulatory enforcement 6.2 0.77 No improper inÙuence 6.3 0.76 No unreasonable delay 6.4 0.67 Respect for due process 6.5 0.76 No expropriation w/out adequate compensation Civil Justice 7.1 0.75 Accessibility and affordability 7.2 0.85 No discrimination 7.3 0.72 No corruption 7.4 0.80 No improper gov't inÙuence 7.5 0.77 No unreasonable delay 7.6 0.85 Effective enforcement 7.7 0.77 Impartial and effective ADRs Criminal Justice 8.1 0.67 Effective investigations 8.2 0.58 Timely and effective adjudication 8.3 0.61 Effective correctional system 8.4 0.79 No discrimination 8.5 0.67 No corruption 8.6 0.79 No improper gov't inÙuence 8.7 0.71 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 114 Luxembourg Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.83 6/31 6/46 6/142 Score Change Rank Change 0.01 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.82 -0.01 12/31 14/46 14/142 Absence of Corruption 0.85 0.00 6/31 8/46 8/142 Open Government 0.82 0.02 6/31 7/46 7/142 Fundamental Rights 0.85 0.00 6/31 6/46 6/142 Order and Security 0.95 0.02 1/31 1/46 1/142 Regulatory Enforcement 0.87 0.01 4/31 4/46 4/142 Civil Justice 0.78 0.01 9/31 10/46 10/142 Criminal Justice 0.73 0.00 10/31 13/46 13/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Luxembourg EU, EFTA, and North America High Constraints on Government Powers 1.1 0.88 Limits by legislature 1.2 0.80 Limits by judiciary 1.3 0.83 Independent auditing 1.4 0.71 Sanctions for ofØcial misconduct 1.5 0.79 Non-governmental checks 1.6 0.89 Lawful transition of power Absence of Corruption 2.1 0.84 In the executive branch 2.2 0.94 In the judiciary 2.3 0.89 In the police/military 2.4 0.74 In the legislature Open Government 3.1 0.82 Publicized laws and gov't data 3.2 0.77 Right to information 3.3 0.83 Civic participation 3.4 0.86 Complaint mechanisms Fundamental Rights 4.1 0.76 No discrimination 4.2 0.95 Right to life and security 4.3 0.86 Due process of law 4.4 0.79 Freedom of expression 4.5 0.86 Freedom of religion 4.6 0.91 Right to privacy 4.7 0.88 Freedom of association 4.8 0.81 Labor rights Order and Security 5.1 0.95 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.91 Absence of violent redress Regulatory Enforcement 6.1 0.80 Effective regulatory enforcement 6.2 0.89 No improper inÙuence 6.3 0.88 No unreasonable delay 6.4 0.89 Respect for due process 6.5 0.87 No expropriation w/out adequate compensation Civil Justice 7.1 0.72 Accessibility and affordability 7.2 0.79 No discrimination 7.3 0.86 No corruption 7.4 0.85 No improper gov't inÙuence 7.5 0.68 No unreasonable delay 7.6 0.73 Effective enforcement 7.7 0.83 Impartial and effective ADRs Criminal Justice 8.1 0.75 Effective investigations 8.2 0.60 Timely and effective adjudication 8.3 0.75 Effective correctional system 8.4 0.59 No discrimination 8.5 0.86 No corruption 8.6 0.72 No improper gov't inÙuence 8.7 0.86 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 115 WJP Rule of Law Index 2023 Madagascar Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.43 20/34 10/18 114/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.43 -0.02 18/34 8/18 106/142 Absence of Corruption 0.27 -0.01 30/34 16/18 134/142 Open Government 0.45 -0.01 12/34 6/18 91/142 Fundamental Rights 0.46 0.00 18/34 10/18 98/142 Order and Security 0.71 0.00 8/34 3/18 80/142 Regulatory Enforcement 0.38 0.00 26/34 11/18 127/142 Civil Justice 0.43 0.00 22/34 8/18 113/142 Criminal Justice 0.32 -0.01 26/34 12/18 117/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Madagascar Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.44 Limits by legislature 1.2 0.39 Limits by judiciary 1.3 0.44 Independent auditing 1.4 0.36 Sanctions for of cial misconduct 1.5 0.47 Non-governmental checks 1.6 0.46 Lawful transition of power Absence of Corruption 2.1 0.31 In the executive branch 2.2 0.26 In the judiciary 2.3 0.34 In the police/military 2.4 0.16 In the legislature Open Government 3.1 0.33 Publicized laws and gov't data 3.2 0.43 Right to information 3.3 0.49 Civic participation 3.4 0.56 Complaint mechanisms Fundamental Rights 4.1 0.58 No discrimination 4.2 0.42 Right to life and security 4.3 0.34 Due process of law 4.4 0.47 Freedom of expression 4.5 0.55 Freedom of religion 4.6 0.25 Right to privacy 4.7 0.53 Freedom of association 4.8 0.54 Labor rights Order and Security 5.1 0.66 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.46 Absence of violent redress Regulatory Enforcement 6.1 0.34 Effective regulatory enforcement 6.2 0.39 No improper in ence 6.3 0.39 No unreasonable delay 6.4 0.32 Respect for due process 6.5 0.47 No expropriation w/out adequate compensation Civil Justice 7.1 0.40 Accessibility and affordability 7.2 0.50 No discrimination 7.3 0.31 No corruption 7.4 0.32 No improper gov't in ence 7.5 0.47 No unreasonable delay 7.6 0.44 Effective enforcement 7.7 0.55 Impartial and effective ADRs Criminal Justice 8.1 0.33 Effective investigations 8.2 0.42 Timely and effective adjudication 8.3 0.19 Effective correctional system 8.4 0.38 No discrimination 8.5 0.30 No corruption 8.6 0.26 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 116 Malawi Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.52 8/34 2/18 69/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.57 0.00 7/34 2/18 55/142 Absence of Corruption 0.42 0.00 12/34 6/18 90/142 Open Government 0.46 0.00 10/34 4/18 86/142 Fundamental Rights 0.58 0.00 6/34 1/18 65/142 Order and Security 0.68 0.00 14/34 7/18 92/142 Regulatory Enforcement 0.46 0.00 14/34 6/18 90/142 Civil Justice 0.56 0.01 7/34 2/18 56/142 Criminal Justice 0.44 -0.01 9/34 3/18 73/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Malawi Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.60 Limits by legislature 1.2 0.64 Limits by judiciary 1.3 0.49 Independent auditing 1.4 0.46 Sanctions for ofØcial misconduct 1.5 0.59 Non-governmental checks 1.6 0.66 Lawful transition of power Absence of Corruption 2.1 0.38 In the executive branch 2.2 0.53 In the judiciary 2.3 0.40 In the police/military 2.4 0.35 In the legislature Open Government 3.1 0.16 Publicized laws and gov't data 3.2 0.47 Right to information 3.3 0.62 Civic participation 3.4 0.57 Complaint mechanisms Fundamental Rights 4.1 0.55 No discrimination 4.2 0.61 Right to life and security 4.3 0.39 Due process of law 4.4 0.59 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.55 Right to privacy 4.7 0.71 Freedom of association 4.8 0.52 Labor rights Order and Security 5.1 0.70 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.34 Absence of violent redress Regulatory Enforcement 6.1 0.42 Effective regulatory enforcement 6.2 0.40 No improper inÙuence 6.3 0.39 No unreasonable delay 6.4 0.51 Respect for due process 6.5 0.61 No expropriation w/out adequate compensation Civil Justice 7.1 0.50 Accessibility and affordability 7.2 0.56 No discrimination 7.3 0.50 No corruption 7.4 0.69 No improper gov't inÙuence 7.5 0.42 No unreasonable delay 7.6 0.63 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.34 Effective investigations 8.2 0.47 Timely and effective adjudication 8.3 0.32 Effective correctional system 8.4 0.47 No discrimination 8.5 0.45 No corruption 8.6 0.65 No improper gov't inÙuence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 117 WJP Rule of Law Index 2023 Malaysia Region: East Asia and PaciØc Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.57 7/15 11/41 55/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.57 0.02 7/15 11/41 56/142 Absence of Corruption 0.57 0.01 7/15 10/41 51/142 Open Government 0.42 0.00 12/15 34/41 103/142 Fundamental Rights 0.49 0.00 9/15 30/41 90/142 Order and Security 0.78 0.00 8/15 10/41 50/142 Regulatory Enforcement 0.55 0.00 8/15 10/41 55/142 Civil Justice 0.62 0.00 7/15 6/41 42/142 Criminal Justice 0.56 0.00 7/15 5/41 43/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Malaysia East Asia and PaciØc Upper-Middle Constraints on Government Powers 1.1 0.58 Limits by legislature 1.2 0.62 Limits by judiciary 1.3 0.58 Independent auditing 1.4 0.56 Sanctions for ofØcial misconduct 1.5 0.48 Non-governmental checks 1.6 0.60 Lawful transition of power Absence of Corruption 2.1 0.52 In the executive branch 2.2 0.77 In the judiciary 2.3 0.60 In the police/military 2.4 0.38 In the legislature Open Government 3.1 0.34 Publicized laws and gov't data 3.2 0.39 Right to information 3.3 0.49 Civic participation 3.4 0.48 Complaint mechanisms Fundamental Rights 4.1 0.49 No discrimination 4.2 0.52 Right to life and security 4.3 0.54 Due process of law 4.4 0.48 Freedom of expression 4.5 0.44 Freedom of religion 4.6 0.41 Right to privacy 4.7 0.46 Freedom of association 4.8 0.58 Labor rights Order and Security 5.1 0.77 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.58 Absence of violent redress Regulatory Enforcement 6.1 0.49 Effective regulatory enforcement 6.2 0.67 No improper inÙuence 6.3 0.54 No unreasonable delay 6.4 0.54 Respect for due process 6.5 0.50 No expropriation w/out adequate compensation Civil Justice 7.1 0.59 Accessibility and affordability 7.2 0.54 No discrimination 7.3 0.70 No corruption 7.4 0.55 No improper gov't inÙuence 7.5 0.65 No unreasonable delay 7.6 0.64 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.62 Effective investigations 8.2 0.58 Timely and effective adjudication 8.3 0.60 Effective correctional system 8.4 0.47 No discrimination 8.5 0.65 No corruption 8.6 0.46 No improper gov't inÙuence 8.7 0.54 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 118 Mali Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.40 24/34 12/18 121/142 Score Change Rank Change -0.02 -5 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.41 -0.02 19/34 9/18 108/142 Absence of Corruption 0.31 0.00 24/34 12/18 122/142 Open Government 0.45 0.00 11/34 5/18 88/142 Fundamental Rights 0.50 -0.01 13/34 6/18 87/142 Order and Security 0.37 -0.13 34/34 17/18 140/142 Regulatory Enforcement 0.47 -0.01 12/34 4/18 87/142 Civil Justice 0.41 0.00 27/34 13/18 120/142 Criminal Justice 0.28 -0.01 32/34 17/18 129/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Mali Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.51 Limits by legislature 1.2 0.36 Limits by judiciary 1.3 0.33 Independent auditing 1.4 0.26 Sanctions for of cial misconduct 1.5 0.53 Non-governmental checks 1.6 0.50 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.20 In the judiciary 2.3 0.31 In the police/military 2.4 0.36 In the legislature Open Government 3.1 0.25 Publicized laws and gov't data 3.2 0.46 Right to information 3.3 0.54 Civic participation 3.4 0.57 Complaint mechanisms Fundamental Rights 4.1 0.57 No discrimination 4.2 0.38 Right to life and security 4.3 0.35 Due process of law 4.4 0.53 Freedom of expression 4.5 0.69 Freedom of religion 4.6 0.28 Right to privacy 4.7 0.60 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.59 Absence of crime 5.2 0.17 Absence of civil con ct 5.3 0.34 Absence of violent redress Regulatory Enforcement 6.1 0.41 Effective regulatory enforcement 6.2 0.50 No improper in ence 6.3 0.37 No unreasonable delay 6.4 0.47 Respect for due process 6.5 0.60 No expropriation w/out adequate compensation Civil Justice 7.1 0.43 Accessibility and affordability 7.2 0.44 No discrimination 7.3 0.12 No corruption 7.4 0.40 No improper gov't in ence 7.5 0.57 No unreasonable delay 7.6 0.45 Effective enforcement 7.7 0.50 Impartial and effective ADRs Criminal Justice 8.1 0.23 Effective investigations 8.2 0.29 Timely and effective adjudication 8.3 0.26 Effective correctional system 8.4 0.36 No discrimination 8.5 0.24 No corruption 8.6 0.26 No improper gov't in ence 8.7 0.35 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 119 WJP Rule of Law Index 2023 Malta Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.68 22/31 29/46 30/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.64 0.00 26/31 34/46 38/142 Absence of Corruption 0.68 -0.01 21/31 30/46 31/142 Open Government 0.64 0.01 23/31 30/46 31/142 Fundamental Rights 0.74 0.00 20/31 25/46 26/142 Order and Security 0.91 -0.01 7/31 11/46 11/142 Regulatory Enforcement 0.59 0.01 26/31 39/46 43/142 Civil Justice 0.60 0.00 25/31 37/46 48/142 Criminal Justice 0.63 0.00 20/31 27/46 27/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Malta EU, EFTA, and North America High Constraints on Government Powers 1.1 0.53 Limits by legislature 1.2 0.66 Limits by judiciary 1.3 0.61 Independent auditing 1.4 0.55 Sanctions for ofØcial misconduct 1.5 0.66 Non-governmental checks 1.6 0.83 Lawful transition of power Absence of Corruption 2.1 0.59 In the executive branch 2.2 0.86 In the judiciary 2.3 0.82 In the police/military 2.4 0.44 In the legislature Open Government 3.1 0.69 Publicized laws and gov't data 3.2 0.48 Right to information 3.3 0.66 Civic participation 3.4 0.70 Complaint mechanisms Fundamental Rights 4.1 0.65 No discrimination 4.2 0.85 Right to life and security 4.3 0.72 Due process of law 4.4 0.66 Freedom of expression 4.5 0.70 Freedom of religion 4.6 0.82 Right to privacy 4.7 0.73 Freedom of association 4.8 0.79 Labor rights Order and Security 5.1 0.86 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.88 Absence of violent redress Regulatory Enforcement 6.1 0.65 Effective regulatory enforcement 6.2 0.73 No improper inÙuence 6.3 0.46 No unreasonable delay 6.4 0.59 Respect for due process 6.5 0.55 No expropriation w/out adequate compensation Civil Justice 7.1 0.69 Accessibility and affordability 7.2 0.50 No discrimination 7.3 0.76 No corruption 7.4 0.70 No improper gov't inÙuence 7.5 0.27 No unreasonable delay 7.6 0.59 Effective enforcement 7.7 0.71 Impartial and effective ADRs Criminal Justice 8.1 0.62 Effective investigations 8.2 0.50 Timely and effective adjudication 8.3 0.54 Effective correctional system 8.4 0.51 No discrimination 8.5 0.77 No corruption 8.6 0.77 No improper gov't inÙuence 8.7 0.72 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 120 Mauritania Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.36 32/34 31/37 133/142 Score Change Rank Change -0.01 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.33 0.00 31/34 31/37 129/142 Absence of Corruption 0.29 -0.01 27/34 31/37 128/142 Open Government 0.28 0.00 34/34 34/37 138/142 Fundamental Rights 0.38 -0.01 28/34 27/37 123/142 Order and Security 0.65 0.01 18/34 22/37 104/142 Regulatory Enforcement 0.28 0.00 34/34 35/37 139/142 Civil Justice 0.40 0.00 30/34 29/37 126/142 Criminal Justice 0.27 -0.02 33/34 29/37 130/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Mauritania Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.35 Limits by legislature 1.2 0.24 Limits by judiciary 1.3 0.33 Independent auditing 1.4 0.28 Sanctions for of cial misconduct 1.5 0.33 Non-governmental checks 1.6 0.44 Lawful transition of power Absence of Corruption 2.1 0.29 In the executive branch 2.2 0.41 In the judiciary 2.3 0.25 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.20 Publicized laws and gov't data 3.2 0.32 Right to information 3.3 0.35 Civic participation 3.4 0.26 Complaint mechanisms Fundamental Rights 4.1 0.39 No discrimination 4.2 0.45 Right to life and security 4.3 0.32 Due process of law 4.4 0.33 Freedom of expression 4.5 0.33 Freedom of religion 4.6 0.32 Right to privacy 4.7 0.48 Freedom of association 4.8 0.45 Labor rights Order and Security 5.1 0.62 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.32 Absence of violent redress Regulatory Enforcement 6.1 0.24 Effective regulatory enforcement 6.2 0.32 No improper in ence 6.3 0.40 No unreasonable delay 6.4 0.16 Respect for due process 6.5 0.30 No expropriation w/out adequate compensation Civil Justice 7.1 0.44 Accessibility and affordability 7.2 0.33 No discrimination 7.3 0.28 No corruption 7.4 0.28 No improper gov't in ence 7.5 0.49 No unreasonable delay 7.6 0.34 Effective enforcement 7.7 0.63 Impartial and effective ADRs Criminal Justice 8.1 0.24 Effective investigations 8.2 0.45 Timely and effective adjudication 8.3 0.25 Effective correctional system 8.4 0.21 No discrimination 8.5 0.28 No corruption 8.6 0.15 No improper gov't in ence 8.7 0.32 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 121 WJP Rule of Law Index 2023 Mauritius Region: Sub-Saharan Africa Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.61 3/34 5/41 46/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.58 -0.01 6/34 9/41 51/142 Absence of Corruption 0.58 0.00 2/34 8/41 48/142 Open Government 0.53 0.00 4/34 17/41 61/142 Fundamental Rights 0.64 0.00 2/34 7/41 45/142 Order and Security 0.75 0.00 2/34 17/41 60/142 Regulatory Enforcement 0.62 0.00 1/34 2/41 35/142 Civil Justice 0.63 0.00 2/34 3/41 34/142 Criminal Justice 0.54 0.00 4/34 9/41 47/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Mauritius Sub-Saharan Africa Upper-Middle Constraints on Government Powers 1.1 0.62 Limits by legislature 1.2 0.67 Limits by judiciary 1.3 0.40 Independent auditing 1.4 0.50 Sanctions for of cial misconduct 1.5 0.59 Non-governmental checks 1.6 0.70 Lawful transition of power Absence of Corruption 2.1 0.59 In the executive branch 2.2 0.80 In the judiciary 2.3 0.61 In the police/military 2.4 0.30 In the legislature Open Government 3.1 0.56 Publicized laws and gov't data 3.2 0.41 Right to information 3.3 0.59 Civic participation 3.4 0.55 Complaint mechanisms Fundamental Rights 4.1 0.59 No discrimination 4.2 0.73 Right to life and security 4.3 0.53 Due process of law 4.4 0.59 Freedom of expression 4.5 0.80 Freedom of religion 4.6 0.60 Right to privacy 4.7 0.65 Freedom of association 4.8 0.65 Labor rights Order and Security 5.1 0.83 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.44 Absence of violent redress Regulatory Enforcement 6.1 0.67 Effective regulatory enforcement 6.2 0.82 No improper in ence 6.3 0.49 No unreasonable delay 6.4 0.55 Respect for due process 6.5 0.60 No expropriation w/out adequate compensation Civil Justice 7.1 0.63 Accessibility and affordability 7.2 0.71 No discrimination 7.3 0.77 No corruption 7.4 0.72 No improper gov't in ence 7.5 0.30 No unreasonable delay 7.6 0.60 Effective enforcement 7.7 0.69 Impartial and effective ADRs Criminal Justice 8.1 0.40 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.54 Effective correctional system 8.4 0.60 No discrimination 8.5 0.66 No corruption 8.6 0.64 No improper gov't in ence 8.7 0.53 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 122 Mexico Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.42 27/32 38/41 116/142 Score Change Rank Change -0.01 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.44 0.00 26/32 33/41 104/142 Absence of Corruption 0.26 0.00 30/32 40/41 136/142 Open Government 0.58 -0.01 7/32 7/41 45/142 Fundamental Rights 0.48 0.00 26/32 32/41 93/142 Order and Security 0.50 -0.01 31/32 41/41 133/142 Regulatory Enforcement 0.44 0.00 24/32 33/41 102/142 Civil Justice 0.37 0.00 27/32 39/41 131/142 Criminal Justice 0.26 -0.02 26/32 40/41 132/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Mexico Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.48 Limits by legislature 1.2 0.41 Limits by judiciary 1.3 0.40 Independent auditing 1.4 0.27 Sanctions for of cial misconduct 1.5 0.48 Non-governmental checks 1.6 0.62 Lawful transition of power Absence of Corruption 2.1 0.31 In the executive branch 2.2 0.30 In the judiciary 2.3 0.31 In the police/military 2.4 0.13 In the legislature Open Government 3.1 0.65 Publicized laws and gov't data 3.2 0.57 Right to information 3.3 0.50 Civic participation 3.4 0.60 Complaint mechanisms Fundamental Rights 4.1 0.35 No discrimination 4.2 0.41 Right to life and security 4.3 0.34 Due process of law 4.4 0.48 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.49 Right to privacy 4.7 0.59 Freedom of association 4.8 0.48 Labor rights Order and Security 5.1 0.38 Absence of crime 5.2 0.90 Absence of civil con ct 5.3 0.22 Absence of violent redress Regulatory Enforcement 6.1 0.53 Effective regulatory enforcement 6.2 0.51 No improper in ence 6.3 0.36 No unreasonable delay 6.4 0.32 Respect for due process 6.5 0.48 No expropriation w/out adequate compensation Civil Justice 7.1 0.39 Accessibility and affordability 7.2 0.25 No discrimination 7.3 0.31 No corruption 7.4 0.40 No improper gov't in ence 7.5 0.25 No unreasonable delay 7.6 0.37 Effective enforcement 7.7 0.61 Impartial and effective ADRs Criminal Justice 8.1 0.19 Effective investigations 8.2 0.27 Timely and effective adjudication 8.3 0.22 Effective correctional system 8.4 0.17 No discrimination 8.5 0.28 No corruption 8.6 0.36 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 123 WJP Rule of Law Index 2023 Moldova Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.53 6/15 20/41 68/142 Score Change Rank Change 0.01 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.51 0.01 4/15 18/41 74/142 Absence of Corruption 0.38 0.01 12/15 34/41 106/142 Open Government 0.57 0.00 3/15 11/41 49/142 Fundamental Rights 0.59 0.01 6/15 18/41 63/142 Order and Security 0.81 0.00 4/15 4/41 40/142 Regulatory Enforcement 0.46 0.00 8/15 31/41 96/142 Civil Justice 0.50 0.00 9/15 24/41 79/142 Criminal Justice 0.41 0.02 8/15 23/41 78/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Moldova Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.64 Limits by legislature 1.2 0.42 Limits by judiciary 1.3 0.51 Independent auditing 1.4 0.36 Sanctions for ofØcial misconduct 1.5 0.55 Non-governmental checks 1.6 0.60 Lawful transition of power Absence of Corruption 2.1 0.37 In the executive branch 2.2 0.45 In the judiciary 2.3 0.53 In the police/military 2.4 0.16 In the legislature Open Government 3.1 0.63 Publicized laws and gov't data 3.2 0.56 Right to information 3.3 0.55 Civic participation 3.4 0.54 Complaint mechanisms Fundamental Rights 4.1 0.56 No discrimination 4.2 0.72 Right to life and security 4.3 0.49 Due process of law 4.4 0.55 Freedom of expression 4.5 0.66 Freedom of religion 4.6 0.53 Right to privacy 4.7 0.64 Freedom of association 4.8 0.54 Labor rights Order and Security 5.1 0.84 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.59 Absence of violent redress Regulatory Enforcement 6.1 0.56 Effective regulatory enforcement 6.2 0.45 No improper inÙuence 6.3 0.54 No unreasonable delay 6.4 0.30 Respect for due process 6.5 0.43 No expropriation w/out adequate compensation Civil Justice 7.1 0.55 Accessibility and affordability 7.2 0.52 No discrimination 7.3 0.36 No corruption 7.4 0.42 No improper gov't inÙuence 7.5 0.42 No unreasonable delay 7.6 0.57 Effective enforcement 7.7 0.66 Impartial and effective ADRs Criminal Justice 8.1 0.29 Effective investigations 8.2 0.47 Timely and effective adjudication 8.3 0.43 Effective correctional system 8.4 0.44 No discrimination 8.5 0.39 No corruption 8.6 0.36 No improper gov't inÙuence 8.7 0.49 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 124 Mongolia Region: East Asia and PaciØc Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.53 8/15 3/37 64/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.53 -0.01 9/15 7/37 71/142 Absence of Corruption 0.42 -0.01 12/15 9/37 86/142 Open Government 0.49 0.00 8/15 9/37 74/142 Fundamental Rights 0.56 0.00 7/15 4/37 69/142 Order and Security 0.76 0.00 10/15 3/37 58/142 Regulatory Enforcement 0.48 0.00 10/15 12/37 79/142 Civil Justice 0.54 0.00 8/15 4/37 66/142 Criminal Justice 0.48 0.00 8/15 1/37 58/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Mongolia East Asia and PaciØc Lower-Middle Constraints on Government Powers 1.1 0.66 Limits by legislature 1.2 0.50 Limits by judiciary 1.3 0.44 Independent auditing 1.4 0.43 Sanctions for ofØcial misconduct 1.5 0.60 Non-governmental checks 1.6 0.54 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.52 In the judiciary 2.3 0.62 In the police/military 2.4 0.15 In the legislature Open Government 3.1 0.42 Publicized laws and gov't data 3.2 0.50 Right to information 3.3 0.59 Civic participation 3.4 0.46 Complaint mechanisms Fundamental Rights 4.1 0.55 No discrimination 4.2 0.67 Right to life and security 4.3 0.49 Due process of law 4.4 0.60 Freedom of expression 4.5 0.60 Freedom of religion 4.6 0.41 Right to privacy 4.7 0.65 Freedom of association 4.8 0.55 Labor rights Order and Security 5.1 0.84 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.58 Effective regulatory enforcement 6.2 0.47 No improper inÙuence 6.3 0.54 No unreasonable delay 6.4 0.34 Respect for due process 6.5 0.45 No expropriation w/out adequate compensation Civil Justice 7.1 0.54 Accessibility and affordability 7.2 0.59 No discrimination 7.3 0.52 No corruption 7.4 0.47 No improper gov't inÙuence 7.5 0.57 No unreasonable delay 7.6 0.46 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.41 Effective investigations 8.2 0.55 Timely and effective adjudication 8.3 0.51 Effective correctional system 8.4 0.52 No discrimination 8.5 0.54 No corruption 8.6 0.37 No improper gov't inÙuence 8.7 0.49 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 125 WJP Rule of Law Index 2023 Montenegro Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.56 2/15 13/41 57/142 Score Change Rank Change Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.52 3/15 16/41 72/142 Absence of Corruption 0.48 3/15 15/41 62/142 Open Government 0.53 5/15 16/41 59/142 Fundamental Rights 0.67 1/15 4/41 41/142 Order and Security 0.82 3/15 2/41 36/142 Regulatory Enforcement 0.48 5/15 23/41 82/142 Civil Justice 0.54 — 3/15 17/41 67/142 Criminal Justice 0.47 2/15 15/41 60/142 Indicates statistically signi cant change at the 10 percent level Low Medium High Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 2023 Score 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Montenegro Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.57 Limits by legislature 1.2 0.39 Limits by judiciary 1.3 0.62 Independent auditing 1.4 0.40 Sanctions for of cial misconduct 1.5 0.61 Non-governmental checks 1.6 0.56 Lawful transition of power Absence of Corruption 2.1 0.48 In the executive branch 2.2 0.60 In the judiciary 2.3 0.54 In the police/military 2.4 0.31 In the legislature Open Government 3.1 0.47 Publicized laws and gov't data 3.2 0.46 Right to information 3.3 0.57 Civic participation 3.4 0.61 Complaint mechanisms Fundamental Rights 4.1 0.68 No discrimination 4.2 0.82 Right to life and security 4.3 0.55 Due process of law 4.4 0.61 Freedom of expression 4.5 0.74 Freedom of religion 4.6 0.53 Right to privacy 4.7 0.67 Freedom of association 4.8 0.72 Labor rights Order and Security 5.1 0.94 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.53 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.68 No improper in ence 6.3 0.48 No unreasonable delay 6.4 0.26 Respect for due process 6.5 0.50 No expropriation w/out adequate compensation Civil Justice 7.1 0.65 Accessibility and affordability 7.2 0.61 No discrimination 7.3 0.53 No corruption 7.4 0.40 No improper gov't in ence 7.5 0.40 No unreasonable delay 7.6 0.46 Effective enforcement 7.7 0.72 Impartial and effective ADRs Criminal Justice 8.1 0.47 Effective investigations 8.2 0.37 Timely and effective adjudication 8.3 0.35 Effective correctional system 8.4 0.73 No discrimination 8.5 0.51 No corruption 8.6 0.34 No improper gov't in ence 8.7 0.55 Due process of law — — — — — — — The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 126 Morocco Region: Middle East and North Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.48 6/9 14/37 92/142 Score Change Rank Change 0.00 4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.51 0.00 4/9 8/37 76/142 Absence of Corruption 0.42 0.00 6/9 10/37 88/142 Open Government 0.43 0.00 3/9 18/37 101/142 Fundamental Rights 0.43 0.01 7/9 22/37 114/142 Order and Security 0.69 0.01 6/9 14/37 89/142 Regulatory Enforcement 0.50 -0.01 4/9 5/37 66/142 Civil Justice 0.52 0.01 5/9 6/37 72/142 Criminal Justice 0.37 0.01 6/9 16/37 95/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Morocco Middle East and North Africa Lower-Middle Constraints on Government Powers 1.1 0.56 Limits by legislature 1.2 0.50 Limits by judiciary 1.3 0.48 Independent auditing 1.4 0.51 Sanctions for ofØcial misconduct 1.5 0.45 Non-governmental checks 1.6 0.57 Lawful transition of power Absence of Corruption 2.1 0.47 In the executive branch 2.2 0.44 In the judiciary 2.3 0.46 In the police/military 2.4 0.31 In the legislature Open Government 3.1 0.38 Publicized laws and gov't data 3.2 0.44 Right to information 3.3 0.43 Civic participation 3.4 0.48 Complaint mechanisms Fundamental Rights 4.1 0.62 No discrimination 4.2 0.34 Right to life and security 4.3 0.39 Due process of law 4.4 0.45 Freedom of expression 4.5 0.40 Freedom of religion 4.6 0.21 Right to privacy 4.7 0.46 Freedom of association 4.8 0.57 Labor rights Order and Security 5.1 0.66 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.41 Absence of violent redress Regulatory Enforcement 6.1 0.54 Effective regulatory enforcement 6.2 0.55 No improper inÙuence 6.3 0.45 No unreasonable delay 6.4 0.42 Respect for due process 6.5 0.55 No expropriation w/out adequate compensation Civil Justice 7.1 0.51 Accessibility and affordability 7.2 0.51 No discrimination 7.3 0.42 No corruption 7.4 0.42 No improper gov't inÙuence 7.5 0.61 No unreasonable delay 7.6 0.48 Effective enforcement 7.7 0.68 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.31 Effective correctional system 8.4 0.34 No discrimination 8.5 0.42 No corruption 8.6 0.24 No improper gov't inÙuence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 127 WJP Rule of Law Index 2023 Mozambique Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.38 29/34 14/18 128/142 Score Change Rank Change -0.01 -4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.41 -0.01 20/34 10/18 109/142 Absence of Corruption 0.35 0.00 21/34 11/18 113/142 Open Government 0.36 0.00 25/34 13/18 122/142 Fundamental Rights 0.39 -0.01 27/34 14/18 122/142 Order and Security 0.43 -0.05 32/34 16/18 138/142 Regulatory Enforcement 0.41 0.00 23/34 9/18 120/142 Civil Justice 0.42 -0.01 25/34 11/18 116/142 Criminal Justice 0.31 -0.02 28/34 14/18 122/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Mozambique Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.52 Limits by legislature 1.2 0.42 Limits by judiciary 1.3 0.38 Independent auditing 1.4 0.37 Sanctions for of cial misconduct 1.5 0.38 Non-governmental checks 1.6 0.40 Lawful transition of power Absence of Corruption 2.1 0.34 In the executive branch 2.2 0.39 In the judiciary 2.3 0.35 In the police/military 2.4 0.33 In the legislature Open Government 3.1 0.19 Publicized laws and gov't data 3.2 0.31 Right to information 3.3 0.41 Civic participation 3.4 0.54 Complaint mechanisms Fundamental Rights 4.1 0.55 No discrimination 4.2 0.19 Right to life and security 4.3 0.25 Due process of law 4.4 0.38 Freedom of expression 4.5 0.59 Freedom of religion 4.6 0.20 Right to privacy 4.7 0.44 Freedom of association 4.8 0.46 Labor rights Order and Security 5.1 0.61 Absence of crime 5.2 0.42 Absence of civil con ct 5.3 0.25 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.50 No improper in ence 6.3 0.43 No unreasonable delay 6.4 0.32 Respect for due process 6.5 0.34 No expropriation w/out adequate compensation Civil Justice 7.1 0.44 Accessibility and affordability 7.2 0.52 No discrimination 7.3 0.39 No corruption 7.4 0.30 No improper gov't in ence 7.5 0.32 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.58 Impartial and effective ADRs Criminal Justice 8.1 0.26 Effective investigations 8.2 0.31 Timely and effective adjudication 8.3 0.20 Effective correctional system 8.4 0.41 No discrimination 8.5 0.37 No corruption 8.6 0.34 No improper gov't in ence 8.7 0.25 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 128 Myanmar Region: East Asia and Paci c Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.35 14/15 33/37 135/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.29 -0.02 14/15 34/37 135/142 Absence of Corruption 0.42 -0.01 11/15 8/37 85/142 Open Government 0.30 -0.01 14/15 33/37 137/142 Fundamental Rights 0.20 -0.02 15/15 36/37 141/142 Order and Security 0.58 -0.03 15/15 31/37 125/142 Regulatory Enforcement 0.42 0.00 14/15 24/37 115/142 Civil Justice 0.32 -0.01 14/15 34/37 138/142 Criminal Justice 0.25 0.00 15/15 32/37 136/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Myanmar East Asia and Paci c Lower-Middle Constraints on Government Powers 1.1 0.24 Limits by legislature 1.2 0.37 Limits by judiciary 1.3 0.22 Independent auditing 1.4 0.32 Sanctions for of cial misconduct 1.5 0.24 Non-governmental checks 1.6 0.35 Lawful transition of power Absence of Corruption 2.1 0.48 In the executive branch 2.2 0.26 In the judiciary 2.3 0.45 In the police/military 2.4 0.50 In the legislature Open Government 3.1 0.16 Publicized laws and gov't data 3.2 0.38 Right to information 3.3 0.27 Civic participation 3.4 0.39 Complaint mechanisms Fundamental Rights 4.1 0.38 No discrimination 4.2 0.05 Right to life and security 4.3 0.14 Due process of law 4.4 0.24 Freedom of expression 4.5 0.10 Freedom of religion 4.6 0.07 Right to privacy 4.7 0.14 Freedom of association 4.8 0.51 Labor rights Order and Security 5.1 0.72 Absence of crime 5.2 0.50 Absence of civil con ct 5.3 0.52 Absence of violent redress Regulatory Enforcement 6.1 0.47 Effective regulatory enforcement 6.2 0.58 No improper in ence 6.3 0.48 No unreasonable delay 6.4 0.25 Respect for due process 6.5 0.32 No expropriation w/out adequate compensation Civil Justice 7.1 0.32 Accessibility and affordability 7.2 0.17 No discrimination 7.3 0.30 No corruption 7.4 0.16 No improper gov't in ence 7.5 0.41 No unreasonable delay 7.6 0.35 Effective enforcement 7.7 0.55 Impartial and effective ADRs Criminal Justice 8.1 0.38 Effective investigations 8.2 0.34 Timely and effective adjudication 8.3 0.28 Effective correctional system 8.4 0.16 No discrimination 8.5 0.39 No corruption 8.6 0.07 No improper gov't in ence 8.7 0.14 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 129 WJP Rule of Law Index 2023 Namibia Region: Sub-Saharan Africa Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.61 2/34 4/41 44/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.66 0.01 1/34 2/41 33/142 Absence of Corruption 0.52 0.01 5/34 13/41 58/142 Open Government 0.58 0.00 2/34 8/41 46/142 Fundamental Rights 0.66 0.01 1/34 6/41 43/142 Order and Security 0.75 0.00 3/34 18/41 63/142 Regulatory Enforcement 0.58 -0.01 4/34 5/41 46/142 Civil Justice 0.62 0.00 3/34 5/41 39/142 Criminal Justice 0.55 0.00 3/34 7/41 45/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Namibia Sub-Saharan Africa Upper-Middle Constraints on Government Powers 1.1 0.74 Limits by legislature 1.2 0.73 Limits by judiciary 1.3 0.49 Independent auditing 1.4 0.60 Sanctions for of cial misconduct 1.5 0.71 Non-governmental checks 1.6 0.71 Lawful transition of power Absence of Corruption 2.1 0.43 In the executive branch 2.2 0.81 In the judiciary 2.3 0.55 In the police/military 2.4 0.28 In the legislature Open Government 3.1 0.33 Publicized laws and gov't data 3.2 0.53 Right to information 3.3 0.71 Civic participation 3.4 0.75 Complaint mechanisms Fundamental Rights 4.1 0.54 No discrimination 4.2 0.76 Right to life and security 4.3 0.58 Due process of law 4.4 0.71 Freedom of expression 4.5 0.79 Freedom of religion 4.6 0.61 Right to privacy 4.7 0.75 Freedom of association 4.8 0.55 Labor rights Order and Security 5.1 0.71 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.54 Absence of violent redress Regulatory Enforcement 6.1 0.48 Effective regulatory enforcement 6.2 0.67 No improper in ence 6.3 0.50 No unreasonable delay 6.4 0.62 Respect for due process 6.5 0.64 No expropriation w/out adequate compensation Civil Justice 7.1 0.42 Accessibility and affordability 7.2 0.62 No discrimination 7.3 0.78 No corruption 7.4 0.72 No improper gov't in ence 7.5 0.45 No unreasonable delay 7.6 0.66 Effective enforcement 7.7 0.70 Impartial and effective ADRs Criminal Justice 8.1 0.40 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.41 Effective correctional system 8.4 0.69 No discrimination 8.5 0.66 No corruption 8.6 0.74 No improper gov't in ence 8.7 0.58 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 130 Nepal Region: South Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.52 1/6 5/37 71/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.59 0.00 1/6 3/37 48/142 Absence of Corruption 0.40 0.00 2/6 13/37 94/142 Open Government 0.52 0.00 2/6 4/37 65/142 Fundamental Rights 0.52 -0.01 1/6 6/37 80/142 Order and Security 0.73 0.00 1/6 7/37 70/142 Regulatory Enforcement 0.49 -0.02 1/6 8/37 73/142 Civil Justice 0.44 -0.02 1/6 21/37 107/142 Criminal Justice 0.45 0.00 1/6 4/37 68/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Nepal South Asia Lower-Middle Constraints on Government Powers 1.1 0.73 Limits by legislature 1.2 0.59 Limits by judiciary 1.3 0.48 Independent auditing 1.4 0.48 Sanctions for ofØcial misconduct 1.5 0.62 Non-governmental checks 1.6 0.64 Lawful transition of power Absence of Corruption 2.1 0.46 In the executive branch 2.2 0.37 In the judiciary 2.3 0.59 In the police/military 2.4 0.19 In the legislature Open Government 3.1 0.31 Publicized laws and gov't data 3.2 0.54 Right to information 3.3 0.60 Civic participation 3.4 0.62 Complaint mechanisms Fundamental Rights 4.1 0.51 No discrimination 4.2 0.42 Right to life and security 4.3 0.41 Due process of law 4.4 0.62 Freedom of expression 4.5 0.62 Freedom of religion 4.6 0.39 Right to privacy 4.7 0.64 Freedom of association 4.8 0.54 Labor rights Order and Security 5.1 0.78 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.40 Absence of violent redress Regulatory Enforcement 6.1 0.47 Effective regulatory enforcement 6.2 0.57 No improper inÙuence 6.3 0.43 No unreasonable delay 6.4 0.46 Respect for due process 6.5 0.50 No expropriation w/out adequate compensation Civil Justice 7.1 0.43 Accessibility and affordability 7.2 0.36 No discrimination 7.3 0.38 No corruption 7.4 0.49 No improper gov't inÙuence 7.5 0.47 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.53 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.56 Timely and effective adjudication 8.3 0.45 Effective correctional system 8.4 0.36 No discrimination 8.5 0.51 No corruption 8.6 0.35 No improper gov't inÙuence 8.7 0.41 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 131 WJP Rule of Law Index 2023 Netherlands Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.83 7/31 7/46 7/142 Score Change Rank Change 0.00 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.85 0.00 6/31 7/46 7/142 Absence of Corruption 0.87 0.00 5/31 7/46 7/142 Open Government 0.83 0.00 5/31 5/46 5/142 Fundamental Rights 0.84 -0.01 9/31 9/46 9/142 Order and Security 0.85 0.00 19/31 26/46 28/142 Regulatory Enforcement 0.85 0.00 5/31 6/46 6/142 Civil Justice 0.84 0.00 3/31 3/46 3/142 Criminal Justice 0.74 -0.01 8/31 10/46 10/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Netherlands EU, EFTA, and North America High Constraints on Government Powers 1.1 0.81 Limits by legislature 1.2 0.86 Limits by judiciary 1.3 0.87 Independent auditing 1.4 0.81 Sanctions for ofØcial misconduct 1.5 0.83 Non-governmental checks 1.6 0.91 Lawful transition of power Absence of Corruption 2.1 0.83 In the executive branch 2.2 0.98 In the judiciary 2.3 0.90 In the police/military 2.4 0.79 In the legislature Open Government 3.1 0.78 Publicized laws and gov't data 3.2 0.77 Right to information 3.3 0.83 Civic participation 3.4 0.93 Complaint mechanisms Fundamental Rights 4.1 0.80 No discrimination 4.2 0.93 Right to life and security 4.3 0.82 Due process of law 4.4 0.83 Freedom of expression 4.5 0.81 Freedom of religion 4.6 0.81 Right to privacy 4.7 0.86 Freedom of association 4.8 0.82 Labor rights Order and Security 5.1 0.90 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.64 Absence of violent redress Regulatory Enforcement 6.1 0.77 Effective regulatory enforcement 6.2 0.89 No improper inÙuence 6.3 0.85 No unreasonable delay 6.4 0.80 Respect for due process 6.5 0.93 No expropriation w/out adequate compensation Civil Justice 7.1 0.79 Accessibility and affordability 7.2 0.84 No discrimination 7.3 0.93 No corruption 7.4 0.90 No improper gov't inÙuence 7.5 0.70 No unreasonable delay 7.6 0.88 Effective enforcement 7.7 0.83 Impartial and effective ADRs Criminal Justice 8.1 0.54 Effective investigations 8.2 0.68 Timely and effective adjudication 8.3 0.76 Effective correctional system 8.4 0.65 No discrimination 8.5 0.89 No corruption 8.6 0.86 No improper gov't inÙuence 8.7 0.82 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 132 New Zealand Region: East Asia and PaciØc Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.83 1/15 8/46 8/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.85 0.00 1/15 6/46 6/142 Absence of Corruption 0.87 0.00 2/15 6/46 6/142 Open Government 0.82 0.00 1/15 6/46 6/142 Fundamental Rights 0.82 0.00 1/15 12/46 12/142 Order and Security 0.88 0.00 4/15 21/46 22/142 Regulatory Enforcement 0.84 -0.01 2/15 7/46 7/142 Civil Justice 0.78 0.00 2/15 11/46 11/142 Criminal Justice 0.73 0.00 3/15 11/46 11/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 New Zealand East Asia and PaciØc High Constraints on Government Powers 1.1 0.85 Limits by legislature 1.2 0.87 Limits by judiciary 1.3 0.84 Independent auditing 1.4 0.81 Sanctions for ofØcial misconduct 1.5 0.83 Non-governmental checks 1.6 0.93 Lawful transition of power Absence of Corruption 2.1 0.86 In the executive branch 2.2 0.96 In the judiciary 2.3 0.92 In the police/military 2.4 0.75 In the legislature Open Government 3.1 0.86 Publicized laws and gov't data 3.2 0.75 Right to information 3.3 0.83 Civic participation 3.4 0.83 Complaint mechanisms Fundamental Rights 4.1 0.71 No discrimination 4.2 0.93 Right to life and security 4.3 0.78 Due process of law 4.4 0.83 Freedom of expression 4.5 0.83 Freedom of religion 4.6 0.84 Right to privacy 4.7 0.84 Freedom of association 4.8 0.76 Labor rights Order and Security 5.1 0.87 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.78 Absence of violent redress Regulatory Enforcement 6.1 0.76 Effective regulatory enforcement 6.2 0.94 No improper inÙuence 6.3 0.81 No unreasonable delay 6.4 0.87 Respect for due process 6.5 0.84 No expropriation w/out adequate compensation Civil Justice 7.1 0.73 Accessibility and affordability 7.2 0.71 No discrimination 7.3 0.94 No corruption 7.4 0.86 No improper gov't inÙuence 7.5 0.71 No unreasonable delay 7.6 0.70 Effective enforcement 7.7 0.80 Impartial and effective ADRs Criminal Justice 8.1 0.61 Effective investigations 8.2 0.72 Timely and effective adjudication 8.3 0.67 Effective correctional system 8.4 0.58 No discrimination 8.5 0.90 No corruption 8.6 0.87 No improper gov't inÙuence 8.7 0.78 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 133 WJP Rule of Law Index 2023 Nicaragua Region: Latin America and Caribbean Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.35 30/32 35/37 137/142 Score Change Rank Change -0.02 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.23 0.00 31/32 37/37 141/142 Absence of Corruption 0.30 -0.02 27/32 30/37 127/142 Open Government 0.32 -0.03 31/32 32/37 133/142 Fundamental Rights 0.30 -0.02 32/32 33/37 136/142 Order and Security 0.70 -0.01 14/32 13/37 88/142 Regulatory Enforcement 0.36 -0.02 30/32 33/37 134/142 Civil Justice 0.32 -0.03 30/32 35/37 139/142 Criminal Justice 0.25 0.00 28/32 33/37 137/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Nicaragua Latin America and Caribbean Lower-Middle Constraints on Government Powers 1.1 0.36 Limits by legislature 1.2 0.30 Limits by judiciary 1.3 0.08 Independent auditing 1.4 0.12 Sanctions for of cial misconduct 1.5 0.32 Non-governmental checks 1.6 0.20 Lawful transition of power Absence of Corruption 2.1 0.35 In the executive branch 2.2 0.23 In the judiciary 2.3 0.43 In the police/military 2.4 0.18 In the legislature Open Government 3.1 0.32 Publicized laws and gov't data 3.2 0.20 Right to information 3.3 0.32 Civic participation 3.4 0.46 Complaint mechanisms Fundamental Rights 4.1 0.34 No discrimination 4.2 0.16 Right to life and security 4.3 0.23 Due process of law 4.4 0.32 Freedom of expression 4.5 0.48 Freedom of religion 4.6 0.03 Right to privacy 4.7 0.32 Freedom of association 4.8 0.49 Labor rights Order and Security 5.1 0.69 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.40 Absence of violent redress Regulatory Enforcement 6.1 0.41 Effective regulatory enforcement 6.2 0.58 No improper in ence 6.3 0.39 No unreasonable delay 6.4 0.08 Respect for due process 6.5 0.33 No expropriation w/out adequate compensation Civil Justice 7.1 0.46 Accessibility and affordability 7.2 0.26 No discrimination 7.3 0.30 No corruption 7.4 0.07 No improper gov't in ence 7.5 0.29 No unreasonable delay 7.6 0.34 Effective enforcement 7.7 0.49 Impartial and effective ADRs Criminal Justice 8.1 0.32 Effective investigations 8.2 0.35 Timely and effective adjudication 8.3 0.22 Effective correctional system 8.4 0.24 No discrimination 8.5 0.38 No corruption 8.6 0.00 No improper gov't in ence 8.7 0.23 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 134 Niger Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.44 17/34 7/18 109/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.39 0.00 23/34 12/18 113/142 Absence of Corruption 0.39 -0.01 15/34 7/18 100/142 Open Government 0.37 0.00 23/34 11/18 120/142 Fundamental Rights 0.47 -0.01 17/34 9/18 95/142 Order and Security 0.62 0.00 20/34 9/18 113/142 Regulatory Enforcement 0.47 0.00 11/34 3/18 85/142 Civil Justice 0.46 0.00 16/34 6/18 99/142 Criminal Justice 0.35 0.00 20/34 7/18 102/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Niger Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.51 Limits by legislature 1.2 0.39 Limits by judiciary 1.3 0.30 Independent auditing 1.4 0.31 Sanctions for of cial misconduct 1.5 0.41 Non-governmental checks 1.6 0.45 Lawful transition of power Absence of Corruption 2.1 0.38 In the executive branch 2.2 0.46 In the judiciary 2.3 0.45 In the police/military 2.4 0.26 In the legislature Open Government 3.1 0.21 Publicized laws and gov't data 3.2 0.37 Right to information 3.3 0.46 Civic participation 3.4 0.43 Complaint mechanisms Fundamental Rights 4.1 0.54 No discrimination 4.2 0.39 Right to life and security 4.3 0.40 Due process of law 4.4 0.41 Freedom of expression 4.5 0.65 Freedom of religion 4.6 0.21 Right to privacy 4.7 0.55 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.69 Absence of crime 5.2 0.63 Absence of civil con ct 5.3 0.55 Absence of violent redress Regulatory Enforcement 6.1 0.48 Effective regulatory enforcement 6.2 0.56 No improper in ence 6.3 0.45 No unreasonable delay 6.4 0.36 Respect for due process 6.5 0.51 No expropriation w/out adequate compensation Civil Justice 7.1 0.48 Accessibility and affordability 7.2 0.42 No discrimination 7.3 0.47 No corruption 7.4 0.27 No improper gov't in ence 7.5 0.45 No unreasonable delay 7.6 0.51 Effective enforcement 7.7 0.60 Impartial and effective ADRs Criminal Justice 8.1 0.30 Effective investigations 8.2 0.37 Timely and effective adjudication 8.3 0.31 Effective correctional system 8.4 0.41 No discrimination 8.5 0.41 No corruption 8.6 0.27 No improper gov't in ence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 135 WJP Rule of Law Index 2023 Nigeria Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.41 23/34 24/37 120/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.50 0.00 14/34 13/37 85/142 Absence of Corruption 0.32 0.01 23/34 27/37 121/142 Open Government 0.42 0.00 14/34 19/37 104/142 Fundamental Rights 0.42 0.00 23/34 23/37 116/142 Order and Security 0.37 0.01 33/34 36/37 139/142 Regulatory Enforcement 0.41 0.00 22/34 26/37 119/142 Civil Justice 0.46 0.00 17/34 17/37 100/142 Criminal Justice 0.39 0.01 13/34 12/37 86/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Nigeria Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.64 Limits by legislature 1.2 0.49 Limits by judiciary 1.3 0.47 Independent auditing 1.4 0.42 Sanctions for of cial misconduct 1.5 0.48 Non-governmental checks 1.6 0.46 Lawful transition of power Absence of Corruption 2.1 0.30 In the executive branch 2.2 0.45 In the judiciary 2.3 0.34 In the police/military 2.4 0.18 In the legislature Open Government 3.1 0.20 Publicized laws and gov't data 3.2 0.38 Right to information 3.3 0.50 Civic participation 3.4 0.60 Complaint mechanisms Fundamental Rights 4.1 0.48 No discrimination 4.2 0.28 Right to life and security 4.3 0.34 Due process of law 4.4 0.48 Freedom of expression 4.5 0.49 Freedom of religion 4.6 0.30 Right to privacy 4.7 0.56 Freedom of association 4.8 0.45 Labor rights Order and Security 5.1 0.53 Absence of crime 5.2 0.17 Absence of civil con ct 5.3 0.43 Absence of violent redress Regulatory Enforcement 6.1 0.39 Effective regulatory enforcement 6.2 0.49 No improper in ence 6.3 0.33 No unreasonable delay 6.4 0.38 Respect for due process 6.5 0.46 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.49 No discrimination 7.3 0.44 No corruption 7.4 0.44 No improper gov't in ence 7.5 0.23 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.42 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.30 Effective correctional system 8.4 0.51 No discrimination 8.5 0.36 No corruption 8.6 0.44 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 136 North Macedonia Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.53 5/15 19/41 67/142 Score Change Rank Change 0.00 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.46 -0.01 6/15 28/41 97/142 Absence of Corruption 0.45 0.00 7/15 23/41 75/142 Open Government 0.50 0.00 7/15 21/41 70/142 Fundamental Rights 0.60 0.01 4/15 15/41 56/142 Order and Security 0.80 0.00 6/15 6/41 43/142 Regulatory Enforcement 0.46 -0.02 7/15 29/41 94/142 Civil Justice 0.52 -0.01 7/15 22/41 75/142 Criminal Justice 0.44 -0.01 7/15 20/41 72/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 North Macedonia Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.53 Limits by legislature 1.2 0.32 Limits by judiciary 1.3 0.54 Independent auditing 1.4 0.34 Sanctions for ofØcial misconduct 1.5 0.53 Non-governmental checks 1.6 0.52 Lawful transition of power Absence of Corruption 2.1 0.39 In the executive branch 2.2 0.49 In the judiciary 2.3 0.63 In the police/military 2.4 0.29 In the legislature Open Government 3.1 0.49 Publicized laws and gov't data 3.2 0.49 Right to information 3.3 0.50 Civic participation 3.4 0.52 Complaint mechanisms Fundamental Rights 4.1 0.65 No discrimination 4.2 0.73 Right to life and security 4.3 0.56 Due process of law 4.4 0.53 Freedom of expression 4.5 0.71 Freedom of religion 4.6 0.48 Right to privacy 4.7 0.60 Freedom of association 4.8 0.57 Labor rights Order and Security 5.1 0.78 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.62 Absence of violent redress Regulatory Enforcement 6.1 0.41 Effective regulatory enforcement 6.2 0.50 No improper inÙuence 6.3 0.46 No unreasonable delay 6.4 0.38 Respect for due process 6.5 0.54 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.62 No discrimination 7.3 0.36 No corruption 7.4 0.39 No improper gov't inÙuence 7.5 0.41 No unreasonable delay 7.6 0.58 Effective enforcement 7.7 0.68 Impartial and effective ADRs Criminal Justice 8.1 0.39 Effective investigations 8.2 0.45 Timely and effective adjudication 8.3 0.38 Effective correctional system 8.4 0.52 No discrimination 8.5 0.46 No corruption 8.6 0.33 No improper gov't inÙuence 8.7 0.56 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 137 WJP Rule of Law Index 2023 Norway Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.89 2/31 2/46 2/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.92 -0.01 2/31 2/46 2/142 Absence of Corruption 0.94 0.00 2/31 2/46 2/142 Open Government 0.88 0.00 1/31 1/46 1/142 Fundamental Rights 0.91 0.00 2/31 2/46 2/142 Order and Security 0.93 0.00 4/31 5/46 5/142 Regulatory Enforcement 0.88 0.01 2/31 2/46 2/142 Civil Justice 0.86 0.02 1/31 1/46 1/142 Criminal Justice 0.83 -0.01 3/31 3/46 3/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Norway EU, EFTA, and North America High Constraints on Government Powers 1.1 0.94 Limits by legislature 1.2 0.95 Limits by judiciary 1.3 0.83 Independent auditing 1.4 0.88 Sanctions for ofØcial misconduct 1.5 0.95 Non-governmental checks 1.6 0.97 Lawful transition of power Absence of Corruption 2.1 0.92 In the executive branch 2.2 0.97 In the judiciary 2.3 0.96 In the police/military 2.4 0.89 In the legislature Open Government 3.1 0.88 Publicized laws and gov't data 3.2 0.86 Right to information 3.3 0.92 Civic participation 3.4 0.85 Complaint mechanisms Fundamental Rights 4.1 0.83 No discrimination 4.2 0.96 Right to life and security 4.3 0.89 Due process of law 4.4 0.95 Freedom of expression 4.5 0.87 Freedom of religion 4.6 0.90 Right to privacy 4.7 0.96 Freedom of association 4.8 0.90 Labor rights Order and Security 5.1 0.96 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.82 Absence of violent redress Regulatory Enforcement 6.1 0.83 Effective regulatory enforcement 6.2 0.99 No improper inÙuence 6.3 0.81 No unreasonable delay 6.4 0.86 Respect for due process 6.5 0.90 No expropriation w/out adequate compensation Civil Justice 7.1 0.72 Accessibility and affordability 7.2 0.77 No discrimination 7.3 0.96 No corruption 7.4 0.94 No improper gov't inÙuence 7.5 0.81 No unreasonable delay 7.6 0.93 Effective enforcement 7.7 0.92 Impartial and effective ADRs Criminal Justice 8.1 0.62 Effective investigations 8.2 0.74 Timely and effective adjudication 8.3 0.92 Effective correctional system 8.4 0.74 No discrimination 8.5 0.95 No corruption 8.6 0.91 No improper gov't inÙuence 8.7 0.89 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 138 Pakistan Region: South Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.38 5/6 29/37 130/142 Score Change Rank Change -0.01 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.47 -0.01 4/6 17/37 95/142 Absence of Corruption 0.31 0.00 5/6 28/37 123/142 Open Government 0.41 -0.01 4/6 20/37 105/142 Fundamental Rights 0.38 0.00 4/6 28/37 125/142 Order and Security 0.33 -0.03 5/6 37/37 141/142 Regulatory Enforcement 0.38 0.00 5/6 31/37 128/142 Civil Justice 0.38 -0.01 4/6 30/37 129/142 Criminal Justice 0.36 0.00 4/6 19/37 99/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Pakistan South Asia Lower-Middle Constraints on Government Powers 1.1 0.56 Limits by legislature 1.2 0.47 Limits by judiciary 1.3 0.45 Independent auditing 1.4 0.33 Sanctions for of cial misconduct 1.5 0.52 Non-governmental checks 1.6 0.49 Lawful transition of power Absence of Corruption 2.1 0.36 In the executive branch 2.2 0.39 In the judiciary 2.3 0.29 In the police/military 2.4 0.22 In the legislature Open Government 3.1 0.26 Publicized laws and gov't data 3.2 0.35 Right to information 3.3 0.52 Civic participation 3.4 0.51 Complaint mechanisms Fundamental Rights 4.1 0.37 No discrimination 4.2 0.30 Right to life and security 4.3 0.34 Due process of law 4.4 0.52 Freedom of expression 4.5 0.39 Freedom of religion 4.6 0.22 Right to privacy 4.7 0.54 Freedom of association 4.8 0.34 Labor rights Order and Security 5.1 0.58 Absence of crime 5.2 0.10 Absence of civil con ct 5.3 0.30 Absence of violent redress Regulatory Enforcement 6.1 0.40 Effective regulatory enforcement 6.2 0.48 No improper in ence 6.3 0.33 No unreasonable delay 6.4 0.20 Respect for due process 6.5 0.49 No expropriation w/out adequate compensation Civil Justice 7.1 0.38 Accessibility and affordability 7.2 0.35 No discrimination 7.3 0.38 No corruption 7.4 0.43 No improper gov't in ence 7.5 0.28 No unreasonable delay 7.6 0.30 Effective enforcement 7.7 0.54 Impartial and effective ADRs Criminal Justice 8.1 0.32 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.35 Effective correctional system 8.4 0.26 No discrimination 8.5 0.40 No corruption 8.6 0.44 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 139 WJP Rule of Law Index 2023 Panama Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.51 15/32 46/46 74/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.53 0.00 15/32 44/46 67/142 Absence of Corruption 0.40 0.00 19/32 46/46 98/142 Open Government 0.56 0.00 10/32 38/46 53/142 Fundamental Rights 0.62 0.00 12/32 40/46 49/142 Order and Security 0.70 0.00 12/32 44/46 84/142 Regulatory Enforcement 0.49 -0.01 14/32 44/46 68/142 Civil Justice 0.46 -0.01 21/32 45/46 95/142 Criminal Justice 0.34 0.00 18/32 45/46 106/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Panama Latin America and Caribbean High Constraints on Government Powers 1.1 0.56 Limits by legislature 1.2 0.47 Limits by judiciary 1.3 0.45 Independent auditing 1.4 0.32 Sanctions for ofØcial misconduct 1.5 0.66 Non-governmental checks 1.6 0.74 Lawful transition of power Absence of Corruption 2.1 0.42 In the executive branch 2.2 0.46 In the judiciary 2.3 0.56 In the police/military 2.4 0.14 In the legislature Open Government 3.1 0.39 Publicized laws and gov't data 3.2 0.54 Right to information 3.3 0.63 Civic participation 3.4 0.68 Complaint mechanisms Fundamental Rights 4.1 0.49 No discrimination 4.2 0.73 Right to life and security 4.3 0.49 Due process of law 4.4 0.66 Freedom of expression 4.5 0.66 Freedom of religion 4.6 0.60 Right to privacy 4.7 0.69 Freedom of association 4.8 0.69 Labor rights Order and Security 5.1 0.67 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.43 Absence of violent redress Regulatory Enforcement 6.1 0.48 Effective regulatory enforcement 6.2 0.55 No improper inÙuence 6.3 0.48 No unreasonable delay 6.4 0.41 Respect for due process 6.5 0.54 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.52 No discrimination 7.3 0.46 No corruption 7.4 0.41 No improper gov't inÙuence 7.5 0.23 No unreasonable delay 7.6 0.46 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.32 Effective investigations 8.2 0.34 Timely and effective adjudication 8.3 0.17 Effective correctional system 8.4 0.34 No discrimination 8.5 0.47 No corruption 8.6 0.27 No improper gov't inÙuence 8.7 0.49 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 140 Paraguay Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.46 24/32 34/41 99/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.50 0.01 20/32 22/41 80/142 Absence of Corruption 0.29 -0.02 28/32 38/41 131/142 Open Government 0.56 0.00 9/32 13/41 52/142 Fundamental Rights 0.51 -0.01 23/32 28/41 83/142 Order and Security 0.70 0.00 11/32 25/41 81/142 Regulatory Enforcement 0.46 -0.01 22/32 26/41 91/142 Civil Justice 0.42 0.00 24/32 35/41 117/142 Criminal Justice 0.26 -0.01 25/32 39/41 131/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Paraguay Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.65 Limits by legislature 1.2 0.42 Limits by judiciary 1.3 0.34 Independent auditing 1.4 0.34 Sanctions for of cial misconduct 1.5 0.62 Non-governmental checks 1.6 0.66 Lawful transition of power Absence of Corruption 2.1 0.35 In the executive branch 2.2 0.38 In the judiciary 2.3 0.29 In the police/military 2.4 0.14 In the legislature Open Government 3.1 0.47 Publicized laws and gov't data 3.2 0.55 Right to information 3.3 0.59 Civic participation 3.4 0.62 Complaint mechanisms Fundamental Rights 4.1 0.42 No discrimination 4.2 0.54 Right to life and security 4.3 0.39 Due process of law 4.4 0.62 Freedom of expression 4.5 0.67 Freedom of religion 4.6 0.29 Right to privacy 4.7 0.65 Freedom of association 4.8 0.47 Labor rights Order and Security 5.1 0.65 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.45 Effective regulatory enforcement 6.2 0.55 No improper in ence 6.3 0.41 No unreasonable delay 6.4 0.38 Respect for due process 6.5 0.52 No expropriation w/out adequate compensation Civil Justice 7.1 0.55 Accessibility and affordability 7.2 0.38 No discrimination 7.3 0.32 No corruption 7.4 0.41 No improper gov't in ence 7.5 0.23 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.28 Effective investigations 8.2 0.19 Timely and effective adjudication 8.3 0.05 Effective correctional system 8.4 0.40 No discrimination 8.5 0.23 No corruption 8.6 0.29 No improper gov't in ence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 141 WJP Rule of Law Index 2023 Peru Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 21/32 28/41 88/142 Score Change Rank Change 0.00 4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.59 0.00 10/32 7/41 49/142 Absence of Corruption 0.33 0.00 25/32 37/41 116/142 Open Government 0.54 0.00 13/32 15/41 58/142 Fundamental Rights 0.60 -0.01 15/32 16/41 57/142 Order and Security 0.62 0.00 23/32 33/41 115/142 Regulatory Enforcement 0.48 0.00 19/32 22/41 81/142 Civil Justice 0.41 0.00 25/32 37/41 121/142 Criminal Justice 0.33 0.01 20/32 33/41 112/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Peru Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.68 Limits by legislature 1.2 0.52 Limits by judiciary 1.3 0.57 Independent auditing 1.4 0.42 Sanctions for ofØcial misconduct 1.5 0.64 Non-governmental checks 1.6 0.67 Lawful transition of power Absence of Corruption 2.1 0.37 In the executive branch 2.2 0.46 In the judiciary 2.3 0.40 In the police/military 2.4 0.10 In the legislature Open Government 3.1 0.37 Publicized laws and gov't data 3.2 0.60 Right to information 3.3 0.60 Civic participation 3.4 0.59 Complaint mechanisms Fundamental Rights 4.1 0.45 No discrimination 4.2 0.71 Right to life and security 4.3 0.41 Due process of law 4.4 0.64 Freedom of expression 4.5 0.71 Freedom of religion 4.6 0.68 Right to privacy 4.7 0.70 Freedom of association 4.8 0.51 Labor rights Order and Security 5.1 0.54 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.31 Absence of violent redress Regulatory Enforcement 6.1 0.52 Effective regulatory enforcement 6.2 0.55 No improper inÙuence 6.3 0.27 No unreasonable delay 6.4 0.41 Respect for due process 6.5 0.63 No expropriation w/out adequate compensation Civil Justice 7.1 0.44 Accessibility and affordability 7.2 0.36 No discrimination 7.3 0.41 No corruption 7.4 0.48 No improper gov't inÙuence 7.5 0.15 No unreasonable delay 7.6 0.38 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.27 Effective investigations 8.2 0.26 Timely and effective adjudication 8.3 0.24 Effective correctional system 8.4 0.37 No discrimination 8.5 0.35 No corruption 8.6 0.41 No improper gov't inÙuence 8.7 0.41 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 142 Philippines Region: East Asia and Paci c Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.46 13/15 16/37 100/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.47 -0.01 10/15 15/37 93/142 Absence of Corruption 0.43 -0.01 10/15 7/37 83/142 Open Government 0.47 -0.02 10/15 11/37 79/142 Fundamental Rights 0.40 0.00 12/15 25/37 120/142 Order and Security 0.67 0.01 13/15 17/37 94/142 Regulatory Enforcement 0.47 0.00 11/15 14/37 84/142 Civil Justice 0.45 0.00 12/15 18/37 103/142 Criminal Justice 0.31 -0.02 13/15 27/37 120/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Philippines East Asia and Paci c Lower-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.50 Limits by judiciary 1.3 0.46 Independent auditing 1.4 0.37 Sanctions for of cial misconduct 1.5 0.50 Non-governmental checks 1.6 0.47 Lawful transition of power Absence of Corruption 2.1 0.45 In the executive branch 2.2 0.39 In the judiciary 2.3 0.49 In the police/military 2.4 0.39 In the legislature Open Government 3.1 0.41 Publicized laws and gov't data 3.2 0.59 Right to information 3.3 0.48 Civic participation 3.4 0.42 Complaint mechanisms Fundamental Rights 4.1 0.45 No discrimination 4.2 0.18 Right to life and security 4.3 0.27 Due process of law 4.4 0.50 Freedom of expression 4.5 0.58 Freedom of religion 4.6 0.22 Right to privacy 4.7 0.49 Freedom of association 4.8 0.48 Labor rights Order and Security 5.1 0.69 Absence of crime 5.2 0.83 Absence of civil con ct 5.3 0.51 Absence of violent redress Regulatory Enforcement 6.1 0.49 Effective regulatory enforcement 6.2 0.59 No improper in ence 6.3 0.37 No unreasonable delay 6.4 0.41 Respect for due process 6.5 0.50 No expropriation w/out adequate compensation Civil Justice 7.1 0.52 Accessibility and affordability 7.2 0.46 No discrimination 7.3 0.51 No corruption 7.4 0.33 No improper gov't in ence 7.5 0.34 No unreasonable delay 7.6 0.37 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.41 Effective investigations 8.2 0.32 Timely and effective adjudication 8.3 0.19 Effective correctional system 8.4 0.25 No discrimination 8.5 0.46 No corruption 8.6 0.26 No improper gov't in ence 8.7 0.27 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 143 WJP Rule of Law Index 2023 Poland Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.64 26/31 35/46 36/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.53 -0.01 29/31 43/46 66/142 Absence of Corruption 0.72 0.00 18/31 25/46 25/142 Open Government 0.58 -0.02 29/31 37/46 44/142 Fundamental Rights 0.61 0.00 30/31 41/46 54/142 Order and Security 0.86 0.00 18/31 25/46 26/142 Regulatory Enforcement 0.63 -0.01 23/31 33/46 34/142 Civil Justice 0.61 0.00 24/31 36/46 45/142 Criminal Justice 0.58 0.00 24/31 34/46 37/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Poland EU, EFTA, and North America High Constraints on Government Powers 1.1 0.45 Limits by legislature 1.2 0.50 Limits by judiciary 1.3 0.46 Independent auditing 1.4 0.50 Sanctions for of cial misconduct 1.5 0.58 Non-governmental checks 1.6 0.70 Lawful transition of power Absence of Corruption 2.1 0.63 In the executive branch 2.2 0.87 In the judiciary 2.3 0.87 In the police/military 2.4 0.50 In the legislature Open Government 3.1 0.57 Publicized laws and gov't data 3.2 0.48 Right to information 3.3 0.60 Civic participation 3.4 0.68 Complaint mechanisms Fundamental Rights 4.1 0.60 No discrimination 4.2 0.71 Right to life and security 4.3 0.62 Due process of law 4.4 0.58 Freedom of expression 4.5 0.54 Freedom of religion 4.6 0.52 Right to privacy 4.7 0.63 Freedom of association 4.8 0.66 Labor rights Order and Security 5.1 0.94 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.63 Absence of violent redress Regulatory Enforcement 6.1 0.65 Effective regulatory enforcement 6.2 0.88 No improper in ence 6.3 0.49 No unreasonable delay 6.4 0.49 Respect for due process 6.5 0.63 No expropriation w/out adequate compensation Civil Justice 7.1 0.64 Accessibility and affordability 7.2 0.71 No discrimination 7.3 0.78 No corruption 7.4 0.47 No improper gov't in ence 7.5 0.34 No unreasonable delay 7.6 0.52 Effective enforcement 7.7 0.80 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.54 Timely and effective adjudication 8.3 0.57 Effective correctional system 8.4 0.58 No discrimination 8.5 0.80 No corruption 8.6 0.42 No improper gov't in ence 8.7 0.62 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 144 Portugal Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.68 21/31 28/46 28/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.76 -0.01 15/31 18/46 19/142 Absence of Corruption 0.71 -0.01 19/31 26/46 26/142 Open Government 0.64 -0.01 22/31 29/46 30/142 Fundamental Rights 0.76 0.00 18/31 22/46 23/142 Order and Security 0.78 0.00 28/31 38/46 49/142 Regulatory Enforcement 0.60 -0.01 25/31 38/46 42/142 Civil Justice 0.65 -0.01 20/31 30/46 32/142 Criminal Justice 0.56 -0.01 26/31 37/46 41/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Portugal EU, EFTA, and North America High Constraints on Government Powers 1.1 0.83 Limits by legislature 1.2 0.72 Limits by judiciary 1.3 0.72 Independent auditing 1.4 0.60 Sanctions for ofØcial misconduct 1.5 0.79 Non-governmental checks 1.6 0.89 Lawful transition of power Absence of Corruption 2.1 0.64 In the executive branch 2.2 0.86 In the judiciary 2.3 0.86 In the police/military 2.4 0.48 In the legislature Open Government 3.1 0.55 Publicized laws and gov't data 3.2 0.59 Right to information 3.3 0.75 Civic participation 3.4 0.68 Complaint mechanisms Fundamental Rights 4.1 0.63 No discrimination 4.2 0.90 Right to life and security 4.3 0.63 Due process of law 4.4 0.79 Freedom of expression 4.5 0.82 Freedom of religion 4.6 0.81 Right to privacy 4.7 0.85 Freedom of association 4.8 0.65 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.62 Effective regulatory enforcement 6.2 0.80 No improper inÙuence 6.3 0.41 No unreasonable delay 6.4 0.52 Respect for due process 6.5 0.64 No expropriation w/out adequate compensation Civil Justice 7.1 0.65 Accessibility and affordability 7.2 0.72 No discrimination 7.3 0.77 No corruption 7.4 0.70 No improper gov't inÙuence 7.5 0.43 No unreasonable delay 7.6 0.51 Effective enforcement 7.7 0.74 Impartial and effective ADRs Criminal Justice 8.1 0.47 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.45 Effective correctional system 8.4 0.45 No discrimination 8.5 0.77 No corruption 8.6 0.75 No improper gov't inÙuence 8.7 0.63 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 145 WJP Rule of Law Index 2023 Romania Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.63 27/31 39/46 40/142 Score Change Rank Change 0.00 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.61 0.00 27/31 37/46 43/142 Absence of Corruption 0.56 0.01 27/31 41/46 52/142 Open Government 0.63 0.00 25/31 32/46 33/142 Fundamental Rights 0.67 0.00 27/31 36/46 39/142 Order and Security 0.83 0.00 23/31 31/46 34/142 Regulatory Enforcement 0.59 0.00 27/31 40/46 44/142 Civil Justice 0.63 -0.02 21/31 31/46 35/142 Criminal Justice 0.52 0.00 27/31 40/46 52/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Romania EU, EFTA, and North America High Constraints on Government Powers 1.1 0.70 Limits by legislature 1.2 0.58 Limits by judiciary 1.3 0.47 Independent auditing 1.4 0.52 Sanctions for of cial misconduct 1.5 0.67 Non-governmental checks 1.6 0.72 Lawful transition of power Absence of Corruption 2.1 0.51 In the executive branch 2.2 0.71 In the judiciary 2.3 0.73 In the police/military 2.4 0.29 In the legislature Open Government 3.1 0.59 Publicized laws and gov't data 3.2 0.55 Right to information 3.3 0.61 Civic participation 3.4 0.76 Complaint mechanisms Fundamental Rights 4.1 0.70 No discrimination 4.2 0.80 Right to life and security 4.3 0.58 Due process of law 4.4 0.67 Freedom of expression 4.5 0.71 Freedom of religion 4.6 0.55 Right to privacy 4.7 0.67 Freedom of association 4.8 0.69 Labor rights Order and Security 5.1 0.92 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.57 Absence of violent redress Regulatory Enforcement 6.1 0.59 Effective regulatory enforcement 6.2 0.70 No improper in ence 6.3 0.53 No unreasonable delay 6.4 0.43 Respect for due process 6.5 0.70 No expropriation w/out adequate compensation Civil Justice 7.1 0.58 Accessibility and affordability 7.2 0.70 No discrimination 7.3 0.63 No corruption 7.4 0.66 No improper gov't in ence 7.5 0.47 No unreasonable delay 7.6 0.63 Effective enforcement 7.7 0.74 Impartial and effective ADRs Criminal Justice 8.1 0.54 Effective investigations 8.2 0.45 Timely and effective adjudication 8.3 0.39 Effective correctional system 8.4 0.53 No discrimination 8.5 0.62 No corruption 8.6 0.56 No improper gov't in ence 8.7 0.58 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 146 Russian Federation Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.44 14/15 37/41 113/142 Score Change Rank Change -0.01 -4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.32 -0.01 13/15 36/41 132/142 Absence of Corruption 0.41 0.00 11/15 30/41 92/142 Open Government 0.45 -0.02 11/15 30/41 89/142 Fundamental Rights 0.38 -0.02 13/15 37/41 124/142 Order and Security 0.66 -0.02 14/15 27/41 101/142 Regulatory Enforcement 0.46 0.00 6/15 27/41 92/142 Civil Justice 0.51 0.00 8/15 23/41 78/142 Criminal Justice 0.29 -0.01 15/15 38/41 127/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Russian Federation Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.34 Limits by legislature 1.2 0.31 Limits by judiciary 1.3 0.32 Independent auditing 1.4 0.31 Sanctions for of cial misconduct 1.5 0.31 Non-governmental checks 1.6 0.34 Lawful transition of power Absence of Corruption 2.1 0.40 In the executive branch 2.2 0.52 In the judiciary 2.3 0.52 In the police/military 2.4 0.22 In the legislature Open Government 3.1 0.56 Publicized laws and gov't data 3.2 0.37 Right to information 3.3 0.31 Civic participation 3.4 0.57 Complaint mechanisms Fundamental Rights 4.1 0.56 No discrimination 4.2 0.34 Right to life and security 4.3 0.34 Due process of law 4.4 0.31 Freedom of expression 4.5 0.45 Freedom of religion 4.6 0.15 Right to privacy 4.7 0.33 Freedom of association 4.8 0.58 Labor rights Order and Security 5.1 0.78 Absence of crime 5.2 0.85 Absence of civil con ct 5.3 0.35 Absence of violent redress Regulatory Enforcement 6.1 0.54 Effective regulatory enforcement 6.2 0.53 No improper in ence 6.3 0.57 No unreasonable delay 6.4 0.31 Respect for due process 6.5 0.36 No expropriation w/out adequate compensation Civil Justice 7.1 0.62 Accessibility and affordability 7.2 0.48 No discrimination 7.3 0.48 No corruption 7.4 0.25 No improper gov't in ence 7.5 0.74 No unreasonable delay 7.6 0.47 Effective enforcement 7.7 0.50 Impartial and effective ADRs Criminal Justice 8.1 0.19 Effective investigations 8.2 0.34 Timely and effective adjudication 8.3 0.32 Effective correctional system 8.4 0.33 No discrimination 8.5 0.46 No corruption 8.6 0.07 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 147 WJP Rule of Law Index 2023 Rwanda Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.63 1/34 1/18 41/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.60 -0.01 5/34 1/18 45/142 Absence of Corruption 0.68 0.01 1/34 1/18 32/142 Open Government 0.57 0.01 3/34 1/18 51/142 Fundamental Rights 0.52 0.00 9/34 3/18 76/142 Order and Security 0.85 0.00 1/34 1/18 27/142 Regulatory Enforcement 0.60 0.00 2/34 1/18 41/142 Civil Justice 0.65 -0.02 1/34 1/18 31/142 Criminal Justice 0.56 0.01 2/34 1/18 42/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Rwanda Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.53 Limits by legislature 1.2 0.55 Limits by judiciary 1.3 0.65 Independent auditing 1.4 0.76 Sanctions for of cial misconduct 1.5 0.45 Non-governmental checks 1.6 0.67 Lawful transition of power Absence of Corruption 2.1 0.74 In the executive branch 2.2 0.66 In the judiciary 2.3 0.70 In the police/military 2.4 0.60 In the legislature Open Government 3.1 0.56 Publicized laws and gov't data 3.2 0.61 Right to information 3.3 0.52 Civic participation 3.4 0.58 Complaint mechanisms Fundamental Rights 4.1 0.69 No discrimination 4.2 0.40 Right to life and security 4.3 0.53 Due process of law 4.4 0.45 Freedom of expression 4.5 0.56 Freedom of religion 4.6 0.32 Right to privacy 4.7 0.56 Freedom of association 4.8 0.67 Labor rights Order and Security 5.1 0.73 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.81 Absence of violent redress Regulatory Enforcement 6.1 0.59 Effective regulatory enforcement 6.2 0.69 No improper in ence 6.3 0.51 No unreasonable delay 6.4 0.52 Respect for due process 6.5 0.71 No expropriation w/out adequate compensation Civil Justice 7.1 0.65 Accessibility and affordability 7.2 0.71 No discrimination 7.3 0.54 No corruption 7.4 0.60 No improper gov't in ence 7.5 0.71 No unreasonable delay 7.6 0.69 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.48 Effective investigations 8.2 0.56 Timely and effective adjudication 8.3 0.49 Effective correctional system 8.4 0.77 No discrimination 8.5 0.71 No corruption 8.6 0.40 No improper gov't in ence 8.7 0.53 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 148 Senegal Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.55 6/34 1/37 60/142 Score Change Rank Change -0.01 -2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.55 -0.02 9/34 5/37 63/142 Absence of Corruption 0.55 0.00 4/34 1/37 54/142 Open Government 0.44 0.00 13/34 15/37 98/142 Fundamental Rights 0.59 -0.01 4/34 2/37 60/142 Order and Security 0.70 0.00 9/34 11/37 82/142 Regulatory Enforcement 0.56 -0.01 5/34 2/37 49/142 Civil Justice 0.56 0.00 8/34 2/37 57/142 Criminal Justice 0.47 -0.01 7/34 2/37 64/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Senegal Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.45 Limits by judiciary 1.3 0.50 Independent auditing 1.4 0.51 Sanctions for ofØcial misconduct 1.5 0.60 Non-governmental checks 1.6 0.68 Lawful transition of power Absence of Corruption 2.1 0.48 In the executive branch 2.2 0.61 In the judiciary 2.3 0.69 In the police/military 2.4 0.44 In the legislature Open Government 3.1 0.35 Publicized laws and gov't data 3.2 0.47 Right to information 3.3 0.59 Civic participation 3.4 0.34 Complaint mechanisms Fundamental Rights 4.1 0.68 No discrimination 4.2 0.56 Right to life and security 4.3 0.47 Due process of law 4.4 0.60 Freedom of expression 4.5 0.75 Freedom of religion 4.6 0.36 Right to privacy 4.7 0.68 Freedom of association 4.8 0.67 Labor rights Order and Security 5.1 0.67 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.44 Absence of violent redress Regulatory Enforcement 6.1 0.56 Effective regulatory enforcement 6.2 0.67 No improper inÙuence 6.3 0.45 No unreasonable delay 6.4 0.52 Respect for due process 6.5 0.63 No expropriation w/out adequate compensation Civil Justice 7.1 0.55 Accessibility and affordability 7.2 0.64 No discrimination 7.3 0.53 No corruption 7.4 0.41 No improper gov't inÙuence 7.5 0.56 No unreasonable delay 7.6 0.58 Effective enforcement 7.7 0.67 Impartial and effective ADRs Criminal Justice 8.1 0.55 Effective investigations 8.2 0.50 Timely and effective adjudication 8.3 0.27 Effective correctional system 8.4 0.57 No discrimination 8.5 0.60 No corruption 8.6 0.30 No improper gov't inÙuence 8.7 0.47 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 149 WJP Rule of Law Index 2023 Serbia Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.48 11/15 30/41 93/142 Score Change Rank Change -0.01 -8 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.35 -0.02 12/15 35/41 127/142 Absence of Corruption 0.42 -0.01 10/15 29/41 91/142 Open Government 0.45 0.00 12/15 31/41 90/142 Fundamental Rights 0.55 0.00 9/15 23/41 72/142 Order and Security 0.76 -0.01 11/15 15/41 57/142 Regulatory Enforcement 0.46 0.00 9/15 32/41 97/142 Civil Justice 0.47 -0.02 12/15 30/41 92/142 Criminal Justice 0.39 -0.01 10/15 28/41 90/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Serbia Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.36 Limits by legislature 1.2 0.32 Limits by judiciary 1.3 0.37 Independent auditing 1.4 0.23 Sanctions for ofØcial misconduct 1.5 0.44 Non-governmental checks 1.6 0.41 Lawful transition of power Absence of Corruption 2.1 0.40 In the executive branch 2.2 0.47 In the judiciary 2.3 0.55 In the police/military 2.4 0.24 In the legislature Open Government 3.1 0.53 Publicized laws and gov't data 3.2 0.46 Right to information 3.3 0.44 Civic participation 3.4 0.38 Complaint mechanisms Fundamental Rights 4.1 0.62 No discrimination 4.2 0.58 Right to life and security 4.3 0.51 Due process of law 4.4 0.44 Freedom of expression 4.5 0.67 Freedom of religion 4.6 0.41 Right to privacy 4.7 0.55 Freedom of association 4.8 0.63 Labor rights Order and Security 5.1 0.88 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.41 Absence of violent redress Regulatory Enforcement 6.1 0.47 Effective regulatory enforcement 6.2 0.52 No improper inÙuence 6.3 0.40 No unreasonable delay 6.4 0.37 Respect for due process 6.5 0.52 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.66 No discrimination 7.3 0.44 No corruption 7.4 0.33 No improper gov't inÙuence 7.5 0.20 No unreasonable delay 7.6 0.51 Effective enforcement 7.7 0.61 Impartial and effective ADRs Criminal Justice 8.1 0.39 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.45 Effective correctional system 8.4 0.36 No discrimination 8.5 0.42 No corruption 8.6 0.16 No improper gov't inÙuence 8.7 0.51 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 150 Sierra Leone Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.44 18/34 8/18 110/142 Score Change Rank Change -0.01 -3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.48 -0.02 16/34 6/18 92/142 Absence of Corruption 0.36 0.00 18/34 9/18 110/142 Open Government 0.39 0.00 19/34 9/18 114/142 Fundamental Rights 0.48 -0.02 15/34 8/18 92/142 Order and Security 0.67 0.00 17/34 8/18 98/142 Regulatory Enforcement 0.37 -0.01 28/34 13/18 130/142 Civil Justice 0.43 -0.01 23/34 9/18 114/142 Criminal Justice 0.35 -0.01 21/34 8/18 103/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Sierra Leone Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.56 Limits by legislature 1.2 0.40 Limits by judiciary 1.3 0.45 Independent auditing 1.4 0.34 Sanctions for of cial misconduct 1.5 0.53 Non-governmental checks 1.6 0.57 Lawful transition of power Absence of Corruption 2.1 0.35 In the executive branch 2.2 0.36 In the judiciary 2.3 0.35 In the police/military 2.4 0.36 In the legislature Open Government 3.1 0.10 Publicized laws and gov't data 3.2 0.42 Right to information 3.3 0.53 Civic participation 3.4 0.49 Complaint mechanisms Fundamental Rights 4.1 0.47 No discrimination 4.2 0.39 Right to life and security 4.3 0.40 Due process of law 4.4 0.53 Freedom of expression 4.5 0.74 Freedom of religion 4.6 0.32 Right to privacy 4.7 0.56 Freedom of association 4.8 0.47 Labor rights Order and Security 5.1 0.60 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.39 Absence of violent redress Regulatory Enforcement 6.1 0.38 Effective regulatory enforcement 6.2 0.38 No improper in ence 6.3 0.34 No unreasonable delay 6.4 0.39 Respect for due process 6.5 0.38 No expropriation w/out adequate compensation Civil Justice 7.1 0.47 Accessibility and affordability 7.2 0.49 No discrimination 7.3 0.33 No corruption 7.4 0.26 No improper gov't in ence 7.5 0.41 No unreasonable delay 7.6 0.52 Effective enforcement 7.7 0.51 Impartial and effective ADRs Criminal Justice 8.1 0.44 Effective investigations 8.2 0.31 Timely and effective adjudication 8.3 0.28 Effective correctional system 8.4 0.45 No discrimination 8.5 0.37 No corruption 8.6 0.20 No improper gov't in ence 8.7 0.40 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 151 WJP Rule of Law Index 2023 Singapore Region: East Asia and PaciØc Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.78 4/15 17/46 17/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.67 0.00 5/15 30/46 31/142 Absence of Corruption 0.91 0.00 1/15 3/46 3/142 Open Government 0.61 -0.01 6/15 33/46 35/142 Fundamental Rights 0.67 -0.01 5/15 37/46 40/142 Order and Security 0.93 0.00 1/15 3/46 3/142 Regulatory Enforcement 0.86 0.01 1/15 5/46 5/142 Civil Justice 0.79 0.00 1/15 9/46 9/142 Criminal Justice 0.77 0.00 1/15 7/46 7/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Singapore East Asia and PaciØc High Constraints on Government Powers 1.1 0.50 Limits by legislature 1.2 0.73 Limits by judiciary 1.3 0.68 Independent auditing 1.4 0.88 Sanctions for ofØcial misconduct 1.5 0.47 Non-governmental checks 1.6 0.75 Lawful transition of power Absence of Corruption 2.1 0.89 In the executive branch 2.2 0.94 In the judiciary 2.3 0.94 In the police/military 2.4 0.88 In the legislature Open Government 3.1 0.79 Publicized laws and gov't data 3.2 0.52 Right to information 3.3 0.51 Civic participation 3.4 0.62 Complaint mechanisms Fundamental Rights 4.1 0.81 No discrimination 4.2 0.73 Right to life and security 4.3 0.75 Due process of law 4.4 0.47 Freedom of expression 4.5 0.79 Freedom of religion 4.6 0.55 Right to privacy 4.7 0.47 Freedom of association 4.8 0.75 Labor rights Order and Security 5.1 0.98 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.82 Absence of violent redress Regulatory Enforcement 6.1 0.82 Effective regulatory enforcement 6.2 0.94 No improper inÙuence 6.3 0.90 No unreasonable delay 6.4 0.87 Respect for due process 6.5 0.75 No expropriation w/out adequate compensation Civil Justice 7.1 0.62 Accessibility and affordability 7.2 0.80 No discrimination 7.3 0.89 No corruption 7.4 0.69 No improper gov't inÙuence 7.5 0.87 No unreasonable delay 7.6 0.90 Effective enforcement 7.7 0.75 Impartial and effective ADRs Criminal Justice 8.1 0.83 Effective investigations 8.2 0.74 Timely and effective adjudication 8.3 0.87 Effective correctional system 8.4 0.76 No discrimination 8.5 0.91 No corruption 8.6 0.55 No improper gov't inÙuence 8.7 0.75 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 152 Slovak Republic Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.66 25/31 33/46 34/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.67 0.00 22/31 28/46 29/142 Absence of Corruption 0.53 0.02 29/31 43/46 56/142 Open Government 0.69 0.00 19/31 26/46 27/142 Fundamental Rights 0.73 0.00 22/31 28/46 29/142 Order and Security 0.90 0.00 9/31 13/46 13/142 Regulatory Enforcement 0.62 0.00 24/31 35/46 37/142 Civil Justice 0.55 0.00 29/31 44/46 64/142 Criminal Justice 0.58 0.00 23/31 33/46 35/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Slovak Republic EU, EFTA, and North America High Constraints on Government Powers 1.1 0.71 Limits by legislature 1.2 0.59 Limits by judiciary 1.3 0.64 Independent auditing 1.4 0.49 Sanctions for of cial misconduct 1.5 0.72 Non-governmental checks 1.6 0.89 Lawful transition of power Absence of Corruption 2.1 0.59 In the executive branch 2.2 0.64 In the judiciary 2.3 0.59 In the police/military 2.4 0.29 In the legislature Open Government 3.1 0.65 Publicized laws and gov't data 3.2 0.66 Right to information 3.3 0.69 Civic participation 3.4 0.75 Complaint mechanisms Fundamental Rights 4.1 0.59 No discrimination 4.2 0.91 Right to life and security 4.3 0.69 Due process of law 4.4 0.72 Freedom of expression 4.5 0.66 Freedom of religion 4.6 0.74 Right to privacy 4.7 0.80 Freedom of association 4.8 0.74 Labor rights Order and Security 5.1 0.88 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.83 Absence of violent redress Regulatory Enforcement 6.1 0.67 Effective regulatory enforcement 6.2 0.69 No improper in ence 6.3 0.50 No unreasonable delay 6.4 0.59 Respect for due process 6.5 0.67 No expropriation w/out adequate compensation Civil Justice 7.1 0.57 Accessibility and affordability 7.2 0.45 No discrimination 7.3 0.59 No corruption 7.4 0.67 No improper gov't in ence 7.5 0.33 No unreasonable delay 7.6 0.56 Effective enforcement 7.7 0.64 Impartial and effective ADRs Criminal Justice 8.1 0.52 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.71 Effective correctional system 8.4 0.46 No discrimination 8.5 0.50 No corruption 8.6 0.76 No improper gov't in ence 8.7 0.69 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 153 WJP Rule of Law Index 2023 Slovenia Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.69 20/31 27/46 27/142 Score Change Rank Change 0.01 4 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.65 0.02 25/31 32/46 36/142 Absence of Corruption 0.67 0.02 22/31 32/46 35/142 Open Government 0.66 0.01 21/31 28/46 29/142 Fundamental Rights 0.75 0.01 19/31 23/46 24/142 Order and Security 0.89 0.00 13/31 17/46 18/142 Regulatory Enforcement 0.65 0.00 21/31 30/46 31/142 Civil Justice 0.67 0.02 18/31 27/46 27/142 Criminal Justice 0.56 0.01 25/31 36/46 40/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Slovenia EU, EFTA, and North America High Constraints on Government Powers 1.1 0.68 Limits by legislature 1.2 0.60 Limits by judiciary 1.3 0.66 Independent auditing 1.4 0.50 Sanctions for ofØcial misconduct 1.5 0.69 Non-governmental checks 1.6 0.80 Lawful transition of power Absence of Corruption 2.1 0.62 In the executive branch 2.2 0.79 In the judiciary 2.3 0.78 In the police/military 2.4 0.47 In the legislature Open Government 3.1 0.69 Publicized laws and gov't data 3.2 0.65 Right to information 3.3 0.69 Civic participation 3.4 0.60 Complaint mechanisms Fundamental Rights 4.1 0.74 No discrimination 4.2 0.88 Right to life and security 4.3 0.74 Due process of law 4.4 0.69 Freedom of expression 4.5 0.77 Freedom of religion 4.6 0.71 Right to privacy 4.7 0.75 Freedom of association 4.8 0.74 Labor rights Order and Security 5.1 0.96 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.72 Absence of violent redress Regulatory Enforcement 6.1 0.70 Effective regulatory enforcement 6.2 0.75 No improper inÙuence 6.3 0.60 No unreasonable delay 6.4 0.57 Respect for due process 6.5 0.62 No expropriation w/out adequate compensation Civil Justice 7.1 0.69 Accessibility and affordability 7.2 0.76 No discrimination 7.3 0.70 No corruption 7.4 0.63 No improper gov't inÙuence 7.5 0.45 No unreasonable delay 7.6 0.63 Effective enforcement 7.7 0.80 Impartial and effective ADRs Criminal Justice 8.1 0.44 Effective investigations 8.2 0.50 Timely and effective adjudication 8.3 0.58 Effective correctional system 8.4 0.53 No discrimination 8.5 0.66 No corruption 8.6 0.50 No improper gov't inÙuence 8.7 0.74 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 154 South Africa Region: Sub-Saharan Africa Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.57 5/34 12/41 56/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.62 -0.01 3/34 4/41 40/142 Absence of Corruption 0.47 -0.01 7/34 18/41 69/142 Open Government 0.62 -0.01 1/34 2/41 34/142 Fundamental Rights 0.63 0.00 3/34 9/41 48/142 Order and Security 0.60 -0.01 22/34 35/41 119/142 Regulatory Enforcement 0.52 -0.01 7/34 14/41 60/142 Civil Justice 0.59 -0.01 5/34 11/41 49/142 Criminal Justice 0.50 0.00 5/34 13/41 55/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 South Africa Sub-Saharan Africa Upper-Middle Constraints on Government Powers 1.1 0.58 Limits by legislature 1.2 0.67 Limits by judiciary 1.3 0.58 Independent auditing 1.4 0.46 Sanctions for ofØcial misconduct 1.5 0.70 Non-governmental checks 1.6 0.71 Lawful transition of power Absence of Corruption 2.1 0.37 In the executive branch 2.2 0.71 In the judiciary 2.3 0.56 In the police/military 2.4 0.22 In the legislature Open Government 3.1 0.54 Publicized laws and gov't data 3.2 0.54 Right to information 3.3 0.69 Civic participation 3.4 0.72 Complaint mechanisms Fundamental Rights 4.1 0.49 No discrimination 4.2 0.60 Right to life and security 4.3 0.53 Due process of law 4.4 0.70 Freedom of expression 4.5 0.70 Freedom of religion 4.6 0.61 Right to privacy 4.7 0.73 Freedom of association 4.8 0.67 Labor rights Order and Security 5.1 0.48 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.33 Absence of violent redress Regulatory Enforcement 6.1 0.42 Effective regulatory enforcement 6.2 0.57 No improper inÙuence 6.3 0.48 No unreasonable delay 6.4 0.55 Respect for due process 6.5 0.59 No expropriation w/out adequate compensation Civil Justice 7.1 0.51 Accessibility and affordability 7.2 0.47 No discrimination 7.3 0.68 No corruption 7.4 0.66 No improper gov't inÙuence 7.5 0.48 No unreasonable delay 7.6 0.57 Effective enforcement 7.7 0.74 Impartial and effective ADRs Criminal Justice 8.1 0.38 Effective investigations 8.2 0.49 Timely and effective adjudication 8.3 0.30 Effective correctional system 8.4 0.52 No discrimination 8.5 0.58 No corruption 8.6 0.72 No improper gov't inÙuence 8.7 0.53 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 155 WJP Rule of Law Index 2023 Spain Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.72 18/31 24/46 24/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.72 -0.01 18/31 22/46 23/142 Absence of Corruption 0.73 -0.01 16/31 23/46 23/142 Open Government 0.70 0.00 17/31 22/46 22/142 Fundamental Rights 0.79 0.00 14/31 16/46 17/142 Order and Security 0.83 0.00 24/31 32/46 35/142 Regulatory Enforcement 0.69 0.00 19/31 27/46 27/142 Civil Justice 0.65 -0.01 19/31 29/46 30/142 Criminal Justice 0.66 -0.01 18/31 25/46 25/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Spain EU, EFTA, and North America High Constraints on Government Powers 1.1 0.74 Limits by legislature 1.2 0.65 Limits by judiciary 1.3 0.74 Independent auditing 1.4 0.63 Sanctions for ofØcial misconduct 1.5 0.71 Non-governmental checks 1.6 0.84 Lawful transition of power Absence of Corruption 2.1 0.67 In the executive branch 2.2 0.88 In the judiciary 2.3 0.89 In the police/military 2.4 0.47 In the legislature Open Government 3.1 0.71 Publicized laws and gov't data 3.2 0.65 Right to information 3.3 0.70 Civic participation 3.4 0.75 Complaint mechanisms Fundamental Rights 4.1 0.71 No discrimination 4.2 0.88 Right to life and security 4.3 0.79 Due process of law 4.4 0.71 Freedom of expression 4.5 0.77 Freedom of religion 4.6 0.86 Right to privacy 4.7 0.82 Freedom of association 4.8 0.75 Labor rights Order and Security 5.1 0.87 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.61 Absence of violent redress Regulatory Enforcement 6.1 0.67 Effective regulatory enforcement 6.2 0.82 No improper inÙuence 6.3 0.54 No unreasonable delay 6.4 0.74 Respect for due process 6.5 0.69 No expropriation w/out adequate compensation Civil Justice 7.1 0.70 Accessibility and affordability 7.2 0.67 No discrimination 7.3 0.72 No corruption 7.4 0.65 No improper gov't inÙuence 7.5 0.49 No unreasonable delay 7.6 0.54 Effective enforcement 7.7 0.80 Impartial and effective ADRs Criminal Justice 8.1 0.58 Effective investigations 8.2 0.55 Timely and effective adjudication 8.3 0.73 Effective correctional system 8.4 0.63 No discrimination 8.5 0.77 No corruption 8.6 0.61 No improper gov't inÙuence 8.7 0.79 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 156 Sri Lanka Region: South Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.50 2/6 7/37 77/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.50 0.00 3/6 10/37 81/142 Absence of Corruption 0.47 0.00 1/6 2/37 65/142 Open Government 0.51 0.01 3/6 5/37 67/142 Fundamental Rights 0.50 0.00 2/6 9/37 86/142 Order and Security 0.68 0.00 2/6 15/37 91/142 Regulatory Enforcement 0.48 0.00 2/6 9/37 75/142 Civil Justice 0.43 0.00 2/6 23/37 109/142 Criminal Justice 0.41 -0.01 2/6 10/37 80/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Sri Lanka South Asia Lower-Middle Constraints on Government Powers 1.1 0.50 Limits by legislature 1.2 0.55 Limits by judiciary 1.3 0.39 Independent auditing 1.4 0.38 Sanctions for of cial misconduct 1.5 0.55 Non-governmental checks 1.6 0.63 Lawful transition of power Absence of Corruption 2.1 0.43 In the executive branch 2.2 0.61 In the judiciary 2.3 0.55 In the police/military 2.4 0.28 In the legislature Open Government 3.1 0.45 Publicized laws and gov't data 3.2 0.46 Right to information 3.3 0.57 Civic participation 3.4 0.58 Complaint mechanisms Fundamental Rights 4.1 0.53 No discrimination 4.2 0.38 Right to life and security 4.3 0.37 Due process of law 4.4 0.55 Freedom of expression 4.5 0.60 Freedom of religion 4.6 0.34 Right to privacy 4.7 0.63 Freedom of association 4.8 0.59 Labor rights Order and Security 5.1 0.80 Absence of crime 5.2 0.92 Absence of civil con ct 5.3 0.33 Absence of violent redress Regulatory Enforcement 6.1 0.55 Effective regulatory enforcement 6.2 0.56 No improper in ence 6.3 0.38 No unreasonable delay 6.4 0.40 Respect for due process 6.5 0.51 No expropriation w/out adequate compensation Civil Justice 7.1 0.46 Accessibility and affordability 7.2 0.37 No discrimination 7.3 0.61 No corruption 7.4 0.46 No improper gov't in ence 7.5 0.19 No unreasonable delay 7.6 0.36 Effective enforcement 7.7 0.59 Impartial and effective ADRs Criminal Justice 8.1 0.41 Effective investigations 8.2 0.35 Timely and effective adjudication 8.3 0.36 Effective correctional system 8.4 0.42 No discrimination 8.5 0.61 No corruption 8.6 0.34 No improper gov't in ence 8.7 0.37 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 157 WJP Rule of Law Index 2023 St. Kitts and Nevis Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.63 6/32 38/46 39/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.61 0.01 6/32 35/46 41/142 Absence of Corruption 0.65 0.01 6/32 35/46 39/142 Open Government 0.45 -0.01 26/32 43/46 94/142 Fundamental Rights 0.69 0.00 8/32 33/46 36/142 Order and Security 0.78 0.00 4/32 39/46 54/142 Regulatory Enforcement 0.62 -0.01 6/32 37/46 39/142 Civil Justice 0.68 -0.01 2/32 25/46 25/142 Criminal Justice 0.59 0.01 2/32 30/46 31/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 St. Kitts and Nevis Latin America and Caribbean High Constraints on Government Powers 1.1 0.59 Limits by legislature 1.2 0.70 Limits by judiciary 1.3 0.62 Independent auditing 1.4 0.44 Sanctions for of cial misconduct 1.5 0.64 Non-governmental checks 1.6 0.69 Lawful transition of power Absence of Corruption 2.1 0.50 In the executive branch 2.2 0.92 In the judiciary 2.3 0.78 In the police/military 2.4 0.39 In the legislature Open Government 3.1 0.22 Publicized laws and gov't data 3.2 0.42 Right to information 3.3 0.62 Civic participation 3.4 0.54 Complaint mechanisms Fundamental Rights 4.1 0.62 No discrimination 4.2 0.85 Right to life and security 4.3 0.62 Due process of law 4.4 0.64 Freedom of expression 4.5 0.67 Freedom of religion 4.6 0.78 Right to privacy 4.7 0.67 Freedom of association 4.8 0.67 Labor rights Order and Security 5.1 0.83 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.50 Absence of violent redress Regulatory Enforcement 6.1 0.51 Effective regulatory enforcement 6.2 0.76 No improper in ence 6.3 0.57 No unreasonable delay 6.4 0.63 Respect for due process 6.5 0.61 No expropriation w/out adequate compensation Civil Justice 7.1 0.71 Accessibility and affordability 7.2 0.63 No discrimination 7.3 0.81 No corruption 7.4 0.82 No improper gov't in ence 7.5 0.55 No unreasonable delay 7.6 0.47 Effective enforcement 7.7 0.76 Impartial and effective ADRs Criminal Justice 8.1 0.44 Effective investigations 8.2 0.53 Timely and effective adjudication 8.3 0.38 Effective correctional system 8.4 0.65 No discrimination 8.5 0.76 No corruption 8.6 0.77 No improper gov't in ence 8.7 0.62 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 158 St. Lucia Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.62 8/32 3/41 43/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.60 0.01 8/32 6/41 46/142 Absence of Corruption 0.64 0.00 9/32 5/41 44/142 Open Government 0.50 0.01 19/32 22/41 72/142 Fundamental Rights 0.66 0.00 9/32 5/41 42/142 Order and Security 0.73 -0.01 7/32 20/41 67/142 Regulatory Enforcement 0.61 0.02 7/32 3/41 40/142 Civil Justice 0.65 -0.01 4/32 1/41 29/142 Criminal Justice 0.54 0.00 8/32 8/41 46/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 St. Lucia Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.65 Limits by legislature 1.2 0.69 Limits by judiciary 1.3 0.42 Independent auditing 1.4 0.48 Sanctions for of cial misconduct 1.5 0.63 Non-governmental checks 1.6 0.74 Lawful transition of power Absence of Corruption 2.1 0.56 In the executive branch 2.2 0.90 In the judiciary 2.3 0.69 In the police/military 2.4 0.39 In the legislature Open Government 3.1 0.28 Publicized laws and gov't data 3.2 0.54 Right to information 3.3 0.62 Civic participation 3.4 0.56 Complaint mechanisms Fundamental Rights 4.1 0.76 No discrimination 4.2 0.77 Right to life and security 4.3 0.53 Due process of law 4.4 0.63 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.58 Right to privacy 4.7 0.64 Freedom of association 4.8 0.67 Labor rights Order and Security 5.1 0.75 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.44 Absence of violent redress Regulatory Enforcement 6.1 0.49 Effective regulatory enforcement 6.2 0.74 No improper in ence 6.3 0.45 No unreasonable delay 6.4 0.68 Respect for due process 6.5 0.67 No expropriation w/out adequate compensation Civil Justice 7.1 0.64 Accessibility and affordability 7.2 0.72 No discrimination 7.3 0.82 No corruption 7.4 0.78 No improper gov't in ence 7.5 0.46 No unreasonable delay 7.6 0.39 Effective enforcement 7.7 0.75 Impartial and effective ADRs Criminal Justice 8.1 0.36 Effective investigations 8.2 0.38 Timely and effective adjudication 8.3 0.48 Effective correctional system 8.4 0.65 No discrimination 8.5 0.71 No corruption 8.6 0.72 No improper gov't in ence 8.7 0.53 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 159 WJP Rule of Law Index 2023 St. Vincent and the Grenadines Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.63 7/32 2/41 42/142 Score Change Rank Change -0.01 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.58 0.00 11/32 8/41 50/142 Absence of Corruption 0.68 0.00 4/32 1/41 29/142 Open Government 0.51 0.00 18/32 20/41 66/142 Fundamental Rights 0.70 0.00 6/32 2/41 34/142 Order and Security 0.76 -0.01 5/32 16/41 59/142 Regulatory Enforcement 0.53 0.00 10/32 11/41 56/142 Civil Justice 0.64 0.00 5/32 2/41 33/142 Criminal Justice 0.60 -0.02 1/32 1/41 30/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 St. Vincent and the Grenadines Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.59 Limits by judiciary 1.3 0.58 Independent auditing 1.4 0.50 Sanctions for of cial misconduct 1.5 0.57 Non-governmental checks 1.6 0.72 Lawful transition of power Absence of Corruption 2.1 0.61 In the executive branch 2.2 0.87 In the judiciary 2.3 0.77 In the police/military 2.4 0.47 In the legislature Open Government 3.1 0.24 Publicized laws and gov't data 3.2 0.60 Right to information 3.3 0.60 Civic participation 3.4 0.62 Complaint mechanisms Fundamental Rights 4.1 0.64 No discrimination 4.2 0.87 Right to life and security 4.3 0.63 Due process of law 4.4 0.57 Freedom of expression 4.5 0.76 Freedom of religion 4.6 0.79 Right to privacy 4.7 0.68 Freedom of association 4.8 0.66 Labor rights Order and Security 5.1 0.78 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.49 Absence of violent redress Regulatory Enforcement 6.1 0.48 Effective regulatory enforcement 6.2 0.71 No improper in ence 6.3 0.45 No unreasonable delay 6.4 0.45 Respect for due process 6.5 0.57 No expropriation w/out adequate compensation Civil Justice 7.1 0.68 Accessibility and affordability 7.2 0.63 No discrimination 7.3 0.76 No corruption 7.4 0.65 No improper gov't in ence 7.5 0.46 No unreasonable delay 7.6 0.46 Effective enforcement 7.7 0.83 Impartial and effective ADRs Criminal Justice 8.1 0.44 Effective investigations 8.2 0.58 Timely and effective adjudication 8.3 0.58 Effective correctional system 8.4 0.55 No discrimination 8.5 0.72 No corruption 8.6 0.68 No improper gov't in ence 8.7 0.63 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 160 Sudan Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.36 31/34 16/18 132/142 Score Change Rank Change -0.03 -5 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.29 -0.10 34/34 18/18 136/142 Absence of Corruption 0.35 -0.01 20/34 10/18 112/142 Open Government 0.35 -0.02 27/34 15/18 127/142 Fundamental Rights 0.33 -0.01 33/34 16/18 132/142 Order and Security 0.56 -0.07 27/34 12/18 130/142 Regulatory Enforcement 0.33 0.00 33/34 18/18 138/142 Civil Justice 0.36 -0.01 34/34 17/18 134/142 Criminal Justice 0.34 -0.01 23/34 10/18 105/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Sudan Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.24 Limits by legislature 1.2 0.29 Limits by judiciary 1.3 0.20 Independent auditing 1.4 0.28 Sanctions for of cial misconduct 1.5 0.39 Non-governmental checks 1.6 0.33 Lawful transition of power Absence of Corruption 2.1 0.35 In the executive branch 2.2 0.39 In the judiciary 2.3 0.36 In the police/military 2.4 0.31 In the legislature Open Government 3.1 0.23 Publicized laws and gov't data 3.2 0.24 Right to information 3.3 0.46 Civic participation 3.4 0.45 Complaint mechanisms Fundamental Rights 4.1 0.33 No discrimination 4.2 0.20 Right to life and security 4.3 0.34 Due process of law 4.4 0.39 Freedom of expression 4.5 0.45 Freedom of religion 4.6 0.16 Right to privacy 4.7 0.47 Freedom of association 4.8 0.29 Labor rights Order and Security 5.1 0.61 Absence of crime 5.2 0.55 Absence of civil con ct 5.3 0.51 Absence of violent redress Regulatory Enforcement 6.1 0.30 Effective regulatory enforcement 6.2 0.51 No improper in ence 6.3 0.37 No unreasonable delay 6.4 0.17 Respect for due process 6.5 0.28 No expropriation w/out adequate compensation Civil Justice 7.1 0.30 Accessibility and affordability 7.2 0.31 No discrimination 7.3 0.36 No corruption 7.4 0.29 No improper gov't in ence 7.5 0.30 No unreasonable delay 7.6 0.47 Effective enforcement 7.7 0.47 Impartial and effective ADRs Criminal Justice 8.1 0.41 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.32 Effective correctional system 8.4 0.27 No discrimination 8.5 0.41 No corruption 8.6 0.27 No improper gov't in ence 8.7 0.34 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 161 WJP Rule of Law Index 2023 Suriname Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 18/32 24/41 81/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.48 0.00 24/32 26/41 90/142 Absence of Corruption 0.43 -0.01 17/32 26/41 80/142 Open Government 0.38 0.00 29/32 38/41 115/142 Fundamental Rights 0.51 -0.01 22/32 27/41 81/142 Order and Security 0.65 0.00 18/32 28/41 103/142 Regulatory Enforcement 0.46 0.00 23/32 30/41 95/142 Civil Justice 0.49 -0.01 19/32 27/41 86/142 Criminal Justice 0.53 0.00 12/32 11/41 51/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Suriname Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.57 Limits by judiciary 1.3 0.37 Independent auditing 1.4 0.37 Sanctions for ofØcial misconduct 1.5 0.51 Non-governmental checks 1.6 0.54 Lawful transition of power Absence of Corruption 2.1 0.43 In the executive branch 2.2 0.60 In the judiciary 2.3 0.54 In the police/military 2.4 0.17 In the legislature Open Government 3.1 0.34 Publicized laws and gov't data 3.2 0.32 Right to information 3.3 0.52 Civic participation 3.4 0.36 Complaint mechanisms Fundamental Rights 4.1 0.53 No discrimination 4.2 0.57 Right to life and security 4.3 0.43 Due process of law 4.4 0.51 Freedom of expression 4.5 0.61 Freedom of religion 4.6 0.33 Right to privacy 4.7 0.61 Freedom of association 4.8 0.51 Labor rights Order and Security 5.1 0.76 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.18 Absence of violent redress Regulatory Enforcement 6.1 0.44 Effective regulatory enforcement 6.2 0.59 No improper inÙuence 6.3 0.44 No unreasonable delay 6.4 0.34 Respect for due process 6.5 0.47 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.52 No discrimination 7.3 0.61 No corruption 7.4 0.52 No improper gov't inÙuence 7.5 0.32 No unreasonable delay 7.6 0.48 Effective enforcement 7.7 0.46 Impartial and effective ADRs Criminal Justice 8.1 0.44 Effective investigations 8.2 0.58 Timely and effective adjudication 8.3 0.37 Effective correctional system 8.4 0.58 No discrimination 8.5 0.59 No corruption 8.6 0.71 No improper gov't inÙuence 8.7 0.43 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 162 Sweden Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.85 4/31 4/46 4/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.86 0.00 5/31 5/46 5/142 Absence of Corruption 0.90 0.00 3/31 4/46 4/142 Open Government 0.84 -0.01 4/31 4/46 4/142 Fundamental Rights 0.87 0.00 4/31 4/46 4/142 Order and Security 0.92 0.00 5/31 7/46 7/142 Regulatory Enforcement 0.83 0.00 7/31 9/46 9/142 Civil Justice 0.82 -0.01 5/31 5/46 5/142 Criminal Justice 0.79 -0.01 4/31 4/46 4/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Sweden EU, EFTA, and North America High Constraints on Government Powers 1.1 0.82 Limits by legislature 1.2 0.83 Limits by judiciary 1.3 0.95 Independent auditing 1.4 0.80 Sanctions for ofØcial misconduct 1.5 0.84 Non-governmental checks 1.6 0.93 Lawful transition of power Absence of Corruption 2.1 0.86 In the executive branch 2.2 0.97 In the judiciary 2.3 0.97 In the police/military 2.4 0.81 In the legislature Open Government 3.1 0.76 Publicized laws and gov't data 3.2 0.90 Right to information 3.3 0.84 Civic participation 3.4 0.87 Complaint mechanisms Fundamental Rights 4.1 0.75 No discrimination 4.2 0.98 Right to life and security 4.3 0.90 Due process of law 4.4 0.84 Freedom of expression 4.5 0.83 Freedom of religion 4.6 0.98 Right to privacy 4.7 0.87 Freedom of association 4.8 0.80 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.87 Absence of violent redress Regulatory Enforcement 6.1 0.78 Effective regulatory enforcement 6.2 0.92 No improper inÙuence 6.3 0.79 No unreasonable delay 6.4 0.78 Respect for due process 6.5 0.87 No expropriation w/out adequate compensation Civil Justice 7.1 0.75 Accessibility and affordability 7.2 0.72 No discrimination 7.3 0.91 No corruption 7.4 0.89 No improper gov't inÙuence 7.5 0.78 No unreasonable delay 7.6 0.88 Effective enforcement 7.7 0.80 Impartial and effective ADRs Criminal Justice 8.1 0.51 Effective investigations 8.2 0.71 Timely and effective adjudication 8.3 0.77 Effective correctional system 8.4 0.77 No discrimination 8.5 0.92 No corruption 8.6 0.94 No improper gov't inÙuence 8.7 0.90 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 163 WJP Rule of Law Index 2023 Tanzania Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.47 12/34 15/37 98/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.50 0.00 11/34 9/37 79/142 Absence of Corruption 0.41 0.00 13/34 12/37 93/142 Open Government 0.39 0.01 18/34 23/37 113/142 Fundamental Rights 0.44 0.01 21/34 20/37 112/142 Order and Security 0.70 0.01 10/34 12/37 83/142 Regulatory Enforcement 0.44 0.01 17/34 17/37 101/142 Civil Justice 0.47 0.00 13/34 14/37 90/142 Criminal Justice 0.37 0.00 15/34 15/37 94/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Tanzania Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.57 Limits by legislature 1.2 0.53 Limits by judiciary 1.3 0.46 Independent auditing 1.4 0.51 Sanctions for of cial misconduct 1.5 0.43 Non-governmental checks 1.6 0.52 Lawful transition of power Absence of Corruption 2.1 0.45 In the executive branch 2.2 0.42 In the judiciary 2.3 0.37 In the police/military 2.4 0.39 In the legislature Open Government 3.1 0.26 Publicized laws and gov't data 3.2 0.43 Right to information 3.3 0.49 Civic participation 3.4 0.36 Complaint mechanisms Fundamental Rights 4.1 0.56 No discrimination 4.2 0.29 Right to life and security 4.3 0.32 Due process of law 4.4 0.43 Freedom of expression 4.5 0.56 Freedom of religion 4.6 0.23 Right to privacy 4.7 0.52 Freedom of association 4.8 0.57 Labor rights Order and Security 5.1 0.70 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.40 Absence of violent redress Regulatory Enforcement 6.1 0.58 Effective regulatory enforcement 6.2 0.49 No improper in ence 6.3 0.34 No unreasonable delay 6.4 0.33 Respect for due process 6.5 0.48 No expropriation w/out adequate compensation Civil Justice 7.1 0.45 Accessibility and affordability 7.2 0.53 No discrimination 7.3 0.44 No corruption 7.4 0.41 No improper gov't in ence 7.5 0.37 No unreasonable delay 7.6 0.51 Effective enforcement 7.7 0.60 Impartial and effective ADRs Criminal Justice 8.1 0.38 Effective investigations 8.2 0.39 Timely and effective adjudication 8.3 0.25 Effective correctional system 8.4 0.48 No discrimination 8.5 0.43 No corruption 8.6 0.36 No improper gov't in ence 8.7 0.32 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 164 Thailand Region: East Asia and PaciØc Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 10/15 25/41 82/142 Score Change Rank Change -0.01 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.45 0.00 11/15 30/41 100/142 Absence of Corruption 0.45 -0.01 9/15 21/41 72/142 Open Government 0.48 0.00 9/15 24/41 77/142 Fundamental Rights 0.46 -0.01 10/15 34/41 101/142 Order and Security 0.74 -0.01 11/15 19/41 65/142 Regulatory Enforcement 0.44 -0.01 12/15 34/41 103/142 Civil Justice 0.49 0.00 10/15 28/41 87/142 Criminal Justice 0.41 -0.01 11/15 22/41 76/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Thailand East Asia and PaciØc Upper-Middle Constraints on Government Powers 1.1 0.49 Limits by legislature 1.2 0.57 Limits by judiciary 1.3 0.36 Independent auditing 1.4 0.44 Sanctions for ofØcial misconduct 1.5 0.49 Non-governmental checks 1.6 0.37 Lawful transition of power Absence of Corruption 2.1 0.43 In the executive branch 2.2 0.67 In the judiciary 2.3 0.45 In the police/military 2.4 0.27 In the legislature Open Government 3.1 0.43 Publicized laws and gov't data 3.2 0.45 Right to information 3.3 0.49 Civic participation 3.4 0.55 Complaint mechanisms Fundamental Rights 4.1 0.48 No discrimination 4.2 0.35 Right to life and security 4.3 0.42 Due process of law 4.4 0.49 Freedom of expression 4.5 0.62 Freedom of religion 4.6 0.34 Right to privacy 4.7 0.44 Freedom of association 4.8 0.53 Labor rights Order and Security 5.1 0.79 Absence of crime 5.2 0.94 Absence of civil conÙict 5.3 0.49 Absence of violent redress Regulatory Enforcement 6.1 0.50 Effective regulatory enforcement 6.2 0.52 No improper inÙuence 6.3 0.44 No unreasonable delay 6.4 0.27 Respect for due process 6.5 0.47 No expropriation w/out adequate compensation Civil Justice 7.1 0.59 Accessibility and affordability 7.2 0.51 No discrimination 7.3 0.71 No corruption 7.4 0.48 No improper gov't inÙuence 7.5 0.31 No unreasonable delay 7.6 0.33 Effective enforcement 7.7 0.48 Impartial and effective ADRs Criminal Justice 8.1 0.37 Effective investigations 8.2 0.41 Timely and effective adjudication 8.3 0.25 Effective correctional system 8.4 0.32 No discrimination 8.5 0.56 No corruption 8.6 0.56 No improper gov't inÙuence 8.7 0.42 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 165 WJP Rule of Law Index 2023 Togo Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.45 14/34 5/18 102/142 Score Change Rank Change -0.01 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.38 0.01 27/34 15/18 118/142 Absence of Corruption 0.43 -0.01 10/34 4/18 82/142 Open Government 0.32 0.00 32/34 17/18 134/142 Fundamental Rights 0.46 0.00 19/34 11/18 104/142 Order and Security 0.70 -0.03 11/34 4/18 86/142 Regulatory Enforcement 0.51 -0.01 9/34 2/18 65/142 Civil Justice 0.48 0.00 12/34 4/18 88/142 Criminal Justice 0.36 -0.01 16/34 6/18 96/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Togo Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.46 Limits by legislature 1.2 0.34 Limits by judiciary 1.3 0.35 Independent auditing 1.4 0.31 Sanctions for of cial misconduct 1.5 0.41 Non-governmental checks 1.6 0.39 Lawful transition of power Absence of Corruption 2.1 0.47 In the executive branch 2.2 0.44 In the judiciary 2.3 0.45 In the police/military 2.4 0.38 In the legislature Open Government 3.1 0.21 Publicized laws and gov't data 3.2 0.29 Right to information 3.3 0.40 Civic participation 3.4 0.38 Complaint mechanisms Fundamental Rights 4.1 0.56 No discrimination 4.2 0.41 Right to life and security 4.3 0.37 Due process of law 4.4 0.41 Freedom of expression 4.5 0.72 Freedom of religion 4.6 0.17 Right to privacy 4.7 0.49 Freedom of association 4.8 0.51 Labor rights Order and Security 5.1 0.75 Absence of crime 5.2 0.89 Absence of civil con ct 5.3 0.45 Absence of violent redress Regulatory Enforcement 6.1 0.50 Effective regulatory enforcement 6.2 0.64 No improper in ence 6.3 0.40 No unreasonable delay 6.4 0.50 Respect for due process 6.5 0.49 No expropriation w/out adequate compensation Civil Justice 7.1 0.53 Accessibility and affordability 7.2 0.57 No discrimination 7.3 0.34 No corruption 7.4 0.34 No improper gov't in ence 7.5 0.51 No unreasonable delay 7.6 0.43 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.30 Effective investigations 8.2 0.43 Timely and effective adjudication 8.3 0.25 Effective correctional system 8.4 0.55 No discrimination 8.5 0.37 No corruption 8.6 0.29 No improper gov't in ence 8.7 0.37 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 166 Trinidad and Tobago Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.52 14/32 44/46 70/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.53 -0.01 16/32 45/46 68/142 Absence of Corruption 0.49 -0.01 13/32 45/46 61/142 Open Government 0.53 -0.01 14/32 40/46 60/142 Fundamental Rights 0.59 0.00 16/32 43/46 61/142 Order and Security 0.65 0.01 17/32 46/46 102/142 Regulatory Enforcement 0.49 -0.01 16/32 45/46 72/142 Civil Justice 0.56 -0.01 12/32 43/46 59/142 Criminal Justice 0.31 0.00 23/32 46/46 121/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Trinidad and Tobago Latin America and Caribbean High Constraints on Government Powers 1.1 0.62 Limits by legislature 1.2 0.64 Limits by judiciary 1.3 0.30 Independent auditing 1.4 0.31 Sanctions for ofØcial misconduct 1.5 0.61 Non-governmental checks 1.6 0.71 Lawful transition of power Absence of Corruption 2.1 0.43 In the executive branch 2.2 0.73 In the judiciary 2.3 0.56 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.39 Publicized laws and gov't data 3.2 0.48 Right to information 3.3 0.64 Civic participation 3.4 0.60 Complaint mechanisms Fundamental Rights 4.1 0.60 No discrimination 4.2 0.61 Right to life and security 4.3 0.39 Due process of law 4.4 0.61 Freedom of expression 4.5 0.76 Freedom of religion 4.6 0.38 Right to privacy 4.7 0.72 Freedom of association 4.8 0.69 Labor rights Order and Security 5.1 0.71 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.24 Absence of violent redress Regulatory Enforcement 6.1 0.42 Effective regulatory enforcement 6.2 0.63 No improper inÙuence 6.3 0.32 No unreasonable delay 6.4 0.48 Respect for due process 6.5 0.58 No expropriation w/out adequate compensation Civil Justice 7.1 0.59 Accessibility and affordability 7.2 0.56 No discrimination 7.3 0.77 No corruption 7.4 0.70 No improper gov't inÙuence 7.5 0.25 No unreasonable delay 7.6 0.40 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.32 Effective investigations 8.2 0.21 Timely and effective adjudication 8.3 0.15 Effective correctional system 8.4 0.18 No discrimination 8.5 0.52 No corruption 8.6 0.38 No improper gov't inÙuence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 167 WJP Rule of Law Index 2023 Tunisia Region: Middle East and North Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.52 4/9 6/37 72/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.54 -0.02 3/9 6/37 64/142 Absence of Corruption 0.47 0.00 4/9 4/37 68/142 Open Government 0.48 0.00 1/9 10/37 75/142 Fundamental Rights 0.51 -0.01 1/9 7/37 84/142 Order and Security 0.72 0.01 5/9 8/37 71/142 Regulatory Enforcement 0.50 -0.01 5/9 6/37 67/142 Civil Justice 0.49 0.01 7/9 10/37 81/142 Criminal Justice 0.41 0.01 4/9 8/37 77/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Tunisia Middle East and North Africa Lower-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.45 Limits by judiciary 1.3 0.54 Independent auditing 1.4 0.45 Sanctions for ofØcial misconduct 1.5 0.59 Non-governmental checks 1.6 0.69 Lawful transition of power Absence of Corruption 2.1 0.49 In the executive branch 2.2 0.49 In the judiciary 2.3 0.56 In the police/military 2.4 0.33 In the legislature Open Government 3.1 0.35 Publicized laws and gov't data 3.2 0.50 Right to information 3.3 0.57 Civic participation 3.4 0.52 Complaint mechanisms Fundamental Rights 4.1 0.56 No discrimination 4.2 0.48 Right to life and security 4.3 0.45 Due process of law 4.4 0.59 Freedom of expression 4.5 0.55 Freedom of religion 4.6 0.31 Right to privacy 4.7 0.65 Freedom of association 4.8 0.45 Labor rights Order and Security 5.1 0.77 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.40 Absence of violent redress Regulatory Enforcement 6.1 0.49 Effective regulatory enforcement 6.2 0.58 No improper inÙuence 6.3 0.37 No unreasonable delay 6.4 0.43 Respect for due process 6.5 0.61 No expropriation w/out adequate compensation Civil Justice 7.1 0.55 Accessibility and affordability 7.2 0.60 No discrimination 7.3 0.38 No corruption 7.4 0.49 No improper gov't inÙuence 7.5 0.41 No unreasonable delay 7.6 0.41 Effective enforcement 7.7 0.62 Impartial and effective ADRs Criminal Justice 8.1 0.40 Effective investigations 8.2 0.44 Timely and effective adjudication 8.3 0.34 Effective correctional system 8.4 0.42 No discrimination 8.5 0.47 No corruption 8.6 0.36 No improper gov't inÙuence 8.7 0.45 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 168 Türkiye Region: Eastern Europe and Central Asia Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.41 15/15 39/41 117/142 Score Change Rank Change 0.00 1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.28 0.01 14/15 39/41 137/142 Absence of Corruption 0.44 -0.01 8/15 25/41 77/142 Open Government 0.40 0.00 13/15 35/41 107/142 Fundamental Rights 0.30 0.00 15/15 39/41 133/142 Order and Security 0.72 -0.01 13/15 23/41 75/142 Regulatory Enforcement 0.42 0.01 14/15 38/41 116/142 Civil Justice 0.41 -0.01 15/15 36/41 119/142 Criminal Justice 0.34 -0.01 13/15 31/41 107/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Türkiye Eastern Europe and Central Asia Upper-Middle Constraints on Government Powers 1.1 0.38 Limits by legislature 1.2 0.32 Limits by judiciary 1.3 0.18 Independent auditing 1.4 0.25 Sanctions for of cial misconduct 1.5 0.25 Non-governmental checks 1.6 0.32 Lawful transition of power Absence of Corruption 2.1 0.43 In the executive branch 2.2 0.55 In the judiciary 2.3 0.64 In the police/military 2.4 0.15 In the legislature Open Government 3.1 0.48 Publicized laws and gov't data 3.2 0.44 Right to information 3.3 0.28 Civic participation 3.4 0.39 Complaint mechanisms Fundamental Rights 4.1 0.38 No discrimination 4.2 0.33 Right to life and security 4.3 0.41 Due process of law 4.4 0.25 Freedom of expression 4.5 0.11 Freedom of religion 4.6 0.25 Right to privacy 4.7 0.30 Freedom of association 4.8 0.39 Labor rights Order and Security 5.1 0.80 Absence of crime 5.2 0.81 Absence of civil con ct 5.3 0.54 Absence of violent redress Regulatory Enforcement 6.1 0.41 Effective regulatory enforcement 6.2 0.62 No improper in ence 6.3 0.40 No unreasonable delay 6.4 0.18 Respect for due process 6.5 0.47 No expropriation w/out adequate compensation Civil Justice 7.1 0.55 Accessibility and affordability 7.2 0.28 No discrimination 7.3 0.49 No corruption 7.4 0.20 No improper gov't in ence 7.5 0.24 No unreasonable delay 7.6 0.49 Effective enforcement 7.7 0.65 Impartial and effective ADRs Criminal Justice 8.1 0.43 Effective investigations 8.2 0.36 Timely and effective adjudication 8.3 0.33 Effective correctional system 8.4 0.23 No discrimination 8.5 0.55 No corruption 8.6 0.05 No improper gov't in ence 8.7 0.41 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 169 WJP Rule of Law Index 2023 Uganda Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.39 28/34 13/18 125/142 Score Change Rank Change 0.00 5 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.39 0.01 24/34 13/18 114/142 Absence of Corruption 0.26 0.00 31/34 17/18 135/142 Open Government 0.39 0.00 17/34 8/18 112/142 Fundamental Rights 0.35 0.00 31/34 15/18 129/142 Order and Security 0.57 -0.01 26/34 11/18 128/142 Regulatory Enforcement 0.43 0.00 20/34 7/18 111/142 Civil Justice 0.42 0.01 24/34 10/18 115/142 Criminal Justice 0.32 -0.01 27/34 13/18 119/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Uganda Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.41 Limits by legislature 1.2 0.39 Limits by judiciary 1.3 0.43 Independent auditing 1.4 0.39 Sanctions for of cial misconduct 1.5 0.36 Non-governmental checks 1.6 0.35 Lawful transition of power Absence of Corruption 2.1 0.26 In the executive branch 2.2 0.34 In the judiciary 2.3 0.26 In the police/military 2.4 0.20 In the legislature Open Government 3.1 0.16 Publicized laws and gov't data 3.2 0.42 Right to information 3.3 0.43 Civic participation 3.4 0.55 Complaint mechanisms Fundamental Rights 4.1 0.43 No discrimination 4.2 0.18 Right to life and security 4.3 0.30 Due process of law 4.4 0.36 Freedom of expression 4.5 0.60 Freedom of religion 4.6 0.05 Right to privacy 4.7 0.45 Freedom of association 4.8 0.41 Labor rights Order and Security 5.1 0.54 Absence of crime 5.2 0.94 Absence of civil con ct 5.3 0.24 Absence of violent redress Regulatory Enforcement 6.1 0.39 Effective regulatory enforcement 6.2 0.36 No improper in ence 6.3 0.40 No unreasonable delay 6.4 0.41 Respect for due process 6.5 0.58 No expropriation w/out adequate compensation Civil Justice 7.1 0.40 Accessibility and affordability 7.2 0.41 No discrimination 7.3 0.36 No corruption 7.4 0.44 No improper gov't in ence 7.5 0.29 No unreasonable delay 7.6 0.49 Effective enforcement 7.7 0.59 Impartial and effective ADRs Criminal Justice 8.1 0.31 Effective investigations 8.2 0.35 Timely and effective adjudication 8.3 0.41 Effective correctional system 8.4 0.30 No discrimination 8.5 0.28 No corruption 8.6 0.28 No improper gov't in ence 8.7 0.30 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 170 Ukraine Region: Eastern Europe and Central Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 9/15 12/37 89/142 Score Change Rank Change -0.01 -11 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.47 0.00 5/15 16/37 94/142 Absence of Corruption 0.33 0.00 14/15 24/37 118/142 Open Government 0.55 -0.01 4/15 3/37 55/142 Fundamental Rights 0.59 -0.01 5/15 1/37 59/142 Order and Security 0.61 -0.08 15/15 29/37 118/142 Regulatory Enforcement 0.43 0.00 12/15 21/37 108/142 Civil Justice 0.53 -0.01 5/15 5/37 69/142 Criminal Justice 0.36 0.00 12/15 17/37 97/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Ukraine Eastern Europe and Central Asia Lower-Middle Constraints on Government Powers 1.1 0.54 Limits by legislature 1.2 0.32 Limits by judiciary 1.3 0.43 Independent auditing 1.4 0.29 Sanctions for ofØcial misconduct 1.5 0.60 Non-governmental checks 1.6 0.66 Lawful transition of power Absence of Corruption 2.1 0.30 In the executive branch 2.2 0.49 In the judiciary 2.3 0.43 In the police/military 2.4 0.07 In the legislature Open Government 3.1 0.61 Publicized laws and gov't data 3.2 0.49 Right to information 3.3 0.57 Civic participation 3.4 0.54 Complaint mechanisms Fundamental Rights 4.1 0.65 No discrimination 4.2 0.61 Right to life and security 4.3 0.45 Due process of law 4.4 0.60 Freedom of expression 4.5 0.76 Freedom of religion 4.6 0.43 Right to privacy 4.7 0.64 Freedom of association 4.8 0.62 Labor rights Order and Security 5.1 0.76 Absence of crime 5.2 0.50 Absence of civil conÙict 5.3 0.58 Absence of violent redress Regulatory Enforcement 6.1 0.42 Effective regulatory enforcement 6.2 0.41 No improper inÙuence 6.3 0.53 No unreasonable delay 6.4 0.41 Respect for due process 6.5 0.39 No expropriation w/out adequate compensation Civil Justice 7.1 0.63 Accessibility and affordability 7.2 0.66 No discrimination 7.3 0.42 No corruption 7.4 0.36 No improper gov't inÙuence 7.5 0.49 No unreasonable delay 7.6 0.46 Effective enforcement 7.7 0.69 Impartial and effective ADRs Criminal Justice 8.1 0.27 Effective investigations 8.2 0.38 Timely and effective adjudication 8.3 0.38 Effective correctional system 8.4 0.49 No discrimination 8.5 0.32 No corruption 8.6 0.27 No improper gov't inÙuence 8.7 0.45 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 171 WJP Rule of Law Index 2023 United Arab Emirates Region: Middle East and North Africa Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.64 1/9 36/46 37/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.57 0.00 1/9 40/46 54/142 Absence of Corruption 0.78 0.00 1/9 19/46 19/142 Open Government 0.34 -0.01 7/9 46/46 129/142 Fundamental Rights 0.45 0.00 5/9 46/46 108/142 Order and Security 0.91 0.01 1/9 10/46 10/142 Regulatory Enforcement 0.70 0.00 1/9 26/46 26/142 Civil Justice 0.66 0.00 1/9 28/46 28/142 Criminal Justice 0.68 0.00 1/9 24/46 24/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 United Arab Emirates Middle East and North Africa High Constraints on Government Powers 1.1 0.52 Limits by legislature 1.2 0.55 Limits by judiciary 1.3 0.75 Independent auditing 1.4 0.73 Sanctions for of cial misconduct 1.5 0.31 Non-governmental checks 1.6 0.58 Lawful transition of power Absence of Corruption 2.1 0.76 In the executive branch 2.2 0.83 In the judiciary 2.3 0.82 In the police/military 2.4 0.71 In the legislature Open Government 3.1 0.37 Publicized laws and gov't data 3.2 0.28 Right to information 3.3 0.31 Civic participation 3.4 0.39 Complaint mechanisms Fundamental Rights 4.1 0.60 No discrimination 4.2 0.47 Right to life and security 4.3 0.72 Due process of law 4.4 0.31 Freedom of expression 4.5 0.49 Freedom of religion 4.6 0.33 Right to privacy 4.7 0.25 Freedom of association 4.8 0.40 Labor rights Order and Security 5.1 0.92 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.82 Absence of violent redress Regulatory Enforcement 6.1 0.61 Effective regulatory enforcement 6.2 0.87 No improper in ence 6.3 0.73 No unreasonable delay 6.4 0.64 Respect for due process 6.5 0.64 No expropriation w/out adequate compensation Civil Justice 7.1 0.60 Accessibility and affordability 7.2 0.59 No discrimination 7.3 0.80 No corruption 7.4 0.61 No improper gov't in ence 7.5 0.66 No unreasonable delay 7.6 0.68 Effective enforcement 7.7 0.66 Impartial and effective ADRs Criminal Justice 8.1 0.69 Effective investigations 8.2 0.74 Timely and effective adjudication 8.3 0.82 Effective correctional system 8.4 0.60 No discrimination 8.5 0.75 No corruption 8.6 0.43 No improper gov't in ence 8.7 0.72 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 172 United Kingdom Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.78 12/31 15/46 15/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.81 -0.01 13/31 15/46 15/142 Absence of Corruption 0.83 0.00 8/31 11/46 11/142 Open Government 0.79 -0.01 10/31 12/46 12/142 Fundamental Rights 0.80 -0.01 13/31 15/46 15/142 Order and Security 0.84 0.00 20/31 28/46 30/142 Regulatory Enforcement 0.79 -0.01 13/31 18/46 18/142 Civil Justice 0.71 0.00 13/31 19/46 19/142 Criminal Justice 0.70 0.00 13/31 18/46 18/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 United Kingdom EU, EFTA, and North America High Constraints on Government Powers 1.1 0.83 Limits by legislature 1.2 0.82 Limits by judiciary 1.3 0.80 Independent auditing 1.4 0.75 Sanctions for ofØcial misconduct 1.5 0.82 Non-governmental checks 1.6 0.86 Lawful transition of power Absence of Corruption 2.1 0.84 In the executive branch 2.2 0.96 In the judiciary 2.3 0.86 In the police/military 2.4 0.66 In the legislature Open Government 3.1 0.91 Publicized laws and gov't data 3.2 0.68 Right to information 3.3 0.82 Civic participation 3.4 0.78 Complaint mechanisms Fundamental Rights 4.1 0.68 No discrimination 4.2 0.91 Right to life and security 4.3 0.77 Due process of law 4.4 0.82 Freedom of expression 4.5 0.84 Freedom of religion 4.6 0.80 Right to privacy 4.7 0.86 Freedom of association 4.8 0.69 Labor rights Order and Security 5.1 0.90 Absence of crime 5.2 0.92 Absence of civil conÙict 5.3 0.71 Absence of violent redress Regulatory Enforcement 6.1 0.76 Effective regulatory enforcement 6.2 0.94 No improper inÙuence 6.3 0.69 No unreasonable delay 6.4 0.82 Respect for due process 6.5 0.75 No expropriation w/out adequate compensation Civil Justice 7.1 0.52 Accessibility and affordability 7.2 0.59 No discrimination 7.3 0.91 No corruption 7.4 0.82 No improper gov't inÙuence 7.5 0.63 No unreasonable delay 7.6 0.70 Effective enforcement 7.7 0.79 Impartial and effective ADRs Criminal Justice 8.1 0.67 Effective investigations 8.2 0.70 Timely and effective adjudication 8.3 0.52 Effective correctional system 8.4 0.57 No discrimination 8.5 0.86 No corruption 8.6 0.81 No improper gov't inÙuence 8.7 0.77 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 173 WJP Rule of Law Index 2023 United States Region: EU, EFTA, and North America Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.70 19/31 26/46 26/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.69 0.00 21/31 27/46 28/142 Absence of Corruption 0.73 0.00 15/31 22/46 22/142 Open Government 0.76 0.00 13/31 15/46 15/142 Fundamental Rights 0.68 0.00 26/31 35/46 38/142 Order and Security 0.83 -0.01 22/31 30/46 33/142 Regulatory Enforcement 0.72 0.00 16/31 22/46 22/142 Civil Justice 0.62 -0.01 22/31 33/46 38/142 Criminal Justice 0.60 -0.01 22/31 29/46 29/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 United States EU, EFTA, and North America High Constraints on Government Powers 1.1 0.71 Limits by legislature 1.2 0.70 Limits by judiciary 1.3 0.70 Independent auditing 1.4 0.58 Sanctions for of cial misconduct 1.5 0.71 Non-governmental checks 1.6 0.75 Lawful transition of power Absence of Corruption 2.1 0.72 In the executive branch 2.2 0.90 In the judiciary 2.3 0.83 In the police/military 2.4 0.48 In the legislature Open Government 3.1 0.77 Publicized laws and gov't data 3.2 0.72 Right to information 3.3 0.74 Civic participation 3.4 0.80 Complaint mechanisms Fundamental Rights 4.1 0.48 No discrimination 4.2 0.76 Right to life and security 4.3 0.62 Due process of law 4.4 0.71 Freedom of expression 4.5 0.70 Freedom of religion 4.6 0.81 Right to privacy 4.7 0.79 Freedom of association 4.8 0.55 Labor rights Order and Security 5.1 0.83 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.66 Absence of violent redress Regulatory Enforcement 6.1 0.69 Effective regulatory enforcement 6.2 0.87 No improper in ence 6.3 0.58 No unreasonable delay 6.4 0.73 Respect for due process 6.5 0.71 No expropriation w/out adequate compensation Civil Justice 7.1 0.45 Accessibility and affordability 7.2 0.36 No discrimination 7.3 0.82 No corruption 7.4 0.69 No improper gov't in ence 7.5 0.60 No unreasonable delay 7.6 0.68 Effective enforcement 7.7 0.77 Impartial and effective ADRs Criminal Justice 8.1 0.66 Effective investigations 8.2 0.63 Timely and effective adjudication 8.3 0.51 Effective correctional system 8.4 0.34 No discrimination 8.5 0.77 No corruption 8.6 0.66 No improper gov't in ence 8.7 0.62 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 174 Uruguay Region: Latin America and Caribbean Income Group: High Overall Score Regional Rank Income Rank Global Rank 0.72 1/32 25/46 25/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.76 0.01 2/32 17/46 18/142 Absence of Corruption 0.73 0.00 1/32 21/46 21/142 Open Government 0.73 0.00 1/32 18/46 18/142 Fundamental Rights 0.80 0.01 1/32 14/46 14/142 Order and Security 0.71 -0.01 10/32 43/46 78/142 Regulatory Enforcement 0.71 0.00 1/32 25/46 25/142 Civil Justice 0.72 0.00 1/32 18/46 18/142 Criminal Justice 0.58 0.01 6/32 35/46 38/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Uruguay Latin America and Caribbean High Constraints on Government Powers 1.1 0.78 Limits by legislature 1.2 0.73 Limits by judiciary 1.3 0.68 Independent auditing 1.4 0.69 Sanctions for ofØcial misconduct 1.5 0.78 Non-governmental checks 1.6 0.91 Lawful transition of power Absence of Corruption 2.1 0.66 In the executive branch 2.2 0.91 In the judiciary 2.3 0.83 In the police/military 2.4 0.54 In the legislature Open Government 3.1 0.72 Publicized laws and gov't data 3.2 0.65 Right to information 3.3 0.77 Civic participation 3.4 0.78 Complaint mechanisms Fundamental Rights 4.1 0.76 No discrimination 4.2 0.87 Right to life and security 4.3 0.63 Due process of law 4.4 0.78 Freedom of expression 4.5 0.85 Freedom of religion 4.6 0.81 Right to privacy 4.7 0.85 Freedom of association 4.8 0.83 Labor rights Order and Security 5.1 0.69 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.44 Absence of violent redress Regulatory Enforcement 6.1 0.68 Effective regulatory enforcement 6.2 0.81 No improper inÙuence 6.3 0.55 No unreasonable delay 6.4 0.67 Respect for due process 6.5 0.81 No expropriation w/out adequate compensation Civil Justice 7.1 0.78 Accessibility and affordability 7.2 0.79 No discrimination 7.3 0.80 No corruption 7.4 0.81 No improper gov't inÙuence 7.5 0.51 No unreasonable delay 7.6 0.62 Effective enforcement 7.7 0.73 Impartial and effective ADRs Criminal Justice 8.1 0.40 Effective investigations 8.2 0.54 Timely and effective adjudication 8.3 0.30 Effective correctional system 8.4 0.62 No discrimination 8.5 0.77 No corruption 8.6 0.77 No improper gov't inÙuence 8.7 0.63 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 175 WJP Rule of Law Index 2023 Uzbekistan Region: Eastern Europe and Central Asia Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.50 8/15 8/37 78/142 Score Change Rank Change 0.00 2 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.38 -0.01 11/15 25/37 119/142 Absence of Corruption 0.47 0.00 6/15 3/37 66/142 Open Government 0.37 0.01 14/15 25/37 119/142 Fundamental Rights 0.45 0.00 12/15 17/37 109/142 Order and Security 0.90 0.00 1/15 1/37 17/142 Regulatory Enforcement 0.45 0.01 10/15 15/37 99/142 Civil Justice 0.52 0.01 6/15 7/37 74/142 Criminal Justice 0.44 -0.01 6/15 6/37 71/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Uzbekistan Eastern Europe and Central Asia Lower-Middle Constraints on Government Powers 1.1 0.30 Limits by legislature 1.2 0.36 Limits by judiciary 1.3 0.34 Independent auditing 1.4 0.43 Sanctions for of cial misconduct 1.5 0.39 Non-governmental checks 1.6 0.43 Lawful transition of power Absence of Corruption 2.1 0.44 In the executive branch 2.2 0.46 In the judiciary 2.3 0.43 In the police/military 2.4 0.55 In the legislature Open Government 3.1 0.26 Publicized laws and gov't data 3.2 0.35 Right to information 3.3 0.35 Civic participation 3.4 0.53 Complaint mechanisms Fundamental Rights 4.1 0.59 No discrimination 4.2 0.47 Right to life and security 4.3 0.39 Due process of law 4.4 0.39 Freedom of expression 4.5 0.58 Freedom of religion 4.6 0.31 Right to privacy 4.7 0.32 Freedom of association 4.8 0.54 Labor rights Order and Security 5.1 0.91 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.78 Absence of violent redress Regulatory Enforcement 6.1 0.57 Effective regulatory enforcement 6.2 0.53 No improper in ence 6.3 0.62 No unreasonable delay 6.4 0.26 Respect for due process 6.5 0.29 No expropriation w/out adequate compensation Civil Justice 7.1 0.50 Accessibility and affordability 7.2 0.58 No discrimination 7.3 0.46 No corruption 7.4 0.27 No improper gov't in ence 7.5 0.77 No unreasonable delay 7.6 0.44 Effective enforcement 7.7 0.61 Impartial and effective ADRs Criminal Justice 8.1 0.50 Effective investigations 8.2 0.60 Timely and effective adjudication 8.3 0.61 Effective correctional system 8.4 0.34 No discrimination 8.5 0.45 No corruption 8.6 0.22 No improper gov't in ence 8.7 0.39 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 176 Venezuela, RB Region: Latin America and Caribbean Income Group: Upper-Middle Overall Score Regional Rank Income Rank Global Rank 0.26 32/32 41/41 142/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.18 0.01 32/32 41/41 142/142 Absence of Corruption 0.28 0.00 29/32 39/41 132/142 Open Government 0.28 0.00 32/32 41/41 139/142 Fundamental Rights 0.30 0.00 31/32 40/41 134/142 Order and Security 0.51 0.01 30/32 40/41 132/142 Regulatory Enforcement 0.19 0.01 32/32 41/41 142/142 Civil Justice 0.26 -0.01 32/32 41/41 141/142 Criminal Justice 0.12 0.00 32/32 41/41 142/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Venezuela, RB Latin America and Caribbean Upper-Middle Constraints on Government Powers 1.1 0.33 Limits by legislature 1.2 0.13 Limits by judiciary 1.3 0.08 Independent auditing 1.4 0.11 Sanctions for ofØcial misconduct 1.5 0.25 Non-governmental checks 1.6 0.19 Lawful transition of power Absence of Corruption 2.1 0.29 In the executive branch 2.2 0.15 In the judiciary 2.3 0.32 In the police/military 2.4 0.34 In the legislature Open Government 3.1 0.18 Publicized laws and gov't data 3.2 0.25 Right to information 3.3 0.26 Civic participation 3.4 0.43 Complaint mechanisms Fundamental Rights 4.1 0.46 No discrimination 4.2 0.09 Right to life and security 4.3 0.17 Due process of law 4.4 0.25 Freedom of expression 4.5 0.59 Freedom of religion 4.6 0.02 Right to privacy 4.7 0.28 Freedom of association 4.8 0.53 Labor rights Order and Security 5.1 0.33 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.20 Absence of violent redress Regulatory Enforcement 6.1 0.31 Effective regulatory enforcement 6.2 0.40 No improper inÙuence 6.3 0.10 No unreasonable delay 6.4 0.01 Respect for due process 6.5 0.14 No expropriation w/out adequate compensation Civil Justice 7.1 0.44 Accessibility and affordability 7.2 0.49 No discrimination 7.3 0.17 No corruption 7.4 0.05 No improper gov't inÙuence 7.5 0.06 No unreasonable delay 7.6 0.16 Effective enforcement 7.7 0.44 Impartial and effective ADRs Criminal Justice 8.1 0.12 Effective investigations 8.2 0.12 Timely and effective adjudication 8.3 0.04 Effective correctional system 8.4 0.11 No discrimination 8.5 0.25 No corruption 8.6 0.02 No improper gov't inÙuence 8.7 0.17 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 177 WJP Rule of Law Index 2023 Vietnam Region: East Asia and PaciØc Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.49 11/15 11/37 87/142 Score Change Rank Change 0.00 -1 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.45 0.00 12/15 19/37 102/142 Absence of Corruption 0.42 0.00 13/15 11/37 89/142 Open Government 0.45 -0.02 11/15 14/37 96/142 Fundamental Rights 0.45 0.00 11/15 18/37 110/142 Order and Security 0.78 0.01 9/15 2/37 51/142 Regulatory Enforcement 0.44 0.00 13/15 18/37 104/142 Civil Justice 0.45 -0.02 13/15 19/37 104/142 Criminal Justice 0.46 0.01 9/15 3/37 66/142 Indicates statistically signiØcant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Vietnam East Asia and PaciØc Lower-Middle Constraints on Government Powers 1.1 0.38 Limits by legislature 1.2 0.37 Limits by judiciary 1.3 0.50 Independent auditing 1.4 0.60 Sanctions for ofØcial misconduct 1.5 0.40 Non-governmental checks 1.6 0.45 Lawful transition of power Absence of Corruption 2.1 0.48 In the executive branch 2.2 0.40 In the judiciary 2.3 0.45 In the police/military 2.4 0.34 In the legislature Open Government 3.1 0.52 Publicized laws and gov't data 3.2 0.34 Right to information 3.3 0.42 Civic participation 3.4 0.51 Complaint mechanisms Fundamental Rights 4.1 0.55 No discrimination 4.2 0.47 Right to life and security 4.3 0.45 Due process of law 4.4 0.40 Freedom of expression 4.5 0.30 Freedom of religion 4.6 0.41 Right to privacy 4.7 0.39 Freedom of association 4.8 0.62 Labor rights Order and Security 5.1 0.92 Absence of crime 5.2 1.00 Absence of civil conÙict 5.3 0.43 Absence of violent redress Regulatory Enforcement 6.1 0.60 Effective regulatory enforcement 6.2 0.41 No improper inÙuence 6.3 0.45 No unreasonable delay 6.4 0.34 Respect for due process 6.5 0.39 No expropriation w/out adequate compensation Civil Justice 7.1 0.49 Accessibility and affordability 7.2 0.56 No discrimination 7.3 0.34 No corruption 7.4 0.32 No improper gov't inÙuence 7.5 0.48 No unreasonable delay 7.6 0.40 Effective enforcement 7.7 0.54 Impartial and effective ADRs Criminal Justice 8.1 0.50 Effective investigations 8.2 0.53 Timely and effective adjudication 8.3 0.45 Effective correctional system 8.4 0.48 No discrimination 8.5 0.53 No corruption 8.6 0.29 No improper gov't inÙuence 8.7 0.45 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 178 Zambia Region: Sub-Saharan Africa Income Group: Low Overall Score Regional Rank Income Rank Global Rank 0.45 15/34 6/18 105/142 Score Change Rank Change 0.00 0 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.46 0.00 17/34 7/18 98/142 Absence of Corruption 0.38 0.00 17/34 8/18 104/142 Open Government 0.39 0.00 15/34 7/18 110/142 Fundamental Rights 0.40 0.00 25/34 13/18 119/142 Order and Security 0.69 0.01 13/34 6/18 90/142 Regulatory Enforcement 0.43 0.00 21/34 8/18 113/142 Civil Justice 0.46 -0.01 15/34 5/18 98/142 Criminal Justice 0.40 -0.01 12/34 5/18 84/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Zambia Sub-Saharan Africa Low Constraints on Government Powers 1.1 0.49 Limits by legislature 1.2 0.36 Limits by judiciary 1.3 0.47 Independent auditing 1.4 0.52 Sanctions for of cial misconduct 1.5 0.41 Non-governmental checks 1.6 0.51 Lawful transition of power Absence of Corruption 2.1 0.41 In the executive branch 2.2 0.51 In the judiciary 2.3 0.33 In the police/military 2.4 0.26 In the legislature Open Government 3.1 0.25 Publicized laws and gov't data 3.2 0.45 Right to information 3.3 0.44 Civic participation 3.4 0.43 Complaint mechanisms Fundamental Rights 4.1 0.50 No discrimination 4.2 0.30 Right to life and security 4.3 0.36 Due process of law 4.4 0.41 Freedom of expression 4.5 0.61 Freedom of religion 4.6 0.23 Right to privacy 4.7 0.41 Freedom of association 4.8 0.40 Labor rights Order and Security 5.1 0.62 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.43 Absence of violent redress Regulatory Enforcement 6.1 0.49 Effective regulatory enforcement 6.2 0.40 No improper in ence 6.3 0.45 No unreasonable delay 6.4 0.34 Respect for due process 6.5 0.43 No expropriation w/out adequate compensation Civil Justice 7.1 0.42 Accessibility and affordability 7.2 0.43 No discrimination 7.3 0.42 No corruption 7.4 0.42 No improper gov't in ence 7.5 0.42 No unreasonable delay 7.6 0.58 Effective enforcement 7.7 0.50 Impartial and effective ADRs Criminal Justice 8.1 0.45 Effective investigations 8.2 0.46 Timely and effective adjudication 8.3 0.30 Effective correctional system 8.4 0.47 No discrimination 8.5 0.36 No corruption 8.6 0.38 No improper gov't in ence 8.7 0.36 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 179 WJP Rule of Law Index 2023 Zimbabwe Region: Sub-Saharan Africa Income Group: Lower-Middle Overall Score Regional Rank Income Rank Global Rank 0.40 26/34 26/37 123/142 Score Change Rank Change 0.01 3 Factor Score Score Change Regional Rank Income Rank Global Rank Constraints on Government Powers 0.33 0.00 32/34 32/37 130/142 Absence of Corruption 0.32 0.01 22/34 26/37 120/142 Open Government 0.33 0.00 29/34 29/37 130/142 Fundamental Rights 0.34 0.00 32/34 31/37 130/142 Order and Security 0.67 0.00 15/34 18/37 95/142 Regulatory Enforcement 0.35 0.00 31/34 34/37 135/142 Civil Justice 0.46 0.01 14/34 16/37 94/142 Criminal Justice 0.36 0.00 17/34 18/37 98/142 Indicates statistically signi cant change at the 10 percent level Low Medium High 2023 Score 2022 Score Constraints on Government Powers Absence of Corruption Open Government Fundamental Rights Order and Security Regulatory Enforcement Civil Justice Criminal Justice 0 0.5 1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 8.1 8.2 8.3 8.4 8.5 8.6 8.7 Zimbabwe Sub-Saharan Africa Lower-Middle Constraints on Government Powers 1.1 0.33 Limits by legislature 1.2 0.31 Limits by judiciary 1.3 0.43 Independent auditing 1.4 0.36 Sanctions for of cial misconduct 1.5 0.26 Non-governmental checks 1.6 0.27 Lawful transition of power Absence of Corruption 2.1 0.30 In the executive branch 2.2 0.45 In the judiciary 2.3 0.31 In the police/military 2.4 0.23 In the legislature Open Government 3.1 0.18 Publicized laws and gov't data 3.2 0.39 Right to information 3.3 0.28 Civic participation 3.4 0.48 Complaint mechanisms Fundamental Rights 4.1 0.42 No discrimination 4.2 0.32 Right to life and security 4.3 0.35 Due process of law 4.4 0.26 Freedom of expression 4.5 0.47 Freedom of religion 4.6 0.17 Right to privacy 4.7 0.27 Freedom of association 4.8 0.47 Labor rights Order and Security 5.1 0.61 Absence of crime 5.2 1.00 Absence of civil con ct 5.3 0.40 Absence of violent redress Regulatory Enforcement 6.1 0.41 Effective regulatory enforcement 6.2 0.44 No improper in ence 6.3 0.32 No unreasonable delay 6.4 0.31 Respect for due process 6.5 0.27 No expropriation w/out adequate compensation Civil Justice 7.1 0.44 Accessibility and affordability 7.2 0.46 No discrimination 7.3 0.43 No corruption 7.4 0.27 No improper gov't in ence 7.5 0.50 No unreasonable delay 7.6 0.63 Effective enforcement 7.7 0.53 Impartial and effective ADRs Criminal Justice 8.1 0.40 Effective investigations 8.2 0.49 Timely and effective adjudication 8.3 0.36 Effective correctional system 8.4 0.38 No discrimination 8.5 0.34 No corruption 8.6 0.24 No improper gov't in ence 8.7 0.35 Due process of law The scores range from 0 to 1, where 1 signifies the highest possible score and 0 signifies the lowest possible score. 180 180 181 WJP Rule of Law Index 2023 182 Methodology Snapshot 183 Methodology 190 Contributing Experts 215 Acknowledgements 216 About the WJP 218 More from the WJP 182 1. The WJP developed the conceptual framework summarized in the Index’s nine factors and 47 sub-factors, in consultation with academics, practitioners, and community leaders from around the world. 2. The Index team developed a set of five questionnaires based on the Index’s conceptual framework to be administered to experts and the general public. Questionnaires were translated into several languages and adapted to reflect commonly used terms and expressions. 3. The Index team identified, on average, more than 300 potential local experts per country to respond to the expert surveys, or Qualified Respondents’ Questionnaires (QRQs). The team engaged the services of leading local polling companies to implement the household surveys, or General Population Poll (GPP). 4. Polling companies conducted pilot tests of the GPP in consultation with the Index team, and launched the final survey for full fieldwork. 5. The Index team sent the QRQ questionnaires to local experts and engaged in continual interaction with them. 6. The Index team collected and mapped the data onto the 44 sub-factors with global comparability. 7. The Index team constructed the final scores using a five-step process: a. Codified the questionnaire items as numeric values; b. Produced raw country scores by aggregating the responses from several individuals (experts and/or general public); c. Normalized the raw scores; d. Aggregated the normalized scores into sub-factors and factors using simple averages; e. Produced the normalized scores, which are rounded to two decimal points, and the final rankings. 8. The data was subject to a series of tests to identify possible biases and errors. For example, the Index team cross-checked all sub-factors against more than 70 third-party sources, including quantitative data and qualitative assessments drawn from local and international organizations. 9. A sensitivity analysis was conducted by the Econometrics and Applied Statistics Unit of the European Commission’s Joint Research Centre, in collaboration with the Index team, to assess the statistical reliability of the results. 10. To illustrate whether the rule of law in a country significantly changed over the course of the past year, a measure of change over time was produced based on the annual difference in the country-level factor scores, the standard errors of these scores (estimated from a set of 100 bootstrap samples), and the results of the corresponding t-tests. 11. The data was organized into country reports, tables, and figures to facilitate its presentation and interpretation. For tables organized by income group, the WJP follows the World Bank income classifications. Methodology Snapshot: Steps to Produce the WJP Rule of Law Index The production of the WJP Rule of Law Index® can be summarized in 11 steps: 183 WJP Rule of Law Index 2023 The WJP Rule of Law Index 2023 report presents information on eight composite factors that are further disaggregated into 44 specific sub-factors (see page 16). Factor 9: Informal Justice, is included in the conceptual framework but has been excluded from the aggregated scores and rankings in order to provide meaningful cross-country comparisons. The country scores and rankings presented in this report are built from more than 500 variables drawn from the assessments of over 149,000 households and 3,400 legal practitioners and experts in 142 countries and jurisdictions, making it the most accurate portrayal of the factors that contribute to shaping the rule of law in a country or jurisdiction. Data Sources To present an image that accurately portrays the rule of law as experienced by ordinary people, each score of the Index is calculated using a large number of questions drawn from two original data sources collected by the World Justice Project in each country: a General Population Poll (GPP) and a series of Qualified Respondents’ Questionnaires (QRQs). These two data sources collect up-to-date firsthand information that is not available at the global level and constitute the world’s most comprehensive dataset of its kind. They capture the experiences and perceptions of ordinary citizens and in-country professionals concerning the performance of the state and its agents and the actual operation of the legal framework in their country. The GPP surveys provide firsthand information on the experiences and the perceptions of ordinary people regarding a range of pertinent rule of law information, including their dealings with the government, the ease of interacting with state bureaucracy, the extent of bribery and corruption, the availability of dispute resolution systems, and the prevalence of common crimes to which they are exposed. The GPP questionnaire includes 127 perception-based questions and 213 experience-based questions, along with socio-demographic information on all respondents. The questionnaire is translated into local languages, adapted to common expressions, and administered by leading local polling companies using a probability sample of 1,000 respondents.1 In previous editions of the Index, the poll was conducted in the three largest cities of each country. However, the World Justice Project’s goal was to update its methodology to include nationally representative polls. Towards this end, nationally representative polls have been conducted in 83 countries and jurisdictions covered in the 2023 WJP Rule of Law Index. Nationally representative polls will be conducted in the remaining countries in future editions of the Index. Depending on the particular situation of each country, one of three different polling methodologies is used: face-to-face, telephone, or online. The GPP has been carried out in each country every few years. The polling data used in this year’s report was collected during summer 2023 (for three countries), fall 2021 through summer 2022 (for 21 countries), fall 2020 through summer 2021 (for 16 countries), fall 2019 (for five countries), fall 2018 (for 57 countries), fall 2017 (for 42 countries), fall 2016 (for four countries), fall 2014 (for three countries), fall 2012 (for one country), and fall 2011 (for two countries). Detailed information regarding the country coverage (cities covered or nationally representative), the polling companies contracted to administer the questionnaire, and the polling methodology employed in each of the 142 countries and jurisdictions is presented on page 186. The QRQs complement the household data with assessments from in-country practitioners and academics with expertise in civil and commercial law; constitutional law, civil liberties, and criminal law; labor law; and public health. These questionnaires gather timely input on a range of topics from practitioners who frequently interact with state institutions. Such topics include information on the efficacy of courts, the strength of regulatory enforcement, and the reliability of accountability mechanisms. The questionnaires contain closed-ended perception questions and several hypothetical scenarios with highly detailed factual assumptions aimed at ensuring comparability across countries. The QRQ surveys are conducted annually, and the questionnaires are completed by respondents selected from directories of law firms, universities and colleges, research organizations, and non-governmental organizations (NGOs), as well as through referrals from the WJP global network of practitioners, and all are vetted by WJP staff based on their expertise. The expert surveys are administered in six languages: Arabic, English, French, Portuguese, Russian, and Spanish. The QRQ data for this report includes more than 3,400 surveys, which represents an average of 24 respondents per country. This data was collected from February 2023 through June 2023. 1. Due to small populations or obstacles to data collection in certain countries and jurisdictions, the sampling plan was adjusted in some cases. For more information on specific countries and jurisdictions and sample sizes, see pages 186-189 Methodology The WJP Rule of Law Index is the first attempt to systematically and comprehensively quantify the rule of law around the world and remains unique in its operationalization of rule of law dimensions into concrete questions. 184 2. Botero, J. and Ponce, A. (2011) “Measuring the Rule of Law”: WJP Working Paper No. 1, available at worldjusticeproject.org/publications. Data Cleaning and Score Computation Once collected, the data is carefully processed to arrive at country-level scores. As a first step, the respondent level data is edited to exclude partially completed surveys, suspicious data, and outliers (which are detected using the Z-score method). Individual answers are then mapped onto the 44 sub-factors of the Index (or onto the intermediate categories that make up each sub-factor), codified so that all values fall between 0 (weakest adherence to the rule of law) and 1 (strongest adherence to the rule of law), and aggregated at the country level using the simple (or unweighted) average of all respondents. This year, to allow for an easier comparison across years, the resulting 2023 scores have been normalized using the Min-Max method with a base year of 2015. These normalized scores were then successively aggregated from the variable level all the way up to the factor level to produce the final country scores, rounded to two decimal points, and rankings. In most cases, the GPP and QRQ questions are equally weighted in the calculation of the scores of the intermediate categories (sub-factors and sub-sub-factors) A full picture of how questions are mapped onto indicators and how they are weighted is available on the WJP Rule of Law Index web page at https:/ /worldjusticeproject.org/index. Data Validation As a final step, data is validated and cross-checked against qualitative and quantitative third-party sources to provide an additional layer of analysis and to identify possible mistakes or inconsistencies within the data. Most of the third-party data sources used to cross-check the Index scores are described in Botero and Ponce (2011).2 Methodological Changes to this Year’s Report Every year, the WJP reviews the methods of data collection to ensure that the information produced is valid, useful, and continues to capture the status of the rule of law in the world. To maintain consistency with previous editions and to facilitate tracking changes over time, this year’s questionnaires and data maps are closely aligned with those administered in the past. In order to improve the accuracy of the QRQ results and reduce respondent burden, proactive dependent interviewing techniques were used to remind respondents who participated in last year’s survey of their responses in the previous year. This year, no new questions or indicators were added to the Index. Overall, 100 percent of questions remained the same between the 2022 and 2023 editions of the Index. A description of the variables is available at worldjusticeproject.org. Tracking Changes Over Time This year’s report includes a measure to illustrate whether the rule of law in a country, as measured through the factors of the WJP Rule of Law Index, has experienced a statistically significant change since the previous year. This measure is marked with an asterisk and represents a summary of rigorous statistical testing based on the use of bootstrapping procedures (see below). For each factor, the change in score will be marked with an asterisk and shaded in green if there was a statistically significant improvement in the score while statistically significant deteriorations in score are marked with an asterisk and shaded in red. This measure complements the numerical scores and rankings presented in this report, which benchmark each country’s current performance on the factors and sub-factors of the Index against that of other countries. The measure of change over time is constructed in three steps: 1. First, last year’s scores are subtracted from this year’s to obtain, for each country and each factor, the annual difference in scores. 2. To test whether the annual changes are statistically significant, a bootstrapping procedure is used to estimate standard errors. To calculate these errors, 100 sample sizes of respondent-level observations (of equal size to the original sample) are randomly selected with replacement for each country from the pooled set of respondents for last year and this year. These samples are used to produce a set of 100 country-level scores for each factor and each country, which are utilized to calculate the final standard errors. These errors--which measure uncertainty associated with picking a particular sample of respondents--are then employed to conduct pair-wise t-tests for each country and each factor. 3. Finally, to illustrate the annual change, a measure of change over time is produced based on the value of the annual difference and its statistical significance (at the 10% level). Strengths and Limitations The Index methodology has both strengths and limitations. Among its strengths is the inclusion of both expert and household surveys to ensure that the findings reflect the conditions experienced by the population. Another strength is that it approaches the measurement of rule of law from various angles by triangulating information across data sources and types of questions. This approach not only enables accounting for different perspectives on the rule of law, but it also helps to reduce possible bias that might be introduced by any other particular data collection method. Finally, it relies on statistical testing to determine the significance of the changes in the factor scores over the last year. 185 WJP Rule of Law Index 2023 With the aforementioned methodological strengths come a number of limitations. First, the data sheds light on rule of law dimensions that appear comparatively strong or weak, but is not specific enough to establish causation. Thus, it will be necessary to use the Index in combination with other analytical tools to provide a full picture of causes and possible solutions. Second, in previous editions of the Index, the methodology has only been applied in three major urban areas in each of the indexed countries for the General Population Poll. However, the World Justice Project’s goal was to update its methodology to include nationally representative polls. Towards this end, nationally representative polls have been conducted in 83 countries and jurisdictions covered in the 2023 WJP Rule of Law Index. Nationally representative polls will be conducted in the remaining countries in future editions of the Index. Third, given the rapid changes to the rule of law occurring in some countries, scores for some countries may be sensitive to the specific points in time when the data was collected. To address this, the WJP is piloting test methods of moving averages to account for short-term fluctuations. Fourth, the QRQ data may be subject to problems in measurement error due to the limited number of experts in some countries, resulting in less precise estimates. To address this, the WJP works constantly to expand its network of in-country academic and practitioner experts who contribute their time and expertise to this endeavor. Finally, due to the limited number of experts in some countries (which implies higher standard errors) and the fact that the GPP is carried out in each country every few years (which implies that for some countries, some variables do not change from one year to another), it is possible that the test described above fails to detect small changes in a country’s situation over time. Other Methodological Considerations A detailed presentation of the methodology, including a table and description of the more than 500 variables used to construct the Index scores, is available at: worldjusticeproject. org and in Botero, J. and Ponce, A. (2011) “Measuring the Rule of Law”: WJP Working Paper No.1, available at: worldjusticeproject.org/publications. Using the WJP Rule of Law Index The WJP Rule of Law Index has been designed to offer a reliable and independent data source for policy makers, businesses, non-governmental organizations (NGOs), and other constituencies to assess a country’s adherence to the rule of law as perceived and experienced by the average person, identify a country’s strengths and weaknesses in comparison to similarly situated countries, and track changes over time. The Index has been designed to include several features that set it apart from other indices and make it valuable for a large number of countries, thus providing a powerful resource that can inform policy debates both within and across countries. However, the Index’s findings must be interpreted in light of certain inherent limitations. 1. The WJP Rule of Law Index does not identify priorities for reform and is not intended to establish causation or to ascertain the complex relationship among different rule of law dimensions in various countries. 2. The Index’s scores and rankings are the product of a rigorous data collection and aggregation methodology. Nonetheless, as with all measurements, they are subject to measurement error. 3. Given the uncertainty associated with picking a particular sample of respondents, standard errors have been calculated using bootstrapping methods to test whether the annual changes in the factor scores are statistically significant. 4. Indices and indicators are subject to potential abuse and misinterpretation. Once released to the public, they can take on a life of their own and be used for purposes unanticipated by their creators. If data is taken out of context, it can lead to unintended or erroneous policy decisions. 5. Rule of law concepts measured by the Index may have different meanings across countries. Users are encouraged to consult the specific definitions of the variables employed in the construction of the Index, which are discussed in greater detail in the methodology section of the WJP Rule of Law Index website. 6. The Index is generally intended to be used in combination with other instruments, both quantitative and qualitative. Just as in the areas of health or economics, no single index conveys a full picture of a country’s situation. Policy-making in the area of rule of law requires careful consideration of all relevant dimensions—which may vary from country to country—and a combination of sources, instruments, and methods. 7. Pursuant to the sensitivity analysis of the Index data conducted in collaboration with the Econometrics and Applied Statistics Unit of the European Commission’s Joint Research Centre, confidence intervals have been calculated for all figures included in the WJP Rule of Law Index. These confidence intervals and other relevant considerations regarding measurement error are reported in Saisana and Saltelli (2015) and Botero and Ponce (2011). The following pages (186-189) list the coverage and polling methodology for the GPP in the 142 indexed countries and jurisdictions. 186 Afghanistan Nationally representative D3: Designs, Data, Decisions & ACSOR Surveys Face-to-face 3019 2019 Albania Nationally representative IDRA Research & Consulting Face-to-face 1000 2018 Algeria Nationally representative WJP in collaboration with local partner Face-to-face 1000 2018 Angola Nationally representative Marketing Support Consultancy Face-to-face 1010 2018 Antigua and Barbuda Nationally representative DMR Insights Ltd. Face-to-face 513/500 2018/2022 Argentina Nationally representative StatMark Group Face-to-face 759 2022 Australia Nationally representative Big Picture Marketing Strategy & Research Online 1067 2018 Austria Vienna, Graz, Linz YouGov Online 1008 2017 The Bahamas Nationally representative DMR Insights Ltd. Face-to-face 500 2022 Bangladesh Dhaka, Chittagong, Khulna Org-Quest Research Ltd. Face-to-face 1000 2016 Barbados Nationally representative DMR Insights Ltd. Face-to-face 513/500 2018/2022 Belarus Minsk, Gomel, Mogilev Market Research & Polls - EURASIA (MRP-EURASIA)/WJP in collaboration with local partner Face-to-face 1000/401 2014/2017 Belgium Nationally representative YouGov Online 1007 2018 Belize Nationally representative CID Gallup Face-to-face 2004 2021 Benin Nationally representative Liaison Marketing Face-to-face 1010 2018 Bolivia Nationally representative Captura Consulting Face-to-face 1000 2022 Bosnia and Herzegovina Sarajevo, Banja Luka, Tuzla Kantar TNS MIB Face-to-face 1000 2017 Botswana Nationally representative BJKA Consulting Face-to-face 1000 2018 Brazil Nationally representative About Brazil Market Research Face-to-face 1109 2022 Bulgaria Sofia, Plovdiv, Varna Alpha Research Ltd. Face-to-face 1001 2018 Burkina Faso Ouagadougou, Bobo Dioulasso, Koudougou Kantar TNS Face-to-face 1029 2017 Cambodia Phnom Penh, Battambang, Kampong Cham Indochina Research Face-to-face 1000 2014 Cameroon Nationally representative Liaison Marketing Face-to-face 1006 2018 Canada Toronto, Montreal, Calgary YouGov Online 1000 2017 Chile Santiago, Valparaíso/Viña del Mar, Antofagasta Datum Internacional S.A./Cadem S.A. Face-to-face 1011 2017 China Shanghai, Beijing, Guangzhou WJP in collaboration with local partner Face-to-face 508 2018 Colombia Nationally representative Tempo Group SA Face-to-face 1000 2022 Congo, Dem. Rep. Kinshasa, Lubumbashi, Mbuji-Mayi Kantar Public at TNS RMS Senegal Face-to-face 1083 2018 Congo, Rep. Nationally representative Liaison Marketing Face-to-face 517 2021 Costa Rica Nationally representative CID Gallup Face-to-face 1005 2022 Côte d'Ivoire Abidjan, Bouaké, Daloa Liaison Marketing Face-to-face 1011 2017 Croatia Nationally representative Ipsos Face-to-face 1010 2018 Cyprus Nationally representative Pulse Market Research Online 504 2021 Czechia Prague, Brno, Ostrava YouGov Online 1013 2017 Denmark Copenhagen, Aarhus, Aalborg YouGov Online 1016 2017 Dominica Nationally representative DMR Insights Ltd. Face-to-face 500 2022 Dominican Republic Nationally representative CID Gallup Face-to-face 1002 2022 Ecuador Nationally representative StatMark Group Face-to-face 1005 2022 Egypt, Arab Rep. Cairo, Alexandria, Giza WJP in collaboration with local partner Face-to-face 1000 2017 El Salvador Nationally representative CID Latinoamerica Face-to-face 1000 2018 Estonia Tallinn, Tartu, Narva Norstat Eesti Online 1010 2017 Ethiopia Addis Ababa, Gondar, Nazret Infinite Insight Ltd. Face-to-face 1037 2017 Coverage Polling Company Methodology Country/Jurisdiction Sample Year 187 WJP Rule of Law Index 2023 Finland Helsinki, Espoo, Tampere YouGov Online 1014 2017 France Nationally representative YouGov Online 1040 2018 Gabon Nationally representative Marketing Support Consultancy Ltd. Face-to-face 513 2022 The Gambia Nationally representative Infinite Insight Ltd. Face-to-face 1030 2019 Georgia Tbilisi, Batumi, Kutaisi ACT Market Research and Consulting Company Face-to-face 1000 2017 Germany Nationally representative YouGov Online 1048 2018 Ghana Nationally representative Infinite Insight Ltd. Face-to-face 1103 2018 Greece Athens, Thessaloniki, Patras YouGov Online 1015 2017 Grenada Nationally representative DMR Insights Ltd. Face-to-face 500 2022 Guatemala Nationally representative Mercaplan Central America & Caribbean Face-to-face 2508 2021 Guinea Conakry, Nzerekore, Kankan Kantar Public at TNS RMS Senegal Face-to-face 1065 2018 Guyana Nationally representative StatMark Group Face-to-face 527/500 2018/2022 Haiti Nationally representative CID Gallup Face-to-face 507 2022 Honduras Nationally representative CID Gallup Face-to-face 3003 2021 Hong Kong SAR, China Hong Kong WJP in collaboration with local partner Face-to-face 1004 2017 Hungary Budapest, Debrecen, Szeged Ipsos Hungary Face-to-face 1000 2017 India Nationally representative Market Xcel Data Matrix Pvt. Ltd. Face-to-face 1059 2018 Indonesia Jakarta, Surabaya, Bandung MRI (Marketing Research Indonesia) Face-to-face 1004 2017 Iran, Islamic Rep. Tehran, Mashhad, Isfahan BJKA consulting with local partner MHA Research Face-to-face 1010 2018 Ireland Nationally representative Dynata Online 1027 2021 Italy Rome, Milan, Naples YouGov Online 1004 2017 Jamaica Nationally representative StatMark Group Face-to-face 1001 2022 Japan Nationally representative Acorn Marketing & Research Consultant (M) Sdn Bhd Online 1000 2018 Jordan Nationally representative WJP in collaboration with local partner Face-to-face 1000 2018 Kazakhstan Almaty, Astana, Shymkent WJP in collaboration with local partner Face-to-face 1000 2017 Kenya Nationally representative Infinite Insight Ltd. Face-to-face 1099 2018 Korea, Rep. Nationally representative Acorn Marketing & Research Consultant (M) Sdn Bhd Online 1000 2018 Kosovo Nationally representative IDRA Research & Consulting Face-to-face 1000 2019 Kuwait Nationally representative D3: Designs, Data, Decisions Face-to-face 50 2023 Kyrgyz Republic Nationally representative Ipsos Face-to-face 1000 2018 Latvia Nationally representative YouGov Online 1049 2021 Lebanon Beirut, Tripoli, Sidon REACH SAL Face-to-face 1000 2017 Liberia Monrovia, Gbarnga and Buchanan Infinite Insight Ltd. Face-to-face 1113 2018 Lithuania Nationally representative YouGov Online 1066 2021 Luxembourg Nationally representative TNS Ilres Online 651 2021 Madagascar Antananarivo, Toamasina, Antsirabe DCDM Research Face-to-face 1000 2017 Malawi Lilongwe, Blantyre, Mzuzu Infinite Insight Ltd. Face-to-face 1039 2017 Malaysia Klang Valley, Johor Bahru, Ipoh Acorn Marketing & Research Consultant (M) Sdn Bhd Face-to-face 1000 2017 Mali Nationally representative Marketing Support Consultancy Face-to-face 1012 2018 Malta Nationally representative MISCO International Limited Face-to-face 500 2021 Mauritania Nationally representative Liaison Marketing Face-to-face 1000 2018 Mauritius Nationally representative DCDM Research Face-to-face 1000 2018 Coverage Polling Company Methodology Country/Jurisdiction Sample Year 188 Mexico Mexico City, Guadalajara, Monterrey Data Opinión Pública y Mercados Face-to-face 1000 2017 Moldova Chisinau, Balti, Cahul Georgian Opinion Research Business International (GORBI) in collaboration with local partner Face-to-face 1043 2017 Mongolia Ulaanbaatar, Erdenet, Darkhan Mongolian Marketing Consulting Group LLC Face-to-face 1000 2017 Montenegro Nationally representative Indago Face-to-face 1000 2023 Morocco Casablanca, Fes, Tangier WJP in collaboration with local partner Face-to-face 1000 2017 Mozambique Nationally representative Quest Research Services Face-to-face 1009 2018 Myanmar Yangon, Mandalay, Nay Pyi Taw Myanmar Survey Research Co., Ltd (MSR) Face-to-face 1000 2018 Namibia Nationally representative Quest Research Services Face-to-face 1001 2018 Nepal Kathmandu, Pokhara, Lalitpur Solutions Consultant Face-to-face 1000 2017 Netherlands Nationally representative YouGov Online 1113 2018 New Zealand Auckland, Wellington, Christchurch Big Picture Marketing Strategy & Research Online 1000 2017 Nicaragua Nationally representative CID Gallup Face-to-face 1000 2019 Niger Niamey, Zinder, Maradi Liaison Marketing Face-to-face 1011 2018 Nigeria Nationally representative Infinite Insight Ltd. Face-to-face 1083 2018 North Macedonia Nationally representative Ipsos dooel Skopje Face-to-face 1594 2023 Norway Oslo, Bergen, Trondheim YouGov Online 1007 2017 Pakistan Nationally representative Gallup Pakistan Face-to-face 1000 2019 Panama Nationally representative CID Gallup Face-to-face 2502 2021 Paraguay Nationally representative Datum Internacional/BM Business Partners Face-to-face 1000 2021 Peru Nationally representative Datum Internacional S.A. Face-to-face 1029 2022 Philippines Manila, Cebu, Davao APMI Partners Face-to-face 1008 2016 Poland Warsaw, Krakow, Lodz IQS Sp. z o.o. Face-to-face 1000 2018 Portugal Lisbon, Porto, Amadora YouGov Online 1016 2017 Romania Nationally representative Alpha Research Ltd. in collaboration with local partner Face-to-face 1000 2018 Russian Federation Moscow, St. Petersburg, Novosibirsk/ Nationally representative WJP in collaboration with local partner Face-to-face 1000/1000 2016/2018 Rwanda Kigali Infinite Insight Ltd. Face-to-face 316 2018 Senegal Pikine, Dakar, Thiès Kantar TNS Face-to-face 1012 2017 Serbia Belgrade, Novi Sad, Niš Ipsos Strategic Marketing d.o.o. Face-to-face 1002 2017 Sierra Leone Nationally representative Infinite Insight Ltd. Face-to-face 1165 2018 Singapore Singapore Survey Sampling International Online 1000 2017 Slovak Republic Nationally representative WJP in collaboration with local partner Online 1022 2021 Slovenia Ljubljana, Maribor, Celje Ipsos d.o.o. Face-to-face 1006 2017 South Africa Nationally representative Quest Research Services Face-to-face 1014 2018 Spain Nationally representative YouGov Online 1051 2018 Sri Lanka Colombo, Kaduwela, Maharagama Kantar LMRB Face-to-face 1010 2017 St. Kitts and Nevis Nationally representative DMR Insights Ltd. Face-to-face 500 2018 St. Lucia Nationally representative DMR Insights Ltd. Face-to-face 500 2022 St. Vincent and the Grenadines Nationally representative DMR Insights Ltd. Face-to-face 500/500 2018/2022 Sudan Nationally representative Sudan Polling and Statistics Center Face-to-face 500 2021 Suriname Nationally representative D3: Designs, Data, Decisions Face-to-face 502 2022 Sweden Nationally representative YouGov Online 1049 2018 Coverage Polling Company Methodology Country/Jurisdiction Sample Year 189 WJP Rule of Law Index 2023 Coverage Polling Company Methodology Country/Jurisdiction Sample Year Tanzania Dar es Salaam, Mwanza, Arusha Infinite Insight Ltd. Face-to-face 1037 2018 Thailand Bangkok, Nakhon Ratchasima, Udon Thani Infosearch Limited Face-to-face 1000 2018 Togo Nationally representative Marketing Support Consultancy Face-to-face 1005 2018 Trinidad and Tobago Nationally representative CID Gallup Face-to-face 1001 2022 Tunisia Big Tunis, Sfax, Sousse BJKA Consulting Face-to-face 1001 2017 Türkiye İstanbul, Ankara, İzmir Kantar Insights Face-to-face 1039 2018 Uganda Kampala, Nansana, Kira Kantar Public East Africa Face-to-face 1062 2018 Ukraine Kyiv, Kharkiv, Odessa GfK Ukraine Face-to-face 1079 2017 United Arab Emirates Dubai, Abu Dhabi, Sharjah WJP in collaboration with local partner Face-to-face 1011/200 2011/2017 United Kingdom Nationally representative YouGov Online 1056 2018 United States Nationally representative YouGov Online 1258 2021 Uruguay Nationally representative BM Business Partners Face-to-face 1000 2018 Uzbekistan Tashkent, Namangan, Samarkand/Nationally representative Market Research & Polls - EURASIA/ Ipsos/Info Sapiens International LLC Face-to-face 1000/300/507 2014/2018/2021 Venezuela, RB Nationally representative StatMark Group Face-to-face 1000/1015 2016/2018 Vietnam Ho Chi Minh City, Hanoi, Hai Phong Indochina Research (Vietnam) Ltd. Face-to-face 1000/1000 2011/2017 Zambia Lusaka, Kitwe, Chipata SIS International Research/Intraspace Market Consultancy Ltd. Face-to-face 1004/1014 2012/2017 Zimbabwe Nationally representative Quest Research Services Face-to-face 1001 2018 190 Afghanistan Baryalai Hakimi Kabul University Hashmatullah Hoshmand Masnad Law Firm Ihsanullah Himmat Khalid Massoudi Masnad Law Firm Khalid Sekander Mahir Hazim Arizona State University Mohammad Imran Malikzai Anti-Corruption Watch Organization Mohammad Tareq Eqtedary Generation Positive Nesar Ahmad Noori Afghanistan Independent Bar Association Noorulhuda Niazi Masnad Law Firm Sayed Reza Hussaini University of Hertfordshire Shabnam Salehi Shabnam Salehi AIHRC Thomas Kraemer Kakar Advocates Anonymous Contributors Albania Adi Muja Muja Law Firm Albana Fona LPA Legal Anteo Papa Bledar Cenameri CLO Legal Solutions Blerta Kalavace IDRA Research & Consulting Brunilda Subashi Universiteti “Ismail Qemali” Vlore Dorant Ekmekçiu Hoxha, Memi & Hoxha Dudi Ilias Open Akte Endri Mykaj Leiden University Eno Muja Muja Law Firm Erion Fejzulla Eris Hoxha Hoxha, Memi & Hoxha Gentiana Tirana Tirana Law Firm Isuf Shehu Shehu & Partners Law Firm Jonada Zyberaj Universiteti “Ismail Qemali” Vlore Jonida Melani Braja Wolf Theiss Katerina Kaçani LPA Legal Krenare Muja Muja Law Firm Xhet Hushi Wolf Theiss Anonymous Contributors Algeria Adel Messaoudi Ligue Algerienne des Droits de L’homme Farouk Yaya Fatiha Aouam Hezam Abosuraima Nadia Zouani Samia Goudjil Goudjil Law Firm Anonymous Contributors Angola Ana Leitão de Matos ALC Advogados António José Ventura Universidade Lusíada de Angola Arlete Amaralmaia CBAM - Advogados Associados Cristiano Santana Agostinho Sanda Paciência CKA & Associados Sociedade Advogados, RL Eduardo Afonso Elsa Tchicana Joni Garcia Orlando Buta Tomás Yessu D’Oliveira Chiquito Anonymous Contributors Antigua and Barbuda Charlotte Jeremy-Cuffy Allthings Medical Chevanev Charles Temple Stoke Craig L. Jacas Stapleton Chambers Luann M. De Costa Lighthouse Law Anonymous Contributors Argentina Adrian Fernando Otero Abogados Patagonicos SH Alberto Gonzalez Torres Baker & McKenzie Alfredo M. Vitolo FORES Analia Duran MBB Balado Bevilacqua Abogados Carlos Ignacio Salvadores de Arzuaga Universidad del Salvador Claudio Jesus Santagati Defensoria General de Lomas de Zamora Dante Omar Graña Fundación Avedis Donabedian Argentina Denise Schalet World Health Organization Diego Silva Ortiz Silva Ortiz, Alfonso, Pavic & Louge Fernando Rebaudi Estudio Rebaudi Francisco Castex Universidad de Buenos Aires Guillermo Michelson Irusta Estudio Michelson Joaquín Emilio Zappa J. P. O’Farrell Abogados Manuel Gonzalo Casas Universidad de San Pablo-Tucumán Marcelo Octavio de Jesús FORES Maria de las Mercedes Balado Bevilacqua MBB Balado Bevilacqua Abogados Marina Mansilla UNPSJB Martin G. Langsam Fundación Bunge & Born; New York University Matías Tonon Universidad del Salvador Contributing Experts The WJP Rule of Law Index 2023 was made possible by the generous contributions of academics and practitioners who contributed their time and expertise. The names of those wishing to be acknowledged individually are listed below. This report was also made possible by the work of the polling companies who conducted fieldwork, and the thousands of individuals who have responded to the General Population Poll around the world. 191 WJP Rule of Law Index 2023 Pablo Zan Bisignani Sandra Guillan Moriondo, Guillan & Asociados Anonymous Contributors Australia Anne Cregan Gilbert + Tobin Breen Creighton RMIT University Brendan Ashdown John Toohey Chambers David Gill Hewlett Packard Enterprise Edouard Tursan d’Espaignet University of New England, Australia Fiona McDonald Australian Centre for Health Law Research, Queensland University of Technology James Gillespie University of Sydney Merrilyn Walton University of Sydney Nicholas Cowdery University of Sydney Patricia Cahill Francis Burt Chambers Peter Cashman University of New South Wales Sean Baron Levi State Chambers; University of New England, Australia Sonia Allan Griffith University; Sonia Allan Consulting; University of Sydney Stephen Blanks New South Wales Council for Civil Liberties Teresa Farmer John Toohey Chambers Anonymous Contributors Austria Christian Klausegger Binder Grösswang Rechtsanwälte GmbH Claudia Habl Austrian National Public Health Institute Daniela Haluza Medical University of Vienna Doris Wydra University of Salzburg Gerhard Jarosch RGJ-Partner Inc. Igor Grabovac Medical University of Vienna Karl Stöger University of Vienna Martin Gruber-Risak University of Vienna Martin Reinisch Brauneis Rechtsanwälte GmbH Sebastian Huter Austrian Society for General Practice and Family Medicine Thomas Frad KWR Karasek Wietrzyk Rechtsanwälte GmbH Anonymous Contributors The Bahamas Arthur Seligman Lennox Paton Catherine Conliffe Princess Margaret Hospital, Public Hospitals Authority Joseph Darville Save The Bays, Waterkeepers Bahamas Leah A. Rolle University of the Bahamas Sharanda Humes-Forbes Vann P. Gaitor Higgs & Johnson Anonymous Contributors Bangladesh A. B. M. Asrafuzzaman University of Dhaka Abdur Rahman Junaid Rahman’s Chambers Badhan Roy Roy & Associates Darras Abdullah Akhtar Imam & Associates Gazi Md. Rokib Bin Hossain The Legal Circle Khandaker Mashfique Ahmed Rahman’s Chambers Md. Tajul Islam Mohammad Habibur Rahman Mohammad Rafiqul Islam Chowdhury M. R. I. Chowdhury & Associates Tanim Hussain Shawon Dr. Kamal Hossain & Associates Anonymous Contributors Barbados Alexandria Thomas Anya Lorde Ayo Barnard Rawlins Harridyal-Sodha & Associates Heather Walker Chancery Chambers Jaydene Thomas Emineo Caribbean Leslie F. Haynes Marvalee Franklyn Franklyn Law Chambers Noah Haynes Sharon R. Harris Zarina Khan Prospect Chambers Anonymous Contributors Belarus Andrei Famenka Vadzim Samaryn Belarusian State University Valentina Komova Stepanovski, Papakul and Partners Anonymous Contributors Belgium Aloys Muberanziza Brussels United Lawyers (BUL) Ann Witters Christoph Van der Elst Ghent University; Tilburg University Didier Ledoux Centre Hospitalier Universitaire de Liège Freek Louckx Universiteit Antwerpen; Vrije Universiteit Brussel Jean-Marc Gollier UCLouvain Jerome Aubertin Stibbe Patrick Goffaux Université Libre de Bruxelles Patrick Henry Avocats Sans Frontières Pieter J. De Koster Bird & Bird Sofie Bekaert Ghent University Walter P. Verstrepen Elegis Brussels Anonymous Contributors Belize Adler G. L. Waight Barrow & Williams LLP Darinka Munoz Barrow & Williams LLP Lissette V. Staine Barrow & Williams LLP Natalia Bevans The Law Firm of Bevans Consultancy Inc. Wayne A. Piper Flores Piper LLP Anonymous Contributors Benin A. Jacqueline Adabra Dedjinou DDS du Littoral Adounlohoun Gisèle Obognon épouse Nanako Plan International Bénin Aline Odje Cabinet d’Avocats Aline Odje Chris A. M. Balogoun Barreau du Bénin Gervais C. Houédété Justine Françoise Houzanme ROAFEM Luciano Hounkponou Société Civile Professionnelle d’Avocats Hounkponou-Kounou et Associés Ludovic Guedje Autorite de Regulation des Marches Publics Marius Kedote COPES-AOC Ogoudjé César Guegni Cabinet d’Avocats Charles Badou & Partners Anonymous Contributors Bolivia Alan Elliott Vargas Lima Academia Boliviana de Estudios Constitucionales Alfonso M. Dorado Escobar 192 Arturo Yañez Cortes Ilustre Colegio de Abogados de Chuquisaca Asdruval Martin Universidad Mayor de San Andrés Betty Carolina Ortuste Tellería Academia Boliviana de Estudios Constitucionales Cesar Burgoa Rodriguez Bufete Burgoa Javier Mir Peña Mir & Asociados Abogados Laborales Jorge Antonio Asbun Estudio Jurídico Asbun Associados Jorge Omar Mostajo Barrios Universidad Mayor de San Andrés José Antonio Rivera Santivañez Universidad Mayor de San Simón Juan Pablo Sánchez Orsini Kathia Saucedo Paz María Micaela Alarcón Gambarte Academia Boliviana de Estudios Constitucionales Nicolás Soliz Peinado ACF Abogados Raul A. Baldivia Baldivia Unzaga & Asociados Yerko Ilijic Crosa Proyecto Acceso CWSL Anonymous Contributors Bosnia and Herzegovina Bahrija Umihanić University of Tuzla Boris Stojanović Denis Pajić University “Džemal Bijedić” Duraković Adnan University of Zenica Dzeneta Omerdic University of Tuzla Emir Sudžuka Hana Korać University of Travnik Harun Išerić University of Sarajevo Jasminka Gradaščević-Sijerčić University of Sarajevo Kanita Pruščanović University of Sarajevo Lana Bubalo University of Stavanger Mehmed Ganic International University of Sarajevo Mirjana Šarkinović Attorney’s Office Mirjana Šarkinović Samir Lemes Eko Forum Zenica Šefket Goletić University of Zenica Slaven Dizdar Anonymous Contributors Botswana Buhle Ncube Letshabo Legal Group Kabo Godfrey Motswagole Motswagole & Company Kagiso Jani Tshekiso Ditiro & Jani Legal Practice Kwadwo Osei-Ofei Osei-Ofei Legal Mabedi Cedella G. Masedi Piyush Sharma Attorneys & Co. Motsomi Ndala Marobela University of Botswana Neo Thelma Moatlhodi Chobe District Council Olebile Daphney Muzila Bookbinder Business Law Piyush Sharma Piyush Sharma Attorneys & Co. Precious Gondwe Precious and Partners Legal Practice Tachilisa Balule University of Botswana Tatenda Caroline Dumba Minchin & Kelly Thuto Senwedi Non-Bank Financial Institutions Tribunal Tshekiso Tshekiso TDJ Legal Practice Unoda Mack Mack, Bahuma Attorneys Anonymous Contributors Brazil Ana Paula de Barcellos Universidade do Estado do Rio de Janeiro Carolina Giesbrecht Forte Korbage de Castro Korbage de Castro Sociedade de Advogados Daniel Bushatsky Advocacia Bushatsky Daniel de Pádua Andrade Universidade Federal de Viçosa Debora Regina Pastana Universidade Federal de Uberlândia Elival da Silva Ramos Universidade de São Paulo Fabio Queiroz Pereira Universidade Federal de Minas Gerais Fábio Ulhoa Coelho Pontifícia Universidade Católica de São Paulo Fernanda Vargas Terrazas Conselho Nacional de Secretarias Municipais de Saúde (CONASEMS) Fernando Aith Universidade de São Paulo Gerson Branco Universidade Federal do Rio Grande do Sul Guilherme Bier Barcelos RMMG Advogados Guilherme de Jesus France Instituto de Estudos Sociais e Políticos, Universidade do Estado do Rio de Janeiro Heloisa Estellita Fundação Getúlio Vargas Heloisa Uelze Trench Rossi Watanabe João Augusto Gameiro Trench Rossi Watanabe Jorge Magalhães Fiocruz Ligia Bahia Universidade Federal do Rio de Janeiro Luciana Zaffalon Plataforma JUSTA Luciano Feldens Feldens Advogados; Pontifícia Universidade Católica do Rio de Janeiro Luis Eduardo Serra Netto Duarte Garcia, Serra Netto e Terra Advogados Luiz Felipe Monteiro Seixas Universidade Federal de Pernambuco Luiz G. P. Dellore Universidade Presbiteriana Mackenzie Maria Celina Bodin de Moraes Pontifícia Universidade Católica do Rio de Janeiro; Universidade do Estado do Rio de Janeiro Maria Valeria Junho Penna Federal University of Rio de Janeiro Matheus Cherulli Alcantara Viana Viana e Azevedo Advogados Michael Freitas Mohallem Pontifícia Universidade Católica do Rio de Janeiro Ordélio Azevedo Sette Azevedo Sette Advogados Paulo Sérgio João Pontifícia Universidade Católica de São Paulo Rafael Lamera Giesta Cabral Universidade Federal Rural do Semiárido Raoni Bielschowsky Universidade Federal de Uberlândia Raquel Betty de Castro Pimenta Tribunal Regional do Trabalho da 3a Região Raquel Lima Scalcon Cavalcanti Sion Advogados Roberta de Freitas Campos Fundação Oswaldo Cruz Rodrigo Ghiringhelli de Azevedo Pontifícia Universidade Católica do Rio Grande do Sul Rubens Hofmeister Neto Feldens Advogados Sara Carvalho Matanzaz Ferreira de Melo Advogados Sérgio Cruz Arenhart Ministério Público Federal Sergio Mannheimer Mannheimer, Perez e Lyra Advogados Soraia Saleh Saleh Advogados Associados Suzana de Queiroz Alves Defensoria Pública da União Teresa Ancona Lopez Universidade de São Paulo 193 WJP Rule of Law Index 2023 Thiago Bottino Fundação Getulio Vargas Ulisses Levy Silvério dos Reis Universidade Federal Rural do Semiárido Victor Hugo Criscuolo Boson Universidade Federal do Sul da Bahia Anonymous Contributors Bulgaria Ana-Mari Eremieva Dimitrov, Petrov & Co. Gergana Ilieva Lachezar Raichev Penkov, Markov & Partners Nikolai Hristov Medical University of Sofia Nikolay Zisov Boyanov & Co. Pavel Tsanov Penkov, Markov & Partners Petko Salchev National Center of Public Health and Analyses Stanislav Malchev Air4Health Association Vania Todorova Stoeva Tchompalov & Znepolski Attorney Partnership Victor Gugushev Gugushev and Partners Anonymous Contributors Burkina Faso Abdoulaye Soma Drabo K. Maxime IRSS-CNRST Hervé Hien Institut National de Santé Publique (INSP) K. Frédéric Hermann Minoungou Legalis Advisory Paulin Küssome Somda Institut National de Santé Publique (INSP); Ministère de la Santé et de l’Hygiène Publique (MSHP) S. Ibrahim Guitanga SCPA Birba-Guitanga & Associes Sièlma Salimata Sou-Kone Société Burkinabè de Droit Constitutionnel Anonymous Contributors Cambodia Darwin (Naryth) Hem BNG Legal Fil B. 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Individual Advisors Mark David Agrast; Maurits Barendrecht, Tilburg University; Tim Besley, London School of Economics; Juan Carlos Botero, Pontificia Universidad Javeriana; Paul Brest, Stanford University; Jose Caballero, IMD Business School; Giulio Caperna, EU-JRC; Thomas Carothers, Carnegie Endowment for International Peace; Eduardo Cifuentes, Universidad de los Andes; Christine M. Cole, Crime & Justice Institute; Mariano-Florentino Cuellar, Carnegie Endowment for International Peace; Nicolas Dassen, Inter-American Development Bank; Larry Diamond, Stanford University; Sandra Elena; Brad Epperly, University of British Columbia; Julio Faundez, Warwick University; Hazel Feigenblatt, University of Maryland; Todd Foglesong, Munk School of Global Affairs at the University of Toronto; Tom Ginsburg, University of Chicago; Joseph Foti, Open Government Partnership; James Goldston, Open Society Foundation; Jorge Gonzalez, Universidad Javeriana; Jon Gould, Arizona State University; Martin Gramatikov, HiiL; Brendan Halloran, Transparency and Accountability Initiative; Linn Hammergren; Tim Hanstad; Wassim Harb, Arab Center for the Development of Rule of Law and Integrity; Nathaniel Heller; Susan Hirsch, George Mason University; Ronald Janse, University of Amsterdam Law School; Erik G. Jensen, Stanford University; Haroon Khadim, PAE; Rachel Kleinfeld, Carnegie Endowment for International Peace; Jack Knight, Duke University; Harold H. Koh, Yale University; Matija Kovacic, EU-JRC; Aart Kraay, The World Bank; Margaret Levi, Stanford University; Iris Litt, Stanford University; Clare Lockhart, The Institute for State Effectiveness; Zsuzsanna Lonti, OECD; Diego Lopez, Universidad de los Andes; William T. Loris, Loyola University; Lauren E. Loveland, National Democratic Institute (NDI); Paul Maassen, Open Government Partnership; Beatriz Magaloni, Stanford University; Jenny S. Martinez, Stanford University; Toby McIntosh, FreedomInfo.org; Toby Mendel, Centre for Law and Democracy; Nicholas Menzies, The World Bank; Ghada Moussa, Cairo University; Sam Muller, HiiL; Robert L. Nelson, American Bar Foundation and Northwestern University; Harris Pastides, University of South Carolina; Randal Peerenboom, La Trobe University and Oxford University; Angela Pinzon, Universidad del Rosario; Pascoe Pleasence, University College London; Shannon Portillo, George Mason University; Michael H. Posner, New York University; Roy L. Prosterman, University of Washington; Anita Ramasastry, University of Washington; Audrey Sacks, The World Bank; Lutforahman Saeed, Kabul University; Michaela Saisana, EU-JRC; Andrea Saltelli, EU-JRC; Moises Sanchez, Alianza Regional por la Libertad de Expresión; Andrei Shleifer, Harvard University; Jorge Luis Silva, Secretaria de Economia; Christopher Stone, Oxford University; Stefan Voigt, University of Hamburg; Barry Weingast, Stanford University; Michael Woolcock, The World Bank. Institutional Contributors and Advisors Altus Global Alliance; APCO Worldwide; Fleishman-Hillard; The Center for Advanced Study in the Behavioral Sciences, Stanford University; The Center on Democracy, Development, and the Rule of Law, Stanford University; The German Bar Association in Brussels; Governance Data Alliance; Google Inc.; The Hague Institute for Innovation of Law (HiiL); Investigative Reporting Program, UC Berkeley Graduate School of Journalism; The Legal Department of Hewlett Packard Enterprise; The Legal Department of Microsoft Corporation; The Whitney and Betty MacMillan Center for International and Area Studies, Yale University; Rule of Law Collaborative, University of South Carolina; The University of Chicago Law School; Vera Institute of Justice; The Wright Center for the Study of Computation and Just Communities. 216 About the World Justice Project The World Justice Project (WJP) is an independent, multidisciplinary organization working to create knowledge, build awareness, and stimulate action to advance the rule of law worldwide. Effective rule of law reduces corruption, combats poverty and disease, and protects people from injustices large and small. It underpins development, accountable government, and respect for fundamental rights, and it is the foundation for communities of justice, opportunity, and peace. Our Approach T raditionally, the rule of law has been viewed as the domain of lawyers and judges. However, everyday issues of safety, rights, justice, and governance affect us all; everyone is a stakeholder in the rule of law. Based on this, the WJP builds and supports a global, multi-disciplinary movement for the rule of law by: Collecting, organizing, and analyzing original, independent rule of law data, including the WJP Rule of Law Index; Supporting research, scholarship, and teaching about the importance of rule of law, its relationship to development, and effective strategies to strengthen it; and Connecting and building an engaged global network of policy-makers and advocates to advance the rule of law through strategic partnerships, convenings, coordinated advocacy, and support for locally-led initiatives, including through the World Justice Forum and the World Justice Challenge, a competition to identify, recognize, and promote good practices and successful solutions for strengthening the rule of law worldwide. Honorary Chairs The World Justice Project has the support of outstanding leaders representing a range of disciplines around the world. The Honorary Chairs of the World Justice Project are: Giuliano Amato; Robert Badinter; James A. Baker III; Cherie Blair; Stephen G. Breyer; Sharan Burrow; David Byrne; Jimmy Carter; Maria Livanos Cattaui; Hans Corell; Hilario G. Davide, Jr.; Hernando de Soto; Adama Dieng; Richard J. Goldstone; Kunio Hamada; Lee H. Hamilton; Mohamed Ibrahim; Tassaduq Hussain Jillani; Anthony M. Kennedy; Beverley McLachlin; George J. Mitchell; Sandra Day O’Connor; Ana Palacio; Roy L. Prosterman; Richard W. Riley; Mary Robinson; Antonio Vitorino; Harold Woolf; and Andrew Young. Board of Directors Sheikha Abdulla Al-Misnad; Kamel Ayadi; Adam Bodnar; Michael Chu; William C. Hubbard; Hassan Bubacar Jallow; Suet-Fern Lee; Mondli Makhanya; M. Margaret McKeown; John Nery; William H. Neukom; Ellen Gracie Northfleet; and James R. Silkenat. Directors Emeriti Ashraf Ghani Ahmadzai Emil Constantinescu Petar Stoyanov Rule of Law Leadership Council Beverley McLachlin, Co-Chair; Judy Perry Martinez, Co-Chair; Kerry Abrams; Paige Alexander; Bernard Amyot; Donald Ayer; Cherie Blair; Nicola Bonucci; Nouzha Chekrouni; Antony Cook; Halimah DeLaine Prado; Tom Dery; Adam Goldenberg; Karen F. Green; Tim Hanstad; Mo Ibrahim; Jeffrey Jowell; Paul Lawrence; Ian McDougall; Judith A. Miller; Ali H. Mokdad; Michael Posner; Regan Ralph; David S. Steuer; and David K.Y. Tang. 217 WJP Rule of Law Index 2023 Officers and Staff Officers: William C. Hubbard, Co-Founder and Chairman of the Board; William H. Neukom, Co-Founder and CEO; Mark D. Agrast, Vice President; Deborah Enix-Ross, Vice President; Judy Perry Martinez, Vice President; Nancy Ward, Vice President; James R. Silkenat, Director and Treasurer; and Gerold W. Libby, General Counsel and Secretary. Staff, Consultants, and Interns: Elizabeth Andersen, Executive Director; Amy Gryskiewicz, Chief of Staff and Operations; Mark Lewis, Chief of Public Sector Partnerships; Alejandro Ponce, Chief Research Officer; Richard Schorr, Chief Financial Officer; James van der Klok, Chief of Philanthropic Partnerships; Tanya Weinberg, Chief Communications Officer; Said Aarji; Sally Aldrich; Daniela Barba; Marta Basystiuk; Hailey Bouwman; Abigail Boyce; Erin Campbell; Ana Cárdenas; Estefany Caudillo; Lilian Chapa Koloffon; Maria Chavarria; Lloyd Cleary; Avery Comar; Miguel Contreras; John Cullen; Shallum David; James Davis; Christine Detz; Giacomo D’Urbano; Alicia Evangelides; Allyse Feitzinger; Vianney Fernández; Marco Ivan Figueroa; Renae Ford; Joshua Fuller; Amir Galván; Alejandro González Arreola; Lucía Estefanía González Medel; Eréndira González Portillo; Kirssy González; Issa Guerra; Fernando Gutierrez O.; Dalia Habiby; Shakhlo Hasanova; Irene Heras; Roberto Hernández; Grace Hulseman; Skye Jacobs; Natalia Jardon; Verónica Jaso; Osvaldo Jiménez; Alphina Kamara; Lauren Kitz; Ally Knapp; Lauren Littlejohn; Mariana López; Debby Manley; Gabriela Marquez; Olimpia Martínez; Raquel Medina; Ignacio Miranda; María José Montiel; Ana María Montoya; Gustavo Núñez Peralta; Fernando Omedé; María Fernanda Ortega; Horacio Ortiz; Santiago Pardo; Enrique Paulin; Artha Pillai; Srirak Plipat; Stephanie Presch; Tanya Primiani; Hannah Rigazzi; Marien Rivera; Mario Rodríguez; Natalia Rodríguez Cajamarca; Juan Salgado; Alicia Segovia; Nikta Shahbaz; Leslie Solís Saravia; Helen Souki Reyes; Victoria Thomaides; Carlos Toruño; Paulina Vega; Sergio Villanueva; Katrina Wanner; and Moss Woodbury. World Justice Project Funders The World Justice Project thanks the following major 2022-2023 donors whose support makes our work possible: Institutional Donors 2022-2023 American Bar Association Section of Environment, Energy & Resources Carnegie Mellon University The Chandler Foundation The Charles Stewart Mott Foundation The Hewlett Foundation The Lawrence Foundation Luminate Mo Ibrahim Foundation Neukom Family Foundation The Open Society Foundations Private Sector Partnership for the Rule of Law 2022-2023 Arnold & Porter Google Hewlett Packard Enterprise (HPE) K&L Gates LLP LexisNexis Rule of Law Foundation Microsoft Corporation Shell Global Uber Wilson Sonsini Foundation Additional Private Sector Support 2022-2023 Cooley Fredrikson & Byron Jupitice Pacifica Law Group Perkins Coie LLP Zuber Lawler Major Individual Donors 2022-2023 Mark Agrast and David Hollis Sheikha Abdulla Al-Misnad Bernard Amyot Elizabeth Andersen and Stephen Pomper Anne Avis Elizabeth F. Bagley Margaret Breen and Stewart Landefeld Cindy and Bob Carlson Patricia Casale and Gary Gut Stephan Coonrod and Cheryl Clark Kelly Cullinane Eileen and John Donahoe Deborah Enix-Ross Barbara and R. Bradford Evans Suzanne E. Gilbert Karen Green The Haldeman Family Foundation David A. Heiner Samuel D. Heins Ruth Holzer and Michael Byowitz Lynn Hubbard and David A. Zapolsky Kappy and William C. Hubbard R. William Ide Jean Johnson and Peter Miller Cynthia Jones and Paul Lawrence P.K. Kilty and Stoddard Lambertson Linda Klein and Michael S. Neuren The Lawrence Foundation Suet-Fern Lee Gerold W. Libby Julie and Tom Lombardo Karla Kurbjun Mathews M. Margaret McKeown Sally and William H. Neukom Ellen Gracie Northfleet Judy Perry Martinez Michelle and Karl Quackenbush Suzanne Ragen Lucy Reed Lesley and Ted Rosenthal Paula Selis and Jonathan Fine Barbara Shufro and David Steuer James R. Silkenat John H. Stout Tonia and Robert B. Strassler David K.Y. Tang Lucy Vance James van der Klok Nancy Ward and Tobias Bright David Weiner Stephen Zack A current list of funders can be found at worldjusticeproject.org. 218 2022 Forum Outcome Report Latin America and the Caribbean Country Reports 2023 WJP Rule of Law Index 2023 Insights Highlights and data trends from the WJP Rule of Law Index 2023 WJP Mexico States Rule of Law Index 2022-2023 Perceptions and experiences in 32 states Corruption in the Caribbean 2023 WJP Justice Data Graphical Report I 2023 More from the World Justice Project For more information or to read these reports, visit worldjusticeproject.org/our-work Washington, DC Singapore Mexico City worldjusticeproject.org facebook.com/thewjp x.com/theWJP Scan the QR code or visit worldjusticeproject.org/index to view our interactive data portal.
2215	https://study.com/academy/lesson/what-is-an-element-in-math.html	Elements of a Set \| Definition & Examples - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses / Math for Kids Course Elements of a Set \| Definition & Examples Lesson Transcript Antonette Dela Cruz, Allison Petrovic Author Antonette Dela Cruz Antonette Dela Cruz is a veteran teacher of Mathematics with 25 years of teaching experience. She has a bachelor’s degree in Chemical Engineering (cum laude) and a graduate degree in Business Administration (magna cum laude) from the University of the Philippines. She’s currently teaching Analysis of Functions and Trigonometry Honors at Volusia County Schools in Florida. View bio Instructor Allison Petrovic Allison has experience teaching high school and college mathematics and has a master's degree in mathematics education. View bio Learn to define what an element is in math. Discover the elements of a set. Find out the size of a set and the notation of a set. See examples of elements of a set. Updated: 11/21/2023 Table of Contents Elements in Math What is A Set? Elements of a Set Size of Set Notation Element of Set Examples Lesson Summary Show Frequently Asked Questions How do you count elements in a set? The elements in a set may be counted by counting the commas and adding one or by counting the items that are separated by commas. Set V = { red, blue, yellow, green, white, brown }, for example, has 6 elements. How do you find an element in math? The element in math is found in the list of items inside the curly bracket of a set notation, separated by commas. Each of the items is an element. What is an element in math and an example? In math, an element is a member of a set. A set is a collection of elements. An example is the set of counting numbers: C = {1, 2, 3, 4, ...} Create an account Table of Contents Elements in Math What is A Set? Elements of a Set Size of Set Notation Element of Set Examples Lesson Summary Show Elements in Math ---------------- Element is a word that has different meanings for mathematics and science. In science, an element is any substance in its simplest form and cannot be broken down into more basic substances by ordinary chemical methods. They are all presented and organized on a periodic table in chemistry. In mathematics, however, elements are members of a set. A set is any grouping of a collection of objects that share a common attribute or classification. For example, a set of colors is made up of different colors for its members or elements. The elements of this particular set are the colors such as red, blue, yellow, green, etc. While science has elements known to exist as substances and uniquely identified by the number of protons in the nuclei of their atoms, the elements in math definition are variable, infinite in number, and mainly depend on a defined set. To unlock this lesson you must be a Study.com Member. Create your account An error occurred trying to load this video. Try refreshing the page, or contact customer support. You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. 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Your next lesson will play in 10 seconds 0:05 An Element in Science 0:34 An Element in Math 1:33 States & People as Elements 2:28 Writing a Set With Elements 2:48 Lesson Summary View Video OnlySaveTimeline 30K views Video Quiz Course Video Only 30K views What is A Set? -------------- A set is a group of defined objects that share common attributes. In mathematics, it is denoted by a capital letter. Sets are not limited to numbers only. It can be letters, names, or symbols. Some examples of various sets include: Set of numbers: N = { all even real numbers } Set of colors: C = { red, yellow, blue, green,... } Set of female students in the class: F = { Sue, Cindy, Lea, Nona, Mildred } Set of operators in arithmetic: O = { +, -, x, ÷ } Sets are not limited to being presented as a list. Sometimes they contain expressions or descriptions such as: Z = { set of positive integers } D = { all values of x for which x > 5 } To unlock this lesson you must be a Study.com Member. Create your account Elements of a Set ----------------- In set notation, the capital letter is followed by an equal sign, and then curly brackets list down the members or the elements. For example, a set of whole numbers may be represented as W={0, 1, 2, 3, 4, 5, ...}. The triple dots at the end of the list denote a continuation of the list to infinity or forever. The elements are separated by the commas within the bracket. Note that an element can be just about anything. Its existence primarily depends on the set. Each item that is mentioned when the set is defined is an element. Sometimes when sets do not enumerate but instead describe, any object that falls under the description is an element of the set. All others objects that cannot be classified under the description are not part of the set and are not elements of the set. To unlock this lesson you must be a Study.com Member. Create your account Size of Set Notation -------------------- The size of a set notation can be finite or infinite. If the elements are listed and shown inside the curly bracket or their number can be determined, the size of the set is the number of elements present within a set. \| Size of a Set is the number of elements belonging to the set. \| For infinite sets, the elements are infinitely many to list down, so the triple dots are used instead to represent continuous listing. Examples of Finite Sets Finite sets have elements that can be counted. A = { 1, 2, 5, 8, 9 } Size of Set: 5 B = { Sanford, Lake Mary, Orlando, Heathrow, Deland, Orange City } Size of Set: 6 C = { red, blue, yellow } Size of Set: 3 Examples of Infinite Sets Infinite sets have elements that are too many or impossible to count. D = { set of all negative integers } E = { set of all the stars in the universe } F = { 5, 10, 15, 20, 25, 30, ... } To unlock this lesson you must be a Study.com Member. Create your account Element of Set Examples ----------------------- Example 1 Set G is a set of countries in Asia. Write the proper set notation for set G and give five elements. SOLUTION There are many countries in Asia, and naming five of them will be the elements for this set. G = { China, Korea, Japan, Philippines, Singapore } Example 2 What is an element in a set? Which of the following is not an element of Set G from Example 1? A) Indonesia B) Laos C) Thailand D) Switzerland SOLUTION D is the answer. Switzerland is NOT a country in Asia. Therefore, it is NOT an element of the set. Example 3 Which of the following is a correct set notation for a set containing the different states of the USA? A) Florida, New York, California, Wyoming B) S = Florida, New York, California, Wyoming C) \|Florida\|New York\|California\|Wyoming D) s = { Florida, New York, California, Wyoming } E) S = { Florida, New York, California, Wyoming } SOLUTION Checking each of the choices: A) is not a set because there it is just a list of the four states B) is not a set because there is no curly bracket C) is not a set because this choice is in table form and no curly brackets and commas D) is not a set because the letter s is not in a capital format E) is the correct set notation for this example To unlock this lesson you must be a Study.com Member. Create your account Lesson Summary -------------- An element in math is an item that belongs to a set. A set is a collection of elements. Anything described by the set may be included as part of its list of elements. The set is denoted by a capital letter followed by an equal sign and curly brackets containing the elements. The proper way to write all elements in a set is to remember two things: The elements must always be listed inside a set of curly brackets, and the elements must be separated by commas. Sets may either be shown as a list with enumerated elements or described with no list of elements. The number of elements in a set may be finite or infinite. The finite sets have a countable number of elements. This means that there is a limit to the number of elements in a set. On the other hand, an infinite set has an endless list of elements. This is impossible to count and list down. For most cases, the triple dots are used after the last element on the list. These dots mean that the list will continue and go on indefinitely. The number of elements is also the size of the set. For a finite set, the size is determined by counting the number of elements. To unlock this lesson you must be a Study.com Member. Create your account Video Transcript An Element in Science Maybe you heard the word ''element'' in your science class when you learned about the Periodic Table. There are many elements on the Periodic Table. Oxygen (O), helium (He), and silver (Ag) are just a few. My favorite element is probably gold (Au). Do any of those science elements sound familiar? Like in science, an element is also a word you will hear in math class. Yes, it is. The word ''element'' in math actually means something similar to what it means in science. Let's take a look. An Element in Math In math, we have what is called a set. A set is basically a collection of things that typically have something in common. Each item in a set is called an element. An element can be a number, letter, color, food, piece of clothing, person, animal, or just about anything. It really just depends on the set. So, a set is made up of elements. Remember when I said that the word ''element'' in math means something similar to what it means in science? You may be wondering how that is. Let's talk about the similarities. Well, the Periodic Table is a set of all of the elements in science. Oxygen is an element in the Periodic Table. Helium is an element in the Periodic Table, and gold, my favorite, is an element in the Periodic Table. There are many more elements in the Periodic Table. That's why we can say that the Periodic Table is a set of all elements in science. Let's look at more examples of elements. States and People as Elements Like I said, an element can be just about anything. All the states in the United States can be a set. Each state would then be an element in that set. Ohio would be an element. Florida would be an element. Hawaii would be an element. These are just a few. The state where you live would be an element, if you live in the United States like me. There would be a total of 50 elements in that set. That is because there are 50 states in the United States. Take out your phone. You probably never hear that in class. If you don't have a phone near you, just imagine a list of all of your friends and family members that you would have as contacts in your phone. If you have a phone with you, open your contacts list. All the contacts in this list make up a set. Each person in your contacts list is an element. So next time you talk to your grandmother, tell her that she is an element in your contact list (and then explain to her what that means). Writing a Set With Elements When we write a set on a sheet of paper, we first need to write two curly brackets. Then, we list each element inside the set of curly brackets. Each element is separated by a comma. Here we have a list of all the elements in the colors of the rainbow. And no, the colors of the rainbow do not have to be written in color. Lesson Summary A set is a group of items that usually have something in common. Each item in a set is called an element. To properly show a set, place all of the elements between two curly brackets and use commas to separate each element. Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a Member Already a member? Log In Back Resources created by teachers for teachers Over 30,000 video lessons& teaching resources‐all in one place. Video lessons Quizzes & Worksheets Classroom Integration Lesson Plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. 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2216	https://www.youtube.com/watch?v=Tap0e-XC6fc	17-Step Response of RL and RC Circuits Enjoy Your Universe 1150 subscribers 94 likes Description 5179 views Posted: 23 Mar 2020 Online lecture for ENGR 2305, Linear Circuits, discussing the step response for RL and RC circuits. This is for a 16-week course taught to community college sophomores entering into an engineering program. The prerequisites for the course are second-semester calculus and differential equations. 4 comments Transcript: we're going to start with our L and remember what the step response is the step response is where energy is being added to the magnetic field so energy build up but after a long period of time the inductor behaves just like a piece of wire so let's say our circuit looks like this here's the resistor here's the inductor there is some voltage across the inductor that's going to be a function of time the switch is initially open but it is closed at T equals zero and this is our voltage source so let's say it has a voltage of V sub s over here so initially there is no energy inside the magnetic field of the inductor so at t equals zero when the switch is closed that's when current begins to flow now if you know how inductors behave if you remember that part the inductor is initially going to create a back EMF to oppose the change in current through the inductor so that is going to oppose the current flow at least initially however over time that's going to die down the change in current is going to slowly the current is going to slowly change until it is moving continuously through here and at that point all of the current is moving through the inductor and that current is just I is equal to V over R so that's N and the inductor at that point is just behaving like it's a piece of wire okay so if we think in terms of KVL here the voltage across the source there is equal to I R plus the voltage across the inductor which is L di DT okay now this the at least the right hand portion over here this looks very similar to something we've seen before however the left hand portion here is is very different it's no longer zero previously this was a first-order ordinary differential equation we had L di DT plus I R equals zero but now we don't have that now it's equal to some nonzero thing so this is now a non-homogeneous it's a little bit harder to solve it's not that difficult you can still solve this one with separation of variables because the V sub s is actually a constant here okay I'm not going to go through the steps remember this isn't the differential equations class but we use differential equations in here but you can go through the steps to solve that and whenever you do you're going to end up with AI as a function of time is equal to and here's where you get a slightly different answer V sub s over R minus V sub s over R e to the negative T over tau now tau is exactly the same thing it is L over R okay but look at this we can simplify this a little bit let's factor out the V sub s over R it's going to be 1 minus e to the negative T over tau like that ok so if you were to graph this let's think about what it looks like here if we do I of T on that axis T down here let's think about what's going on here at T equals 0 switches close remember I said the current is going to be opposed by the back EMF generated by the inductor and that back EMF is generated because we have a di DT you cannot have an instantaneous change in current so it's this occurrence got to slowly change here so the current initially is going to be down here at 0 at t equals zero okay but over time it's going to rise up it's going to approach a maximum value and then it's going to level off and reach that maximum value in the limit here now if we were to trace that maximum value back to the vertical axis here that is V sub s over R like that okay so that's how this is going to behave now now if you'll think about it that also should be exactly the way you expected it to turn out because the current even though it initially starts out at zero it slowly builds up and after a long time how long is a long time v towel after v towel this inductor it behaves just as if it wasn't wasn't there so the circuit behaves as if the inductor wasn't there there's still energy stored in the magnetic field inside the inductor but the current is just going to go straight through so after a long time v towel the current is essentially these of s over R ok so that is the step response for an RL circuit for the current at least now let's think in terms of let's think in terms of how that 1 minus e to the negative T over tau behaves because we've already made this chart once where let's say that we have a chart here and we've said T and we did e to the negative T over tau well now let's expand that and do 1 minus e to the negative T over tau and after 1 tau of time we said this is down to like point 3 6 7 something but this is 1 minus that so this is point 6 3 something ok and then after 2 tau this is down to point 1 3 5 this is going to be like 0.8 6 3 tau I'm just rounding these numbers here and I could round them all just keep that in mind point zero four nine so that's actually close to point zero five so that makes this one like point nine five so it's 95 percent up to its maximum value by three towel after four towel this is point zero one eight which means it's point ninety eight percent and after five tau we're below one percent so it's point zero zero six which makes this point nine nine four approximately so it's 99 percent done after that so it's it's less than one percent left to finishing so once again five tau is the time for this term that's in the parenthesis the one minus e to the negative T over tau it takes five towel for that to complete its process as well okay now let's turn our attention to the voltage let's say that we wanted to figure out what these voltages we got the current so what is the sub T in other words well it's going to follow the same general pattern but you gotta use your circuits knowledge to figure this out how is that inductor going to behave initially it's got a back EMF right at T equals zero that opposes the current that doesn't allow the current through but then after a long time all of the current is going through it there so if all of the current is going through it and it's behaving like a piece of wire what's the potential difference across a piece of wire and the answer is zero so you expect this to behave a little differently than what the current function looks like you expect it actually to decay to be a decaying exponential in other words we're kind of expecting this based upon our circuits knowledge that the V of T versus T is going to start at some maximum value and then decay over time like that okay but how do you find the V of T well V is just L di DT okay this is what the eye looks like down here so you take the derivative of that so it's just L times the derivative with respect to time of V sub s over R times one minus e to the negative T over tau so we need to take the derivative of that let's see this first term take the derivative I'm imagining this V sub s over R being distributed through there so this first term with the wine is going to go to zero then we're going to have the derivative of this over here so we're going to have V sub s over R now I'm taking that negative sign put it right there and we're gonna have a factor that's going to be negative one over tau times e to the negative T over tau so right just like that now if you plug in for tau remember what tau is tau is L over R if you were to plug in for tau L over R you're going to discover that the elves will cancel since this is 1 over tau and you're going to end up with something that looks like this the negatives are also going to cancel you're going to end up with something that the oh by the way the R will also cancel here because you've got an R there so this is L over R you're it's going to cancel out this is going to end up being V sub s e to the negative T over tau which using your circuits knowledge you figured out yeah that's what it ought to look like so having a familiarity with these circuit elements how they behave when you turn the switch on and what to expect like they're either going to rise up and level off or they're going to decay like that those are really the only two possibilities here either the quantity you're after voltage or current is going to rise up and level off or it's going to decay over time so that's that's really the only two possibilities let's very quickly look at an RC step response as well now that we've got all that under our belt here okay so let's say that we've got a circuit current source we've got resistor and switch and capacitor here okay so this is our capacitor this is our resistor the switch is initially open but the switch is closed at T equals zero and this has current source size of s okay initially this is an incomplete circuit so there's nothing no energy going over here to the capacitor initially all the current is just over here in the resistor right here you could figure out what the voltage is if you knew what har was and what I said that so as you can figure out what the voltage was across that resistor right away okay but there's no there's no energy stored in the capacitor so nothing is there before t equals zero however at t equals zero the switch is closed now at t equals zero then when you get to this junction when the current comes around and it gets to this Junction then it has a choice to go down this way or to go that way after that switch is closed well all of that current is going to go down this way and the reason is because there's nothing opposing it over here remember how the capacitor creates a displacement current across its plates the current comes in builds up a charge on one plate which induces the charge on the other plate so it looks like that current is passing right straight through so at least for the fraction of a moment right after T equals zero all of that currents going to be flowing over here now you can do that you can have an instantaneous change in current for the capacitor can I have an instantaneous change in voltage okay so you've got to keep that in mind as well so let's go through the steps here of looking at what this equation is going to look like and try to find the equation of state the first order differential equation for this circuit okay what we're going to do the way that we're going to do it is we're going to sum the currents away from this top essential node so I've kind of messed it up here so let's say there's the top essential node there I'm going to imagine a current going that way of current going down this way and a current going that way so this is essentially no voltage at that point well the one going to the left that's going to be negative high submiss okay the one going down the bottom I'm going to call this node voltage here well the node voltage here is going to be the voltage across the capacitor because if I call this my ground down here then that node is going to have the same voltage as the capacitor so that one is V sub C over R that's the current going down the middle and now over to the right-hand side that current is going to be C D V sub C DT and now all that's equal to zero so that came about because of a node voltage equation I'm going to rearrange the terms just so it looks a little easier to read D V sub C DT plus V sub C over R C tell what I've done is I've divided through by the C so go RC on bottom there and then I'm gonna put the hi on the other side it's constant so it's going to be I sub s over C wave down now this is our first-order ordinary differential equation that's the equation of State for this circuit here with this RC circuit this is again a step response so how do you solve this one well same sort of way you can go through the possibilities or through the process of solving this if you want I'm just going to jump to the solution here V sub C as a function of time is equal to I so s are plus points this V not that's V at 0 minus I sub s are e to the negative T over tau and since this is an RC circuit tau is just equal to RC like that okay so once again we've got something that looks a little similar now you see that V naught right there that came about when you tried to solve the circuit here you that's the initial voltage that's on the capacitor and you may have seen something similar that's you know if we did like I DT plus V naught that's the voltage you know which you've seen equations like that okay well that's that V naught that's where that came from that's the initial voltage across the capacitor but for our example here that is 0 so V naught is 0 for our capacitor so this reduces down to I sub s R - hi sub s are a did this room ignore that it's going to be I sub s are minus I so F are e to the negative T over tau okay maybe I did it right in the first place but that's what it looks like now you can factor out the I sub s R and I've got 1 minus e to the negative T over tau once again so this one is going to look very similar I've got my V sub C versus T starts off at 0 rises up levels off to some maximum value in the limit and that maximum value is going to be high so the s or there okay so that's the voltage across the capacitor now what if I I want to find the current what if I want to find this current going through the capacitor there okay well I equals C DV DT and that voltage is a voltage across the capacitor this current is current through the capacitor so you just take this you take the derivative of it and you've got the equation for that now what do you expect that to look like remember what we talked about at the beginning all of the current is going to pass through here initially which is going to build up a very slow increase in the potential difference between these capacitors when I say very slow I mean within v towel it's going to build that up so it's it's going to quickly increase the voltage across this but the current initially is going to be very high so what do you expect the current through that capacitor to look like will I expect the current to look like I sub C of T starting high and going low like that in fact I expect that to be I sub s all of the current to be going through that capacitor at least at T equals 0 well I'm going to leave it as an exercise to you to take this equation take your V sub C that you've got down here at the bottom and take the derivative of it multiply it by C and you're going to discover exactly what you end up is I sub C is going to be I sub s e to the negative T over tau that should be exactly what you end up with all right all right so what we've done now is we've gone through all of these natural and step responses for both RL and RC circuits now what I want to do is turn our attention to a more general approach because you have probably noticed that there are patterns present in every one of these and those patterns are going to allow us to figure out more general solutions that are going to apply you may have also noticed that the the solutions are very similar to each other because the differential equations are very similar to each other now let's really there's really only four possible ways to arrange these simple RC circuits and RL circuits let's say that in fact let's draw all four all four of them you can have an R and an L like that so you can have an R and an L with some voltage there you could also have a situation like this where we've got an L here there's your R and this is some eye now these two circuits you may notice are equivalent to each other if you think of that being the terminals then what's on the left-hand portion there of the pin is a than an equivalent so you've got a V th and an R th that and the load here is the inductor but you could do a transformation a source transformation on that and this is a SC which is the same as the th / r th so that is the Norton equivalent and here the resistor would be our attention once again these would be the terminals so you've got the inductor as the load on either one of those two cases right there now let's say that you've got what's a capacitor now so capacitor once again you got some voltage you got a resistor and got your capacitor once again this is a demeaning equivalent or with source transformation you can get the Norton out of it and that's once again is is C and this is our th okay so really you've got four possibilities here now in each of these cases you're probably going to be interested in whatever the voltages across the inductor or whatever this current is or the voltage across that and whatever this current is here or the voltage across this guy and whatever that current is or the voltage across the capacitor there and whatever that current is so you're going to be interested in those currents or those voltages in all of those cases there okay for each of these equations then there is a pattern that is evident all of the all of the equations of state reduced down to this let me show you DX DT and X is either going to be the voltage or the current that you're interested in plus X over tau equals and that's supposed to be a Kappa Greek letter Kappa okay now you might say well I don't I'm not seeing that well think back you had DV DT plus R over L V zero okay that was the equation for the natural response for an RL circuit okay so the V is X the V is X here the Tau you notice that's town bottom tau is L over R for an RL circuits so it's one over that so we got har over L here and the K happens to be zero the K happens to be zero because it's the natural response however if we did that same sort of thing for the for the coupon sorry for the capacitor then you're going to have di DT plus I over RC that's equal to zero so there's the natural response for the RC and then the step response equations looked I guess pretty similar except you're not going to have the zero over here on the right hand side for each one of these so if you're talking about the RL we set it up so that we had this equation as our equation of state plus R over L PI equals the sub Sol or for for the capacitor the step response was DV DT plus V over R C equals I Sub s over C so down here on bottom these are the step responses these are the natural responses and this is for the inductor and these are for these two over here for the capacitor so we've got a towel value and we've got a K value here or a Kappa value for each one of those now if you pay attention to this and you spend some time with these equations and you start seeing patterns in here in each case the X is going to be either V or high I think that part is obvious but in each case that X approaches a constant value so in the end in the limit the X is going to be zero or it's going to approach the constant value remember in some cases we had the rising leveling off exponential and in other cases we had the decaying exponential so it's going to be one of those it's either going to in the limit go to 0 or in the limit level off to some maximum value here so that final value that let's call that X sub F that's X final in each case if you know which one of these cases you've got the X final is always going to be the kappa x the towel okay so that's the final value always okay that means that in each of these cases you can solve these equations in fact you could even use this general form right here and go through the process of solving it and here's what you would end up with I'm gonna get a another sheet of paper here I'll just use the backside okay you're going to end up with a generic solution that looks like this X final plus X hat T naught and T naught is the first time when they when the switch is closed minus X final times e to the negative t minus T naught over tau okay so this is the final value this is the initial value and once again that's final and T naught is the time the switch is thrown that could be T equals zero but that's the time the switch is thrown that's when you start making the observation and then tau here is the time constant okay so everything that we have just done we've spent what three videos going through then or maybe it's four videos by this point I don't know going through the natural and step response for all of these all of these various RL and RC circuits but it all boils down to this is your solution so really this is all you need to know so if you want to find the natural and step response for an RL and RC circuit then all you have to do is identify which what variable are you interested in are you interested in AI or are you interested in V okay figure out which one of those you're interested in because that's going to be your X all right then determine the initial values so determine X at T naught now that could be X at 0 we've sometimes called it X naught or V naught or I naught or I sub s whatever it is that's the initial value whenever the switch is thrown that's the initial value right there that X sub F is going to be Kappa times tau though you're going to need to calculate tau so tau is going to be either RC or L over R but then once you've gotten these things figured out once you've got these initial and final values figured out then you're just plug it in and oftentimes you don't even have to do any calculations here you can like I said using your circuits knowledge you can almost just by inspection figure out what the final value of X is going to be or what the initial value of X is going to be and then almost immediately jump to the solution now let me show you an example of of us doing being able to do that so that's called the general approach and we're going to use the general approach here alright here is our problem current source resistor in parallel with it switch and capacitor okay we're interested in the voltage across that capacitor so there's our capacitors switches open switch is closed at T equals zero there's our R and this is going to be high so that's so that's our our source there okay now previous previously we had to go through finding the the differential equation the first order differential equation then we had to go through the steps of solving it and then we figured out and I will just show you what the answer it was in those cases V sub C of T is going to be I sub s times R times 1 minus e to the negative T over tau and tau in this case it's just going to be RC okay so we know that that's the answer right there but let's use our general approach and let's try and figure out what the what the answer should look of it remember here's the general approach X of T is equal to X sub F plus X let's call it t naught minus X sub F times e to the negative T over tau so this is our general general solution we need to figure out what what we're interested in well we're interested in X sub C so I'm sorry V sub C so V sub C is what X is okay that's the first step now let's figure out what the sub C at 0 is that's the initial value well V sub C at 0 since the switch is open before t equals 0 that's going to be 0 okay okay so that's the next step what about V sub C at infinity at time infinity as T goes to infinity in other words what is V sub C final okay well that's going to be Kappa Tau all right well what is capa here well in order for you to remember what Kappa is you're gonna have to look back at the differential equations and it turns out that it's I sub s over C multiplied by tau which is RC so it's just I sub s R in other words the final value for the capacitor is just going to be whatever the voltages across this thing right here that's that's the final voltage okay now that now you did not have to go through this to figure that out your circuits knowledge should have been able to tell you that this final voltage across this is going to be this final voltage across that resistor there well what's the final voltage across that resistor it's just I times R its I sub s times R so that's your final value okay so that's your your X final or your V sub C final okay then what about tau well tau is going to be RC in this case so we're ready to apply that right there V sub C at T is equal to X final let's see where was X well that's just I sub s are sorry plus let's see what is the initial value the initial value was 0 minus the final again I sub s R times e to the negative T over tau which is just RC simplifying it I sub s R is 1 minus e to the negative T over tau where tau equals R see there there's our answer which matches that right there okay so you can use the general approach here to get to the same solutions that we did so that equation right there if I were you I would commit that to memory because you're probably going to use it quite a bit now let's follow through on that let's say that I wanted to find the current this current here okay well you use the same equation and you don't use a different equation you use exactly the same one all right I want the current so that's going to be I sub C that's what I'm interested in so I sub C of T is what X is so let's write that down I sub C of T is equal to I final now what's i final what's the current final current going to be over here well the final current is going to be 0 after this thing gets charged up so this is going to be 0 plus okay and what about the initial current the initial current well that's that's just going to be I sub s miss miss - and we're back to the final again 0 times e to the negative T over tau so what does that tell me that tells me that I sub C T is equal to I sub s e to the negative T over tau there so you can use this same technique on any component where you have or any type of circuit where you have one R and one C however in reality you're going to have multiple lawyers and I have multiple C's so what we need to do is we need to go through an approach so that we can figure it so that we can I guess we knit down so that we only have one R in one see if we can do that then we can use this general approach right here and with no problem ok and we'll we'll do that in just a minute okay in the meantime I want to show you something else let's say that we've got op-amp I'm betting you know what this one else one feedback resistor this is the resistor that's in series with a source here what is that that's an inverting op-amp so it's an inverting op-amp now here's what I'm going to do I'm going to take that and I'm going to replace it with a capacitor okay and I'll redraw that on a clean sheet of paper so the only difference here is that now we have a capacitor where the feedback resistor used to be so this is C so if that's a seen way so that our output over here I mean exactly the same way okay let's solve this one well using the techniques that we know the the initial what am I trying to say the op-amp constraint so that's what I'm trying to say the ideal op-amp constraints visa P equals V sub N and I suppose I sub n equals 0 well this is visa P and you see it's going straight to ground so we know that V sub P is going to be 0 so that means this V sub n is also going to be 0 here okay we know that we also know that these currents going in here that's that's hi sub n this is I so P that those are going to be 0 so we just know that by inspection by inspection thus Appeals V sub N equals 0 all right now at the V sub n node then we use our regular technique of node voltage equation so V sub n minus V sub s over V sub S Plus now the current going this way is C sub F times DV DT but this V here this V is equal to base of n minus V output like that ok so this is the derivative of that right there so at the V sub n that's zero this V sub n that's zero and what are we left with minus V sub s over R sub s how did that get on bottom that's a mistake R sub s that should be our savez thank you for pointing that out I just got that message okay plus I'm screwing up here minus C sub F D V not DT and that's all equal to zero in fact all that's equal to zero as well okay so we got minus signs on everything and get rid of that minus sign just rewrite it one more time V sub s r sub s + c sub F DV not DT equals zero okay now we've got it once again a first-order differential equation we should be able to solve that but really remember what we want to do here we want to write the output in terms of the input and once we've done that we've solved this op-amp problem so let's do separation of variables for this here I'm gonna leave D V not DT on one side by itself and let's put everything else on the other side so I'm going to have a minus V sub s over R soda and I'm going to what divided by the C sub F so C so that so you notice what I've got over here that's a V sub s over what is effectively the time constant here now what am I going to do I'm going to integrate both sides and let me just move up here for that whenever you do that on the left-hand side here you're going to end up with me not as a function of time now we're going from that initial time T naught to some later time T so I'm it's actually going to be V naught minus V naught at T naught and that's equal to this stuff over here on the right hand side which is minus the integral of V sub s over R sub s C sub F DT plus know that that's all that's all right there so when it comes to V naught of T that's equal to minus the integral V sub s over R so C sub F DT plus V naught at T naught okay now look at what we've got here this says that the output is the integral of the input now it's a scaled integral because of these constants down here R sub s and C sub F you could factor those out of the integral though and this is just the integral of the input plus whatever the initial voltage was at t equals zero okay now that output may have been zero there may have been no voltage initially over here on the output in which case that would have been zero right there but what do you think this one is called this is an integrator op-amp it's an integrating amplifier because you're going to get out the integral of whatever the input was and well you might say well how does that work well if you let's say you input a step function so let's say the V sub s versus time you do some kind of step function and it makes a transition down at a point at time t1 and then it goes over here it goes back up and let's say the entire period of time is 2 t1 so it starts here at 0 it goes up to the maximum value of V sub m and then it goes to some negative value down here which is 20 minus V sub M okay so that's that's what the source is doing okay well what should the integral function look like in other words what's the output going to look like for something like that okay well this is the output just bringing that time t1 down here there's that time - t1 right there this is the T axis okay well what's what's that going to look like well remember this is the it's going to be inverted to begin with so it's going to look something like that because we're taking the integral of that thing and it's the reason it's down here is because it's it's negative but then right here it changes and does something like that so I would hope that you would be able to figure that out if you if you need to do the math to see that then you know the the output of T is going to be minus 1 over RC of the integral of whatever the source function is DT plus V naught at T naught and you're going from T naught to time T here so what are you doing you take this function here between 0 and T 1 and figure out take the integral of it ok well it's a constant so when you integrate that you're going to get a variable in it the negative sign is going to bring it down here so it's going to be like that then over here it's going to be like that I'm assuming that this is zero my advice would be spend a little time with us see if you can work your way backwards here this one has a slope that is downwards so the derivative of this is going to be a constant so it's a linear downward here the only reason that it's below the line here in the negative region has to do with that negative sign right there okay we'll stop right there for today and then we'll work some more problems with using the general approach and maybe we'll talk about some other things as well okay
2217	https://brainly.com/question/20532908	[FREE] A(3,0) and B(-3,0) are the two fixed points. Find the locus of a point P at which AB subtends a right angle - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +60,3k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +28,4k Ace exams faster, with practice that adapts to you Practice Worksheets +8,2k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified A(3,0) and B(-3,0) are the two fixed points. Find the locus of a point P at which AB subtends a right angle at P. 1 See answer Explain with Learning Companion NEW Asked by sthapresan06 • 01/11/2021 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 20065657 people 20M 5.0 2 Upload your school material for a more relevant answer y 2+x 2=9 Explanation Let's take a general point P(x,y). If there is a right angle at P then the slopes of the lines passing through AP and BP must be perpendicular. Two lines of slopes m1 and m2 are perpendicular if: m1m2=-1 The slope of the line passing through P(x,y) and A(3,0) is: m 1=x−3 y−0=x−3 y The slope of the line passing through P(x,y) and B(-3,0) is: m 2=x+3 y−0=x+3 y Substituting in the equation: x−3 y⋅x+3 y=−1 Operating: (x−3)(x+3)y 2=−1 y 2=−(x−3)(x+3) y 2=−(x 2−9) y 2=−x 2+9 y 2+x 2=9 The locus is a circle centered at the origin with radius 3 Answered by elcharly64 •3.2K answers•20.1M people helped Thanks 2 5.0 (3 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 20065657 people 20M 5.0 2 Non-Euclidean Geometry - Henry Manning Surveying and Mapping - Christian Tiberius The Foundations of Geometry - David Hilbert Upload your school material for a more relevant answer The locus of the point P such that line segment AB subtends a right angle at P is defined by the equation y 2+x 2=9. This describes a circle centered at the origin with a radius of 3. Thus, point P will lie on this circle. Explanation To find the locus of a point P such that the line segment AB subtends a right angle at P, we will follow a systematic approach. Identify the Points: We have two fixed points A(3,0) and B(-3,0). General Point: Let's consider a general point P with coordinates (x, y). Determine the Slopes: The slope of line AP is given by: m 1=x−3 y−0=x−3 y The slope of line BP is given by: m 2=x+3 y−0=x+3 y Condition for Right Angle: For the lines AB to subtend a right angle at point P, the product of their slopes must equal -1. Thus, we have: m 1⋅m 2=−1 Substituting the expressions for m_1 and m_2 gives: (x−3 y)(x+3 y)=−1 This simplifies to: (x−3)(x+3)y 2=−1 Rearranging the Equation: Multiply both sides by (x - 3)(x + 3): y 2=−(x 2−9) Thus, we get: y 2=9−x 2 Locus Equation: Reorganizing this equation gives us: x 2+y 2=9 This represents a circle. Interpretation: The locus of point P is a circle centered at the origin (0,0) with a radius of 3. This means that as point P traces this circle, the angle subtended by line AB will always be a right angle. In conclusion, the locus is described by the equation y 2+x 2=9. Examples & Evidence For example, if we choose point P at (0, 3) or (0, -3), both points will lie on the circle described by the equation and will ensure that the angle subtended by AB is a right angle. This conclusion follows from the property of circles that states if a point P lies on the circle defined by the segment A and B as diameter, then angle APB will always be a right angle. Thanks 2 5.0 (2 votes) Advertisement sthapresan06 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 19 What is the molar mass of magnesium sulfate heptahydrate (MgSO4⋅ 7H2O) ? Be sure to include the mass of all elements in the formula including the elements in water. 73 g/mol 120 g/mol 138 g/mol 246 g/mol Community Answer 4.6 17 Unit 1, Lesson 6 2. The point (-4,1) is rotated 180 degrees counterclockwise using center (-3,0). What are the coordinates of the image? A. (-5,-2) B. (-4,-1) C. (-2, -1) D. (4,-1) 3. Describe a sequence of transformations for which Triangle B is the image of Triangle A. Community Answer 11 All atoms of uranium have the same (1) mass number (2) atomic number (3) number of neutrons plus protons (4) number of neutrons plus electrons Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Compare the average rate of change for f(x)=3 x and g(x)=3 x+5 for 0≤x≤4. A. The average rate of change of g(x) is greater than that of f(x). B. The average rate of change of f(x) is greater than that of g(x). C. The average rates of change of f(x) and g(x) are the same. Solve for z. z+−6.4=11.6 −5 1⋅9 2 Write your answer in simplest form. Solve for d. $\begin{array}{l} 2 d=9.6 \ d=\square \end{array}$ Select the tables that show a proportional relationship between x and y. \| x \| y \| :---: \| \| 5 \| 1 \| \| 10 \| 2 \| \| 15 \| 3 \| Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
2218	https://books.google.com/books/about/Fundamentals_of_Statistical_Signal_Proce.html?id=vA9LAQAAIAAJ	Fundamentals of Statistical Signal Processing: Detection theory - Steven M. Kay - Google Books Sign in Hidden fields Try the new Google Books Books Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Get print book No eBook available Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» ### Get Textbooks on Google Play Rent and save from the world's largest eBookstore. Read, highlight, and take notes, across web, tablet, and phone. Go to Google Play Now » My library My History Fundamentals of Statistical Signal Processing: Detection theory Steven M. Kay Prentice-Hall PTR, 1998 - Computers - 560 pages The most comprehensive overview of signal detection available. This is a thorough, up-to-date introduction to optimizing detection algorithms for implementation on digital computers. It focuses extensively on real-world signal processing applications, including state-of-the-art speech and communications technology as well as traditional sonar/radar systems. Start with a quick review of the fundamental issues associated with mathematical detection, as well as the most important probability density functions and their properties. Next, review Gaussian, Chi-Squared, F, Rayleigh, and Rician PDFs, quadratic forms of Gaussian random variables, asymptotic Gaussian PDFs, and Monte Carlo Performance Evaluations. Three chapters introduce the basics of detection based on simple hypothesis testing, including the Neyman-Pearson Theorem, handling irrelevant data, Bayes Risk, multiple hypothesis testing, and both deterministic and random signals. The author then presents exceptionally detailed coverage of composite hypothesis testing to accommodate unknown signal and noise parameters. These chapters will be especially useful for those building detectors that must work with real, physical data. Other topics covered include: Detection in nonGaussian noise, including nonGaussian noise characteristics, known deterministic signals, and deterministic signals with unknown parameters Detection of model changes, including maneuver detection and time-varying PSD detection Complex extensions, vector generalization, and array processing The book makes extensive use of MATLAB, and program listings are included wherever appropriate. Designed for practicing electrical engineers, researchers, and advanced students, it is an ideal complement to Steven M. Kay's Fundamentals of Statistical Signal Processing, Vol. 1: Estimation Theory (Prentice Hall PTR, 1993, ISBN: 0-13-345711-7). More » From inside the book Contents Introduction 1 Detection based on simple hypothesis testing is described in Chapters 3 parameters is the subject of Chapters 69 Other chapters address detection 13 Copyright 17 other sections not shown Common terms and phrases Appendixassumeasymptotic performanceChaptercorrelationcovariance matrixDC leveldecide H₁denoteddetection performancedetection problemdetector decides H₁determinedeterministic signaleigenvaluesenergy detectorequivalentestimatorestimator-correlatorexampleFisher information matrixfrequencyGaussian noiseGaussian randomgivenGLRTH₁ if N-1H₁ if p(xHencehypothesis testinghypothesis testing problemKay-Iknown signallarge data recordslevel in WGNlinear modelln LG(xln p(xmatched filtern=nononGaussian noiseNP detectorNP testnuisance parametersoptimalparameter testperiodogramprior probabilitiesPss(frandom processrandom signalrandom variableRao testRayleigh fadingright-tail probabilitysensorshown in FigureSignal Processingsinusoidtest statistictheoremthresholdTR(Xunknown amplitudeunknown parametersvectorWald testWGN with varianceWiener filterΗοθοθςθτοσ² About the author(1998) STEVEN M. KAY is Professor of Electrical Engineering at the University of Rhode Island and a leading expert in signal processing. Bibliographic information Title Fundamentals of Statistical Signal Processing: Detection theory Volume 2 of Fundamentals of Statistical Signal Processing, Steven M. Kay Prentice-Hall signal processing series, ISSN 1050-2769 AuthorSteven M. Kay Edition reprint Publisher Prentice-Hall PTR, 1998 Original from the University of California Digitized Jun 20, 2011 ISBN 013504135X, 9780135041352 Length 560 pages SubjectsComputers › Design, Graphics & Media › Graphics Tools Computers / Design, Graphics & Media / Graphics Tools Technology & Engineering / Electrical Technology & Engineering / Engineering (General) Technology & Engineering / General Technology & Engineering / Signals & Signal Processing Technology & Engineering / Telecommunications Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
2219	https://www.acep.org/criticalcare/newsroom/newsroom-articles/april-2025/a-review-of-difficult-airway-management-strategies-in-the-emergency-department	American College of Emergency Physicians 'Critical Care Home Home Join SectionContact Us Critical Care Section Newsroom April 14, 2025 A Review of Difficult Airway Management Strategies in the Emergency Department Gabrielle Michaeli, MDIndiana University School of Medicine, Department of Emergency Medicine In the emergency department, one of the most critical skills is airway management and performing intubation when necessary. The reasons for intubation vary but typically are a result of decreased or insufficient respiratory drive, an inability for the patient to clear secretions, or concern for impending airway compromise. Nearly 400,000 intubations occur in the emergency department each year.1 A large body of literature exists comparing different intubation techniques and the effects of adjuncts on successful intubation. One definitive finding is that higher attempt rates resulting in longer periods of apneic time are known to have deleterious effects on patients. For this reason, first pass success is a common surrogate for better clinical outcomes in many of these studies. This article will focus on several different methods for airway management that have been studied to improve first pass success. Video Laryngoscopy By far, the most robust evidence exists for comparing direct laryngoscopy (DL) to video laryngoscopy (VL). Direct laryngoscopy involves direct visualization by the intubator during the procedure. Video laryngoscopy is an indirect method of visualization via a camera present at the end of the laryngoscope blade. Video laryngoscopes exist as standard geometry (curved or straight) blades or hyperangulated. One study by Prekker et al. used an observation analysis of data gathered from the BOUGIE and PREPARE II trials analyzing the rate of first past success in DL versus VL. This study demonstrated that VL use produced superior views of the vocal cords and thus increased the rate of first pass success.2 A second study by Prekker et al. known as the DEVICE trial aimed to answer the same question. This prospective randomized control trial enrolled 1420 patients to either DL or VL over an 8-month period. Intubators were either emergency medicine residents or critical care fellows. First pass success was 85.1% in the VL group and 70.8% in DL. One weakness reported in this study is that the intubators had a higher rate of experience utilizing VL, possibly skewing results. Overall, however, this study further strengthened prior published data reporting higher first pass success in VL.3 A third study by Brown et al. utilizing data from the National Emergency Airway Registry adds that even when adding augmented positioning maneuvers such as ramped up body positioning, bougie, or cricoid pressure, VL without augmentation still outperforms DL with these maneuvers.4 Bougie The bougie is a semi rigid stylet that is 60 cm long, 5 mm wide with a coudé tip. It is designed to easily pass through the vocal cords and is able to provide tactile feedback during intubations. There are many different versions of bougies that exist. To utilize the bougie, best practice is to insert your laryngoscope and pass the bougie as if it were an endotracheal tube. When passing the bougie, if you are able to continue passing it past the 50 cm mark, you are likely in the esophagus. If you are in the trachea, you will feel bumps from the tracheal rings and you will get tactile feedback when you hit the carina. When loading the endotracheal tube over the bougie, the laryngoscope should be kept in place to maintain tongue displacement.7 The BEAM trial was a single center RCT that randomized 757 patients to “bougie-first” intubation vs. standard intubation. First pass success occurred in 96% of the bougie-first patients compared to 82% of those undergoing standard intubation.8 The BOUGIE trial by Driver et al. was a multicenter randomized control trial that similarly randomized 1102 patients into a bougie-first intubation vs. standard intubation. Intubations were largely performed by fellows and residents. In this trial, first pass success was achieved 80.4% of the time in the bougie-first group and 83% in the standard intubation.9 This is quite a bit lower than in the BEAM trial, which begs the question if higher success occurred in the BEAM trial due to greater experience and comfort with the use of bougie during intubation. Ultrasound The use of ultrasound in the emergency setting continues to grow. There are numerous ways to incorporate the ultrasound into daily practice. One use being investigated is for confirmation of endotracheal tube (ETT) placement. The current gold standard for tracheal confirmation is end-tidal capnography. However, in certain clinical scenarios, such as cardiac arrest, end-tidal is unreliable. Additionally, during an arrest, you are not able to obtain radiographic confirmation of ETT depth. As such, ultrasound is an ideal way to confirm ETT placement during critical situations. An article by Gottlieb et al. discusses different ways to utilize point-of-care ultrasound (POCUS) for intubation. It assists with determining appropriate ETT depth by measuring the distance from the thyrohyoid membrane to the hyoid bone or epiglottis, as well as dynamically confirming endotracheal intubation or using multiple windows to confirm ETT placement. To use ultrasound during intubation, more than one operator is needed. While intubating, you look for the “snowstorm sign.” This is described as a flutter of the tracheal rings that occurs when the tube makes contact with the trachea. Following intubation, you can look for the “bullet sign” or the “double track sign.” The “bullet sign” shows rounding of the vocal cords due to the presence of the ETT. The “double track sign” is seen if there is esophageal intubation. The double track is seen because when the ETT is in the esophagus, there are two air filled lumens present. Other surrogates for ETT placement include evaluation of lung sliding and diaphragmatic elevation, which can help evaluate for mainstem intubation. ETT depth has been evaluated by placing the probe at the sternal notch and looking for the ETT cuff. Studies have evaluated filling the cuff with saline to better visualize the cuff, although this is not standard practice in adults. Ultrasound has also been utilized to better visualize the cricothyroid membrane to assist in cricothyrotomies.10 A randomized control trial by Hossein-Nejad et al. assessed the ability of emergency medicine residents to determine esophageal versus endotracheal intubation in cadavers utilizing ultrasound. Correct locations were determined 89.4% of the time.11 Long, Koyfman and Gottlieb performed a meta-analysis that showed ultrasound is 98.7% sensitive and 97.1% specific for determining endotracheal intubation. The meta-analysis included 17 studies. There was no demonstrated benefit in linear when compared to curvilinear probe choice. Ultimately, they report that this is user-dependent, and there is a learning curve to achieve the highest sensitivity and specificity.12 SALAD Method An important indication for intubation is large-volume emesis or hematemesis in an altered patient. Intubating in these scenarios can be challenging, given the presence of thick products that can obscure the view of the vocal cords. A proposed method to help combat this is known as the suction-assisted laryngoscopy assisted decontamination technique, otherwise known as the SALAD method. This was conceptualized by Dr. Jim Ducanto. This method incorporates the use of a large bore suction catheter while performing intubation. During this technique, the suction catheter is placed into the mouth along with the laryngoscope. Once an adequate view is obtained, the suction is removed and subsequently replaced to the left of the blade into the esophagus known as the “SALAD Park Maneuver,” allowing for space for the ETT. A bougie or ETT is then inserted through the cords.14 Fiberoptic Fiberoptic laryngoscopy is becoming more prevalent in the emergency department, and single-use, disposable endoscopes are more readily available. Fiberoptic nasal intubation is beneficial in cases of angioedema, burns, trauma, foreign body aspiration, or other instances where intubating through the mouth is difficult. It is also useful in cases where awake intubation is preferred.17 In these cases, pharmacologic pretreatment is beneficial. These include: Glycopyrrolate or atropine to reduce secretions Ondansetron to reduce gag reflex Viscous lidocaine applied intranasally Nebulized lidocaine to anesthetize posterior oropharynx Ketamine or midazolam for anxiolysis To perform nasal fiberoptic intubation, you will pretreat with the above medications. Next, you should evaluate both nostrils to determine patency, followed by preloading the endotracheal tube, typically a 7.0 or smaller, onto the endoscope. After the endoscope is inserted into the chosen nare, it is directed backwards and parallel to the transverse anatomic plane. Additional topicalization with aerosolized lidocaine may be required. Once past the nasopharynx, you will ideally have a direct view of the vocal cords. Once at the cords, spraying additional lidocaine onto the cords is often needed. At this point, the provider can then decide whether to administer parenteral sedation and/or a paralytic agent. When ready, the laryngoscope can be advanced through the cords. At this point, the endoscope may be used as a bougie, and the ETT can be passed through the nostril and then through the cords. Once past the cords, the laryngoscope can confirm the positioning, and the scope is then removed.18 In summary, there are a multitude of techniques that can be utilized for intubation. Overwhelmingly, literature favors the use of VL over DL. One important takeaway is there is a higher first pass success rate in method the operator has most practice with. Thus, in airways that you anticipate being difficult, use the methods and adjuncts with which you feel most comfortable. References Maguire S, Schmitt PR, Sternlicht E, et al. Endotracheal Intubation of Difficult Airways in Emergency Settings: A Guide for Innovators. Med Devices (Auckl). 2023 Jul 18;16:183-99. Prekker ME, Trent SA, Lofrano A, et al. Laryngoscopy and tracheal intubation: does use of a video laryngoscope facilitate both steps of the procedure? Ann Emerg Med. 2023 Oct;82(4):425-31. doi:10.1016/j.annemergmed.2023.02.016. Epub 2023 Apr 5. Prekker ME, Driver BE, Trent SA, et al. Video versus direct laryngoscopy for tracheal intubation of critically ill adults. N Engl J Med. 2023;389(5):418-429. Brown CA 3rd, Kaji AH, Fantegrossi A, et al. Video Laryngoscopy Compared to Augmented Direct Laryngoscopy in Adult Emergency Department Tracheal Intubations: A National Emergency Airway Registry (NEAR) Study. Acad Emerg Med. 2020 Feb;27(2):100-108. doi: 10.1111/acem.13851. Epub 2020 Jan 20. PMID: 31957174. Azam S, Khan ZZ, Shahbaz H, et al. Video Versus Direct Laryngoscopy for Intubation: Updated Systematic Review and Meta-Analysis. Cureus. 2024 Jan 5;16(1):e51720. doi: 10.7759/cureus.51720. PMID: 38322075; PMCID: PMC10846758. Ruderman BT, Mali M, Kaji AH, et al. Direct vs Video Laryngoscopy for Difficult Airway Patients in the Emergency Department: A National Emergency Airway Registry Study. West J Emerg Med. 2022 Aug 19;23(5):706-715. doi: 10.5811/westjem.2022.6.55551. PMID: 36205675; PMCID: PMC9541990. Strayer, R. (2016b, February 1). Bougie every intubation. EM. Driver BE, Prekker ME, Klein LR, et al. Effect of Use of a Bougie vs Endotracheal Tube and Stylet on First-Attempt Intubation Success Among Patients With Difficult Airways Undergoing Emergency Intubation: A Randomized Clinical Trial. JAMA. 2018;319(21):2179-89. PMID: 29800096 Driver BE, Semler MW, Self WH, et al. Effect of Use of a Bougie vs Endotracheal Tube With Stylet on Successful Intubation on the First Attempt Among Critically Ill Patients Undergoing Tracheal Intubation: A Randomized Clinical Trial.JAMA. 2021 Dec 28;326(24):2488-2497. doi: 1001/jama.2021.22002 PMID: 34879143 Gottlieb M, Holladay D, Burns K, et al. Ultrasound for airway management: An evidence-based review for the emergency clinician. J Emerg Med. 2020 May;38(5)1007-13. doi: 10.1016/j.ajem.2019.12.019. Hossein-Nejad H, Mehrjerdi MS, Abdollahi A, et al. Ultrasound for Intubation Confirmation: A Randomized Controlled Study among Emergency Medicine Residents. Ultrasound Med Biol. 2021 Feb;47(2):230-235. doi: 10.1016/j.ultrasmedbio.2020.10.012. Epub 2020 Nov 18. PMID: 33218839. Long B, Koyfman A, Gottlieb M. Diagnostic Accuracy of Ultrasound for Confirmation of Endotracheal Tube Placement. Acad Emerg Med. 2019 Sep;26(9)1096-1098. doi.org/10.11/acem.13773 Chen W, Chen J, Wang H, et al. Application of bedside real-time tracheal ultrasonography for confirmation of emergency endotracheal intubation in patients in the Intensive Care Unit. J Int Med Res. 2020;48(4):030006051989477. Epub 2019 Dec 27. Root CW, Mitchell OJL, Brown R, et al. Suction Assisted Laryngoscopy and Airway Decontamination (SALAD): A technique for improved emergency airway management. Resusc Plus. 2020 May 21;1-2:100005. doi: 10.1016/j.resplu.2020.100005. PMID: 34223292; PMCID: PMC8244406. Lin LW, Huang CC, Ong JR, et al. The suction-assisted laryngoscopy assisted decontamination technique toward successful intubation during massive vomiting simulation: A pilot before-after study. Medicine (Baltimore). 2019 Nov;98(46):e17898. doi: 10.1097/MD.0000000000017898. PMID: 31725637; PMCID: PMC6867733. Fiore MP, Marmer SL, Steuerwald MT, et al. Three Airway Management Techniques for Airway Decontamination in Massive Emesis: A Manikin Study. West J Emerg Med. 2019 Aug 6;20(5):784-90. doi: 10.5811/westjem.2019.6.42222. PMID: 31539335; PMCID: PMC6754197. Pirotte A, Panchananam V, Finley M, et al. Current Considerations in Emergency Airway Management. Curr Emerg Hosp Med Rep. 2022;10(4):73-86. doi: 10.1007/s40138-022-00255-y. Epub 2022 Dec 3. PMID: 36531125; PMCID: PMC9734887. Nickson, C. (2020, November 3). Awake intubation. Life in the Fast Lane LITFL. Salad Method Figure 1: Introduce suction into patient oropharynx Figure 2: Insert laryngoscope Figure 3: Switch suction to the left of the laryngoscope to allow room for the endotracheal tube or bougie Figure 4: Suction tip at the proximal esophagus known as the "Park Maneuver" Figure 5: Suction in the "Park Maneuver" with appropriately placed endotracheal tube Other Option for SALAD Method Photos Fiberoptic Intubation Photos Figure 6: Loading the endotracheal tube onto the fiberoptic laryngoscope Figure 7: Introducing the fiberoptic laryngoscope into the nostril, directing the scope directly posterior Figure 8: View when exiting the nasopharynx Figure 9: Advancing the endotracheal tube through the nostril Ultrasound Figure 10: The trachea (T) and collapsed esophagus (E)13 Figure 11: Trachea (T) with air filled esophagus (E) indicated esophageal intubation13 Critical Care Section News Section Related Articles Hyperviscosity Syndrome Leadership and Advancement in EM-CCM: Can We Change the Landscape Considering the Past, Looking to the Future [ Feedback → ]
2220	https://sidbanerjee.orie.cornell.edu/docs/ORIE4520/files/HW3_solns.pdf	ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) Problem 1: (Practice with Chebyshev and Chernoﬀbounds) When using concentration bounds to analyze randomized algorithms, one often has to approach the problem in diﬀerent ways depending on the speciﬁc bound being used. Typically, Chebyshev is useful when dealing with more complicated random variables, and in particular, when they are pairwise independent; Chernoﬀbounds are usually used along with the union bound for events which are easier to analyze. We’ll now go back and look at a few of our older examples using both these techniques. Part (a) (Number of collisions) Recall we showed that if we throw m balls in n bins, the average number of collisions Xm,n to be µm,n = m 2 1 n. Use Chebyshev’s inequality to show that: P[\|Xm,n −µm,n\| ≥c√µm,n] ≤1/c2. Next suppose we choose m = 2√n, then µm,n ≤1. Use Chernoﬀbounds plus the union bound to bound the probability that no bin has more than 1 ball. Compare this to the more exact analysis you did in homework 1. Solution: Let σm,n be the standard deviation of Xm,n. As we did before, for every pair of balls (i, j), let Yi,j be the indicator that the two balls collide. Note that since Yi,j are {0, 1} valued r.vs, we have E[Y 2 i,j] = E[Yi,j], and thus V ar(Yi,j) ≥E[Yi,j]. Moreover, observe that Yi,j are pairwise independent, i.e., for any (i, j) ̸= (i′, j′), the r.v.s Yi,j and Yi′,j′ are independent (in particular, note that Yi,j, Yi,k and Yj,k are pairwise independent – however they are not mutually independent). Now we have Xm,n = P i,j Yi,j, and thus µm,n = P i,j E[Yi,j] and V ar(Xm,n) = P i,j V ar(Yi,j) ≤ P i,j E[Yi,j] = µm,n. Thus, by Chebyshev’s inequality, we have: P \|Xm,n −µm,n\| ≥c√µm,n ≤1/c2. To use Chernoﬀbounds, we instead consider the number of balls in each bin. Let Zb,i be the indicator that ball i fell in bin b – these are now mutually independent. Further, let Bb = P i Zb,i be the number of balls in bin b – then we know that E[Bb] = m/n, and moreover using our standard Chernoﬀbound, we have: P [Bb ≥2] = P [Bb ≥(1 + (2n/m −1)) E[Bb]] ≤exp −(2n/m −1)2(m/n) 2 + (2n/m −1) Using P[X ≥(1 + ϵ)µ] ≤exp −ϵ2µ 2 + ϵ = exp −(2n −m)2 n(m + 2n) Finally, by the union bound, we have P[Some bin has ≥2 balls] ≤n exp −(2n−m)2 n(m+2n) . Now if m = 2√n, we get exp −(2n−m)2 n(m+2n) = Θ(1), and thus the bound on P[Some bin has ≥2 balls] is not 1 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) useful (as it grows with n). On the other hand, in the ﬁrst homework, we did a direct calculation to show that P[No collisions] ≤exp (−E[Xm,n]), and thus if m = 2√n, we have P[No collisions] ≤1/2. Thus, the Chernoﬀbound does not give us the tightest scaling in this case. Part (b) (Coupon collector) For n bins, recall that we deﬁned Ti to be the ﬁrst time when i unique bins were ﬁlled, and used these random variables to show that Tn, i.e., the number of balls we need to throw before every bin has at least one ball, satisﬁes E[Tn] = nHn = Θ(n log n). Using the same random variables, show that P[\|Tn −E[Tn]\| ≥cE[Tn]] ≤ π2 6c2H2 n . Next, suppose we throw in m = n log n+cn balls – using Chernoﬀbounds plus the union bound, choose c such that no bin is empty with probability greater than 1 −δ. Hint: Use P∞ i=1 1/i2 = π2/6 Solution: First, using the independence of Tis let’s calculate V ar[Tn]: V ar[Tn] = n−1 X i=0 V ar[Ti] = n−1 X i=0 1 −pi pi2 = n−1 X i=0 ni (n −i)2 = n X i=1 n(n −i) i2 ≤n2 n X i=1 1 i2 ≤π2n2 6 . Now, using Chebyshev’s inequality, we get: P[\|Tn −E[Tn]\| ≥cE[Tn]] = P[\|Tn −E[Tn]\| ≥cnHn] ≤V ar[Tn] (cnHn)2 ≤ π2 6c2Hn2 . Next, let m = n log n + cn, and let Bb be the number of balls in bin b. We know E[Bb] = log n + c. Then by the Chernoﬀbound, we have: P[Bb ≤0] = P[Bb ≤(1 −1)E[Bb]] ≤exp −E[Bb] 2 =≤exp −(log n + c) 2 Finally, using the union bound, we have P[Some bin is empty] ≤nP[Bb ≤0] = exp(log n−(log n+c) 2 ), and we can set c = log n + 2 log(1/δ) to get this less than δ. Here, using a direct calculation is better than the Chernoﬀbound. In particular, we have: P[Bb ≤0] = 1 −1 n m ≤e−m/n = e−c/n By the union bound, we have P[Some bin is empty] ≤e−c, and thus we need c = log(1/δ) to ensure this is less than δ. The main takeaway again is that Chernoﬀbounds are ﬁne when probabilities are small and we do not want the tightest bounds, but may be weak when probabilities are larger and we want tighter bounds. 2 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) Problem 2: (The Hoeﬀding Extension) In class, we saw that for a r.v. Xi ∼Bernoulli(pi), we have: E[eθX] = 1 −p + peθ Plugging this into the Chernoﬀbound and optimizing over θ, we obtained a variety of bounds – in particular, for independent r.vs {Xi} with X = P i Xi and µ = E[X] = P i pi, we showed for any ϵ > 0: P[\|X −µ\| ≥ϵµ] ≤2 exp −ϵ2µ 2 + ϵ We now extend this to more general bounded r.vs. Part (a) First, for any θ, argue that the function f(x) = eθx for x ∈[0, 1] is bounded above by the line joining (0, 1) and (1, eθ). Using this, ﬁnd constants α, β such that ∀x ∈[0, 1]: eθX ≤αx + β Solution: Recall that f(x) = eθx is a convex function. Now, note that f(0) = 1 and f(1) = eθ. Therefore, for x ∈[0, 1], it is bounded above by the line joining (0, 1) and (1, eθ), which means: eθx ≤(eθ −1)x + 1. Part (b) Next, for any random variable Xi taking values in [0, 1] such that E[Xi] = µi, show that: E[eθXi] ≤1 −µi + µieθ. Using this, for independent r.vs Xi taking values in [0, 1] with E[Xi] = µi, and deﬁning X = P i Xi, µ = P i µi, show that: P[\|X −µ\| ≥ϵµ] ≤2 exp −ϵ2µ 2 + ϵ (Note: You can directly use the inequality for Bernoulli r.v.s – no need to show the optimization.) Solution: Let’s take the expectations of both sides of the inequality in the solution of the part a): E[eθXi] ≤E[(eθ −1)Xi + 1] = (eθ −1)E[Xi] + 1 = (eθ −1)µ1 + 1 = 1 −µi + µieθ. Now, let X = P i Xi, µ = P i µi. Using exactly the same method is we did in class for Bernoulli r.vs, we get: P[\|X −µ\| ≥ϵµ] ≤2 exp −ϵ2µ 2 + ϵ 3 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) Part (c) (Optional) Next, for any random variable Xi taking values in [ai, bi] such that E[Xi] = µi, a similar bounding technique as above can be used to show: E h eθ(Xi−µi)i ≤exp 1 8θ2(b −a)2 (This is sometimes referred to as Hoeﬀding’s lemma – for the proof, see the wikipedia article) Now consider independent r.vs Xi taking values in [ai, bi] with E[Xi] = µi, and as before, let X = P i Xi, µ = P i µi. Using the above inequality, optimize over θ to show that: P[(X −µ) ≥ϵµ] ≤exp −2ϵ2µ2 Pn i=1(bi −ai)2 Solution: As in the standard Chernoﬀinequality, we have: P[X −µ ≥ϵµ] = P h eθ(X−µ) ≥eθϵµi ≤min θ>0 E[eθ(X−µ)] eθϵµ = min θ>0 ΠiE[eθ(Xi−µi)] eθϵµ ≤min θ>0 exp θ2 8 X i (bi −ai)2 −θϵµ ! To optimize, we choose θ∗= 4ϵµ/ P i(bi −ai)2, to get: P[X −µ ≥ϵµ] ≤exp − 2ϵ2µ2 P i(bi −ai)2 Problem 3: (A Weaker Sampling Theorem: Adapted from MU Ex 4.9) In class we saw the following ‘sampling theorem’ for estimating the mean of a {0, 1}-valued random variable: In order to get an estimate within ±ϵ with conﬁdence 1 −δ, we need n ≥2+ϵ ϵ ln 2 δ. A crucial component in this proof was using the Chernoﬀbound for Bernoulli(p) random variables. Suppose instead we want to estimate a more general random variable X (for example, the average number of hours of TV watched by a random person) – we may not be able to use a Chernoﬀbound if we do not know the moment generating function We now show how to to get a similar sampling theorem which only uses knowledge of the mean and variance of aX. We want to estimate a r.v. X with mean E[X] and variance V ar[X], given i.i.d samples X1, X2, . . .. Let r = p V ar[X]/E[X] – we now show that we can estimate it up to accuracy ±ϵE[X] and conﬁdence 1 −δ using O ϵ2 r2 ln 1 δ samples. 4 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) Part (a) Given n samples X1, . . . , Xn, suppose we use the estimator b X = (Pn i=1 Xi) /n. Show that n = O(r2/ϵ2δ) is suﬃcient to ensure: P h \| b X −E[X]\| ≥ϵE[X] i ≤δ Solution: By linearity of expectation, we have that E[ b X] = Pn i=1 E[Xi]/n = E[X]. Moreover, since Xis are i.i.d., we have that V ar b X = Pn i=1 V ar(Xi)/n2 = V ar(X)/n = r2E[X]2/n. Now, using Chebyshev’s inequality we get: P h \| b X −E[X]\| ≥ϵE[X] i ≤V ar( b X) ϵ2E[X]2 = r2 nϵ2 To ensure P h \| b X −E[X]\| ≥ϵE[X] i ≤δ, we can choose n such that r2 nϵ2 ≤δ. Using n = O(r2/ϵ2δ) samples is thus suﬃcient. Part (b) We say an estimator is a weak estimator if it satisﬁes that P h \| b X −E[X]\| ≥ϵE[X] i ≤1/4 – using part (a), show, that we need O(r2/ϵ2) samples to obtain a weak estimator. Now suppose we are given m weak estimates b X1, b X2, . . . , b Xm, and we deﬁne a new estimator e X to be the median of these weak estimates. Show that using m = O(ln(1/δ) weak estimates gives us an estimate e X that satisﬁes: P h \| e X −E[X]\| ≥ϵE[X] i ≤δ What could go wrong if we used the mean of b X1, b X2, . . . , b Xm instead of the median? NOTE: I got the deﬁnition of weak estimator inverted in the Homework - this is the correct deﬁni-tion. Essentially, we want the probability of samples being close to the mean to be > 1/2. Solution: If we take δ = 1 4 in part a), we see that b X is a weak estimator if we use n = O(r2/ϵ2) samples. Now we want to show that we need O(log 1/δ) such weak estimators to ensure that the median of these estimators is within (1 ± ϵ)E[X]. Let us introduce new r.vs Yi s.t. Yi = 1, if \| b Xi −E[X]\| ≤ϵE[X], and Yi = 0, if \| b Xi −E[X]\| ≥ ϵE[X]. Y = Pm i=i Yi is the number of weak estimates for which P h \| b X −E[X]\| ≤ϵE[X] i ≤δ; to ensure that the median is also within (1 ± ϵ)E[X], we need Y > m/2. In other words, we have: P h \| e X −E[X]\| ≥ϵE[X] i = P [Y ≤m/2] . 5 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) Moreover, P[Yi = 0] ≤3/4, and hence E[Y ] ≥3m/4. Now for independent random variables Zi ∼ Bernoulli(pi), with Z = Pn i=1 Zi, we saw the following ChernoﬀBound: P[Z ≤(1 −ϵ)E[Z]] ≤exp −ϵ2 2 E[Z] Using this, we can write: P [Y ≤m/2] = P [Y ≤(1 −1/3)3m/4] ≤P[Y ≤(1 −1/3)E[Y ]] ≤e −(1/9)E[Y ] 2 ≤e−m/18. Now, if we take m ≥18 ln(1/δ) = O(log 1/δ), we get P h \| b X −E[X]\| ≤ϵE[X] i ≤δ. Note though that we only needed to count the weak estimators that fell outside (1 ± ϵ)E[X] – we did not need to assume they are bounded, or have bounded variance, etc. If instead we had used the mean of the weak estimators, we could have a problem if these bad estimates happened to take very large values. Using the median made our estimate robust to such outliers. Problem 4: (Randomized Set-Cover) In this problem, we’ll look at randomized rounding, which is a very powerful technique for solving large-scale combinatorial optimization problems. The main idea is that given a problem which can be written as an optimization problem with integer constraints, we can sometimes solve the relaxed problem with non-integer constraints, and then round the solutions to get a good assignment. We will highlight this technique for the Minimum Set-Cover problem. We are given a collection of m subsets {S1, S2, . . . , Sm} which are subsets of some large set U of n elements, such that S i Si = U. The Minimum Set-Cover problem is that of selecting the smallest number of sets C from the collection {S1, S2, . . . , Sm} such that they cover U, i.e., such that each element in U lies in at least one of the sets in C. Part (a) Argue that the minimum-set cover problem is equivalent to the following integer program: Minimize x X i xi subject to X i\|e∈Si xi ≥1, e ∈U xi ∈{0, 1}, i ∈{1, 2, . . . , m} Let the solution to this problem, i.e., the minimum set-cover, be denoted OPT. Next, argue that if we solve the same problem, but now change the last constraint to xi ∈[0, 1] for all i, then the resulting solution OPTLP of this relaxed problem obeys OPTLP ≤OPT. Note that the relaxed problem is an LP and hence can be solved eﬃciently. 6 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) Solution: For each set Si, we associate a variable xi ∈{0, 1} that indicates of we want to choose Si or not. We can then write the solutions for Minimum Set-Cover problem as a vector x ∈{0, 1}m. The objective is clearly to minimize the sum of these variables – moreover, since the sets we select must cover each element in U, we need one constraint to ensure this for each element. If we replace each constraint xi ∈{0, 1} with xi ∈[0, 1] for all i, then the resulting problem can be solved in a polynomial time as it is a Linear Program (LP). However, by replacing integer constraints with continuous domains, we have expanded the set of feasible solutions – thus, the resulting minimization problem must give a smaller value (as all feasible integer solutions are within our new feasible region), and hence we have that OPTLP ≤OPT. Part (b) Given a solution z to the relaxed LP, we now round the values to obtain a feasible solution for the original minimum set-cover problem. For each set Si, we generate k = c log n i.i.d Bernoulli(zi) random variables Xi,1, Xi,2, . . . , Xi,k – if any of them is 1, then we set xi = 1, i.e., we add Si to our cover C. Prove that the resulting set-cover obeys E[\|C\|] ≤c log n · OPT. Solution: First, from the deﬁnition of the rounding process, for any j ∈{1, 2, . . . , k}, we have: X i E[Xi,j] = X i P[Xi,j = 1] = X i zi = OPTLP Therefore, by linearity of expectation, we have: E[\|C\|] ≤ X i c log n X j=1 E[Xi,j] = c log n · OPTLP ≤c log n · OPT. Part (c) Finally, choose c to ensure that the probability that the resulting set-cover C does not cover any element e ∈U is less than 1/n2. Solution: For any element e ∈U, and for any j ∈{1, 2, . . . , c log n}, let Ye,j = P i\|e∈Si Xi,j, and let Ye = Pc log n j=1 Ye,j. Note that for the sets Si containing element e, if any of the Xi,j = 1, then e is covered – in other words, P[e is covered] = P[Ye ≥1]. Moreover, E[Ye,j] = P i\|e∈Si E[Xi,j] = P i\|e∈Si zi. Let ke = \|{i\|e ∈Si}\|; since e is fractionally covered in the LP, we have z1 +· · ·+zke ≥1, and thus E[Ye] ≥c log n. Now using our basic Chernoﬀbound, we have: P[Ye ≤0] = P [Ye ≤(1 −1)E[Ye]] ≤e −E[Ye] 2 ≤e −c log n 2 = 1 nc/2 Thus P[e is not covered] ≤n−c/2. Moreover, by the union bound, we have that: P[Any element e ∈U is not covered] ≤n1−c/2. 7 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) If c ≥6, we get 1 −c/2 = −2, and hence P[Every element e ∈U is covered] ≥1 −n−2. (Note: You can actually get a tighter bound of c ≥3 using the following argument: First, we need to observe that probability of e being covered is minimized when zi are all equal, i.e., z1 = · · · = zke = 1/ke (this is not obvious - try to prove it. . .). Thus we get for any j: P  X i\|e∈Si Yi,j = 0  ≤(1 −z1) · · · (1 −zke) ≤ 1 −1 ke ke ≤1 e. Therefore, each element is covered with probability at least 1 −1/e if we draw only one sample of each Xi. Since we draw c log n samples, the probability that element e is not covered is: P[e is not covered] ≤ 1 e c log n = 1 nc . Now via the union bound, we see that if we take c = 3, then we’ll have: P[Any element e ∈U is not covered] ≤1 n2 . Problem 5: (Papadimitrou’s 2SAT Algorithm) (Optional) The 2SAT problem (Wikipedia entry), and more generally, the boolean satisﬁability (SAT) prob-lem (Wikipedia entry) are one of the cornerstones of theoretical algorithms, and also a very useful modeling tool for a variety of optimization problems. In the general SAT problem, we want to ﬁnd a satisfying assignment for a given a Boolean expression in n Boolean (i.e., {0, 1}, or FALSE/TRUE) variables {X1, X2, . . . , Xn} typically involving conjunctions (i.e., logical AND, denoted as ∧), dis-junctions (i.e., logical OR, denoted as ∨) and negations (logical NOT; typically X denotes the negation of a variable X). In 2SAT, the expression is restricted to being a conjunction (AND) of several clauses, where each clause is the disjunction (OR) of two literals (either a variable or its negation). For example, the expression (X1 ∨X2) ∧(X1 ∨X3) ∧(X2 ∨X3) is a 2SAT formula, which has several satisfying assignments including (X1 = 1, X2 = 1, X3 = 1). Note that in order to ﬁnd a satisfying assignment, we need to set the variables such that each clause in the formula has at least one literal which is TRUE. Although the general SAT problem is known to be NP-complete, 2SAT can be solved in polynomial time – we will now see a simple randomized algorithm that demonstrates this fact: Papapdimitrou’s 2SAT Algorithm): Given a 2CNF formula F involving n Boolean variables, and an arbitrary assignment τ, we check if τ satisﬁes F. If not, we pick an arbitrary unsatisﬁed clause, pick one of its literals uniformly at random, and ﬂip it to get a new assignment τ ′. We then repeat this until we ﬁnd a satisfying assignment. Part (a) Assume that F has a unique satisfying assignment τ ∗, and for any assignment τ, let N(τ) be the number of literals in τ which agree (i.e., have the same value) as the corresponding literal in 8 ORIE 4520: Stochastics at Scale Fall 2015 Homework 3: Solutions Sid Banerjee (sbanerjee@cornell.edu) τ. Argue that each time we execute an iteration of Papadimitrou’s 2SAT algorithm with input assignment τ, the new assignment τ ′ satisﬁes: N(τ ′) = ( N(τ) + 1 with probability at least 1 2 N(τ) −1 with probability at most 1 2 Solution: Given any unsatisﬁed clause, we know that in τ ∗at least one of the literals is ﬂipped (it could be both are ﬂipped). Since we choose to ﬂip a uniform random literal out of the two, we are guaranteed that we increase the agreement between our guess τ and τ ∗by 1 with probability at least 1/2. Part (b) Based on the above, argue that the running time Tn of the algorithm is upper bounded by the ﬁrst time that a symmetric random walk starting from 0 hits n or −n (equivalently, Tn = arg minK>0 n \| PK i=1 Xi\| = n o , where Xi are i.i.d Rademacher random variables). Next, show that E[Tn] = n2, and thus, prove that after O(n3) iterations, Papadimitrou’s algorithm terminates with probability greater than 1 −1/n. Solution: See this lecture (and the next) from Tim Roughgarden’s algorithms course for a beau-tiful exposition of this proof. One thing you should note: although running the algorithm till O(n3) steps gives the desired probability, one can get much better bounds if we instead stop after O(n2) steps, and then restart the process. Markov’s tells us that after 2n2 steps, we ﬁnd the solution with probability at least 1/2 – now from previous assignments, you know that doing O(log n) independent trials is suﬃcient to amplify the probability to 1 −1/n. However, the analysis works for any starting point – so this does suggest that O(n2 log n) steps of Papadimitrou’s algorithm was suﬃcient... Is this surprising? Not really – basically what this says is that the failure probability does follow a ‘Chernoﬀ-style’ exponential decay. We could not show this directly as the random variables were not independent – however, Markov’s inequality still works, and we can exploit that in a clever way to get our result. 9
2221	https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems?srsltid=AfmBOopyFShtCsXFw1Fo1K1dUy4ucT8XR0erFmxU_nj2uc64ODgj11_M	Art of Problem Solving 2022 AMC 10A Problems - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. 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CHECK SCHEDULE 2022 AMC 10A Problems 2022 AMC 10A (Answer Key) Printable versions: Wiki • AoPS Resources • PDF Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator). Figures are not necessarily drawn to scale. You will have 75 minutes working time to complete the test. 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 Contents [hide] 1 Problem 1 2 Problem 2 3 Problem 3 4 Problem 4 5 Problem 5 6 Problem 6 7 Problem 7 8 Problem 8 9 Problem 9 10 Problem 10 11 Problem 11 12 Problem 12 13 Problem 13 14 Problem 14 15 Problem 15 16 Problem 16 17 Problem 17 18 Problem 18 19 Problem 19 20 Problem 20 21 Problem 21 22 Problem 22 23 Problem 23 24 Problem 24 25 Problem 25 26 See also Problem 1 What is the value of Solution Problem 2 Mike cycled laps in minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first minutes? Solution Problem 3 The sum of three numbers is The first number is times the third number, and the third number is less than the second number. What is the absolute value of the difference between the first and second numbers? Solution Problem 4 In some countries, automobile fuel efficiency is measured in liters per kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals miles, and gallon equals liters. Which of the following gives the fuel efficiency in liters per kilometers for a car that gets miles per gallon? Solution Problem 5 Square has side length . Points , , , and each lie on a side of such that is an equilateral convex hexagon with side length . What is ? Solution Problem 6 Which expression is equal to for Solution Problem 7 The least common multiple of a positive integer and is , and the greatest common divisor of and is . What is the sum of the digits of ? Solution Problem 8 A data set consists of (not distinct) positive integers: , , , , , and . The average (arithmetic mean) of the numbers equals a value in the data set. What is the sum of all possible values of ? Solution Problem 9 A rectangle is partitioned into regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible? Solution Problem 10 Daniel finds a rectangular index card and measures its diagonal to be centimeters. Daniel then cuts out equal squares of side cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be centimeters, as shown below. What is the area of the original index card? Solution Problem 11 Ted mistakenly wrote as What is the sum of all real numbers for which these two expressions have the same value? Solution Problem 12 On Halloween, children walked into the principal's office asking for candy. They can be classified into three types: Some always lie; some always tell the truth; and some alternately lie and tell the truth. The alternaters arbitrarily choose their first response, either a lie or the truth, but each subsequent statement has the opposite truth value from its predecessor. The principal asked everyone the same three questions in this order. "Are you a truth-teller?" The principal gave a piece of candy to each of the children who answered yes. "Are you an alternater?" The principal gave a piece of candy to each of the children who answered yes. "Are you a liar?" The principal gave a piece of candy to each of the children who answered yes. How many pieces of candy in all did the principal give to the children who always tell the truth? Solution Problem 13 Let be a scalene triangle. Point lies on so that bisects The line through perpendicular to intersects the line through parallel to at point Suppose and What is Solution Problem 14 How many ways are there to split the integers through into pairs such that in each pair, the greater number is at least times the lesser number? Solution Problem 15 Quadrilateral with side lengths is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form where and are positive integers such that and have no common prime factor. What is Solution Problem 16 The roots of the polynomial are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by units. What is the volume of the new box? Solution Problem 17 How many three-digit positive integers are there whose nonzero digits and satisfy (The bar indicates repetition, thus is the infinite repeating decimal ) Solution Problem 18 Let be the transformation of the coordinate plane that first rotates the plane degrees counterclockwise around the origin and then reflects the plane across the -axis. What is the least positive integer such that performing the sequence of transformations returns the point back to itself? Solution Problem 19 Define as the least common multiple of all the integers from to inclusive. There is a unique integer such that What is the remainder when is divided by ? Solution Problem 20 A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are , , and . What is the fourth term of this sequence? Solution Problem 21 A bowl is formed by attaching four regular hexagons of side to a square of side . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? Solution Problem 22 Suppose that cards numbered are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards are picked up on the first pass, and on the second pass, on the third pass, on the fourth pass, and on the fifth pass. For how many of the possible orderings of the cards will the cards be picked up in exactly two passes? Solution Problem 23 Isosceles trapezoid has parallel sides and with and There is a point in the plane such that and What is Solution Problem 24 How many strings of length formed from the digits , , , , are there such that for each , at least of the digits are less than ? (For example, satisfies this condition because it contains at least digit less than , at least digits less than , at least digits less than , and at least digits less than . The string does not satisfy the condition because it does not contain at least digits less than .) Solution Problem 25 Let , , and be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the -axis. The left edge of and the right edge of are on the -axis, and contains as many lattice points as does . The top two vertices of are in , and contains of the lattice points contained in See the figure (not drawn to scale). The fraction of lattice points in that are in is times the fraction of lattice points in that are in . What is the minimum possible value of the edge length of plus the edge length of plus the edge length of ? Solution See also 2022 AMC 10A (Problems • Answer Key • Resources) Preceded by 2021 Fall AMC 10B ProblemsFollowed by 2022 AMC 10B Problems 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 10 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 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2222	http://medianetlab.ee.ucla.edu/SocalNEGT2014/Barman.pdf	An Approximate Version of Carath´ eodory’s Theorem and Its Algorithmic Applications Siddharth Barman California Institute of Technology Carath´ eodory’s Theorem Any vector in the convex hull of a set V in Rd can be expressed as a convex combination of at most d + 1 vectors of V . v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd w v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd w v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors Approx. Carath´ eodory’s Theorem Given set V in the p-unit ball with norm p ≥2, for every vector in the convex hull of V there exists an ε-close (under p-norm distance) vector that is a convex combination of O p ε2 vectors of V . Application: Algorithm for Approximate Nash Equilibria      2 7 · · · 1 8 2 · · · 8 . . . . . . ... . . . 18 28 · · · 4     ,      3 1 · · · 4 1 5 · · · 9 . . . . . . ... . . . 26 5 · · · 35      Payoﬀs Nash Equilibrium Prob. Player 1 Prob. Player 2 Algorithm . .      2 7 · · · 1 8 2 · · · 8 . . . . . . ... . . . 18 28 · · · 4     ,      3 1 · · · 4 1 5 · · · 9 . . . . . . ... . . . 26 5 · · · 35      Payoﬀs Nash Equilibrium Prob. Player 1 Prob. Player 2 Algorithm . . Nash equilibrium in two-player games is PPAD-hard [GP06, DGP06, CD06, CDT09].      2 7 · · · 1 8 2 · · · 8 . . . . . . ... . . . 18 28 · · · 4     ,      3 1 · · · 4 1 5 · · · 9 . . . . . . ... . . . 26 5 · · · 35      Payoﬀs Nash Equilibrium Prob. Player 1 Prob. Player 2 Algorithm Hard even in two-player games [DGP06, CDT09] . .      2 7 · · · 1 8 2 · · · 8 . . . . . . ... . . . 18 28 · · · 4     ,      3 1 · · · 4 1 5 · · · 9 . . . . . . ... . . . 26 5 · · · 35      Payoﬀs Approx. Nash Eq. Prob. Player 1 Prob. Player 2 Algorithm Hard even in two-player games [DGP06, CDT09] . .      2 7 · · · 1 8 2 · · · 8 . . . . . . ... . . . 18 28 · · · 4     ,      3 1 · · · 4 1 5 · · · 9 . . . . . . ... . . . 26 5 · · · 35      Payoﬀs Approx. Nash Eq. Prob. Player 1 Prob. Player 2 Algorithm Hard even in two-player games [DGP06, CDT09] . . Focus: Two-Player Games      2 7 · · · 1 8 2 · · · 8 . . . . . . ... . . . 18 28 · · · 4     ,      3 1 · · · 4 1 5 · · · 9 . . . . . . ... . . . 26 5 · · · 35      Payoﬀs Approx. Nash Eq. Prob. Player 1 Prob. Player 2 Algorithm Hard even in two-player games [DGP06, CDT09] . . Focus: Two-Player Games Payoﬀmatrices A and B of size n × n      1 2 · · · n 1 A11 A12 · · · A1n 2 A21 A22 · · · A2n . . . . . . . . . ... . . . n An1 An2 · · · Ann           1 2 · · · n 1 B11 B12 · · · B1n 2 B21 B22 · · · B2n . . . . . . . . . ... . . . n Bn1 Bn2 · · · Bnn      Payoﬀmatrices A and B of size n × n Probability vectors over [n]: x and y Payoﬀmatrices A and B of size n × n Probability vectors over [n]: x and y Nash equilibrium (x, y): No player can beneﬁt by unilateral deviation eT i Ay ≤xT Ay ∀i ∈[n] and xT Bej ≤xT By ∀j ∈[n] Payoﬀmatrices A and B of size n × n Probability vectors over [n]: x and y Nash equilibrium (x, y): No player can beneﬁt by unilateral deviation eT i Ay ≤xT Ay ∀i ∈[n] and xT Bej ≤xT By ∀j ∈[n] Approximate Nash equilibrium (x, y): No player can beneﬁt more than ε by unilateral deviation eT i Ay ≤xT Ay + ε ∀i ∈[n] and xT Bej ≤xT By + ε ∀j ∈[n] Computation of Eq. in Two-Player Games Nash Equilibria General Games: Exp. time [Lemke & Howson 1964] Zero-Sum Games: Poly. time [von Neumann 1928, Dantzig 1951] Computation of Eq. in Two-Player Games Nash Equilibria General Games: Exp. time [Lemke & Howson 1964] Zero-Sum Games: Poly. time [von Neumann 1928, Dantzig 1951] Approximate Nash Equilibria General Games: nO(log n) [Lipton et al. 2003] Low-Rank Games: (1/ε)rank [Alon et al. 2013] Computation of Eq. in Two-Player Games Nash Equilibria General Games: Exp. time [Lemke & Howson 1964] Zero-Sum Games: Poly. time [von Neumann 1928, Dantzig 1951] Approximate Nash Equilibria General Games: nO(log n) [Lipton et al. 2003] Low-Rank Games: (1/ε)rank [Alon et al. 2013] This Talk: Sparsity Deﬁnition (Sparsity of a Game) The sparsity of a game (A, B) is deﬁned to be the maximum number of non-zero entries in any column of A + B. Deﬁnition (Sparsity of a Game) The sparsity of a game (A, B) is deﬁned to be the maximum number of non-zero entries in any column of A + B. • Sparsity = 0 in zero-sum games Deﬁnition (Sparsity of a Game) The sparsity of a game (A, B) is deﬁned to be the maximum number of non-zero entries in any column of A + B. • Sparsity = 0 in zero-sum games • In general, sparsity is at most n Deﬁnition (Sparsity of a Game) The sparsity of a game (A, B) is deﬁned to be the maximum number of non-zero entries in any column of A + B. Theorem In a two-player s-sparse game an ε-Nash equilibrium can be computed in time nO(log s/ε2). Payoﬀmatrices normalized A, B ∈[−1, 1]n×n. Deﬁnition (Sparsity of a Game) The sparsity of a game (A, B) is deﬁned to be the maximum number of non-zero entries in any column of A + B. Theorem In a two-player s-sparse game an ε-Nash equilibrium can be computed in time nO(log s/ε2). Implications: • When s is a ﬁxed constant we get a polynomial-time algorithm • For general games (s ≤n) the running time matches the best-known upper bound: nO(log n/ε2) [LMM’03]. Deﬁnition (Sparsity of a Game) The sparsity of a game (A, B) is deﬁned to be the maximum number of non-zero entries in any column of A + B. Theorem In a two-player s-sparse game an ε-Nash equilibrium can be computed in time nO(log s/ε2). Note: Sparse Games are Hard [CDT 2006]. Nash eq: eT i Ay ≤xT Ay ∀i and xT Bej ≤xT By ∀j Bilinear Program for Nash Eq. [MS’64] maximize xT (A + B)y −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] Nash eq: eT i Ay ≤xT Ay ∀i and xT Bej ≤xT By ∀j Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] Say (x∗, y∗) is a Nash eq. Given u∗= Cy∗we get an LP. maximize xT u∗−π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] Cy = u∗ Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] Say (x∗, y∗) is a Nash eq. Given u∗= Cy∗we get an LP. maximize xT u∗−π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] Cy = u∗ A vector close to Cy∗is suﬃcient to ﬁnd an approx. Nash eq. Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] C1 C2 C3 C4 C5 C6 C7 . . Matrix C Cy∗ Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] C1 C2 C3 C4 C5 C6 C7 . . Matrix C Cy∗ w′ ε Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] C1 C2 C3 C4 C5 C6 C7 . . Matrix C Cy∗ w′ ε Idea: Exhaustively search for w′ Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] C1 C2 C3 C4 C5 C6 C7 . . Matrix C Cy∗ w′ ε Idea: Exhaustively search for w′, by enumerating subsets of columns of C. Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] C1 C2 C3 C4 C5 C6 C7 . . Matrix C Cy∗ w′ ε Theorem In a two-player s-sparse game an ε-Nash equilibrium can be computed in time nO(log s/ε2). Bilinear Program for Nash Eq. [MS’64] maximize xT Cy −π1 −π2 subject to xT B ≤π2 and Ay ≤π1 x, y ∈∆n and π1, π2 ∈[−1, 1] C1 C2 C3 C4 C5 C6 C7 . . Matrix C Cy∗ w′ ε General Result We can eﬃciently approximate any sparse bilinear or quadratic form over the simplex. ✓Application: Algorithm for Approximate Nash Equilibria v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors Approx. Carath´ eodory’s Theorem Given set V in the p-unit ball with norm p ≥2, for every vector in the convex hull of V there exists an ε-close (under p-norm distance) vector that is a convex combination of O p ε2 vectors of V . v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors Approx. Carath´ eodory’s Theorem Given set V in the p-unit ball with norm p ≥2, for every vector in the convex hull of V there exists an ε-close (under p-norm distance) vector that is a convex combination of O p ε2 vectors of V . Proof via the Probabilistic Method and Khintchine-Kahane Inequality Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm Approx. Birkhoﬀ-von Neumann Theorem For every d × d doubly stochastic matrix D there exists an ε-close (under the entrywise ∞-norm) doubly stochastic matrix D′ such that D′ is a convex combination of O log d ε2 permutation matrices. Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem . . d + 1colors in Rd Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem . . d + 1colors in Rd Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem . . d + 1colors in Rd Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem . . d + 1colors in Rd w Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem . . d + 1colors in Rd w Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem . . d + 1colors in Rd w ε w′ Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem Barman, An Approximate Version of Carath´ eodory’s Theorem with Applications to Approximating Nash Equilibria and Dense Bipartite Subgraphs, 2014 Extensions • Convex hull of matrices with entrywise norm and Schatten p-norm • Colorful Carath´ eodory Theorem Barman, An Approximate Version of Carath´ eodory’s Theorem with Applications to Approximating Nash Equilibria and Dense Bipartite Subgraphs, 2014 Thank You! v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors Approx. Carath´ eodory’s Theorem Given set V in the p-unit ball with norm p ≥2, for every vector in the convex hull of V there exists an ε-close (under p-norm distance) vector that is a convex combination of O p ε2 vectors of V . v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors Proof via the Probabilistic Method v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors w = P i αivi where P i αi = 1 and αi ≥0 v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors w = P i αivi where P i αi = 1 and αi ≥0 Sample vi with probability αi v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors w = P i αivi where P i αi = 1 and αi ≥0 Sample vi with probability αi Empirical mean of m i.i.d. samples s1, s2, . . . , sm w = E " 1 m m X i=1 si # v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors w = P i αivi where P i αi = 1 and αi ≥0 Sample vi with probability αi Consider g(s1, s2, . . . , sm) := 1 m m X i=1 si −w p With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m (proof via Mcdiarmid’s inequality) With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m • Bounded expectation: E[g] ≤ε, for m = O(p/ε2) With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m • Bounded expectation: E[g] ≤ε, for m = O(p/ε2) (proof via Khintchine inequality) With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m • Bounded expectation: E[g] ≤ε, for m = O(p/ε2) • Overall: Pr (g > 2ε) ≤e−ε2m With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m • Bounded expectation: E[g] ≤ε, for m = O(p/ε2) • Overall: Pr (g > 2ε) ≤e−ε2m v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m • Bounded expectation: E[g] ≤ε, for m = O(p/ε2) • Overall: Pr (g > 2ε) ≤e−ε2m v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors w′ := 1 m Pm i=1 si With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m • Bounded expectation: E[g] ≤ε, for m = O(p/ε2) • Overall: Pr(∥w′ −w∥p > 2ε) ≤e−ε2m v1 v2 v3 v4 v5 v6 v7 . . V ⊂Rd Norm p ≥2 w w′ ε O p ε2 vectors w′ := 1 m Pm i=1 si With i.i.d. samples s1, . . . , sm ∼w, g(s1, . . . , sm) := 1 m Pm i=1 si −w p • g concentrates: Pr (\|g −E[g]\| > ε) ≤e−ε2m • Bounded expectation: E[g] ≤ε, for m = O(p/ε2) • Overall: Pr(∥w′ −w∥p > 2ε) ≤e−ε2m Approx. Carath´ eodory’s Theorem Given set V in the p-unit ball with norm p ≥2, for every vector in the convex hull of V there exists an ε-close (under p-norm distance) vector that is a convex combination of O p ε2 vectors of V . Khintchine-Kahane Inequality [TJ74,S11] Let r1, r2, . . . , rm be a sequence of i.i.d. random variables with Pr(ri = ±1) = 1 2 In addition, let u1, u2, . . . , um ∈Rd be a deterministic se-quence of vectors. Then, for 2 ≤p < ∞ E m X i=1 riui p ≤√p m X i=1 ∥ui∥2 p ! 1 2
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2224	http://www.chemguideforcie.co.uk/section93/learningd.html	chemguide: CIE A level chemistry support: Learning outcome 9.3(d) Chemguide: Support for CIE A level Chemistry Learning outcome 9.3(d) This statement covers a lot of chemistry of the oxides of Group 4 elements. Before you go on, you should find and read the statement in your copy of the syllabus. In all honesty, this is boring stuff - but it is an area where CIE tend to ask questions. However boring it is, you must take the trouble to learn as much of it as you can. You should check as many past papers as you can so that you can see which are the things most likely to come up, and how much you need to learn as opposed to working it out if you have to. The oxides The elements form two ranges of oxides - monoxides and dioxides. That is because the elements have two possible oxidation states, +2 and +4. Monoxides include carbon monoxide, CO, tin(II) oxide, SnO, and lead(II) oxide, PbO. Dioxides include CO 2, SiO 2, SnO 2 and PbO 2. The two non-metal oxides would normally be called carbon dioxide and silicon dioxide. The metal oxides would normally be called tin(IV) oxide and lead(IV) oxide. But you may find the names used fairly interchangeably, so that lead(IV) oxide is often called lead dioxide, for example. Lead also has another oxide with the formula Pb 3 O 4, called trilead tetroxide. This is a bright red solid known popularly as "red lead". Pb 3 O 4 behaves chemically as if it were a mixture of PbO and PbO 2 in the ratio 2PbO.PbO 2. You will come across this again later on this page. Bonding in the oxides In the non-metallic elements from carbon to germanium, the oxides are covalent. For CIE purposes, you can think of the metal oxides (of tin and lead) as being ionic. The CIE teacher support material says that SnO 2 and PbO 2 are most easily described as ionic, but with a degree of covalency - and so that is what you need to learn. It is worth looking separately at a few of the oxides in more detail. The oxides of carbon The fact that the two oxides of carbon are gases must mean that they are simple molecular substances. Carbon monoxide can be thought of as having two ordinary covalent bonds between the carbon and the oxygen plus a co-ordinate bond using a lone pair on the oxygen atom. In carbon dioxide, carbon forms simple molecules with oxygen because of the possibility of carbon-oxygen double bonds. None of the other elements in Group 4 form double bonds with oxygen, and so that forces completely different structures on them. Silicon and germanium dioxides Neither of these elements form double bonds with oxygen. Mainly for interest:Silicon atoms are bigger than carbon atoms. That means that silicon-oxygen bonds will be longer than carbon-oxygen bonds. To make a double bond, you have to have sideways overlap between p orbitals on adjacent atoms. With the longer silicon-oxygen bonds, the p orbitals on the silicon and the oxygen aren't quite close enough together to allow enough sideways overlap to give a stable pi bond. So, silicon bonds with oxygen in such a way that only single bonds are formed. Similarly with germanium. There are various different structures for silicon dioxide. The easiest to remember and draw is: This is based on a diamond structure with each of the silicon atoms being bridged to its other four neighbours via an oxygen atom. This means that silicon dioxide is a giant covalent structure. The strong bonds in three dimensions make it a hard, high melting point solid. Exactly the same is true of germanium dioxide. The tin and lead oxides If you think of these as ionic, then SnO and PbO will have giant ionic lattices consisting of Sn 2+ or Pb 2+ ions and oxide ions. SnO 2 and PbO 2 will have giant ionic lattices consisting of Sn 4+ or Pb 4+ ions and oxide ions. The acid-base behaviour of the oxides Before we get bogged down in detail, here is a summary of the important points. The oxides of the elements at the top of Group 4 are slightly acidic, but the acidity of the oxides falls as you go down the Group. Towards the bottom of the Group, the oxides have basic properties as well as acidic ones - in other words, they are amphoteric. The trend is therefore from acidic oxides at the top of the Group towards amphoteric ones at the bottom. Carbon and silicon dioxides are acidic. (There is a question mark over carbon monoxide - see later.) The germanium, tin and lead oxides are all amphoteric. The dioxides of germanium, tin and lead tend to have more acidic character than the monoxides. The oxides of carbon Carbon monoxide is only slightly soluble in water and doesn't react with it. It is normally counted as a neutral oxide. CIE describe it as neutral in their teacher support material - and so that is what you must learn. Note:Carbon monoxide does in fact react with hot concentrated sodium hydroxide solution to produce sodium methanoate. The fact that the carbon monoxide reacts with the basic hydroxide ion shows that it must be acidic. Carbon dioxide is weakly acidic. With water Carbon dioxide reacts with water to a slight extent to produce hydrogen ions (strictly, hydroxonium ions) and hydrogencarbonate ions. Overall, this reaction is: The solution of carbon dioxide in water is sometimes known as carbonic acid, but in fact only about 0.1% of the carbon dioxide has actually reacted. The position of equilibrium is well to the left-hand side. With bases Carbon dioxide reacts with sodium hydroxide solution in the cold to give either sodium carbonate or sodium hydrogencarbonate solution - depending on the reacting proportions. Silicon dioxide Silicon dioxide doesn't react with water, because of the difficulty of breaking up the giant covalent structure. Silicon dioxide does react with sodium hydroxide solution, but only if it is hot and concentrated. Sodium silicate solution is formed. Silicon dioxide is weakly acidic. Germanium, tin and lead oxides The monoxides All of these oxides are amphoteric - they show both basic and acidic properties. The basic nature of the oxides is shown by their reaction with acids to form salts. For example, they all react with hydrochloric acid. This can be summarised as: For example, with tin(II) oxide, you get tin(II) chloride and water. The acidic nature of the oxides is shown by their reaction with bases like sodium hydroxide solution. In general terms: For example, with tin(II) oxide, you get sodium stannate(II) solution and water. The stannate name is given to ions where tin is a part of the negative ion. The (II), of course, shows the tin in the +2 oxidation state. Note:There are various versions of the formulae for compounds like sodium stannate(II). This is the simplest one, and is acceptable to CIE. The dioxides These dioxides are again amphoteric - showing both basic and acidic properties. The basic nature of the dioxides is shown by their reaction with concentrated hydrochloric acid to give compounds of the type XCl 4: Note:The XCl 4 actually goes on to react with excess chloride ions in the hydrochloric acid to give complexes such as XCl 6 2-, but this isn't required by CIE. In the case of lead(IV) oxide, the reaction has to be done with ice-cold hydrochloric acid. If the reaction is done any warmer, the lead(IV) chloride decomposes to give lead(II) chloride and chlorine gas. This is an effect of the preferred oxidation state of lead being +2 rather than +4. The reaction with hydrochloric acid isn't typical of lead(IV) oxide, which doesn't react with most acids. The acidic nature of the dioxides is shown by their reaction with hot concentrated sodium hydroxide solution or molten sodium hydroxide. In general terms: For example, with tin(IV) oxide, you get sodium stannate(IV) and water. Note:As in the case of the similar compound formed with tin(II) oxide, there are various formulae for the product in this reaction. This is the simplest, and is acceptable to CIE. The special case of Pb 3 O 4 You may remember that Pb 3 O 4 can be thought of chemically as a mixture of 2PbO.PbO 2. If you react this with an acid, then the two oxides which can be thought of as making up the Pb 3 O 4 behave independently. For example, the PbO will react with nitric acid to give lead(II) nitrate solution and water: However, the lead(IV) oxide doesn't react with nitric acid. Thinking of Pb 3 O 4 as 2PbO.PbO 2, the PbO would react with the nitric acid, and the PbO 2 would be left over. Care!This equation came up in a CIE question. The examiner's report said that many candidates forgot to multiply the lead(II) oxide part of the equation by two. That would be an annoying mistake to make, because you have already done the hard part by working out or remembering that it can be thought of as 2PbO.PbO 2. The thermal stability of the oxides This is about the way the oxides are affected by heating. The monoxides CO, SiO and GeO Apart from carbon monoxide, these oxides are unimportant for A level chemistry purposes. They all disproportionate on heating - that is, one molecule undergoes both oxidation and reduction. For example, in the case of SiO, one molecule is reduced to silicon while the other is oxidised to silicon dioxide. They all undergo the change: In the case of carbon monoxide, a catalyst is needed even at relatively high temperatures. These changes reflect the fact that the +4 oxidation state (in the dioxides) is more stable than the +2 state at the top of Group 4. SnO and PbO These are stable to heat. The dioxides The dioxides are stable to heat except for PbO 2. This decomposes on heating to give lead(II) oxide and oxygen. This reflects the greater stability of the +2 oxidation state at the bottom of Group 4. Pb 3 O 4 (thought of as 2PbO.PbO 2) will, of course, also give off oxygen on heating to give simple PbO. Go to the Section 9.3 Menu . . . To return to the list of learning outcomes in Section 9.3 Go to the CIE Main Menu . . . To return to the list of all the CIE sections Go to Chemguide Main Menu . . . This will take you to the main part of Chemguide. © Jim Clark 2011 (modified August 2013)
2225	https://www.leonschools.net/cms/lib7/FL01903265/Centricity/Domain/4275/7th%20Homework%202-1.pdf	Math 7 Name_______ Chapter 6 Test - Percents REVIEW WORKSHEET #1 Date__ Score___ Write each as a percent. 1. 0.34 _ 2. 0.05 ___________ 3. 0.00625 ___________ 4. 3.5 ___________ Write each as a decimal and reduced fraction. 5. 21% ________ _____ 6. 15% ________ _____ 7. 650% ________ _____ 8. .8% ________ _____ Write the fraction as a decimal and as a percent. _3 31 9. 16 __ _ 10. 80 ___________ ___________ _53 7 11. 25 __ _ 12. 140 __ _ Solve each problem using the equation method. Please show the equations and work for each problem in the space provided. 13. What number is 45% of 580? 14. What percent of 160 is 62? 13. __ 14. __ Solve each problem using the proportion method. Please show the proportion and work for each problem in the space provided. 15. 60% of what number is 348? 16. 320% of 69 is what number? 15. __ 16. ___ Solve each of the remaining problems using either the equation or the proportion method. Be sure to show your equation or proportion in the space provided. 17. 48 is what percent of 642? 18. 34 is 68% of what number? 17. __ 18. __ 19. 110% of 124 is what number? 20. What is 34% of 375? 19. __ 20.__ Find the percent of change (rounded to the nearest tenth of a percent). Show all proportions or equations. 21. 125 is increased to 210. 22. 50 is decreased to 14. 23. 45 is increased to 180. 21.__ 22.___ 23.__ 24. Mrs. Rogers sold 49 out of 78 crafts at the fair. What percent of her crafts did she sell? Show your proportion or equation set-up. 24.__ 25. Bill bought a shirt for $35.80 plus he paid an additional 7% in sales tax. What was the final cost? Show your proportion or equation set-up. 25.___ 26. Kathy bought a pair of jeans on sale for 30% off of the regular price. If the jeans regularly cost $75.00. How much did she pay? Show your proportion or equation set-up. 26. __ 27. There are 625 students at Median Middle School. If there 8% of the students were absent on Tuesday, how many students were absent? 27.___ 28. A store buys cell phones from the manufacturer for $34.40 and marks them up by 85%. How much do they charge for the phones (what is the retail price)? 28. __ 29. A new top-selling DVD is marked 20% off at Discount DVD, where it normally sells for $29.90. The same disc sells for $28.60 at Value Video, where you have a coupon for 15% off anything in the store. Where would you buy the DVD? Explain how you decided and be sure to include work to support your answer. 30. You put $500 in the bank, where it earns 1.5% simple interest each year. How long until your money will triple? 31. 25% off Sale A) In a sale, all the prices are reduced by 25%. Jane sees a jacket before the sale. What is the sale price? In the second week of the sale, the prices are reduced by 25% more. In the third week, by 25% more. In the fourth week, the prices are again reduced by 25% more. B) Jane thinks this will mean that the prices will be reduced to $0 after the four reductions because 4 x 25% = 100%. Explain why Jane is wrong. C) If Jane is able to buy her jacket after the four reductions. a. How much will she have to pay? b. What percentage of the original price does she save? 32. T-shirt Sale A) Tom bought these three T-shirts at the sale price of $14.50. How much money did he save compared to the original total price of the T-shirts? B) What percentage of the total price did Tom save?
2226	https://www.onlinemathlearning.com/area-triangle.html	OnlineMathLearning.com - Do Not Process My Personal Information If you wish to opt-out of the sale, sharing to third parties, or processing of your personal or sensitive information for targeted advertising by us, please use the below opt-out section to confirm your selection. Please note that after your opt-out request is processed you may continue seeing interest-based ads based on personal information utilized by us or personal information disclosed to third parties prior to your opt-out. You may separately opt-out of the further disclosure of your personal information by third parties on the IAB’s list of downstream participants. This information may also be disclosed by us to third parties on the IAB’s List of Downstream Participants that may further disclose it to other third parties. Area Of Triangles - Formulas Related Pages Types Of Triangles Word Problems - Area of Triangles More Geometry Lessons In these lessons, we will look at different formulas that can be used to calculate the area of triangles when we are given Base and Height the Length of Three Sides (Heron’s formula) Side, Angle, Side an Equilateral Triangle a Triangle drawn on a Grid Three Vertices on the Coordinate Plane Two Vectors from One Vertex Share this page to Google Classroom Area Of Triangles There are several ways to calculate the area of a triangle, depending on the information you are given. The following table gives the most common and useful formulas. Scroll down the page if you need more explanations about the formulas, how to use them as well as worksheets to practice. Printable & Online Triangle Worksheets Printable Triangle Sum Theorem Exterior Angle Theorem Triangle Correspondence Explore Triangle Congruence Explore SSA Triangle Congruence Identify Congruent Triangles Online Angles in a Triangle 1 Angles in a Triangle 2 Given Base And Height If we are given the base of the triangle and the perpendicular height then we can use the formula. A=12bh Area of a triangle is equal to half of the product of its base and height. The height of a triangle is the perpendicular distance from a vertex to the base of the triangle. Any of the 3 sides of a triangle can be used as a base. It all depends on where the height is drawn. If you are given the sides of an isosceles or equilateral triangle, you can use the Pythagorean Theorem to find the height of the triangle and then use the above formula to find the area. How to calculate the area of a triangle when given the base and height? Show Video Lesson How to use the Pythagorean theorem to get the height of an isosceles triangle and then calculate the area of the triangle? Show Video Lesson Worksheets Worksheet to calculate the areas of triangles Finding areas of triangles and parallelograms Finding the height given the area of the triangle Problems involving the base, height and area of a triangle Given Length Of Three Sides If we are given three sides of a triangle, we can use Heron’s formula: where a, b, and c are the lengths of the sides and s is half the perimeter. Use Heron’s Formula to determine the area of a triangle while only knowing the lengths of the sides Show Video Lesson Given Side Angle Side If we are given the lengths of two sides of a triangle and the size of angle between them we can use the formula: Area=12absinC The area of a triangle is equal to half the product of two sides times the sine of the included angle. This is also known as the sine rule for the area of a triangle. By considering sin A and sin B in a similar way, we obtain Area=12bcsinA or Area=12acsinB We may use any of the above formulas depending on which two sides and angle are given. Remember that the given angle must be between the two given sides. Example: Find the area of triangle PQR if p = 6.5 cm, r = 4.3 cm and Q = 39˚. Give your answer correct to 2 decimal places. Solution: Area of triangle PQR = 1/2 pr sin Q = 1/2 × 6.5 × 4.3 × sin 39˚ = 8.79 cm2 How to find the area of an oblique triangle when given two sides and an angle? Show Video Lesson Given Equilateral Triangle To find the area of an equilateral triangle, we can use the Pythagorean Theorem to get the height of the triangle and then use formula A=12bh or we can use the following formula: The formula for the area of an equilateral triangle (with all sides congruent) is equal to A=s23√4 where s is the length of any side of the triangle. How to find the area of an equilateral triangle with side s? Show Video Lesson Given A Triangle Drawn On A Grid If the triangle is drawn on a grid then we can use the "box" method to calculate the area of the triangle. This method involves drawing a smallest box that will enclose the triangle. Make sure that the box follows the grid of the graph paper. The space between the triangle and box is subdivided into right triangles and rectangles and the total area of the space is calculated. The area of the required triangle is then the area of the space subtracted from the area of the box. How to use the box method to calculate the area of a triangle drawn on a grid? Show Video Lesson Given Three Vertices On The Coordinate Plane When we are given three vertices of a triangle on the coordinate plane, we should first check whether the three vertices form a right triangle. If it is a right triangle then we can use the formula of half the product of its base and height to calculate the area. If it not a right triangle then we can either use Heron’s formula or the determinant of a matrix. To use Heron’s formula, we need to first get the length of each side by using the distance formula. Then plug the values into the Heron’s formula. where a, b, and c are the lengths of the sides and s is half the perimeter. If you are familiar with matrices and determinants, then you can use the determinant of a matrix to get the area of the triangle. The area of a triangle is equal to ±12∣∣∣∣x1x2x3y1y2y3111∣∣∣∣ where (x1, y1), (x2, y2), (x3, y3) are the coordinates of the three vertices. The plus/minus (±) sign is meant to take whichever sign is needed to make the answer positive. If the answer is zero, then the three points are collinear (forms a straight line). Using the determinant of a matrix we can find the area of a triangle whose coordinates are on the coordinate plane. Example: Black-necked stilts are birds that live throughout Florida and surrounding areas but breed mostly in the triangular region shown on the map. Estimate the area of this region. The coordinates given are measured in miles. Show Video Lesson How to find the area of a triangle with the determinant of a matrix? Show Video Lesson Given Two Vectors From One Vertex If we are given the three vertices of a triangle in space, we can use cross products to find the area of the triangle. If a triangle is specified by vectors u and v originating at one vertex, then the area is half the magnitude of their cross product. A=12∥∥u→×v→∥∥ How to compute the area of a triangle given its three vertices in space, using cross products? Example: A triangle in R3 has vertices at points (1, 2, 3), (5, 5, 5) and (7, 8, 9). What is the area? Show Video Lesson Try out our new and fun Fraction Concoction Game. Add and subtract fractions to make exciting fraction concoctions following a recipe. 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2227	https://www.youtube.com/watch?v=cZU7AxUKTxA	Law of Cosines for Vectors - Linear Algebra Made Easy (2016) Course Grinder 25400 subscribers 68 likes Description 7546 views Posted: 23 Jan 2016 Transcript: the last video introduced us to the law of cosines and now we want to see the law while K maybe I should spell this right law of cosines for vectors so we're gonna try to you know translate that similar concept over to vectors let's go ahead and write two vectors out over here I don't want to I think I'm gonna you know treat myself to some straight lines because we did that last video and it worked out pretty nicely look at that ooh that's fresh okay I [ __ ] up that last little edge there okay that's fresh that's awesome I'm always trying to draw it out with the stylus and it just doesn't work oh yeah look at that [ __ ] that sexy okay now okay I'll have to go to the there okay so we have those set up over here this will be X this will be Y and this will be the angle that's between them later so we have these in this is in space we'll just call these these are in space R and right so they're in any space right now and we want to see if we can translate those over to you know a formula where we'll have the lengths of the vectors and the angle involved which is essentially what we saw with the law of cosines right we wanted to have the three edges of the triangle as well as the angle involved in some type of formula that we can use when he when the triangle did not contain a right angle so we want to go ahead and draw this and essentially how we're going to write this is we'll have this line over here and this is gonna act as like our C right this is EC that we had from before is going to be here and essentially is going to be X minus y so the length of X minus y squared is going to be that one and so in this case B squared is going to be B squared is going to be over here B squared B squared so this is remember we're relating back to the last video so this B squared is length of x squared and a squared over here is the length of Y squared right do I need to write out the the cosine triangle again let's do it just in case right just in case you're a bit lost of what's going on I'm relating it to this that we saw in the last video so I'm treating myself to some straight lines again because I can do it look at that man that's sexy okay so problem is I have to scroll down to go to my layers give me a second yeah cuz I'm doing all this in Photoshop so I have to go through the layers okay so we have that here and then we'll have we had some height H we had a we had B we had C this was 2 right-angled triangles we had the angle theta over here is that all we need yeah that was all we need and we got the formula well here was with C squared is equal to a squared plus B squared minus 2a B coasts of theta you can go ahead and check the earlier video in the playlist I'll put the playlist in the description and go one video earlier and you'll see how I went and got this formula over here so we're trying to look at how we can find that with these vectors so this is you know I've did the replacements over here these things over here which we're gonna be replacing the values that we have in here so what we can do is we can simply just go and plug those values into this formula that we had so we went and said that C squared was going to be the length of X minus y squared so we have that right there that will be equal to what we had was a squared was going to be this alright whatever it doesn't matter the actual way that I wrote it so I think I had what would it have for y squared I said B squared was x squared so you know you could essentially reverse those but it doesn't make a difference here it's gonna be the same thing because we're adding them so the order doesn't really matter then we have subtracted by 2 then we have subtracted by 2 we don't have these squared values here so we'll just have lengths of X like why and we'll have the angle cosine so we've essentially written it in the exact same way that we had before right we wrote it in we just replaced those values that we had for those two vectors and we had an angle between those two vectors and we essentially replaced it for the triangle now we can actually take this a little bit further I'm gonna show you how we do that give me a second actually I'm gonna end this off here the one I put too much in this video and we will resume from this exact spot in the next one we're going to relate this to the formula for the dot product and we're gonna see if those two cosines are going to be the same the angles for the cosine so we're going to see that in the next video
2228	https://www.youtube.com/watch?v=7oGKTiJUOlQ	prove cos(x + pi) = -cos x MSolved Tutoring 66100 subscribers 19 likes Description 1735 views Posted: 6 Sep 2019 prove cos(x + pi) = -cos x 1 comments Transcript: identity so what we'll do here is cosine of X plus PI it's gonna be cosine X times cosine PI minus sine of X times sine of Pi so cosine of cosine of pi is equal to negative 1 so we get cosine x times negative 1 minus the sine of X sine of pi is equal to 0 so this just becomes a 0 negative 1 times cosine of X becomes negative cosine of X which matches up with our identity up here and that's how you prove that ok I hope that this video helped you out appreciate you watching it you have a great
2229	https://www.mathworksheets4kids.com/3by2-division.php	Dividing 3-digit by 2-digit Numbers Worksheets [x] Login Child Login Main MenuMathLanguage ArtsScienceSocial StudiesInteractive WorksheetsBrowse By GradeBecome a Member Become a Member Math Interactive Worksheets Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Worksheets by Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Number Sense and Operations Number Recognition Counting Skip Counting Place Value Number Lines Addition Subtraction Multiplication Division Word Problems Comparing Numbers Ordering Numbers Odd and Even Prime and Composite Roman Numerals Ordinal Numbers Properties Patterns Rounding Estimation In and Out Boxes Number System Conversions More Number Sense Worksheets Measurement Size Comparison Time Calendar Money Measuring Length Weight Capacity Metric Unit Conversion Customary Unit Conversion Temperature More Measurement Worksheets Financial Literacy Writing Checks Profit and Loss Discount Sales Tax Simple Interest Compound Interest Statistics and Data Analysis Tally Marks Pictograph Line Plot Bar Graph Line Graph Pie Graph Scatter Plot Mean, Median, Mode, Range Mean Absolute Deviation Stem-and-leaf Plot Box-and-whisker Plot Factorial Permutation and Combination Probability Venn Diagram More Statistics Worksheets Geometry Positions Shapes - 2D Shapes - 3D Lines, Rays and Line Segments Points, Lines and Planes Angles Symmetry Transformation Area Perimeter Rectangle Triangle Circle Quadrilateral Polygon Ordered Pairs Midpoint Formula Distance Formula Slope Parallel, Perpendicular and Intersecting Lines Scale Factor Surface Area Volume Pythagorean Theorem More Geometry Worksheets Pre-Algebra Factoring GCF LCM Fractions Decimals Converting between Fractions and Decimals Integers Significant Figures Percents Convert between Fractions, Decimals, and Percents Ratio Proportions Direct and Inverse Variation Order of Operations Exponents Radicals Squaring Numbers Square Roots Scientific Notations Logarithms Speed, Distance, and Time Absolute Value More Pre-Algebra Worksheets Algebra Translating Algebraic Phrases Evaluating Algebraic Expressions Simplifying Algebraic Expressions Algebraic Identities Equations Quadratic Equations Systems of Equations Functions Polynomials Inequalities Sequence and Series Matrices Complex Numbers Vectors More Algebra Worksheets Trigonometry Calculus Math Workbooks English Language Arts Summer Review Packets Science Social Studies Holidays and Events Support Worksheets> Math> Number Sense> Division> 3-digit by 2-digit Numbers Dividing 3-digit by 2-digit Numbers Worksheets Sharpen skills with our 3-digit by 2-digit division worksheets. Learn to solve division problems following the basic steps of long division, find the quotient and remainder of standard division problems, and real-life word problems. Complete the steps to divide numbers, check your answers, and decode riddles! Try some of these worksheets for free! Standard Division Problems \| No Remainder Instruct children to divide 3-digit numbers by 2-digit divisors step by step until they end up with a zero. Download the set Division Using Grids The problems involving 3-digit dividends and 2-digit divisors are aligned on grids for children to divide them effortlessly. Download the set Standard Division Problems \| With Remainder Each PDF worksheet comprises 12 problems involving 3-digit by 2-digit numbers. Calculate the quotient and remainder and write the answers in the space provided. Download the set Filling in Missing Numbers Multiply, divide, or subtract to determine the missing parts in dividing numbers such as the divisor, dividend, quotient, or remainder. Download the set Standard Division with Word Problems \| With or Without Remainder Incorporate this blend of standard division problems and word problems to recapitulate skills in finding the quotient and the remainder. Download the set Division Riddles Let learners rattle their brains to solve each division sentence, map the answers to the letters, and decode the riddles. Download the set Standard Division Problems \| With or Without Remainder Test skills with these worksheets on dividing 3-digit numbers by 2-digit whole numbers. The problems may or may not result in a remainder. Download the set Division Using Area Models Relate the 2-digit divisor and quotient to the width and length of the rectangle and the 3-digit dividend to the area of the rectangle while dividing. Without Remainder With Remainder Download the set Division with Word Problems \| No Remainder Connect division to day-to-day life while solving standard problems and word problems. The numbers divide equally, leaving no remainder. Download the set Division Timed Tests without Remainders Build speed and accuracy with these timed division quizzes. Assess if kids can solve 25 division problems in 4 minutes and 50 in 8 minutes. 25 Problems 50 Problems Download the set Horizontal Division Problems \| Without Remainder Reaffirm skills with 15 horizontal 3-digit by 2-digit division problems that leave no remainders in each worksheet. Download the set Divide and Check Evaluate your answer by plugging the values in the formula - quotient divisor + remainder = dividend, and check if the equation is balanced. Download the set Division with Word Problems \| With Remainder Read the scenario carefully, use your comprehension skills, and apply the concept to solve each 3-digit by 2-digit division problem. Download the set Horizontal Division \| With Remainder Write the quotient along with what is left over while dividing the 3-digit dividends by double-digit divisors in these printable worksheets. Download the set Division Timed Tests with Remainders Navigate through these division drills, apply the rules for division, and calculate the quotients while writing the remainder next to them. 25 Problems 50 Problems Download the set Related Worksheets »3-digit by 1-digit Division »Division Word Problems »In and Out Boxes for Division Home Become a Member Membership Information Login Printing Help FAQ How to Use Interactive Worksheets How to Use Printable Worksheets About Us Privacy Policy Terms of Use Contact Us Follow us Copyright © 2025 - Math Worksheets 4 Kids × This is a members-only feature! 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2230	https://openstax.org/books/biology-2e/pages/16-chapter-summary	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Biology 2e Chapter Summary Biology 2eChapter Summary Search for key terms or text. ## 16.1 Regulation of Gene Expression While all somatic cells within an organism contain the same DNA, not all cells within that organism express the same proteins. Prokaryotic organisms express most of their genes most of the time. However, some genes are expressed only when they are needed. Eukaryotic organisms, on the other hand, express only a subset of their genes in any given cell. To express a protein, the DNA is first transcribed into RNA, which is then translated into proteins, which are then targeted to specific cellular locations. In prokaryotic cells, transcription and translation occur almost simultaneously. In eukaryotic cells, transcription occurs in the nucleus and is separate from the translation that occurs in the cytoplasm. Gene expression in prokaryotes is mostly regulated at the transcriptional level (some epigenetic and post-translational regulation is also present), whereas in eukaryotic cells, gene expression is regulated at the epigenetic, transcriptional, post-transcriptional, translational, and post-translational levels. ## 16.2 Prokaryotic Gene Regulation The regulation of gene expression in prokaryotic cells occurs at the transcriptional level. There are two majors kinds of proteins that control prokaryotic transcription: repressors and activators. Repressors bind to an operator region to block the action of RNA polymerase. Activators bind to the promoter to enhance the binding of RNA polymerase. Inducer molecules can increase transcription either by inactivating repressors or by activating activator proteins. In the trp operon, the trp repressor is itself activated by binding to tryptophan. Therefore, if tryptophan is not needed, the repressor is bound to the operator and transcription remains off. The lac operon is activated by the CAP (catabolite activator protein), which binds to the promoter to stabilize RNA polymerase binding. CAP is itself activated by cAMP, whose concentration rises as the concentration of glucose falls. However, the lac operon also requires the presence of lactose for transcription to occur. Lactose inactivates the lac repressor, and prevents the repressor protein from binding to the lac operator. With the repressor inactivated, transcription may proceed. Therefore glucose must be absent and lactose must be present for effective transcription of the lac operon. ## 16.3 Eukaryotic Epigenetic Gene Regulation In eukaryotic cells, the first stage of gene-expression control occurs at the epigenetic level. Epigenetic mechanisms control access to the chromosomal region to allow genes to be turned on or off. Chromatin remodeling controls how DNA is packed into the nucleus by regulating how tightly the DNA is wound around histone proteins. The DNA itself may be methylated to selectively silence genes. The addition or removal of chemical modifications (or flags) to histone proteins or DNA signals the cell to open or close a chromosomal region. Therefore, eukaryotic cells can control whether a gene is expressed by controlling accessibility to the binding of RNA polymerase and its transcription factors. ## 16.4 Eukaryotic Transcription Gene Regulation To start transcription, general transcription factors, such as TFIID, TFIIB, and others, must first bind to the TATA box and recruit RNA polymerase to that location. Additional transcription factors may also bind to other regulatory elements at the promoter to increase or prevent transcription. In addition to promoter sequences, enhancer regions help augment transcription. Enhancers can be upstream, downstream, within a gene itself, or on other chromosomes. Specific transcription factors bound to enhancer regions may either increase or prevent transcription. ## 16.5 Eukaryotic Post-transcriptional Gene Regulation Post-transcriptional control can occur at any stage after transcription, including RNA splicing and RNA stability. Once RNA is transcribed, it must be processed to create a mature RNA that is ready to be translated. This involves the removal of introns that do not code for protein. Spliceosomes bind to the signals that mark the exon/intron border to remove the introns and ligate the exons together. Once this occurs, the RNA is mature and can be translated. Alternative splicing can produce more than one mRNA from a given transcript. Different splicing variants may be produced under different conditions. RNA is created and spliced in the nucleus, but needs to be transported to the cytoplasm to be translated. RNA is transported to the cytoplasm through the nuclear pore complex. Once the RNA is in the cytoplasm, the length of time it resides there before being degraded, called RNA stability, can also be altered to control the overall amount of protein that is synthesized. The RNA stability can be increased, leading to longer residency time in the cytoplasm, or decreased, leading to shortened time and less protein synthesis. RNA stability is controlled by RNA-binding proteins (RPBs) and microRNAs (miRNAs). These RPBs and miRNAs bind to the 5' UTR or the 3' UTR of the RNA to increase or decrease RNA stability. MicroRNAs associated with RISC complexes may repress translation or lead to mRNA breakdown. ## 16.6 Eukaryotic Translational and Post-translational Gene Regulation Changing the status of the RNA or the protein itself can affect the amount of protein, the function of the protein, or how long it is found in the cell. To translate the protein, a protein initiator complex must assemble on the RNA. Modifications (such as phosphorylation) of proteins in this complex can prevent proper translation from occurring. Once a protein has been synthesized, it can be modified (phosphorylated, acetylated, methylated, or ubiquitinated). These post-translational modifications can greatly impact the stability, degradation, or function of the protein. ## 16.7 Cancer and Gene Regulation Cancer can be described as a disease of altered gene expression. Changes at every level of eukaryotic gene expression can be detected in some form of cancer at some point in time. In order to understand how changes to gene expression can cause cancer, it is critical to understand how each stage of gene regulation works in normal cells. By understanding the mechanisms of control in normal, non-diseased cells, it will be easier for scientists to understand what goes wrong in disease states including complex ones like cancer. 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2231	https://www.oxfordreference.com/abstract/10.1093/acref/9780199661350.001.0001/acref-9780199661350-e-5954	Thrive - Oxford Reference Update Jump to Content Personal Profile About News Subscriber Services Contact Us Help For Authors Oxford Reference PublicationsPages Publications Pages Help Subject Archaeology Art & Architecture Bilingual dictionaries Classical studies Encyclopedias English Dictionaries and Thesauri History Language reference Law Linguistics Literature Media studies Medicine and health Music Names studies Performing arts Philosophy Quotations Religion Science and technology Social sciences Society and culture Browse All Reference Type Overview Pages Subject Reference Timelines Quotations English Dictionaries Bilingual Dictionaries Browse All My Content (1) Recently viewed (1) thrive My Searches (0)### Recently viewed (0) Close Fowler’s Dictionary of Modern English Usage (4 ed.) Edited by: Jeremy Butterfield Publisher:Oxford University Press Print Publication Date:2015 Print ISBN-13:9780199661350 Published online:2015 Current Online Version:2015 eISBN:9780191744532 Find at OUP.com Google Preview Read More Highlight search term - [x] Cite Save Share ThisFacebook LinkedIn Twitter Email this contentShare Link Copy this link, or click below to email it to a friend Email this contentor copy the link directly: The link was not copied. Your current browser may not support copying via this button. Link copied successfully Copy link Sign inGet help with access You could not be signed in, please check and try again. Username Please enter your Username Password Please enter your Password Forgot password? Don't have an account? Sign in via your Institution You could not be signed in, please check and try again. Sign in with your library card Please enter your library card number Search within work Related Content In this work Key to the Pronunciation Abbreviations and Symbols Bibliographical Abbreviations Show Summary Details Publishing Information Dedication Some Quotations A revision for the twenty-first century Acknowledgements Dedication, 1926 Key to the Pronunciation Abbreviations and Symbols Bibliographical Abbreviations Page of PRINTED FROM OXFORD REFERENCE (www.oxfordreference.com). (c) Copyright Oxford University Press, 2023. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single entry from a reference work in OR for personal use (for details see Privacy Policy and Legal Notice). date: 29 September 2025 thrive (verb). Source:Fowler’s Dictionary of Modern English Usage Author(s):Jeremy ButterfieldJeremy Butterfield The regular or ‘weak’ form thrived has won the centuries-old battle between irregular or ‘strong’ throv e as past tense and ... ... Access to the complete content on Oxford Reference requires a subscription or purchase. Public users are able to search the site and view the abstracts and keywords for each book and chapter without a subscription. Please subscribe or login to access full text content. If you have purchased a print title that contains an access token, please see the token for information about how to register your code. For questions on access or troubleshooting, please check our FAQs, and if you can''t find the answer there, please contact us. Publishing Information Dedication Some Quotations A revision for the twenty-first century Acknowledgements Dedication, 1926 Key to the Pronunciation Abbreviations and Symbols Bibliographical Abbreviations Oxford University Press Copyright © 2025. All rights reserved. PRINTED FROM OXFORD REFERENCE (www.oxfordreference.com). (c) Copyright Oxford University Press, 2023. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single entry from a reference work in OR for personal use (for details see Privacy Policy and Legal Notice). date: 29 September 2025 Cookie Policy Privacy Policy Legal Notice Credits Accessibility [34.96.49.197] 34.96.49.197 Sign in to annotate Close Edit Character limit 500/500 Delete Cancel Save @! Character limit 500/500 Cancel Save Oxford University Press uses cookies to enhance your experience on our website. By selecting ‘accept all’ you are agreeing to our use of cookies. 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2232	https://cran.r-project.org/web/packages/weibulltools/vignettes/Life_Data_Analysis_Part_II.html	Life Data Analysis Part II - Estimation Methods for Parametric Lifetime Models Life Data Analysis Part II - Estimation Methods for Parametric Lifetime Models Rank Regression and Maximum Likelihood Tim-Gunnar Hensel David Barkemeyer 2023-04-04 This document introduces two methods for the parameter estimation of lifetime distributions. Whereas Rank Regression (RR) fits a straight line through transformed plotting positions (transformation is described precisely in vignette(topic = "Life_Data_Analysis_Part_I", package = "weibulltools")), Maximum likelihood (ML) strives to maximize a function of the parameters given the sample data. If the parameters are obtained, a cumulative distribution function (CDF) can be computed and added to a probability plot. In the theoretical part of this vignette the focus is on the two-parameter Weibull distribution. The second part is about the application of the provided estimation methods in {weibulltools}. All implemented models can be found in the help pages of rank_regression() and ml_estimation(). The Weibull Distribution The Weibull distribution is a continuous probability distribution, which is specified by the location parameter μ μ and the scale parameter σ σ. Its CDF and PDF (probability density function) are given by the following formulas: F(t)=Φ S E V(log(t)−μ σ)F(t)=Φ S E V(log⁡(t)−μ σ) f(t)=1 σ t ϕ S E V(log(t)−μ σ)f(t)=1 σ t ϕ S E V(log⁡(t)−μ σ) The practical benefit of the Weibull in the field of lifetime analysis is that the common profiles of failure rates, which are observed over the lifetime of a large number of technical products, can be described using this statistical distribution. In the following, the estimation of the specific parameters μ μ and σ σ is explained. Rank Regression (RR) In RR the CDF is linearized such that the true, unknown population is estimated by a straight line which is analytically placed among the plotting pairs. The lifetime characteristic, entered on the x-axis, is displayed on a logarithmic scale. A double-logarithmic representation of the estimated failure probabilities is used for the y-axis. Ordinary Least Squares (OLS) determines a best-fit line in order that the sum of squared deviations between this fitted regression line and the plotted points is minimized. In reliability analysis, it became prevalent that the line is placed in the probability plot in the way that the horizontal distances between the best-fit line and the points are minimized 1. This procedure is called x on y rank regression. The formulas for estimating the slope and the intercept of the regression line according to the described method are given below. Slope: b^=∑n i=1(x i−x¯)⋅(y i−y¯)∑n i=1(y i−y¯)2 b^=∑i=1 n(x i−x¯)⋅(y i−y¯)∑i=1 n(y i−y¯)2 Intercept: a^=x¯−b^⋅y¯a^=x¯−b^⋅y¯ With x i=log(t i);x¯=1 n⋅∑i=1 n log(t i);x i=log⁡(t i);x¯=1 n⋅∑i=1 n log⁡(t i); as well as y i=Φ−1 S E V[F(t)]=log{−log[1−F(t i)]}a n d y¯=1 n⋅∑i=1 n log{−log[1−F(t i)]}.y i=Φ S E V−1[F(t)]=log⁡{−log⁡[1−F(t i)]}a n d y¯=1 n⋅∑i=1 n log⁡{−log⁡[1−F(t i)]}. The estimates of the intercept and slope are equal to the Weibull parameters μ μ and σ σ, i.e. μ^=a^μ^=a^ and σ^=b^.σ^=b^. In order to obtain the parameters of the shape-scale parameterization the intercept and the slope need to be transformed 2. η^=exp(a^)=exp(μ^)η^=exp⁡(a^)=exp⁡(μ^) and β^=1 b^=1 σ^.β^=1 b^=1 σ^. Maximum Likelihood (ML) The ML method of Ronald A. Fisher estimates the parameters by maximizing the likelihood function. Assuming a theoretical distribution, the idea of ML is that the specific parameters are chosen in such a way that the plausibility of obtaining the present sample is maximized. The likelihood and log-likelihood are given by the following equations: L=∏i=1 n{1 σ t i ϕ S E V(log(t i)−μ σ)}L=∏i=1 n{1 σ t i ϕ S E V(log⁡(t i)−μ σ)} and log L=∑i=1 n log{1 σ t i ϕ S E V(log(t i)−μ σ)}log⁡L=∑i=1 n log⁡{1 σ t i ϕ S E V(log⁡(t i)−μ σ)} Deriving and nullifying the log-likelihood function according to parameters results in two formulas that have to be solved numerically in order to obtain the estimates. In large samples, ML estimators have optimality properties. In addition, the simulation studies by Genschel and Meeker3 have shown that even in small samples it is difficult to find an estimator that regularly has better properties than ML estimators. Data To apply the introduced parameter estimation methods the shock and alloy datasets are used. Shock Absorber In this dataset kilometer-dependent problems that have occurred on shock absorbers are reported. In addition to failed items the dataset also contains non-defective (censored) observations. The data can be found in Statistical Methods for Reliability Data4. For consistent handling of the data, {weibulltools} introduces the function reliability_data() that converts the original dataset into a wt_reliability_data object. This formatted object allows to easily apply the presented methods. ``` shock_tbl <- reliability_data(data = shock, x = distance, status = status) shock_tbl > Reliability Data with characteristic x: 'distance': > # A tibble: 38 × 3 > x status id > > 1 6700 1 ID1 > 2 6950 0 ID2 > 3 7820 0 ID3 > 4 8790 0 ID4 > 5 9120 1 ID5 > 6 9660 0 ID6 > 7 9820 0 ID7 > 8 11310 0 ID8 > 9 11690 0 ID9 > 10 11850 0 ID10 > # ℹ 28 more rows ``` Alloy T7989 The dataset alloy in which the cycles until a fatigue failure of a special alloy occurs are inspected. The data is also taken from Meeker and Escobar 5. Again, the data have to be formatted as a wt_reliability_data object: ``` Data: alloy_tbl <- reliability_data(data = alloy, x = cycles, status = status) alloy_tbl > Reliability Data with characteristic x: 'cycles': > # A tibble: 72 × 3 > x status id > > 1 300 0 ID1 > 2 300 0 ID2 > 3 300 0 ID3 > 4 300 0 ID4 > 5 300 0 ID5 > 6 291 1 ID6 > 7 274 1 ID7 > 8 271 1 ID8 > 9 269 1 ID9 > 10 257 1 ID10 > # ℹ 62 more rows ``` RR and ML with {weibulltools} rank_regression() and ml_estimation() can be applied to complete data as well as failure and (multiple) right-censored data. Both methods can also deal with models that have a threshold parameter γ γ. In the following both methods are applied to the dataset shock. RR for two-parameter Weibull distribution ``` rank_regression needs estimated failure probabilities: shock_cdf <- estimate_cdf(shock_tbl, methods = "johnson") Estimating two-parameter Weibull: rr_weibull <- rank_regression(shock_cdf, distribution = "weibull") rr_weibull > Rank Regression > Coefficients: > mu sigma > 10.2596 0.3632 Probability plot: weibull_grid <- plot_prob( shock_cdf, distribution = "weibull", title_main = "Weibull Probability Plot", title_x = "Mileage in km", title_y = "Probability of Failure in %", title_trace = "Defectives", plot_method = "ggplot2" ) Add regression line: weibull_plot <- plot_mod( weibull_grid, x = rr_weibull, title_trace = "Rank Regression" ) weibull_plot ``` Figure 1: RR for a two-parametric Weibull distribution. ML for two-parameter Weibull distribution ``` Again estimating Weibull: ml_weibull <- ml_estimation( shock_tbl, distribution = "weibull" ) ml_weibull > Maximum Likelihood Estimation > Coefficients: > mu sigma > 10.2299 0.3164 Add ML estimation to weibull_grid: weibull_plot2 <- plot_mod( weibull_grid, x = ml_weibull, title_trace = "Maximum Likelihood" ) weibull_plot2 ``` Figure 2: ML for a two-parametric Weibull distribution. ML for two- and three-parameter log-normal distribution Finally, two- and three-parametric log-normal distributions are fitted to the alloy data using maximum likelihood. ``` Two-parameter log-normal: ml_lognormal <- ml_estimation( alloy_tbl, distribution = "lognormal" ) ml_lognormal > Maximum Likelihood Estimation > Coefficients: > mu sigma > 5.1278 0.3276 Three-parameter Log-normal: ml_lognormal3 <- ml_estimation( alloy_tbl, distribution = "lognormal3" ) ml_lognormal3 > Maximum Likelihood Estimation > Coefficients: > mu sigma gamma > 4.5015 0.6132 72.0727 ``` ``` Constructing probability plot: tbl_cdf_john <- estimate_cdf(alloy_tbl, "johnson") lognormal_grid <- plot_prob( tbl_cdf_john, distribution = "lognormal", title_main = "Log-normal Probability Plot", title_x = "Cycles", title_y = "Probability of Failure in %", title_trace = "Failed units", plot_method = "ggplot2" ) Add two-parametric model to grid: lognormal_plot <- plot_mod( lognormal_grid, x = ml_lognormal, title_trace = "Two-parametric log-normal" ) lognormal_plot ``` Figure 3: ML for a two-parametric log-normal distribution. ``` Add three-parametric model to lognormal_plot: lognormal3_plot <- plot_mod( lognormal_grid, x = ml_lognormal3, title_trace = "Three-parametric log-normal" ) lognormal3_plot ``` Figure 4: ML for a three-parametric log-normal distribution. Berkson, J.: Are There Two Regressions?, Journal of the American Statistical Association 45 (250), DOI: 10.2307/2280676, 1950, pp.164-180↩︎ ReliaSoft Corporation: Life Data Analysis Reference Book, online: ReliaSoft, accessed 19 December 2020↩︎ Genschel, U.; Meeker, W. Q.: A Comparison of Maximum Likelihood and Median-Rank Regression for Weibull Estimation, in: Quality Engineering 22 (4), DOI: 10.1080/08982112.2010.503447, 2010, pp.236-255↩︎ Meeker, W. Q.; Escobar, L. A.: Statistical Methods for Reliability Data, New York, Wiley series in probability and statistics, 1998, p.630↩︎ Meeker, W. Q.; Escobar, L. A.: Statistical Methods for Reliability Data, New York, Wiley series in probability and statistics, 1998, p.131↩︎
2233	https://www.atlas.org/solution/b9a0e041-eb8c-4044-81eb-33134424b0a5/explain-the-difference-between-a-co-ordinate-dative-covalent-bond-and-a-polar-covalent-bond-give-an-example-of-each	BlogSign inSign up free Question Explain the difference between a co-ordinate (dative covalent) bond and a polar covalent bond. Give an example of each. Answer 􀊀100% (1 rated) A co-ordinate (dative covalent) bond is a bond in which both electrons come from one atom (e.g., the bond formed between ammonia and boron trifluoride, where nitrogen donates a pair of electrons). A polar covalent bond involves unequal sharing of electrons between two atoms with different electronegativities (e.g., the bond in HCl, where electrons are closer to chlorine than hydrogen). Steps 1. To explain the difference between a co-ordinate (dative covalent) bond and a polar covalent bond, we first need to define each type of bond clearly. 2. A co-ordinate bond, also known as a dative covalent bond, is a type of covalent bond where both electrons in the bond come from the same atom. For example, in the reaction between ammonia (NH3) and boron trifluoride (BF3), the lone pair of electrons from nitrogen in ammonia forms a co-ordinate bond with the electron-deficient boron atom. 3. In contrast, a polar covalent bond occurs when two atoms share a pair of electrons unequally due to a difference in their electronegativities. This results in a partial negative charge (δ-) on the more electronegative atom and a partial positive charge (δ+) on the less electronegative atom. An example of a polar covalent bond is the bond between hydrogen and chlorine in hydrogen chloride (HCl); here, chlorine is more electronegative than hydrogen, leading to an unequal sharing of electrons. 4. In summary, the key difference is that a coordinate bond involves the sharing of both electrons from one atom, while in a polar covalent bond, the electrons are shared unequally between two different atoms. 􀐅 􀉿 Helpful 􀊁 Not Helpful Related Simplify the expression 7 3×7−3 3×3−4\frac{7^3 \times 7^{-3}}{3 \times 3^{-4}}3×3−4 7 3×7−3. 􀊀 63%(5 rated)What is the result of the expression 3x3-3:3+3? 􀊀 83%(5 rated)Evaluate, ( 5/7 × 2/3) + (5/6-8/9) ÷ 7/15 of 5/6 (3 marks) 􀊀 100%(2 rated)There are 7 characteristics of living things? 􀊀 80%(4 rated)Find the (a) first (b) second (c) third and (d) fourth moments of the set 2, 3, 7 , 8, and 10. 􀊀 100%(2 rated)Calculate the number of positive integers not exceeding 1200 that are divisible by 2, 3, 5 or by 7 􀊀 100%(2 rated)Given that tan x degrees = 3/7, find cos(90-x) degrees, giving the answer to 4 significant figures 􀊀 100%(2 rated)Given p x( ) = −4(x − 15)2 + 2, what is the value of p(7)? 􀊀 100%(3 rated) We're a team of current and former students and teachers on a mission to make learning accessible and engaging for everyone. © Triangle Labs · LA & NYC DiscordTikTokTwitterInstagramLinkedIn ChangelogBlogContactiOS AppAndroid App ResourcesPrivacyTermsAffiliate ProgramAI Detector © Triangle Labs · LA & NYC Solved: Explain the difference between a co-ordinate (dative covalent) bond and a polar covalent bond. Give an example of each. ===============
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Extras Teacher's Guides The Physics Classroom » Physics Tutorial » 1-D Kinematics » Slope of v-t Graphs 1-D Kinematics - Lesson 4 - Describing Motion with Velocity vs. Time Graphs Determining the Slope on a v-t Graph Meaning of Shape for a v-t Graph Meaning of Slope for v-t Graphs Relating Shape to Motion for v-t Graphs Slope of v-t Graphs Area Under v-t Graphs Getting your Trinity Audio player ready... Hold down the T key for 3 seconds to activate the audio accessibility mode, at which point you can click the K key to pause and resume audio. Useful for the Check Your Understanding and See Answers. Report This Ad It was learned earlier in Lesson 4 that the slope of the line on a velocity versus time graph is equal to the acceleration of the object. If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s. If the object is moving with an acceleration of -8 m/s/s, then the slope of the line will be -8 m/s/s. If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. Because of its importance, a student of physics must have a good understanding of how to calculate the slope of a line. In this part of the lesson, the method for determining the slope of a line on a velocity-time graph will be discussed. Let's begin by considering the velocity versus time graph below. The line is sloping upwards to the right. But mathematically, by how much does it slope upwards for every 1 second along the horizontal (time) axis? To answer this question we must use the slope equation. Using the Slope Equation The slope equation says that the slope of a line is found by determining the amount of rise of the line between any two points divided by the amount of run of the line between the same two points. A method for carrying out the calculation is Pick two points on the line and determine their coordinates. Determine the difference in y-coordinates for these two points (rise). Determine the difference in x-coordinates for these two points (run). Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope). The calculations below shows how this method can be applied to determine the slope of the line. Note that three different calculations are performed for three different sets of two points on the line. In each case, the result is the same: the slope is 10 m/s/s. For points (5 s, 50 m/s) and (0 s, 0 m/s): Slope = (50 m/s - 0 m/s) / (5 s - 0 s) = 10 m/s/s For points (5 s, 50 m/s) and (2 s, 20 m/s): Slope = (50 m/s - 20 m/s) / (5 s - 2 s) = 10 m/s/s For points (4 s, 40 m/s) and (3 s, 30 m/s): Slope = (40 m/s - 30 m/s) / (4 s - 3 s) = 10 m/s/s Observe that regardless of which two points on the line are chosen for the slope calculation, the result remains the same - 10 m/s/s. Check Your Understanding Consider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as portrayed by the graph. Use the button to view the answer. See Answer The acceleration (i.e., slope) is 4 m/s/s. If you think the slope is 5 m/s/s, then you're making a common mistake. You are picking one point (probably 5 s, 25 m/s) and dividing y/x. Instead you must pick two points (as in the discussed in this part of the lesson) and divide the change in y by the change in x. 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2235	https://demonstrations.wolfram.com/SumOfATelescopingSeries/	Sum of a Telescoping Series \| Wolfram Demonstrations Project 13,000+ Open Interactive Demonstrations From the makers of Wolfram Language and Mathematica Wolfram Demonstrations Project Topics Latest Random Authoring Notebook Sum of a Telescoping Series n2 y=1/x e e 3 e 4 e 1.0 1.5 2.0 2.5 Initializing live version Open Notebook in CloudCopy Manipulate to Clipboard Source Code For each n=1,2,… ,the area of the corresponding colored region is 1 n - 1 n+1 = 1 n(n+1) ,therefore ∞ ∑ n=1 1 n(n+1) =1 . Contributed by: Soledad Mª Sáez Martínez, Félix Martínez de la Rosa(2011) Open content licensed under CC BY-NC-SA Snapshots Details Reference: Füsun Akman, "Proofs without Words under the Magic Curve," The College Mathematics Journal, 33(1), 2002 pp. 42–46. External Links Series (Wolfram MathWorld)» Permanent Citation Cite this as Soledad Mª Sáez Martínez, Félix Martínez de la Rosa (2011), "Sum of a Telescoping Series" Wolfram Demonstrations Project. demonstrations.wolfram.com/SumOfATelescopingSeries/ Related Demonstrations Sum of a Telescoping Series (II) Soledad Mª Sáez Martínez, Félix Martínez de la Rosa Sum of a Geometric Series Soledad Mª Sáez Martínez, Félix Martínez de la Rosa Sum of the Alternating Harmonic Series (I) Soledad Mª Sáez Martínez, Félix Martínez de la Rosa Sum of the Alternating Harmonic Series (II) Soledad Mª Sáez Martínez, Félix Martínez de la Rosa Visual Proof of Two Integrals Soledad Mª Sáez Martínez, Félix Martínez de la Rosa Visual Computation of an Integral Soledad Mª Sáez Martínez, Félix Martínez de la Rosa Sum of Reciprocals of Triangular Numbers Soledad Mª Sáez Martínez, Félix Martínez de la Rosa Strips of Equal Width on a Sphere Have Equal Surface Areas Mito Are, Daniel Relix Series Expansion of 1/(1-x) Jon Perry Power Series Interval of Convergence Olivia M. Carducci Plot of a Geometric Sequence and Its Partial Sums Aaron Dunigan AtLee Numerical Inversion of the Laplace Transform: The Fourier Series Approximation Housam Binous Graphical Representation of Geometric Series Jonathan Shih Common Methods of Estimating the Area under a Curve Scott Liao Average Value of a Function Michael Largey, Samuel Leung Accuracy of Series Approximations Fred E. Moolekamp III, Kevin L. Stokes Zeros of Truncated Series of Elementary Functions Michael Trott Volumes of Solids of Revolution: Shell Method Helen Papadopoulos Volumes of Revolution Using Cylindrical Shells Stephen Wilkerson Related Topics High School Calculus and Analytic Geometry Calculus Integrals Series Browse all topics Wolfram LanguageMathematicaWolfram UCommunityMore» ©2025 Wolfram Demonstrations Project &Contributors \| Terms of Use \| Privacy Policy \| Give feedback AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA
2236	https://tmedweb.tulane.edu/pharmwiki/doku.php/zaleplon	zaleplon [TUSOM \| Pharmwiki] skip to content Site Tools TUSOM \| Pharmwiki Search Recent Changes Media Manager Sitemap User Tools Log In Trace:•zaleplon zaleplon Zaleplon Trade Name: Sonata ® Drug Class: Non-benzodiazepine hypnotic Mechanism of Action: Its effect is believed to result from its interaction with GABA-receptor complexes at binding domains located close to or allosterically coupled to benzodiazepine receptors Zaleplon binds selectively to the brain alpha-1 subunit of the GABA-A/chloride receptor/channel complex Chemically unrelated to benzodiazepines & barbiturates (it is a member of the pyrazolopyrimidine class of hypnotics) Indications: for the short-term treatment of insomnia Sonata should be taken immediately before bedtime or after the patient has gone to bed and has experienced difficulty falling asleep Sonata has been shown to decrease the time to sleep onset for up to 30 days in controlled clinical studies It has not been shown to increase total sleep time or decrease the number of awakenings Hypnotics should generally be limited to 7 to 10 days of use, and reevaluation of the patient is recommended if they are to be taken for more than 2 to 3 weeks. Sonata should not be prescribed in quantities exceeding a 1-month supply Sonata should be taken immediately before bedtime or after the patient has gone to bed and has experienced difficulty falling asleep Pharmacokinetics: rapidly eliminated with a half life of ~1 hr primarily metabolized by aldehyde oxidase. Side Effects: increased incidence of headache with higher doses Drug Interactions: Zaleplon's effects can be reversed by flumazenil if side effects are severe or life threatening Warnings: some patients may exhibit abnormal thinking or behavioral changes (either depressed actions, or disinhibiting actions) some depressed patients may experience an increase of suicidal ideation sleep driving & other complex behaviors, such as making phone calls & preparing food have been reported. Notes: its shorter duration of action causes less confusion & tiredness on awakening; it may be a safer drug to take if you have to ambulate (get up and walk around) at night Pronunciation: Zal e plahn References: Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0) Vanderah TW (2024): Sedative-Hypnotic Drugs. Chapter 22. In: Basic & Clinical Pharmacology. 16th Ed. Vanderah TW (Editor). McGraw Hill / Lange. rxlist.com (Sonata ®) Keywords zaleplon, Sonata zaleplon.txt · Last modified: 2024/02/07 17:23 by cclarks Page Tools Show pagesource Old revisions Backlinks Back to top TUSOM SOM \| Pharmacology Department \| TMedWeb
2237	https://en.wikipedia.org/wiki/Polar	Jump to content Polar Català Čeština Deutsch Español Euskara فارسی Français 한국어 Italiano עברית Magyar Nederlands 日本語 Norsk bokmål Norsk nynorsk Polski Português Suomi Svenska Türkçe Edit links From Wikipedia, the free encyclopedia Look up polar in Wiktionary, the free dictionary. Polar(s) may refer to: Geography [edit] Geographical pole, either of the two points on Earth where its axis of rotation intersects its surface Polar climate, the climate common in polar regions Polar regions of Earth, locations within the polar circles, referred to as the Arctic and Antarctic Places [edit] Polar, Wisconsin, town in Langlade County, Wisconsin, United States Polar (community), Wisconsin, unincorporated community in Langlade County, Wisconsin, United States Arts, entertainment, and media [edit] Music [edit] Polar (album), second album by the High Water Marks Polar Music, a record label Polar Studios, music studio of ABBA in Sweden Polars (album), an album by the Dutch metal band, Textures Other uses in arts and entertainment [edit] Polar, a 2002 novel by T. R. Pearson Polar (webcomic), a webcomic and series of graphic novels by Víctor Santos Polar (film), a 2019 Netflix film adaption of the comic series Brands and enterprises [edit] Polar Air Cargo, an American airline Polar Airlines, a Russian airline Polar Beverages, an American soft drink company Polar Electro, a Finnish manufacturer of sports training computers Sisu Polar, a truck model series produced by the Finnish heavy vehicle producer Sisu Auto Linguistics [edit] Grammatical polarity, a grammatical category of affirmative vs. negative Polar question, a question that can be answered yes or no Mathematics [edit] Polar point group, a symmetry in geometry and crystallography Pole and polar (a point and a line), a construction in geometry Polar cone Polar coordinate system, uses a central point and angles Polar curve (a point and a curve), a generalization of a point and a line Polar set, with respect to a bilinear pairing of vector spaces Science and technology [edit] Chemical polarity, a concept in chemistry which describes how equally bonding electrons are shared between atoms Polar (satellite), a satellite launched by NASA in 1996 Polar (star), a strongly magnetic cataclysmic variable star system POLAR III and POLAR II, a pedestrian test dummy created by Honda, used to study pedestrian injuries in road traffic accidents Polars (software), an open-source software library for data manipulation Other uses [edit] Polar (musician), Norwegian electronic music producer See also [edit] All pages with titles containing Polar Festival Polar de Cognac, a French festival focused on crime fiction Polar curve (aviation), a diagram that depicts the gliding performance of an aircraft Polar fleece, an insulating synthetic wool fabric Polar organelle, a specialised region of the cell membrane found surrounding the flagella base(s) in some bacteria Polar overdominance a form of genetic mutation Polarity (disambiguation) Polarization (disambiguation) Pole (disambiguation) Topics referred to by the same term Retrieved from " Category: Disambiguation pages Hidden categories: Short description is different from Wikidata All article disambiguation pages All disambiguation pages
2238	https://www.geeksforgeeks.org/dsa/generate-all-permutations-of-a-string-that-follow-given-constraints/	Generate all permutations of a string that follow given constraints Last Updated : 11 Jul, 2025 Suggest changes 13 Likes Given a string, generate all permutations of it that do not contain 'B' after 'A', i.e., the string should not contain "AB" as a substring. Examples: Input : str = "ABC" Output : ACB, BAC, BCA, CBA Out of 6 permutations of "ABC", 4 follow the given constraint and 2 ("ABC" and "CAB") do not follow. Input : str = "BCD" Output : BCD, BDC, CDB, CBD, DCB, DBC A simple solution is to generate all permutations. For every permutation, check if it follows the given constraint. C++ ```` // Simple C++ program to print all permutations // of a string that follow given constraint include using namespace std; void permute(string& str, int l, int r) { // Check if current permutation is // valid if (l == r) { if (str.find("AB") == string::npos) cout << str << " "; return; } // Recursively generate all permutation for (int i = l; i <= r; i++) { swap(str[l], str[i]); permute(str, l + 1, r); swap(str[l], str[i]); } } // Driver Code int main() { string str = "ABC"; permute(str, 0, str.length() - 1); return 0; } ```` // Simple C++ program to print all permutations // Simple C++ program to print all permutations // of a string that follow given constraint // of a string that follow given constraint ``` include #include ``` using namespace std; using namespace std void permute(string& str, int l, int r) void permute string& str int l int r { // Check if current permutation is // Check if current permutation is // valid // valid if (l == r) {if l == r if (str.find("AB") == string::npos) if str find "AB" ==string::npos cout << str << " "; cout<< str<< " " return; return } // Recursively generate all permutation // Recursively generate all permutation for (int i = l; i <= r; i++) {for int i = l i<= r i ++ swap(str[l], str[i]); swap str l str i permute(str, l + 1, r); permute str l + 1 r swap(str[l], str[i]); swap str l str i } } // Driver Code // Driver Code int main() int main { string str = "ABC"; string str = "ABC" permute(str, 0, str.length() - 1); permute str 0 str length - 1 return 0; return 0 } Java ```` // Simple Java program to print all // permutations of a String that // follow given constraint import java.util.; class GFG{ static void permute(char[] str, int l, int r) { // Check if current permutation is // valid if (l == r) { if (!String.valueOf(str).contains("AB")) System.out.print(String.valueOf(str) + " "); return; } // Recursively generate all permutation for(int i = l; i <= r; i++) { char tmp = str[l]; str[l] = str[i]; str[i] = tmp; permute(str, l + 1, r); tmp = str[l]; str[l] = str[i]; str[i] = tmp; } } // Driver Code public static void main(String[] args) { String str = "ABC"; permute(str.toCharArray(), 0, str.length() - 1); } } // This code is contributed by Amit Katiyar ```` Python ```` Simple Python program to print all permutations of a string that follow given constraint def permute(str, l, r): # Check if current permutation is # valid if (l == r): if "AB" not in ''.join(str): print(''.join(str), end=" ") return # Recursively generate all permutation for i in range(l, r + 1): str[l], str[i] = str[i], str[l] permute(str, l + 1, r) str[l], str[i] = str[i], str[l] Driver Code str = "ABC" permute(list(str), 0, len(str) - 1) This code is contributed by SHUBHAMSINGH10 ```` C# ```` // Simple C# program to print all permutations // of a string that follow given constraint using System; using System.Text; class GFG { static void permute(StringBuilder str, int l, int r) { // Check if current permutation is // valid if (l == r) { if (str.ToString().IndexOf("AB") == -1) { Console.Write(str.ToString() + " "); } return; } // Recursively generate all permutation for (int i = l; i <= r; i++) { char tmp = str[l]; str[l] = str[i]; str[i] = tmp; permute(str, l + 1, r); tmp = str[l]; str[l] = str[i]; str[i] = tmp; } } // Driver code static void Main(string[] arg) { string str = "ABC"; StringBuilder s = new StringBuilder(str); permute(s, 0, str.Length - 1); } } // This code is contributed by rutvik_56 ```` JavaScript ```` // JavaScript code to implement the above approach function permute(str, l, r) { // Check if current permutation is // valid if (l == r) { if (str.indexOf("AB") == -1) document.write(str," "); return; } // Recursively generate all permutation for (let i = l; i <= r; i++) { let a = str.split(""); let temp = a[l]; a[l] = a[i]; a[i] = temp; permute(a.join(""), l + 1, r); temp = a[l]; a[l] = a[i]; a[i] = temp; } } // Driver Code let str = "ABC"; permute(str, 0, str.length - 1); // This code is contributed by shinjanpatra ```` Output ACB BAC BCA CBA Time Complexity: O(n! X n) where n! is the number of permutations generated and O(n) times to print a permutation. Auxiliary Space: O(n! X n ) The above solution first generates all permutations, then for every permutation, it checks if it follows given constraint or not. An efficient solution is to use Backtracking. We cut down the recursion tree whenever we see that substring "AB" is formed. How do we do this? we add a isSafe() function. Before doing a swap, we check if previous character is 'A' and current character is 'B'. Below is the implementation of the above code: C++ ```` // Backtracking based CPP program to print all // permutations of a string that follow given // constraint include using namespace std; bool isSafe(string& str, int l, int i, int r) { // If previous character was 'A' and character // is 'B', then do not proceed and cut down // the recursion tree. if (l != 0 && str[l - 1] == 'A' && str[i] == 'B') return false; // This condition is explicitly required for // cases when last two characters are "BA". We // do not want them to swapped and become "AB" if (r == l + 1 && str[i] == 'A' && str[l] == 'B' \|\| r == l + 1 && l == i && str[r] == 'B' && str[l] == 'A') return false; return true; } void permute(string& str, int l, int r) { // We reach here only when permutation // is valid if (l == r) { cout << str << " "; return; } // Fix all characters one by one for (int i = l; i <= r; i++) { // Fix str[i] only if it is a // valid move. if (isSafe(str, l, i, r)) { swap(str[l], str[i]); permute(str, l + 1, r); swap(str[l], str[i]); } } } // Driver Code int main() { string str = "ABC"; // Function call permute(str, 0, str.length() - 1); return 0; } ```` // Backtracking based CPP program to print all // Backtracking based CPP program to print all // permutations of a string that follow given // permutations of a string that follow given // constraint // constraint ``` include #include ``` using namespace std; using namespace std bool isSafe(string& str, int l, int i, int r) bool isSafe string& str int l int i int r { // If previous character was 'A' and character // If previous character was 'A' and character // is 'B', then do not proceed and cut down // is 'B', then do not proceed and cut down // the recursion tree. // the recursion tree. if (l != 0 && str[l - 1] == 'A' && str[i] == 'B') if l!= 0&& str l - 1 == 'A'&& str i == 'B' return false; return false // This condition is explicitly required for // This condition is explicitly required for // cases when last two characters are "BA". We // cases when last two characters are "BA". We // do not want them to swapped and become "AB" // do not want them to swapped and become "AB" if (r == l + 1 && str[i] == 'A' && str[l] == 'B' if r == l + 1&& str i == 'A'&& str l == 'B' \|\| r == l + 1 && l == i && str[r] == 'B'\|\| r == l + 1&& l == i&& str r == 'B' && str[l] == 'A')&& str l == 'A' return false; return false return true; return true } void permute(string& str, int l, int r) void permute string& str int l int r { // We reach here only when permutation // We reach here only when permutation // is valid // is valid if (l == r) {if l == r cout << str << " "; cout<< str<< " " return; return } // Fix all characters one by one // Fix all characters one by one for (int i = l; i <= r; i++) {for int i = l i<= r i ++ // Fix str[i] only if it is a // Fix str[i] only if it is a // valid move. // valid move. if (isSafe(str, l, i, r)) {if isSafe str l i r swap(str[l], str[i]); swap str l str i permute(str, l + 1, r); permute str l + 1 r swap(str[l], str[i]); swap str l str i } } } // Driver Code // Driver Code int main() int main { string str = "ABC"; string str = "ABC" ``` ``` // Function call // Function call permute(str, 0, str.length() - 1); permute str 0 str length - 1 return 0; return 0 } Java ```` // Backtracking based JAVA program // to print all permutations of a // string that follow given constraint public class GFG { public boolean isSafe(String str, int l, int i, int r) { // If previous character was 'A' // and character is 'B', then // do not proceed and cut down the // recursion tree. if (l != 0 && str.charAt(l - 1) == 'A' && str.charAt(i) == 'B') return false; // This condition is explicitly required // for cases when last two characters // are "BA". We do not want them to // swapped and become "AB" if (r == l + 1 && str.charAt(i) == 'A' && str.charAt(l) == 'B' \|\| r == l + 1 && l == i && str.charAt(r) == 'B' && str.charAt(l) == 'A') return false; return true; } / permutation function @param str string to calculate permutation for @param l starting index @param r end index / private void permute(String str, int l, int r) { // We reach here only when permutation // is valid if (l == r) System.out.print(str + " "); else { // Fix all characters one by one for (int i = l; i <= r; i++) { // Fix str[i] only if it is a // valid move. if (isSafe(str, l, i, r)) { str = swap(str, l, i); permute(str, l + 1, r); str = swap(str, l, i); } } } } / Swap Characters at position @param a string value @param i position 1 @param j position 2 @return swapped string / public String swap(String a, int i, int j) { char temp; char[] charArray = a.toCharArray(); temp = charArray[i]; charArray[i] = charArray[j]; charArray[j] = temp; return String.valueOf(charArray); } // Driver Code public static void main(String[] args) { String str = "ABC"; int n = str.length(); GFG permutation = new GFG(); // Function call permutation.permute(str, 0, n - 1); } } ```` Python ```` Backtracking based Python3 program to print all permutations of a string that follow given constraint def isSafe(str, l, i, r): # If previous character was 'A' and character # is 'B', then do not proceed and cut down # the recursion tree. if (l != 0 and str[l - 1] == 'A' and str[i] == 'B'): return False # This condition is explicitly required for # cases when last two characters are "BA". We # do not want them to swapped and become "AB" if (r == l + 1 and str[i] == 'A' and str[l] == 'B' or r == l + 1 and l == i and str[r] == 'B' and str[l] == 'A'): return False return True def permute(str, l, r): # We reach here only when permutation # is valid if (l == r): print(str, sep="", end=" ") return # Fix all characters one by one for i in range(l, r + 1): # Fix str[i] only if it is a # valid move. if (isSafe(str, l, i, r)): str[l], str[i] = str[i], str[l] permute(str, l + 1, r) str[l], str[i] = str[i], str[l] Driver Code str = "ABC" Function call permute(list(str), 0, len(str) - 1) This code is contributed by SHUBHAMSINGH10 ```` C# ```` // Backtracking based C# program to print all // permutations of a string that follow given // constraint using System; public class GFG { // Backtracking based C# program // to print all permutations of a // string that follow given constraint using System; using System.Text; class GFG { static bool isSafe(StringBuilder str, int l, int i, int r) { // If previous character was 'A' // and character is 'B', then do not // proceed and cut down the recursion tree. if (l != 0 && str[l - 1] == 'A' && str[i] == 'B') return false; // This condition is explicitly // required for cases when last two // characters are "BA". We do not want // them to swapped and become "AB" if (r == l + 1 && str[i] == 'A' && str[l] == 'B' \|\| r == l + 1 && l == i && str[r] == 'B' && str[l] == 'A') return false; return true; } static void permute(StringBuilder str, int l, int r) { // We reach here only when permutation // is valid if (l == r) { Console.Write(str + " "); return; } // Fix all characters one by one for (int i = l; i <= r; i++) { // Fix str[i] only if it is a // valid move. if (isSafe(str, l, i, r)) { char temp = str[l]; str[l] = str[i]; str[i] = temp; permute(str, l + 1, r); temp = str[l]; str[l] = str[i]; str[i] = temp; } } } // Driver code static void Main() { string str = "ABC"; StringBuilder s = new StringBuilder(str); // Function call permute(s, 0, str.Length - 1); } } // This code is contributed by divyeshrabadiya07 ```` JavaScript ```` const GFG = { isSafe(str, l, i, r) { // If previous character was 'A' // and character is 'B', then // do not proceed and cut down the // recursion tree. if (l !== 0 && str[l - 1] === 'A' && str[i] === 'B') return false; // This condition is explicitly required // for cases when last two characters // are "BA". We do not want them to // swapped and become "AB" if ((r === l + 1 && str[i] === 'A' && str[l] === 'B') \|\| (r === l + 1 && l === i && str[r] === 'B' && str[l] === 'A')) { return false; } return true; }, permute(str, l, r) { // We reach here only when permutation // is valid if (l === r) console.log(${str}); else { // Fix all characters one by one for (let i = l; i <= r; i++) { // Fix str[i] only if it is a // valid move. if (this.isSafe(str, l, i, r)) { str = this.swap(str, l, i); this.permute(str, l + 1, r); str = this.swap(str, l, i); } } } }, swap(a, i, j) { let temp; const charArray = a.split(''); temp = charArray[i]; charArray[i] = charArray[j]; charArray[j] = temp; return charArray.join(''); }, }; // Driver Code const str = 'ABC'; const n = str.length; // Function call GFG.permute(str, 0, n - 1); ```` Output ACB BAC BCA CBA K kartik Improve Article Tags : Misc Strings Backtracking Combinatorial DSA Explore DSA Fundamentals Logic Building Problems 2 min readAnalysis of Algorithms 1 min read Data Structures Array Data Structure 3 min readString in Data Structure 2 min readHashing in Data Structure 2 min readLinked List Data Structure 2 min readStack Data Structure 2 min readQueue Data Structure 2 min readTree Data Structure 2 min readGraph Data Structure 3 min readTrie Data Structure 15+ min read Algorithms Searching Algorithms 2 min readSorting Algorithms 3 min readIntroduction to Recursion 14 min readGreedy Algorithms 3 min readGraph Algorithms 3 min readDynamic Programming or DP 3 min readBitwise Algorithms 4 min read Advanced Segment Tree 2 min readBinary Indexed Tree or Fenwick Tree 15 min readSquare Root (Sqrt) Decomposition Algorithm 15+ min readBinary Lifting 15+ min readGeometry 2 min read Interview Preparation Interview Corner 3 min readGfG160 3 min read Practice Problem GeeksforGeeks Practice - Leading Online Coding Platform 6 min readProblem of The Day - Develop the Habit of Coding 5 min read Improvement Suggest Changes Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal. Create Improvement Enhance the article with your expertise. Contribute to the GeeksforGeeks community and help create better learning resources for all. Suggest Changes min 4 words, max Words Limit:1000 Thank You! Your suggestions are valuable to us. What kind of Experience do you want to share? Interview Experiences Admission Experiences Career Journeys Work Experiences Campus Experiences Competitive Exam Experiences
2239	https://www.365jz.com/article/29124	常见的收集数据的方法有哪些 - 365建站网首页网站模板整站模板站群模板应用中心站群模板插件整站建站软件解决方案帮助中心教程中心建站论坛您现在的位置：365建站网>365文章> 常见的收集数据的方法有哪些常见的收集数据的方法有哪些文章来源：365jz.com 点击数：425 更新时间：2023-11-12 23:08 参与评论在现代信息时代，数据已经成为一种非常重要的资源。无论是企业、政府还是学术界，都需要收集数据来进行决策和分析。下面将介绍一些常见的数据收集方法。问卷调查：问卷调查是最常见的数据收集方法之一。通过设计一系列问题，可以收集到大量的定量和定性数据。问卷调查可以通过面对面、电话、邮件或在线形式进行。优点是可以一次性收集到大量数据，而缺点是可能受到回答者主观偏见的影响。访谈：访谈是一种深入了解个体或群体观点和经验的方法。可以通过面对面、电话或视频等方式进行。访谈可以提供详细和全面的信息，但是受访者可能会受到访谈者的影响而回答不真实的问题。观察：观察是一种通过观察和记录行为、事件和现象来收集数据的方法。观察可以是有目的的，也可以是无目的的。观察可以提供客观的数据，但是可能会受到观察者主观偏见的影响。实验：实验是一种通过控制和操纵变量来收集数据的方法。实验可以用于确定因果关系和验证假设。实验需要设计实验组和对照组，并确保实验过程的可重复性和可靠性。文献研究：文献研究是通过收集和分析已有的文献和资料来收集数据的方法。可以通过阅读书籍、期刊论文、报纸文章、互联网等途径进行。文献研究可以提供历史数据和理论支持，但是可能受到文献来源的限制和偏见的影响。网络爬虫：网络爬虫是一种自动化的收集网络上信息的方法。可以通过编写程序来自动抓取网页上的数据。网络爬虫可以高效地收集大量的数据，但是需要注意合法性和道德问题。数据挖掘：数据挖掘是一种通过使用统计和机器学习等方法来发现和提取隐藏在大量数据中的模式和关联的方法。数据挖掘可以帮助发现未知的信息和规律，但是需要具备相关的技术和专业知识。总结起来，常见的收集数据的方法包括问卷调查、访谈、观察、实验、文献研究、网络爬虫和数据挖掘等。不同的方法适用于不同的研究目的和数据类型。在进行数据收集时，需要注意合法性、可靠性和道德问题，以确保数据的质量和可信度。同时，还需要根据具体情况选择合适的方法和工具，以提高数据收集的效率和准确性。如对本文有疑问，请提交到交流论坛，广大热心网友会为你解答！！点击进入论坛您可能感兴趣的文章: 数据错误循环冗余检查大数据的应用有哪些 php 几种常见的快速排序代码大数据技术就业和发展前景数据库设计规范有哪些发表评论(425 人查看， 0 条评论) 请自觉遵守互联网相关的政策法规，严禁发布色情、暴力、反动的言论。昵称：提交评论最新评论 ------分隔线---------------------------- 上一篇：ps羽化在哪怎么用下一篇：bonjour是什么软件大家感兴趣的内容 1 数据错误循环冗余检查 2 大数据的应用有哪些 3 php 几种常见的快速排序代码 4 大数据技术就业和发展前景 5 数据库设计规范有哪些 6 如何做数据分析 7 Oracle数据库报错ORA-00904: 标识符无效问题解决办法 8 数据库三范式是指什么 9 mysql数据库管理系统软件有哪些 10 数据挖掘的意义及价值实例最近更新的内容 1 jquery获取元素的所有宽高(包括边框内边距和外边距) 2 复制移动整个文件夹php代码函数三种写法 3 cefsharp devtools 截取节点屏幕截图功能 vb.net代码 4 vb.net Application.ThreadException捕获程序自动退出事件代码 5 http请求 412 Precondition Failed是什么 6 vb.net调用Microsoft.Office.Interop.Excel.dll 超过65536行报错解决方法 7 网站建设什么公司好 8 电脑画图软件有哪些 9 什么是ipc 10 epic是什么平台公司信息 · 网站介绍· 联系我们· 加入我们· 服务条款快速入口 · 365软件· 杰创官网· 建站工具· 网站大全其它栏目 · 建站教程· 365学习业务咨询 · 技术支持· 服务时间：9:00-18:00 Powered by 365建站网RSS地图HTML地图 copyright © 2013-2024 版权所有鄂ICP备17013400号
2240	http://fractal.math.unr.edu/~ejolson/311-06/handout/sumpart.pdf	Abel’s Summation by Parts Formula Given two sequences (an)n∈N and (bn)n∈N of real numbers consider the partial sums An = n X k=1 ak and Bn = n X k=1 bk. Further deﬁne A0 = 0 and B0 = 0. Thus, Ak −Ak−1 = ak and Bk −Bk−1 = bk for k = 1, 2, . . . . Now consider the diﬀerence AkBk −Ak−1Bk−1 of the products. Thus, AkBk −Ak−1Bk−1 = AkBk −AkBk−1 + AkBk−1 −Ak−1Bk−1 = Ak(Bk −Bk−1) + (Ak −Ak−1)Bk−1 = Akbk + akBk−1. Given m, n ∈N with m < n, sum both sides from k = m to n. Since the left side telescopes we have n X k=m AkBk −Ak−1Bk−1 = AnBn −Am−1Bm−1. Therefore AnBn −Am−1Bm−1 = n X k=m Akbk + n X k=m akBk−1. This formula should be compared to the usual integration by parts formula. Let F and G be diﬀerentiable functions on [a, b] with derivatives F ′ = f and G′ = g. Then by the product rule (FG)′ = FG′ + F ′G = Fg + fG. Integrate both sides from a to b. The Fundamental Theorem of Calculus Part I implies that Z b a (FG)′ = F(b)G(b) −F(a)G(a). Therefore, F(b)G(b) −F(a)G(a) = Z b a F(x)g(x) dx + Z b a f(x)G(x) dx. Comparing this formula to one above we see that sums play the role of integrals, ak plays the role of the derivative of Ak, and similarly bk is like the derivative of Bk. Finally, the summation by parts formula in the book follows by taking m = 1 and using A0 = 0 and B0 = 0 to obtain AnBn = n X k=1 Akbk + n X k=1 akBk−1 and then identifying the books notation bn+1 with our Bn.
2241	https://www.calculatoratoz.com/en/critical-depth-for-rectangular-channel-calculator/Calc-18293	Critical Depth for Rectangular Channel Calculator \| \| \| \| \| --- --- \| \| ✖The Discharge per unit Width is the ratio of total discharge in the channel to the width considered.ⓘ Discharge per unit Width [q] \| \| CentistokesKilostokesMegastokesSquare Centimeter per SecondSquare Foot per SecondSquare Meter per HourSquare Meter per SecondSquare Millimeter per SecondStokes \| +10% -10% \| \| \| \| \| \| --- --- \| \| ✖The Critical Depth of Rectangular Channel is defined as the depth of flow where energy is at a minimum for a particular discharge.ⓘ Critical Depth for Rectangular Channel [hr] \| \| Angstrom Astronomical Unit Centimeter Decimeter Earth Equatorial Radius Fermi Foot Inch Kilometer Light Year Meter Microinch Micrometer Micron Mile Millimeter Nanometer Picometer Yard \| ⎘ Copy \| Critical Depth for Rectangular Channel Solution Credits < Critical Flow and its Computation Calculators Critical Depth for Rectangular Channel Formula What is Critical Depth? The concept of critical depth is conventionally defined in open-channel hydraulics (Chow 1959; Montes 1998; Chanson 2004) as the depth at which the specific energy reaches a minimum value, considering the mean specific energy Hm within the whole flow section in flows with parallel streamlines. How to Calculate Critical Depth for Rectangular Channel? Critical Depth for Rectangular Channel calculator uses Critical Depth of Rectangular Channel = ((Discharge per unit Width^2)/([g]))^(1/3) to calculate the Critical Depth of Rectangular Channel, The Critical Depth for Rectangular Channel is defined as the minimum flow depth at which specific flow conditions occur, influencing flow velocity and discharge in hydraulic systems. Critical Depth of Rectangular Channel is denoted by hr symbol. How to calculate Critical Depth for Rectangular Channel using this online calculator? To use this online calculator for Critical Depth for Rectangular Channel, enter Discharge per unit Width (q) and hit the calculate button. Here is how the Critical Depth for Rectangular Channel calculation can be explained with given input values -> 2.182934 = ((10.1^2)/([g]))^(1/3). FAQ Home
2242	https://sg.indeed.com/career-advice/career-development/client-vs-customer	Client vs. Customer (Definitions, Differences and Examples) \| Indeed.com Singapore Skip to main content Home Company reviews Salary guide Sign in Sign in Employers / Post Job 1 new update Home Company reviews Salary guide Employers Create your CV Change country 🇸🇬 Singapore Help Privacy Centre Start of main content Career Guide Finding a Job Resumes & Cover Letters Resumes & Cover Letters articles Resume samples Cover letter samples Interviewing Pay & Salary Career Development Career Development Client vs. Customer (Definitions, Differences and Examples) Client vs. Customer (Definitions, Differences and Examples) Written by Indeed Editorial Team Updated 5 June 2025 Clients and customers are people who engage in business transactions with companies. They're buyers who purchase goods or services from sellers. Learning the differences between these two groups may help you tailor successful sales and marketing strategies that meet their needs. In this article, we explore the differences between clients vs. customers, provide examples of businesses that rely on them and offer tips on how to turn customers into clients. Related jobs on Indeed Part-time jobs Full-time jobs Remote jobs Urgently needed jobs View more jobs on Indeed Definition of client vs. customer To understand the distinction between a client vs. a customer, it's important to examine their definitions. Clients refer to people who purchase professional services from a business or company to satisfy a need or solve a problem. Customers are people who pay for goods and services from companies or stores without going into a formal business relationship with these establishments.Clients typically buy solutions or advice, while customers usually buy products and services. Clients may have long-term relationships with the businesses they purchase from. In contrast, customers often end their engagements with sellers once they receive the goods or services they want to buy. Examples of clients Many companies that offer professional services often refer to the people they do business with as clients. Clients usually require services that satisfy a personal or commercial need on a recurring basis. Here are some examples of industries and businesses that handle clients: law firms marketing and advertising agencies accounting services design studios insurance firms real estate agencies Related:What Is Client Servicing? (With Definition and Tips) Examples of customers Customers perform monetary transactions to buy products or services from businesses and stores. These transactions may be non-recurring, but there may be some customers who make purchases repeatedly. Here are some examples of businesses that handle customers: retail stores food and beverage establishments hotels banks hair salons supermarkets Related:How to Handle Difficult Customers in 10 Steps (With Tips) Differences between client and customer While both clients and customers are important sources of revenue for businesses, there are differences between them in terms of their behaviours and requirements. Successful businesses understand how to serve clients or customers according to their needs. Here are some differences between them: Level of personalisation Businesses serving clients can differentiate their services according to the clients' requirements. For example, an HR consultancy firm may offer a wide range of services. If a client requires professional services in payroll processing and recruitment, the consultancy firm may customise its offerings to target these two areas. The level of personalisation is lower for customers, as they usually purchase ready-made products or services with no customisation. Level of exclusivity Clients typically seek professional services from a single provider. For instance, clients may rely exclusively on a private bank for both their banking and investment needs. There's typically a lower level of exclusivity with customers. For example, a restaurant's customers are likely to patronise other restaurants too.Related: Client Services Resume Example (With Tips) Expectations of service support Businesses that handle clients can expect to provide higher levels of service support. For example, a law consultancy firm for a manufacturing company may receive frequent support requests relating to legal aspects of the company's affairs. Customers may also require service support, such as requesting product refunds or exchanges from retail stores. The support for customers is usually only for that purchase, and the engagement typically ends after customer service resolves the issue. Marketing and sales strategies Companies may adopt different marketing and sales strategies depending on whether they serve clients or customers. Client-based companies may have a smaller target market, but focus on the retention of their clients. For instance, a law firm may rely on a small base of long-term key clients who provide their business revenue for a long period. They may attract new business by emphasising their service levels and client testimonials.In contrast, companies that rely on customers may use marketing strategies that attract a wider range of customers and encourage them to spend more. They typically aim to generate a larger base of transactional customers, as they require a higher number of purchases to achieve revenue targets.Related:What Does a Marketing and Sales Professional Do? (With Skills) Duration of the sales process It often takes longer to sell to clients than to customers. The sales process starts from initial client contact to closing the deal and follow-ups. As client-based businesses usually involve larger financial commitments, the sales cycle may be longer as clients spend more time selecting the best provider that suits their requirements. Customers may have shorter sales cycles as they perform quick transactions that usually involve a single purchase. Presence of contractual obligations Clients often sign formal contracts when they enter sales relationships with companies. For example, if a client engages the services of an accounting consultancy firm, both parties may sign a contract that includes the scope of service, fees and withdrawal clauses. There are usually no contractual obligations for customers, as the value of products and services are usually lower. Style and depth of relationship Client-centric companies may provide more personal attention to their individual clients. These companies may invest more resources to develop strong relationships with their clients, such as employing dedicated client service representatives. Customers may base their purchasing decisions on convenience, location and price. Relationships with a particular business and their service staff may be a less important factor for customers to consider.Related:What Does a Customer Service Representative Do? (Plus Salary) Differences between customer and consumer Consumer is another term that's interchangeable with customer, but there are differences between them. Consumers refer to people who are the end users of products and services. Customers may buy the products and services, but they may not be the ones using them.Knowing this distinction is useful for businesses who may target both consumers and customers. Successful businesses understand that a consumer can influence the purchasing decisions of a customer. These businesses may tailor their sales and marketing strategies to suit the preferences of consumers. Search jobs and companies hiring now Job title, keywords or company Location Search Tips on how to turn customers into clients Companies that serve customers and consumers usually perform single transactions and rarely develop deep relationships with those who pay for their products and services. There are some companies that aim to turn short-term customers into long-term clients. Loyal customers become clients when they patronise a certain business regularly, perform repeated transactions and develop relationships with the business owners or staff.Companies that invest resources to turn customers into clients realise it may be easier to keep a returning customer than to find a new one. Serving a group of repeat clients may lead to a more stable revenue than relying on new customers. Here are some tips on how to turn customers into clients: Provide attractive incentives To attract existing customers to return and make recurring purchases, you can offer incentives such as free consultations and customised trials. For instance, a beauty salon may turn their one-time customers into clients by offering complimentary manicure services. This may encourage existing customers to come back regularly and successfully convert them into clients. Provide excellent customer service Companies that provide high levels of service may do recurring businesses with their customers. When they receive excellent service, they're more likely to return for more purchases and recommend these businesses to others. For example, a grocery store manager who greets customers by their names is likely to develop good relationships with them and turn them into regular clients.Related:What Is a Customer Relations Manager? (With Average Salary) Maintain regular communication Customers are more likely to transact again with businesses that listen to their opinions and cater to their needs. Communicate regularly with customers as if you're serving clients and address their concerns. Show that you understand their requirements and offer personalised offers or discounts. Successful businesses maintain personal connections with their customers to develop them into clients. Ensure product or service quality The quality of your product or service is a direct factor that influences your customers' purchasing decisions. Ensure that you offer products and services that are consistent in quality. When customers feel confident that they can expect good quality products and services every time they patronise your business, the likelihood of them making recurring purchases is higher. Adopt a relational selling model A relational selling model aims to build long-term relationships with the people who provide revenue for your business. This differs from a transactional selling model, where the focus is on closing short-term sales. When you use relational selling, you put your customer's interests ahead of your self-interest. Companies that succeed at relational selling have a better chance of turning customers into clients. The information on this site is provided as a courtesy and for informational purposes only. Indeed is not a career or legal advisor and does not guarantee job interviews or offers. Share: Related Articles How to Manage Customer Relationships (With Importance) 9 Types of Customer Needs (Plus Steps on How to Meet Them) What Is Customer Relations? 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2243	https://fiveable.me/key-terms/combinatorics/restricted-permutations	printables 🧮combinatorics review key term - Restricted Permutations Citation: MLA Definition Restricted permutations refer to arrangements of a set of items where certain conditions limit how those items can be ordered. These conditions can include constraints like specific items needing to be together, certain items being excluded from specific positions, or maintaining a particular sequence among some of the elements. Understanding restricted permutations is essential for solving combinatorial problems where limitations are placed on the arrangement of objects. 5 Must Know Facts For Your Next Test When calculating restricted permutations, it's common to first consider the total number of unrestricted permutations and then subtract cases that violate the restrictions. Common restrictions might include keeping certain elements together as a block or ensuring that no two identical items are adjacent. The principle of complementary counting is often applied in restricted permutations by counting the total arrangements and then subtracting invalid cases. When handling groups or blocks within restricted permutations, treating these blocks as single entities simplifies calculations. Restricted permutations can often be solved using recursive methods or generating functions to account for varying conditions. Review Questions How do restricted permutations differ from unrestricted ones, and what methods can be used to solve problems involving them? Restricted permutations differ from unrestricted ones in that they have specific conditions that limit how elements can be arranged. To solve problems involving restricted permutations, one can use methods like complementary counting, where you first calculate the total number of unrestricted arrangements and then subtract the invalid configurations. Additionally, treating groups or blocks as single entities helps simplify complex arrangements into manageable calculations. What role does the Inclusion-Exclusion Principle play in calculating restricted permutations? The Inclusion-Exclusion Principle plays a critical role in calculating restricted permutations by providing a systematic approach to account for overlaps between restricted sets. When certain arrangements violate multiple restrictions, simply subtracting invalid cases would lead to double-counting. The principle helps in accurately determining valid arrangements by ensuring all intersections are appropriately included or excluded based on their counts, thereby refining the total count of valid restricted permutations. Evaluate how the concept of factorials is integrated into understanding restricted permutations, particularly in complex arrangement scenarios. Factorials are integral to understanding restricted permutations as they provide the foundational framework for calculating total arrangements without restrictions. In complex scenarios where restrictions apply, factorial calculations help establish a baseline from which adjustments can be made. For instance, if certain elements must remain together as a block, factorials help determine the total arrangements by treating this block as a single unit while still considering the internal arrangement possibilities within that block. Thus, factorials facilitate both basic and modified calculations essential for grasping the nuances of restricted permutations. Related terms Factorial: A mathematical function that multiplies a given number by every positive integer less than that number, denoted as n!. Combinations: A selection of items from a larger set where the order does not matter, contrasting with permutations where order is significant. Inclusion-Exclusion Principle: A counting technique used to calculate the size of the union of multiple sets by including the sizes of individual sets and excluding the sizes of their intersections.
2244	https://www.math.uh.edu/~irina/MATH3336/3336Notes/3336S11.pdf	Page 1 of 9 © 2020, I. Perepelitsa Discrete Mathematics Propositional Logic What is a proposition? Definition: A proposition (or a statement) is a sentence that is either true or false, but not both. Examples of Propositions: a. Austin is the capital of Texas. b. Texas is the largest state of the United States. c. 1 + 0 = 1 Examples that are NOT Propositions: a. Watch out! b. What time is it? c. 𝑥+ 3 = 5  Letters are used to denote propositions: 𝑝, 𝑞, 𝑟, 𝑠…  The truth value of a proposition that is always true denoted by 𝑇, the truth value of a proposition that is always false denoted by 𝐹. New propositions (compound propositions) can be formed from existing propositions using logical operators. Definition: Let 𝑝 be a proposition. The negation of 𝑝, denoted by ¬𝑝, the statement “It is not the case that 𝑝.” Examples: a. Proposition: A triangle has three sides. Negation: It is not the case that triangle has three sides. Negation in simple English: A triangle does not have three sides. b. Proposition: All fish can swim. Negation: It is not the case that all fish can swim. Negation in simple English: Some fish cannot swim. c. Proposition: 2 + 3 > 5 Negation: Page 2 of 9 © 2020, I. Perepelitsa A proposition and its negation have OPPOSITE truth values! Construct a truth table for the negation of 𝑝. 𝒑 ¬𝒑 Definition: Let 𝑝 and 𝑞 be propositions. The conjunction of 𝑝 and 𝑞, denoted by 𝑝∧𝑞, is the proposition “𝑝 and 𝑞.” The conjunction is true when BOTH 𝑝 and 𝑞 are true and is false otherwise. Example: Construct a truth table for the conjunction. 𝒑 𝒒 𝒑∧𝒒 Example: Find the conjunction of the following propositions and determine its truth value. a. 𝑝: All birds can fly. 𝑞: 2 + 3 = 5 Conjunction: Page 3 of 9 © 2020, I. Perepelitsa Definition: Let 𝑝 and 𝑞 be propositions. The disjunction of 𝑝 and 𝑞, denoted by 𝑝∨𝑞, is the proposition “𝑝 or 𝑞.” The disjunction is false when BOTH 𝑝 and 𝑞 are false and is true otherwise. Example: Construct the truth table for the disjunction. 𝒑 𝒒 𝒑∨𝒒 T T T F F T F F Example: Find the disjunction of the following propositions and determine its truth value. a. 𝑝: Triangles are square. 𝑞: Circles are round. Disjunction: Definition: Let 𝑝 and 𝑞 be propositions. The exclusive of 𝑝 and 𝑞, denoted by 𝑝⨁𝑞, is the proposition “𝑝 or 𝑞, but not both.” The exclusive is true when ONE of 𝑝 and 𝑞 is true and is false otherwise. Example: Students who have taken calculus or computer science can take this class. Soup or salad comes with this entrée. Page 4 of 9 © 2020, I. Perepelitsa Example: Construct the truth table for the exclusive. 𝒑 𝒒 𝒑⨁𝒒 T T T F F T F F Definition: Let 𝑝 and 𝑞 be propositions. The conditional statement (implication) 𝑝→𝑞 is the proposition “if 𝑝, then 𝑞.” The conditional statement 𝑝→𝑞 is false then 𝑝 is true and 𝑞 is false, and true otherwise. In the conditional statement 𝑝→𝑞, 𝑝 is called hypothesis and 𝑞 is called conclusion. Example: The Truth Table for the Conditional Statement 𝑝→𝑞. 𝒑 𝒒 𝒑→𝒒 T T T T F F F T T F F T  Connection between the hypothesis and conclusion is NOT necessary.  Think: Implication = Contract. Example: a. If you get 100% on the final, then you will get an A. b. If the Moon made of cheese, then 1 + 1 = 2. Page 5 of 9 © 2020, I. Perepelitsa Different Ways of Expressing 𝒑→𝒒 if p, then q p implies q if p, q p only if q q unless ¬p q when p q if p q whenever p p is sufficient for q q follows from p q is necessary for p a necessary condition for p is q a sufficient condition for q is p Example: Write the statement in the “If…, then…” form. a. It is hot whenever it is sunny. b. To get a good grade it is necessary that you study. Page 6 of 9 © 2020, I. Perepelitsa Definitions: The proposition 𝑞→𝑝 is called converse. The proposition ¬𝑝→¬𝑞 is called inverse. The proposition ¬𝑞→¬𝑝 is called contrapositive. Example: Write the converse, inverse, and contrapositive for the following statement. a. If 3 ≥5, then 7 > 7. Converse: Inverse: Contrapositive: b. I come to class whenever there is going to be a quiz. (Hint: Rewrite the proposition in the “if, then” form) Converse: Inverse: Contrapositive: Page 7 of 9 © 2020, I. Perepelitsa Definition: Let 𝑝 and 𝑞 be propositions. The biconditional statement 𝑝↔𝑞 is the proposition “𝑝 if and only if 𝑞.” The biconditional statement 𝑝↔𝑞 is true when p and q have the SAME truth values, and is false otherwise. “if and only if” = “iff” Example: You can drive a car if and only if your gas tank is not empty. Example: The Truth Table for the Biconditional Statement ↔𝑞 . 𝒑 𝒒 𝒑↔𝒒 T T T F F T F F Expressing the Biconditional p is necessary and sufficient for q if p then q , and conversely p iff q Truth Tables for Compound Propositions Construction of a truth table: 1. Rows  Need a row for every possible combination of values for the atomic propositions. 2. Columns  Need a column for the compound proposition (usually at far right)  Need a column for the truth value of each expression that occurs in the compound proposition as it is built up. Page 8 of 9 © 2020, I. Perepelitsa Example: Construct a truth table for (𝑝∨¬𝑞) →(𝑝∧𝑞) (𝑝∨¬𝑞) →(𝑝∧𝑞) Equivalent Propositions Definition: Two propositions are equivalent if they always have the same truth value. Example: Show using a truth table that the conditional is equivalent to the contrapositive. Page 9 of 9 © 2020, I. Perepelitsa Precedence of Logical Operators Operator Precedence ¬ 1 ∧ 2 ∨ 3 → 4 ↔ 5 Example: 𝑝∨𝑞→¬𝑟 is equivalent to If the intended meaning is 𝑝∨(𝑞→¬𝑟) then parentheses must be used.
2245	https://www.albert.io/blog/right-triangle-trigonometry-special-right-triangles-a-guide/	Skip to content ➜ Right Triangle Trigonometry & Special Right Triangles: A Guide The Albert Team Last Updated On: What We Review Introduction Right triangles show up on nearly every SAT® Math exam. When a problem involves ladders, ramps, or slanted roofs, right triangle trigonometry is usually the hidden tool. By the end of this guide, students will be able to: Apply the Pythagorean Theorem with confidence Use SOH-CAH-TOA in both calculator and non-calculator sections Recognize and exploit 30-60-90 and 45-45-90 special right triangles Therefore, score-boosting shortcuts will feel natural. Start practicing SAT® Math (Digital) on Albert now! Pythagorean Theorem Refresher A. Formula and Quick Facts For any right triangle, the squares of the legs add to the square of the hypotenuse: Key points: The hypotenuse is always opposite the 90^\circ angle. If two sides are known, the third follows quickly. Common SAT® leg lengths include 3-4-5, 5-12-13, and their multiples. B. Common SAT® Twists Legs with radicals: 5\sqrt{2} often pairs with 5\sqrt{2} in isosceles right triangles. Missing hypotenuse in coordinate plane distance questions. Hidden right angles are created by altitude lines or slopes. C. Example Points A (–2, 3) and B (4, –1) form two vertices of a right triangle with the right angle at A. Find the length of the hypotenuse \overline{AB}. Solution Identify the legs using the distance in x and y directions. Change in x = \|4 – (–2)\| = 6 Change in y = \|–1 – 3\| = 4 Because the right angle is at A, those lengths are legs. Therefore: a=6,b=4 Apply the theorem: 6^{2}+4^{2}=c^{2} 36+16=c^{2} 52=c^{2} c=\sqrt{52}=2\sqrt{13} Meet SOH-CAH-TOA: The Core of Right Triangle Trigonometry A. Defining Sine, Cosine, and Tangent For an acute angle θ in a right triangle: Sine: \sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}} Cosine: \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} Tangent: \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}} The mnemonic “SOH-CAH-TOA” summarizes these ratios. B. Setting Up Ratios First, label the triangle relative to angle θ. Next, assign each given length to the opposite, adjacent, or hypotenuse. Finally, substitute into the correct ratio. C. Calculator vs. Non-Calculator Sections Non-calculator: angles of 30^\circ, 45^\circ, 60^\circ dominate. Memorize their exact values. Calculator: Any angle may appear, yet set the device to degree mode. D. Example Problem In right triangle RST, m\angle T = 90^\circ and leg RS = 8 cm. If m\angle R = 37^\circ, find RT to the nearest tenth. Solution Use the cosine function: \cos(\angle R) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{RT}{RS} Substitute the known values: \cos(37^\circ) = \dfrac{RT}{8} Now solve for RT: RT = 8 \times \cos(37^\circ) RT \approx 8 \times 0.7986 = 6.4 \text{ cm} Complementary Angles: The Sine-Cosine Link Because the acute angles in a right triangle are complementary, \theta + (90^{\circ}-\theta)=90^{\circ}. Consequently, \sin\theta=\cos(90^{\circ}-\theta)\cos\theta=\sin(90^{\circ}-\theta) Quick Example If \cos25^{\circ}=0.9063, then \sin65^{\circ} must also equal 0.9063. No calculator required! Ready to boost your SAT® Math (Digital) scores? Explore our plans and pricing here! Special Right Triangles Cheat Sheet A. 45-45-90 Triangles Pattern: sides are x,,x,,x\sqrt{2} (legs, leg, hypotenuse). Derivation: An isosceles right triangle splits a square diagonally. Example Problem The legs of an isosceles right triangle measure 7 cm. Find the hypotenuse. Solution c=7\sqrt{2}\text{ cm}. B. 30-60-90 Triangles Pattern: x,;x\sqrt{3},;2x (short leg, long leg, hypotenuse). Derivation: drop an altitude in an equilateral triangle of side 2x. Example Problem The long leg of a 30-60-90 triangle is 9\sqrt{3}. Determine the other two sides. Solution Long leg = x\sqrt{3} → x=9. Short leg = x=9; hypotenuse = 2x=18. Using Similarity to Build Exact Trig Values By placing special right triangles on the unit circle, exact ratios follow. \| \| \| \| \| --- --- \| \| Angles \| Sine \| Cosine \| Tangent \| \| 30^\circ \| \dfrac{1}{2} \| \dfrac{\sqrt{3}}{2} \| \dfrac{\sqrt{3}}{3} \| \| 45^\circ \| \dfrac{\sqrt{2}}{2} \| \dfrac{\sqrt{2}}{2} \| 1 \| \| 60^\circ \| \dfrac{\sqrt{3}}{2} \| \dfrac{1}{2} \| \sqrt{3} \| Example Find \sin45^{\circ}. Solution: In a 45-45-90 triangle with legs 1, hypotenuse \sqrt{2}, \sin45^{\circ}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2} Strategy Section: Which Tool When? Decision bullets guide quick choices: Need a missing side and have the other two? → Use Pythagorean Theorem. Know one side and one non-right angle? → Apply SOH-CAH-TOA. Angle equal to 30^\circ, 45^\circ, 60^\circ? → Special triangle shortcuts save time. Asked for exact value, not decimal? → Prefer special triangles or similarity. Timed-test tips: Write the SOH-CAH-TOA letters on scrap paper immediately; it prevents mix-ups under pressure. Real-World & SAT®-Style Applications 1. Angle of Elevation Problem A surveyor stands 40 m from a tower. The line of sight to the top forms a 33^\circ angle of elevation. Estimate the tower’s height to the nearest meter. Solution Opposite = height, adjacent = 40 m. \tan33^{\circ}=\dfrac{h}{40} → h=40\tan33^{\circ}\approx40(0.649)=26\text{ m} 2. Coordinate Plane Tie-In Problem Triangle PQR has right angle at Q with Q(3, –2) and R(7, 5). If PQ = 5, find PR. Solution Find QR using distance formula: Δx = 7 – 3 = 4, Δy = 5 – (–2) = 7 → QR=\sqrt{4^{2}+7^{2}}=\sqrt{65}. Apply the Pythagorean Theorem to the triangle: pq^{2}+(QR)^{2}=(PR)^{2} 5^{2}+65=(PR)^{2} 25+65=90 → PR=\sqrt{90}=3\sqrt{10}. Quick Reference Vocabulary Chart \| \| \| --- \| \| Term \| Definition \| \| Hypotenuse \| The side opposite the right angle; longest side. \| \| Opposite Side \| The side across from a given acute angle. \| \| Adjacent Side \| The side next to a given acute angle (not the hypotenuse). \| \| Altitude \| A perpendicular segment from a vertex to the opposite side. \| \| Angle of Elevation \| The angle formed by a horizontal line and a line of sight upward. \| \| Angle of Depression \| The angle formed by a horizontal line and a line of sight downward. \| \| Complementary Angles \| Two angles whose measures add to 90^\circ. \| \| Similar Triangles \| Triangles with equal corresponding angles and proportional sides. \| Practice Questions A right triangle has legs 9 cm and 12 cm. Find the hypotenuse. In △XYZ, m\angle Y = 90^\circ, XY = 10 m, m\angle X = 28^\circ. Find YZ to the nearest tenth. Find \cos60^{\circ} using a special triangle. The hypotenuse of a 30-60-90 triangle is 14. What is the shorter leg? \angle A and \angle B are complementary. If \sin A=0.8, what is \cos B? Answer Key with Brief Explanations 15 10\sin28^{\circ}\approx4.7\text{ m} \cos60^{\circ}=\dfrac{1}{2}. 7. 0.8. Key Takeaways & Next Steps Memorize the Pythagorean triple patterns (3-4-5, 5-12-13). Keep SOH-CAH-TOA written out until it sticks. Special right triangles convert angles 30^\circ, 45^\circ, 60^\circ into speedy exact values. The sine of an angle equals the cosine of its complement. Practice daily; short bursts of study reinforce retention more than cramming. Next, tackle official SAT® practice sets and time every attempt. With the strategies above, right triangle trigonometry will become a reliable strength. Good luck! Sharpen Your Skills for SAT® Math (Digital) Are you preparing for the SAT® Math (Digital) test? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world SAT® Math (Digital) problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now! Need help preparing for your SAT® Math (Digital) exam? Albert has hundreds of SAT® Math (Digital) practice questions, free response, and full-length practice tests to try out. Start practicing SAT® Math (Digital) on Albert now! Interested in a school license? Bring Albert to your school and empower all teachers with the world's best question bank for: ➜ SAT® & ACT® ➜ AP® ➜ ELA, Math, Science, & Social Studies ➜ State assessments Options for teachers, schools, and districts. EXPLORE OPTIONS Popular Posts AP® Score Calculators Simulate how different MCQ and FRQ scores translate into AP® scores AP® Review Guides The ultimate review guides for AP® subjects to help you plan and structure your prep. Core Subject Review Guides Review the most important topics in Physics and Algebra 1. 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2246	https://thecollegepanda.com/books/excerpts/act-math-logarithms-chapter.pdf	&IALCNBGM Warning: Logarithms rarely show up on the exam. This chapter is only for those who want to cover all their bases (haha!). At this point, we know how to solve an equation like x3 = 8 We take the cube root of both sides to get x = 2. But what about 3x = 8 Do we take the xth root of both sides? That doesn’t really work. Instead, we have to use logs (short for logarithms). Here’s how they work. Given an equation like 3x = 8, we can isolate x like so: x = log3 8 The little “3” is called the base. The “8” is called the argument or the power. Notice where those two numbers came from and where they were placed. It’s important that you know how to go from one form to the other: 3x = 8 x = log3 8 Now to evaluate log3 8, we need a calculator, which would give you x = log3 8 ≈1.89 As you can see, log3 8 is just a number. That’s all it is. Don’t be scared by it. Now we can state that 31.89 = 8 We used a calculator for this one, but don’t worry about any calculator steps. The ACT will never ask you to evaluate a logarithm that you can’t do by hand. Lastly, a log with a base of 10 is typically written without the base. For example, log 7 = log10 7. 183 CHAPTER 25 LOGARITHMS EXAMPLE 1: What is the value of log2 Å1 8 ã ? A. −3 B. −1 3 C. 1 4 D. 1 3 E. 3 Let log2 Å1 8 ã = x. Then by deﬁnition, 2x = 1 8 Since 2−3 = 1 23 = 1 8, x = −3. Answer (A) . EXAMPLE 2: Given that 52x = 9, which of the following gives the value of x ? A. log2 9 5 B. log5 9 2 C. 2 log5 9 D. 5 log2 9 E. 9 log2 5 The equation 52x = 9 is equivalent to 2x = log5 9 x = log5 9 2 Answer (B) . &;QMI@&IALCNBGM There are several laws of logarithms you should know: • loga 1 = 0 • loga a = 1 • loga(xy) = loga x + loga y • loga(xy) = y loga x • loga Åx y ã = loga x −loga y • loga b = logc b logc a The ﬁrst two don’t really need to be memorized because they stem from the basic deﬁnition of a logarithm. You should memorize the rest. The last one allows you to change the base of a logarithm. For example, if you wanted to convert log2 5 to a logarithm expression in base 10, which might be easier to input on your calculator, then log2 5 = log 5 log 2. 184 THE COLLEGE PANDA EXAMPLE 3: Which of the following is equivalent to log3 5 + log3 7 ? A. log3 12 B. log3 35 C. 2 log3 6 D. 123 E. 312 log3 5 + log3 7 = log3(5 · 7) = log3 35 Answer (B) . EXAMPLE 4: If p = log2 x, what is the value of log2(2x3) in terms of p ? A. 6p B. 2p3 C. 1 + 3p D. 3 + 3p E. 1 + p3 log2(2x3) = log2 2 + log2 x3 = 1 + 3 log2 x = 1 + 3p Answer (C) . Note that we cannot move the exponent “3” to the front ﬁrst because it applies only to x, not the entire argument. EXAMPLE 5: If log3 b −log3 4 = 2, then b = ? A. 6 B. 8 C. 13 D. 24 E. 36 log3 b −log3 4 = 2 log3 Åb 4 ã = 2 By deﬁnition, 32 = b 4 b = 32 · 4 = 36 Answer (E) . 185 CHAPTER 25 LOGARITHMS CHAPTER EXERCISE: Answers for this chapter start on page 262. 1. What is the value of x if log4 x = 3 ? A. 4 √ 3 B. 3 √ 4 C. 12 D. 64 E. 81 2. If a = log5 3 and b = log5 4, which of the following expressions is equal to 12? A. ab B. a + b C. 5a + 5b D. 5a+b E. 25ab 3. If c is a positive number such that logc 9 = 1 2, then c = ? A. 1 3 B. √ 3 C. 3 D. 18 E. 81 4. If log5 4 = x, what is the value of 52−x ? A. 1 25 B. 4 25 C. 25 4 D. 21 E. 25 5. Given that log3 x2 = c, what is the value of log3 x in terms of c ? A. c 2 B. c −2 C. 2c D. c2 E. √c 6. If log2 7 = p and log2 3 = q, what is the value of log2 63 in terms of p and q ? A. 3pq B. p + 2q C. p + 3q D. 3p + 3q E. p + q2 7. For all x > 0, which of the following expressions is equivalent to log !(3x)2" ? A. log 3 + 2 log x B. 2 log 3 + log x C. 2 log 3 + 2 log x D. 2(log 3)(log x) E. log 6 + log 2x 8. If a, b, and c are positive numbers such that ax = b and ay = c, then xy = ? A. a √ bc B. loga(bc) C. logb a + logc a D. (loga b)(loga c) E. (logb a)(logc a) 9. If x > 0 and log4(x + 3) + log4(x −3) = 2, then x = ? A. 5 B. √ 7 C. 8 D. √ 17 E. 25 10. If logb 40 −logb 5 = 3, then b = ? A. 2 B. 5 C. 8 D. 15 E. 512 186
2247	https://www.kaptest.com/study/gre/gre-text-completion-tips-strategies/?srsltid=AfmBOoqaq1iRjN38TL5HAGC2jlOwdsd8pHkrpLs4GmURWSkT0zcGTexT	GRE Text Completion: Tips and Strategies – Kaplan Test Prep ☰ About the GRE® ▼ GRE Exam Info What's Tested on the GRE? What's a Good GRE Score? Taking the GRE Studying for the GRE ▼ GRE Study Plans GRE Tips and Tricks GRE Books Applying to Grad School Free GRE Practice Questions ▼ GRE Question of the Day GRE Pop Quiz GRE 20-Minute Workout Free GRE Practice Test GRE Prep ▼ GRE Qbank GRE Practice Packs GRE Classes GRE Text Completion: Tips and Strategies August 2, 2019/in GRE/by admin GRE Text Completion questions feature a brief passage (from one to five sentences) in which one to three words have been left blank. You are given a menu of words for each blank, and you choose the answers for all blanks concurrently in order to reconstruct a sensible and meaningful passage. In questions with multiple blanks, you need to select the correct choice for every blank in order to get credit. The basic strategy for this type of question is 1) Read the whole passage actively (noting signpost words and key phrases) 2) Ask “How can I make sense of this passage?” If possible, come up with your own words that fit 3) Find the answer choice(s) that matches the sense of the passage arrived at in step two 4) Reread the passage with your answer and confirm that it works logically, grammatically and stylistically These tasks become a little more complicated in a multiple blank question, and we’ll focus on that in another entry. Let’s get started with a simpler one-blank ETS sample question: “Dramatic literature often _____ the history of a culture in that it takes as its subject matter the important events that have shaped and guided the culture.” – confounds – repudiates – recapitulates – anticipates – polarizes We’ll use the basic strategy outlined above. First, read the whole sentence; note the construction “in that it takes.” This is a signpost indicating to the reader that the second part of the sentence is an explanation or restatement of the first part. That means we want a word that would make the first part and second part equivalent or similar. In step two we come up with our own words such as “reflects,” “reveals,” or “traces.” Those might not appear in the answers, but this process insures that we really do have an understanding of the passage. We turn to the answer choices in step three. “Confounds,” “repudiates,” and “polarizes” do not fit the basic sense of the passage we’ve come up with. So let’s focus on “recapitulates” and “anticipates.” Both of these seem plausible, and neither are particularly obscure words. This is an example of how the test focuses more on distinctions of context. Does “anticipates” work? This would require literature to anticipate the history of the culture, in a sense predicting it or happening before the events themselves–that’s an interesting hypothesis but not the one suggested by the tense of the passage, where the subject matter are the events that “have shaped” the culture. We’re left with “recapitulates.” Rereading the sentence with this word, we find that the first part of the sentence now indeed reflects the second part in both the connection of literature and history and their causal order. GRE Text Completion: Signpost Words The relative importance of individual words in the sentence to understanding its meaning varies. One way to think about this is to consider which words or phrases you would underline, or even underline twice. These words and phrases act as signposts that indicate both the logical direction and the essence of the sentence. Two basic such categories are those that indicate continuation or similitude and those that indicate contrast or a change in direction. Similitude: because of, like, similar to, since, consequently, so, as a result, for example, and, also, in so far as, etc. Contrast: whereas, however, unlike, even though, although, but, alternatively, nevertheless, etc. These are by no means complete lists, and it is more important that you actively pay attention to the role of such words in creating the structure of a passage than that you memorize them. You will already have an intuitive grasp of such clues; it is more a matter of being able to draw on them in the test situation with ambiguous stimuli. Consider this ETS sample question: “Since she believed him to be both candid and trustworthy, she refused to consider the possibility that his statement had been ______.” What words would you underline here? How about “since she believed him” and “she refused to consider?” “Since” is a word from our list, but notice that ‘believed” and “refused” are key words in this sentence also. These two small segments give us enough information to induce that the second half of the sentence must be a continuing statement of her belief in him; therefore, the missing word must be something like “false” (because she refuses to think he might have been lying). – irrelevant – facetious – mistaken – critical – insincere “Insincere” and “facetious” and “mistaken” fit the general sense of falseness, but as we try the words and refine our grasp of the tone, we see that “insincere” most directly addresses the idea of trust indicated by her belief in him. “Facetious” has a connonation more of unserious, while “mistaken” misses the issue of intent. GRE Text Completion: Multiple Blanks The good news is that in multiple blank questions, instead of five choices per blank, there are only three choices per blank. The not-so-good news (and try on your quantitative skills here) are that you now have 27 (333) possible answer permutations on a three-blank question, only one of which is right! That means your chances of guessing correctly are only 1 in 27, instead of 1 in 5. Getting these questions right requires an attentive application of the skills you use on the simpler questions, plus the confidence to keep your cool when faced with a longer passage that has multiple question marks. One of the key strategies for these questions is to find the hinge of the passage: one of the three blanks will often be the most informative to address first because it is the most strongly keyed to the main meaning of the sentence and will reveal the most about the other choices. In order to target in on this, you’ll apply the same overall strategy that we examined earlier. Steps one and two (getting the gist and providing your own words) are especially important, because it’s easy to get lost in the sea of answer choices if you look at them before you understand what’s happening in the passage. Therefore, let’s revise the second of the four steps. 2a) Ask “How can I make sense of this passage?”and identify your point of attack. 2b) Come up with your own words that fit Let’s apply this to an ETS sample question: “To the untutored eye the tightly forested Ardennes hills around Sedan look quite (i)__, (ii)_ place through which to advance a modern army, even with today’s more numerous and better roads and bridges, the woods and the river Meuse form a significant (iii) _____.” On first read, the beginning of this passage may seem ambiguous with its double blanks right next to each other. Do not fear though, the question will always give us a feasible approach. We know that we’re talking about geography and the capacity of an army to move in it. Here the second piece of the passage will be easier to resolve first. It sets up a contrast with the signpost word “even.” “Even with today’s.. better roads.. the woods and river form a significant _.” Better roads would tend to make something easier to navigate, so the contrast implies we’re looking for something like “obstacle.” The choices for blank (iii) are: -resource -impediment -passage. Neither “resource” nor “passage” are like an obstacle, so we have “impediment.” With this clarified, we can turn with more confidence towards the beginning. Notice that the second half of the passage elaborates the first part. There are no indications of a change of direction (something like “To the untutored eye…,but to the military expert the woods and river form a barrier.”) That means that we need words that will be aligned with the sense of the area posing a challenge. “To the untutored eye the tightly forested Ardennes hills around Sedan look quite (i)__, (ii)_____ place through which to advance a modern army…” Blank (i) Blank (ii) impenetrable a makeshift inconsiderable an unpropitious uncultivated an unremarkable Notice that blank (ii) has two words for each selection; sometimes the blanks may even be more complex. This can make it more challenging to come up with your own words, but if you’ve done the work of analyzing the gist, you’ll still be prepared to look at the answer choices. In this case, we see that “impenetrable” fits the sense of challenge we’ve identified for blank (i), and “an unpropitious” is the best match for blank (ii). Even if you don’t know what “unpropitious” means (unfavorable, poor omen), you can probably see that the other two choices are not a great fit–makeshift meaning “thrown together, temporary” and “unremarkable” reversing the sense that the Ardennes are being described as remarkable for the challenges posed. Get The Test Take a free practice test How would you do if you took the GRE® today? Get The Test Take a free practice test How would you do if you took the GRE® today? Tags:GRE, GRE strategy, GRE test prep, GRE text completions, gre verbal, new GRE, revised GRE verbal Share this entry Share on Facebook Share on Twitter 0 0 admin admin 2019-08-02 00:00:06 2020-09-11 20:41:31 GRE Text Completion: Tips and Strategies You might also like Making the Most of My Summers in College Quantitative Comparison Strategies for the GRE GRE Reading Comprehension: Types of Questions and Tips What Should You Do After Taking A GRE Practice Test? 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2248	https://en.wikipedia.org/wiki/Nitrogen_fixation	Jump to content Search Contents 1 History 2 Biological 2.1 Importance of nitrogen 2.2 Nitrogenase 2.3 Evolution of nitrogenase 2.4 Microorganisms 2.5 Algae 2.6 Root nodule symbioses 2.6.1 Legume family 2.6.2 Non-leguminous 2.7 Other plant symbionts 3 Industrial processes 3.1 Historical 3.2 Haber process 3.3 Homogeneous catalysis 4 Lightning 5 See also 6 References 7 External links Nitrogen fixation Afrikaans العربية বাংলা Bosanski Català Čeština Dansk Deutsch Eesti Español Esperanto Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն हिन्दी Bahasa Indonesia Íslenska Italiano עברית Қазақша Kreyòl ayisyen Lietuvių Македонски മലയാളം Bahasa Melayu Монгол 日本語 Norsk nynorsk ଓଡ଼ିଆ Oʻzbekcha / ўзбекча Polski Português Romnă Русский සිංහල Simple English Slovenčina Српски / srpski Suomi Svenska தமிழ் Türkçe Українська Tiếng Việt 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Conversion of molecular nitrogen into biologically accessible nitrogen compounds See also: Marine nitrogen fixation Nitrogen fixation is a chemical process by which molecular dinitrogen (N2) is converted into ammonia (NH3). It occurs both biologically and abiologically in chemical industries. Biological nitrogen fixation or diazotrophy is catalyzed by enzymes called nitrogenases. These enzyme complexes are encoded by the Nif genes (or Nif homologs) and contain iron, often with a second metal (usually molybdenum, but sometimes vanadium). Some nitrogen-fixing bacteria have symbiotic relationships with plants, especially legumes, mosses and aquatic ferns such as Azolla. Looser non-symbiotic relationships between diazotrophs and plants are often referred to as associative, as seen in nitrogen fixation on rice roots. Nitrogen fixation occurs between some termites and fungi. It occurs naturally in the air by means of NOx production by lightning. Fixed nitrogen is essential to life on Earth. Organic compounds such as DNA and proteins contain nitrogen. Industrial nitrogen fixation underpins the manufacture of all nitrogenous industrial products, which include fertilizers, pharmaceuticals, textiles, dyes and explosives. History [edit] Biological nitrogen fixation was discovered by Jean-Baptiste Boussingault in 1838. Later, in 1880, the process by which it happens was discovered by German agronomist Hermann Hellriegel and Hermann Wilfarth [de] and was fully described by Dutch microbiologist Martinus Beijerinck. "The protracted investigations of the relation of plants to the acquisition of nitrogen begun by de Saussure, Ville, Lawes, Gilbert and others, and culminated in the discovery of symbiotic fixation by Hellriegel and Wilfarth in 1887." "Experiments by Bossingault in 1855 and Pugh, Gilbert & Lawes in 1887 had shown that nitrogen did not enter the plant directly. The discovery of the role of nitrogen fixing bacteria by Herman Hellriegel and Herman Wilfarth in 1886–1888 would open a new era of soil science." In 1901, Beijerinck showed that Azotobacter chroococcum was able to fix atmospheric nitrogen. This was the first species of the azotobacter genus, so-named by him. It is also the first known diazotroph, species that use diatomic nitrogen as a step in the complete nitrogen cycle. Biological [edit] Biological nitrogen fixation (BNF) occurs when atmospheric nitrogen is converted to ammonia by a nitrogenase enzyme. The overall reaction for BNF is: N2 + 16ATP + 16H2O + 8e− + 8H+ → 2NH3 +H2 + 16ADP + 16Pi The process is coupled to the hydrolysis of 16 equivalents of ATP and is accompanied by the co-formation of one equivalent of H2. The conversion of N2 into ammonia occurs at a metal cluster called FeMoco, an abbreviation for the iron-molybdenum cofactor. The mechanism proceeds via a series of protonation and reduction steps wherein the FeMoco active site hydrogenates the N2 substrate. In free-living diazotrophs, nitrogenase-generated ammonia is assimilated into glutamate through the glutamine synthetase/glutamate synthase pathway. The microbial nif genes required for nitrogen fixation are widely distributed in diverse environments. Nitrogenases are rapidly degraded by oxygen. For this reason, many bacteria cease production of the enzyme in the presence of oxygen. Many nitrogen-fixing organisms exist only in anaerobic conditions, respiring to draw down oxygen levels, or binding the oxygen with a protein such as leghemoglobin. Importance of nitrogen [edit] \| \| \| Part of a series on \| \| Biogeochemical cycles \| \| Water cycle Water cycle + deep water cycle \| \| Carbon cycle Global + atmospheric + terrestrial + oceanic Sequestration + carbon sink + deep carbon cycle + soil carbon + mycorrhizal fungi Boreal forests \| \| Nutrient cycle Hydrogen cycle Nitrogen cycle + human impact + nitrification + nitrogen and lichens + fixation + assimilation Oxygen cycle Phosphorus cycle + assimilation Sulfur cycle + assimilation \| \| Rock cycle Calcium cycle Silica cycle Carbonate–silicate cycle \| \| Marine cycle Marine biogeochemical cycles Biological pump + microbial loop + viral shunt Calcareous ooze Siliceous ooze \| \| Methane cycle Atmospheric methane Methane clathrate + clathrate gun hypothesis + Arctic methane emissions \| \| Other cycles aluminum arsenic boron bromine cadmium chlorine chromium copper fluorine gold iodine iron lead lithium manganese mercury ozone–oxygen selenium vanadium zinc \| \| Related topics Biogeochemistry + geochemical cycle + chemical cycling + environmental chemistry Biosequestration Deep biosphere Ocean acidification + acid rain Biogeochemical planetary boundaries \| \| Research groups DAAC GEOTRACES IMBER NOBM SOLAS \| \| Category \| \| v t e \| Atmospheric nitrogen cannot be metabolized by most organisms, because its triple covalent bond is very strong. Most take up fixed nitrogen from various sources. For every 100 atoms of carbon, roughly 2 to 20 atoms of nitrogen are assimilated. The atomic ratio of carbon (C) : nitrogen (N) : phosphorus (P) observed on average in planktonic biomass was originally described by Alfred Redfield, who determined the stoichiometric relationship between C:N:P atoms, The Redfield Ratio, to be 106:16:1. Nitrogenase [edit] Main article: Nitrogenase The protein complex nitrogenase is responsible for catalyzing the reduction of nitrogen gas (N2) to ammonia (NH3). In cyanobacteria, this enzyme system is housed in a specialized cell called the heterocyst. The production of the nitrogenase complex is genetically regulated, and the activity of the protein complex is dependent on ambient oxygen concentrations, and intra- and extracellular concentrations of ammonia and oxidized nitrogen species (nitrate and nitrite). Additionally, the combined concentrations of both ammonium and nitrate are thought to inhibit NFix, specifically when intracellular concentrations of 2-oxoglutarate (2-OG) exceed a critical threshold. The specialized heterocyst cell is necessary for the performance of nitrogenase as a result of its sensitivity to ambient oxygen. Nitrogenase consist of two proteins, a catalytic iron-dependent protein, commonly referred to as MoFe protein and a reducing iron-only protein (Fe protein). Three iron-dependent proteins are known: molybdenum-dependent, vanadium-dependent, and iron-only, with all three nitrogenase protein variations containing an iron protein component. Molybdenum-dependent nitrogenase is most common. The different types of nitrogenase can be determined by the specific iron protein component. Nitrogenase is highly conserved. Gene expression through DNA sequencing can distinguish which protein complex is present in the microorganism and potentially being expressed. Most frequently, the nifH gene is used to identify the presence of molybdenum-dependent nitrogenase, followed by closely related nitrogenase reductases (component II) vnfH and anfH representing vanadium-dependent and iron-only nitrogenase, respectively. In studying the ecology and evolution of nitrogen-fixing bacteria, the nifH gene is the biomarker most widely used. nifH has two similar genes anfH and vnfH that also encode for the nitrogenase reductase component of the nitrogenase complex. Evolution of nitrogenase [edit] Nitrogenase is thought to have evolved sometime between 1.5-2.2 billion years ago (Ga), although there is some isotopic support for nitrogenase evolution as early as around 3.2 Ga. Nitrogenase appears to have evolved from maturase-like proteins, although the function of the preceding protein is currently unknown. Nitrogenase has three different forms (Nif, Anf, and Vnf) that correspond with the metal found in the active site of the protein (molybdenum, iron, and vanadium respectively). Marine metal abundances over Earth's geologic timeline are thought to have driven the relative abundance of which form of nitrogenase was most common. Currently, there is no conclusive agreement on which form of nitrogenase arose first. Microorganisms [edit] Main article: Diazotroph Diazotrophs are widespread within domain Bacteria including cyanobacteria (e.g. the highly significant Trichodesmium and Cyanothece), green sulfur bacteria, purple sulfur bacteria, Azotobacteraceae, rhizobia and Frankia. Several obligately anaerobic bacteria fix nitrogen including many (but not all) Clostridium spp. Some archaea such as Methanosarcina acetivorans also fix nitrogen, and several other methanogenic taxa, are significant contributors to nitrogen fixation in oxygen-deficient soils. Cyanobacteria, commonly known as blue-green algae, inhabit nearly all illuminated environments on Earth and play key roles in the carbon and nitrogen cycle of the biosphere. In general, cyanobacteria can use various inorganic and organic sources of combined nitrogen, such as nitrate, nitrite, ammonium, urea, or some amino acids. Several cyanobacteria strains are also capable of diazotrophic growth, an ability that may have been present in their last common ancestor in the Archean eon. Nitrogen fixation not only naturally occurs in soils but also aquatic systems, including both freshwater and marine. Indeed, the amount of nitrogen fixed in the ocean is at least as much as that on land. The colonial marine cyanobacterium Trichodesmium is thought to fix nitrogen on such a scale that it accounts for almost half of the nitrogen fixation in marine systems globally. Marine surface lichens and non-photosynthetic bacteria belonging in Proteobacteria and Planctomycetes fixate significant atmospheric nitrogen. Species of nitrogen fixing cyanobacteria in fresh waters include: Aphanizomenon and Dolichospermum (previously Anabaena). Such species have specialized cells called heterocytes, in which nitrogen fixation occurs via the nitrogenase enzyme. Algae [edit] One type of organelle, originating from cyanobacterial endosymbionts called UCYN-A2, can turn nitrogen gas into a biologically available form. This nitroplast was discovered in algae, particularly in the marine algae Braarudosphaera bigelowii. Diatoms in the family Rhopalodiaceae also possess cyanobacterial endosymbionts called spheroid bodies or diazoplasts. These endosymbionts have lost photosynthetic properties, but have kept the ability to perform nitrogen fixation, allowing these diatoms to fix atmospheric nitrogen. Other diatoms in symbiosis with nitrogen-fixing cyanobacteria are among the genera Hemiaulus, Rhizosolenia and Chaetoceros. Root nodule symbioses [edit] Main article: Root nodule Legume family [edit] Plants that contribute to nitrogen fixation include those of the legume family—Fabaceae— with taxa such as kudzu, clover, soybean, alfalfa, lupin, peanut and rooibos. They contain symbiotic rhizobia bacteria within nodules in their root systems, producing nitrogen compounds that help the plant to grow and compete with other plants. When the plant dies, the fixed nitrogen is released, making it available to other plants; this helps to fertilize the soil. The great majority of legumes have this association, but a few genera (e.g., Styphnolobium) do not. In many traditional farming practices, fields are rotated through various types of crops, which usually include one consisting mainly or entirely of clover. Fixation efficiency in soil is dependent on many factors, including the legume and air and soil conditions. For example, nitrogen fixation by red clover can range from 50 to 200 lb/acre (56 to 224 kg/ha). Non-leguminous [edit] The ability to fix nitrogen in nodules is present in actinorhizal plants such as alder and bayberry, with the help of Frankia bacteria. They are found in 25 genera in the orders Cucurbitales, Fagales and Rosales, which together with the Fabales form a nitrogen-fixing clade of eurosids. The ability to fix nitrogen is not universally present in these families. For example, of 122 Rosaceae genera, only four fix nitrogen. Fabales were the first lineage to branch off this nitrogen-fixing clade; thus, the ability to fix nitrogen may be plesiomorphic and subsequently lost in most descendants of the original nitrogen-fixing plant; however, it may be that the basic genetic and physiological requirements were present in an incipient state in the most recent common ancestors of all these plants, but only evolved to full function in some of them. In addition, Trema (Parasponia), a tropical genus in the family Cannabaceae, is unusually able to interact with rhizobia and form nitrogen-fixing nodules. Non-legumious nodulating plants \| Family \| Genera \| Species \| \| Betulaceae \| Alnus (alders) \| Most or all species \| \| Boraginaceae \| Phacelia \| Phacelia tanacetifolia \| \| Cannabaceae \| Trema (Parasponia) \| Trema orientale Trema lamarckiana \| \| Casuarinaceae \| Allocasuarina Casuarina Ceuthostoma Gymnostoma \| \| \| Coriariaceae \| Coriaria \| Coriaria arborea Coriaria myrtifolia \| \| Datiscaceae \| Datisca \| \| \| Elaeagnaceae \| Elaeagnus (silverberries) Hippophae (sea-buckthorns) Shepherdia (buffaloberries) \| \| \| Myricaceae \| Comptonia (sweetfern) Myrica (bayberries) \| \| \| Posidoniaceae \| Posidonia (seagrass) \| \| \| Rhamnaceae \| Ceanothus Colletia Discaria Kentrothamnus Retanilla Talguenea Trevoa \| \| \| Rosaceae \| Cercocarpus (mountain mahoganies) Chamaebatia (mountain miseries) Dryas Purshia/Cowania (bitterbrushes/cliffroses) \| \| Other plant symbionts [edit] Some other plants live in association with a cyanobiont (cyanobacteria such as Nostoc) which fix nitrogen for them: Some lichens such as Lobaria and Peltigera Mosquito fern (Azolla species) Cycads Gunnera Blasia (liverwort) Hornworts Some symbiotic relationships involving agriculturally-important plants are: Sugarcane and unclear endophytes Foxtail millet and Azospirillum brasilense Kallar grass and Azoarcus sp. strain BH72 Rice and Herbaspirillum seropedicae Wheat and Klebsiella pneumoniae Maize landrace 'Sierra Mixe' / 'olotón' and various Bacteroidota and Pseudomonadota Industrial processes [edit] Historical [edit] A method for nitrogen fixation was first described by Henry Cavendish in 1784 using electric arcs reacting nitrogen and oxygen in air. This method was implemented in the Birkeland–Eyde process of 1903. The fixation of nitrogen by lightning is a very similar natural occurring process. The possibility that atmospheric nitrogen reacts with certain chemicals was first observed by Desfosses in 1828. He observed that mixtures of alkali metal oxides and carbon react with nitrogen at high temperatures. With the use of barium carbonate as starting material, the first commercial process became available in the 1860s, developed by Margueritte and Sourdeval. The resulting barium cyanide reacts with steam, yielding ammonia. In 1898 Frank and Caro developed what is known as the Frank–Caro process to fix nitrogen in the form of calcium cyanamide. The process was eclipsed by the Haber process, which was discovered in 1909. Haber process [edit] Main article: Haber process The dominant industrial method for producing ammonia is the Haber process also known as the Haber-Bosch process in 1909. Fertilizer production is now the largest source of human-produced fixed nitrogen in the terrestrial ecosystem. Ammonia is a required precursor to fertilizers, explosives, and other products. The Haber process requires high pressures (around 200 atm) and high temperatures (at least 400 °C), which are routine conditions for industrial catalysis. This process uses natural gas as a hydrogen source and air as a nitrogen source. The ammonia product has resulted in an intensification of nitrogen fertilizer globally and is credited with supporting the expansion of the human population from around 2 billion in the early 20th century to roughly 8 billion people now. Homogeneous catalysis [edit] Main article: Abiological nitrogen fixation Much research has been conducted on the discovery of catalysts for nitrogen fixation, often with the goal of lowering energy requirements. However, such research has thus far failed to approach the efficiency and ease of the Haber process. Many compounds react with atmospheric nitrogen to give dinitrogen complexes. The first dinitrogen complex to be reported was Ru(NH3)5(N2)2+. Some soluble complexes do catalyze nitrogen fixation. Lightning [edit] Nitrogen can be fixed by lightning converting nitrogen gas (N2) and oxygen gas (O2) in the atmosphere into NOx (nitrogen oxides). The N2 molecule is highly stable and nonreactive due to the triple bond between the nitrogen atoms. Lightning produces enough energy and heat to break this bond allowing nitrogen atoms to react with oxygen, forming NOx. These compounds cannot be used by plants, but as this molecule cools, it reacts with oxygen to form NO2, which in turn reacts with water to produce HNO2 (nitrous acid) or HNO3 (nitric acid). When these acids seep into the soil, they make NO3− (nitrate), which is of use to plants. See also [edit] Birkeland–Eyde process: an industrial fertilizer production process Carbon fixation Denitrification: an organic process of nitrogen release George Washington Carver: an American botanist Heterocyst Nitrification: biological production of nitrogen Nitrogen cycle: the flow and transformation of nitrogen through the environment Nitrogen deficiency Nitrogen fixation package for quantitative measurement of nitrogen fixation by plants Nitrogenase: enzymes used by organisms to fix nitrogen Ostwald process: a chemical process for making nitric acid (HNO3) Electrification of catalytic processes: electrochemical reduction of N2 References [edit] ^ a b c d Einsle O, Rees DC (2020). "Structural Enzymology of Nitrogenase Enzymes". Chemical Reviews. 120 (12): 4969–5004. doi:10.1021/acs.chemrev.0c00067. PMC 8606229. PMID 32538623. ^ Burris RH, Wilson PW (June 1945). "Biological Nitrogen Fixation". Annual Review of Biochemistry. 14 (1): 685–708. doi:10.1146/annurev.bi.14.070145.003345. ISSN 0066-4154. ^ Wagner SC (2011). "Biological Nitrogen Fixation". Nature Education Knowledge. 3 (10): 15. Archived from the original on 13 September 2018. Retrieved 29 January 2019. ^ Zahran HH (December 1999). "Rhizobium-legume symbiosis and nitrogen fixation under severe conditions and in an arid climate". Microbiology and Molecular Biology Reviews. 63 (4): 968–89, table of contents. doi:10.1128/MMBR.63.4.968-989.1999. PMC 98982. PMID 10585971. ^ Sapountzis P, de Verges J, Rousk K, Cilliers M, Vorster BJ, Poulsen M (2016). "Potential for Nitrogen Fixation in the Fungus-Growing Termite Symbiosis". Frontiers in Microbiology. 7: 1993. doi:10.3389/fmicb.2016.01993. PMC 5156715. PMID 28018322. ^ Slosson E (1919). Creative Chemistry. New York, NY: The Century Co. pp. 19–37. ^ Hill RD, Rinker RG, Wilson HD (1979). "Atmospheric Nitrogen Fixation by Lightning". J. Atmos. Sci. 37 (1): 179–192. 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"Proposed Process for the Fixation of Atmospheric Nitrogen", historical perspective, Scientific American, 13 July 1878, p. 21 A global ocean snapshot of nitrogen fixers by matching sequences to cells in the Tara Ocean \| Authority control databases \| \| International \| \| \| National \| United States France BnF data Japan Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Nitrogen cycle Metabolism Plant physiology Soil biology Hidden categories: Pages with reference errors CS1 French-language sources (fr) CS1 German-language sources (de) CS1 maint: article number as page number CS1 Korean-language sources (ko) Pages with broken reference names Articles with short description Short description is different from Wikidata Use dmy dates from March 2021 Nitrogen fixation Add topic
2249	https://en.wikipedia.org/wiki/Bipartite_graph	Jump to content Bipartite graph العربية Català Čeština Dansk Deutsch Eesti Español Esperanto فارسی Français Galego 한국어 Hrvatski Íslenska Italiano עברית Magyar 日本語 Polski Português Romnă Русский Slovenčina Slovenščina Српски / srpski Svenska தமிழ் ไทย Türkçe Українська Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Graph divided into two independent sets Example of a bipartite graph without cycles In the mathematical field of graph theory, a bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint and independent sets and , that is, every edge connects a vertex in to one in . Vertex sets and are usually called the parts of the graph. Equivalently, a bipartite graph is a graph that does not contain any odd-length cycles. The two sets and may be thought of as a coloring of the graph with two colors: if one colors all nodes in blue, and all nodes in red, each edge has endpoints of differing colors, as is required in the graph coloring problem. In contrast, such a coloring is impossible in the case of a non-bipartite graph, such as a triangle: after one node is colored blue and another red, the third vertex of the triangle is connected to vertices of both colors, preventing it from being assigned either color. One often writes to denote a bipartite graph whose partition has the parts and , with denoting the edges of the graph. If a bipartite graph is not connected, it may have more than one bipartition; in this case, the notation is helpful in specifying one particular bipartition that may be of importance in an application. If , that is, if the two subsets have equal cardinality, then is called a balanced bipartite graph. If all vertices on the same side of the bipartition have the same degree, then is called biregular. Examples [edit] When modelling relations between two different classes of objects, bipartite graphs very often arise naturally. For instance, a graph of football players and clubs, with an edge between a player and a club if the player has played for that club, is a natural example of an affiliation network, a type of bipartite graph used in social network analysis. Another example where bipartite graphs appear naturally is in the (NP-complete) railway optimization problem, in which the input is a schedule of trains and their stops, and the goal is to find a set of train stations as small as possible such that every train visits at least one of the chosen stations. This problem can be modeled as a dominating set problem in a bipartite graph that has a vertex for each train and each station and an edge for each pair of a station and a train that stops at that station. A third example is in the academic field of numismatics. Ancient coins are made using two positive impressions of the design (the obverse and reverse). The charts numismatists produce to represent the production of coins are bipartite graphs. More abstract examples include the following: Every tree is bipartite. Cycle graphs with an even number of vertices are bipartite. Every planar graph whose faces all have even length is bipartite. Special cases of this are grid graphs and squaregraphs, in which every inner face consists of 4 edges and every inner vertex has four or more neighbors. The complete bipartite graph on m and n vertices, denoted by Kn,m is the bipartite graph , where U and V are disjoint sets of size m and n, respectively, and E connects every vertex in U with all vertices in V. It follows that Km,n has mn edges. Closely related to the complete bipartite graphs are the crown graphs, formed from complete bipartite graphs by removing the edges of a perfect matching. Hypercube graphs, partial cubes, and median graphs are bipartite. In these graphs, the vertices may be labeled by bitvectors, in such a way that two vertices are adjacent if and only if the corresponding bitvectors differ in a single position. A bipartition may be formed by separating the vertices whose bitvectors have an even number of ones from the vertices with an odd number of ones. Trees and squaregraphs form examples of median graphs, and every median graph is a partial cube. Properties [edit] Characterization [edit] Bipartite graphs may be characterized in several different ways: An undirected graph is bipartite if and only if it does not contain an odd cycle. A graph is bipartite if and only if it is 2-colorable, (i.e. its chromatic number is less than or equal to 2). A graph is bipartite if and only if every edge belongs to an odd number of bonds, minimal subsets of edges whose removal increases the number of components of the graph. A graph is bipartite if and only if the spectrum of the graph is symmetric. Kőnig's theorem and perfect graphs [edit] In bipartite graphs, the size of minimum vertex cover is equal to the size of the maximum matching; this is Kőnig's theorem. An alternative and equivalent form of this theorem is that the size of the maximum independent set plus the size of the maximum matching is equal to the number of vertices. In any graph without isolated vertices the size of the minimum edge cover plus the size of a maximum matching equals the number of vertices. Combining this equality with Kőnig's theorem leads to the facts that, in bipartite graphs, the size of the minimum edge cover is equal to the size of the maximum independent set, and the size of the minimum edge cover plus the size of the minimum vertex cover is equal to the number of vertices. Another class of related results concerns perfect graphs: every bipartite graph, the complement of every bipartite graph, the line graph of every bipartite graph, and the complement of the line graph of every bipartite graph, are all perfect. Perfection of bipartite graphs is easy to see (their chromatic number is two and their maximum clique size is also two) but perfection of the complements of bipartite graphs is less trivial, and is another restatement of Kőnig's theorem. This was one of the results that motivated the initial definition of perfect graphs. Perfection of the complements of line graphs of perfect graphs is yet another restatement of Kőnig's theorem, and perfection of the line graphs themselves is a restatement of an earlier theorem of Kőnig, that every bipartite graph has an edge coloring using a number of colors equal to its maximum degree. According to the strong perfect graph theorem, the perfect graphs have a forbidden graph characterization resembling that of bipartite graphs: a graph is bipartite if and only if it has no odd cycle as a subgraph, and a graph is perfect if and only if it has no odd cycle or its complement as an induced subgraph. The bipartite graphs, line graphs of bipartite graphs, and their complements form four out of the five basic classes of perfect graphs used in the proof of the strong perfect graph theorem. It follows that any subgraph of a bipartite graph is also bipartite because it cannot gain an odd cycle. Degree [edit] For a vertex, the number of adjacent vertices is called the degree of the vertex and is denoted . The degree sum formula for a bipartite graph states that The degree sequence of a bipartite graph is the pair of lists each containing the degrees of the two parts and . For example, the complete bipartite graph K3,5 has degree sequence . Isomorphic bipartite graphs have the same degree sequence. However, the degree sequence does not, in general, uniquely identify a bipartite graph; in some cases, non-isomorphic bipartite graphs may have the same degree sequence. The bipartite realization problem is the problem of finding a simple bipartite graph with the degree sequence being two given lists of natural numbers. (Trailing zeros may be ignored since they are trivially realized by adding an appropriate number of isolated vertices to the digraph.) Relation to hypergraphs and directed graphs [edit] The biadjacency matrix of a bipartite graph is a (0,1) matrix of size that has a one for each pair of adjacent vertices and a zero for nonadjacent vertices. Biadjacency matrices may be used to describe equivalences between bipartite graphs, hypergraphs, and directed graphs. A hypergraph is a combinatorial structure that, like an undirected graph, has vertices and edges, but in which the edges may be arbitrary sets of vertices rather than having to have exactly two endpoints. A bipartite graph may be used to model a hypergraph in which U is the set of vertices of the hypergraph, V is the set of hyperedges, and E contains an edge from a hypergraph vertex v to a hypergraph edge e exactly when v is one of the endpoints of e. Under this correspondence, the biadjacency matrices of bipartite graphs are exactly the incidence matrices of the corresponding hypergraphs. As a special case of this correspondence between bipartite graphs and hypergraphs, any multigraph (a graph in which there may be two or more edges between the same two vertices) may be interpreted as a hypergraph in which some hyperedges have equal sets of endpoints, and represented by a bipartite graph that does not have multiple adjacencies and in which the vertices on one side of the bipartition all have degree two. A similar reinterpretation of adjacency matrices may be used to show a one-to-one correspondence between directed graphs (on a given number of labeled vertices, allowing self-loops) and balanced bipartite graphs, with the same number of vertices on both sides of the bipartition. For, the adjacency matrix of a directed graph with n vertices can be any (0,1) matrix of size , which can then be reinterpreted as the adjacency matrix of a bipartite graph with n vertices on each side of its bipartition. In this construction, the bipartite graph is the bipartite double cover of the directed graph. Algorithms [edit] Testing bipartiteness [edit] It is possible to test whether a graph is bipartite, and to return either a two-coloring (if it is bipartite) or an odd cycle (if it is not) in linear time, using depth-first search (DFS). The main idea is to assign to each vertex the color that differs from the color of its parent in the DFS forest, assigning colors in a preorder traversal of the depth-first-search forest. This will necessarily provide a two-coloring of the spanning forest consisting of the edges connecting vertices to their parents, but it may not properly color some of the non-forest edges. In a DFS forest, one of the two endpoints of every non-forest edge is an ancestor of the other endpoint, and when the depth first search discovers an edge of this type it should check that these two vertices have different colors. If they do not, then the path in the forest from ancestor to descendant, together with the miscolored edge, form an odd cycle, which is returned from the algorithm together with the result that the graph is not bipartite. However, if the algorithm terminates without detecting an odd cycle of this type, then every edge must be properly colored, and the algorithm returns the coloring together with the result that the graph is bipartite. Alternatively, a similar procedure may be used with breadth-first search in place of DFS. Again, each node is given the opposite color to its parent in the search forest, in breadth-first order. If, when a vertex is colored, there exists an edge connecting it to a previously-colored vertex with the same color, then this edge together with the paths in the breadth-first search forest connecting its two endpoints to their lowest common ancestor forms an odd cycle. If the algorithm terminates without finding an odd cycle in this way, then it must have found a proper coloring, and can safely conclude that the graph is bipartite. For the intersection graphs of line segments or other simple shapes in the Euclidean plane, it is possible to test whether the graph is bipartite and return either a two-coloring or an odd cycle in time , even though the graph itself may have up to edges. Odd cycle transversal [edit] Main article: Odd cycle transversal Odd cycle transversal is an NP-complete algorithmic problem that asks, given a graph G = (V,E) and a number k, whether there exists a set of k vertices whose removal from G would cause the resulting graph to be bipartite. The problem is fixed-parameter tractable, meaning that there is an algorithm whose running time can be bounded by a polynomial function of the size of the graph multiplied by a larger function of k. The name odd cycle transversal comes from the fact that a graph is bipartite if and only if it has no odd cycles. Hence, to delete vertices from a graph in order to obtain a bipartite graph, one needs to "hit all odd cycle", or find a so-called odd cycle transversal set. In the illustration, every odd cycle in the graph contains the blue (the bottommost) vertices, so removing those vertices kills all odd cycles and leaves a bipartite graph. The edge bipartization problem is the algorithmic problem of deleting as few edges as possible to make a graph bipartite and is also an important problem in graph modification algorithmics. This problem is also fixed-parameter tractable, and can be solved in time , where k is the number of edges to delete and m is the number of edges in the input graph. Matching [edit] Main article: Matching (graph theory) A matching in a graph is a subset of its edges, no two of which share an endpoint. Polynomial time algorithms are known for many algorithmic problems on matchings, including maximum matching (finding a matching that uses as many edges as possible), maximum weight matching, and stable marriage. In many cases, matching problems are simpler to solve on bipartite graphs than on non-bipartite graphs, and many matching algorithms such as the Hopcroft–Karp algorithm for maximum cardinality matching work correctly only on bipartite inputs. As a simple example, suppose that a set of people are all seeking jobs from among a set of jobs, with not all people suitable for all jobs. This situation can be modeled as a bipartite graph where an edge connects each job-seeker with each suitable job. A perfect matching describes a way of simultaneously satisfying all job-seekers and filling all jobs; Hall's marriage theorem provides a characterization of the bipartite graphs which allow perfect matchings. The National Resident Matching Program applies graph matching methods to solve this problem for U.S. medical student job-seekers and hospital residency jobs. The Dulmage–Mendelsohn decomposition is a structural decomposition of bipartite graphs that is useful in finding maximum matchings. Additional applications [edit] Bipartite graphs are extensively used in modern coding theory, especially to decode codewords received from the channel. Factor graphs and Tanner graphs are examples of this. A Tanner graph is a bipartite graph in which the vertices on one side of the bipartition represent digits of a codeword, and the vertices on the other side represent combinations of digits that are expected to sum to zero in a codeword without errors. A factor graph is a closely related belief network used for probabilistic decoding of LDPC and turbo codes. In computer science, a Petri net is a mathematical modeling tool used in analysis and simulations of concurrent systems. A system is modeled as a bipartite directed graph with two sets of nodes: A set of "place" nodes that contain resources, and a set of "event" nodes which generate and/or consume resources. There are additional constraints on the nodes and edges that constrain the behavior of the system. Petri nets utilize the properties of bipartite directed graphs and other properties to allow mathematical proofs of the behavior of systems while also allowing easy implementation of simulations of the system. In projective geometry, Levi graphs are a form of bipartite graph used to model the incidences between points and lines in a configuration. Corresponding to the geometric property of points and lines that every two lines meet in at most one point and every two points be connected with a single line, Levi graphs necessarily do not contain any cycles of length four, so their girth must be six or more. See also [edit] Bipartite dimension, the minimum number of complete bipartite graphs whose union is the given graph Bipartite double cover, a way of transforming any graph into a bipartite graph by doubling its vertices Bipartite hypergraph, a generalization of bipartiteness to hypergraphs. Bipartite matroid, a class of matroids that includes the graphic matroids of bipartite graphs Bipartite network projection, a weighting technique for compressing information about bipartite networks Convex bipartite graph, a bipartite graph whose vertices can be ordered so that the vertex neighborhoods are contiguous Multipartite graph, a generalization of bipartite graphs to more than two subsets of vertices Parity graph, a generalization of bipartite graphs in which every two induced paths between the same two points have the same parity Quasi-bipartite graph, a type of Steiner tree problem instance in which the terminals form an independent set, allowing approximation algorithms that generalize those for bipartite graphs Split graph, a graph in which the vertices can be partitioned into two subsets, one of which is independent and the other of which is a clique Zarankiewicz problem on the maximum number of edges in a bipartite graph with forbidden subgraphs References [edit] ^ Diestel, Reinard (2005), Graph Theory, Graduate Texts in Mathematics, Springer, ISBN 978-3-642-14278-9, archived from the original on 2011-04-09, retrieved 2012-02-27 ^ Asratian, Armen S.; Denley, Tristan M. 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This result has sometimes been called the "two color theorem"; Soifer credits it to a famous 1879 paper of Alfred Kempe containing a false proof of the four color theorem. ^ Bandelt, H.-J.; Chepoi, V.; Eppstein, D. (2010), "Combinatorics and geometry of finite and infinite squaregraphs", SIAM Journal on Discrete Mathematics, 24 (4): 1399–1440, arXiv:0905.4537, doi:10.1137/090760301, S2CID 10788524. ^ Asratian, Denley & Häggkvist (1998), p. 11. ^ Archdeacon, D.; Debowsky, M.; Dinitz, J.; Gavlas, H. (2004), "Cycle systems in the complete bipartite graph minus a one-factor", Discrete Mathematics, 284 (1–3): 37–43, doi:10.1016/j.disc.2003.11.021. ^ Ovchinnikov, Sergei (2011), Graphs and Cubes, Universitext, Springer. See especially Chapter 5, "Partial Cubes", pp. 127–181. ^ Asratian, Denley & Häggkvist (1998), Theorem 2.1.3, p. 8. Asratian et al. attribute this characterization to a 1916 paper by Dénes Kőnig. 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(1990), "A proof of McKee's Eulerian-bipartite characterization", Discrete Mathematics, 84 (2): 217–220, doi:10.1016/0012-365X(90)90380-Z, MR 1071664 ^ Biggs, Norman (1994), Algebraic Graph Theory, Cambridge Mathematical Library (2nd ed.), Cambridge University Press, p. 53, ISBN 9780521458979. ^ Kőnig, Dénes (1931), "Gráfok és mátrixok", Matematikai és Fizikai Lapok, 38: 116–119. ^ Gross, Jonathan L.; Yellen, Jay (2005), Graph Theory and Its Applications, Discrete Mathematics And Its Applications (2nd ed.), CRC Press, p. 568, ISBN 9781584885054. ^ Chartrand, Gary; Zhang, Ping (2012), A First Course in Graph Theory, Courier Dover Publications, pp. 189–190, ISBN 9780486483689. ^ Béla Bollobás (1998), Modern Graph Theory, Graduate Texts in Mathematics, vol. 184, Springer, p. 165, ISBN 9780387984889. ^ Chudnovsky, Maria; Robertson, Neil; Seymour, Paul; Thomas, Robin (2006), "The strong perfect graph theorem", Annals of Mathematics, 164 (1): 51–229, arXiv:math/0212070, CiteSeerX 10.1.1.111.7265, doi:10.4007/annals.2006.164.51, S2CID 119151552. ^ DeVos, Matt, "Matchings" (PDF), Lecture notes: Introduction to Graph Theory, Math 345, Simon Fraser University ^ Lovász, László (2014), Combinatorial Problems and Exercises (2nd ed.), Elsevier, p. 281, ISBN 9780080933092 ^ Asratian, Denley & Häggkvist (1998), p. 17. ^ A. A. Sapozhenko (2001) , "Hypergraph", Encyclopedia of Mathematics, EMS Press ^ Brualdi, Richard A.; Harary, Frank; Miller, Zevi (1980), "Bigraphs versus digraphs via matrices", Journal of Graph Theory, 4 (1): 51–73, doi:10.1002/jgt.3190040107, MR 0558453. Brualdi et al. credit the idea for this equivalence to Dulmage, A. L.; Mendelsohn, N. S. (1958), "Coverings of bipartite graphs", Canadian Journal of Mathematics, 10: 517–534, doi:10.4153/CJM-1958-052-0, MR 0097069, S2CID 123363425. ^ Sedgewick, Robert (2004), Algorithms in Java, Part 5: Graph Algorithms (3rd ed.), Addison Wesley, pp. 109–111. ^ Kleinberg, Jon; Tardos, Éva (2006), Algorithm Design, Addison Wesley, pp. 94–97. ^ Eppstein, David (2009), "Testing bipartiteness of geometric intersection graphs", ACM Transactions on Algorithms, 5 (2): Art. 15, arXiv:cs.CG/0307023, doi:10.1145/1497290.1497291, MR 2561751, S2CID 60496. ^ Yannakakis, Mihalis (1978), "Node-and edge-deletion NP-complete problems", Proceedings of the 10th ACM Symposium on Theory of Computing (STOC '78), pp. 253–264, doi:10.1145/800133.804355, S2CID 363248 ^ Reed, Bruce; Smith, Kaleigh; Vetta, Adrian (2004), "Finding odd cycle transversals", Operations Research Letters, 32 (4): 299–301, CiteSeerX 10.1.1.112.6357, doi:10.1016/j.orl.2003.10.009, MR 2057781. ^ Guo, Jiong; Gramm, Jens; Hüffner, Falk; Niedermeier, Rolf; Wernicke, Sebastian (2006), "Compression-based fixed-parameter algorithms for feedback vertex set and edge bipartization", Journal of Computer and System Sciences, 72 (8): 1386–1396, doi:10.1016/j.jcss.2006.02.001 ^ Ahuja, Ravindra K.; Magnanti, Thomas L.; Orlin, James B. (1993), "12. Assignments and Matchings", Network Flows: Theory, Algorithms, and Applications, Prentice Hall, pp. 461–509. ^ Ahuja, Magnanti & Orlin (1993), p. 463: "Nonbipartite matching problems are more difficult to solve because they do not reduce to standard network flow problems." ^ Hopcroft, John E.; Karp, Richard M. (1973), "An n5/2 algorithm for maximum matchings in bipartite graphs", SIAM Journal on Computing, 2 (4): 225–231, doi:10.1137/0202019. ^ Ahuja, Magnanti & Orlin (1993), Application 12.1 Bipartite Personnel Assignment, pp. 463–464. ^ Robinson, Sara (April 2003), "Are Medical Students Meeting Their (Best Possible) Match?" (PDF), SIAM News (3): 36, archived from the original (PDF) on 2016-11-18, retrieved 2012-08-27. ^ Dulmage & Mendelsohn (1958). ^ Moon, Todd K. (2005), Error Correction Coding: Mathematical Methods and Algorithms, John Wiley & Sons, p. 638, ISBN 9780471648000. ^ Moon (2005), p. 686. ^ Cassandras, Christos G.; Lafortune, Stephane (2007), Introduction to Discrete Event Systems (2nd ed.), Springer, p. 224, ISBN 9780387333328. ^ Grünbaum, Branko (2009), Configurations of Points and Lines, Graduate Studies in Mathematics, vol. 103, American Mathematical Society, p. 28, ISBN 9780821843086. External links [edit] "Graph, bipartite", Encyclopedia of Mathematics, EMS Press, 2001 Information System on Graph Classes and their Inclusions: bipartite graph Weisstein, Eric W., "Bipartite Graph", MathWorld Bipartite graphs in systems biology and medicine Retrieved from " Categories: Graph families Bipartite graphs Parity (mathematics) Hidden categories: Articles with short description Short description is different from Wikidata Pages using multiple image with auto scaled images
2250	https://oeis.org/A000009/internal	The OEIS is supported by the many generous donors to the OEIS Foundation. Expansion of Product_{m >= 1} (1 + x^m); number of partitions of n into distinct parts; number of partitions of n into odd parts. (Formerly M0281 N0100) %I M0281 N0100 #566 Jul 28 2025 10:06:31 %S 1,1,1,2,2,3,4,5,6,8,10,12,15,18,22,27,32,38,46,54,64,76,89,104,122, %T 142,165,192,222,256,296,340,390,448,512,585,668,760,864,982,1113, %U 1260,1426,1610,1816,2048,2304,2590,2910,3264,3658,4097,4582,5120,5718,6378 %N Expansion of Product_{m >= 1} (1 + x^m); number of partitions of n into distinct parts; number of partitions of n into odd parts. %C Partitions into distinct parts are sometimes called "strict partitions". %C Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700). %C The result that number of partitions of n into distinct parts = number of partitions of n into odd parts is due to Euler. %C Bijection: given n = L1 1 + L23 + L35 + L77 + ..., a partition into odd parts, write each Li in binary, Li = 2^a1 + 2^a2 + 2^a3 + ... where the aj's are all different, then expand n = (2^a1 1 + ...)1 + ... by removing the brackets and we get a partition into distinct parts. For the reverse operation, just keep splitting any even number into halves until no evens remain. %C Euler transform of period 2 sequence [1,0,1,0,...]. - _Michael Somos_, Dec 16 2002 %C Number of different partial sums 1+[1,2]+[1,3]+[1,4]+..., where [1,x] indicates a choice. E.g., a(6)=4, as we can write 1+1+1+1+1+1, 1+2+3, 1+2+1+1+1, 1+1+3+1. - _Jon Perry_, Dec 31 2003 %C a(n) is the sum of the number of partitions of x_j into at most j parts, where j is the index for the j-th triangular number and n-T(j)=x_j. For example; a(12)=partitions into <= 4 parts of 12-T(4)=2 + partitions into <= 3 parts of 12-T(3)=6 + partitions into <= 2 parts of 12-T(2)=9 + partitions into 1 part of 12-T(1)=11 = (2)(11) + (6)(51)(42)(411)(33)(321)(222) + (9)(81)(72)(63)(54)+(11) = 2+7+5+1 = 15. - _Jon Perry_, Jan 13 2004 %C Number of partitions of n where if k is the largest part, all parts 1..k are present. - _Jon Perry_, Sep 21 2005 %C Jack Grahl and Franklin T. Adams-Watters prove this claim of Jon Perry's by observing that the Ferrers dual of a "gapless" partition is guaranteed to have distinct parts; since the Ferrers dual is an involution, this establishes a bijection between the two sets of partitions. - _Allan C. Wechsler_, Sep 28 2021 %C The number of connected threshold graphs having n edges. - Michael D. Barrus (mbarrus2(AT)uiuc.edu), Jul 12 2007 %C Starting with offset 1 = row sums of triangle A146061 and the INVERT transform of A000700 starting: (1, 0, 1, -1, 1, -1, 1, -2, 2, -2, 2, -3, 3, -3, 4, -5, ...). - _Gary W. Adamson_, Oct 26 2008 %C Number of partitions of n in which the largest part occurs an odd number of times and all other parts occur an even number of times. (Such partitions are the duals of the partitions with odd parts.) - _David Wasserman_, Mar 04 2009 %C Equals A035363 convolved with A010054. The convolution square of A000009 = A022567 = A000041 convolved with A010054. A000041 = A000009 convolved with A035363. - _Gary W. Adamson_, Jun 11 2009 %C Considering all partitions of n into distinct parts: there are A140207(n) partitions of maximal size which is A003056(n), and A051162(n) is the greatest number occurring in these partitions. - _Reinhard Zumkeller_, Jun 13 2009 %C Equals left border of triangle A091602 starting with offset 1. - _Gary W. Adamson_, Mar 13 2010 %C Number of symmetric unimodal compositions of n+1 where the maximal part appears once. Also number of symmetric unimodal compositions of n where the maximal part appears an odd number of times. - _Joerg Arndt_, Jun 11 2013 %C Because for these partitions the exponents of the parts 1, 2, ... are either 0 or 1 (j^0 meaning that part j is absent) one could call these partitions also 'fermionic partitions'. The parts are the levels, that is the positive integers, and the occupation number is either 0 or 1 (like Pauli's exclusion principle). The 'fermionic states' are denoted by these partitions of n. - _Wolfdieter Lang_, May 14 2014 %C The set of partitions containing only odd parts forms a monoid under the product described in comments to A047993. - _Richard Locke Peterson_, Aug 16 2018 %C Ewell (1973) gives a number of recurrences. - _N. J. A. Sloane_, Jan 14 2020 %C a(n) equals the number of permutations p of the set {1,2,...,n+1}, written in one line notation as p = p_1p_2...p_(n+1), satisfying p_(i+1) - p_i <= 1 for 1 <= i <= n, (i.e., those permutations that, when read from left to right, never increase by more than 1) whose major index maj(p) := Sum_{p_i > p_(i+1)} i equals n. For example, of the 16 permutations on 5 letters satisfying p_(i+1) - p_i <= 1, 1 <= i <= 4, there are exactly two permutations whose major index is 4, namely, 5 3 4 1 2 and 2 3 4 5 1. Hence a(4) = 2. See the Bala link in A007318 for a proof. - _Peter Bala_, Mar 30 2022 %C Conjecture: Each positive integer n can be written as a_1 + ... + a_k, where a_1,...,a_k are strict partition numbers (i.e., terms of the current sequence) with no one dividing another. This has been verified for n = 1..1350. - _Zhi-Wei Sun_, Apr 14 2023 %C Conjecture: For each integer n > 7, a(n) divides none of p(n), p(n) - 1 and p(n) + 1, where p(n) is the number of partitions of n given by A000041. This has been verified for n up to 10^5. - _Zhi-Wei Sun_, May 20 2023 [Verified for n <= 210^6. - _Vaclav Kotesovec_, May 23 2023] %C The g.f. Product_{k >= 0} 1 + x^k = Product_{k >= 0} 1 - x^k + 2x^k == Product_{k >= 0} 1 - x^k == Sum_{k in Z} (-1)^kx^(k(3k-1)/2) (mod 2) by Euler's pentagonal number theorem. It follows that a(n) is odd iff n = k(3k - 1)/2 for some integer k, i.e., iff n is a generalized pentagonal number A001318. - _Peter Bala_, Jan 07 2025 %D Mohammad K. 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Sills, and Peter Zimmer, Rogers-Ramanujan-Slater Type Identities, Electronic J. Combinatorics, DS15, 1-59, May 31, 2008; see also arXiv version, arXiv:1901.00946 [math.NT], 2019. %H Günter Meinardus, Über Partitionen mit Differenzenbedingungen, Mathematische Zeitschrift (1954/55), Volume 61, page 289-302. %H Mircea Merca, Combinatorial interpretations of a recent convolution for the number of divisors of a positive integer, Journal of Number Theory, Volume 160, March 2016, Pages 60-75, function q(n). %H Mircea Merca, The Lambert series factorization theorem, The Ramanujan Journal, Vol. 44, No. 2 (2017), pp. 417-435; alternative link. %H István Mező, Several special values of Jacobi theta functions arXiv:1106.2703v3 [math.CA], 2011-2013. %H Donald J. Newman, A Problem Seminar, pp. 18;93;102-3 Prob. 93 Springer-Verlag NY 1982. %H Hieu D. Nguyen and Douglas Taggart, Mining the OEIS: Ten Experimental Conjectures, 2013. Mentions this sequence. %H Kimeu Arphaxad Ngwava, Nick Gill, and Ian Short, Nilpotent covers of symmetric groups, arXiv:2005.13869 [math.GR], 2020. %H Matthew Parker, The first 100K terms (7-Zip compressed file). %H Marko Riedel, Proof of recurrence by V. Jovovic. %H Ed Sandifer, How Euler Did It, Philip Naude's problem. %H Michael Somos, Introduction to Ramanujan theta functions. %H Zhi-Wei Sun, A representation problem involving strict partition numbers, Question 444761 at MathOverflow, April 14, 2023. %H Eric Weisstein's World of Mathematics, Partition Function P, Partition Function Q, Partition Function bk, Euler Identity, Ramanujan Theta Functions, q-Pochhammer Symbol. %H Wikipedia, Glaisher's Theorem. %H Wolfram Research, Generating functions for q(n). %H Mingjia Yang and Doron Zeilberger, Systematic Counting of Restricted Partitions, arXiv:1910.08989 [math.CO], 2019. %H Michael P. Zaletel and Roger S. K. Mong, Exact matrix product states for quantum Hall wave functions, Physical Review B, Vol. 86, No. 24 (2012), 245305; arXiv preprint, arXiv:1208.4862 [cond-mat.str-el], 2012. - From _N. J. A. Sloane_, Dec 25 2012 %H Index entries for "core" sequences %F G.f.: Product_{m>=1} (1 + x^m) = 1/Product_{m>=0} (1-x^(2m+1)) = Sum_{k>=0} Product_{i=1..k} x^i/(1-x^i) = Sum_{n>=0} x^(n(n+1)/2) / Product_{k=1..n} (1-x^k). %F G.f.: Sum_{n>=0} x^nProduct_{k=1..n-1} (1+x^k) = 1 + Sum_{n>=1} x^nProduct_{k>=n+1} (1+x^k). - _Joerg Arndt_, Jan 29 2011 %F Product_{k>=1} (1+x^(2k)) = Sum_{k>=0} x^(k(k+1))/Product_{i=1..k} (1-x^(2i)) - Euler (Hardy and Wright, Theorem 346). %F Asymptotics: a(n) ~ exp(Pi l_n / sqrt(3)) / ( 4 3^(1/4) l_n^(3/2) ) where l_n = (n-1/24)^(1/2) (Ayoub). %F For n > 1, a(n) = (1/n)Sum_{k=1..n} b(k)a(n-k), with a(0)=1, b(n) = A000593(n) = sum of odd divisors of n; cf. A000700. - _Vladeta Jovovic_, Jan 21 2002 %F a(n) = t(n, 0), t as defined in A079211. %F a(n) = Sum_{k=0..n-1} A117195(n,k) = A117192(n) + A117193(n) for n>0. - _Reinhard Zumkeller_, Mar 03 2006 %F a(n) = A026837(n) + A026838(n) = A118301(n) + A118302(n); a(A001318(n)) = A051044(n); a(A090864(n)) = A118303(n). - _Reinhard Zumkeller_, Apr 22 2006 %F Expansion of 1 / chi(-x) = chi(x) / chi(-x^2) = f(-x) / phi(x) = f(x) / phi(-x^2) = psi(x) / f(-x^2) = f(-x^2) / f(-x) = f(-x^4) / psi(-x) in powers of x where phi(), psi(), chi(), f() are Ramanujan theta functions. - _Michael Somos_, Mar 12 2011 %F G.f. is a period 1 Fourier series which satisfies f(-1 / (1152 t)) = 2^(-1/2) / f(t) where q = exp(2 Pi i t). - _Michael Somos_, Aug 16 2007 %F Expansion of q^(-1/24) eta(q^2) / eta(q) in powers of q. %F Expansion of q^(-1/24) 2^(-1/2) f2(t) in powers of q = exp(2 Pi i t) where f2() is a Weber function. - _Michael Somos_, Oct 18 2007 %F Given g.f. A(x), then B(x) = x A(x^3)^8 satisfies 0 = f(B(x), B(x^2)) where f(u, v) = v - u^2 + 16uv^2 . - _Michael Somos_, May 31 2005 %F Given g.f. A(x), then B(x) = x A(x^8)^3 satisfies 0 = f(B(x), B(x^3)) where f(u, v) = (u^3 - v) (u + v^3) - 9 u^3 v^3. - _Michael Somos_, Mar 25 2008 %F From Evangelos Georgiadis, Andrew V. Sutherland, Kiran S. Kedlaya (egeorg(AT)mit.edu), Mar 03 2009: (Start) %F a(0)=1; a(n) = 2(Sum_{k=1..floor(sqrt(n))} (-1)^(k+1) a(n-k^2)) + sigma(n) where sigma(n) = (-1)^j if (n=(j(3j+1))/2 OR n=(j(3j-1))/2) otherwise sigma(n)=0 (simpler: sigma = A010815). (End) %F From _Gary W. Adamson_, Jun 13 2009: (Start) %F The product g.f. = (1/(1-x))(1/(1-x^3))(1/(1-x^5))...; = (1,1,1,...) %F (1,0,0,1,0,0,1,0,0,1,...)(1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,...) ...; = %F abc... where a, ab, abc, ... converge to A000009: %F 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, ... = ab %F 1, 1, 1, 2, 2, 3, 4, 4, 5, 6, ... = abc %F 1, 1, 1, 2, 2, 3, 4, 5, 6, 7, ... = abcd %F 1, 1, 1, 2, 2, 3, 4, 5, 6, 8, ... = abcde %F 1, 1, 1, 2, 2, 3, 4, 5, 6, 8, ... = abcdef %F ... (cf. analogous example in A000041). (End) %F a(A004526(n)) = A172033(n). - _Reinhard Zumkeller_, Jan 23 2010 %F a(n) = P(n) - P(n-2) - P(n-4) + P(n-10) + P(n-14) + ... + (-1)^m P(n-2p_m) + ..., where P(n) is the partition function (A000041) and p_m = m(3m-1)/2 is the m-th generalized pentagonal number (A001318). - _Jerome Malenfant_, Feb 16 2011 %F a(n) = A054242(n,0) = A201377(n,0). - _Reinhard Zumkeller_, Dec 02 2011 %F G.f.: 1/2 (-1; x)_{inf} where (a; q)_k is the q-Pochhammer symbol. - _Vladimir Reshetnikov_, Apr 24 2013 %F More precise asymptotics: a(n) ~ exp(Pisqrt((n-1/24)/3)) / (43^(1/4)(n-1/24)^(3/4)) (1 + (Pi^2-27)/(24Pisqrt(3(n-1/24))) + (Pi^4-270Pi^2-1215)/(3456Pi^2(n-1/24))). - _Vaclav Kotesovec_, Nov 30 2015 %F a(n) = A067661(n) + A067659(n). _Wolfdieter Lang_, Jan 18 2016 %F From _Vaclav Kotesovec_, May 29 2016: (Start) %F a(n) ~ exp(Pisqrt(n/3))/(43^(1/4)n^(3/4)) (1 + (Pi/(48sqrt(3)) - (3sqrt(3))/(8Pi))/sqrt(n) + (Pi^2/13824 - 5/128 - 45/(128Pi^2))/n). %F a(n) ~ exp(Pisqrt(n/3) + (Pi/(48sqrt(3)) - 3sqrt(3)/(8Pi))/sqrt(n) - (1/32 + 9/(16Pi^2))/n) / (43^(1/4)n^(3/4)). %F (End) %F a(n) = A089806(n)A010815(floor(n/2)) + a(n-1) + a(n-2) - a(n-5) - a(n-7) + a(n-12) + ... + A057077(m-1)a(n-A001318(m)) + ..., where n > A001318(m). - _Gevorg Hmayakyan_, Jul 07 2016 %F a(n) ~ PiBesselI(1, Pisqrt((n+1/24)/3)) / sqrt(24n+1). - _Vaclav Kotesovec_, Nov 08 2016 %F a(n) = A000041(n) - A047967(n). - _R. J. Mathar_, Nov 20 2017 %F Sum_{n>=1} 1/a(n) = A237515. - _Amiram Eldar_, Nov 15 2020 %F From _Peter Bala_, Jan 15 2021: (Start) %F G.f.: (1 + x)Sum_{n >= 0} x^(n(n+3)/2)/Product_{k = 1..n} (1 - x^k) = %F (1 + x)(1 + x^2)Sum_{n >= 0} x^(n(n+5)/2)/Product_{k = 1..n} (1 - x^k) = (1 + x)(1 + x^2)(1 + x^3)Sum_{n >= 0} x^(n(n+7)/2)/Product_{k = 1..n} (1 - x^k) = .... %F G.f.: (1/2)Sum_{n >= 0} x^(n(n-1)/2)/Product_{k = 1..n} (1 - x^k) = %F (1/2)(1/(1 + x))Sum_{n >= 0} x^((n-1)(n-2)/2)/Product_{k = 1..n} (1 - x^k) = (1/2)(1/((1 + x)(1 + x^2)))Sum_{n >= 0} x^((n-2)(n-3)/2)/Product_{k = 1..n} (1 - x^k) = .... %F G.f.: Sum_{n >= 0} x^n/Product_{k = 1..n} (1 - x^(2k)) = (1/(1 - x)) Sum_{n >= 0} x^(3n)/Product_{k = 1..n} (1 - x^(2k)) = (1/((1 - x)(1 - x^3))) Sum_{n >= 0} x^(5n)/Product_{k = 1..n} (1 - x^(2k)) = (1/((1 - x)(1 - x^3)(1 - x^5))) Sum_{n >= 0} x^(7n)/Product_{k = 1..n} (1 - x^(2k)) = .... (End) %F From _Peter Bala_, Feb 02 2021: (Start) %F G.f.: A(x) = Sum_{n >= 0} x^(n(2n-1))/Product_{k = 1..2n} (1 - x^k). (Set z = x and q = x^2 in Mc Laughlin et al. (2019 ArXiv version), Section 1.3, Identity 7.) %F Similarly, A(x) = Sum_{n >= 0} x^(n(2n+1))/Product_{k = 1..2n+1} (1 - x^k). (End) %F a(n) = A001227(n) + A238005(n) + A238006(n). - _R. J. Mathar_, Sep 08 2021 %F G.f.: A(x) = exp ( Sum_{n >= 1} x^n/(n(1 - x^(2n))) ) = exp ( Sum_{n >= 1} (-1)^(n+1)x^n/(n(1 - x^n)) ). - _Peter Bala_, Dec 23 2021 %F Sum_{n>=0} a(n)/exp(Pin) = exp(Pi/24)/2^(1/8) = A292820. - _Simon Plouffe_, May 12 2023 [Proof: Sum_{n>=0} a(n)/exp(Pin) = phi(exp(-2Pi)) / phi(exp(-Pi)), where phi(q) is the Euler modular function. We have phi(exp(-2Pi)) = exp(Pi/12) Gamma(1/4) / (2 Pi^(3/4)) and phi(exp(-Pi)) = exp(Pi/24) Gamma(1/4) / (2^(7/8) Pi^(3/4)), see formulas (14) and (13) in I. Mező, 2013. - _Vaclav Kotesovec_, May 12 2023] %F a(2n) = Sum_{j=1..n} p(n+j, 2j) and a(2n+1) = Sum_{j=1..n+1} p(n+j,2j-1), where p(n, s) is the number of partitions of n having exactly s parts. - _Gregory L. Simay_, Aug 30 2023 %e G.f. = 1 + x + x^2 + 2x^3 + 2x^4 + 3x^5 + 4x^6 + 5x^7 + 6x^8 + 8x^9 + ... %e G.f. = q + q^25 + q^49 + 2q^73 + 2q^97 + 3q^121 + 4q^145 + 5q^169 + ... %e The partitions of n into distinct parts (see A118457) for small n are: %e 1: 1 %e 2: 2 %e 3: 3, 21 %e 4: 4, 31 %e 5: 5, 41, 32 %e 6: 6, 51, 42, 321 %e 7: 7, 61, 52, 43, 421 %e 8: 8, 71, 62, 53, 521, 431 %e ... %e From _Reinhard Zumkeller_, Jun 13 2009: (Start) %e a(8)=6, A140207(8)=#{5+2+1,4+3+1}=2, A003056(8)=3, A051162(8)=5; %e a(9)=8, A140207(9)=#{6+2+1,5+3+1,4+3+2}=3, A003056(9)=3, A051162(9)=6; %e a(10)=10, A140207(10)=#{4+3+2+1}=1, A003056(10)=4, A051162(10)=4. (End) %p N := 100; t1 := series(mul(1+x^k,k=1..N),x,N); A000009 := proc(n) coeff(t1,x,n); end; %p spec := [ P, {P=PowerSet(N), N=Sequence(Z,card>=1)} ]: [ seq(combstructcount, n=0..58) ]; %p spec := [ P, {P=PowerSet(N), N=Sequence(Z,card>=1)} ]: combstructallstructs; # to get the actual partitions for n=10 %p A000009 := proc(n) %p local x,m; %p product(1+x^m,m=1..n+1) ; %p expand(%) ; %p coeff(%,x,n) ; %p end proc: # _R. J. Mathar_, Jun 18 2016 %p lim := 99; # Enlarge if more terms are needed. %p simplify(expand(QDifferenceEquations:-QPochhammer(-1, x, lim)/2, x)): %p seq(coeff(%, x, n), n=0..55); # _Peter Luschny_, Nov 17 2016 %p # Alternative: %p a:= proc(n) option remember; if(n=0, 1, add(a(n-j)add( %p if(d::odd, d, 0), d=numtheorydivisors), j=1..n)/n) %p end: %p seq(a(n), n=0..55); # _Alois P. Heinz_, Jun 24 2025 %t PartitionsQ[Range[0, 60]] ( _Harvey Dale_, Jul 27 2009 ) %t a[ n_] := SeriesCoefficient[ Product[ 1 + x^k, {k, n}], {x, 0, n}]; ( _Michael Somos_, Jul 06 2011 ) %t a[ n_] := SeriesCoefficient[ 1 / Product[ 1 - x^k, {k, 1, n, 2}], {x, 0, n}]; ( _Michael Somos_, Jul 06 2011 ) %t a[ n_] := With[ {t = Log[q] / (2 Pi I)}, SeriesCoefficient[ q^(-1/24) DedekindEta[2 t] / DedekindEta[ t], {q, 0, n}]]; ( _Michael Somos_, Jul 06 2011 ) %t a[ n_] := SeriesCoefficient[ 1 / QPochhammer[ x, x^2], {x, 0, n}]; ( _Michael Somos_, May 24 2013 ) %t a[ n_] := SeriesCoefficient[ Series[ QHypergeometricPFQ[ {q}, {q x}, q, - q x], {q, 0, n}] /. x -> 1, {q, 0, n}]; ( _Michael Somos_, Mar 04 2014 ) %t a[ n_] := SeriesCoefficient[ QHypergeometricPFQ[{}, {}, q, -1] / 2, {q, 0, n}]; ( _Michael Somos_, Mar 04 2014 ) %t nmax = 60; CoefficientList[Series[Exp[Sum[(-1)^(k+1)/kx^k/(1-x^k), {k, 1, nmax}]], {x, 0, nmax}], x] ( _Vaclav Kotesovec_, Aug 25 2015 ) %t nmax = 100; poly = ConstantArray[0, nmax + 1]; poly = 1; poly = 1; Do[Do[poly += poly, {j, nmax, k, -1}];, {k, 2, nmax}]; poly ( _Vaclav Kotesovec_, Jan 14 2017 ) %o (PARI) {a(n) = if( n<0, 0, polcoeff( prod( k=1, n, 1 + x^k, 1 + x O(x^n)), n))}; / _Michael Somos_, Nov 17 1999 / %o (PARI) {a(n) = my(A); if( n<0, 0, A = x O(x^n); polcoeff( eta(x^2 + A) / eta(x + A), n))}; %o (PARI) {a(n) = my(c); forpart(p=n, if( n<1 \|\| p<2, c++; for(i=1, #p-1, if( p[i+1] > p[i]+1, c--; break)))); c}; / _Michael Somos_, Aug 13 2017 / %o (PARI) lista(nn) = {q='q+O('q^nn); Vec(eta(q^2)/eta(q))} \ _Altug Alkan_, Mar 20 2018 %o (Magma) Coefficients(&[1+x^m:m in [1..100]])[1..100] where x is PolynomialRing(Integers()).1; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006 %o (Haskell) %o import Data.MemoCombinators (memo2, integral) %o a000009 n = a000009_list !! n %o a000009_list = map (pM 1) [0..] where %o pM = memo2 integral integral p %o p _ 0 = 1 %o p k m \| m < k = 0 %o \| otherwise = pM (k + 1) (m - k) + pM (k + 1) m %o -- _Reinhard Zumkeller_, Sep 09 2015, Nov 05 2013 %o (Maxima) num_distinct_partitions(60,list); / _Emanuele Munarini_, Feb 24 2014 / %o (Maxima) %o h(n):=if oddp(n)=true then 1 else 0; %o S(n,m):=if n=0 then 1 else if n %o makelist(S(n,1),n,0,27); / _Vladimir Kruchinin_, Sep 07 2014 / %o (SageMath) # uses[EulerTransform from A166861] %o a = BinaryRecurrenceSequence(0, 1) %o b = EulerTransform(a) %o print([b(n) for n in range(56)]) # _Peter Luschny_, Nov 11 2020 %o (Python) # uses A010815 %o from functools import lru_cache %o from math import isqrt %o @lru_cache(maxsize=None) %o def A000009(n): return 1 if n == 0 else A010815(n)+2sum((-1)(k+1)A000009(n-k2) for k in range(1,isqrt(n)+1)) # _Chai Wah Wu_, Sep 08 2021 %o (Python) %o import numpy as np %o n = 1000 %o arr = np.zeros(n,dtype=object) %o arr = 1 %o for i in range(1,n): %o arr[i:] += arr[:n-i] %o print(arr) # _Yigit Oktar_, Jul 12 2025 %o (Julia) # uses A010815 %o using Memoize %o @memoize function A000009(n) %o n == 0 && return 1 %o s = sum((-1)^kA000009(n - k^2) for k in 1:isqrt(n)) %o A010815(n) - 2s %o end # _Peter Luschny_, Sep 09 2021 %Y Apart from the first term, equals A052839-1. The rows of A053632 converge to this sequence. When reduced modulo 2 equals the absolute values of A010815. The positions of odd terms given by A001318. %Y a(n) = Sum_{n=1..m} A097306(n, m), row sums of triangle of number of partitions of n into m odd parts. %Y Cf. A001318, A000041, A000700, A003724, A004111, A007837, A010815, A035294, A068049, A078408, A081360, A088670, A109950, A109968, A132312, A146061, A035363, A010054, A057077, A089806, A091602, A237515, A118457 (the partitions), A118459 (partition lengths), A015723 (total number of parts), A230957 (boustrophedon transform). %Y Cf. A167377 (complement). %Y Cf. A067659 (odd number of parts), A067661 (even number of parts). %Y Number of r-regular partitions for r = 2 through 12: A000009, A000726, A001935, A035959, A219601, A035985, A261775, A104502, A261776, A328545, A328546. %K nonn,core,easy,nice %O 0,4 %A _N. J. A. 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2251	https://www.investopedia.com/ask/answers/122214/what-difference-between-revenue-and-income.asp	Revenue vs. Income: What's the Difference? 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By Claire Boyte-White Full Bio Claire Boyte-White is the lead writer for NapkinFinance.com, co-author of I Am Net Worthy, and an Investopedia contributor. Claire's expertise lies in corporate finance & accounting, mutual funds, retirement planning, and technical analysis. Learn about our editorial policies Updated December 19, 2024 Reviewed by David Kindness Reviewed by David Kindness Full Bio David Kindness is a Certified Public Accountant (CPA) and an expert in the fields of financial accounting, corporate and individual tax planning and preparation, and investing and retirement planning. David has helped thousands of clients improve their accounting and financial systems, create budgets, and minimize their taxes. Learn about our Financial Review Board Fact checked by Amanda Bellucco-Chatham Fact checked by Amanda Bellucco-Chatham Full Bio Amanda Bellucco-Chatham is an editor, writer, and fact-checker with years of experience researching personal finance topics. Specialties include general financial planning, career development, lending, retirement, tax preparation, and credit. Learn about our editorial policies Revenue vs. Income: An Overview Revenue is the total amount of money generated from a business's primary operations.It's also referred to as gross sales or "the top line"because it's the first line on an income statement. It's calculated by multiplying a company's average sales price by the number of units sold. Income is a company's total earnings after all expenses and earnings that aren't counted as revenue are deducted.It's calculated by subtracting expenses, interest, cost of sales or goods sold, and taxes from total revenues. Key Takeaways Revenue is the total amount of money generated by the sale of goods or services related to the company's primary operations. Revenue is calculated before any expenses are taken out. Income or net income is a company's total earnings after deducting expenses. Both revenue and net income are useful in determining the financial strength of a company but they aren't interchangeable. Revenue Revenue is the money a company generates before any expenses are subtracted.It only indicates how effective a company is at generating sales. It doesn't consider operating efficiencies which could have a dramatic impact on the bottom line. Revenue can come from a variety of sources. These include but aren't limited to: The sale of goods, services, and assets Advertising Licensing agreements Fees and service charges Subscriptions Rental income Open a New Bank Account Advertiser Disclosure × The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace. Companies recognize and record revenue differently so it isn't always the same even for companies within the same industry. You can find out how a specific company defines it in its financial statements if you're unsure. Income Income is earnings left after all expenses and non-revenue, additional income are deducted.It's more commonly called net income because it's the net result after the deductions. There may be several line items subtracted from revenue to arrive at net income. Income can be broken up into different categories just like revenue: Gross Income:Gross income is the total income recorded before any taxes and expenses are deducted. Gross income may also be referred to as gross profit or gross margin. It's found on the income statement. Net Income: Net income is calculated by subtracting the costs of doing business such asdepreciation, interest, taxes,and other expenses from revenues. The bottom line or net income describes how efficient a company is with its spending and managing itsoperating costs. This figure appears on a company's income statement and is an important measure of its profitability. Income can be used to analyze and determine whether a company is operating efficiently. Important Common financial ratios that use data from the income statement include profit margin, operating margin, earnings per share (EPS), price-to-earnings ratio, and return on stockholders' equity. Key Differences: An Example Consider Apple, one of the largest tech companies on the market, to grasp how significant the difference between revenue and income can be. From a net sales (total revenue) of $119.5 billion in Q4 of 2023, Apple deducted its: Total cost of sales: $64.7 billion Total operating expenses: $14.4 billion Other income (expense), net: $50 million Income taxes: 7.2 million 1 This gave the company a net income of $40.3 billion in Q4 2023. Can Income Be Higher Than Revenue? Income can generally never be higher than revenue because income is derived from revenue after subtracting all costs. Revenue is the starting point and income is the endpoint. The business will have received income from an outside source that isn't operating income such as from a specific transaction or investment in cases where income is higher than revenue. Is Revenue or Income More Important? Both measures are important and income is derived from revenue but income is generally considered more important. Income is profit that shows that a business can cover its expenses and use that profit to grow the business. It won't have to rely on outside sources such as debt to continue operating. Strong revenues indicate that a business can sell its product or service but strong profits indicate that a business is in good financial health. What Are the Advantages of Revenue Management? Revenue management allows a company to better manage its sales tactics and its costs such as the need for raw materials. It can help it to offer a better price point to customers, run operations more efficiently, and keep inventory slim. The Bottom Line Companies report several important financial metrics each quarter including revenue and income. These two figures are often used interchangeably because they refer to the money a company earns. Revenue refers to money earned from a variety of sources, however. Income is any money that's left over after all expenses are accounted for including taxes and other costs. Sponsored Maximize Your Retirement Income Taking a strategic approach to investing can help you maximize your retirement income. Vanguard Personal Advisor® will work with you to build a customized financial plan that aligns with your goals. You’ll also benefit from innovative service at a low cost, and ongoing access to advisors. 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2252	https://www2.math.upenn.edu/~pemantle/1400b.pdf	v1.02 The Penn Calc Companion Part II: Integration and Applications About this Document This material is taken from the wiki-format Penn Calc Wiki, created to accompany Robert Ghrist’s Single-Variable Calculus class, as presented on Coursera beginning January 2013. All material is copyright 2012-2013 Robert Ghrist. Past editors and contributors include: Prof. Robert Ghrist, Mr. David Lonoﬀ, Dr. Subhrajit Bhattachayra, Dr. Alberto Garcia-Raboso, Dr. Vidit Nanda, Ms. Lee Ling Tan, Mr. Brett Bernstein, Prof. Antonella Grassi, and Prof. Dennis DeTurck. Contents 17 Antidiﬀerentiation 2 17.1 Diﬀerential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 17.2 Initial value problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 17.3 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 17.4 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 18 Exponential Growth Examples 8 18.1 Radioactive decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 18.2 Population growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 18.3 Interest accumulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 18.4 Linguistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 18.5 Zombies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 18.6 Newton’s law of cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 18.7 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 18.8 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 19 More Diﬀerential Equations 13 19.1 Autonomous diﬀerential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 19.2 Separable diﬀerential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 19.3 Linear 1st order diﬀerential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 19.4 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 19.5 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1 20 ODE Linearization 21 20.1 Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 20.2 Simple Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 20.3 Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 20.4 Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 20.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 20.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 21 Integration By Substitution 27 21.1 Integration rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 21.2 Substitution: the chain rule in reverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 21.3 Perspective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 21.4 Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 21.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 21.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 22 Integration By Parts 34 22.1 LIPET: A tip for choosing u and dv . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 22.2 Repeated use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 22.3 Reduction formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 22.4 Additional examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 22.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 22.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 23 Trigonometric Substitution 46 23.1 Typical substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 23.2 Forms with other constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 23.3 Completing the square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 23.4 Hyperbolic trigonometric substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 23.5 Blow-ups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 23.6 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 23.7 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 24 Partial Fractions 56 24.1 Partial fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 24.2 Other technicalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 24.3 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 24.4 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 2 25 Deﬁnite Integrals 65 25.1 Partitions and Riemann sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 25.2 The deﬁnite integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 25.3 Properties of deﬁnite integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 25.4 More examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 25.5 Odd and even functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 25.6 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 25.7 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 26 Fundamental Theorem Of Integral Calculus 78 26.1 Limits of integration and substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 26.2 Additional examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 26.3 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 26.4 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 27 Improper Integrals 88 27.1 Dealing with improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 27.2 Bounds at inﬁnity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 27.3 The p-integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 27.4 Converge or diverge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 27.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 27.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 28 Trigonometric Integrals 97 28.1 Product of sines and cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 28.2 Product of tangents and secants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 28.3 Product of sine and cosine with constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 28.4 Additional examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 28.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 28.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 29 Tables And Computers 107 29.1 Tables of integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 29.2 Mathematical software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 29.3 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 30 Simple Areas 110 30.1 Length of an interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 30.2 Parallelogram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 3 30.3 Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 30.4 Disc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 30.5 The area between two curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 30.6 Gini Index (An application of area formula) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 30.7 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 30.8 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 31 Complex Areas 118 31.1 Complex regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 31.2 Horizontal strips . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 31.3 Polar shapes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 31.4 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 31.5 Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 32 Volumes 128 32.1 Finding the volume element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 32.2 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 32.3 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 33 Volumes Of Revolution 139 33.1 Volume element for solid of revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 33.2 Cylindrical shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 33.3 Washers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 33.4 Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 33.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 33.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 34 Volumes In Arbitrary Dimension 147 34.1 The cube in dimension n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 34.2 Simplex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 34.3 Volume of spheres in arbitrary dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 34.4 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 35 Arclength 150 35.1 Parametric curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 35.2 Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 35.3 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 35.4 Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 36 Surface Area 159 4 36.1 Surface area of a cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 36.2 Surface area element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 36.3 Rotations about the y-axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 36.4 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 36.5 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 37 Work 168 37.1 Work element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 37.2 Work element by slices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 37.3 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 37.4 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 38 Elements 178 38.1 Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 38.2 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 38.3 Hydrostatic force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 38.4 Present value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 38.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 38.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 39 Averages 186 39.1 Average value of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 39.2 Root mean square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 39.3 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 39.4 Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 40 Centroids And Centers Of Mass 190 40.1 The area element revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 40.2 Centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 40.3 Center of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 40.4 Centroids using point masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 40.5 Application: Pappus’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 40.6 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 40.7 Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 41 Moments And Gyrations 203 41.1 Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 41.2 Radius of gyration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 41.3 Higher mass moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 5 41.4 Additivity of moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 41.5 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 41.6 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 42 Fair Probability 212 42.1 Uniform distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 42.2 Buﬀon needle problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 42.3 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 43 Probability Densities 224 43.1 Random variable and probability density function (PDF) . . . . . . . . . . . . . . . . . . . . . 224 43.2 Properties of a probability density function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 43.3 Several speciﬁc density functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 43.4 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 43.5 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 44 Expectation And Variance 230 44.1 Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 44.2 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 44.3 Standard deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 44.4 Interpretations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 44.5 The normal distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 44.6 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 44.7 Answers to Selected Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 6 17 Antidiﬀerentiation This module begins our study of integration. Integration, or anti-diﬀerentiation, can be thought of as running diﬀerentiation in reverse, or undoing the derivative. This motivates the following deﬁnition: The Indeﬁnite Integral The indeﬁnite integral of f (t), denoted R f (t) dt, is the class of functions whose derivative is f (t). R f (t) dt is also referred to as the anti-derivative of f . The act of taking the indeﬁnite integral is an operator which is referred to as anti-diﬀerentiation or integration. Note The indeﬁnite integral of a function is only deﬁned up to an added constant, called the constant of integration. In other words, if F(x) is an anti-derivative of f (x), then G(x) = F(x) + C, C a constant, is also an anti-derivative of f , because C disappears when diﬀerentiated. Conversely, any two indeﬁnite integrals of f (x) diﬀer only by some constant. Any of the known derivatives from the previous chapter can be rephrased as an integral. For example, just as there was a power rule for diﬀerentiating monomials, there is a corresponding power rule for integrating monomials. And any anti-derivative can easily be checked by taking the derivative and seeing that the result gives back the original function. Example Give the integral of each of the following functions: xn, 1 x , sin x, cos x, ex. (See Answer 1) There are other functions which are harder to integrate by merely using one of the derivatives we already know. Some of these can be integrated using other techniques from upcoming modules, but there are also functions whose anti-derivative cannot be expressed in terms of simple functions. 17.1 Diﬀerential equations The motivating problem for the study of anti-diﬀerentiation is solving a diﬀerential equation. A diﬀerential equation is an equation involving a function and its derivative. In this course, we deal with ordinary diﬀerential equations, ODEs, which are diﬀerential equations involving only functions of one variable and the derivative 7 with respect to that variable (future courses deal with partial diﬀerential equations, which involve functions of several variables and partial derivatives). Solving a diﬀerential equation means ﬁnding the function (or class of functions, usually) which satisfy the diﬀerential equation. A Simple ODE The simplest diﬀerential equation is of the form dx dt = f (t). Here, the goal is to ﬁnd the function x(t) whose derivative with respect to t is f (t). But this is precisely what the integral is. And so, the solution of the diﬀerential equation dx dt = f (t) is given by x(t) = R f (t)dt. Using the interpretation of the derivative as slope, one can think of the function f (t) as describing the slope of the function x(t): Thus, x(t) is a function which ﬁts the slopes prescribed by f (t). Note that any constant vertical shift of a solution x(t) will still have the same slope at each point. This is one interpretation of the integration constant: it represents the potential vertical shifts to a solution of the diﬀerential equation. Example Consider a falling object. Let x(t) be the height of the object at time t, v(t) be the velocity of the object, and assume that acceleration is the constant −g (negative because gravity pulls down). Express the height of the object as a function of t, v0, and x0; here, v0 and x0 are the velocity and height of the object, respectively, at time t = 0. (See Answer 2) The Next Simplest ODE Another slightly more complex ODE is of the form dx dt = f (x). 8 Before we discuss how to solve this in general, we consider a speciﬁc example, which is one of the most famous diﬀerential equations: dx dt = ax, where a is a constant. We solve this diﬀerential equation in three diﬀerent ways: 1. (Guess) Solve this diﬀerential equation by ﬁrst observing that x = Cet satisﬁes dx dt = x and then adjusting the exponent so that an extra factor of a comes out when diﬀerentiating. Hint: remember the chain rule. (See Answer 3) 2. (Series) Solve the diﬀerential equation by assuming x(t) = c0 + c1t + c2t2 + c3t3 + · · · and then determining what the constants ci must be to satisfy the diﬀerential equation. (See Answer 4) 3. (Integration) Rearrange the diﬀerential equation into the form dx x = a dt and integrate both sides to solve the diﬀerential equation. (See Answer 5) The diﬀerential equation from this example is sometimes used as a simple model of population growth. In words, the diﬀerential equation says that the growth of a population is proportional to the size of the population. As the solution above slows, this model implies the population has exponential growth. This is not a very good model for most populations because of competition for resources and overcrowding. But under certain conditions and for short periods of time, some populations (for instance, bacteria with an abundant food supply) do exhibit exponential growth. For more examples of exponential growth, see the next module. 17.2 Initial value problems Although a general diﬀerential equation’s solution often depends on a constant (sometimes several), an addi-tional condition called an initial value or initial condition can specify a speciﬁc solution. This condition is usually of the form y(t0) = y0. A diﬀerential equation with such an initial condition is called an initial value problem. To solve such a problem, ﬁrst ﬁnd the general solution and then use the initial value to ﬁnd the speciﬁc constant of integration which satisﬁes the initial condition. In the context of population growth, the initial value is typically the size of the population at time 0. This is particularly nice in the exponential growth model, because the solution is of the form P(t) = DeAt. So if P(0) = P0 is given, then plugging this in gives P(t) = P0eAt. 17.3 EXERCISES • Z (4x3 + 3x2 + 2x + 1) dx = • d dx Z ln tan x dx = • Z d dx e−x dx = 9 • Find the general solution of the diﬀerential equation dx dt = t2 • Find the general solution of the diﬀerential equation dx dt = x2 • There is a large class of diﬀerential equations – the so-called “linear” ones – for which we can ﬁnd solutions using the Taylor series method discussed in the Lecture. One such diﬀerential equation is t d2x dt2 + dx dt + tx = 0 This is a particular case of the more general “Bessel diﬀerential equation,” and one solution of it is given by the Bessel function J0(t) that we saw earlier. Notice that this involves not only the ﬁrst derivative but also the second derivative. For this reason, it is said to be a “second order” diﬀerential equation. In this problem we will content ourselves with ﬁnding a relationship (speciﬁcally, a “recurrence relation”) on the coeﬃcients of a Taylor series expansion about t = 0 of a solution to our equation. Hence consider the Taylor series x(t) = ∞ X k=0 cktk Substituting this into the diﬀerential equation above will give you two conditions. The ﬁrst one is c1 = 0. What is the other one? “Note:” this problem involves some nontrivial manipulation of indices in summation notation. Do not get discouraged if it feels more diﬃcult than other problems: it is! • Find the general solution of the diﬀerential equation dx dt = t3 + x2t3 17.4 Answers to Selected Examples 1. Z xndx = 1 n + 1xn+1 + C Z 1 x dx = ln \|x\| + C Z sin xdx = −cos x + C Z cos xdx = sin x + C Z exdx = ex + C (Don’t forget the constant!) (Return) 10 2. We know from an earlier module that dv dt = a = −g. Beginning with the second of these equations, we ﬁnd that v(t) = Z (−g) dt = −gt + C. We can determine C by plugging in t = 0. This cancels that −gt and leaves us with C = v(0) = v0, the initial velocity. Thus, v(t) = −gt + v0 Now, using the fact that dx dt = v, we ﬁnd that x(t) = Z v(t) dt = Z (−gt + v0) dt = −1 2gt2 + v0t + C. Again, we can ﬁnd C by plugging in t = 0. This leaves us with x(0) = C, and so C = x0, the initial height. Thus, x(t) = −1 2gt2 + v0t + x0. (Return) 3. Observe that x = Ceat will get an extra factor of a when diﬀerentiated by the chain rule. That is, d dt Ceat = aCeat. And so x(t) = Ceat is a solution of the diﬀerential equation. (Return) 4. Assuming that x(t) = c0 + c1t + c2t2 + c3t3 + · · · , and then taking the derivative of this series, term by term, we ﬁnd dx dt = 0 + c1 + 2c2t + 3c3t2 + · · · . On the other hand, from the original diﬀerential equation we have dx dt = ax = a c0 + c1t + c2t2 + c3t3 + · · · = ac0 + ac1t + ac2t2 + ac3t3 + · · · . 11 Because these two series both equal dx dt , they must be equal to each other. But two series are equal if and only if their corresponding coeﬃcients are equal. Therefore, c1 = ac0 2c2 = ac1 3c3 = ac2, and so on. Solving these equations one by one gives c1 = ac0 c2 = 1 2ac1 = 1 2a2c0 c3 = 1 3ac2 = 1 6a3c0 And, generally, cn = 1 n!anc0 (this can be proven using a method called induction). Doing a little bit of factoring and grouping of factors, we ﬁnd x(t) = c0 + ac0t + 1 2!a2c0t2 + 1 3!a3c0t3 + · · · = c0 1 + (at) + 1 2!(at)2 + 1 3!(at)3 + · · · = c0eat, which is, again, of the form Ceat. (Return) 5. Using the chain rule and substituting according to the diﬀerential equation, we have dx = dx dt dt dx = ax dt dx x = a dt. Integrating both sides of the equation gives ln x = at + C (only one constant of integration is necessary here, because a constant on the left side could be subtracted from both sides and absorbed into C). Now, exponentiating the equation gives x = eat+C. By exponential rules, eat+C = eateC, and the eC is often rewritten as a new constant, often written C again. Thus, the solution to dx dt = ax is x(t) = Ceat, where C is any constant. (Return) 12 18 Exponential Growth Examples Recall from the last module that the diﬀerential equation dx dt = ax has solution x = Ceat, where C is some constant. The constant C can be thought of as an initial condition, the value of the function at time t = 0. When a > 0, the function has exponential growth. When a < 0, the function has exponential decay: This module is devoted to several examples of exponential growth and decay. 18.1 Radioactive decay Carbon-14 is a radioactive isotope of carbon which exists in organic materials. It is known that the rate at which carbon-14 atoms decay is proportional to the number of carbon-14 atoms present. If I represents the number of atoms, then the diﬀerential equation is dI dt = −λI, where λ is positive (and so the number of atoms is decreasing). 13 18.2 Population growth For bacteria with an abundant food supply, the population P satisﬁes dP dt = bP, for some positive b. But as the food supply dwindles, or overcrowding occurs, the population growth will necessarily slow (or else the bacteria would eventually consume the earth). Thus, this is not usually an accurate population model. 18.3 Interest accumulation Consider a bank account with initial deposit (also called the principal) P, annual interest rate r, and which is compounded n times a year (so n = 1 gives simple interest, n = 4 is quarterly interest, etc.). Then the value of the account at the end of k years is P 1 + r n nk. What happens as n gets bigger and bigger? Recall that lim n→∞ 1 + α n n = eα. Then it follows that lim n→∞P 1 + r n nk = lim n→∞P h 1 + r n nik = Perk. This is called continuous compounding. So an account with continuous compounding is worth Perk after k years, where P is the initial investment, and r is the annual interest rate. Rule of 70 The Rule of 70 is a mental math trick that approximates the number of years it takes for a continuously compounded account to double in value. The rule says that # years for the account to double = 70 100r . In other words, the number of years is 70 divided by the annual percentage. Verify the Rule of 70. (See Answer 1) 18.4 Linguistics Historical linguists study (among other things) word usage and the rate at which words fall out of use. Let W(t) be the number of words which are in active use in English after t years. One model predicts that dW dt = −λW, hence W = Ce−λt, and C is the number of common words at time 0. Example Suppose the writing of John Milton (from around 1667) consists of 20% words which are unfamiliar to us today. Use the above model to estimate the fraction of words that Shakespeare’s audience (from around 1600) recognized of Chaucer’s writing (from around 1400). (See Answer 2) This model ignores the creation of new words as well as the fact that deﬁnitions of existing words can evolve over time. So it is probably not the most accurate model. It is worth noting that even if the mathematics are correct, if the underlying model is not very good, then the resulting answer will not be very good either. 14 18.5 Zombies Suppose in the zombie apocalypse that the rate of change of the infected population Z(t) is proportional to the uninfected population U(t). Let P be the total population (assumed to be constant). Then dZ dt = rU = r(P −Z), since U = P −Z (the number of uninfected is the total population less the infected). Taking the derivative of U = P −Z gives dU dt = dP dt −dZ dt = −dZ dt = −rU, since P was assumed to be constant. Thus, U(t) = U0e−rt, where U0 is the initial uninfected population. And so the zombie population is given by Z(t) = P −U0e−rt. Note the population is doomed, as Z(t) →P as t →∞. Although a zombie apocalypse is unlikely, this model is still useful for other phenomena involving how quickly something spreads. This includes the spread of disease, propaganda, and technology. Another example is heat transfer, discussed in detail below. 18.6 Newton’s law of cooling Newton’s law of cooling says that the rate of change of the temperature of a body is proportional to the diﬀerence of the temperatures of the body and ambient environment. Let T be the temperature of the body and A be the ambient temperature. Then dT dt = k(A −T), for some positive constant k. Separating gives dT T −A = −k dt, and integrating and exponentiating gives T −A = Ce−kt. Thus the solution is T = A + Ce−kt. Note that k must be positive for this to model to make sense. Indeed, as t →∞, the temperature of the body should approach the ambient temperature, which will only happen if k > 0. Also, note that C = T0 −A is the initial diﬀerence in temperature. 18.7 EXERCISES • After drinking a cup of coﬀee, the amount C of caﬀeine in a person’s body obeys the diﬀerential equation dC dt = −αC where the constant α has an approximate value of 0.14 hours−1 How many hours will it take a human body to metabolize half of the initial amount of caﬀeine? 15 • The amount I of a radioactive substance in a given sample will decay in time according to the following equation: dI dt = −λI Nuclear engineers and scientists tend to be concerned with the “half-life” of a substance, that is, the time it takes for the amount of radioactive material to be halved. Find the half-life of a substance in terms of its decay constant λ. • In a highly viscous ﬂuid, a falling spherical object of radius r decelerates right before reaching the bottom of the container. A simple model for this behavior is provided by the equation dh dt = −α r h, where h is the height of the object measured from the bottom, and α is a constant that depends on the viscosity of the ﬂuid. Find the time it would take the object to drop from h = 6r to h = 2r in terms of α and r. • On a cold day you want to brew a nice hot cup of tea. You pour boiling water (at a temperature of 212◦ F) into a mug and drop a tea bag in it. The water cools down in contact with the cold air according to Newton’s law of cooling: dT dt = κ(A −T) where T is the temperature of the water, A = 32◦F the ambient temperature, and κ = 0.36 min−1. The threshold for human beings to feel pain when entering in contact with something hot is around 107◦ F. How many seconds do you have to wait until you can safely take a sip? • On the night of April 14, 1912, the British passenger liner RMS Titanic collided with an iceberg and sank in the North Atlantic Ocean. The ship lacked enough lifeboats to accommodate all of the passengers, and many of them died from hypothermia in the cold sea waters. Hypothermia is the condition in which the temperature of a human body drops below normal operating levels (around 36◦C). When the core body temperature drops below 28◦C, the hypothermia is said to have become severe: major organs shut down and eventually the heart stops. If the water temperature that night was -2◦C, how long did it take for passengers of the Titanic to enter severe hypothermia? Recall from lecture that heat transfer is described by Newton’s law of cooling: dT dt = κ(A −T) where T is the body temperature of a passenger, A the water temperature, and κ = 0.016 min−1. • The birthrate of a population (number of births per year × 100 / number of population) is 20%, and the mortality rate (number of deaths per year × 100/ number of population) is 5%. If the initial population is 10,000, ﬁnd the function P(t), the population as a function of time. How long does it take for the population to double? • Let y(t) denote the number of atoms of a particular radioactive isotope of carbon at year t. We know that the rate at which y(t) decreases is proportional to y itself. (a) What diﬀerential equation does y(t) satisfy? (b) If it takes 5 years for the number to decrease to half of its initial number, what is the constant involved in your answer of part (a)? 16 18.8 Answers to Selected Examples 1. The problem asks for k when the balance has doubled. That is, ﬁnd k such that 2P = Perk. Dividing by P and taking logarithms gives rk = ln(2), so k = ln(2) r . Since ln(2) ≈.7, this shows that k = ln(2) r ≈.7 r ≈ 70 100r , as desired. (Return) 2. Let t denote time measured in years. Let WM(t) denote the number of words from Milton’s time which are in common use at time t. Then WM(t) = WM(1667)e−λ(t−1667) (the t −1667 is used in the exponent because we are interested in the number of years since Milton). Let WC(t) denote the number of words from Chaucer’s time which are in common use at time t. Then WC(t) = WC(1400)e−λ(t−1400). From the given information, we have that WM(2013) WM(1667) = 4 5. And according to the above formula, we have WM(1667)e−λ·346 WM(1667) = 4 5. The factors of WM(1667) cancel, leaving e−λ·346 = 4 5. Taking the logarithm of both sides and dividing by −346 gives λ = −ln(4/5) 346 ≈6.5 × 10−4. Knowing λ allows us to compute the fraction of words from Chaucer’s time that would be recognized in Shakespeare’s time: WC(1600) WC(1400) = WC(1400)e−λ(200) WC(1400) = e−λ(200) ≈.88 So, according to this model, approximately 88% of words from Chaucer’s work would have been understood in Shakespeare’s time. (Return) 17 19 More Diﬀerential Equations Recall that an ordinary diﬀerential equation is an equation involving a function and its derivatives. The solution to a diﬀerential equation is a function which satisﬁes the equation. An earlier module introduced a few basic diﬀerential equations. This module deals with a few diﬀerent families of diﬀerential equations and the methods of solving them. 19.1 Autonomous diﬀerential equations A diﬀerential equation is called autonomous if the derivative of the function x(t) is independent of t, i.e. the equation is of the form dx dt = f (x). Logically, a nonautonomous diﬀerential equation is one where the derivative equals a function of both x and t: dx dt = f (x, t). In general, nonautonomous diﬀerential equations can be very diﬃcult, but certain types yield to a little algebra and integration. These include separable diﬀerential equations and linear ﬁrst order diﬀerential equations, which are covered here. 19.2 Separable diﬀerential equations A separable diﬀerential equation is one where, with a little algebra, we are able to express the diﬀerential equation in the form dx dt = f (x)g(t). This may involve some algebra. Note in particular that any autonomous equation is separable (think of g(t) = 1). Once a diﬀerential equation is factored this way, it can be solved by using the chain rule and some algebra: dx = dx dt dt dx = f (x)g(t) dt dx f (x) = g(t) dt. 18 (This is why these equations are called separable–the variables can be separated to opposite sides of the equation). To solve this equation, one must ﬁnd the functions whose derivatives are 1 f (x) and g(t), respectively. In other words, one integrates both sides. Example Solve the diﬀerential equation y ′ = 3yx2. (See Answer 1) Example Solve the diﬀerential equation dx dt = et−x. (See Answer 2) Example Solve the initial value problem dy dx = xy + x, with y(0) = 3. (See Answer 3) Example Assume that a falling body with mass m has a drag force proportional to velocity v(t). Then the downward acceleration mg is being counteracted by the upward acceleration κv, for some constant κ. Thus, mdv dt = mg −κv = −κ v −mg κ . which is separable. Solve this diﬀerential equation. (See Answer 4) 19.3 Linear 1st order diﬀerential equations The product rule gives a technique (integration by parts) for seemingly diﬃcult integrals; the product rule also gives a technique for solving a certain class of non-separable diﬀerential equations called linear 1st order diﬀerential equations. This is a diﬀerential equation which can be written in the form dx dt = A(t)x + B(t) (as for separable diﬀerential equations, this may involve a little algebra). This form gives the reason for calling these equations linear, since dropping the t’s gives dx dt = Ax + B, which is reminiscent of the equation of a line. 1st order means that the equation only involves the function x and its derivative dx dt (and no higher derivatives), along with functions of t. 19 The standard form of a linear 1st order diﬀerential equation is achieved by bringing all the terms involving x to the left side, which gives dx dt −A(t)x = B. Example Identify which of these is a linear 1st order diﬀerential equation, and put it in standard form if it is. In the cases that are not, identify which condition is violated. Are all separable equations also linear 1st order? 1. tx′ + x = 0 2. x′ −etx2 = 0 3. x′ = x sin(t) 4. x′′ −t2 dx dt = 0 (See Answer 5) Integrating factors The method for solving linear 1st order diﬀerential equations is to use the product rule to factor the sum of two derivatives into the derivative of a product. It is best explained by example. Example In part 1 from the previous example, note that tx′ + x = (tx)′ by the product rule. So that diﬀerential equation can be written as (tx)′ = 0. Integrating both sides gives tx = C for some constant C. Thus the solution is x = C t . However, not all linear 1st order diﬀerential equations are expressed so nicely. For instance, in example 3 above, one cannot rewrite x′ −sin(t)x as the derivative of a product of functions. This is where an integrating factor is used. The integrating factor, denoted by I(t) in this course, is a function which is multiplied through the entire diﬀerential equation, giving I dx dt −IAx = IB. I(t) is chosen so that the left side of this equation can be factored as a derivative of a product using the product rule. Symbolically, the goal is to choose I(t) so that I dx dt −IAx = d dt (Ix). To ﬁnd I, expand the product d dt (Ix) = I dx dt + dI dt x. For this to equal the left side of the above equation, it must be that −IA = dI dt . This diﬀerential equation is separable, and one ﬁnds that dI I = −A dt. Integrating and exponentiating gives that I(t) = e R −A(t) dt. One need not work through all this algebra every time but can jump straight to writing down the integrating factor. Multiplying through by the integrating factor allows the left side to be rewritten by the product rule, and integrating both sides ﬁnishes the problem. To summarize the method: 20 1. Get the diﬀerential equation into standard form dx dt −A(t)x = B(t). 2. Compute the integrating factor I(t) = e− R A(t) dt. 3. Multiply the entire equation by I(t), which gives I dx dt −IAx = IB. 1. Rewrite the left side as the derivative of a product (this works because of the way I(t) was chosen): d dt (Ix) = IB. 2. Integrate both sides and then divide by I. 3. The ﬁnal answer, then, is given by x(t) = e R A · Z Be− R A. Example Solve the diﬀerential equation tx′ + tx = t2. Hint: R tet dt = tet −et + C. (See Answer 6) Example Suppose a 1000 gallon tank is 90% full. An additive is is pumped into the tank at a rate of 10 gallons per minute. The mixture is well stirred and drained at a rate of 5 gallons per minute. What is the concentration of the additive when the tank is full? (See Answer 7) 19.4 EXERCISES • Solve the diﬀerential equation dx dt = x t . • Solve the diﬀerential equation dx dt = √ 1 −x2 √ 1 −t2 . • Given that x(0) = 0 and dx dt = tex, compute x(1). • What integrating factor should be used to solve the linear diﬀerential equation t2 dx dt = 4t −t5x • Solve the diﬀerential equation dx dt −5x = 3. • Solve the diﬀerential equation dx dt = x 1 + t + 2. 21 • Suppose that, in order to buy a house, you obtain a mortgage. If the lender advertises an annual interest rate r, your debt D(t) will increase exponentially according to the simple O.D.E. dD dt = rD If you pay your debt at a rate of P (annual rate, paid continuously), the evolution of your debt will then (under assumptions of continual compounding and payment) obey the linear diﬀerential equation dD dt = rD −P Using this model, answer the following question: if initial amount of the mortgage is for $400,000, the annual interest rate is 5%, and you pay at a rate of $40,000 every year, how many years will it take you to pay oﬀthe debt? • German physician Ernst Heinrich Weber (1795-1878) is considered one of the fathers of experimental psychology. In his study of perception, he noticed that the perceived diﬀerence between two almost-equal stimuli is proportional to the percentual diﬀerence between them. In terms of diﬀerentials, we can express Weber’s law as dp = k dS S , where p is the perceived intensity of a stimulus and S its actual strength. Observe the relative rate of change on the right hand side. In what way must the magnitude of a stimulus change in time for a human being to perceive a linear growth? Linearly? Logarithmically? Polynomially? • Some nonlinear diﬀerential equations can be reduced to linear ones by a clever change of variables. Bernouilli equations dx dt + p(t)x = q(t)xα, α ∈R constitute the most important case. Notice that for α = 0 or α = 1 the above equation is already linear. For other values of α, the substitution u = x1−α yields a linear diﬀerential equation in the variable u. Apply the above change of variables in the case dx dt + 2tx = x3 What diﬀerential equation on u do you get? 19.5 Answers to Selected Examples 1. Separating gives dy y = 3x2 dx. Integrating both sides, we have Z dy y = Z 3x2 dx ln y = x3 + C. 22 Now exponentiating both sides gives y = ex3+C = Cex3, for some constant C (remember that the C is not the same in the ﬁrst and second line above, but we just rewrite it for convenience). (Return) 2. First, using a law of exponents on the right side, we have et−x = ete−x. Now, separating gives ex dx = et dt. We might be tempted at this point to say x = t because of the symmetry of this equation. But we must integrate both sides, which introduces an integration constant: ex = et + C. Now taking the logarithm gives x = ln(et + C). Note We must have the integration constant contained within the natural logarithm. In general, it is best to introduce the integration constant as soon as the integration occurs. A common mistake is to forget the constant and then at the very end of the problem add it. This is frequently incorrect, as in this case. (Return) 3. First, this diﬀerential equation does not look like it is of the form of a separable diﬀerential equation. However, with a little factoring, one ﬁnds that dy dx = x(y + 1). Thus, dy y + 1 = x dx. Anti-diﬀerentiating gives ln \|y + 1\| = 1 2x2 + C. Then, exponentiating gives \|y + 1\| = ex2/2+C = Dex2/2. Apply the initial condition by plugging in x = 0 and y = 3, which gives that D = 4. Thus, the solution to the initial value problem is y = 4ex2/2 −1. One should double check that this satisﬁes the diﬀerential equation. (Return) 4. Separating gives dv v −mg/κ = −κ mdt. Integrating both sides gives ln(v −mg/κ) = −κt/m + C, and so exponentiating and solving for v gives v = Ce−κt/m + mg κ 23 (here C is replacing the constant eC from exponentiating). Note that as t →∞, the exponential term goes to 0, and so v(t) →mg κ , which is the terminal velocity of the falling body (when the force of gravity and drag cancel each other). (Return) 5. 1. Linear 1st order. Standard form is x′ + 1 t x = 0. 2. Not linear because of the x2. Note that this is separable though. 3. Linear 1st order. Standard form is x′ −sin(t)x = 0. 4. Not 1st order because of the presence of x′′. Number 2 shows that a diﬀerential equation can be separable even though it is not linear 1st order. (Return) 6. Divide through by t to get the equation in standard form: x′ + x = t. Compute the integrating factor I = e R dt = et. Multiplying through gives etx′ + etx = ett. Rewriting this using the product rule gives (etx)′ = tet. Integrating both sides and using the hint gives etx = tet −et + C. Finally, dividing by et gives x = t −1 + C et as a ﬁnal answer. (Return) 7. Begin by setting up a few variables to help make sense of what is happening. Let V (t) be the volume of the total mixture at time t. Let Q(t) be the total amount of additive in the mixture at time t. Let C(t) be the concentration of the mixture, i.e. C(t) = Q(t) V (t). The volume is not too diﬃcult to compute. Since there are 10 gallons per minute entering the tank, and 5 gallons per minute leaving the tank, the net amount of ﬂuid entering the tank is 5 gallons per minute. The tank begins at 90% full, which is 900 gallons. So V (t) = 900 + 5t, and is full at t = 20. Next, consider the rate at which the quantity Q(t) of additive in the tank is changing. There is 10 gallons of pure additive entering per minute and 5 gallons of mixture leaving. Therefore the amount of additive leaving is 5C. Putting this together gives d dt Q = 10 −5C = 10 − 5Q 900 + 5t . Rearranging this gives dQ dt + 1 180 + t Q = 10. 24 This is a 1st order linear diﬀerential equation. Computing the integrating factor we ﬁnd I = exp Z 1 180 + t dt = exp(ln(180 + t)) = 180 + t. Multiplying through gives (180 + t)dQ dt + Q = 10(180 + t). Now, as always, the left side can be rewritten using the product rule to give d dt [Q(180 + t)] = 1800 + 10t. Integrating both sides gives Q · (180 + t) = 1800t + 5t2 + K (using K here to avoid confusion with concentration C). We can ﬁnd K by setting t = 0 in this equation. Since Q(0) = 0 (there is no additive in the tank initially), we ﬁnd K = 0. Now, solving for Q and evaluating at t = 20 gives Q = 1800t + 5t2 180 + t = 190. So the ﬁnal concentration of the full tank is 190 1000 = 19%. (Return) 25 20 ODE Linearization We have seen techniques for solving two types of diﬀerential equations: separable and linear. Unfortunately, there are a lot of diﬀerential equations which do not ﬁt into these categories. In some of these cases, we can use linearization to determine the behavior of such diﬀerential equations. 20.1 Oscillation How does one model oscillation? It turns out that a ﬁrst order diﬀerential equation will not work, but a second order (i.e. involving the second derivative) equation will: d2x dt2 = −a2x. Solving such an equation is beyond the scope of this course, but in a course on diﬀerential equations one ﬁnds the pair of solutions x = C1 cos(at) x = C2 sin(at). For this course we will look at a simpler way to model oscillation. 20.2 Simple Oscillator Consider a spinner where θ(t) represents the angle the arrow makes with the positive x-axis at time t. Then θ increases linearly with t and whenever θ gets to 2π, it goes back to 0: (Click Here: Simple Oscillator Animated GIF) This can be modeled by dθ dt = a θ = at + θ0 mod 2π where mod 2π means ”take the remainder when divided by 2π”. Here a can be thought of as the frequency of the spinner (e.g. how many revolutions per minute it makes). 26 20.3 Coupled Oscillators Now consider two simple oscillators, θ1 and θ2, with the same frequency a, but which are slightly out of phase with each other (i.e. one arrow is slightly ahead of the other): (Click Here: Two Oscillators Animated GIF) Now suppose these oscillators are coupled so that each exerts a small inﬂuence on the other (e.g. by connecting their axles with a rod). One way to represent this mathematically is to adjust the rates of change of the oscillators so that they are aﬀected by the diﬀerence in angles: dθ1 dt = a + ϵ sin(θ2 −θ1) dθ2 dt = a −ϵ sin(θ2 −θ1). Here, ϵ is some small constant which represents the strength of the eﬀect of the coupling. When θ2 is bigger than θ1, the above diﬀerential equations speed up θ1 slightly and slow down θ2 slightly. One can ﬁnd by simulation that this coupling eﬀect causes the oscillators to synchronize relatively quickly, depending on how big the phase is between them and how big ϵ is: (Click Here: Coupled Oscillators Animated GIF) Synchronization To analyze the synchronization eﬀect mathematically, consider the phase ϕ between the two oscillators: ϕ = θ2 −θ1. Looking at how the phase ϕ changes with respect to time gives dϕ dt = d dt (θ2 −θ1) = dθ2 dt −dθ1 dt = (a −ϵ sin(θ2 −θ1)) −(a + ϵ sin(θ2 −θ1)) = −2ϵ sin(θ2 −θ1) = −2ϵ sin(ϕ). This is a separable diﬀerential equation, but solving it to ﬁnd ϕ as an explicit function of t is not so easy, and does not really help us understanding the synchronization phenomenon. But linearization will help us understand the synchronization eﬀect and how quickly it occurs. Linearization Going back to the diﬀerential equation for the phase, suppose we replace sin ϕ with its linearization: dϕ dt = −2ϵ sin ϕ = −2ϵ ϕ + O(ϕ3) ≈−2ϵϕ. This will be a good approximation assuming the phase is small (the oscillators are not too far out of sync). This is a familiar diﬀerential equation, which gives us the approximate solution ϕ(t) ≈ϕ0e−2ϵt, 27 where ϕ0 is the initial phase. This is called the linearized solution to the original diﬀerential equation. Here, the linearized solution predicts that the phase decays exponentially, which is consistent with the above simulation. 20.4 Equilibria Another way to study diﬀerential equations and predict their behavior, is to study the equilibria of the equation. An equilibrium of the equation ˙ x = f (x) (here, ˙ x = dx dt ), is a solution x(t) = C, C a constant, such that ˙ x = 0. In other words, an equilibrium is a root of f . In terms of the diﬀerential equation, an equilibrium is a steady state where the quantity x does not change. One way to ﬁnd the equilibria of a diﬀerential equation is to plot the derivative of a function versus the function itself. From the phase diﬀerential equation above, we plot ˙ ϕ on the y-axis and ϕ on the x-axis and look for roots: The roots of this equation are the values of ϕ for which sin ϕ = 0. For the range of values in which we are interested, the roots are ϕ = −π, 0, π. The equilibrium at 0 is familiar, because that is the state of synchronization to which the above simulation converged. The other two correspond to a phase of π, which means the oscillators are completely opposite one another (it is the same for −π since these angles are co-terminal). Stable and Unstable A logical question at this point is why did the above coupled oscillator simulation eventually synchronize rather than ending up in opposite directions? In general, some equilibria are attractive in the sense that if the quantity x gets near such an equilibrium, it will be drawn towards it and stay at it. Some equilibria are repellent in the sense that even if x is very close to such an equilibrium, it will be pushed away from it. Formally, 28 Stable and Unstable Equilibria An equilibrium C of the diﬀerential equation dx dt = f (x) is stable if f ′(C) < 0 and is unstable if f ′(C) > 0. It is best to make sense of these deﬁnitions visually. Plot ˙ x versus x. Then each root of this equation is an equilibrium. If the graph crosses from positive to negative (going from left to right), then the equilibrium is stable. If the graph crosses from negative to positive (again, from left to right), then the equilibrium is unstable: Another way to think of stable and unstable equilibria is to visualize one ball sitting in a bowl, and another ball sitting on top of an inverted bowl: Each of these balls is in equilibrium (it will stay where it is as long as it is not disturbed). But the ball in the bowl is stable because if we nudge it in either direction, it will return to its equilibrium. However, the ball on the inverted bowl is unstable because if it is nudged in either direction it will roll oﬀthe bowl. 29 Example Find and classify the equilibria of the equation dx dt = x2 −4x + 3. (See Answer 1) 20.5 EXERCISES • The diﬀerential equation dx dt = (ex −1)(x −1) has an equilibrium at x = 0. What is the linearized equation at this equilibrium? Hint: Taylor-expand the right hand side about zero. • There is also an equilibrium at x = 1 in the equation above. What is the linearized equation at this equilibirum? Hint: let h = x −1 be a local coordinate and compute ˙ h = ˙ x = · · · by Taylor expanding the right hand side at x = 1. • Recall from Lecture 18, Newton’s Law of Heat Transfer, which states that dT dt = κ(A −T), where κ > 0 is a thermal conductivity constant and A is the (constant) ambient temperature. Find and classify the equilibria in this system (using the derivative of the right hand side at the equilibria, recall...). Wasn’t that easy? • Find and classify all the equilibria of the ODE dy dt = −2y + y 2 + y 3 • Recall from Lecture 19 how we computed the terminal velocity of a falling body with linear drag given by mdv dt = mg −κv, where, of course, m is mass, g is gravitation, v is velocity, and κ > 0 is the drag coeﬃcient. Can you see how easily one can solve for the equilibrium v∞= mg/κ? Do it! • Very good. Now, let’s use a more realistic model of drag that is quadratic as opposed to linear: mdv dt = mg −λv 2, where λ > 0 is a constant drag coeﬃcient. This diﬀerential equation is not as easy to solve (but soon you will learn how). Is there is terminal velocity? What is it? 30 • Recall that with continuous compounding at an interest rate of r > 0, an investment I(t) with initial investment I0 = I(0) is I(t) = I0ert. What happens if you wish to withdraw funds from the investment at a rate of spending S, where S > 0 is constant? The diﬀerential equation is: dI dt = rI −S. Your goals are as follows. You have an initial investment I0, and you cannot change it or the rate r. You want to be able to spend as much as possible but you also don’t want to ever spend all your money. What amount of spending rate S can you bear? Hint: if you’re not sure what to do, ﬁnd and classify the equilibria in this model and think about which initial conditions lead to which long-term behaviors. • In our lesson, we looked at two oscillators with ”sinusoidal” coupling. Other types of coupling are possible as well. Consider the system of two oscillators modeled by dθ1 dt = 2 + ϵ(eθ1−θ2 −1) ; dθ2 dt = 2 + ϵ(1 −eθ1−θ2) Consider the phase diﬀerence ϕ = θ2 −θ1. Note that ϕ = 0 (where the oscillators are coupled) is an equilibrium. What is the linearized equation for ϕ about 0? This looks intimidating, but is very straightforward. If you’re not sure how to start, compute dϕ dt . Then linearize this about ϕ = 0. 20.6 Answers to Selected Examples 1. Here, f (x) = x2 −4x + 3. Factoring, one ﬁnds dx dt = (x −1)(x −3). So the roots (and hence the equilibria) are x = 1 and x = 3. Taking the derivative, we ﬁnd f ′(x) = 2x −4 f ′(1) = −2 < 0 f ′(3) = 2 > 0. Thus x = 1 is a stable equilibrium, and x = 3 is an unstable equilibrium. (Return) 31 21 Integration By Substitution The previous modules gave some of the motivation for integration as a method of solving diﬀerential equations. In this and the next few modules, we turn to techniques of integration. 21.1 Integration rules Since integration is the inverse of diﬀerentiation, one can turn diﬀerentiation rules into integration rules. For example, by the linearity of the derivative, we have linearity of the integral: Z (u + v) dx = Z u dx + Z v dx Z (cu) dx = c Z u dx where c is a constant. In other words, integration is a linear operator. The rest of this module deals with turning the chain rule for diﬀerentiation into a rule for integration. This rule is called substitution, or u-substitution traditionally. 21.2 Substitution: the chain rule in reverse Recall the chain rule, which says that if u = u(x) is a function of x, then du = du dx dx. Now if f = f (u) is a function of u, then we ﬁnd Z f (u) du = Z f (u(x))du dx dx (To get from the left side to the right, all we have done is replace u and du by u(x) and du dx dx, respectively). This is the formula for substitution, or u-substitution. Substitution is a useful technique but is not always easy to apply. In a typical problem, one encounters the right side of the above equation, but without knowing what f and u are. If one can ﬁnd the correct f and u so that the integral can be expressed as above, then one can switch over to the left side of the above equation, which is usually easier to evaluate. 32 Example Compute Z esin x cos x dx. (See Answer 1) Example Compute Z 2xex2 dx. (See Answer 2) It is not always so easy to see the ideal choice of u, and sometimes it might take a few tries to ﬁnd the right substitution. Usually, a good strategy is to look for the inner function of a composition of functions and let that be u. Another idea is to look for a function whose derivative is also a factor of the integrand. Example Compute Z x √ x −1 dx (See Answer 3) Another general tip for integration by substitution is to try to simplify the integrand as much as possible before integrating. Example Compute Z cot θ csc θ dθ. (See Answer 4) Example The Gompertz model for the size N(t) of a tumor at time t is dN dt = −aN ln(bN), where a > 0 and 0 < b < 1 are constants. Solve this diﬀerential equation. Hint: it is separable. Then ﬁnd the limit behavior lim t→∞N(t). Finally, ﬁnd the equilibria of the original diﬀerential equation and classify them as stable or unstable. (See Answer 5) 33 Example Compute Z 4(2x + 5)4dx. (See Answer 6) 21.3 Perspective The big idea of this module is that a change of variables (a substitution of one variable for a function of another) can change a diﬃcult integral into an easier one. After computing the easier integral, we can change the variables back again. This idea will come up again in this course and in multivariable calculus. 21.4 Additional Examples Example Compute Z (ln x)2 x dx. (See Answer 7) Example Compute Z tan θ dθ. (See Answer 8) Example Compute Z x5p 1 + x3 dx. (See Answer 9) 21.5 EXERCISES • Compute the integral Z 3 cos x dx • Compute the integral Z x sec2 x2 dx • Compute the integral Z 4x (x2 −1)3 dx 34 • Compute the integral Z e √x √x dx • Compute the integral Z ln(15x5) x dx • Compute the integral x dx √x + 3 dx • Compute the integral dx x √ x2 −1 dx using the substitution u = √ x2 −1 • Now do the same integral using the substitution u = x−1 What is going on here? • Compute the integral dx x √ x2 + 1 dx 21.6 Answers to Selected Examples 1. Let u = sin x and f (u) = eu. Then du = du dx dx = cos x dx, and Z esin x cos x dx = Z f (sin x)d(sin x) = Z f (u) du = Z eu du = eu + C = esin x + C. Note that after evaluating the integral in terms of u, we usually replace u with its function of x, since the original integral was with respect to x. We can check that this is the correct antiderivative by diﬀerentiating (remembering to apply the chain rule) and seeing that we get back the function which we were integrating originally. Typically, we need not write out all the details of what f is. It is suﬃcient to identify u and du and then make the necessary substitutions. (Return) 2. The inner function here looks like u = x2. Then du = 2x dx, which is the remaining factor in the integrand. Thus, Z 2xex2 dx = Z eu du = eu + C = ex2 + C. (Return) 35 3. Here, the inner function seems to be x −1. So let u = x −1. Then du = dx. This takes care of what is under the square root, and the diﬀerential, but what about the extra factor of x? We can take care of this factor by noting that u = x −1 implies x = u + 1. So the integral becomes Z x √ x −1 dx = Z (u + 1)√u du = Z (u + 1)u1/2 du = Z u3/2 + u1/2 du = 2 5u5/2 + 2 3u3/2 + C = 2 5(x −1)5/2 + 2 3(x −1)3/2 + C. From the third to the fourth line above, we used the power rule on each term of the integrand. Again, we can diﬀerentiate to make sure we get back to our original integrand (though it might require a little algebra to show that they are in fact equal). (Return) 4. Since cotangent and cosecant are not very familiar functions, it is helpful to rewrite them in terms of sine and cosine. This gives Z cot θ csc θ dθ = Z cos θ sin θ · 1 sin θ dθ = Z cos θ sin2 θ dθ. It is easier now to see that u = sin θ is a good choice, since its derivative du = cos θ dθ is in the numerator. Making this substitution shows Z cos θ sin2 θ dθ = Z 1 u2 du = Z u−2 du = −u−1 + C = −1 sin θ + C = −csc θ + C. Of course, if one happened to remember the fact that d dθ csc θ = −csc θ cot θ, then we would not require a substitution. But substitution allows us to do these integrals (and harder ones) without needing to memorize a lot of information. (Return) 5. Separating variables and integrating both sides gives Z dN N ln(bN) = Z −a dt. 36 The right side is easy, but the left side requires some work. Looking for a function whose derivative is a factor in the integrand, we see that u = ln(bN) is a good choice. In this case du = 1 bN · b dN = dN N . And so, the integral on the left above becomes Z dN N ln(bN) = Z 1 u du = ln u + C = ln ln(bN) + C. Putting this together with the integral on the right above (and combining the integration constants to one integration constant on the right), we have ln ln(bN) = −at + C Exponentiating twice and then dividing by b gives N = 1 bee−at+C = 1 beCe−at. By plugging in t = 0, we ﬁnd that C = ln(bN0), where N0 is the initial size of the tumor. In the long run, the exponential e−at →0, since a > 0 by assumption. Therefore, the entire exponent is going to 0, and so lim t→∞N(t) = 1 b. Note that N(t) = 1 b is an equilibrium solution to the original diﬀerential equation, since it gives dN dt = 0. It is a stable equilibrium since the graph of of −aN ln(bN) goes from positive to negative as N goes from less than 1 b to greater than 1 b. Another equilibrium is N = 0. This is unstable, since N > 0 means dN dt > 0. Intuitively, even if the tumor is very tiny, it will grow according to this model. (Return) 6. In this case, the inner function is u = 2x + 5, and one ﬁnds that du = 2dx. Thus dx = du 2 , which gives Z 4(2x + 5)4dx = Z 4u4 du 2 = Z 2u4du = 2 5u5 + C = 2 5(2x + 5)5 + C. (Return) 37 7. Here, a good inner function is u = ln x, because the derivative du = 1 x dx. Thus Z (ln x)2 x dx = Z u2 du = u3 3 + C = (ln x)3 3 + C. (Return) 8. First, rewrite tangent in terms of sine and cosine: Z tan θ dθ = Z sin θ cos θ dθ Now, note that u = sin θ would not work, because its derivative, cos θ is in the denominator. On the other hand, u = cos θ is a good substitution because its derivative (up to a constant) is in in the numerator. That is, du = −sin θ dθ. Therefore, Z sin θ cos θ dθ = Z −1 u du = −ln(u) + C = −ln(cos θ) + C. (Return) 9. The logical choice of inner function is u = 1 + x3, which gives du = 3x2 dx and so dx = du 3x2 Substituting in, we ﬁnd Z x5p 1 + x3 dx = Z x5√u du 3x2 = 1 3 Z x3√u du. This seems problematic, because we haven’t been able to get everything in terms of u. But we can use our original substitution to help. Since u = 1 + x3, we have that x3 = u −1, and so 1 3 Z x3√u du = 1 3 Z (u −1)√u du = 1 3 Z u3/2 −u1/2 du = 1 3 2 5u5/2 −2 3u3/2 + C = 2 15(1 + x3)5/2 −2 9(1 + x3)3/2 + C. (Return) 38 22 Integration By Parts This module uses the product rule to derive another useful integration technique: integration by parts. Recall the product rule: d(u · v) = u · dv + v · du. Integrating both sides gives Z d(u · v) = Z u dv + Z v du. Solving for R u dv gives Integration by parts If u = u(x) and v = v(x) are two functions of x, then Z u dv = uv − Z v du. Intuitively, we are given a diﬃcult integral R u dv. By breaking the integrand into u and dv and applying the above formula, we are hopefully able to wind up with an easier integral R v du. Like with the substitution technique, it requires a little bit of thought to choose suitable u and dv. Once u and dv are picked, it is a fairly mechanical process to apply the formula (assuming a good choice of u and dv). Note that the selection is constrained by the fact that u dv must be the entire integrand. So whatever choice is made for u, whatever factors are left over become dv. Note also that the formula involves ﬁnding v, and so dv must be integrable. Ideally, dv should be easy to integrate, which can help guide the selection. Example Compute Z xexdx. (See Answer 1) Example Compute Z ln(x)dx. 39 (See Answer 2) Example Try to compute Z sin x x dx Hint: it cannot be done using integration by parts. (See Answer 3) 22.1 LIPET: A tip for choosing u and dv It is not always obvious how to choose u and dv. The mnemonic LIPET gives a suggestion for how to select u, and then whatever is left over becomes dv. 1. Logarithm 2. Inverse function 3. Polynomial 4. Exponential 5. Trigonometric. When picking u, go down the list until some factor of the integrand ﬁrst matches something from the list. So in the ﬁrst example above, there was no logarithm, no inverse function, but there was a polynomial, x, which was chosen for u. In the second example, there was a logarithm, so that became u. This will not always work perfectly, because (as the above example showed) some integrals simply cannot be computed using integration by parts. But in most examples where integration by parts works, the above mnemonic will help give the correct selection of u and dv. Example Compute Z ln x x2 dx. (See Answer 4) 22.2 Repeated use Sometimes integration by parts requires repeated use, if the integral R v du is not easy to compute. It is not always easy to tell when repeating integration by parts will help, but with practice it becomes easier. Example Compute Z ex cos(x)dx. 40 (See Answer 5) Example Compute Z e2x sin(3x) dx. (See Answer 6) There are integrals that require several applications of integration by parts before they are ﬁnished. Unfortu-nately, it is not always clear when it will work and when it will not. Doing a lot of practice can help develop the intuition to tell the diﬀerence. As the next example shows, sometimes an integral that looks like a perfect candidate for integration by parts does not yield to this method. Example Compute Z ex cosh x dx. (See Answer 7) 22.3 Reduction formulae A ﬁnal application of integration by parts is to prove what are known as reduction formulae. These formulae express one integral in terms of another slightly simpler integral. One can use a reduction formula to repeatedly simplify an integral, eventually reaching a known integral. These formulae are invariably derived by using integration by parts and some algebra. Example For a ﬁxed integer n ≥0, show that Z xn cos x dx = xn sin x + nxn−1 cos x −n(n −1) Z xn−2 cos x dx. Use this formula to ﬁnd Z x2 cos xdx. (See Answer 8) Example Similar algebra as in the above example shows that for n ≥0 Z xn sin x dx = −xn cos x + nxn−1 sin x −n(n −1) Z xn−2 sin x dx. 41 Example Find a reduction formula for Z xnex dx. Use it to evaluate Z x2ex dx. (See Answer 9) Example Show that for n ≥2, Z secn(x) dx = 1 n −1 secn−2(x) tan(x) + n −2 n −1 Z secn−2(x) dx. (See Answer 10) 22.4 Additional examples Example Compute Z x sin(x) dx. (See Answer 11) Example Compute Z arctan x dx. Hint: recall that d dx arctan x = 1 1 + x2 . (See Answer 12) 22.5 EXERCISES Compute the following integrals: • Z xex/2 dx • Z x2ex/2 dx 42 • Z 3x ln x dx • Z 3x2 ln x dx • Z x2 cos x 2 dx • Z e2x sin 3x dx • Z ln x dx • Z ln2 x dx • Z sin(ln x) dx • Z arcsin(2x) dx To solve the integral Z ex cos x dx , we used the method of integration by parts twice. Based on how we solved the integral of ex cosh x, we can try the same with the cosine version, using the fact that cos x = 1 2(eix +e−ix). Try! The integration is the easy part...the hard part is getting the algebra to work out (hello again, Euler’s formula...) • Compute Z sin(2x) cos(3x) dx 22.6 Answers to Selected Examples 1. Letting u = x and dv = exdx (we see that these factors together make up our integrand), one ﬁnds by diﬀerentiating u and integrating dv that du = dx and v = ex. Many students ﬁnd it helps to organize this information in a grid: u = x du = dx dv = ex dx v = ex. Then, from the formula it follows that Z xexdx = xex − Z exdx = xex −ex + C = (x −1)ex + C. One can check that the derivative of this gives back the original integrand, as desired. (Return) 43 2. The selection of u and dv that works is u = ln(x) du = 1 x dx dv = dx v = x. Note that the only other possibility would be dv = ln(x)dx. But this choice would mean that to ﬁnd v we would need to ﬁnd the integral of ln(x), which is the problem at hand. Applying the formula, we ﬁnd Z ln(x)dx = x ln(x) − Z x · 1 x dx = x ln(x) − Z dx = x ln(x) −x + C = x(ln(x) −1) + C. Again, we can check that the derivative of this function gives back ln x, our original integrand. (Return) 3. One choice we might try is u = sin x du = cos x dx dv = 1 x dx v = ln x. However, this requires us to compute the integral Z v du = Z ln x cos x dx, which is no better than the original integral. Another possible choice is u = 1 x du = −1 x2 dx dv = sin x dx v = −cos x. This leads to the integral Z v du = Z cos x x2 dx, which is again no better than the original integral. It turns out no selection will work. Some integrals cannot be computed using integration by parts. And some integrals (like this one) have no elementary answer (i.e. some combination of trigonometric functions, polynomials, exponentials, etc.). That said, we could expand sin(x) as a Taylor series, divide by x and integrate term by term, which gives a series solution. This gives a perfectly suitable solution provided that x is not too far from 0. (Return) 44 4. A good choice is u = ln x du = 1 x dx dv = 1 x2 dx v = −1 x . Then Z ln x x2 dx = −1 x ln x − Z −1 x2 dx = −1 x ln x −1 x + C. (Return) 5. We can take u = ex du = ex dx dv = cos x dx v = sin x. (it turns out that this would work equally well if we reversed these). Then the formula says that Z ex cos(x)dx = ex sin(x) − Z sin(x)exdx. This does not seem much better than the original problem. However, with some persistence and algebra, this will work. Let I = R ex cos(x)dx be the original integral, and let J = R sin(x)exdx be the new integral. So the above calculation shows I = ex sin(x) −J Using integration by parts on J, we pick u = ex du = ex dx dv = sin x dx v = −cos x. Then it follows that J = Z sin(x)exdx = ex(−cos(x)) − Z (−cos(x))exdx = −ex cos(x) + Z ex cos(x)dx = −ex cos(x) + I. So the problem has come back to the original integral I. This might seem like cause for despair, but putting together the previous calculations shows I = ex sin(x) −J = ex sin(x) −(−ex cos(x) + I) = ex sin(x) + ex cos(x) −I. 45 Now, solving for I gives 2I = ex sin(x) + ex cos x + C I = 1 2(ex sin(x) + ex cos(x)) + C. (Return) 6. The algebra is similar to the above example, but care must be taken with the constants. Let I = Z e2x sin(3x) dx. Let u = e2x du = 2e2x dx dv = sin(3x) dx v = −1 3 cos(3x). Then I = −1 3e2x cos(3x) − Z −1 3 cos(3x) 2e2x dx = −1 3e2x cos(3x) + 2 3 Z e2x cos(3x) dx. Now, let J = Z e2x cos(3x) dx. Selecting u = e2x du = 2e2x dx dv = cos(3x) dx v = 1 3 sin(3x), we have J = 1 3e2x sin(3x) −2 3 Z e2x sin(3x) dx = 1 3e2x sin(3x) −2 3I. Putting this all together, we have I = −1 3e2x cos(3x) + 2 3J = −1 3e2x cos(3x) + 2 3 1 3e2x sin(3x) −2 3I = −1 3e2x cos(3x) + 2 9e2x sin(3x) −4 9I. 46 Solving for I gives 13 9 I = −1 3e2x cos(3x) + 2 9e2x sin(3x) I = 9 13 −1 3e2x cos(3x) + 2 9e2x sin(3x) = 1 13e2x (−3 cos(3x) + 2 sin(3x)) . Remember that any indeﬁnite integral has an integration constant, so the ﬁnal answer is Z e2x sin(3x) dx = 1 13e2x (−3 cos(3x) + 2 sin(3x)) + C. (Return) 7. This looks so similar to the above examples, that it is reasonable to expect that two applications of integration by parts will allow us to algebraically ﬁnd this integral. Unfortunately, there is a problem that will soon present itself. Let I = Z ex cosh x dx. If we set u = ex du = ex dx dv = cosh x dx v = sinh x, we ﬁnd I = ex sinh x − Z ex sinh x = ex sinh x −J, where we have set J = Z ex sinh x dx. Letting u = ex du = ex dx dv = sinh x dx v = cosh x, we ﬁnd J = ex cosh x − Z ex cosh x dx = ex cosh x −I. Putting it all together, I = ex sinh x −J = ex sinh x −(ex cosh x −I) = ex sinh x −ex cosh x + I. 47 Here is where our problem arises. We cannot solve for I because there is a positive I on both sides. This problem is due to the fact that (unlike sine and cosine), the hyperbolic sine and cosine do not introduce negative signs when integrated or diﬀerentiated, respectively. So what do we do? Rewrite our integral using the deﬁnition of cosh x and it becomes easy: Z ex cosh x dx = Z ex · ex + e−x 2 = 1 2 Z e2x + 1 dx = 1 2 1 2e2x + x + C = 1 4e2x + 1 2x + C. (Return) 8. Let u = xn du = nxn−1 dx dv = cos x dx v = sin x. Then according to the formula, Z xn cos x dx = xn sin x − Z nxn−1 sin x dx. Now, since we want to get our integral in terms of an integral involving cos x and a power of x, we can apply integration by parts to Z xn sin x dx. Here, we let u = xn du = nxn−1 dx dv = sin x dx v = −cos x. This gives Z xn sin x dx = −xn cos x + Z nxn−1 cos x dx. Now, using this in our earlier equation (though with n replaced by n −1), we ﬁnd Z xn cos x dx = xn sin x − Z nxn−1 sin x dx = xn sin x −n Z xn−1 sin x dx = xn sin x −n −xn−1 cos x + Z (n −1)xn−2 cos x dx = xn sin x + nxn−1 cos x −n(n −1) Z xn−2 cos x dx. 48 The formula says that Z x2 cos x dx = x2 sin x + 2x cos x −2 Z cos x dx = x2 sin x + 2x cos x −2 sin x + C = x2 sin x + 2x cos x −2 sin x + C. (Return) 9. Letting u = xn du = nxn−1 dx dv = ex dx v = ex, we ﬁnd that Z xnex = xnex −n Z xn−1ex dx. Applying this when n = 2 (then applying it again) gives Z x2ex = x2ex −2 Z xex dx = x2ex −2 xex − Z ex dx = x2ex −2xex + 2ex + C. (Return) 10. In integrating a power of a trigonometric function, it can be hard to pick how many factors become u and how many become dv. The fact that dv is supposed to be easy to integrate can guide this selection. Since R sec2 xdx = tan x, letting dv = sec2 xdx should work well. Thus, u = secn−2(x) and dv = sec2 xdx, which means du = (n −2) secn−3(x) sec(x) tan(x)dx (by the chain rule), and v = tan x. Recalling the Pythagorean identity tan2 x = sec2 x −1, one ﬁnds that Z secn(x)dx = secn−2(x) tan(x) − Z (n −2) secn−2(x) tan2(x)dx = secn−2(x) tan(x) −(n −2) Z secn−2(x)(sec2(x) −1)dx = secn−2(x) tan(x) −(n −2) Z (secn(x) −secn−2(x))dx = secn−2(x) tan(x) −(n −2) Z secn(x)dx + (n −2) Z secn−2(x)dx Now, solving for R secn(x)dx gives Z secn(x)dx = 1 n −1 secn−2(x) tan(x) + n −2 n −1 Z secn−2(x)dx, as desired. (Return) 49 11. The logical choice (either by the LIPET mnemonic, or by picking u to be something which gets simpler when diﬀerentiated) for parts is u = x du = dx dv = sin(x) dx v = −cos(x) Therefore, Z x sin(x) dx = −x cos(x) − Z −cos(x) dx = −x cos(x) + Z cos(x) dx = −x cos(x) + sin(x) + C. (Return) 12. As in some earlier examples, the only choice we have is to set u = arctan x du = 1 1 + x2 dx dv = dx v = x. Therefore, Z arctan x dx = x arctan x − Z x 1 + x2 dx. This second integral can be solved with a substitution of u = 1 + x2 du = 2x dx. So dx = du 2x . Making the substitution gives Z x 1 + x2 dx = Z x u · du 2x = 1 2 Z du u = 1 2 ln u + C = 1 2 ln(1 + x2) + C. Putting it all together, we ﬁnd Z arctan x dx = x arctan x − Z x 1 + x2 dx = x arctan x −1 2 ln(1 + x2) + C. (Return) 50 23 Trigonometric Substitution There is another class of integrals which usually do not involve trigonometric functions, but which can be solved by substituting the variable with a trigonometric function. This can be thought of as using the substitution formula, from Integration By Substitution, in the other direction. That is, going from the left side to right side in the equality Z f (x) dx = Z f (x(θ))dx dθ dθ, where we have made the substitution x = x(θ). We often use θ when making a trigonometric substitution. Example Compute Z dx 1 + x2 . (See Answer 1) 23.1 Typical substitutions Trigonometric substitution makes use of the Pythagorean identities. In general, the basic trigonometric substi-tutions are: Form Substitution Identity used 1 + x2 x = tan θ 1 + tan2 θ = sec2 θ 1 −x2 x = sin θ 1 −sin2 θ = cos2 θ x2 −1 x = sec θ sec2 θ −1 = tan2 θ Caveat The form x2 −1 often leads to a messy integral involving sec(θ). This can often be avoided using a hyperbolic trigonometric substitution (see below). After a substitution has been made, the resulting integral will often involve a product of trigonometric functions, possibly raised to powers. These types of integrals are covered in more detail in Trigonometric Integrals. For now, here are a few of the useful identities in evaluating these integrals: 51 Power reduction sin2(θ) = 1−cos(2θ) 2 cos2(θ) = 1+cos(2θ) 2 Double angle sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ −sin2 θ Example Compute Z p 1 −x2 dx. (See Answer 2) Example Compute Z dx √ 1 −x2 . (See Answer 3) 23.2 Forms with other constants There are other forms which are similar to the above forms but have diﬀerent constants involved. These are dealt with using similar substitutions which make the constants cancel and factor so that the same identities can be used. Example Compute Z dx x2√ x2 + 4 (See Answer 4) Example Compute Z dx 4 + 9x2 . (See Answer 5) The following table summarizes the substitutions to be made when other constants are involved. The identities used are the same Pythagorean identities given in the above table. Form Substitution a2x2 + b2 x = b a tan θ b2 −a2x2 x = b a sin θ a2x2 −b2 x = b a sec θ 52 23.3 Completing the square Sometimes it is not obvious at ﬁrst that an integral is of the form where a trigonometric substitution is helpful. It may take a little bit of algebra to see what the right substitution is. This common algebraic tool is known as completing the square, which simply rewrites a quadratic expression as the square of a binomial plus a constant. To review the algebra involved in this process, check Wikipedia:Completing the square. Example Compute Z dx √ 3 + 2x −x2 (See Answer 6) 23.4 Hyperbolic trigonometric substitutions Recall that the hyperbolic trigonometric functions sinh(x) and cosh(x) are deﬁned by sinh(θ) = eθ −e−θ 2 cosh(θ) = eθ + e−θ 2 . These functions satisfy the Pythagorean identity cosh2(θ)−sinh2(θ) = 1. Also, note that d dθ cosh(θ) = sinh(θ), and d dθ sinh(θ) = cosh(θ). This means hyperbolic substitutions are another option for dealing with the following forms: Form Substitution Identity used 1 + x2 x = sinh θ 1 + sinh2 θ = cosh2 θ x2 −1 x = cosh θ cosh2 θ −1 = sinh2 θ This often gives a simpler answer than the x = sec θ substitution suggested above, but the trade-oﬀis that the answer will involve hyperbolic functions. Here are some of the other identities for the hyperbolic functions, which are similar to those for regular trigonometric functions: Double Angle sinh(2θ) = 2 sinh(θ) cosh(θ) cosh(2θ) = cosh2(θ) + sinh2(θ) cosh(2θ) = 2 cosh2(θ) −1 cosh(2θ) = 2 sinh2(θ) + 1 Power reduction sinh2(θ) = cosh(2θ)−1 2 cosh2(θ) = cosh(2θ)+1 2 53 Example Compute Z dx √ 1 + x2 . (See Answer 7) Example Compute Z p 1 + x2 dx. (See Answer 8) 23.5 Blow-ups Sometimes a diﬀerential equation can be solved by using a trigonometric substitution. But this can sometimes lead to an unreasonable solution due to blow-ups or singularities, which exist for many trigonometric functions. Example Consider a ﬁnancial model which predicts that marginal proﬁts equal some positive constant plus something which is proportional to the square of net proﬁts. Mathematically, dP dt = b2 + a2P 2, for constants a and b (we square them to ensure that they are positive). Solve this diﬀerential equation and ﬁnd where it has a blow-up. (See Answer 9) 23.6 EXERCISES Compute the following integrals: • Z x2 √ 4 −x2 dx • Z dx √ x2 −2x • Z √ 1 −x2 x2 dx • Z (1 −x2)−3/2 dx • Z x √ 1 + x2 dx • Z dx x √ x2 −1 54 • Z dx √ x2 −6x + 10 • Z dx √ x2 −2x −8 23.7 Answers to Selected Examples 1. Consider the substitution x = tan(θ). Then one ﬁnds that dx = sec2 θ dθ. Making these substitutions and recalling the Pythagorean identity 1 + tan2 θ = sec2 θ, the integral becomes Z dx 1 + x2 = Z sec2 θ dθ 1 + tan2 θ = Z sec2 θ dθ sec2 θ = Z dθ = θ + C = arctan(x) + C. The last line comes from our original substitution: x = tan θ ⇔ arctan x = θ. (Return) 2. According to the above guide, the substitution to make is x = sin θ. Then dx = cos θ dθ, and it follows that Z p 1 −x2 dx = Z p 1 −sin2 θ cos θ dθ = Z √ cos2 θ cos θ dθ = Z cos2 θ dθ. Now using the power reduction identity for cosine, we have Z cos2 θ dθ = Z 1 2(1 + cos(2θ)) dθ = θ 2 + 1 4 sin(2θ) + C. Finally, we must get this back in terms of x. We know that θ = arcsin x. But to take care of sin 2θ, we must use the double angle formula from above. This gives sin 2θ = 2 sin θ cos θ = 2x p 1 −x2 In the last line above, we knew sin θ = x from the original substitution. We found cos θ by drawing a right triangle which relates x and θ according to the substitution sin θ = x: 55 Putting this all together and doing a little simpliﬁcation, we ﬁnd Z p 1 −x2 = 1 2 arcsin x + 1 2x p 1 −x2 + C. (Return) 3. By the above table, the substitution x = sin θ should be used (hence θ = arcsin(x)). Then dx = cos θ dθ, so the integral becomes Z dx √ 1 −x2 = Z cos θ dθ p 1 −sin2 θ = Z cos θ dθ √ cos2 θ = Z dθ = θ + C = arcsin(x) + C. (Return) 4. The form x2 + 4 in the denominator reminds us of the substitution we made earlier for x2 + 1, which was the substitution x = tan θ. This is the correct impulse, but unfortunately it does not work quite right here since there is no nice simpliﬁcation for tan2 θ + 4. We can ﬁx this by adjusting the coeﬃcients. The idea is that we could factor out a 4 if we had 4 tan2 θ + 4 = 4(tan2 θ + 1). To get that extra factor of 4, we can make the substitution x = 2 tan θ. Then dx = 2 sec2 θ dθ, and the integral becomes Z dx x2√ x2 + 4 = Z 2 sec2 θ dθ 4 tan2 θ √ 4 tan2 θ + 4 = Z 2 sec2 θ dθ 4 tan2 θ p 4(tan2 θ + 1) = 1 4 Z sec2 θ dθ tan2 θ sec θ. 56 The last equality above comes from again using the identity tan2 θ+1 = sec2 θ. Doing a little simpliﬁcation and rewriting in terms of sine and cosine gives 1 4 Z sec θ dθ tan2 θ = 1 4 Z 1 cos θ · cos2 θ sin2 θ dθ = 1 4 Z cos θ dθ sin2 θ . This we can handle with a substitution of u = sin θ and du = cos θ dθ, which gives 1 4 Z cos θ dθ sin2 θ = 1 4 Z du u2 = 1 4 −1 u + C = −1 4u + C = − 1 4 sin θ + C Now, we must do one ﬁnal bit of right triangle trigonometry to get sin θ in terms of x. By the original substitution we have tan θ = x 2, and this can be expressed by the following triangle: It follows that sin θ = x √ x2+4. Putting it all together, we have Z dx x2√ x2 + 4 = − 1 4 sin θ + C = − √ x2 + 4 4x + C. (Return) 5. This is another example which looks like x = tan θ is the right type of substitution to make. However, again we need to adjust the coeﬃcient since 4 + 9 tan2 θ does not simplify nicely. 57 The key is to get the constants to cancel and factor. The substitution x = 2 3 tan θ will work, and in this case θ = arctan( 3 2x). Then dx = 2 3 sec2 θdθ, and the integral becomes Z dx 4 + 9x2 = 2 3 Z sec2 θ dθ 4 + 9(4/9) tan2 θ = 2 3 Z sec2 θ dθ 4(1 + tan2 θ) = 2 3 Z sec2 θ dθ 4 sec2 θ = 2 3 · 1 4 Z dθ = 1 6θ + C = 1 6 arctan(3 2x) + C. (Return) 6. Start by completing the square for the quadratic: 3 + 2x −x2 = −x2 + 2x + 3 = −(x2 −2x) + 3 = −(x2 −2x + 1) + 4 = −(x −1)2 + 4 = 4 −(x −1)2. So we can rewrite the integral as Z dx √ 3 + 2x −x2 = Z dx p 4 −(x −1)2 = Z du √ 4 −u2 , where we substituted u = x −1 and du = dx. This can now be dealt with using a trigonometric substitution of u = 2 sin θ (remember, the extra factor of 2 is there so that the 4 will factor out). So du = 2 cos θ dθ, and the integral becomes Z du √ 4 −u2 = Z 2 cos θ dθ p 4 −4 sin2 θ = Z 2 cos θ dθ √ 4 p 1 −sin2 θ = Z 2 cos θ dθ 2 cos θ = Z dθ = θ + C. 58 Solving our original substitution for θ, we see that θ = arcsin u 2 = arcsin x −1 2 . So the ﬁnal answer is Z dx √ 3 + 2x −x2 = arcsin x −1 2 + C. (Return) 7. Using a regular trigonometric substitution, we would set x = tan θ, and dx = sec2 θ dθ, which, after the usual algebra, gives Z dx √ 1 + x2 = Z sec2 θ dθ √ 1 + tan2 θ = Z sec θ dθ. But the integral of secant is not easy to remember, nor easy to rederive. If instead, we make the hyperbolic trigonometric substitution x = sinh u, so dx = cosh u du, then we have Z dx √ 1 + x2 = Z cosh u du p 1 + sinh2 u = Z cosh u du cosh u = Z du = u + C = arcsinh x + C. So the hyperbolic trigonometric substitution led to a much easier integral to evaluate. The trade-oﬀis that the ﬁnal result involves the inverse hyperbolic trigonometric functions, as opposed to more familiar functions. (Return) 8. Using the hyperbolic trigonometric substitution x = sinh(θ) gives Z p 1 + x2 dx = Z p 1 + sinh2 θ cosh θ dθ = Z p cosh2 θ cosh θ dθ = Z cosh2 θ dθ = 1 2 Z (cosh(2θ) + 1) dθ = 1 2 θ + 1 2 sinh(2θ) + C = 1 2θ + 1 42 sinh(θ) cosh(θ) + C = 1 2 sinh−1 x + 1 2x p 1 + x2 + C. 59 (Return) 9. This is a separable equation. Separating the variables and integrating both sides gives Z dP b2 + a2P 2 = Z dt. On the left, we can use the trigonometric substitution P = b a tan θ dP = b a sec2 θ dθ. Note then that θ = arctan a bP. This gives Z dP b2 + a2P 2 = Z b a sec2 θ dθ b2(1 + tan2 θ) = b a Z sec2 θ dθ b2 sec2 θ = 1 ab Z dθ = 1 abθ = 1 ab arctan a bP. (leaving oﬀthe constant for now). On the right side we get t + C, so 1 ab arctan a bP = t + C. Solving this for P gives P(t) = b a tan(abt + C). If initial proﬁts, at t = 0, are 0, then C = 0, so the ﬁnal answer is P(t) = b a tan(abt). Since tangent blows up at π 2 , this model implies proﬁt goes to inﬁnity at t = π 2ab, which is a sign that this model is not perfect. (Return) 60 24 Partial Fractions So far, the techniques of integration covered in this course have all been derived from diﬀerentiation rules run in reverse. This module gives an algebraic method for integrating rational functions. Recall that a rational function is a function of the form f (x) = P(x) Q(x), where P(x) and Q(x) are polynomials. It turns out that with a little bit of algebraic manipulation, many of these integrals are not too diﬃcult to compute. Example Compute Z 3x2 −5 x −2 dx. (See Answer 1) The rest of this module expands on this method (in particular, when the denominator is of a higher degree), which is known as the method of partial fractions. 24.1 Partial fractions Given a rational function P (x) Q(x), and P has a lower power than Q, the method of partial fractions uses algebra to rewrite the function as a sum of simpler terms which are easy to integrate. While there are some cases to deal with, the basic outline of the method is: 1. Given the integral R P (x) Q(x)dx where P and Q are polynomials. 2. Factor Q(x) = (x −r1)(x −r2) . . . (x −rn). Assume for now that each of these factors is distinct. 3. We use the following fact, that the rational function can be expressed as P(x) Q(x) = A1 x −r1 + A2 x −r2 + . . . + An x −rn . 4. Use algebra to ﬁnd what each of the constants Ai is. This step requires the most work. 5. Then 61 Z P(x) Q(x)dx = Z A1 x −r1 + . . . + An x −rn dx = A1 ln \|x −r1\| + . . . + An ln \|x −rn\| + C. Example Compute Z 3x −1 x2 −2x −3dx. (See Answer 2) Example Compute Z 2x2 −6x −2 x3 −x2 −2x dx. (See Answer 3) Example Compute Z x2 + 2x −1 2x3 + 3x2 −2x dx. (See Answer 4) Example A simple model for the deﬂection x(t) of a thin beam under a load proportional to λ2 is dx dt = λ2x −x3 = x(λ −x)(λ + x). Solve this diﬀerential equation (but do not solve for x(t) explicitly). Then ﬁnd the equilibria of the diﬀerential equation and classify them as stable or unstable. (See Answer 5) Example The logistic model for population dynamics says that the rate of change of a population P with respect to time is dP dt = rP −bP 2 where r and b are positive constants which can be thought of as the reproduction rate and death rate, respectively. Factoring and letting K = r b, we have dP dt = bP(K −P). Solve this diﬀerential equation. What is the long run population behavior? (See Answer 6) 62 24.2 Other technicalities Higher degree numerator For the algebra to work out above, the degree of the numerator, P(x), must be lower than that of the denominator, Q(x). However, it is easy to deal with the case when the numerator has equal or higher degree. One can use long division to rewrite the quotient as a divisor plus a remainder, just like writing an improper fraction as a mixed number in middle school. Repeated factors If the denominator has one or more repeated factors, i.e. P(x) Q(x) = P(x) (x −r1)m1 . . . (x −rk)mk , where one or more of the mi is greater than 1. Then the way to express the function is P(x) (x −r1)m1 . . . (x −rn)mn = A1 x −r1 + A2 (x −r1)2 + · · · + Am1 (x −r1)m1 + B1 x −r2 + B2 (x −r2)2 + · · · + Bm2 (x −r2)m2 + . . . . Now, the algebra proceeds as before to ﬁnd the constants in the numerators. It is easiest to see this through an example. Example Compute Z 2x2 −4x −2 (x + 1)(x −1)2 dx. (See Answer 7) Quadratic factors Suppose one of the factors of the denominator is a quadratic which cannot be factored (e.g. x2 + 1). Then the numerator of this factor in the expansion should be of the form Ax + B. Then the algebra proceeds as before. Example Compute Z 3x2 −2x + 1 (x −1) (x2 + 1)dx. (See Answer 8) 24.3 EXERCISES Compute the following integrals: 63 • Z 5 + x x2 + x −6 dx • Z 2x + 3 6x2 + 5x + 1 dx • Z x (x + 1) (x + 2) dx • Z x2 −x + 5 (x −2) (x −1)(x + 3) dx • Z 2x −1 x3 −x dx • Z x2 −3 x2 −4 dx • Z x3 + 10x2 + 33x + 36 x2 + 4x + 3 dx • Z x + 2 (x −1)2 dx • Z dx x4 −6x3 + 12x2 24.4 Answers to Selected Examples 1. By doing polynomial long division on this ratio, we ﬁnd (For more on polynomial long division, see wikipedia). The above observation, which is entirely based on algebra, allows us to evaluate the integral as Z 3x2 −5 x −2 dx = Z 3x + 6 + 7 x −2 dx = 3 2x2 + 6x + 7 ln \|x −2\| + C. (Return) 64 2. Factoring the denominator gives x2 −2x −3 = (x + 1)(x −3). Thus, the goal is to ﬁnd constants A and B such that 3x −1 (x + 1) (x −3) = A x + 1 + B x −3. There are several methods for ﬁnding the constants, but one of the simplest is to clear denominators, which gives 3x −1 = A(x −3) + B(x + 1). This equation must hold for every value of x. In particular, one can pick convenient values of x which make the algebra easy. In this case, by plugging in x = 3, the ﬁrst term on the right disappears. Thus, the equation becomes 8 = B · 4, and so B = 2. Similarly, picking x = −1 makes the second term on the right disappear. Thus, −4 = A · (−4) so A = 1. It follows that Z 3x −1 x2 −2x −3dx = Z 3x −1 (x + 1) (x −3)dx = Z 1 x + 1 + 2 x −3 dx = ln \|x + 1\| + 2 ln \|x −3\| + C. (Return) 3. Factoring the denominator gives x3 −x2 −2x = x(x + 1)(x −2) So we are looking for constants A, B, and C such that 2x2 −6x −2 x(x + 1)(x −2) = A x + B x + 1 + C x −2. Clearing fractions gives 2x2 −6x −2 = A(x + 1)(x −2) + Bx(x −2) + Cx(x + 1). Now, picking the following convenient values of x allows us to ﬁnd each constant: x = 0 −2 = −2A A = 1 x = −1 6 = 3B B = 2 x = 2 −6 = 6C C = −1. So we have that Z 2x2 −6x −2 x(x + 1)(x −2) = Z 1 x + 2 x + 1 − 1 x −2 dx = ln \|x\| + 2 ln \|x + 1\| −ln \|x −2\| + C. (Return) 65 4. Factoring gives 2x3 + 3x2 −2x = x(2x −1)(x + 2) So we are looking for constants A, B, C such that x2 + 2x −1 2x3 + 3x2 −2x = A x + B 2x −1 + C x + 2. As before, we clear fractions which gives x2 + 2x −1 = A(2x −1)(x + 2) + Bx(x + 2) + Cx(2x −1). Now we pick convenient values of x to make the factors cancel and solve for the constants: x = 0 −1 = −2A A = 1 2 x = 1 2 1 4 = 5 4B B = 1 5 x = −2 −1 = 10C C = −1 10. So we ﬁnd Z x2 + 2x −1 x(2x −1)(x + 2) dx = Z 1/2 x + 1/5 2x −1 + −1/10 x + 2 dx = 1 2 ln \|x\| + 1 5 · 1 2 ln \|2x −1\| −1 10 ln \|x + 2\| + C = 1 2 ln \|x\| + 1 10 ln \|2x −1\| −1 10 ln \|x + 2\| + C. Note the extra factor of 1 2 for the middle term comes from doing a substitution of u = 2x −1, which implies dx = 1 2 du. (Return) 5. Factoring gives dx dt = x(λ −x)(λ + x). Separating and integrating gives Z dx x(λ −x)(λ + x) = Z dt. Now, we use partial fractions on the left side: 1 x(λ −x)(λ + x) = A x + B λ −x + C λ + x Clearing fractions gives 1 = A(λ −x)(λ + x) + Bx(λ + x) + Cx(λ −x). 66 Picking convenient values of x gives x = 0 1 = λ2A A = 1 λ2 x = λ 1 = 2λ2B B = 1 2λ2 x = −λ 1 = −2λ2C C = −1 2λ2 . So we have Z dx x(λ −x)(λ + x) = Z 1/λ2 x + 1/2λ2 λ −x −1/2λ2 λ + x dx = 1 λ2 ln \|x\| − 1 2λ2 ln \|λ −x\| − 1 2λ2 ln \|λ + x\|. All of this equals t + C on the right. The equilibria of the diﬀerential equation are x = 0, x = λ, and x = −λ. The equilibrium at 0 is unstable and the other two are stable, as the graph shows: (Return) 6. Separating gives dP P(K −P) = bdt. (1) Integrating the left side is done using partial fractions, and the denominator is already factored. So the next step is to ﬁnd A and B such that 1 P(K −P) = A P + B K −P . Clearing denominators gives 1 = A(K −P) + BP. Remember, K and A are constants, and this equation must hold for every value of P. Setting P = K cancels the ﬁrst term and gives 1 = BK, so B = 1 K . 67 Setting P = 0 cancels the second term and gives A = 1 K . Thus, Z dP P(K −P) = Z 1 K 1 P + 1 K −P dP = 1 K (ln P −ln(K −P)) = 1 K ln P K −P , by a property of logarithms. Multiplying through by K gives ln P K −P = Z Kb dt = Z r dt = rt + C (recall that Kb = r by the deﬁnition of K). Now, exponentiating gives P K −P = ˜ Cert, for a new constant ˜ C. By plugging in t = 0, we ﬁnd that ˜ C = P0 K −P0 , where P0 is the initial population. Multiplying through by K −P and doing a little algebra gives P = ˜ Cert(K −P) P + P ˜ Cert = ˜ CertK P(1 + ˜ Cert) = ˜ CertK P = ˜ CertK 1 + ˜ Cert . Replacing ˜ C = P0 K−P0 gives P = P0 K −P0 ertK · 1 1 + P0 K−P0 ert = KP0ert K −P0 + P0ert = KP0 (K −P0)e−rt + P0 (From the ﬁrst to the second line, we distributed (K −P0) in the denominator. From the second to third line, we multiplied the top and bottom by e−rt.). Note that if P0 = 0 (i.e. there was no population to begin with), then the population will stay at 0. This is consistent with the above equation. On the other hand, if P0 > 0, then as t →∞, the e−rt in the denominator goes to 0, and so lim t→∞P(t) = KP0 P0 = K. 68 We can think of K as the carrying capacity for the population, a sort of ideal size for the population. Alternatively, by looking at the original diﬀerential equation, we see that K is an equilibrium. It is stable, since populations above K have a negative derivative (hence are decreasing), and populations below K have a positive derivative (hence are increasing). On the other hand 0 is an unstable equilibrium. The model implies that as long as the population is not extinct to begin with, it will grow and eventually equal K. (Return) 7. The denominator is already factored, so write 2x2 −4x −2 (x + 1)(x −1)2 = A x + 1 + B x −1 + C (x −1)2 . Clearing the denominators gives 2x2 −4x −2 = A(x −1)2 + B(x + 1)(x −1) + C(x + 1). Plugging in x = 1 cancels the ﬁrst two terms on the right, leaving −4 = 2C, so C = −2. Plugging in x = −1 cancels the second two terms and leaves 4 = 4A, so A = 1. Now, it seems that there are no more nice values of x to help solve for B. But remember that the equation must hold for any value of x. Picking x = 0 (which is an easy value to use), gives −2 = A −B + C. Knowing A = 1 and C = −2 gives B = 1. Thus, Z 2x2 −4x −2 (x + 1) (x −1)2 dx = Z 1 x + 1 + 1 x −1 + −2 (x −1)2 dx = ln \|x + 1\| + ln \|x −1\| + 2 x −1 + C (Return) 8. Write 3x2 −2x + 1 (x −1) (x2 + 1) = A x −1 + Bx + C x2 + 1 . Clearing fractions gives 3x2 −2x + 1 = A(x2 + 1) + (Bx + C)(x −1). Picking x = 1 gives 2 = 2A, so A = 1. Now, picking any other two values for x will allow ﬁnding B and C. For instance, x = 0 gives 1 = A −C, and so C = 0. Finally, picking x = −1 gives 6 = 2A + (−B)(−2), so B = 2. Thus, Z 3x2 −2x + 1 (x −1) (x2 + 1)dx = Z 1 x −1 + 2x x2 + 1 dx = ln \|x −1\| + ln(x2 + 1) + C. (Return) 69 25 Deﬁnite Integrals This module moves from the indeﬁnite integral, which is a class of functions, to the deﬁnite integral, which is a number. The relationship between these seemingly unrelated topics will be revealed in the next module. The idea underlying the deﬁnite integral is that adding up local increments leads to a global total. Before getting into the details of what this means, consider a simple example. Example Consider n X i=1 i = 1 + 2 + 3 + · · · + n. One can visualize this sum as the area of a triangular stack of 1 × 1 boxes. The ﬁrst column has 1 box, the second column has 2 boxes, and so on through the nth column with n boxes: The area of this roughly triangular region can be found by splitting it into two regions: a right triangle of base and height n, and the half boxes left over: 70 The total area is therefore 1 2n(n + 1), and so we ﬁnd that n X i=1 i = 1 2n(n + 1). The point of this example is to compare the amount of computation (e.g. the number of additions) required to do the sum using local information (adding up the terms one by one), verses the global information (evaluating the product on the right above). It is much easier to simply evaluate the product. The deﬁnite integral takes this type of idea and generalizes it to more diﬃcult sums. Before we can deﬁne it, we need a few deﬁnitions. 25.1 Partitions and Riemann sums Given an interval [a, b], a partition P of [a, b] is a division of the interval [a, b] into subintervals Pi. Visually, think of placing hash marks along the interval [a, b] and then labeling the subintervals P1, P2, . . . from left to right: Let (∆x)i be the width of the ith subinterval, Pi. Choose a sample point xi from the ith subinterval (this can be a point chosen at random from the subinterval or systematically; it does not matter). 71 Given a function f , a partition P for an interval [a, b], and sample points xi, the Riemann sum of f on P is given by N X i=1 f (xi)(∆x)i. The Riemann sum can be interpreted as an approximation of the area under the curve of f from a to b using rectangles. The width and height of the ith rectangle are (∆x)i and f (xi), respectively. Note that in this area interpretation, a rectangle which is below the x-axis has negative area (since f (xi) < 0 in this case). For an example with N = 4 rectangles, consider the following ﬁgure: 25.2 The deﬁnite integral The deﬁnite integral The deﬁnite integral of a function f from a to b, denoted Z b x=a f (x) dx, is deﬁned by Z b x=a f (x) dx = lim ∆x→0 N X i=1 f (xi)(∆x)i. The function f being integrated is called the integrand. In other words, the deﬁnite integral is the limit of the Riemann sums as the lengths of the subintervals approach 0. In the area interpretation, the widths of all the rectangles are getting arbitrarily small, which ultimately gives the area under the curve: (Link to Riemann Sum Limit Animated GIF) Remember that when interpreting the deﬁnite integral as the area under the curve, any region which is below the x-axis contributes negative area to the total. 72 Example Using the deﬁnition of the deﬁnite integral, compute Z 1 x=0 x dx. (See Answer 1) Notation Sums The integral sign R and the summation sign P are both short for sum. The integral sign R looks like a stylized S, and the summation sign is the Greek sigma, short for sum. Limits Including the variable in the limits of integration is not strictly necessary, but is a useful habit to develop for future courses where integration will be happening with respect to several variables. It is also ﬁne to suppress the notation and just have R b a f (x) dx: Z b a f (x) dx = Z b x=a f (x) dx. Variables The variable used in the integrand does not matter; it is sometimes referred to as a dummy variable: Z b x=a f (x) dx = Z b t=a f (t) dt = Z b z=a f (z) dz. However, if there is a variable used in one of the limits of integration (as will happen from time to time), it is important to avoid using that as the dummy variable too. For example, Z x a f (t) dt instead of Z x a f (x) dx. Caveat Note that, although their notation is similar, deﬁnite integrals are not the same as indeﬁnite integrals! The indeﬁnite integral of a function is a class of functions, whereas the deﬁnite integral of a function over an interval is a number. That said, it is no accident that they have similar notations, because of their relationship, which is given by the Fundamental Theorem of Integral Calculus in the next module. 25.3 Properties of deﬁnite integrals Linearity The deﬁnite integral is linear, i.e. Z b x=a (f (x) + g(x)) dx = Z b x=a f (x) dx + Z b x=a g(x) dx. Z b x=a c · f (x) dx = c Z b x=a f (x) dx. (See Justiﬁcation 2) 73 Additivity When integrating the same function over two adjacent intervals, we have additivity: Z b a f (x) dx + Z c b f (x) dx = Z c a f (x) dx. In the area interpretation, this can be thought of as taking the area under the curve from a to b and adding the area under the curve from b to c, which gives the area under the curve from a to c: Another way of thinking about it is adding the intervals [a, b] and [b, c] together to get [a, c]. It is important to note that the orientation of the interval matters, as discussed in the next subsection. Orientation The orientation of the interval over which we integrate matters. Integrating from left to right is positive, and integrating from right to left is negative: Z b a f (x) dx = − Z a b f (x) dx. (See Justiﬁcation 3) Dominance This is another intuitive property. If f (x) ≥0 for all x in the interval [a, b], then Z b a f (x) dx ≥0. Also, if f (x) ≥g(x) for all x in the interval, then Z b a f (x) dx ≥ Z b a g(x) dx (See Justiﬁcation 4) 74 25.4 More examples There are a few deﬁnite integrals that we can compute directly from the deﬁnition. But for most functions, it is not easy to work directly with the deﬁnition. Example Compute Z b a c dx (See Answer 5) Example Compute Z b a x dx (See Answer 6) 25.5 Odd and even functions There are a few ﬁnal cases where certain deﬁnite integrals can be simpliﬁed by using properties of the integrand. Odd and even functions A function f (x) is called odd if f (−x) = −f (x). A function g(x) is called even if g(−x) = g(x). 75 The reason for the terminology comes from Taylor series. A function is odd if and only if every term in its Taylor series has odd power. Similarly, a function is even if and only if every term in its Taylor series has even power. (See Justiﬁcation 7) Example Sine and hyperbolic sine are both odd functions because they only have odd powers in their Taylor series. Cosine and hyperbolic cosine are both even functions because they only have even powers in their Taylor series. Odd function over a symmetric domain If an odd function f is integrated over a domain that is symmetric about the origin (i.e., an interval of the form [−L, L], then Z L x=−L f (x) dx = 0. Formally, any subinterval’s on the left half of the interval will make a contribution to the Riemann sum which is equal and opposite to the contribution of the corresponding subinterval on the right half of the interval. These equal and opposite sums cancel, and so the deﬁnite integral over the entire interval is 0. In terms of the area interpretation, the net area under the curve over the left half of the interval will be equal and opposite in sign to the net area under the curve over the right half of the interval. Therefore, the total area will be 0: 76 Even function over a symmetric domain. If an even function g is integrated over a domain that is symmetric about the origin (i.e., an interval of the form [−L, L]), then Z L x=−L g(x) dx = 2 Z L x=0 g(x) dx. Formally, each subinterval on the left half of the interval has a corresponding subinterval on the right with an equal contribution to the Riemann sum. So one can just take the Riemann sum on the right and double it. Using the area interpretation, one can see that the region under the curve on the left will be the mirror image of the region under the curve on the right, so the total area is just twice the area on the right: 25.6 EXERCISES • One particular choice of partition and sampling that can be used to numerically evaluate deﬁnite integrals is the following. With n ﬁxed, divide the interval [a, b] into n subintervals Pi of common length (∆x)i = (b −a)/n. For the sampling, choose the right endpoint of each Pi; this gives you the formula: xi = a + i b −a n With these choices of partition and sampling, compute the Riemann sums for the integral 77 Z 2 x=1 dx x for n = 1, 2, 3 subdivisions. Note: in the next Lecture we will learn that Z 2 x=1 dx x = ln 2 ≃0.693 How does this compare to the values you obtained from the Riemann sums? • With the same choices of partition and sampling as in the previous problem, evaluate the Riemann sum for the integral Z 3 x=0 x2 dx for an arbitrary number n of subdivisions. You may need to use the following: n X i=1 i = n(n + 1) 2 , n X i=1 i2 = n(n + 1)(2n + 1) 6 , n X i=1 i3 = n2(n + 1)2 4 • The line y = x, the x-axis and the vertical line x = 2 bound a triangle of area 2. Thus, I = Z 2 x=0 x dx = 2 Evaluating the Riemann sum for n subdivisions for the above integral with the same choices of partition and sampling as in the previous problem yields an approximation RS(n) for its value I. The error E(n) we commit by using this approximation is deﬁned to be the diﬀerence E(n) = RS(n) −I Show that E(n) is in O(n−k) for some k > 0. What’s the best value of k? • What is the following integral? Think! Z π/4 x=−π/4 x2 + ln \| cos x\| sin x 2 dx • Using the deﬁnition of deﬁnite integrals, compute Z 1 0 x3 dx. Use a uniform partition and the fact that Pn i=1 i3 = n2(n+1)2 4 . 78 25.7 Answers to Selected Examples 1. Let the partition P divide the interval [0, 1] into N equally sized subintervals. Then the ith subinterval of P is given by [(i −1) 1 N , i 1 N ], and (∆x)i = 1 N . Choose the right endpoint of each subinterval to be its sample point, i.e. xi = i N . Finally, note that as N →∞, ∆x →0. It follows that Z 1 0 x dx = lim ∆x→0 N X i=1 f (xi)(∆x)i = lim N→∞ N X i=1 i N · 1 N = lim N→∞ 1 N2 N X i=1 i = lim N→∞ 1 N2 N(N + 1) 2 = lim N→∞ N2 + N 2N2 = 1 2. We used the fact from earlier that n X i=1 i = 1 2n(n + 1) (Return) 2. The deﬁnite integral is deﬁned as the limit of Riemann sums. Note that for any partition P of the interval, N X i=1 (f + g)(xi)(∆x)i = N X i=1 [f (xi) + g(xi)] (∆x)i = N X i=1 f (xi)(∆x)i + g(xi)(∆x)i = N X i=1 f (xi)(∆x)i + N X i=1 g(xi)(∆x)i, because of linearity of ﬁnite sums. Therefore, as one takes the limit as ∆x →0, one ﬁnds (by the linearity of limits) that Z b a (f (x) + g(x)) dx = Z b a f (x) dx + Z b a g(x) dx. The argument for a constant multiple is almost identical: we can pull a constant out from a sum, and pull a constant out from a limit. (Return) 3. Consider what happens if one computes Z b a f (x) dx + Z a b f (x) dx. 79 By the additivity property (where c has been replaced by a), this is Z a a f (x) dx. But this equals 0, which is intuitive in the area interpretation. (More formally, any partition of an interval with 0 width has subintervals of 0 width, so the Riemann sums equal 0). Therefore, Z b a f (x) dx + Z a b f (x) dx = 0, and rearranging gives Z b a f (x) dx = − Z a b f (x) dx, as desired. (Return) 4. For the ﬁrst part, note that regardless of the partition of [a, b], the Riemann sum N X i=1 f (xi)(∆x)i ≥0, because f (xi) ≥0 by the above assumption. Since each Riemann sum is non-negative, the limit is non-negative. For the second part, note that f (x) ≥g(x) = ⇒f (x) −g(x) ≥0. So applying the ﬁrst part, we have Z b a (f (x) −g(x)) dx ≥0. Then by linearity of the deﬁnite integral (above), Z b a f (x) dx − Z b a g(x) dx ≥0, and rearranging gives Z b a f (x) dx ≥ Z b a g(x) dx. (Return) 5. If we use the partition of [a, b] into n equal intervals, then (∆x)i = b −a n 80 Also, note that f (xi) = c for all i. So Z b a c dx = lim ∆x→0 n X i=1 c b −a n = lim ∆x→0 c · n · b −a n = c · (b −a). We could also see this by interpreting this deﬁnite integral as the area under the curve y = c between x = a and x = b, which is simply a rectangle of base b −a and height c. (Return) 6. Again using a partition into n equal sized subintervals, we have that (∆x)i = b−a n . If we take our sample point to be the right endpoint of each subinterval, then we have xi = a + b−a n i. So Z b x=a x dx = lim n→∞ n X i=1 a + b −a n i b −a n = lim n→∞ n X i=1 ab −a n + n X i=1 i b −a n 2 = lim n→∞n · ab −a n + b −a n 2 n(n + 1) 2 = a(b −a) + (b −a)2 2 = 2ab −2a2 + b2 −2ab + a2 2 = 1 2(b2 −a2). This can also be found by interpreting the deﬁnite integral as the area under the curve y = x, which can be broken into a rectangle with base b −a and height a and a triangle with base and height b −a: 81 (Return) 7. If f only has odd powers in its Taylor series, then f (x) = a1x + a3x3 + a5x5 + · · · for some constants a1, a3, · · · . So evaluating f (−x) and doing a little algebra, we ﬁnd f (−x) = a1(−x) + a3(−x)3 + a5(−x)5 + · · · = −a1x −a3x3 −a5x5 −· · · = − a1x + a3x3 + a5x5 + · · · = −f (x), as desired. Similarly, if g(x) has even powers, then g(x) = a0 + a2x2 + a4x4 + · · · and it follows that g(−x) = a0 + a2(−x)2 + a4(−x)4 + · · · = a0 + a2x2 + a4x4 + · · · = g(x), as desired. (Return) 82 26 Fundamental Theorem Of Integral Calculus Computing deﬁnite integrals from the deﬁnition is diﬃcult, even for fairly simple functions. Fortunately, there is a powerful tool—the Fundamental Theorem of Integral Calculus—which connects the deﬁnite integral with the indeﬁnite integral and makes most deﬁnite integrals easy to compute. The Fundamental Theorem of Integral Calculus (FTIC) Given a continuous function f , it follows that 1. d dx R x a f (t)dt = f (x) and 2. R b a f (x)dx = R f (x)dx b a (where F(x) b a = F(b) −F(a)). In other words, Part 1 says that the function F(x) = Z x t=a f (t)dt is an anti-derivative of f . Part 2 says that the deﬁnite integral can be computed by ﬁnding the indeﬁnite integral of f and subtracting the evaluation at the bottom bound from the evaluation at the top bound. Note that even though the indeﬁnite integral is actually a class of functions that diﬀer by a constant, F(b)−F(a) has the same value for any function F in such a class, so when computing the antiderivative for the purpose of computing a deﬁnite integral, it is allowable (and convenient) to forego the constant of integration. Part 2 can be expressed in a slightly diﬀerent way which is illustrative. For a diﬀerentiable function F we have F b x=a = Z b x=a dF. This says that the net change in quantity (given on the left side) equals the integral of the rate of change (given on the right side). This interpretation will be used in many applications in the next chapter. (See Rough Proof 1) 83 Example Compute Z T x=1 1 x dx. (See Answer 2) Example Compute Z 3 1 x2dx. (See Answer 3) Example Suppose a publisher prints 12000 books per month with expected revenue of $60 per book. The marginal cost of each book is given by MC(x) = 10 + 1 2000x. What would be the change in proﬁt from a 25% increase in production? (See Answer 4) Example Find d dx Z x 0 sin(t)dt . (See Answer 5) Example Find d dx Z x3 x sin(t)dt ! . (See Answer 6) Caveat Note that if the integrand f (t) fails to be deﬁned or continuous at a point in the interval [a, b], then the FTIC does not hold. The following example shows this using the singularities of a rational function. Example Compute Z 4 1 dx x2 −5x + 6. (See Answer 7) 84 26.1 Limits of integration and substitution One must be careful when using Part 2 of the Fundamental Theorem of Integral Calculus along with the method of substitution. The reason this can cause problems is that when a substitution is made, the old limits of integration are still in terms of the original variable. Therefore, one must either get the antiderivative in terms of the original variable before evaluating (this is what we usually did at the end of the substitution anyway), or one can change the limits of integration to reﬂect the new variable. Consider the following example which demonstrates both techniques. Example Compute Z 1 x=0 x(x −1)n dx, where n is some ﬁxed positive constant. (See Answer 8) Another case where one must be careful of the limits of integration is with Integration By Parts. One can compute the indeﬁnite integral completely and then apply the limits of integration, or one can apply them as one goes, as in the following: Z b a u dv = uv b a − Z b a v du. Example Again compute the deﬁnite integral Z 1 x=0 x(x −1)n dx, but this time using integration by parts. (See Answer 9) 26.2 Additional examples Example Compute Z e5 x=e3 ln x x dx. (See Answer 10) Example Compute Z 1 x=−1 1 1 + 3x2 dx. (See Answer 11) 85 26.3 EXERCISES • Evaluate the following integrals: Z 1 x=−1 dx 1 + x2 Z 3 x=0 5x √ x + 1 dx Z π x=−π d dx (x cos x) dx • Compute the following derivatives: d dx Z π x=−π x cos x dx d dx Z x t=0 cos t dt d dx Z x4 t=x2 e−t2 dt d dx Z x t=1 1 √t dt d dx Z arcsin x t=0 ln \|sin t + cos t\| dt d dx Z tan x t=sin x e−t2 dt • What is the leading-order term in the Taylor series about x = 0 of f (x) = Z x 0 ln(cosh(t))dt • We usually use Riemann sums to approximate integrals, but we can go the other way, too, using an antiderivative to approximate a sum. Using only your head (no paper, no calculator), compute an approx-imation for 100 X n=0 n3 Hint: what integral does this resemble? • Compute d dx Z x4 3x e−t2/2dt • Find the critical points of the function f (x) = R x2 e ln(1 + t2) cos(√t)dt 86 26.4 Answers to Selected Examples 1. Part 1: By the deﬁnition of the derivative, and the deﬁnition of the deﬁnite integral, d dx Z x a f (t)dt = lim h→0 R x+h a f (t)dt − R x a f (t)dt h = lim h→0 R x+h x f (t)dt h = lim h→0 f (x∗)∆x h = lim h→0 f (x∗)h h = lim h→0 f (x∗) = f (x), since x ≤x∗≤x + h. Part 2: By Part 1, F(x) = R x a f (t)dt is an anti-derivative of f . Furthermore, we have F(b) −F(a) = Z b a f (t)dt − Z a a f (t)dt = Z b a f (t)dt −0 = Z b a f (t)dt. Let G(x) be some anti-derivative of f . Since anti-derivatives of the same function diﬀer only by a constant, G(x) = F(x) −C for some constant C. Then we have G(b) −G(a) = (F(b) −C) −(F(a) −C) = F(b) −F(a) as desired. (Return) 2. Using Part 2 of the Fundamental Theorem, we ﬁnd that Z T x=1 1 x dx = ln x T x=1 = ln T −ln 1 = ln T. Note that the above deﬁnite integral is sometimes used as the deﬁnition of the natural logarithm. (Return) 3. By Part 2 of FTIC, Z 3 1 x2dx = 1 3x3 3 1 = 1 3(33 −13) = 26 3 . (Return) 87 4. The additional 25% means an extra 3000 books. So the goal is to ﬁnd the change in proﬁt P as x goes from 12000 to 15000. That is, P 15000 x=12000 = Z 15000 x=12000 dP, according to Part 2 of the Fundamental Theorem. Now, since proﬁt is revenue minus cost, it follows that marginal proﬁt is given by dP = dR −dC. Here, dR = MR(x) dx = 60 dx since the marginal revenue from each book is $60. And dC = MC(x) dx = 10 + 1 2000x dx. Putting it all together, we ﬁnd that P 15000 x=12000 = Z 15000 x=12000 dP = Z 15000 x=12000 60 − 10 + 1 2000x dx = Z 15000 x=12000 50 − 1 2000x dx = 50x − 1 4000x2 15000 x=12000 = 50 · 15000 −150002 4000 − 50 · 12000 −120002 4000 = $129750 (Return) 5. By Part 1 of FTIC, d dx R x 0 sin(t)dt = sin(x). (Return) 6. One must be careful with a function of x in one or both bounds. A good way to break the problem down is to write F(x) = R x 0 sin(t)dt. By Part 1 of FTIC, F ′(x) = sin(x). Now, note that Z x3 x sin(t)dt = Z x3 0 sin(t)dt − Z x 0 sin(t)dt = F(x3) −F(x). Next, taking the derivative (and remembering the chain rule) gives d dx Z x3 x sin(t)dt ! = d dx F(x3) −F(x) = F ′(x3)(3x2) −F ′(x)(1) = sin(x3)(3x2) −sin(x). (Return) 88 7. This is a rational function, and the denominator factors as (x −3)(x −2), so use partial fractions to express 1 (x −3) (x −2) = A x −3 + B x −2. Clearing denominators gives 1 = A(x −2) + B(x −3). Setting x = 3 gives A = 1. Setting x = 2 gives B = −1. Thus, Z dx x2 −5x + 6 = Z 1 x −3 − 1 x −2 dx = ln \|x −3\| −ln \|x −2\|. Then trying to apply FTIC would give Z 4 1 dx x2 −5x + 6 = ln \|x −3\| −ln \|x −2\| 4 1 = ln(1) −ln(2) −(ln(2) −ln(1)) = −2 ln(2). However, because 1 (x−3)(x−2) has singularities at x = 2 and x = 3, one must evaluate the improper integral as follows: Z 4 1 dx x2 −5x + 6 = Z 2 1 dx x2 −5x + 6 + Z 3 2 dx x2 −5x + 6 + Z 4 3 dx x2 −5x + 6, and none of these integrals exist, as will be shown in the next section on improper integrals, so the entire integral does not exist. (Return) 8. First, we will try the method of substituting back in before evaluating. We make the substitution u = x −1 du = dx. Then the integral becomes Z 1 x=0 x(x −1)n dx = Z 1 x=0 (u + 1)un du = Z 1 x=0 un+1 + un du = un+2 n + 2 + un+1 n + 1 1 x=0 . This is where a lot of students might make the mistake of plugging in the limits of x when trying to evaluate the integral. This is a reason to include the x = a at the bottom limit of integration: it helps remind us that those limits are in terms of x, even if we have made one or more substitution. 89 To avoid this pitfall, we now ﬁnish getting the antiderivative in terms of x so that we can evaluate: un+2 n + 2 + un+1 n + 1 1 x=0 = (x −1)n+2 n + 2 + (x −1)n+1 n + 1 1 x=0 = 0 + 0 −(−1)n+2 n + 2 −(−1)n+1 n + 1 = (−1)n+2 1 n + 1 − 1 n + 2 = (−1)n (n + 1) (n + 2). On the other hand, as soon as we made the substitution u = x −1, we could have changed the limits of integration to reﬂect our new variable. Namely, at the lower limit x = 0, what is the corresponding value of u? Well, u = x −1, so when x = 0, we have u = −1. Similarly, when x = 1, the corresponding value of u is u = 0. So as we were making our substitution, we could make a corresponding change in the limits of integration so that the new deﬁnite integral is entirely in terms of u: Z 1 x=0 x(x −1)n dx = Z 0 u=−1 (u + 1)un du. Now the calculation proceeds as before, and gives the same answer. Changing the limits can sometimes be easier, especially in a complicated integral which may involve (for example) a u-substitution and a trigonometric substitution. Otherwise, the algebra can get messy as one substitutes back in and then substitutes back in again. (Return) 9. The logical choice of parts is u = x du = dx dv = (x −1)n dx v = (x −1)n+1 n + 1 . Then by the formula for parts, we have Z 1 x=0 x(x −1)n dx = x (x −1)n+1 n + 1 1 x=0 − Z 1 x=0 (x −1)n+1 n + 1 dx = 0 − (x −1)n+2 (n + 1) (n + 2) 1 x=0 = (−1)n+2 (n + 1) (n + 2) = (−1)n (n + 1) (n + 2). Note that from the ﬁrst to second line above, we have x (x−1)n+1 n+1 is 0 at both x = 1 and at x = 0, so that entire term disappears. (Return) 90 10. This integral is best computed with a substitution of u = ln x du = 1 x dx (Integration by parts works too, but it involves a little bit of algebra). Here it is convenient to change the limits of integration as we go, so note that when x = e3, we have u = 3. When x = e5 we have u = 5. Thus, Z e5 x=e3 ln x x dx = Z 5 u=3 u du = 1 2u2 5 u=3 = 1 2 (25 −9) = 1 2 · 16 = 8. (Return) 11. This looks like a good candidate for a trigonometric substitution. Because of the extra factor of 3, the logical substitution is x = 1 √ 3 tan θ dx = 1 √ 3 sec2 θ dθ. Remember the constant 1 √ 3 is there so that when it is squared it will cancel with the factor of 3 and allow us to use the identity 1 + tan2 = sec2. Again, we will change the bounds as we go. Note that x = −1 ⇒ 1 √ 3 tan θ = −1 ⇒ tan θ = − √ 3. The value of θ for which this holds is θ = −π 3 . Similarly, when x = 1 we have θ = π 3 . Proceeding, we 91 have Z 1 x=−1 1 1 + 3x2 = 1 √ 3 Z π/3 θ=−π/3 sec2 θ dθ 1 + 3( 1 √ 3 tan θ)2 = 1 √ 3 Z π/3 θ=−π/3 sec2 θ dθ 1 + tan2 θ = 1 √ 3 Z π/3 θ=−π/3 sec2 θ sec2 θ dθ = 1 √ 3 Z π/3 θ=−π/3 dθ = 1 √ 3θ π/3 −π/3 = 1 √ 3 π 3 −−π 3 = 2π 3 √ 3 (Return) 92 27 Improper Integrals An improper integral is a deﬁnite integral which cannot be evaluated using the Fundamental Theorem of Integral Calculus (FTIC). This situation arises because the integral either 1. has a point in its interval of integration which is not in the domain of the integrand (the function being integrated) or 2. has ∞or −∞as a bound of integration. As an example of the ﬁrst type, consider Z 2 0 dx x . This integral is improper because the left endpoint, 0, is not in the domain of 1 x . Another example of the ﬁrst type is Z 4 0 dx 3 √x −2. This is improper because the point 2 is in the interval of integration but is not in the domain of 1 3 √x−2. For an example of the second type, consider Z ∞ 0 dx 1 + x2 . This is improper because the upper bound is ∞. For an example of the danger of trying to apply the Fundamental Theorem of Integral Calculus when it does not apply, consider Z 1 x=−1 1 x2 dx. This is an improper integral (hence FTIC does not apply) because the point x = 0 is not in the domain of 1 x2 . If we were to try to apply FTIC anyway, we would ﬁnd Z 1 x=−1 1 x2 dx = −1 x 1 x=−1 = −1 −1 = −2. This is problematic since 1 x2 > 0 for all x (hence should have a positive integral by the dominance property). 93 27.1 Dealing with improper integrals To deal with an improper integral of the ﬁrst type, ﬁrst consider the integral R b a f (x) dx, where a is not in the domain of f (x), but f is continuous on the rest of the interval (a, b]. In this situation, one replaces the lower bound with a variable T which is slightly larger than a, and then takes the limit as T approaches a from the right (this is denoted by T →a+). Z b a f (x) dx = lim T →a+ Z b T f (x) dx. This replacement allows the integral R b T f (x) dx to be computed using FTIC, since f is continuous on that interval. After that integral is computed (in terms of T), the limit is computed. If the limit exists, then the original integral exists and equals the result. If the limit does not exist or is inﬁnite, then the original integral does not exist either. Example Determine whether the integral Z 2 0 dx x exists, and if it exists, what its value is. (See Answer 1) If the right endpoint, b, of the integral R b a f (x) dx is not in the domain of f , then b gets replaced with a variable T slightly smaller than b, and again the integral is replaced with a limit, this time as T →b−(that is, the limit as T approaches b from the left). Z b a f (x) dx = lim T →b− Z T a f (x) dx. Again, the integral equals this limit, if it exists. If the limit does not exist, then the integral does not exist. Example Determine whether the integral Z 3 1 dx (x −3)2 exists. If it exists, ﬁnd its value. (See Answer 2) For integrals where a point inside the interval of integration is not in the domain of the integrand, the integral is ﬁrst split at the bad point, and then each integral is evaluated separately using the above techniques. So, consider R b a f (x) dx where the point c is not in the domain of f and a < c < b. Then Z b a f (x) dx = Z c a f (x) dx + Z b c f (x) dx = lim T →c− Z T a f (x) dx + lim U→c+ Z b U f (x) dx. Both of the resulting limits must exist for the original integral to exist. 94 Example Determine if the integral Z 1 −1 dx x4/5 exists. (See Answer 3) 27.2 Bounds at inﬁnity For integrals with one bound at inﬁnity, the integral is deﬁned as follows. Z ∞ a f (x) dx = lim T →∞ Z T a f (x) dx. Similarly, Z a −∞ f (x) dx = lim T →−∞ Z a T f (x) dx. In the case of bounds of ∞and −∞, one can ﬁrst split the integral at any real number c, and then compute each integral as above: Z ∞ −∞ f (x) dx = Z c −∞ f (x) dx + Z ∞ c f (x) dx = lim T →−∞ Z c T f (x) dx + lim U→∞ Z U c f (x) dx. As before, both of these limits must exist for the original integral to exist. It is not equivalent to computing a single limit such as lim T →∞ Z T −T f (x) dx. Example Determine if the integral Z ∞ 0 dx 1 + x2 exists. If it exists, ﬁnd its value. (See Answer 4) 27.3 The p-integral One class of improper integrals is common enough to get its own name: the p-integral. This is the name given to integrals of the form Z 1 xp dx = Z x−p dx. There are two versions of this integral that are of interest to us. First, with a limit at inﬁnity, and second with a limit at 0. 95 Example Show that Z ∞ x=1 x−p dx = ( 1 p−1 if p > 1 ∞ if p ≤1. (See Answer 5) Example Consider the p-integral with a limit at 0: Z 1 x=0 x−p dx. This integral is improper because 0 is not in the domain of x−p. Show that Z 1 x=0 x−p dx = ( ∞ if p ≥1 1 1−p if p < 1. (See Answer 6) 27.4 Converge or diverge In some contexts, it is enough to know whether an integral converges (has a ﬁnite answer) or diverges (goes to inﬁnity or does not exist). Using the knowledge of the above p-integrals, and some asymptotic tools from earlier in the course, can help quickly determine whether certain improper integrals converge or diverge. Example Determine if Z 1 x=0 dx √ x2 + x converges or diverges. (See Answer 7) Example Determine if Z ∞ x=1 dx √ x2 + x converges or diverges. (See Answer 8) Example Determine if Z ∞ x=−∞ 2x 1 + x2 dx converges or diverges. (See Answer 9) 96 27.5 EXERCISES • Use a Talyor expansion of the integrand at x = 0 to determine whether the following integrals converge or diverge: Z 1 x=0 e−x x dx Z 1 x=0 cos2 x √x dx • Use the asymptotics of the integrand as x →∞to determine whether the following integrals converge or diverge: Z +∞ x=1 3 √x + 3 x3 dx Z +∞ x=1 1 −5−x x dx • Compute the following integrals, if they converge, by evaluating a limit. Z +∞ x=0 e−x sin x dx Z 2 x=1 dx √x −1 Z 4 x=0 2 dx √ 16 −x2 • The following integral is improper both at x = 1 and at x →∞: Z +∞ x=1 1 √ x3 −1 dx First, as x →∞, show that the integrand is x−3/2 + O(x−9/2), so that it converges at this limit. Next, as x →1+, show that the integrand is (3(x −1))−1/2 + O((x −1)1/2), so that it converges at this limit also. • Consider the following two integrals: I1 = Z +∞ x=2 dx √ x3 −8, I2 = Z +∞ x=2 1 p (x −2)3 dx One converges and one does not. Which is which and why? 97 • Until now we have used asymptotic analysis to relate an improper integral to a p-integral. But sometimes the leading order term is not a power. Identify the leading order term as x →+∞of the integrand of Z +∞ x=1 1 sinh x dx and determine whether the integral converges or diverges. • For p ≥0 an integer, consider the following integral: Ip = Z +∞ x=1 dx lnp x Show that this diverges for any value of p. Hint 1: think about the growth of lnp x as x →+∞as compared to polynomial growth. Hint 2: recall from Lecture 25 that if g(x) ≥f (x) for every x ∈[a, b], then Z b x=a g(x) dx ≥ Z b x=a f (x) dx This is also true if the domain of integration is unbounded and the integrals are deﬁned... • Determine whether the following integral converges or diverges. Z ∞ 2 1 (x5 −4x3)1/4 dx 27.6 Answers to Selected Examples 1. Since the left endpoint is not in the domain of 1 x , the integral becomes Z 2 0 dx x = lim T →0+ Z 2 T dx x = lim T →0+ ln x 2 T = lim T →0+ (ln(2) −ln(T)) . This limit does not exist because limT →0+ ln(T) diverges. Hence, the original integral does not exist. (Return) 2. The integral becomes Z 3 1 dx (x −3)2 = lim T →3− Z T 1 dx (x −3)2 = lim T →3−− 1 x −3 T 1 = lim T →3−− 1 T −3 −(−1 −2). This limit does not exist since limT →3− 1 T −3 diverges. Hence, the integral does not exist. (Return) 98 3. The only point not in the domain of the function 1 x4/5 is 0. Thus, the integral becomes Z 1 −1 dx x4/5 = Z 0 −1 dx x4/5 + Z 1 0 dx x4/5 = lim T →0− Z T −1 dx x4/5 + lim U→0+ Z 1 u dx x4/5 = lim T →0−5x1/5 T −1 + lim U→0+ 5x1/5 1 u = lim T →0− 5T 1/5 −5(−1)1/5 + lim U→0+ 5 −5u1/5 = 0 + 5 + 5 −0 = 10. So the integral exists and equals 10. (Return) 4. Because the top bound is ∞, the integral becomes Z ∞ 0 dx 1 + x2 = lim T →∞ Z T 0 dx 1 + x2 = lim T →∞arctan(x) T 0 = lim T →∞arctan(T) −arctan(0) = lim T →∞arctan(T) = π 2 . So the integral exists and equals π 2 . (Return) 5. First, if p ̸= 1, then we can use the power rule on x−p. Here we ﬁnd Z ∞ x=1 x−p dx = lim T →∞ x−p+1 −p + 1 T x=1 = lim T →∞ T −p+1 1 −p − 1 1 −p . Note that T −p+1 diverges to inﬁnity if p < 1, since in this case we have a positive power of T, which goes to inﬁnity as T goes to inﬁnity. On the other hand, T −p+1 goes to 0 if p > 1, since in that case it is a negative power of T. Putting this together with the above equations, we have Z ∞ x=1 x−p dx = ( 1 p−1 if p > 1 ∞ if p < 1. Finally, consider the case p = 1. In this case, Z ∞ x=1 1 x dx = lim T →∞ln x T 1 = lim T →∞ln T = ∞. 99 And so the cases are as claimed above. (Return) 6. As in the previous example, we ﬁrst consider when p ̸= 1 and use the power rule: Z 1 x=0 x−p dx = lim T →0+ x1−p 1 −p 1 x=T = lim T →0+ 1 1 −p −T 1−p 1 −p . Now, note that the limit is as T →0+. So we have T 1−p diverges if p > 1 (since a negative power of T diverges as T →0+), and converges to 0 if p < 1. Putting this together with the previous computation gives Z 1 x=0 x−p dx = n 1 1−p if p1. Finally, if p = 1, then Z 1 x=0 1 x dx = lim T →0+ ln x 1 x=T = lim T →0+ 0 −ln T which diverges to inﬁnity. Thus, the original integral diverges for all p ≥1, as claimed. (Return) 7. This is not a function for which we can easily ﬁnd an antiderivative. However, since we are interested in the behavior of the function near 0 (that is where the blow-up occurs), we can do a little bit of algebra to see that 1 √ x2 + x = 1 √x · 1 √x + 1 = 1 √x · (1 + x)−1/2 = 1 √x (1 + O(x)) , by the binomial expansion. Therefore, the leading order term of this function as x →0 is 1 √x . This is a convergent p-integral, because p = 1 2 and from the above example Z 1 x=0 dx xp converges when p < 1. (Return) 8. We must do a slightly diﬀerent analysis for this integral, since we are interested in the behavior as x →∞. Because we want to take advantage of the binomial series again (which requires its argument to be less 100 than 1), we do the following algebra: 1 √ x2 + x = 1 √ x2 · 1 √ 1 + x−1 = 1 x 1 + 1 x −1/2 = 1 x 1 + O 1 x . So for this integral, the leading order term is 1 x , which is the p-integral with p = 1. This diverges, as the above example shows, and so the original integral diverges. (Return) 9. As mentioned above, it is not valid to do this with a single limit such as lim T →∞ Z T x=−T 2x 1 + x2 dx. The integral must be split and treated as two separate integrals with limits at inﬁnity: Z ∞ x=−∞ 2x 1 + x2 dx = lim S→−∞ Z 0 x=S 2x 1 + x2 dx + lim T →∞ Z T x=0 2x 1 + x2 dx. Again, a little bit of asymptotic analysis with the help of the geometric series helps determine the behavior of this function: 2x 1 + x2 = 2x 1 + x2 · x−2 x−2 = 2 x · 1 1 + x−2 = 2 x 1 + O(x−2) . Here, the geometric series was used to write 1 1 + x−2 = 1 −x−2 + x−4 −· · · = 1 + O(x−2), which is justiﬁed since x →∞and so x−2 is very small. So the original integrand behaves like 2 x , which diverges when integrated to inﬁnity. Therefore lim T →∞ Z T x=0 2x 1 + x2 dx diverges, and (but for a sign change) the same reasoning shows lim S→−∞ Z 0 x=S 2x 1 + x2 dx diverges too, so the original integral diverges. (Return) 101 28 Trigonometric Integrals A trigonometric integral is an integral involving products and powers of trigonometric functions: cosine, sine, tangent, secant, cosecant, and cotangent. Many of these integrals can be handled with u-substitution, but there are other methods which are outlined in this module. The three families of integrals discussed in this module are Z sinm θ cosn θ dθ Z tanm θ secn θ dθ Z sin(mθ) cos(nθ) dθ. 28.1 Product of sines and cosines Consider the integral Z sinm θ cosn θ dθ. There are several cases to consider based on whether m and n are odd and even. m is odd If m is odd, then one factor of sin θ can be set aside. This leaves behind an even power of sin θ, which can be expressed in terms of cos θ using the Pythagorean identity. Then the substitution u = cos θ can be made. Example Find Z sin3(x) cos(x) dx. (See Answer 1) n is odd If n is odd, the procedure is very similar. This time, we set aside a factor of cos θ. This leaves an even power of cos θ which can be expressed in terms of sin θ using the Pythagorean identities. 102 Example Find Z sin2(x) cos3(x) dx. (See Answer 2) Both m and n are even If neither m nor n is odd, then both are even. This is a bit more diﬃcult and requires using the power reduction formulas: Power reduction sin2(θ) = 1−cos(2θ) 2 cos2(θ) = 1+cos(2θ) 2 Example Find Z sin2 x dx. (See Answer 3) Example Find Z sin2 θ cos2 θ dθ. (See Answer 4) Example Find Z cos4(x) dx. (See Answer 5) 28.2 Product of tangents and secants Next consider the integral Z tanm θ secn θ dθ. As with the product of sines and cosines, the method will depend on whether m and n are odd or even. 103 m is odd If m is odd, we will set aside a factor of tan θ sec θ. Note that this is the derivative of sec θ and so this sets up a substitution of u = sec θ. After setting aside these factors, we are left with an even power of tan θ, which can be expressed in terms of sec θ using the Pythagorean identity tan2 θ = sec2 θ −1. Now, the integral can be computed using the substitution u = sec θ. Example Compute Z tan3 θ sec θ dθ. (See Answer 6) n is even If n is even, then we can set aside a factor of sec2 θ. Note that this is the derivative of tan θ and therefore sets up the substitution u = tan θ. Setting aside sec2 θ leaves an even power of sec θ, which can be expressed in terms of tan θ using the Pythagorean identity sec2 θ = 1 + tan2 θ. Then the substitution u = tan θ allows the computation of the integral. Example Compute Z tan2 θ sec6 θ dθ. (See Answer 7) m is even, n is odd If neither of the above cases holds, then the integral is a bit more diﬃcult. It typically requires a bit of algebra and several applications of a reduction formula (or integration by parts). A general method is to rewrite the even power of tangent entirely in terms of secant by using the Pythagorean identity tan2 θ = sec2 θ −1. This gives an integral which is sums of powers of sec θ. Each of these can be solved using the reduction formula for secant: Z secn θ dθ = 1 n −1 secn−2 θ tan θ + n −2 n −1 Z secn−2 θ dθ, along with the fact that Z sec θ dθ = ln \| sec θ + tan θ\| + C. 104 Example Compute Z tan2 θ sec θ dθ. (See Answer 8) 28.3 Product of sine and cosine with constants Finally, consider the integral Z sin(mθ) cos(nθ) dθ. This integral requires some algebra to simplify the integrand. One can verify using the sum and diﬀerence formulas for sine that sin(mθ) cos(nθ) = 1 2 (sin((m + n)θ) + sin((m −n)θ)) . This expression can be integrated term by term to ﬁnd Z sin(mθ) cos(nθ) dθ = Z 1 2 (sin((m + n)θ) + sin((m −n)θ)) dθ = 1 2 −cos((m + n)θ) m + n −cos((m −n)θ) m −n + C. There are similar formulas for related integrals: Z sin(mθ) sin(nθ) dθ = −sin((m + n)θ) 2 (m + n) + sin((m −n)θ) 2 (m −n) + C Z cos(mθ) cos(nθ) dθ = sin((m + n)θ) 2 (m + n) + sin((m −n)θ) 2 (m −n) + C. These formulas need not be memorized, but be aware they exist and look them up when necessary. Example Compute Z sin(3θ) cos(4θ) dθ. (See Answer 9) 28.4 Additional examples Example Compute Z π/2 θ=−π/2 cosn θ dθ. 105 Use the fact (which is proven using integration by parts) that Z cosn θ dθ = cosn−1 θ sin θ n + n −1 n Z cosn−2 θ dθ. (See Answer 10) Example Compute Z tan3 θ dθ (See Answer 11) 28.5 EXERCISES Compute the following indeﬁnite integrals. You may need to use reduction formulae or coordinate changes. • Z sin2 x cos2 x dx • Z sin3 x 2 cos3 x 2 dx • Z x3 dx √ 9 −x2 • Z 5 tan5 x sec3 x dx • Z 7 tan4 x sec4 x dx • Z 9 sin3 3x dx • Z cos4 x dx • Z sin x sec x tan x dx • Z tan5 2x sec4 2x dx • Z cos x √ 1 −sin x dx • Z x2 dx √ 1 + x2 • Z tan4(2x) sec4(2x) dx 106 28.6 Answers to Selected Examples 1. Following the above outline, set aside one factor of sin x, which gives Z sin3 x cos x dx = Z (sin2 x)(cos x)(sin x) dx. Now, there is an even power of sine remaining, which can be rewritten using the Pythagorean identity sin2 x = 1 −cos2 x. This gives Z sin3 x cos x dx = Z (sin2 x)(cos x) sin(x) dx = Z (1 −cos2 x)(cos x) sin(x) dx. Now, the integral can be handled by letting u = cos(x) (and du = −sin(x)dx). Z (1 −cos2 x)(cos x)(sin x) dx = Z (1 −u2)u(−du) = Z (u3 −u) du = u4 4 −u2 2 + C = cos4 x 4 −cos2 x 2 + C. (Return) 2. Following the procedure outlined above, we set aside a factor of cosine and use the Pythagorean identity, which gives Z sin2(x) cos3(x) dx = Z sin2(x) cos2(x) cos(x) dx = Z sin2(x)(1 −sin2 x) cos(x) dx. Now, we are ready to make the substitution u = sin(x), du = cos(x) dx. This gives Z sin2(x)(1 −sin2 x) cos(x) dx = Z u2(1 −u2) du = Z u2 −u4 du = u3 3 −u5 5 + C = sin3 x 3 −sin5 x 5 + C. (Return) 107 3. Using the ﬁrst power reduction formula gives Z sin2 x dx = Z 1 2(1 −cos(2x)) dx = 1 2 x −sin(2x) 2 + C. (Return) 4. Using both the power reduction formulas and doing some algebra gives Z sin2 θ cos2 θ dθ = Z 1 2 (1 −cos 2θ) 1 2 (1 + cos 2θ) dθ = 1 4 Z (1 −cos 2θ)(1 + cos 2θ) dθ = 1 4 Z (1 −cos2 2θ) dθ = 1 4 Z sin2 2θ dθ = 1 4 Z 1 2 (1 −cos 4θ) dθ = 1 8 θ −sin 4θ 4 + C. (Return) 5. Using the second power reduction formula (and then again in a later step) gives Z cos4(x) dx = Z (cos2(x))2 dx = Z (1 2(1 + cos(2x))2 dx = 1 4 Z 1 + 2 cos(2x) + cos2(2x) dx = 1 4 Z 1 + 2 cos(2x) + 1 2(1 + cos(4x)) dx = 1 4 x + sin(2x) + 1 2x + 1 8 sin(4x) + C (Return) 6. Since the power of tangent is odd, we set aside a factor of tan θ sec θ, and use the Pythagorean identity to ﬁnd Z tan3 θ sec θ dθ = Z tan2 θ(tan θ sec θ) dθ = Z (sec2 θ −1)(tan θ sec θ)dθ. Now, we can make the substitution u = sec θ du = sec θ tan θ dθ, 108 which gives Z (sec2 θ −1)(tan θ sec θ)dθ = Z (u2 −1) du = 1 3u3 −u + C = 1 3 sec3 θ −sec θ + C. (Return) 7. Because the power of secant is even, we set aside sec2 θ and use the Pythagorean identity to ﬁnd Z tan2 θ sec6 θ dθ = Z tan2 θ sec4 θ(sec2 θ) dθ = Z tan2 θ(1 + tan2 θ)2(sec2 θ) dθ. Now, we are prepared for a substitution of u = tan θ du = sec2 θ dθ. Making this substitution and simplifying gives Z tan2 θ(1 + tan2 θ)2(sec2 θ) dθ = Z u2(1 + u2)2 du = Z u2(1 + 2u2 + u4) du = Z (u2 + 2u4 + u6) du = 1 3u3 + 2 5u5 + 1 7u7 + C = 1 3 tan3 θ + 2 5 tan5 θ + 1 7 tan7 θ + C. (Return) 8. Using the Pythagorean identity gives Z tan2 θ sec θ dθ = Z (sec2 θ −1) sec θ dθ = Z (sec3 θ −sec θ) dθ = Z sec3 θ dθ − Z sec θ dθ. Now, using the reduction formula on the ﬁrst of these integrals gives Z sec3 θ dθ = 1 2 sec θ tan θ + 1 2 Z sec θ dθ. 109 Combining this with the above expression and using the integral of secant, we ﬁnd Z tan2 θ sec θ dθ = Z sec3 θ dθ − Z sec θ dθ = 1 2 sec θ tan θ + 1 2 Z sec θ dθ − Z sec θ dθ = 1 2 sec θ tan θ −1 2 Z sec θ dθ = 1 2 sec θ tan θ −1 2 ln \| sec θ + tan θ\| + C. (Return) 9. Using the formula given, we have Z sin(3θ) cos(4θ) dθ = −cos(7θ) 14 −cos(−θ) −2 + C = −cos(7θ) 14 + cos θ 2 + C, where we have used the fact that cosine is even to simplify the ﬁnal expression. (Return) 10. Applying the limits of integration in the above reduction formula gives Z π/2 θ=−π/2 cosn θ dθ = cosn−1 θ sin θ n π/2 −π/2 + n −1 n Z π/2 θ=−π/2 cosn−2 θ dθ. Now, notice that cosn−1 θ sin θ n π/2 −π/2 = 0 because cosine is 0 at ±π/2. Therefore, Z π/2 θ=−π/2 cosn θ dθ = n −1 n Z π/2 θ=−π/2 cosn−2 θ dθ. This is itself a reduction formula. By computing the base cases n = 0 and n = 1, respectively, we ﬁnd Z π/2 θ=−π/2 dθ = θ π/2 −π/2 = π, and Z π/2 θ=−π/2 cos θ dθ = sin θ π/2 −π/2 = 2. 110 Now the value of the integral for any higher value of n can be found by repeatedly using the above formula until the integral reduces to one of the base cases above. Using induction, one ﬁnds Z π/2 θ=−π/2 cosn θ dθ = ( 1·3·5···(n−1) 2·4·6···n π if n is even 2·4·6···(n−1) 3·5·7···n · 2 if n is odd. (Return) 11. Here the power of tangent is odd, so the method calls for setting aside a factor of sec θ tan θ. However, there is no factor of secant in this integral! It turns out that this is not a problem; we can multiply the top and the bottom by secant to introduce a factor of secant, and the algebra works out: Z tan3 θ dθ = Z tan2 θ sec θ (sec θ tan θ) dθ = Z sec2 θ −1 sec θ (sec θ tan θ) dθ = Z sec θ − 1 sec θ (sec θ tan θ) dθ. Now, proceed as usual with the substitution u = sec θ du = sec θ tan θ dθ. Thus, Z sec θ − 1 sec θ (sec θ tan θ) dθ = Z u −1 u du = 1 2u2 −ln \|u\| + C = 1 2 sec2 θ −ln \| sec θ\| + C. (Return) 111 29 Tables And Computers This module discusses some of the shortcuts available by using tables of integrals and mathematical software. Many integrals can be easily computed by a computer algebra system, but it is still important to know the underlying concepts so as to be able to use these tools eﬃciently and accurately. 29.1 Tables of integrals Most calculus textbooks have an appendix containing one hundred or more integral formulas. All of these formulas can be derived using the techniques of the previous modules (possibly with some additional technical algebra), and it is a good exercise to try to derive some of them. Using the table is sometimes diﬃcult, because ﬁnding the correct form can be tricky. And even with the correct form, an integral may not match the form precisely. It may take some algebra and a u-substitution to match the form. Example Use the formula Z dx x2√ x2 −a2 = √ x2 −a2 a2x + C to evaluate the integral Z dx (4x2 + 4x + 1) √ 4x2 + 4x −3. (See Answer 1) Other table entries are more inductive in nature, like the reduction formulas mentioned in the integration by parts module. Example Use the formulas Z sec ax dx = 1 a ln \| sec ax + tan ax\| + C, and (for n ≥2) Z secn ax dx = secn−2 ax tan ax a(n −1) + n −2 n −1 Z secn−2 ax dx + C 112 to ﬁnd Z sec3(x) dx. (See Answer 2) 29.2 Mathematical software Expensive computer algebra systems such as Maple and Mathematica can quickly and accurately dispense with most integrals that can be done by hand. One free alternative, from the makers of Mathematica, is Wolfram Alpha, which for most purposes is as good as its more costly relatives, and in many cases it can explain the intermediate steps of longer computations (though it now only provides three such explanations per day for a user without a paid subscription). Note that the form of the answer given by computer systems may look diﬀerent from what one gets by hand or by a table, so care should be taken when comparing answers. Example Compute R sec3(x) dx using Wolfram Alpha, or other computer algebra system. Note the syntax of the entry (though it is pretty good at parsing other forms of entry). Also note that the answer given is in a diﬀerent form than that found in the earlier example. Answer Example There are limits to what a computer algebra system can do. Consider the integral Z π/2 x=0 sinn x sinn x + cosn x dx. It turns out that the value of this integral is π 4 for all n, although Wolfram Alpha is not able to compute it. But if we enter small particular values of n into Wolfram Alpha, then it does give the answer, although sometimes only in decimal form. 29.3 Answers to Selected Examples 1. Although it is not exactly in the correct form, completing the square should get it closer. Indeed, factoring and completing the square gives Z dx (4x2 + 4x + 1) √ 4x2 + 4x −3 = Z dx (2x + 1)2p (2x + 1)2 −4 . Now, making the u-substitution u = 2x + 1 (hence dx = 1 2 du) gives Z dx (2x + 1)2p (2x + 1)2 −4 = 1 2 Z du u2√ u2 −4 = 1 2 √ u2 −4 4u + C = 1 2 · q (2x + 1)2 −4 4 (2x + 1) + C. 113 (Return) 2. Reducing using the second formula, and then using the ﬁrst formula gives Z sec3x = sec x tan x 2 + 1 2 Z sec x dx = sec x tan x 2 + 1 2 ln \| sec x + tan x\| + C. (Return) 114 30 Simple Areas We know the basic standard formulae for the area of basic shapes, but why are they true? From the point of view of calculus, area A is the integral of dA, the area element. In this chapter, we will use the following procedure to determine a quantity U: 1. Determine the diﬀerential element dU. 2. Integrate to compute U = R dU. 30.1 Length of an interval Before getting to areas, ﬁrst consider how this method works for computing the length L of the interval from a to b. If the length is denoted L, then the length element will be denoted dL, and L = R dL. In this context, the appropriate length element would be dx if we’re working along the x-axis. So, we want to integrate dx as x goes from a to b. The length, L = Z dL = Z b x=a dx = x b x=a = b −a 30.2 Parallelogram The formula for the area of a parallelogram is base × height (bh). Consider the following rearrangement into diﬀerential elements, where we carve the parallelogram into parallel horizontal strips of width b and height dy, where y is the y-axis. 115 In this case, the area element, dA = b dy, is the area of this inﬁnitesimal rectangle. The limits on y should go from 0 to the height, h of the parallelogram. The area, A = Z dA = Z h y=0 b dy = by h y=0 = bh We have our familiar answer bh. This means that we’ve done a rearrangement in terms of inﬁnitesimal strips. Shearing that parallelogram preserves the area element and hence, the area. That is why a parallelogram has the same area as the corresponding rectangle. 30.3 Triangle The formula for the area of a triangle is 1 2× base × height ( 1 2bh). Let’s think in terms of a diﬀerential area element. Given the fact that we can shear and preserve the area element, and thus the area, let’s present our triangle as having a hypotenuse modeled by the line y = h bx. 116 To compute the area element, let’s use a vertical strip. dA = h bx dx where the height of that vertical strip is h bx and the width is the length element dx. The area, A = Z dA = Z b x=0 h bx dx = h b x2 2 b x=0 = hb2 2b = 1 2bh 30.4 Disc We will use three ways to ﬁnd the area of a circular disc of radius r: 1. Using an angular area element. 2. Using a radial variable. 3. Using a lateral, or a vertical rectangular strip. 1. Angular In this case, we’ll use an angular area element. We will take a wedge with angle dθ. If we look at that close up, it’s modeled fairly well as a triangle. It’s not a perfect triangle, there’s a bit of curvature at the end. This is a triangle with two sides of length r whose included angle is dθ. Such a triangle has area 1 2r 2 sin(dθ) ≈1 2r 2dθ, since dθ is very small. If we model that as a triangle with height r, and width r dθ, we can ignore the higher order terms in the Taylor expansion of that area. We obtain an area element dA = 1 2r(r dθ). Integrating to get the area, θ has to spin all the way around the circle from 0 to 2π. 117 The area, A = Z dA = Z 2π θ=0 1 2r 2 dθ = 1 2r 2 Z 2π θ=0 dθ = 1 2r 2θ 2π θ=0 = 1 2r 2(2π) = πr 2. 2. Radial Let’s consider a radial variable. We can sweep out the area of the circular disk using annuli with a radial coordinate t. Then, we’re looking at an annular strip of width dt. The corresponding area element is the circumference (2πt) × thickness (dt). dA = 2πt dt Integrating this from 0 to the radius r gives us the area. A = Z dA = Z r t=0 2πt dt = πt2 r t=0 = πr 2. 3. Lateral We will use a vertical rectangular strip. Again, it is not a perfect rectangle and there’s a little bit of curvature at the end. But, these are higher order terms, and we just care about the diﬀerential element. So, using a vertical strip with width dx, and knowing that the formula for the boundary circle is x2 + y 2 = r 2, we solve for y along the upper and lower branches. y = ± p r 2 −x2 We then obtain an area element that is the area of this rectangular strip. dA = 2 p r 2 −x2 dx In the case of strips, assume the circle is centered at the origin, and let x keep track of where the strip intersects the x-axis. Thus, x ranges from −r to r. Integrating, and using a trigonometric substitution 118 x = r sin u gives A = Z dA = Z r −r 2 p r 2 −x2 dx = 2 Z π/2 −π/2 q r 2(1 −sin2 u) r cos u du = 2r 2 Z π/2 −π/2 cos2 u du = 2r 2 Z π/2 −π/2 1 2(1 + cos(2u))du = r 2(u + 1 2 sin 2u) π/2 −π/2 = πr 2 30.5 The area between two curves Let’s say the f is on top and the g is below. Then as we sweep a vertical strip from left to right, we obtain the area. In this case, the area element is a vertical rectangle of width dx and of height f (x) −g(x), the length of the interval between the two. dA = (f (x) −g(x))dx 119 The general formula for the area between two curves f (x) and g(x), A = Z dA = Z b x=a (f (x) −g(x))dx Example Find the area of the region bounded above by f (x) = 4 + x −x2 and below by g(x) = 1 −x. (See Answer 1) 30.6 Gini Index (An application of area formula) In economics, this ratio is used to quantify income inequality in a population. Let f (x) = Fraction of total income earned by the lowest x fraction of the populace, 0 < x < 1. The Gini index quantiﬁes how far f is from a “ﬂat” distribution. This means that f(0) = 0, f(1) = 1. f is probably going to be below the ﬂat distribution where y = x, the lowest x fraction earns the lowest x fraction. The Gini index, G(f ) is measuring the diﬀerence between these two distributions, in terms of area. It’s the ratio of the area between the ﬂat distribution y = x and the given population’s income distribution y = f (x). One normalizes that by the area between the ﬂat distribution y = x and y = 0, namely the area of that triangle, or 120 1 2. G(f ) = Area between the y = x and y = f (x) Area between y = x and y = 0 = 2 Z 1 x=0 (x −f (x)) dx. Example Compute G for a power law distribution f (x) = xn. (See Answer 2) The Gini Index doesn’t tell you the income distribution, but we could approximate it in the assumption of a power law. For example, in the year 2010, in the state of New York in USA, the Gini Index was very close to 1 2. If we assume that it went by a power law distribution, that would imply a cubic distribution of income. 30.7 EXERCISES • What is the area between the curve f (x) = sin3 x and the x-axis from x = 0 to x = π 3 ? • Find the area of the bounded region enclosed by the curves y = √x and y = x2. • What is the area between the curve y = sin x and the x-axis for 0 ≤x ≤π ? • Calculate the Gini index of a country where the fraction of total income earned by the lowest x fraction of the populace is given by f (x) = 2 5x2 + 3 5x3. • Compute the area between the curves f (x) = ex sec2 x and g(x) = ex tan2 x for 0 ≤x ≤π. • Consider a cone of height h with base a circular disc of radius r. Let’s compute the “surface area” – the area of the “outside” of the cone, not including the bottom. Following how we computed the area of a circular disc (which is, indeed, such a cone with h = 0 ), we can decompose its area into inﬁnitesimal triangles with base rdθ and height the slant length L = √ h2 + r 2. The area element dA is then the area of this inﬁnitesimal triangle. Integrating dA from θ = 0 to θ = 2π gives the “surface area” of the cone. What is its value? • Compute the area between the curves sin(x) and cos(x) from x = 0 to x = π/2. • Compute the area of a triangle with vertices at (0,0), (2,1), (3,6) 30.8 Answers to Selected Examples 1. The logical choice for area element is a vertical strip: 121 The height of this strip is f (x) −g(x) = 3 + 2x −x2, and the width of the strip is dx. So the area element is dA = (3 + 2x −x2)dx. To ﬁnd the intersection points, set the curves equal, which gives 1 −x = 4 + x −x2. This implies x2 −2x −3 = 0, which factors to (x + 1)(x −3) = 0. Thus, the intersections are x = −1 and x = 3. It follows that A = Z dA = Z 3 −1 (3 + 2x −x2)dx = 3x + x2 −1 3x3 3 −1 = (9 + 9 −9) − −3 + 1 + 1 3 = 32 3 . (Return) 2. G(f ) = 2 Z 1 x=0 (x −f (x)) dx = 2 Z 1 x=0 (x −xn) dx = 2(x2 2 −xn+1 n + 1) 1 x=0 = 1 − 2 n + 1 = n −1 n + 1 (Return) 122 31 Complex Areas 31.1 Complex regions Some regions in the plane are more complicated and cannot be evaluated with a single integral. This happens when the area element is not bounded by the same curves throughout the region. For instance, consider the region bounded by a parabola and two lines: In this case, the only way to ﬁnd the area of the region is to divide it into regions which can be integrated separately: 123 31.2 Horizontal strips Other regions are diﬃcult to integrate using vertical strips as the area element, but work well with horizontal strips as the area element. For example, consider the following region bounded on the left by x = g(y) and on the right by x = f (y): In this case, the area of a horizontal strip is a function of y, namely (f (y) −g(y))dy, where x = f (y) is the curve on the right and x = g(y) is the curve on the left. Example Find the area between the curves y −x = 0 y 2 + x = 2. (See Answer 1) Example Find the area of the region bounded by x = 3y and x = y 2 −4. (See Answer 2) Example Find the area of the region bounded by y = ln(x) and the lines y = 0, y = 1, and x = 0. (See Answer 3) 31.3 Polar shapes A polar shape is the graph of a polar function r = f (θ). Here, the input to the function is θ, which is the angle formed with the positive x-axis (known as the pole). The output r is the distance from the origin (or radial 124 distance). For example, the following shows the graph of the polar function r = 1 + cos θ, which is known as a cardioid: The area of such a region is not usually easy to compute by integrating with respect to x or y (for one thing, the polar equation would need to be expressed in terms of x and y ﬁrst!). Instead, the way to integrate over such regions is to use a polar area element, which is a wedge shaped region. Here are several examples of the polar area element for various values of θ: To compute what the polar area element is in terms of θ, r, and dθ, note that the region is roughly triangular (the curved portion at the base of the triangle can be ignored since it is a higher order term). The angle at the tip of the triangle is dθ, the height of the triangle is r, and the base of the triangle is r dθ: 125 Thus, the polar area element is dA = 1 2(r)(r dθ) = 1 2(f (θ))2 dθ, since r = f (θ). Thus, the area of a polar region deﬁned by r = f (θ), where a ≤θ ≤b, is A = Z b θ=a 1 2(f (θ))2 dθ. Example Compute the area of the cardioid r = 1 + cos θ. (See Answer 4) Example Find the area of a single petal of the polar curve r = sin(3θ): Hint: To ﬁnd the bounds on θ, compute when r = 0. (See Answer 5) Example Find the area inside the circle r = 2 sin θ and outside the circle r = 1: 126 (See Answer 6) 31.4 EXERCISES • Find the area enclosed by the curves y = 1, x = 1, and y = ln x. • Find the area of the bounded region enclosed by the x-axis, the lines x = 1 and x = 2 and the hyperbola xy = 1. • Compute the area in the bounded (i.e., ﬁnite) regions between y = x(x −1)(x −2) and the x-axis. • Find the area of the sector of a circular disc of radius r (centered at the origin) given by 1 ≤θ ≤3 (as usual, θ is in radians). • Use polar coordinates to ﬁnd an area within r = 3 −2cos(θ) and outside r = 3. • Find the area of the overlap between two circles of radius 2 that pass through each others’ centers. You can do so with either cartesian or polar coordinates (though one might be easier than the other!). • Find the area bound by the curves y = cos2 x and y = 8x2 π2 . • Kepler’s First Law states that the orbit of every planet is an ellipse with the Sun at one of its two foci. If we think of the Sun as being situated at the origin, we can describe the orbit with the equation: r = p 1 + ε cos θ The point at which the planet is closest to the Sun (the perihelion) corresponds to θ = 0, while the planet is furthest away from the Sun at θ = π (the aphelion). Knowing the distance between the Sun and the planet at these two points would allow you to ﬁx the values of the constants p and ε. Notice that ε = 0 describes a perfect circle, so that the “eccentricity” ε measures how far the orbit is from being a circle. 127 Kepler’s Second Law states that the line joining a planet and the Sun sweeps out equal areas during equal intervals of time. Another way of expressing this fact is by saying that the “areal velocity” vA = dA dt of that line is constant in time. Express the area element dA in terms of the angle element dθ and use Kepler’s Second Law to deduce the diﬀerential equation governing the time evolution of θ. • Let C1 be the circle given by r = sin(θ). Let C2 be the circle given by r = cos(θ). Find the area of region in C2 that is not in C1. 31.5 Answers to Selected Exercises 1. Expressing these curves as functions of y, we ﬁnd x = y x = 2 −y 2. Graphing these curves, one ﬁnds the bounded region: To ﬁnd the intersections, set the curves equal to one another. This gives y = 2 −y 2. A rearranging and factoring gives y 2 + y −2 = (y −1)(y + 2) = 0, and so we ﬁnd that the intersection points are at y = 1 and y = −2 (the x-coordinates are the same, since they are on the line x = y). Note that using a vertical rectangle as the area element here would not be so easy, because the area element depends on the value of x. Sometimes the strip goes from the parabola below to the line above, as shown in blue, and sometimes the strip goes from parabola to parabola, shown in red: 128 In particular, the area element for a vertical strip is dA = ( (x + √2 −x) dx if −2 ≤x ≤1 2√2 −x dx if 1 ≤x ≤2 But using a horizontal strip as the area element works much better because throughout the region the strip is always going from the line on the left to the parabola on the right. So using a horizontal strip gives the area element dA = ((2 −y 2) −y) dy. Integrating this over the range of −2 ≤y ≤1 gives the area: A = Z dA = Z 1 y=−2 2 −y 2 −y dy = 2y −y 3 3 −y 2 2 1 y=−2 = 2 −1 3 −1 2 − −4 + 8 3 −2 = 9 2. (Return) 2. The region looks roughly as in the following: 129 By setting 3y = y 2 −4, collecting like terms, and factoring, one ﬁnds the intersection points at y = −1 and y = 4, as indicated in the ﬁgure. The area element is a horizontal rectangle, which has area dA = (3y −(y 2 −4))dy. Thus, the area between the curves is A = Z dA = Z 4 −1 (3y −y 2 + 4)dy = 3 2y 2 −1 3y 3 + 4y 4 −1 = 125 6 (Return) 3. The region looks roughly like the following: Note that using vertical rectangles would not be ideal because this would require two integrals (for x from 0 to 1 and from 1 to e). Instead, one can express the curve y = ln x as x = ey. Now, using horizontal rectangles gives an area element of dA = eydy. Thus A = Z dA = Z 1 0 eydy = ey 1 0 = e −1. (Return) 4. In this case, f (θ) = 1 + cos θ, and so the area element is dA = 1 2(1 + cos θ)2 dθ = 1 2(1 + 2 cos θ + cos2 θ) dθ 130 Because θ ranges from 0 to 2π to trace out the entire cardioid, it follows that the area is A = Z dA = 1 2 Z 2π θ=0 (1 + 2 cos θ + cos2 θ) dθ = 1 2 Z 2π θ=0 (1 + 2 cos θ + 1 2(1 + cos(2θ))) dθ = 1 2 θ + 2 sin θ + 1 2θ + 1 4 sin(2θ) 2π θ=0 = 3π 2 . (Return) 5. The area element is dA = 1 2 sin2(3θ) dθ. To ﬁnd the bounds on θ, set r = 0, which gives sin(3θ) = 0. The smallest values of θ for which this occurs is θ = 0 and θ = π 3 : Thus, the area of a single petal is A = Z dA = Z π/3 θ=0 1 2 sin2(3θ) dθ = 1 2 Z π/3 θ=0 1 2(1 −cos(6θ)) dθ = 1 4 θ −1 6 sin(6θ) π/3 θ=0 = π 12 (Return) 6. First, we ﬁnd the intersections by setting the curves equal, which gives 2 sin θ = 1 ⇒ sin θ = 1 2, 131 and so the intersections are at π 6 and 5π 6 . The area element of the region is the polar area element of the circle r = 2 sin θ minus the polar area element of the circle r = 1: So we have that dA = 1 2(2 sin θ)2 −1 2(1)2 dθ. Thus, the area is A = Z dA = 1 2 Z 5π/6 θ=π/6 (4 sin2 θ −1) dθ = 1 2 Z 5π/6 θ=π/6 2(1 −cos(2θ)) −1 dθ = 1 2 (θ −sin(2θ)) 5π/6 π/6 = π 3 + √ 3 2 . From the second to the third line above, we used the power reduction formula for sine: sin2 θ = 1 2(1 −cos(2θ)). (Return) 132 32 Volumes 32.1 Finding the volume element Just as area was computed by ﬁnding the area element and integrating, volume is computed by determining the volume element (i.e. the volume of a slice) and then integrating: V = Z dV. The diﬃculty is in ﬁnding a suitable volume element dV . Once that is chosen, the rest is a matter of evaluating the integral. Example Compute the volume of a cylinder of radius R and height H using several diﬀerent volume elements dV : (See Answer 1) 133 Example Find the volume of a sphere of radius R. First, by using discs as the volume element (shown on left below). Then use cylindrical shells as the volume element (shown in the middle below). Finally, use a spherical shell for the volume element, as shown in the third diagram. (See Answer 2) Example Find the volume of a cone of base radius R and height H. (See Answer 3) Example Find the volume of a square pyramid of base edge S and height H. 134 (See Answer 4) Example Show that the volume of a generalized cone of base area B and height H is 1 3BH. Explain the reason for the factor of 1 3. (See Answer 5) 135 32.2 EXERCISES • Find the volume of the following solid: for 1 ≤x < ∞, the intersection of the this solid with the plane perpendicular to the x-axis is a circular disc of radius e−x. • The base of a solid is given by the region lying between the y-axis, the parabola y = x2, and the line y = 16 in the ﬁrst quadrant. Its cross-sections perpendicular to the y-axis are equilateral triangles. Find the volume of this solid. • The base of a solid is given by the region lying between the y-axis, the parabola y = x2, and the line y = 4. Its cross-sections perpendicular to the y-axis are squares. Find the volume of this solid. • Find the volume of the solid whose base is the region enclosed by the curve y = sin x and the x-axis from x = 0 to x = π and whose cross-sections perpendicular to the x-axis are semicircles. • Consider a cone of height h over a circular base of radius r. We computed the volume by slicing parallel to the base. What happens if instead we slice orthogonal to the base? What is the volume element obtained by taking a wedge at angle θ of thickness dθ ? • Consider the following solid, deﬁned in terms of polar coordinates: 0 ≤r ≤R; 0 ≤θ ≤2π; 0 ≤z ≤r. Can you describe this shape? Compute its volume. • Consider the following solid, deﬁned in terms of polar coordinates: 0 ≤r ≤R; 0 ≤θ ≤2π; 0 ≤z ≤θ. Can you describe this shape? Compute its volume. • Challenge: compute the volume intersection of the (inﬁnite) cylinders of radius R centered along the x and y axes in 3-d. That is, compute the volume of intersection of x2 + z2 ≤R2 x2 + z2 ≤R2 in the 3-dimensional (x, y, z) space. 32.3 Answers to Selected Examples 1. First, consider making a slice perpendicular to the base of the cylinder: 136 This gives a rectangular slice whose height is H, the same as the cylinder. The width of the rectangle can be determined by looking at an overhead view of the cylinder. Let x be the distance of the slice from the center of the cylinder (so x ranges from −R to R as the slice sweeps across the cylinder): Doing a little algebra, we ﬁnd that the width of the rectangle is 2 √ R2 −x2. Finally, the thickness of the slice is dx, and so the volume element in this case is dV = 2H p R2 −x2 dx. Integrating this requires the trigonometric substitution x = R sin θ. There are easier volume elements we could choose, as we shall see. Another way to slice is to make cuts parallel to the base of the cylinder. Let y denote the distance of the slice from the base of the cylinder: Then each slice is a circle of radius R and thickness dy. Thus dV = πR2 dy, 137 and y ranges from 0 to H, so the volume is V = Z dV = Z H y=0 πR2 dy = πR2y H y=0 = πR2H. Another possible choice is a wedge shaped volume element. Let θ be the angle that the wedge forms with a ﬁxed axis (so θ ranges from 0 to 2π): Here, the area of the sector of the circle is 1 2R2 dθ. Thus the volume of the wedge is dV = 1 2R2H dθ. Thus the volume is V = Z dV = Z 2π θ=0 1 2R2H dθ = 1 2R2Hθ 2π θ=0 = πR2H. One ﬁnal option is to use cylindrical shells. Let t be the radius of the shell, so that t ranges from 0 to R as the shells sweep through the cylinder. 138 The height of the cylindrical shell is H and the thickness of the shell is dt. Recalling that the lateral surface area of a cylinder is 2πRH, we have dV = 2πtH dt. Integrating gives that the volumes is V = Z dV = Z R t=0 2πtH dt = 2πH t2 2 R t=0 = πR2H. (Return) 2. Let x be the distance from the center of the disc to the center of the sphere (so x ranges from −R to R as the discs sweep across the sphere). Then drawing a right triangle shows that the radius of the disc is √ R2 −x2 (since the radius of the sphere is R). See the diagram below: 139 The thickness of the disc is dx, and so the volume of the disc is π( √ R2 −x2)2dx (the area of the disc times its thickness), and so the volume of the sphere is V = Z dV = Z R x=−R π(R2 −x2)dx = π R2x −x3 3 R x=−R = π R3 −R3 3 − −R3 + R3 3 = 4 3πR3. For the cylindrical shell, let t be the radius of the cylinder (so t ranges from 0 to R as the cylinders sweep out the sphere). Then by drawing in a right triangle, one ﬁnds that the height of the cylinder is 2 √ R2 −t2: Recall that the lateral surface area of a cylinder with radius r and height h is 2πrh. Thus, the lateral surface area of the cylinder is 4πt √ R2 −t2. The thickness of the shell is dt, and so the volume element is 4πt √ R2 −t2dt. It follows (after making the u-substitution u = R2 −t2) that the volume of the sphere is V = Z dV = Z R t=0 4πt p R2 −t2dt = 4π Z 0 u=R2 −1 2 √udu = −2π 2 3u3/2 0 u=R2 ! = −2π 0 −2 3R3 = 4 3πR3. 140 Finally, for the spherical shell, let ρ denote the radius of the spherical shell: Recall that the surface area of a sphere of radius ρ is 4πρ2. Therefore, the volume of the spherical shell (i.e. our volume element) is dV = 4πρ2 dρ. Note that to sweep over the entire sphere, ρ must range from 0 to R. Therefore, V = Z dV = Z R ρ=0 4πρ2 dρ = 4π 1 3ρ3 R ρ=0 = 4 3πR3. (Return) 3. The easiest choice for volume element is a slice parallel to the base of the cone, which gives a disc. Let y be the distance from the tip of the cone to the center of the disc (so y ranges from 0 to H as the disc sweeps across the cone), and x be the radius of the disc: 141 The volume element is the area of the disc, πx2, times the thickness of the disc, dy. It remains to ﬁnd x in terms of y. In the cutaway in the ﬁgure on the right above, one sees that by similar triangles, x y = R H, and so x = R Hy. Thus, the volume element is dV = πx2dy = π R H y 2 dy. Thus, the volume of the cone is V = Z dV = Z H y=0 π R2 H2 y 2dy = πR2 H2 y 3 3 H y=0 ! = π R2 H2 · H3 3 = 1 3πR2H. (Return) 4. Again, use slices parallel to the base. Let y be the distance from the tip of the cone to the center of the slice (so y ranges from 0 to H), and let x be half of the side length of the slice. As shown in the above cutaway, one ﬁnds by similar triangles that x y = S/2 H , and so x = S 2Hy. Therefore, the area of a slice is (2x)2 = S2 H2 y 2, and the thickness of a slice is dy, so the volume element is dV = S2 H2 y 2dy. 142 And so the volume of the pyramid is V = Z dV = Z H y=0 S2 H2 y 2dy = S2 H2 y 3 3 H y=0 ! = S2 H2 · H3 3 = S2H 3 . (Return) 5. Let y be the distance from the tip of the cone to the slice. Because the linear dimensions of the slice grow proportionally with y (e.g. the length of the slice is y H times the length of the base), the area of the slice will grow proportionally with the square of y. This means that Area of the slice = y H 2 B Thus, the volume element is dV = B y 2 H2 dy, and it follows that the volume of the cone is V = Z dV = Z H y=0 B y 2 H2 dy = B H2 y 3 3 H y=0 ! = B H2 · H3 3 = 1 3BH. The factor of 1 3 comes from the fact that the volume element is proportional to the square of y, hence the integral has a y 2, which produces a factor of 1 3 by the power rule. (Return) 143 33 Volumes Of Revolution 33.1 Volume element for solid of revolution Consider a region R in the plane and a line L. The solid of revolution of R about the axis L is the solid which results from taking the region R and revolving it around the line L: (Solid of Revolution Animated GIF) The result is typically something doughnut shaped. The question of this module is to ﬁnd the volume of the solid: The method is the same as the previous modules: ﬁnd the volume element (the contribution of a small slice of the region to the total volume) and integrate. Just as area can be computed using vertical or horizontal slices, volume can be computed using corresponding methods: shells or washers, respectively. The basic outline of ﬁnding the volume element for a solid of revolution is 1. Find a convenient area element for the region R in the plane 2. Determine the volume as that area element gets revolved around the axis L. 144 33.2 Cylindrical shells When the area element is parallel to the axis of rotation, the volume element is a cylindrical shell. Here, the region is bounded by two parabolas. The natural area element for such a region is a vertical rectangle (shown in red). As the region is revolved about the y-axis, the volume element traces out a cylindrical shell, whose volume becomes the volume element of the solid of revolution. Recall that a cylinder has lateral surface area 2πrh. The thickness of the cylindrical shell is dx (if the axis of rotation is a vertical line) or dy (if the axis of rotation is a horizontal line). Here r and h will generally be functions of x or y (again, depending on whether the axis of rotation is vertical or horizontal). If a horizontal rectangle is the natural area element (for instance, the region between two horizontal parabolas), and the axis of revolution is the x-axis, then cylindrical shells again arise naturally as the volume element: 145 Example Suppose the region bounded by y = 3x −x2 and y = x is revolved about the y-axis. What is the volume of the resulting solid? (See Answer 1) 33.3 Washers When the area element is perpendicular to the axis of rotation, the volume element is a washer. So when the area element is a horizontal rectangle (as in a region bounded by horizontal parabolas) and the axis of revolution is vertical, the region traced out by the rectangle is a washer: A washer is just an annulus (a circle with a circle cut out of it) which has been thickened. The volume of the washer is the area of the annulus times the thickness of the washer. The area of the annulus is πR2 −πr 2, where R is the radius of the outer circle and r is the radius of the inner circle. The thickness of the washer is dx or dy (depending on the orientation of the washer. Thus, the volume element when using washers is dV = π(R2 −r 2) dx or dy. Example Given the region bounded by y = 3x −x2 and y = x, ﬁnd the volume of the solid resulting from revolving the region about the x-axis. (See Answer 2) 33.4 Additional Examples Example Find the volume of a doughnut formed by rotating a disc of radius a about the y-axis. Let the radius of the doughnut be R, as shown in this cutaway: 146 Use a vertical area element (which leads to a cylindrical shell). (See Answer 3) Example Compute the volume of the doughnut again, this time using a horizontal area element (which leads to a washer). (See Answer 4) 33.5 EXERCISES • Let D be the region bounded by the curve y = x3, the x-axis, the line x = 0 and the line x = 2. Find the volume of the region obtained by revolving D about the x-axis. • Let D be the same region as above. What is the volume of the region formed by rotating this D about the line x = 3? • Let D be the region between the curve y = −(x −2)2 + 1 and the x-axis. Find the volume of the region obtained by revolving D about the y-axis. • Find the volume obtained by revolving the region between the curves y = x3 and y = 3 √x in the ﬁrst quadrant about the x-axis. • Let D be the region under the curve y = ln √x and above the x-axis from x = 1 to x = e. Find the volume of the region obtained by revolving D about the x-axis. • Let D be the region bounded by the graph of y = 1 −x4, the x-axis and the y-axis in the ﬁrst quadrant. Which of the following integrals can be used to compute the volume of the region obtained by revolving D around the line x = 5? • Challenge: compute the volume of the region obtained by rotating the disc x2 + y 2 ≤ϵ2 about the axis given by y = 1 −x for ϵ ≤1 2. • Let D be the region under the curve √x −1 above the x-axis from x = 1 to x = 2. Compute the volume of solid obtained by rotating D about the x-axis. Compute the volume twice, using both methods. 33.6 Answers to Selected Examples 1. The ﬁrst step in such a calculation is to draw a decent picture of the region. Then determine whether a vertical or horizontal rectangle would make the best area element. In this case, a vertical rectangle is the best choice. Since a vertical rectangle is being revolved about a vertical axis, the result is a cylindrical shell: 147 The radius of the shell is x (the distance from the y-axis), and the height of the shell is the distance from the top curve to the bottom curve: h = (3x −x2) −x = 2x −x2. The thickness of the shell is dx. Recalling that the surface area of a cylinder is 2πrh, it follows that the volume element is just the surface area multiplied by the thickness dx: dV = 2πrh dx = 2πx(2x −x2) dx. A little algebra shows that the intersections occur at x = 0 and x = 2, so the volume is V = Z dV = Z 2 x=0 2πx(2x −x2) dx = 2π Z 2 x=0 (2x2 −x3) dx = 2π 2 3x3 −1 4x4 2 x=0 = 8 3π. (Return) 2. As in the previous example, the optimal area element is a vertical rectangle. A vertical rectangle revolved about a horizontal axis results in a washer: 148 The outer radius R is the upper curve: R = 3x −x2, and the inner radius r is the inner curve: r = x. The thickness of the washer is dx, and so dV = π(R2 −r 2) dx = π((3x −x2)2 −x2) dx. As in the last example, the intersections are at x = 0 and x = 2, so V = Z dV = Z 2 x=0 π(9x2 −6x3 + x4 −x2) dx = π 8 3x3 −3 2x4 + 1 5x5 2 x=0 = 56 15π. (Return) 3. First, suppose we use a vertical area element. Since we are rotating about a vertical axis, the area element and axis of rotation are parallel, and so the resulting volume element is a cylindrical shell. Let x be the distance of the area element (the rectangle) from the y-axis. This also happens to be the radius of the cylindrical shell: 149 The equation of the circle is (x −R)2 + y 2 = a2, and solving for y gives y = ± p a2 −(x −R)2. Therefore, the height of the area element (and hence the height of the cylindrical shell) is 2 p a2 −(x −R)2. Now, the volume of the cylindrical shell (our volume element) is dV = 2πrh dx = 2πx(2 p a2 −(x −R)2) dx = 4πx p a2 −(x −R)2 dx. Note that x ranges from R −a to R + a as it sweeps across the circle. Therefore the volume is V = Z dV = Z R+a x=R−a 4πx p a2 −(x −R)2 dx. This is a bit messy, but with a substitution of u = x −R du = dx, we ﬁnd Z R+a x=R−a 4πx p a2 −(x −R)2 dx = Z a u=−a 4π(u + R) p a2 −u2 du = Z a u=−a 4πu p a2 −u2 du + Z a u=−a 4πR p a2 −u2 du. Here, we have used linearity to split the integral into two integrals. Notice that the ﬁrst integrand is an odd function of u, and since it is integrated over a symmetric interval, the ﬁrst integral is 0: Z a u=−a 4πu p a2 −u2 du = 0. The second integral can be found by noting that Z a u=−a 2 p a2 −u2 du gives the area of a disc of radius a (this was an integral done in the simple areas module). Therefore, the volume is Z a u=−a 4πR p a2 −u2 du = 2πR Z a u=−a 2 p a2 −u2 du = 2πR(πa2) = 2π2Ra2. (Return) 150 4. Carefully drawing and labeling the outer and inner radii of the washer gives the following diagram: The outer and inner radii can be found by solving the equation of the circle for x: (x −R)2 + y 2 = a2 ⇒ x = R ± p a2 −y 2. Thus, the volume element is the area of the washer times its thickness, dy. Computing this and doing a little algebra gives dV = h π(R + p a2 −y 2)2 −π(R − p a2 −y 2)2i dy = 4πR p a2 −y 2 dy. Note that y ranges from −a to a, and so the volume integral is the same one arrived at above: V = Z dV = Z a y=−a 4πR p a2 −y 2 dy = 2πR(πa2) = 2π2Ra2. (Return) 151 34 Volumes In Arbitrary Dimension The motivation for this module is to ﬁnd the volume (often referred to as hypervolume) of an object in dimension n. This has physical meaning for n ≤3, but what happens for n ≥4? 34.1 The cube in dimension n Consider the unit cube (i.e. the cube of side length 1) in n dimensions, sometimes called the n-hypercube or just the n-cube. Formally, this is deﬁned to be the set of n-tuples (i.e. lists of length n) (x1, x2, . . . , xn) such that 0 ≤xi ≤1 for all 1 ≤i ≤n. For n = 0, 1, 2, 3, these are familiar ﬁgures: the point, line segment, square, and cube, respectively. Now, consider some of the various measurements for each of these cubes. Volume of the cube For n = 0, the cube is just a point, and volume is deﬁned to just be the number of points. So a single point has volume 1. For n = 1, the cube is a line segment. The volume in one dimension is just length, so the one dimension cube has volume 1. For n = 2, the cube is a square of side length 1. In two dimensions, volume is area, so the cube in two dimensions has volume w × h = 1 × 1 = 1. For n = 3, the cube is a (traditional) cube of side length 1, which has (traditional) volume l×w×h = 1×1×1 = 1. For higher values of n, this pattern continues. The intuition is that each additional dimension adds an extra factor of 1, thus the volume of each unit n-cube is 1. 152 Surface area of cubes Consider the surface area of the cube in dimension n. As with volume, this has physical meaning for n = 2 and n = 3. For n = 2, the surface area of a square is really its perimeter, which is 4. For n = 3, the surface area is the total area of the faces which bound the cube. There are 6 faces each with area 1, so the surface area is 6. In general, the n dimension cube will have 2n boundary faces, and each face is a cube of dimension n −1, so the surface area (really the hypervolume of the boundary) is 2n. Other features The diagonal of the n-cube can be deﬁned to be the distance from (0, 0, . . . , 0) to (1, 1, . . . , 1). Using the distance formula, one ﬁnds that the diagonal of the n-cube is √n. The number of corners is fairly easy to count. For n = 0, 1, 2, 3, the number of corners is 1, 2, 4, and 8 respectively. Since the n-cube can be thought of as two copies of the (n −1)-cube, one can show by induction that there are 2n corners in the n-cube. 34.2 Simplex A simplex is a generalization of a triangle or a pyramid. In dimension n, the simplex is deﬁned to be the set of n-tuples (x1, x2, . . . , xn) such that 0 ≤xi ≤1 and P xi ≤1. This can be thought of as the corner of the n dimension cube where the sum of the coordinates is less than 1. Here are the simplices of dimension n = 0, 1, 2, 3: 34.3 Volume of spheres in arbitrary dimension Now, consider a sphere of radius r in n dimensions. This is the set of points (x1, x2, . . . , xn) such that x2 1 + x2 2 + . . . + x2 n ≤r 2. Let Vn(r) be the volume of the sphere of radius r in n dimensions (as above, volume means length, area, volume, hypervolume for n = 1, 2, 3, . . ., respectively). With some careful integration and induction, one ﬁnds that Vn(r) = ( πk k! r n if n = 2k 2nπkk! n! r n if n = 2k + 1 . Now, note that as n →∞(and r stays ﬁxed), the volume goes to 0 (since factorial grows faster than exponentials). 153 34.4 EXERCISES • Consider a four-dimensional box (or “rectangular prism”) with side-lengths 1, 1/2, 1/3, and 1/4. What is the 4-dimensional volume of this box? • What is the “diameter” – i.e., the farthest distance between two points – in this 4-d box? Hint: think in terms of diagonals. • High-dimensional objects are everywhere and all about. Let’s consider a very simple model of the space of digital images. Assume a planar digital image (such as that captured by a digital camera), where each pixel is given values that encode color and intensity of light. Let’s assume that this is done via an RGB (red/green/blue) model. Though there are many RGB model speciﬁcations, let us use one well-suited for mathematics: to each pixel on associates three numbers (R, G, B), each taking a value in [0, 1]. Since the red/green/blue values are independent, each pixel has associated to it a 3-d cube of possible color values. Consider a (fairly standard) 10-megapixel camera. If I were to consider the “space of all images” that my camera can capture, what does the space look like? How many dimensions does it have? Note: there’s no calculus in this problem...just counting! • Consider an n-dimensional “hypercube” C of all side-lengths equal to 1. Its n-dimensional volume is, clearly, 1. Now consider what happens when you shrink the hypercube’s side-lengths by 1 percent (concentrically, so that the shrunken cube has the same center as the original) and remove it from the original cube. By subtracting the n-dimensional volume of this slightly smaller hypercube, conclude how much volume remains in the 1-percent outer “shell.” • In the previous question, what happens to the volume of the 1-percent shell as n →∞? • We have seen that the n-dimensional volume of a unit radius ball in dimension n converges to zero as n →∞. But what about a really large ball? For a ball of radius R = 1010 meters in dimension n, what is the limit as n →∞of its volume? (in unit of meters-to-the-nth) • For the brave: so, as n →∞, the volume of the n-ball all concentrates near the surface shell. OK, you’ve got that. Now answer this: what proportion of the volume is concentrated along the “equatorial plane”? Let’s make that speciﬁc. Recall, we computed the volume Vn as In · Vn−1, where In = Z π 2 θ=−π 2 cosn θ dθ. We can compute the volume of the equatorial slice of thickness 2ϵ (for some small but ﬁxed ϵ > 0) as Vn,2ϵ = Vn−1 Z ϵ −ϵ cosn θ dθ. So, here is the (hard!) problem. Compute the limit as n →∞of the ratio of Vn,2ϵ to Vn: lim n→∞ Vn,2ϵ Vn = lim n→∞ 1 In Z ϵ −ϵ cosn θ dθ. If you can do this (a very big if...) you will get a surprise... 154 35 Arclength Consider the graph of a function y = f (x) for a ≤x ≤b. The purpose of this module is to ﬁnd the length of this piece of the curve, known as the arclength of the function f from a to b. As in previous modules, the basic method is to ﬁnd the arclength element dL and then integrate it: L = Z dL. By zooming in on a portion of the curve, it begins to look like a straight line. Then one can express dL in terms of the inﬁnitesimal horizontal change dx and vertical change dy: Now, by the Pythagorean theorem one ﬁnds that dL = p dx2 + dy 2. A little algebra and the chain rule gives that dL = p dx2 + dy 2 = s dx2 + dy dx dx 2 = s 1 + dy dx 2 dx = p 1 + (f ′(x))2dx. So the arclength of the function f from a to b is given by L = Z b a p 1 + (f ′(x))2dx. 155 Example Find the arclength of the curve y = ln sin x; π 4 ≤x ≤π 2 . Hint: recall the facts that 1 + cot2 x = csc2 x Z csc x dx = −ln \| csc x + cot x\| + C. (See Answer 1) Example Find the arclength of the curve y = x2 −1 8 ln(x); 1 ≤x ≤4. (See Answer 2) 35.1 Parametric curves If a curve is deﬁned parametrically, i.e. x = x(t) and y = y(t) for a ≤t ≤b, then the arclength element can be written as dL = p dx2 + dy 2 = sdx dt dt 2 + dy dt dt 2 = sdx dt 2 + dy dt 2 dt = p x′(t)2 + y ′(t)2dt. So the arclength of a parametric curve (x(t), y(t)) for a ≤t ≤b is given by L = Z b a p x′(t)2 + y ′(t)2dt. Example Find the arclength for a circle of radius r. (See Answer 3) Example Find the arclength for the spiral x(t) = t cos(t), y(t) = t sin(t) for 0 ≤t ≤2π. (See Answer 4) 156 35.2 Additional Examples Example Compute the arclength of the curve y = 2 3x3/2; 0 ≤x ≤3 (See Answer 5) Example A catenary is the curve that is formed by hanging a cable between two towers. It is a fact that the rate of change of the slope of a hanging cable is proportional to the rate of change of arclength with respect to x. Mathematically, d dx dy dx = κ · dL dx , for some constant κ. Use this fact to ﬁnd the equation of the catenary. Then ﬁnd the length of the catenary for −l ≤x ≤l. (See Answer 6) Example Show that the spiral x = 1 t cos t y = 1 t sin t for 2π ≤t has inﬁnite arclength. (See Answer 7) 35.3 EXERCISES • Compute the arclength of y = x3 3 + 1 4x , from x = 1 to x = 2. 35.4 Answers to Selected Exercises 1. Computing the arclength element from the above formula gives dL = s 1 + dy dx 2 dx = s 1 + 1 sin x cos x 2 dx = p 1 + cot2 x dx = √ csc2 x dx = csc x dx. 157 Therefore, we ﬁnd that the arclength is L = Z dL = Z π/2 x=π/4 csc x dx = −ln \| csc x + cot x\| π/2 x=π/4 = −ln(1 + 0) + ln( √ 2 + 1) = ln(1 + √ 2). (Return) 2. First, one ﬁnds dy dx = 2x −1 8x . So, with some careful algebra one sees that s 1 + dy dx 2 = s 1 + 2x −1 8x 2 = s 1 + (2x)2 −22x 8x + 1 (8x)2 = s 1 + (2x)2 −1 2 + 1 (8x)2 = s (2x)2 + 1 2 + 1 (8x)2 Now note that by reversing the cancellation done in an earlier step when simplifying −2 2x 8x = −1 2, one ﬁnds that 1 2 = 2 2x 8x . And so, continuing the computation, one ﬁnds s 1 + dy dx 2 = s (2x)2 + 22x 8x + 1 (8x)2 = s 2x + 1 8x 2 = 2x + 1 8x . Thus, dL = 2x + 1 8x dx, and it follows that L = Z dL = Z 4 x=1 2x + 1 8x dx = x2 + 1 8 ln(x) 4 x=1 = (16 + 1 8 ln(4)) −(1 + 1 8 ln(1)) = 15 + ln(4) 8 . 158 (Return) 3. A simple parametrization for the circle of radius r is x = r cos t y = r sin t. Note that t ranges from 0 to 2π. Using the above formula, we ﬁnd that the arclength element is dL = sdx dt 2 + dy dt 2 dt = p (−r sin t)2 + (r cos t)2 dt = q r 2(sin2 t + cos2 t) dt = √ r 2 dt = r dt. (we used the Pythagorean identity sin2 t + cos2 t = 1 from line three to line four). Therefore, L = Z dL = Z 2π t=0 r dt = r · t 2π t=0 = 2πr, as desired. (Return) 4. First, compute x′(t) = cos(t) −t sin(t) and y ′(t) = sin(t) + t cos(t). Then p x′(t)2 + y ′(t)2 = p (cos(t) −t sin(t))2 + (sin(t) + t cos(t))2 = q cos2(t) −2t cos(t) sin(t) + t2 sin2(t) + sin2(t) + 2t cos(t) sin(t) + t2 cos2(t) = p 1 + t2. Thus, dL = √ 1 + t2dt. So one ﬁnds that L = Z 2π 0 p 1 + t2dt. This integral was computed in the Trigonometric Substitution module. The answer becomes L = 1 2 sinh−1(t) + 1 2t p 1 + t2 2π 0 = 1 2 sinh−1(2π) + π p 1 + 4π2. (Return) 159 5. Computing the arclength element, we ﬁnd dL = s 1 + dy dx 2 dx = q 1 + √x 2 dx = √ 1 + x dx. Therefore, the arclength is L = Z dL = Z 3 x=0 √ 1 + x dx = 2 3(1 + x)3/2 3 x=0 = 16 3 −2 3 = 14 3 . (Return) 6. Using the formula for the arclength element, the fact tells us that d dx dy dx = κ s 1 + dy dx 2 Now, making a substitution of u = dy dx simpliﬁes the equation to become du dx = κ p 1 + u2. This is a separable diﬀerential equation. Separating and integrating gives Z du √ 1 + u2 = Z κ dx. The left side can be handled with either a trigonometric or hyperbolic trigonometric substitution. We take the latter approach, and let u = sinh t du = cosh t dt. 160 So we have (remembering the Pythagorean identity for hyperbolic trigonometric functions from the trigonometric substitution module) Z du √ 1 + u2 = Z cosh t p 1 + sinh2 t dt = Z cosh t √ cosh2 t dt = Z cosh t cosh t dt = Z dt = t = arcsinh u (we leave the constant of integration oﬀfor now since we will be integrating on the right side as well). On the right side, we have Z κ dx = κx + C. Putting it together, we have u = sinh(κx + C). If we pick our coordinates so that x = 0 occurs at the low point of the catenary, then note that at this point, we have u = dy dx = 0, since the slope of the catenary is 0 at the low point. Using this fact and plugging in x = 0 into the earlier equation gives u = sinh(C) = 0 and so C = 0. This gives u = dy dx = sinh(κx). Now integrating both sides gives y = 1 κ cosh(κx) + C, where C = y0 is the y value of the low point of the catenary. 161 To ﬁnd the length of the catenary, we have L = Z dL = Z l x=−l s 1 + dy dx 2 dx = Z l x=−l p 1 + sinh2 κx dx = Z l x=−l cosh κx dx = 1 κ sinh κx l x=−l = 1 κ (sinh κl −sinh κ(−l)) = 2 κ sinh κl. since hyperbolic sine is an odd function. This grows very quickly as l increases, because 2 κ sinh κl ≈1 κeκl. (Return) 7. Computing dx dt = −t sin t −cos t t2 dy dt = t cos t −sin t t2 . Plugging these into the formula for the arclength of a parametric curve and noting the cancellation of cross terms, we have dL = sdx dt 2 + dy dt 2 dt = s−t sin t −cos t t2 2 + t cos t −sin t t2 2 dt = s t2 · sin2 t + cos2 t t4 + cos2 t + sin2 t t4 dt = r 1 t2 + 1 t4 dt = √ t2 + 1 t2 dt. Therefore, the arclength is Z ∞ t=2π √ t2 + 1 t2 dt. 162 This integral is diﬃcult to compute exactly, but we only want to show it diverges, which is not as diﬃcult. Note that √ t2 + 1 t2 ≥ √ t2 t2 = t t2 = 1 t . And so by the dominance of deﬁnite integrals, Z ∞ t=2π √ t2 + 1 t2 dt ≥ Z ∞ t=2π 1 t dt but the integral on the right diverges to inﬁnity by our earlier discussions of p-integrals. Thus, our integral on the left, being larger, also diverges to inﬁnity. (Return) 163 36 Surface Area This module deals with the surface area of solids of revolution. Consider the portion of a curve y = f (x) for a ≤x ≤b revolved about a horizontal axis to create a solid. In earlier modules the goal was to ﬁnd the volume of such a solid, but now the focus is on ﬁnding the surface area. As always, the method will be to ﬁnd the surface area element and integrate it. The surface area element which works well is the thin band shown here: 36.1 Surface area of a cone The ﬁrst step towards ﬁnding the surface area element is to ﬁnd the lateral surface area of a more simple solid: the cone. Consider a cone whose base has radius r and lateral height R (the lateral height is the distance from the tip of the cone to a point on the circumference of the base; see the left diagram below). 164 To ﬁnd the area, consider cutting the cone along the straight dotted line from base circumference to tip and unrolling the cone. The result is a portion of a circle whose radius is R, as shown on the right in the diagram above. Note that the circumference of the base of the cone, 2πr, becomes the length of arc of the unrolled cone. This means that the unrolled cone is a fraction of the full circle of radius R, and that fraction is 2πr 2πR (the ratio of the circumference of the partial circle to the circumference of the whole circle). Thus the surface area of the cone is 2πr 2πRπR2 = πrR. The surface area of a cone can be used to ﬁnd the area of a frustum of a cone whose top radius is r1, bottom radius is r2, and lateral height l (as in the below diagram). The area of this frustum is π(r1 + r2)l. Expressed another way, the area is 2πrl, where r = r1+r2 2 is the average of the two radii of the frustum. 36.2 Surface area element Now, the surface area element can be found. When the curve is partitioned into suﬃciently small pieces, the surface area element is just the area of the frustum formed by rotating the arclength element about the axis (see the diagram): Thus, the surface area element is dS = 2πrdL, where r is the distance from the curve to the axis of rotation, and dL is the arclength element (i.e. dL = p 1 + (f ′(x))2dx). In the (common) case where the axis of rotation is the x-axis, one ﬁnds that r = f (x). 165 Thus, the surface area resulting from revolving the curve y = f (x) for a ≤x ≤b about the x-axis is given by S = Z b a 2πrdL = 2π Z b a f (x) p 1 + (f ′(x))2dx. Example Consider the sphere of radius r. If the sphere is cut into slices of equal width, which slice has the most surface area? (See Answer 1) Example Consider the surface generated by revolving the curve y = 1 xp for x ≥1 about the x-axis. Find the values of p for which the surface has ﬁnite surface area. Then ﬁnd the values of p for which the solid of revolution has ﬁnite volume. (See Answer 2) 166 36.3 Rotations about the y-axis Suppose we want to know the surface area which results from revolving the curve y = f (x); a ≤x ≤b about the y-axis. There are two main ways one can go about ﬁnding this surface area: 1. Express everything as a function of y (including range of inputs), and then use the above formula but with the roles of x and y switched. 2. Leave things in terms of x, but adjust the formula slightly. The ﬁrst method expresses the curve as x = f −1(y); c ≤y ≤d, where c = f −1(a) and d = f −1(b). Then express the surface area element as dS = 2πr dL = 2πf −1(y) s 1 + dx dy 2 dy. Putting it together, the surface area can be expressed as S = Z dS = 2π Z d y=c f −1(y) s 1 + dx dy 2 dy. Again, this is really just a reuse of the original formula, with the roles of x and y ﬂipped. The second method is sometimes simpler to apply because it involves less algebra. The main observation to make is that the radius in the surface area element is simply x when the curve is revolved around the y-axis: 167 So the surface area element can be written dS = 2πr dL = 2πx s 1 + dy dx 2 dx. This integral is with respect to x, and so it should be integrated over the original range of x: S = 2π Z b x=a x s 1 + dy dx 2 dx. Example Compute the surface area of the surface resulting from revolving the curve y = 1 2x2; 0 ≤x ≤4. about the y-axis: (See Answer 3) 168 36.4 EXERCISES • Compute the surface area resulting from revolving the curve f (x) = cosh(x), 0 ≤x ≤2 about the x-axis. 36.5 Answers to Selected Examples 1. If we center the sphere at the origin, we can think of the sphere as the surface of revolution obtained by revolving the curve y = p r 2 −x2; −r ≤x ≤r about the x-axis. First, we compute the arclength element: dL = s 1 + dy dx 2 dx = s 1 + −x √ r 2 −x2 2 dx = r 1 + x2 r 2 −x2 dx = r r 2 r 2 −x2 dx = r √ r 2 −x2 dx. Plugging this into the surface area element, we ﬁnd dS = 2πy dL = 2π p r 2 −x2 · r √ r 2 −x2 dx = 2πr dx. Note that this is independent of x! This means that every slice of the sphere has equal surface area. For example, if we were to slice the sphere into four slices of equal thickness, then a middle slice goes from x = 0 to x = r 2, and its surface area Z r/2 x=0 2πr dx = 2πrx r/2 x=0 = 2πr · r 2 = πr 2. The end-cap slice, on the other hand, goes from x = r 2 to x = r, so its surface area is Z r x=r/2 2πr dx = 2πrx r x=r/2 = 2πr r −r 2 = 2πr · r 2 = πr 2. 169 So we see that the pieces have equal surface area. (Return) 2. The surface area, in terms of p, is S = 2π Z ∞ 1 1 xp p 1 + (−px−p−1)2dx = 2π Z ∞ 1 1 xp r 1 + p2 x2p+2 dx Unfortunately, this integral is not computable using standard methods, but we can use a binomial expansion to determine the leading order term of the integrand, which will tell us whether the integral converges or not. We see that 1 xp r 1 + p2 x2p+2 = 1 xp 1 + p2 x2p+2 1/2 = 1 xp 1 + 1 2 · p2 x2p+2 + O 1 x4p+4 = 1 xp + O 1 x3p+2 . Therefore, the leading order term in this integral is 1 xp , which we know converges for p > 1 and diverges for p ≤1 (from our study of p-integrals). So this surface of revolution has ﬁnite area if and only if p > 1. Turning to the volume of this solid, it is best to use slices perpendicular to the x-axis, which leads to discs whose radius is y: The volume element is therefore dV = πy 2 dx = π 1 xp 2 dx = π · 1 x2p dx. 170 Thus, the volume is V = Z dV = π Z ∞ x=1 1 x2p dx. We know this is convergent if 2p > 1, i.e. p > 1 2. So the volume of the solid is ﬁnite if p > 1 2. This leads to the surprising fact that for 1 2 < p ≤1, the volume of the solid is ﬁnite, but the surface area is inﬁnite. (Return) 3. Using the ﬁrst method requires some algebra. The curve becomes x = p 2y; 0 ≤y ≤8. So the area element is dS = 2πr dL = 2πr s 1 + dx dy 2 dy = 2π p 2y s 1 + 1 √2y 2 dy = 2π p 2y + 1 dy. So the surface area is S = Z dS = 2π Z 8 y=0 p 2y + 1 dy = 2π(2y + 1)3/2 · 1 3 8 y=0 = 2 3π 173/2 −1 . Using the second method, we have dS = 2πr dL = 2πx s 1 + dy dx dx = 2πx p 1 + x2 dx. 171 So the surface area is S = Z dS = π Z 4 x=0 2x p 1 + x2 dx = π(1 + x2)3/2 · 2 3 4 x=0 = 2 3π 173/2 −1 . So we get the answer with (perhaps) slightly less algebra involved. (Return) 172 37 Work Recall that work is the amount of energy required to perform some action. When the amount of force is constant, work is simply work = force × distance. For example, if a book weighing 22 Newtons (about 5 pounds) is lifted 2 meters, the total work done is 22N × 2m = 44J (J is the Joule, which equals one Newton-meter). Consider a situation where the force is not constant. For instance, if one were to lift a weight using a non-negligible rope, there is less rope being pulled up (and hence less force) as the weight goes further up. It is in situations like these that we need a better formula to compute work. 37.1 Work element Computing work when the force is not constant requires integration. As in previous sections, the ﬁrst step is to determine the work element dW, and then integrate: W = Z dW. Because work arises in a variety of situations, there is not one simple formula for the work element. For diﬀerent applications the work element will look diﬀerent. In some situations, it is best to consider a small movement dx, where the force can be thought of as constant for that small movement, which allows the work element to be expressed as dW = F · dx. Springs The force required to displace a spring varies with the displacement. The further the spring is stretched, the more resistant it becomes to being stretched further. Consider three types of springs: • Linear. A spring is linear if the force of resistance grows linearly with the displacement. That is, F(x) = κx. for some constant κ, which represents the stiﬀness of the spring. • Hard. A spring is hard if the force of resistance grows faster than linearly with the displacement: 173 F(x) = κx + O(x2). • Soft. A spring is soft if the force of resistance grows slower than linearly with the displacement: F(x) = κx −O(x2). Consider for any of these springs what the work element dW is. When the spring is stretched to x, the force of resistance is F(x). For the next inﬁnitesimal amount of stretching dx, the force can be presumed to be constant: (Stretching Spring Animated GIF) Therefore, the work element (i.e. the amount of work to stretch the spring the additional amount dx) is dW = F(x) dx. Example Compute the amount of work it takes to stretch a linear spring from rest (when x = 0) to x = a. (See Answer 1) Example Consider a nonlinear, soft spring which exerts a force of F(x) = 3x −x2 Newtons when the spring is stretched to x meters. Determine how much work is required to stretch the spring from 1 meter to 3 meters. (See Answer 2) Pulling up a rope In some situations, one must do a little work to determine what F(x) is, and then one can integrate, as in the above examples. Example Consider a rope which is 100 feet long and density 1 pound/foot. It hangs from a wall which is 50 feet high (so 50 feet of rope runs down the length of the wall and the remaining 50 feet is coiled at the bottom of the wall). How much work (in foot-pounds) is required to pull the rope to the top of the wall? 174 (See Answer 3) 37.2 Work element by slices In other situations, such as pumping liquid, digging a hole, or piling gravel, a fruitful method for determining the work involved is to consider a slice of the material which is being moved. Determining the weight of the slice, and multiplying by the distance the slice has to be lifted gives the amount of work required for that slice. That is precisely the work element. Integrating over all the slices in the object gives the total amount of work to move that object. Example Pumping Liquid Consider an inverted conical tank (so the tip of the cone points downward) with base radius 5 feet and height 10 feet. Water is pumped into the tank through a valve at the tip of the cone: How much work is required to ﬁll the tank with water? Leave the weight density of water as the constant ρ. (See Answer 4) Example Digging a Hole Consider two workers digging a hole. How deep should the ﬁrst worker dig so that each does the same amount of work? Let the weight density of the dirt be the constant ρ, the depth of the hole is D, and the cross-sectional area of the hole is the constant A (so we assume that the hole does not get any wider or narrower as the workers dig). (See Answer 5) Example Gravel Pyramid Compute the amount of work required to build a pyramid of gravel. Assume the gravel is inﬁnitesimal with weight density ρ, and that the pyramid has a square base of side length s, and height h: 175 (See Answer 6) Example Rope Revisited Consider the rope example from above, but this time suppose l total feet of rope are hanging from a h foot building, where l ≥h, and let ρ be the weight density of the rope. Compute the work required to lift the rope to the top of the building. For a diﬀerent perspective, this time, use a work element which equals the amount of work required to lift an inﬁnitesimal length of rope to the top of the building (this will depend on whether the inﬁnitesimal length of rope is hanging at the beginning or is part of the coil at the bottom of the building). Then integrate along the entire length of rope. (See Answer 7) 37.3 EXERCISES • Consider a conical tank of height 10m. The vertex of the cone is at the bottom, and the base of cone (which is at height 10m) has radius 2m. Let ρ denote the weight density of water. The water inside the tank has height 4m. How much work would it take to pull all the water to the top of the tank? 37.4 Answers to Selected Examples 1. As shown above, the work element (i.e. the amount of work to stretch the spring a short distance dx) is dW = F(x) dx = κx dx. It follows that W = Z dW = Z a x=0 κx dx = κx2 2 a x=0 = κa2 2 . (Return) 2. The work element is dW = F(x) dx = (3x −x2) dx. 176 Thus, the total work to stretch the spring from 1 meter to 3 meters is W = Z dW = Z 3 x=1 (3x −x2) dx = 3 2x2 −1 3x3 3 x=1 = 27 2 −9 − 3 2 −1 3 = 10 3 Joules. (Return) 3. As the ﬁrst 50 feet of rope are brought up, there is always precisely 50 feet of rope hanging from the building (because every foot of rope brought onto the top of the building is replaced by a rope which is coiled below). These 50 feet of rope weigh 50 lbs, so that is the force required to support them. If x denotes the amount of rope which has been taken onto the roof, then F(x) = 50; 0 ≤x ≤50. After the ﬁrst 50 feet of rope have been brought to the roof, there is now 50 feet of rope dangling with nothing left coiled below. Therefore, as we bring up these last 50 feet, there is less and less rope hanging, and so the weight of the rope (and hence the force we exert) is decreasing. It decreases linearly, since the rope has constant density. Each foot of rope we bring up decreases the weight by 1 lb, and so F(x) = 100 −x; 50 ≤x ≤100 (to see that this is right, note that it is linear and matches at the endpoints). We can graph the force as a function of the amount of rope we have brought up: Now, we can ﬁnd the work by integrating the work element W = Z dW = Z 100 x=0 F(x) dx. 177 Note that this is the area under the graph of the force (highlighted above), which is easier to compute than to do it algebraically. Splitting it into a square and a triangle, the area (and hence the work) is 50 lb × 50 ft + 1 250 lb × 50 ft = 3750 ft-lb. (Return) 4. Consider a slice of the water in the tank. Let x be the distance of the slice from the tip of the tank. That is, x is the distance that the slice of water has to be lifted. Let r be the radius of the slice: Above, we said that it is the weight of the slice multiplied by the distance the slice had to be moved. But the weight of a slice is just the volume of the slice times the density of the slice. Letting ρ denote the weight density of the substance (in this case water), we have dW = weight of slice × distanceslicetravels = ρ · dV · distance slice travels . In the problem at hand, we have dV = πr 2 dx, and the distance the slice is lifted is x, by the way we labeled our diagram. To ﬁnish, we must get r in terms of x, which requires a little bit of geometry. If we ﬂatten our cone and look at it from the side, we get similar triangles: 178 Therefore, r x = 5 10 = 1 2, and so r = x 2. Putting this together, we have dW = ρ · dV · distance slice travels = ρ(πr 2 dx)x = πρ1 4x3 dx. Note that x ranges from 0 to 10, so the work required to ﬁll the tank is W = Z dW = Z 10 x=0 πρ1 4x3 dx = πρ 4 x4 4 10 x=0 = 625πρ. (Return) 5. Let x be the distance down to to the layer of dirt currently being dug: This is convenient because this is the distance that the current slice of dirt has to be lifted to get out of the hole. The area of the slice of dirt is A, its thickness is dx, and the density is ρ, so we have dW = weight of slice × distance slice moves = (ρ · dV ) · x = (ρA dx) · x = ρAx dx. 179 Note that x varies from 0 to D as the hole gets dug. Thus, the total work required to dig the hole is W = Z dW = Z D x=0 ρAx dx = 1 2ρAD2. To ﬁnd the depth ˜ D where the work done is half, we set Z ˜ D x=0 ρAx dx = 1 2W = 1 4ρAD2. Computing the integral on the left, we ﬁnd 1 2ρA ˜ D2 = 1 4ρAD2. Solving for ˜ D gives ˜ D = 1 √ 2D. (Return) 6. If we think of building the pyramid slice by slice, let y be the distance from the base of the pyramid to the slice. This is convenient because this is the distance that the slice must be lifted to be put in place. Also, let x be the side length of the slice: Then using similar triangles, as shown on the right above, we ﬁnd that x s = h −y h . So we ﬁnd that x = h −y h s. 180 Thus, the volume of a slice is just the area x2 multiplied by the thickness dy, and so we have dW = ρdV y = ρ(x2 dy) y = ρ h −y h s 2 y dy = ρs2 h2 (h −y)2y dy. Because y ranges from 0 to h, we have W = Z dW = ρs2 h2 Z h y=0 (h −y)2y dy = ρs2 h2 Z h y=0 (h2y −2hy 2 + y 3) dy = ρs2 h2 1 2h2y 2 −2 3hy 3 + 1 4y 4 h y=0 = ρs2 h2 1 2h4 −2 3h4 + 1 4h4 = ρs2 h2 · 1 12h4 = ρs2h2 12 . (Return) 7. Let L be the distance along the rope of the inﬁnitesimal piece being considered: So L is the distance that the inﬁnitesimal piece must be lifted to get to the top of the building. The weight of the inﬁnitesimal piece is density multiplied by length, and so the work element for a piece of rope which is hanging is dW = ρL dL; 0 ≤L ≤h. 181 For a piece of rope which is part of the coil at the bottom, the distance it must be lifted is always h, so the work element there is dW = ρh dL; h ≤L ≤l. So the work can be computed by integrating these work elements over their respective ranges and then adding: W = Z dW = Z h L=0 ρL dL + Z l L=h ρh dL = ρ1 2L2 h L=0 + ρhL l L=h = ρh2 2 + ρh(l −h). Another way to think about this is to treat the coiled rope at the bottom of the wall as a single solid object. The rope in the coil has length l −h, and so its weight is ρ(l −h). The distance the coil (as a unit) must be lifted is h. It follows that the work to lift the coiled portion of the rope is ρh(l −h), the result of the second integral above. (Return) 182 38 Elements This module deals with various problems that can be modeled using integral calculus. As in the previous sections, the problem will be to ﬁnd the total accumulation of some quantity U, and the method will be to determine a slice of the quantity, the U element dU, and integrate. 38.1 Mass Mass of a rod Consider the problem of determining the mass of a rod. Suppose the rod’s density varies along the length of the rod (but the rod is uniform in cross section). Let ρ(x) denote the linear density (i.e. the mass per unit of length) of the rod at position x: Then the mass element dM is the density ρ(x) times the thickness of the slice dx, as shown above, and it follows that the mass of the rod is M = Z dM = Z L x=0 ρ(x)dx. Mass of the earth Consider the problem of ﬁnding the mass of the earth. Suppose the density of the earth ρ(r) is given as a function of the distance from the center of the earth. Assume that there are just three layers (inner core, outer core, and mantle) and that the density is constant within each layer. 183 What is the mass element in this case? It is important to note that in this example we are measuring the contribution of a spherical shell to the mass of the earth. This contribution is the volume of the spherical shell multiplied by the density of the shell. Mathematically, dM = ρ(r) · dV. Recalling that the surface area of a sphere of radius r is 4πr 2, we have that the volume element is dV = 4πr 2 dr, and so the mass element is dM = 4πρ(r)r 2 dr. Example Using the approximate graph of density above, estimate the mass of the earth. (See Answer 1) 38.2 Torque Imagine a rod of variable density which is attached to a hinge. The torque at the hinge depends not just on the weight of the rod but on the distribution of the weight. If there were just a mass-less rod with a single point mass, the torque would be Force × Distance . This can be used to determine the torque element dT by thinking of each slice of the rod as a point mass. What is the torque on such a slice? 184 First, the torque element is the distance from the hinge, x, times the force element dF (the force on the slice). The force element is the mass of the slice dM times the gravitational constant g. Finally, as in the previous example, the mass element dM = ρ(x) dx. Putting it all together, one ﬁnds dT = x · g · ρ(x) dx. Integrating this over the length of the rod gives the torque. 38.3 Hydrostatic force The next application is to compute the total force exerted by a tank of ﬂuid on a surface submerged in the tank, often called the hydrostatic force. For a tangible example, consider a large aquarium with a circular glass viewing window (see the diagram below). If the viewing window has radius r, and the top of the viewing window is at depth h, then the problem is to ﬁnd the total force of the water on the viewing window. As always, the method will be to ﬁnd the force element dF (the force on a small strip of the window), and then use integration to ﬁnd the total force. Recall that if pressure is constant across a surface, the force on the surface is area × pressure. Hydrostatic pressure is given by P = weight density of ﬂuid × depth. Note the units: N m3 · m = N m2 , which is the correct unit for pressure (force per unit of area). Since the density of the ﬂuid is assumed to be constant, the pressure only depends on the depth.Therefore, the most logical choice for the force element is a horizontal strip, since the depth, and hence the pressure, will be constant across the strip. Letting dA denote the area of the strip, we ﬁnd that the force element is given by dF = P dA = ρx dA, where ρ is the weight density of the ﬂuid and x is the depth of the strip. 185 Example Compute the total force exerted on the circular viewing window in the aquarium shown above. (See Answer 2) Example Compute the force on the endcap of a full cylindrical tank of radius R on its side. (See Answer 3) Example Consider a dam in the shape of a trapezoid with height h, top edge l1 and bottom edge l2. Find the total force exerted on the dam by the water: (See Answer 4) 38.4 Present value Consider the problem of determining the present value of some amount of money at a future time. Turning the problem around, ﬁrst consider the value of an initial amount of money P0 at a future time t. Assuming a constant 186 annual nominal interest rate r and continuous compounding, this problem was an example of exponential growth, and had solution P(t) = P0ert, where t is the time in years. Given some amount of money, P, at time t, ﬁnding its present value is a matter of solving P = P0ert for P0. In other words, solving for present value in this simple case is the same as ﬁnding the initial investment P0 which yields P after t years of continuous compounding interest. Solving this equation gives that the present value of a future amount P at time t is given by P0 = Pe−rt. Example Find the present value of $1000000 in 30 years, assuming an interest rate of r = .08. (See Answer 5) Now consider an income stream, say from a job. If I(t) is the rate of income at time t, what is the present value of that income stream? Let PV be the present value. Then consider the income earned over a small amount of time t years in the future: dI = I(t) dt (the income element). This small bit of income at time t contributes e−rtI(t)dt to the present value of the income stream. Thus the present value element is given by dPV = e−rtI(t)dt. Integrating this over the range of values of t (the time period of the income stream) gives the present value of that income stream. Example The Bigbucks lottery has an option of either a single lump sum payment today or an annuity which pays a constant amount each year for 20 years. Suppose the annuity pays $3 million a year (for 20 years), and that the interest rate will remain steady at r = .05. What is the fair lump sum payout today? (See Answer 6) 38.5 EXERCISES • Consider a dam with the shape of an isosceles triangle. The base of the triangle, which is parallel to the ground, is 5m long, and the height of the triangle is 10m. The weight density of water is given by ρ. Compute the force exerted on the dam by water. 38.6 Answers to Selected Examples 1. Note that our volume is being measured in cubic kilometers, but the density ρ(r) is in grams per cubic centimeter. We need a conversion factor C to make sure the units come out correctly. A little unit conversion gives us that C = 1 g cm3 = 1012 kg km3 . So we need to multiply by this so that the units are correct (and the ﬁnal answer will be in kilograms). 187 Splitting the integral based on the values of r for which ρ(r) is constant, we ﬁnd M = Z dM = C Z 6400 r=0 4πr 2ρ(r) dr = 4πC Z 1200 r=0 r 2ρ(r) dr + Z 3400 r=1200 r 2ρ(r) dr + Z 6400 r=3400 r 2ρ(r) dr = 4πC 13 · r 3 3 1200 0 + 10 · r 3 3 3400 1200 + 5 · r 3 3 6400 3400 ! ≈6.3 · 1024 kilograms According to Wolfram Alpha, the mass of the earth is approximately 5.97 · 1024 kilograms, so our rough estimate is not too far oﬀ. (Return) 2. As mentioned above, we will use horizontal strips of the window as the area element. The force element is the amount of force on that strip of the window. Let x be the distance from the center of the window to the horizontal strip. Let up be negative, down be positive (so the top of the window is x = −r and the bottom of the window is x = r: Then the depth of the strip is h+r +x, and the area of the strip is 2 √ r 2 −x2 dx. Thus the force element in this example is dF = (h + r + x)ρ · 2 p r 2 −x2 dx 188 where ρ is the weight density of water. So F = Z dF = 2ρ Z r −r (h + r + x) p r 2 −x2 dx = 2ρ Z r −r (h + r) p r 2 −x2 dx + 2ρ Z r −r x p r 2 −x2 dx. Now, notice that x √ r 2 −x2 dx is an odd function, so its integral from −r to r is 0. Thus F = 2ρ(h + r) Z r −r p r 2 −x2 dx = 2ρ(h + r)πr 2 2 = ρ(h + r)πr 2, since R √ r 2 −x2 dx gives half the area of a circle of radius r. It is worth observing that with a very symmetric window such as the circle in this example, one can take the area of the window, πr 2, and multiply by the pressure at the center of the window ρ(h + r), to ﬁnd the hydrostatic force: F = ρπr 2(h + r). The reason this works is that the pressure on a horizontal strip above the center of the window averages with the pressure on the strip’s mirror image below the center to give the pressure at the center of the window. (Return) 3. Using the knowledge gleaned from the previous example, we can take the area of the endcap, πR2, and multiply by the hydrostatic pressure at the center of the endcap, which is ρR, to ﬁnd that the force is F = πR2 · ρR = ρπR3. We could also note that this is really a special case of the aquarium window example above, by setting h = 0 in that example. (Return) 4. As above, the force element dF is the force exerted on a horizontal strip. Let x be the distance of the horizontal strip from the top of the dam, and l(x) be the length of the strip 189 Since the shape is a trapezoid, l(x) is a linear function of x, and from the top and the bottom of the dam, one ﬁnds that l(0) = l1 and l(h) = l2. It follows from the slope intercept form of a line that l(x) = l1 + l2−l1 h x. So the force acting on the strip is dF = ρx dA, where ρ is the weight density of the water, x is the depth of the strip, and dA = (l1 + l2−l1 h x) dx is the area of the strip. Putting it all together, one ﬁnds F = Z dF = Z h 0 ρx l1 + l2 −l1 h x dx = ρ l1x2 2 + l2 −l1 3h x3 h 0 = ρ l1h2 2 + l2 −l1 3h h3 = ρh2 6 (l1 + 2l2). (Return) 5. From the above equation one ﬁnds that P0 = Pe−rt = 1000000e(−.08)·30 ≈90717. (Return) 6. The income stream I(t) is constant at 3 · 106. Thus, PV = Z dPV = Z 20 t=0 e−rtI(t)dt = 3 · 106 Z 20 t=0 e−.05tdt = 3 · 106 1 −.05e−.05t 20 t=0 ! = 3 · 106 · (−20)(e−1 −1), which is approximately $38 million. (Return) 190 39 Averages Consider the problem of ﬁnding the average test score in a class of 100 students. The answer is to add up all the scores and divide by 100. But what would happen if there were inﬁnitely many students? This module deals with the problem of ﬁnding the average value of a function. 39.1 Average value of a function The deﬁnition of the average value of a function f (x) over the interval [a, b], denoted f , is f = R b a f (x) dx b −a . One way to interpret the average value is to ﬁnd the rectangle of length b −a whose area equals the area under the curve f over the interval [a, b]. The height of this rectangle is f . Put another way, f is the height of the horizontal line such that the area above the line and below f (x) equals the area which is below the line and above f (x). These areas are shown in red and blue, respectively, in the following diagram: A better formulation of the average value, which will be useful in other situations and higher dimensions, is f = R b x=a f dx R b x=a dx . This emphasizes that the average value over a region is the integral of the function over the region divided by the volume of that region (in this case, the 1-dimensional volume is just the length of the interval). This generalizes nicely to higher dimensions. 191 If we go down a dimension to the discrete average, if fi denotes the ith data point out of n, then the average value of the data is f = Pn i=1 fi n = Pn i=1 fi Pn i=1 1 . This shares a common feature with the earlier formula for average value. Namely, it is the sum (integral) of the function values over a range of inputs divided by the sum (integral) of 1 over that range of inputs. Example Compute the average value of sin2 x over the interval [0, 2π]. (See Answer 1) Example Compute the average of xn and ex over 0 ≤x ≤T. Compute the average of ln x over 1 ≤x ≤T. (See Answer 2) Example Suppose we are given the density function ρ(r) for the density of the earth at a distance r from the center. Find a formula for the average density of the earth, but do not try to evaluate the integral. (See Answer 3) 39.2 Root mean square There is another type of average of a function called the root mean square. The root mean square of f , denoted fRMS is deﬁned by fRMS = p f 2. So the root mean square is the square root of the average value of the square of the function. This is a useful metric when the average value of f is uninteresting. Example Compute and compare the average value and the root mean square of f (x) = sin x on the interval [0, 2π]. (See Answer 4) 39.3 EXERCISES • Consider the polar function f (θ) = cos2(θ). Compute the average value of f from θ = 0 to θ = 2π. 192 39.4 Answers to Selected Exercises 1. From the deﬁnition, f = R 2π 0 sin2 xdx 2π = 1 2π Z 2π 0 1 2 (1 −cos(2x)) dx = 1 2π x 2 −1 4 sin(2x) 2π 0 = 1 2π · 2π 2 = 1 2. (Return) 2. For f (x) = xn, one ﬁnds f = 1 T Z T 0 xn dx = 1 T xn+1 n + 1 T 0 = T n n + 1. For f (x) = ex, the average value is f = 1 T Z T 0 ex dx = eT −1 T For f (x) = ln x, recalling the integral using integration by parts, one ﬁnds f = 1 T −1 Z T 1 ln x dx = 1 T −1(x ln x −x) T 1 = 1 T −1(T ln T −T + 1). (Return) 3. Note that we cannot simply integrate the density function and divide by the radius of the earth, for the same reason that we could not integrate the density function to ﬁnd the mass of the earth in the previous module. One way to logically think about it is to note that the average density times the volume of the earth should give the mass of the earth. That is, ρ · V = M. 193 Remember that when we found the mass of the earth, we had dM = ρ dV , where the volume element dV is a spherical shell. So we can write ρ = M V = R ρ dV R dV . Then, remembering that the volume of the spherical shell (i.e. the volume element) is 4πr 2 dr, we have ρ = Z R r=0 4πr 2ρ(r) dr Z R r=0 4πr 2 dr . (Return) 4. The average value of sin x is f = Z 2π x=0 sin x dx 2π = 1 2π (−cos x) 2π x=0 = 1 2π (−1 −(−1)) = 0. Using the result of an example from above, the root mean square of sin x is fRMS = v u u u t Z 2π x=0 sin2 x dx 2π = r 1 2 = 1 √ 2. (Return) 194 40 Centroids And Centers Of Mass The motivation for this module are the questions: • what is the average of several locations (e.g. cities on a map)? • what is the average of an entire region? The centroid and center of mass give answers to these questions. The formulas for the centroid and the center of mass of a region in the plane seem somewhat mysterious for their apparent lack of symmetry. So before giving the formulas, a brief aside is helpful. 40.1 The area element revisited In future courses, the area element of a region will not be a strip of area but a small rectangle with width dx and height dy: The area of the region, then, is the limit of the sum of the areas of all these small rectangles as the rectangles get inﬁnitely small. The notation used to express this is called a double integral, written Area = ZZ R dx dy. Think of the double integral as a nested integral: RR dx dy = R ( R dx) dy. The inner integral is performed ﬁrst, with respect to x (since the dx is left of the dy). Then the result is integrated with respect to y. Conceptually, 195 the inner integral is adding up the contribution of a row of boxes, and then the outer integral is adding up the rows: Double integrals can be computed in the other order too: RR dy dx. First the inner integral is performed with respect to y, which adds up the contribution of a column of boxes. Then the outer integral adds up the contribution of the columns: Example Express the area of the region bounded by the curves y = x2 −4x + 5 and y = x + 1 as a double integral and evaluate the integral. (See Answer 1) 40.2 Centroid The centroid of a region R in the plane is deﬁned to be the point (x, y), where x is the average x-coordinate of R and y is the average y-coordinate of R. One interpretation is that if the region were cut out of a sheet of uniform density metal and a pin were placed at its centroid, the region would balance on the pin. 196 The centroid is best expressed mathematically using double integrals: x = RR R x dx dy RR R dx dy y = RR R y dx dy RR R dx dy . Suppose the region R is bounded above by the curve y = f (x) and below by the curve y = g(x), and the intersection points are at x = a and x = b. Then integration is easier in the dy dx order, and the centroid can be written more explicitly as Centroid of a region The centroid of the region bounded above by y = f (x) and below by y = g(x) is given by x = R b a R f (x) g(x) xdydx R b a R f (x) g(x) dydx = R b a x(f (x) −g(x))dx R b a (f (x) −g(x))dx . Similarly, y = R b a R f (x) g(x) y dy dx R b a R f (x) g(x) dy dx = R b a 1 2(f (x)2 −g(x)2) dx R b a (f (x) −g(x)) dx . Note that the denominator in each case is the area of the region. 197 Example Find the centroid of a triangle with vertices at (a, 0), (b, 0), and (0, c). (See Answer 2) Example Compute the centroid of the upper half circle of radius R. (See Answer 3) Example Compute the centroid of the quarter circle of radius R: (See Answer 4) Symmetry It is important to note that centroids respect symmetry. What that means is that if there is an axis of symmetry (i.e. a line where if we reﬂect the region about the line we get the same region back), then the centroid must lie on the axis of symmetry. If there is more than one axis of symmetry, then the centroid will lie at the intersection of these axes: 198 40.3 Center of mass Now consider a region R of the plane cut from a sheet of metal of variable density ρ(x, y). Again, the problem is to ﬁnd the balancing point (x, y), but in this context it is called the center of mass. Again, it is expressed as a double integral: x = RR R ρ(x, y)x dx dy RR R ρ(x, y) dx dy y = RR R ρ(x, y)y dx dy RR R ρ(x, y) dx dy . The only diﬀerence between these and the centroid formulas is that instead of the area element dA = dx dy, the mass element dM = ρ(x, y) dx dy is used (multiplying the area element by the density of that point gives the mass contributed by that small rectangle). Indeed, if the density is constant, then ρ(x, y) = ρ factors out of both the numerator and denominator and cancel, leaving the formula for centroid. Note that the denominator for both x and y is the mass of the region. Example Compute the center of mass of the region bounded above by y = 4x −x2 and below by the x-axis, where the density function is given by ρ(x, y) = 2x: 199 (See Answer 5) 40.4 Centroids using point masses Given a complex region which consists of the union of simpler regions, there is a method for ﬁnding the centroid: 1. Find the centroid of each simple region. 2. Replace each region with a point mass at its centroid, where the mass is the area of the region. 3. Find the centroid of these point masses (this is done by taking a weighted average of their x and y coordinates). (Centroids and Point Masses Animated GIF) This is easiest to see with an example: Example Find the centroid of a region consisting of a rectangle of width 2R and height H which has a semicircle of radius R on one end: (See Answer 6) 40.5 Application: Pappus’ theorem One application of the centroid is known as Pappus’ theorem, after the Greek mathematician Pappus of Alexan-dria. It uses the centroid to ﬁnd the volume and surface area of a solid of revolution. 200 Pappus’ theorem Consider the solid which results from rotating the plane region R about the axis L. The volume of this solid is equal to the area of R times the distance the centroid travels (as it gets revolved around the axis). The surface area of the solid is equal to the perimeter of R times the distance the centroid travels. Example Find the volume and surface area of a torus (i.e. a doughnut) with cross sectional radius r and main radius R: (See Answer 7) 40.6 EXERCISES • Compute the area of region bounded by curves x = (y −2)2 + 2 and y = x −2 using double integrals. • Consider the region under the graph y = x2, above the x-axis, from x = 0 to x = 1. Let S be the solid obtained by revolving this region about the y-axis. Compute the average height (average y-coordinate) of S. • Let R1 denote the region inside the triangle with vertices at (0,1), (-2,0), (0,-1). Given a unit circle centered at the origin, let R2 denote the region inside the semicircle for x ≥0. Let R denote the union R1 and R2. compute the centroid of R. 40.7 Answers to Selected Exercises 1. 201 The easier order of integration is dy dx because every vertical strip is bounded on top by y = x + 1 and bounded below by y = x2 −4x + 5; whereas a horizontal strip would sometimes be bounded on the left by y = x + 1, and other times be bounded by y = x2 −4x + 5. Setting the curves equal gives the intersections at x = 1 and x = 4. So the area can be found by computing Z x=4 x=1 Z y=x+1 y=x2−4x+5 dy dx = Z x=4 x=1 y x+1 x2−4x+5 ! dx = Z x=4 x=1 (x + 1 −(x2 −4x + 5)) dx = Z x=4 x=1 (−x2 + 5x −4) dx = −x3 3 + 5 2x2 −4x 4 1 = 9 2. (Return) 2. The easier order of integration is dx dy because a horizontal strip is always bounded on the left by x = −b c y + b and on the right by x = −a c y + a (see the diagram below). So one ﬁnds that x = R y=c y=0 R x= −a c y+a x= −b c y+b x dx dy R y=c y=0 R x= −a c y+a x= −b c y+b dx dy = R y=c y=0 R x= −a c y+a x= −b c y+b x dx dy Area Noting that the area of the triangle is 1 2(a −b)c, one ﬁnds x = 2 (a −b)c Z y=c y=0 Z x= −a c y+a x= −b c y+b x dx dy = 2 (a −b)c Z y=c y=0 1 2 (−a c y + a)2 −(−b c y + b)2 dy = 1 (a −b)c · 1 3 (−a c y + a)3 −c a −(−b c y + b)3 −c b c 0 = 1 3(a −b)c (a2c −b2c) = 1 3(a −b)c c(a + b)(a −b) = 1 3(a + b). A similar computation gives that y = c 3. 202 More generally, the centroid of a triangle with coordinates (x0, y0), (x1, y1), and (x2, y2) is (x, y) = x0 + x1 + x2 3 , y0 + y1 + y2 3 . In other words, the centroid of a triangle is the average of the x coordinates and the average of the y coordinates. (Return) 3. By the symmetry about the y-axis, the x-coordinate of the centroid is 0. To ﬁnd the y-coordinate, note that the equation of the curve is y = √ R2 −x2. Also, note that the area of the region is 1 2πR2. Thus, y = 2 πR2 Z R x=−R 1 2( p R2 −x2)2 dx = 1 πR2 R2x −1 3x3 R x=−R = 1 πR2 · 4R3 3 = 4R 3π . (Return) 4. We know that the area of the region is 1 4πR2. So we have that x = 1 A Z x(f (x) −g(x)) dx = 4 πR2 Z R x=0 x( p R2 −x2 −0) dx. Making a substitution of u = R2 −x2 du = −2x dx 203 gives 4 πR2 Z R x=0 x p R2 −x2 dx = 4 πR2 Z 0 u=R2 −1 2 √u du = −2 πR2 2 3u3/2 0 u=R2 = 2 πR2 · 2 3R3 = 4R 3π . Because the region is symmetric about the line y = x, we predict that y = 4R 3π as well. We can verify this by integrating: y = 1 A Z 1 2(f (x)2 −g(x)2) dx = 2 πR2 Z R x=0 (R2 −x2) dx = 2 πR2 R2x −1 3x3 R x=0 = 2 πR2 R3 −1 3R3 = 2 πR2 · 2 3R3 = 4R 3π , as claimed. (Return) 5. Setting y = 0, we ﬁnd that the curve intersects the x-axis at x = 0 and x = 4. First, we compute the mass of the region, which is the denominator for both x and y. It is easier to integrate in the dy dx order, so we will do that here, and in the integrals that follow. M = ZZ R ρ(x, y) dy dx = Z 4 x=0 Z 4x−x2 y=0 2x dy dx = Z 4 x=0 2xy 4x−x2 y=0 ! dx = Z 4 x=0 2x(4x −x2) dx = Z 4 x=0 (8x2 −2x3) dx = 8 3x3 −1 2x4 4 x=0 = 512 3 −128 = 128 3 . 204 So to compute x, we ﬁnd x = 1 M ZZ R ρ(x, y)x dy dx = 3 128 Z 4 x=0 Z 4x−x2 y=0 (2x)x dy dx = 3 128 Z 4 x=0 2x2y 4x−x2 y=0 ! dx = 3 128 Z 4 x=0 2x2(4x −x2) dx = 3 128 2x4 −2 5x5 4 x=0 = 3 128512 −2048 5 = 3 128 · 512 5 = 12 5 . Similarly, y = 1 M ZZ R ρ(x, y)y dy dx = 3 128 Z 4 x=0 Z 4x−x2 y=0 (2x)y dy dx = 3 128 Z 4 x=0 xy 2 4x−x2 y=0 dx = 3 128 Z 4 x=0 x(4x −x2)2 dx = 3 128 Z 4 x=0 (16x3 −8x4 + x5) dx = 3 128 4x4 −8 5x5 + 1 6x6 4 x=0 = 3 128 · 1024 1 −8 5 + 2 3 = 8 5. (Return) 6. From an earlier example, the centroid of the semicircle is (0, 4R 3π ), and the weight (the area of the semicircle) is 1 2πR2. The rectangle is symmetric, so its centroid (as it is drawn in the coordinate plane) is (0, −H 2 ), and its weight is 2RH: 205 By symmetry, x = 0. Taking the weighted average of the y-coordinates of the points gives y = 1 2πR2 · 4R 3π + 2RH · −H 2 1 2πR2 + 2RH = 2 3R3 −H2R 1 2πR2 + 2RH = 4R2 −6H2 3πR + 12H . We can check that this is reasonable by noting that if H = 0 we get the y-coordinate of the centroid of the semicircle, and when R = 0 we get the y-coordinate of the centroid of the line segment from (0, 0) to (0, −H). (Return) 7. Here, the region being rotated is a circle, which is easy to work with because a circle’s centroid is just its center. For convenience, center the circle at (R, 0) and revolve around the y-axis. Then the distance which the centroid travels is 2πR (the path of the centroid is just a circle of radius R). 206 Therefore the surface area of the torus is Surface area = Perimeter · Centroid travel distance = (2πr) · (2πR) = 4π2r · R. And the volume of the torus is Volume = Area · Centroid travel distance = (πr 2) · (2πR) = 2π2r 2R. (Return) 207 41 Moments And Gyrations This module deals with the moment of inertia and the radius of gyration, which are two properties of an object with physical interpretations. 41.1 Moment of inertia The moment of inertia of an object, usually denoted I, measures the object’s resistance to rotation about an axis. To get an intuitive understanding of moment of inertia consider swinging a hammer by its handle (higher moment of inertia, harder to swing) versus swinging a hammer by its head (lower moment of inertia, easier to swing). So moment of inertia depends on both the object being rotated and the axis about which it is being rotated. (Hammer Animated GIF) Consider ﬁrst a particle of mass. The bigger the mass, the more resistant it will be to rotation about an axis. Similarly, the further the particle is from the axis, the more resistant it will be to rotation. For a point mass, the moment of inertia is given by I = r 2 M, where r is the distance of the particle from the axis of rotation, and M is the mass of the particle: (Particle Animated GIF) The next question is how to calculate the moment of inertia when all the mass is not at a single point. As in previous modules, the method will be to break the object into slices of mass, and consider the contribution of each slice to the moment of inertia: 208 Each slice can be thought of as an individual particle of mass which contributes to the moment of inertia. The contribution of the slice becomes the moment of inertia element dI: dI = r 2 dM. Example Consider a solid disc of radius R and constant density ρ rotated about its central vertical axis: (Disk Rotating Around Diameter Animated GIF) Compute its moment of inertia. (See Answer 1) Example Consider a solid disc of radius R and constant density ρ rotated about its center: (Disk Rotating Around Center Animated GIF) Compute its moment of inertia. (See Answer 2) Example Consider a rectangle of length l and height h. Compute the moment of inertia about the vertical axis through its center. Then compute the moment of inertia about the horizontal axis through its center. Hint: use symmetry to ﬁnd the second answer from the ﬁrst. (See Answer 3) 209 41.2 Radius of gyration Another property of an object, radius of gyration, denoted Rg, can be expressed in terms of the moment of inertia. Imagine replacing the object being rotated about an axis by a single point mass being rotated about that same axis. The radius of gyration is the radius at which the point mass has the same moment of inertia as the object. More speciﬁcally, I = MR2 g, and solving for Rg gives Rg = r I M . Note that because I = Z r 2 dM we can write Rg = sR r 2 dM R dM = p r 2 = rRMS So the radius of gyration is really the root mean square of the radius. 41.3 Higher mass moments In the centroid module, we computed R x dM as part of computing the x-coordinate of the center of mass, x.The moment of inertia I from this module is given by R x2 dM. These are respectively known as the ﬁrst mass moment and the second mass moment (ﬁrst and second referring to the powers of x). There are higher mass moments: R xn dM, for n ≥3, as well as the lower mass moment R x0dM, which is just mass. These moments each give more information about how the mass of the object is distributed. This is similar, in a sense, to how knowledge of the derivative of a function at a point leads to an approximation of the function using the Taylor series. The more derivatives one knows, the better the approximation. A logical question, then, is if one knows all the mass moments of an object, can one perfectly describe the distribution of mass? 41.4 Additivity of moments One nice feature of moments is that, being integrals, they are additive. This means that a complex region can be split into simpler regions for which we already know the moment of inertia, and these moments can be added to ﬁnd the moment of inertia for the entire region. Example Compute the moment of inertia for each of the following ﬁgures about a horizontal axis through their centers. 210 Which has the greater moment of inertia, the I-shaped ﬁgure or the H-shaped ﬁgure? (See Answer 4) 41.5 EXERCISES • Consider a right triangle with vertices at (0,0), (5,0), (0,10). Consider rotating the triangle about the y-axis. The density is given by ρ(x, y) = x. Compute the moment of inertia. Compute the radius of gyration. 211 41.6 Answers to Selected Examples 1. Because the distance to the axis is part of the inertia element, a good area element to use is a vertical rectangle, where every point has the same distance to the center axis. Let x be the distance from the central axis to the rectangle (thus, r = x): The area of this rectangle, as has been computed several times previously, is dA = 2 √ R2 −x2 dx. Then the mass element dM = ρ dA, and it follows that dI = r 2 dM = 2x2ρ p R2 −x2 dx. so integrating the inertia element gives I = Z dI = Z R x=−R 2ρx2p R2 −x2 dx = 4ρ Z R x=0 x2p R2 −x2 dx (using the fact that the integrand is an even function allows the ﬁnal step). Now the substitution x = R sin θ, and some of the trig integral methods gives the answer ρπ 4 R4, which can also be written 1 4MR2, where M = πR2ρ is the mass of the disc. (Return) 2. In this example, a good area element to use is a ring (also called an annulus), because every point in a ring has the same distance to the origin, which is the axis of rotation (one can imagine the axis sticking out of the page perpendicular to the center of the disc): 212 As before, dM = ρ dA. In this case, dA is the area of the ring, which is 2πr dr (the circumference of the ring times the width of the ring). It follows that I = Z dI = Z R r=0 r 2ρ2πr dr = 2πρ Z R r=0 r 3 dr = 2πρr 4 4 R r=0 = πρR4 2 . This can be expressed as 1 2MR2, where M is again the mass of the disc. Note that the answer in this example is twice that of the previous example. This can be explained (using the answer from the previous example) by noting that r 2 = x2 + y 2 in this example. Therefore, I = Z r 2 dM = Z (x2 + y 2) dM = Z x2 dM + Z y 2 dM, and these two integrals are, respectively, the moment of inertia about a vertical axis (from the previous example) and the moment of inertia about a horizontal axis. By symmetry, these are equal, which explains why this answer is twice the answer of the previous example. (Return) 3. Center the rectangle at the origin. About the vertical axis, it is again best to use vertical rectangles. Let r denote the distance of this rectangle from the y-axis: 213 Then r = x, and dM = ρh dx. Thus I = Z dI = ρh Z l/2 x=−l/2 x2 dx = ρhx3 3 l/2 x=−l/2 = 1 12ρhl3 = 1 12Ml2, where M = ρlh is the mass of the rectangle. By symmetry, the moment of inertia about a horizontal axis through the center is 1 12Mh2. (Return) 4. For the ﬁrst ﬁgure, we can divide it into two rectangles (in light blue) and a square which are all being rotated about their horizontal center axis: 214 From the above example, we know that the moment of inertia for a rectangle about its horizontal axis is 1 12Mh2 = 1 12lh3, where l is the length and h is the height of the rectangle. So for each of the tall rectangles we have I = 1 12a3 a−b 2 and for the square in the middle we have I = 1 12b4. Putting it together, we have the moment of inertia for the entire region is I = 2 · 1 12a3 a −b 2 + 1 12b3b = 1 12(a4 −a3b) + 1 12b4 = 1 12 a4 + b4 −a3b . For the other region, we cannot divide it up into rectangles in the same exact way, because we do not know the moment of inertia for a rectangle rotated about an axis other than one through its center. Instead, we can take the entire square of side length a, and compute its moment of inertia. Then we can subtract oﬀthe inertia for the small rectangles we do not want to include, shown in red: 215 Again, using the fact from the previous example, the moment for the whole square is 1 12a4, and the moment for each of the smaller rectangles (which we will subtract) is 1 12 · a−b 2 · b3, so the moment of inertia for the whole region is I = 1 12a4 −2 · 1 12 · a −b 2 · b3 = 1 12a4 −1 12(ab3 −b4) = 1 12(a4 + b4 −ab3). So the I-shaped ﬁgure has the greater moment of inertia. This is important when considering whether to use an H-beam or and I-beam in construction. According to a fact mentioned in higher derivatives, the deﬂection u(x) (the amount the beam sags at location x) satisﬁes the equation EI d4u dx4 = q(x), where E is the elasticity of the material (a constant), and q(x) is a static load at location x along the beam. Because the I-beam has the greater moment of inertia, it follows that their deﬂection will be less, and so I-beams are more common in building construction. (Return) 216 42 Fair Probability Probability is the study of the likelihood of certain events occurring in a random experiment. A simple example is a coin ﬂip. There are two outcomes: heads (H) or tails (T). If the coin is fair, then the probability of each outcome is 1 2, written P(H) = P(T) = 1 2. Another example is a roll of a standard die. There are six outcomes: 1 through 6. If the die is fair then the probability of each outcome is 1 6. In these types of problems, one can ﬁnd the probability of an event occurring by counting the number of desired outcomes and dividing by the total number of outcomes. Example What is the probability that a pair of dice sums to seven or eleven? (See Answer 1) Example Alice and Bob play a game where they take turns ﬂipping a fair coin, with Alice going ﬁrst. The ﬁrst player to get heads wins. What is the probability that Alice wins? Hint: ﬁnd the probability that Alice wins on her ﬁrst ﬂip, and the probability that she wins on her second ﬂip, and the probability that she wins on her third ﬂip, etc. Add up all these (inﬁnitely many) probabilities to ﬁnd the probability that she wins. Second hint: For Alice to win on her second ﬂip, it means that both Alice and Bob got tails on their respective ﬁrst ﬂips (otherwise the game would have ended in the ﬁrst round). So the probability of Alice winning on her second ﬂip is P(A got tails) · P(B got tails) · P(A got heads) = 1 2 · 1 2 · 1 2 = 1 2 3 . (See Answer 2) 42.1 Uniform distribution The above examples refer to a fair coin and a fair die. A discrete experiment (i.e. the possible outcomes can be listed) is said to have the uniform distribution if the experiment is fair in the sense that every outcome is equally likely. What about experiments which are not discrete? For instance, a spinner gives a point along the circumference of a circle, and the individual points of the circle cannot be enumerated. Throwing a dart at a circular dartboard likewise has as many outcomes as there are points in the interior of the disk. What does it mean for such an 217 experiment to be fair, i.e., what does the uniform distribution mean in an experiment with continuous outcomes? To answer this question, consider the probability of a range of outcomes of the experiment. So, for instance, what is the probability of the spinner landing in the ﬁrst quarter of the circle? If the experiment is fair, then this probability should be the same as landing in any other quarter of the circle: 1 4: (Spinner Animated GIF) Thus an experiment is fair (i.e. has the uniform distribution) if for any set of outcomes D, P(D) = volume of D total volume of all outcomes. Here “volume” depends on the dimension of the experiment. For instance, the spinner has dimension 1 (where volume is really just the length) since any point on the circumference can be speciﬁed by a single value (say, the angle of the arrow relative to the positive x-axis). So a spinner is considered fair if the probability of the arrow landing in a certain range along the circumference equals the length of that range divided by the total circumference of the circle. Length Example Find the probability that a randomly chosen angle θ has sin θ > 1 2? (See Answer 3) Example Find the probability that a randomly chosen angle θ has tan θ > 0. (See Answer 4) Area In two dimensions, volume is really area, and so when computing the probability that a randomly chosen point in a region R in the plane lies within the region D, we have P = Area of D Area of R . 218 Example A dartboard is circular with radius 9 inches: The bullseye is a small circle at the center of the board. Find the radius of the bullseye so that the probability of hitting it is 1 100 (assuming a throw hits the board uniformly at random). (See Answer 5) Example Find the probability that a randomly chosen point in a square lies within the circle inscribed in the square: (See Answer 6) 219 Example Find the probability that a randomly chosen point in a circle lies in the equilateral triangle inscribed in the circle: Hint: the area of an equilateral triangle of side length s is A = s2√ 3 4 . (See Answer 7) There are some probability problems that do not seem geometric in nature but can be solved by graphing the possible outcomes and taking the ratio of the areas. Example Xander and Yolanda want to meet up to study calculus. Each friend will arrive at the library at some random time between 5 pm and 6pm, wait 20 minutes for the other person, and then leave if the other person does not arrive in that time. Find the probability that the friends successfully meet up. Hint: Let x be the number of minutes after 5 pm that Xander arrives and y be the number of minutes after 5 pm that Yolanda arrives. Now plot the possible arrival times as a region in the plane and determine the region which corresponds to them successfully meeting up. (See Answer 8) Volume Finally, in dimension 3, volume is volume as we traditionally know it. In this case, we imagine picking a point from within a 3D region and know the probability that the point lies within some subset of that region. Example Find the probability that a randomly chosen point from within a cube lies within the inscribed sphere: 220 (See Answer 9) Example What is the probability that a randomly chosen point in a ball lies within 10% of the boundary (as measured by radius)? (See Answer 10) 42.2 Buﬀon needle problem The Buﬀon needle problem, named after the Count of Buﬀon, asks for the probability that a needle of length l, dropped uniformly at random onto a sheet with parallel lines spaced l units apart, will cross a line. To simplify the problem, consider two parameters which determine whether the needle crosses: 1. h, the distance from the left tip of the needle to the next line to its right 2. θ, the angle that the needle makes with a vertical line: 221 Note that 0 ≤h ≤l and 0 ≤θ ≤π. Now, for what values of h and θ is there a crossing? Note that by right triangle trigonometry, the horizontal distance from the left end of the needle to the right end of the needle is l sin θ: Thus, there is a crossing if h ≤l sin θ, and there is no crossing if h > l sin θ. Graphing this inequality shows that the region below the curve (shown in purple) is where a crossing occurs. The region above the curve is where a crossing does not occur. 222 Dropping a needle at random is like randomly picking a point in this rectangle. Thus, the probability of a random needle creating a crossing equals the probability of randomly picking a point below the curve in the above rectangle. That probability is given by dividing the area under the curve by the area of the entire rectangle. P(crossing) = R π 0 l sin θ dθ lπ = 1 π −cos θ π 0 = 2 π . 42.3 Answers to Selected Examples 1. By listing the desired outcomes, one ﬁnds that (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) are the possible pairings which give 7, and (6, 5) and (5, 6) are the possible pairings which give 11. So there are 8 desired outcomes. The total number of outcomes is 6 × 6 (six outcomes for the ﬁrst die paired with each of the six outcomes for the other die). So the probability is # desired # total = 8 36 = 2 9. (Return) 2. Proceeding as the hint suggests, we look for a pattern. P(A wins on 1st ﬂip) = P(A gets heads) = 1 2. And then P(A wins on 2nd ﬂip) = P(A gets tails) · P(B gets tails) · P(A gets heads) = 1 2 · 1 2 · 1 2 = 1 2 3 . Next, P(A wins on 3rd ﬂip) = P(A gets tails) · P(B gets tails) · P(A gets tails) · P(B gets tails) · P(A gets heads) = 1 2 · 1 2 · 1 2 · 1 2 · 1 2 = 1 2 5 . In general, for Alice to win on the nth ﬂip, she must get a head on that ﬂip, and both Alice and Bob must have gotten tails on each of their previous n −1 ﬂips. Thus, there are a total of 2(n −1) + 1 = 2n −1 coin ﬂips that must come out in a precise way, and the probability of each of these is 1 2, so we have P(A wins on nth ﬂip) = 1 2 2n−1 . 223 Adding these up for all n, and using the geometric series, gives P(A wins) = ∞ X n=1 1 2 2n−1 = 1 2 + 1 2 3 + 1 2 5 + · · · = 1 2 · 1 + 1 2 2 + 1 2 4 + · · · ! = 1 2 · 1 + 1 4 + 1 4 2 + · · · ! = 1 2 · 1 1 −1/4 = 2 3. (Return) 3. We can visualize the sine of the angle by considering a unit circle, and noting that sine is the y-coordinate of a point on the circle: Then one ﬁnds that the angles for which sin θ > 1 2 are π 6 < θ < 5π 6 . The length of this portion of the circumference of the circle is 4π 6 , and so the probability of a random angle θ satisfying sin θ > 1 2 is P = 4π/6 2π = 1 3. 224 (Return) 4. Note that tangent is positive when sine and cosine have the same sign, i.e. if sine and cosine are both positive or if sine and cosine are both negative. This corresponds to the ﬁrst and third quadrants of the unit circle: The length of each these arcs is π 2 , and so the probability that tan θ > 0 is P = 2 · π/2 2π = 1 2 (Return) 5. Ignoring the unnecessary detail of the dartboard, let the radius of the bull’s eye be r. Then P (( bullseye ) = A( bullseye ) A( board ) = πr 2 π · 92 = r 2 81. Setting equal to 1 100 and solving gives r = 0.9 inches. (In reality, the bullseye is much smaller, but the numbers worked out nicer in this example). (Return) 6. If the radius of the circle is r, then the side length of the square is 2r. Thus, the area of the circle is πr 2 and the area of the square is (2r)2 = 4r 2. And so the probability that a point chosen at random within the square also lies within the circle is P = πr 2 4r 2 = π 4 . 225 (Return) 7. By doing a little bit of right triangle trigonometry: we ﬁnd that the side length of the triangle is s = r √ 3. Therefore, the area of the triangle is A = s2√ 3 4 = 3r 2√ 3 4 . And so the probability of a point within the circle being within the triangle is the ratio of the areas: Area of triangle Area of circle = 1 πr 2 3r 2√ 3 4 = 3 √ 3 4π (Return) 8. The possible outcomes form a square for 0 ≤x ≤60 and 0 ≤y ≤60. For the friends to meet, we must have that Yolanda arrives no later than 20 minutes after Xander and that Xander arrives no later than 20 minutes after Yolanda arrives. Mathematically, y ≤x + 20 x ≤y + 20 The two will successfully meet if and only if these two conditions are met. Graphing these inequalities, the points they have in common are shown below in dark blue: 226 So the probability of them meeting is the area of the dark blue region divided by the total area of the square. It is easier to determine the area of the region we do not want and subtract. There are two isosceles right triangles of side length 40, so the area of the region we do not want is bad area = 2 · 1 2 · 40 · 40 = 1600 Therefore, the probability of the friends meeting is P = area of dark blue region total area = total area −light blue area total area = 3600 −1600 3600 = 2000 3600 = 5 9. (Return) 9. If r is the radius of the inscribed sphere, then the side length of the cube is 2r. Therefore, the volume of the sphere is 4 3πr 3 and the volume of the cube is (2r)3 = 8r 3. So the probability that a random point within in the cube lies within the sphere is P = (4/3)πr 3 8r 3 = π 6 . (Return) 10. Let r be the radius of the ball. Then the volume of the ball (the volume of all the possible outcomes) is 4 3πr 3. To ﬁnd the volume of the desired outcomes, consider the volume of the undesired outcomes: those points which lie within 90% of the center. These points form a ball of radius 9 10r, hence their volume is 227 4 3π 9 10r 3. So the desirable outcomes have the complementary volume volume of desired outcomes = total volume −volume of undesired outcomes = 4 3πr 3 −4 3π( 9 10r)3 = 4 3πr 3 1 −(9/10)3 . Thus, the probability of a point being within 10% of the boundary is volume of desired outcomes total volume = 4 3πr 3(1 −(9/10)3) 4 3πr 3 = 1 −(9/10)3 = 0.271 (Return) 228 43 Probability Densities The last module dealt with the uniform distribution, where any one outcome is as likely as another. This module deals with experiments whose outcomes have diﬀerent probabilities. For example, consider an unfair coin which has a 2 3 probability of landing heads and a 1 3 probability of landing tails. Another example is time spent on hold with customer service, where it is more likely that the call is answered in the ﬁrst hour than in the second hour. 43.1 Random variable and probability density function (PDF) A random variable X is a function whose output should be thought of as the outcome of an experiment. Associated with a random variable is a probability density function (PDF) ρ(x), which is deﬁned by P(a ≤X ≤ b) = R b a ρ(x)dx. That is, the probability that the random variable falls in a certain range of values is given by integrating the PDF over that range of values. Phrased another way, we can think of probability P as the quantity we want to compute over a certain range of values, and the probability element is given by dP = ρ(x) dx. Example Consider the spinner from the last module. The outcome of a spin is some angle (relative to the positive x-axis) between 0 and 2π. If X is the random variable which gives the output of a spin, then P(a ≤X ≤b) = b −a 2π , since the spinner was assumed to be fair. This holds for all 0 ≤a ≤b ≤2π. Then the associated PDF is ρ(x) = ( 1 2π if 0 ≤x ≤2π 0. otherwise Note Sometimes a PDF ρ(x) is only deﬁned on a certain domain D. D can be thought of as the set of all possible outcomes of the experiment X. In this case, it is assumed that ρ(x) = 0 for x not in that domain. So another way of deﬁning the PDF for the spinner is ρ(x) = 1 2π for 0 ≤x ≤2π. 229 43.2 Properties of a probability density function The following are deﬁning properties of a PDF. In other words, a function ρ(x) is a PDF on the domain D if and only if it satisﬁes these properties. 1. ρ(x) ≥0 for all x ∈D. 2. R D ρ(x) dx = 1. The ﬁrst property is necessary since probabilities must be non-negative. The second property reﬂects the fact that the random variable X associated with ρ(x) must have some outcome in the domain D (since D is the set of all possible outcomes), and so integrating over all of these outcomes should give 1. Note If ρ(x) is deﬁned on some speciﬁc domain D, then the integral over that speciﬁc domain should equal 1. This is because ρ(x) = 0 outside of that domain, as mentioned in the above note. Example Find the value of the constant c so that ρ(x) = c 1+x2 for all x is a PDF. (See Answer 1) 43.3 Several speciﬁc density functions Uniform density Hinted at above and in the previous module, the uniform density function (or uniform distribution) on [a, b] is given by ρ(x) = 1 b−a (and ρ(x) = 0 if x is not in [a, b]): More generally, the uniform distribution on the domain D (whatever the dimension) is given by ρ(x) = 1 Volume of D. In dimension 0, where outcomes are discrete (as in the rolling of a die or the ﬂipping of a coin), remember that volume is just counting. So in this case the probability of a particular outcome is ρ(x) = 1 n, where n is the number of outcomes in the domain D (e.g. n = 6 for the roll of a die; n = 2 for a coin ﬂip). 230 Exponential density Another density function used to model many common experiments is the exponential density function. This is actually a whole family of density functions given by ρ(t) = αe−αt for t ≥0 and α > 0 some constant. The reason a parameter t is used is that the exponential density is often used to model experiments with a time outcome. Example Show that the exponential density ρ(t) = αe−αt (for t ≥0) satisﬁes the properties of a density function. (See Answer 2) Example Consider a call made to customer service at Acme company. The number of minutes spent on hold before the call is answered is often modeled with an exponential density function ρ(t) = αe−αt. Find, in terms of α, the probability that the waiting time for a call is less than 30 minutes. (See Answer 3) Example Again consider customer service call waiting time at Acme company, and again assume an exponential density function ρ(t) = αe−αt. Suppose half of all customers are answered within 5 minutes. Find α and then ﬁnd the probability that a call takes more than 10 minutes to be answered. (See Answer 4) Gaussian density The last probability density function is the ’Gaussian, or normal, density function. This is an important density function and is expanded on in the next module. Like the exponential, the Gaussian density function usually has parameters (see the next module), but in its simplest form, the Gaussian is given by ρ(x) = 1 √ 2π e−x2/2. 231 The Gaussian has all real x as its domain, but because it tails oﬀso quickly in both directions, the probability of getting values far from the center (in this case x = 0) is very small. 43.4 EXERCISES • Which of the following are probability density functions? a. f (x) =1/2 on D = [0, 2] b. f (x) =sin(x) 2 on D = [0, 3π] c. f (x) =5e−2x on D = [0, ∞) d. f (x) = 1 π(1 + x2) on D = (−∞, ∞) e. f (x) =x 2 for 0 ≤x ≤1 1 2 for 1 ≤x ≤2 3 2 −x 2 for 2 ≤x ≤3 43.5 Answers to Selected Examples 1. As long as c ≥0, the ﬁrst property for a PDF will be met, since 1 + x2 > 0 for all x. To satisfy the second property, compute Z ∞ −∞ c 1 + x2 dx = c arctan(x) ∞ −∞ = c π 2 −(−π 2 ) = cπ. Since this integral is supposed to be 1, we ﬁnd that c = 1 π. (Return) 232 2. The exponential function is never negative, so one need only check the integral. One ﬁnds Z ∞ t=0 αe−αt dt = α 1 −αe−αt ∞ t=0 = −(0 −1) = 1, as desired. So the exponential density is in fact a density. (Return) 3. To ﬁnd the probability that 0 ≤X ≤30, use the relationship between probability and the PDF, which is P(0 ≤X ≤30) = Z 30 0 ρ(x) dx = Z 30 0 λe−λx dx = −e−λx 30 0 = −e−30λ −(−1) = 1 −e−30λ. (Return) 4. Since half of all customers are answered within 5 minutes, we have that P(0 ≤X ≤5) = 1 2. On the other hand, we know that this can be expressed as the integral of the density function, so we have 1 2 = Z 5 t=0 ρ(t) dt = Z 5 t=0 αe−αt dt = −e−αt 5 t=0 = −e−5α −(−1) = 1 −e−5α. So we have that e−5α = 1 2. Taking the log of both sides, dividing by −5 and simplifying, we have α = 1 −5 ln 1 2 = 1 −5(−ln 2) = 1 5 ln 2. 233 For the second part, we want to know the probability of waiting more than 10 minutes. This is (leaving α as a constant for now) P(X ≥10) = Z ∞ t=10 ρ(t) dt = Z ∞ t=10 αe−αt dt = −e−αt ∞ t=10 = 0 − −e−α·10 . Now plugging in the value of α, we have P(X ≥10) = e−(ln 2/5)·10 = e−2 ln 2 = 2−2 = 1 4. (Return) 234 44 Expectation And Variance When performing an experiment, it is useful to know what the expected outcome will be as well as how much variation one can expect among the outcomes. The notions of expected outcome and variation are made formal in this module by the terms expectation,variance, and standard deviation. This module will also show some of the connections of these statistical metrics with the applications of the previous modules. 44.1 Expectation Consider a random variable X with probability density function (PDF) ρ(x) deﬁned on some domain D. The expectation of X, denoted by E, is deﬁned by E = Z D xρ(x) dx = Z D x dP, where dP is the probability element. The expectation of X is sometimes called the mean of X, the expected value, or the ﬁrst moment. In some books it is denoted µX. It is best to think of the expectation as the number one gets by repeating the experiment many times and taking the average of the outputs. The notion of expectation is more general than the mean because one can also take the expectation of a function of X. The expectation of f (X) is deﬁned by E[f (X)] = Z D f (x)ρ(x) dx. Example Find the expectation of X, where X is uniformly distributed on the interval [a, b]. (See Answer 1) Example Recall that the random variable X is said to have the exponential distribution if the PDF associated with X is ρ(t) = αe−αt for t ≥0, where α > 0 is some constant. Find the expectation of the exponential distribution (in terms of α). (See Answer 2) 235 44.2 Variance Consider a random variable X with PDF ρ(x). The variance of X, denoted V, is deﬁned by V = E h (X −E[X])2i = E[X2] −E[X]2. In the notation of the lecture, V = Z D (x −E)2 dP = Z D x2 dP −E2. Note: it requires some calculation to show the second equality above holds. Either of the above expressions may be taken as the deﬁnition of variance, and the second one might be slightly simpler for the sake of computation. (See Justiﬁcation 3) Example Compute the variance of the exponential density function ρ(x) = αe−αx. (See Answer 4) 44.3 Standard deviation Consider a random variable X with PDF ρ(x). Then the standard deviation of X, denoted σX, is deﬁned by σX = p V [X] = p E[X2] −E[X]2 = sZ D x2ρ(x) dx − Z D xρ(x) dx 2 . Example Find the standard deviation of X, where X is uniformly distributed over [a, b]. (See Answer 5) 44.4 Interpretations If one interprets the PDF ρ(x) as the density of a rod at location x, then: 1. The mean, µ = R xρ(x) dx, gives the center of mass of the rod. 2. The variance, V = R (x −µ)2ρ(x) dx, gives the moment of inertia about the line x = µ. 3. The standard deviation, σ = √ V , gives the radius of gyration about the line x = µ. 236 44.5 The normal distribution A random variable X is said to have the normal distribution, or to be normally distributed, with mean µ and standard deviation σ if its PDF is of the form ρ(x) = 1 σ √ 2π e−1 2( x−µ σ ) 2 . Due to its ubiquity throughout the sciences, the normal distribution is one of the most well-known probability distributions. However, because its PDF does not have an elementary anti-derivative, it is not easy to calculate exact probabilities associated with the normal distribution. Instead, there are is a rule of thumb which can be used. The 68-95-99.7 rule Given a random variable X which is normally distributed with mean µ and standard deviation σ, the following hold: 1. P(µ −σ ≤X ≤µ + σ) ≈.68. 2. P(µ −2σ ≤X ≤µ + 2σ) ≈.95. 3. P(µ −3σ ≤X ≤µ + 3σ) ≈.997. In other words, 68% of samples will fall within 1 standard deviation of the mean. 95% of samples will fall within 2 standard deviations of the mean. And 99.7% of samples will fall within 3 standard deviations. These rules, along with the symmetry of the normal PDF, can be used to approximate many probabilities relating to the normal distribution: 237 Example The height of men in a certain population is normally distributed with mean µ = 70 inches and standard deviation σ = 2 inches. If a man is chosen at random from the population, what is the probability that he is taller than 72 inches? (See Answer 6) 44.6 EXERCISES • Compute the expected value of normally distributed random variable with probability density function ρ(x) = 1 √ 2πe−x2 2 on −∞< x < ∞. 44.7 Answers to Selected Examples 1. Recall that the PDF associated with X is given by ρ(x) = 1 b−a for a ≤x ≤b. Thus, the mean is given by E = Z b a x · 1 b −a dx = 1 b −a x2 2 b a = 1 b −a · 1 2(b2 −a2) = 1 b −a · 1 2(b + a)(b −a) = 1 2(a + b). (Return) 238 2. From the deﬁnition of expectation, one ﬁnds E = Z ∞ 0 tαe−αt dt = α Z ∞ 0 te−αt dt. Using integration by parts, with u = t du = dt dv = e−αt v = 1 −αe−αt, we ﬁnd that α Z ∞ 0 te−αt dt = α t −αe−αt − Z ∞ 0 1 −αe−αt dt = −te−αt −1 αe−αt ∞ 0 = (0 −0) −(0 −1 α) = 1 α. (Return) 3. Expanding out the expression and using the linearity of the integral, we ﬁnd Z D (x −E)2 dP = Z D (x −E)2ρ(x) dx = Z D x2 dP − Z D 2xE dP + Z E2 dP = Z D x2 dP −2E Z D x dP + E2 Z dP = Z D x2 dP −2E · E + E2 = Z D x2 dP −E2. because R x dP = E and R dP = 1, by the deﬁnition of expectation and the deﬁnition of the probability density function, respectively. (Return) 4. The variance requires us to compute E(X2) = Z D x2 dP = Z ∞ x=0 x2αe−αx dx. 239 Using integration by parts, with u = x2 du = 2x dx dv = αe−αx dx v = −e−αx, we ﬁnd Z ∞ x=0 x2αe−αx dx = −x2e−αx ∞ x=0 + Z ∞ x=0 2xe−αx dx. This second integral can be done with integration by parts again, or we can use the fact that this is almost the integral for the expectation. Namely, we know Z ∞ x=0 xαe−αx dx = 1 α, and so by dividing through by α, we have Z ∞ x=0 xe−αx dx = 1 α2 . Putting this together, we have Z ∞ x=0 x2αe−αx dx = −x2e−αx ∞ x=0 + 2 α2 = (0 −0) + 2 α2 = 2 α2 . Finally, then, the variance is V = Z D x2 dP −E2 = 2 α2 − 1 α 2 = 1 α2 . (Return) 5. Again, recall that the PDF for the uniform distribution is ρ(x) = 1 b−a for a ≤x ≤b. Thus, E[X2] = Z b a x2 1 b −a dx = 1 b −a x3 3 b a = 1 b −a · 1 3(b3 −a3) = 1 b −a · 1 3(b −a)(b2 + ba + a2) = b2 + ab + a2 3 240 From the previous example, E[X] = µX = a+b 2 . Thus, σX = p E[X2] −E[X]2 = r b2 + ba + a2 3 −b2 + 2ba + a2 4 = r b2 −2ab + a2 12 = b −a √ 12 . (Return) 6. Let X be the height of a randomly chosen man. Then P(68 ≤X ≤72) = .68 by the above rule. By symmetry P(68 ≤X ≤70) = P(70 ≤X ≤72) = .34. Also, by symmetry, P(X ≤70) = .5. Thus, P(X ≤72) = P(X ≤70) + P(70 ≤X ≤72) = .5 + .34 = .84. It follows that P(X > 72) = 1 −P(X ≤72) = 1 −.84 = .16. This is best visualized by labeling the various regions under the normal curve with their areas: So the probability that a randomly chosen man from the population is taller than 72 inches is .16. (Return) 241
2253	https://www.merriam-webster.com/grammar/purposely-purposefully-usage	Est. 1828 Popular in Commonly Confused Democracy or Republic: What's the difference? 'Canceled' or 'cancelled'? The Difference Between 'i.e.' and 'e.g.' 'Lose' or 'loose'? Is it 'home in' or 'hone in'? See All Popular in Grammar & Usage Putting Adjectives in the Right Order What's the difference between 'sympathy' and 'empathy'? 'Affect' vs. 'Effect' Is it 'autumn' or 'fall'? Using Bullet Points ( • ) See All Popular in Wordplay 8 Real (and Perceived) Redundant Words Skibidi, Mog & More: Gen Alpha Slang Color Words to Describe Autumn Leaves Ye Olde Nincompoop: Old-Fashioned Words for 'Stupid' Great Big List of Beautiful and Useless Words, Vol. 3 See All Grammar & Usage Commonly Confused Is it 'purposely' or 'purposefully'? Is 'purposely' a real word? Though the first presidential debate seems like a distant memory at this point, there was one word during the course of the debate that threw a bunch of dictionary users for a loop. Not, not bigly or braggadocious. It was something that was a little more opaque: .@realDonaldTrump: Clinton's email server "more than a mistake; that was done purposely." #debates — Nightline (@Nightline) September 27, 2016 The misconstrued word? Purposely. During the next debate I am going to sit here and say "____ is not a word" every time @realDonaldTrump says something like purposely... — forevereve (@forevereve) September 27, 2016 Digging deeper, it appears that many people felt that Trump should have used purposefully instead, since purposely isn't really a word. Except, like bigly and braggadocious, purposely is a word—and a very common one at that. 'Purposely' means "on purpose"; 'purposefully' means "indicating the existence of a purpose." Meaning of Purposefully and Purposely Purposely came into English in the late 1400s or early 1500s, right in the middle of an -ly-adverb boom in English. Its earliest and current meaning is "on purpose, intentionally": It is ordyned...that no man take any Eyre[r], Gossehauke [etc.] nor purposly drive them oute of their covertes.— Acts of Parliament, 1495 In the second game Jeanette starts burying balls off two and three cushions. Even when she purposely misses, she leaves him blocked in like it's 5 p.m. on an L.A. freeway. — Rick Reilly, Sports Illustrated, 4 July 2005 These two uses are typical, and not rare: purposely has had far more use, historically, than purposefully has (and it still has slightly more use in printed English prose). Purposefully, on the other hand, is a relative newcomer. Our earliest evidence for the word currently comes from the mid-1800s, with the meaning "indicating the existence of a purpose or object," or "not meaningless or aimless," as in "We were purposefully taken to inspect them" (Elizabeth Grant, Memoirs of a Highland Lady, 1854). So if purposely is actually more common than purposefully, why did so many people think Trump was making up yet another word? Likely because the context of his comment seemed to imply something that the word purposely wasn't communicating: determined intention. Determined Intention The two words have the same root—purpose—but slightly different meanings. When used in prose, purposefully seems to connote a determination or intentionality that purposely does not—to do something purposefully is to do it guided by a deliberate aim: "I Knew You Were Trouble" audaciously moves into dubstep territory, boasting heavy bass in some parts and vocals that have been purposefully heavily autotuned to sound mechanized. — The Johns Hopkins News-Letter, 25 Oct. 2012 In common use, purposely seems to lack that level of determination: She wore a long knit dress that looked purposely homespun and showed off her growing baby bump, with her hair swept behind her in a curling bun. — Janelle Brown, This Is Where We Live, 2010 In everyday use, purposely is fine to merely show that something was done or said on purpose (as opposed to accidentally). But if that thing was done or said with a deliberate aim or intention, then purposefully is the adverb to use. Share Word of the Day kerfuffle See Definitions and Examples » Get Word of the Day daily email! Games & Quizzes See All Quordle Can you solve 4 words at once? Blossom Pick the best words! The Missing Letter A daily crossword with a twist Challenging Words You Should Know Not a quiz for the pusillanimous See All Popular Commonly Confused ### Democracy or Republic: What's the difference? ### 'Canceled' or 'cancelled'? ### The Difference Between 'i.e.' and 'e.g.' ### 'Lose' or 'loose'? ### Is it 'home in' or 'hone in'? See All Grammar & Usage ### Putting Adjectives in the Right Order ### What's the difference between 'sympathy' and 'empathy'? ### 'Affect' vs. 'Effect' ### Is it 'autumn' or 'fall'? ### Using Bullet Points ( • ) See All Wordplay ### 8 Real (and Perceived) Redundant Words ### Skibidi, Mog & More: Gen Alpha Slang ### Color Words to Describe Autumn Leaves ### Ye Olde Nincompoop: Old-Fashioned Words for 'Stupid' ### Great Big List of Beautiful and Useless Words, Vol. 3 See All
2254	https://www.fcc.gov/general/cell-phones-and-specific-absorption-rate	Skip to search An official website of the United States government Here’s how you know Official websites use .govA .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPSA lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. Home Engineering & Technology Cell Phones and Specific Absorption Rate The SAR is a value that corresponds to the relative amount of RF energy absorbed in the head of a user of a wireless handset. The FCC limit for public exposure from cellular telephones is an SAR level of 1.6 watts per kilogram (1.6 W/kg). Specific Absorption Rate (SAR) for Wireless Phones and Devices Available at various Web sites. The easiest way to ascertain SAR for many cellular phones is via the FCC's links to individual manufacturers' Web sites: On this page you will find links to most manufacturers' Web pages that include SAR information for their phones, along with instructions on how to search each site for SAR information. You can also obtain SAR information on many cellular phones from the FCC's database if you have the FCC ID number of the phone or device and if it was produced and marketed within the last 1-2 years. The FCC ID number is usually shown somewhere on the case of the phone or device. In many cases, you will have to remove the battery pack to find the number. Once you have the number, proceed as follows. Go to the following Web site: Equipment Authorization. Click on the link for “FCC ID Search”. Once you are there you will see instructions for inserting the FCC ID number. Enter the FCC ID number (in two parts as indicated: “Grantee Code” is comprised of the first three characters, the “Equipment Product Code” is the remainder of the FCC ID). Then click on “Start Search.” The grant of equipment authorization for this particular ID number should appear. Look through the grant for the section on SAR compliance, certification of compliance with FCC rules for RF exposure or similar language. This section should contain the value(s) for typical or maximum SAR for your phone. For portable phones and devices authorized since June 2, 2000, maximum SAR levels should be noted on the grant of equipment authorization. For phones and devices authorized between about mid-1998 and June 2000, detailed information on SAR levels is typically found in the “exhibits” associated with the grant of equipment authorization. Therefore, once a grant is accessed these exhibits can be viewed by clicking on the appropriate entry labeled “View Exhibit.” Electronic records for FCC equipment authorization grants were initiated in 1998. Therefore, prior to this date FCC records for grants are in the form of paper records that are not part of our electronic database. At this time, due to staff limitations, we are unable to routinely search through FCC paper records to extract SAR information for grants filed prior to mid- to late-1998. If you want additional consumer information on safety of cell phones and other transmitting devices please consult the information available below at this Web Site. In particular, you may wish to read or download our OET Bulletin 56 (see “RF Safety Bulletins”) entitled: “Questions and Answers about Biological Effects and Potential Hazards of Radiofrequency Electromagnetic Fields.” If you have any problems or additional questions you may contact us at RF Safety (rfsafety@fcc.gov). You may also wish to consult a consumer update on mobile phone safety published by the U.S. Food and Drug Administration (FDA) that can be found at: www.fda.gov/cellphones/. Bureau/Office: Engineering & Technology
2255	https://www.youtube.com/watch?v=PKP790TrLW8	Simplify 40/60 Into Its Simplest Form Nathan Grieve 3090 subscribers 5 likes Description 652 views Posted: 15 Jan 2025 To simplify or reduce a fraction like 40/60, we need to look for the common factors between the numerator and the denominator. 1 comments Transcript: to simplify 40 over 60 we need to look for the common factors between the top number the numerator and the bottom number the denominator first let's list out the factors for each for 40 the factors are 1 2 4 5 8 10 20 and 40 for 60 the factors are 1 2 3 4 5 6 10 12 15 20 30 and 60 so the largest common factor between 40 and 60 is 20 this means we can divide both the numerator and denominator by 20 to simplify the fraction dividing the numerator 40 by 20 gives us 2 dividing the denominator 60 by 20 gives us 3 so after dividing the simplified fraction is 2 over3 and we know that 2 over3 is in its simplest form because the only common factor between 2 and three is 1 and when one is is the largest common factor we know that our fraction has been simplified
2256	https://math.stackexchange.com/questions/20717/how-to-find-solutions-of-linear-diophantine-ax-by-c	elementary number theory - How to find solutions of linear Diophantine ax + by = c? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to find solutions of linear Diophantine ax + by = c? Ask Question Asked 14 years, 7 months ago Modified3 years, 5 months ago Viewed 193k times This question shows research effort; it is useful and clear 135 Save this question. Show activity on this post. I want to find a set of integer solutions of Diophantine equation: a x+b y=c a x+b y=c, and apparently gcd(a,b)\|c gcd(a,b)\|c. Then by what formula can I use to find x x and y y ? I tried to play around with it: x=(c−b y)/a x=(c−b y)/a, hence a\|(c−b y)a\|(c−b y). a a, c c and b b are known. So to obtain integer solution for a a, then c−b y=a k c−b y=a k, and I lost from here, because y=(c−a k)/b y=(c−a k)/b. I kept repeating this routine and could not find a way to get rid of it? Any hint? Thanks, Chan elementary-number-theory diophantine-equations linear-diophantine-equations Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jun 5, 2016 at 6:22 Martin Sleziak 56.3k 20 20 gold badges 211 211 silver badges 391 391 bronze badges asked Feb 6, 2011 at 19:08 roxrookroxrook 12.4k 24 24 gold badges 110 110 silver badges 149 149 bronze badges 2 8 en.wikipedia.org/wiki/Extended_Euclidean_algorithmQiaochu Yuan –Qiaochu Yuan 2011-02-06 19:21:36 +00:00 Commented Feb 6, 2011 at 19:21 1 Read the following paper researchgate.net/publication/…. Hope it helps . best Issam.Kaddoura –Issam.Kaddoura 2016-09-03 14:56:33 +00:00 Commented Sep 3, 2016 at 14:56 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 190 Save this answer. Show activity on this post. The diophantine equation a x+b y=c a x+b y=c has solutions if and only if gcd(a,b)\|c gcd(a,b)\|c. If so, it has infinitely many solutions, and any one solution can be used to generate all the other ones. To see this, note that the greatest common divisor of a a and b b divides both a x a x and b y b y, hence divides c c if there is a solution. This gives the necessity of the condition ~~(which you have backwards).~~(fixed in edits) The converse is actually a constructive proof, that you can find in pretty much every elementary number theory course or book, and which is essentially the same as yunone's answer above (but without dividing through first). From the Extended Euclidean Algorithm, given any integers a a and b b you can find integers s s and t t such that a s+b t=gcd(a,b)a s+b t=gcd(a,b); the numbers s s and t t are not unique, but you only need one pair. Once you find s s and t t, since we are assuming that gcd(a,b)gcd(a,b) divides c c, there exists an integer k k such that gcd(a,b)k=c gcd(a,b)k=c. Multiplying a s+b t=gcd(a,b)a s+b t=gcd(a,b) through by k k you get a(s k)+b(t k)=gcd(a,b)k=c.a(s k)+b(t k)=gcd(a,b)k=c. So this gives one solution, with x=s k x=s k and y=t k y=t k. Now suppose that a x 1+b y 1=c a x 1+b y 1=c is a solution, and a x+b y=c a x+b y=c is some other solution. Taking the difference between the two, we get a(x 1−x)+b(y 1−y)=0.a(x 1−x)+b(y 1−y)=0. Therefore, a(x 1−x)=b(y−y 1)a(x 1−x)=b(y−y 1). That means that a a divides b(y−y 1)b(y−y 1), and therefore a gcd(a,b)a gcd(a,b) divides y−y 1 y−y 1. Therefore, y=y 1+r a gcd(a,b)y=y 1+r a gcd(a,b) for some integer r r. Substituting into the equation a(x 1−x)=b(y−y 1)a(x 1−x)=b(y−y 1) gives a(x 1−x)=r b(a gcd(a,b))a(x 1−x)=r b(a gcd(a,b)) which yields gcd(a,b)a(x 1−x)=r b a gcd(a,b)a(x 1−x)=r b a or x=x 1−r b gcd(a,b)x=x 1−r b gcd(a,b). Thus, if a x 1+b y 1=c a x 1+b y 1=c is any solution, then all solutions are of the form x=x 1−r b gcd(a,b),y=y 1+r a gcd(a,b)x=x 1−r b gcd(a,b),y=y 1+r a gcd(a,b) exactly as yunone said. To give you an example of this in action, suppose we want to find all integer solutions to 258 x+147 y=369.258 x+147 y=369. First, we use the Euclidean Algorithm to find gcd(147,258)gcd(147,258); the parenthetical equation on the far right is how we will use this equality after we are done with the computation. 258 147 111 36=147(1)+111=111(1)+36=36(3)+3=3(12).(equivalently,111=258−147)(equivalently,36=147−111)(equivalently,3=111−3(36))258=147(1)+111(equivalently,111=258−147)147=111(1)+36(equivalently,36=147−111)111=36(3)+3(equivalently,3=111−3(36))36=3(12). So gcd(147,258)=3 gcd(147,258)=3. Since 3\|369 3\|369, the equation has integral solutions. Then we find a way of writing 3 3 as a linear combination of 147 147 and 258 258, using the Euclidean algorithm computation above, and the equalities on the far right. We have: 3=111−3(36)=111−3(147−111)=4(111)−3(147)=4(258−147)−3(147)=4(258)−7(147).3=111−3(36)=111−3(147−111)=4(111)−3(147)=4(258−147)−3(147)=4(258)−7(147). Then, we take 258(4)+147(−7)=3 258(4)+147(−7)=3, and multiply through by 123 123; why 123 123? Because 3×123=369 3×123=369. We get: 258(492)+147(−861)=369.258(492)+147(−861)=369. So one solution is x=492 x=492 and y=−861 y=−861. All other solutions will have the form x y=492−147 r 3=492−49 r,=−861+258 r 3=86 r−861,r∈Z.x=492−147 r 3=492−49 r,y=−861+258 r 3=86 r−861,r∈Z. You can reduce those constants by making a simple change of variable. For example, if we let r=t+10 r=t+10, then x y=492−49(t+10)=2−49 t,=86(t+10)−861=86 t−1,t∈Z.x=492−49(t+10)=2−49 t,y=86(t+10)−861=86 t−1,t∈Z. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jul 22, 2011 at 2:32 answered Feb 6, 2011 at 20:46 Arturo MagidinArturo Magidin 419k 60 60 gold badges 864 864 silver badges 1.2k 1.2k bronze badges 7 1 is it possible to add a little discussion around same equation in higher dimensions ?zinking –zinking 2016-06-25 12:59:42 +00:00 Commented Jun 25, 2016 at 12:59 5 @Cookie: I wrote this eight years ago... As to your question:No, you cannot write a b a b, because you do not know that b b divides a a. That’s precisely the issue. By writing a b a b you are leaving the integers and going into the rationals. Let d=gcd(a,b)d=gcd(a,b). Write a=d a′a=d a′, b=d b′b=d b′. Then it is well known that gcd(a′,b′)=1 gcd(a′,b′)=1. From a\|b(y−y 1)a\|b(y−y 1) you get a′\|b′(y−y 1)a′\|b′(y−y 1). And since gcd(a′,b′)=1 gcd(a′,b′)=1< you deduce a′\|y−y 1 a′\|y−y 1. This is pretty standard fare.Arturo Magidin –Arturo Magidin 2019-03-19 02:57:19 +00:00 Commented Mar 19, 2019 at 2:57 1 @ArturoMagidin How does a(x 1−x)=b(y−y 1)a(x 1−x)=b(y−y 1) mean that a a divides b(y−y 1)b(y−y 1)? How can we be sure a a always divides b(y−y 1)b(y−y 1) and (x 1−x)(x 1−x) does not divide b(y−y 1)b(y−y 1)?alu –alu 2020-09-07 15:00:00 +00:00 Commented Sep 7, 2020 at 15:00 4 @alu: Huh? By definition, a a divides b(y−y 1)b(y−y 1) if and only if there exists an integer k k such that a k=b(y−y 1)a k=b(y−y 1). Here, the integer is x 1−x x 1−x.Arturo Magidin –Arturo Magidin 2020-09-07 15:16:52 +00:00 Commented Sep 7, 2020 at 15:16 4 @alu: Note: both a a and x−x 1 x−x 1 divide b(y−y 1)b(y−y 1). It’s not a “one or the other”. 2×3=6 2×3=6, so both 2 2 and 3 3 divide 6 6.Arturo Magidin –Arturo Magidin 2020-09-07 18:03:19 +00:00 Commented Sep 7, 2020 at 18:03 \|Show 2 more comments This answer is useful 26 Save this answer. Show activity on this post. As others have mentioned one may employ the extended Euclidean algorithm. It deserves to be better known that this is most easily performed via row-reduction on an augmented matrix - analogous to methods used in linear algebra. See this excerpt from one of my old sci.math posts: ``` For example, to solve mx + ny = gcd(x,y) one begins with two rows [m 1 0], [n 0 1], representing the two equations m = 1m + 0n, n = 0m + 1n. Then one executes the Euclidean algorithm on the numbers in the first column, doing the same operations in parallel on the other columns, Here is an example: d = x(80) + y(62) proceeds as: in equation form \| in row form ---------------------+------------ 80 = 1(80) + 0(62) \| 80 1 0 62 = 0(80) + 1(62) \| 62 0 1 row1 - row2 -> 18 = 1(80) - 1(62) \| 18 1 -1 row2 - 3 row3 -> 8 = -3(80) + 4(62) \| 8 -3 4 row3 - 2 row4 -> 2 = 7(80) - 9(62) \| 2 7 -9 row4 - 4 row5 -> 0 = -31(80) +40(62) \| 0 -31 40 ``` Above the row operations are those resulting from applying the Euclidean algorithm to the numbers in the first column, ``` row1 row2 row3 row4 row5 namely: 80, 62, 18, 8, 2 = Euclidean remainder sequence \| \| for example 62-3(18) = 8, the 2nd step in Euclidean algorithm becomes: row2 -3 row3 = row4 on the identity-augmented matrix. In effect we have row-reduced the first two rows to the last two. The matrix effecting the reduction is in the bottom right corner. It starts as the identity, and is multiplied by each elementary row operation matrix, hence it accumulates the product of all the row operations, namely: [ 7 -9] [ 80 1 0] = [2 7 -9] [-31 40] [ 62 0 1] [0 -31 40] The 1st row is the particular solution: 2 = 7(80) - 9(62) The 2nd row is the homogeneous solution: 0 = -31(80) + 40(62), so the general solution is any linear combination of the two: n row1 + m row2 -> 2n = (7n-31m) 80 + (40m-9n) 62 The same row/column reduction techniques tackle arbitrary systems of linear Diophantine equations. Such techniques generalize easily to similar coefficient rings possessing a Euclidean algorithm, e.g. polynomial rings F[x] over a field, Gaussian integers Z[i]. There are many analogous interesting methods, e.g. search on keywords: Hermite / Smith normal form, invariant factors, lattice basis reduction, continued fractions, Farey fractions / mediants, Stern-Brocot tree / diatomic sequence. ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Apr 8, 2022 at 8:09 answered Feb 6, 2011 at 21:43 Bill DubuqueBill Dubuque 284k 42 42 gold badges 339 339 silver badges 1k 1k bronze badges 5 6 @Chan: It irks me that most textbooks in elementary number theory present more obfuscated approaches. If you go on to study algebra you will learn more about the underlying theory when you study Hermite Smith normal forms and other module-theoretic generalizations of linear algebra results.Bill Dubuque –Bill Dubuque 2011-02-06 22:06:04 +00:00 Commented Feb 6, 2011 at 22:06 3 This is actually discussed in Niven, Zuckerman, Montgomery. Just so you have a reference (pages 217-218 in the 5th edition).Arturo Magidin –Arturo Magidin 2011-02-06 22:07:41 +00:00 Commented Feb 6, 2011 at 22:07 1 @Arturo. Thanks for the reference. I'm happy to see that it finally made it into an edition of a popular textbook, but I'm sad that the presentation there leaves much to be desired.Bill Dubuque –Bill Dubuque 2011-02-06 22:21:07 +00:00 Commented Feb 6, 2011 at 22:21 1 I don't mean to be a smart-aleck, or too pedantic, but in your first grey sandbox, you have written 0=-31(80)-40(62) in the "in equation form" column, when I think it should instead be: 0=-31(80)+40(62).Mr Pie –Mr Pie 2022-04-08 05:28:38 +00:00 Commented Apr 8, 2022 at 5:28 @MrPie Yes, the typo was only in the equation (vs. row) form. Now fixed, thanks.Bill Dubuque –Bill Dubuque 2022-04-08 08:10:34 +00:00 Commented Apr 8, 2022 at 8:10 Add a comment\| This answer is useful 21 Save this answer. Show activity on this post. Here's another method. It doesn't require finding an initial solution, but it may require finding a modular multiplicative inverse. (Edit: I'll show this method with an example instead of a generalization. It may make it more clear.) Solve 2 x+3 y=5 2 x+3 y=5. 2 x≡5(mod 3)2 x≡5(mod 3). You could either find a multiplicative inverse: x≡5⋅2−1(mod 3)x≡5⋅2−1(mod 3) Or do it as follows: 2 x≡2(mod 3)2 x≡2(mod 3) Divide both sides by 2 2 (notice gcd(3,2)=1 gcd(3,2)=1). x≡1(mod 3)x≡1(mod 3), x=3 n+1 x=3 n+1, n∈Z n∈Z. Substitute this value of x x into the original equation: 2(3 n+1)+3 y=5 2(3 n+1)+3 y=5, y=1−2 n y=1−2 n. Answer: (x,y)=(3 n+1,1−2 n)(x,y)=(3 n+1,1−2 n), n∈Z n∈Z. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 17, 2017 at 20:17 answered Jun 27, 2016 at 22:35 user236182user236182 13.5k 1 1 gold badge 25 25 silver badges 48 48 bronze badges 1 1 This answer deserves to be on top. Really great answer, made me realize that linear congruences and diophantine equations are inter-convertible.Manish Chandra Joshi –Manish Chandra Joshi 2019-09-11 14:25:10 +00:00 Commented Sep 11, 2019 at 14:25 Add a comment\| This answer is useful 17 Save this answer. Show activity on this post. Do you mean gcd(a,b)gcd(a,b) divides c c? If so, you can divide both sides of the equation to get a g x+b g y=c g a g x+b g y=c g where g=gcd(a,b)g=gcd(a,b). But since gcd(a/g,b/g)=1 gcd(a/g,b/g)=1, you can use the extended Euclidean algorithm to find a solution (x 0,y 0)(x 0,y 0) to the equation a g x+b g y=1.a g x+b g y=1. Once you have that, the solution (X,Y)=(c g⋅x 0,c g⋅y 0)(X,Y)=(c g⋅x 0,c g⋅y 0) is a solution to your original equation. Furthermore, the values x=X+b g t y=Y−a g t x=X+b g t y=Y−a g t give all solutions when t t ranges over Z Z, I believe. Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Follow Follow this answer to receive notifications answered Feb 6, 2011 at 19:33 yunoneyunone 22.9k 10 10 gold badges 85 85 silver badges 160 160 bronze badges 1 @yuone: Yes, that gives all solutions.Arturo Magidin –Arturo Magidin 2011-02-06 20:47:21 +00:00 Commented Feb 6, 2011 at 20:47 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Look can be deceiving. The integer solution to the equation a x+b y=c a x+b y=c is anything but easy. Please endure a rather long derivation. To make it more comprehensible let's first solve the equation for y: a x+b y=c b y=c−a x y=c−a x b a x+b y=c b y=c−a x y=c−a x b To have an integer solution, y y must be an integer, and that is if c−a x c−a x is a multiple of b b, or c−a x=−n b⟺a x=c+n b c−a x=−n b⟺a x=c+n b. This has the same meaning as a x≡c(m o d n)a x≡c(m o d n). To continue, we need this Theorem 1: The congruence a x≡c(m o d n)a x≡c(m o d n) has a solution iff g c d(a,n)\|c g c d(a,n)\|c. And this Lemma 2: If g c d(p,q)=1 g c d(p,q)=1, then p x≡r(m o d q)p x≡r(m o d q) has a solution modulo q q. To keep this answer manageable, I would like to skip the proof of Theorem 1 and Lemma 2 (which can be found by googling). Just post a question and comment me if you encounter some trouble with them. Let's define d=g c d(a,n)d=g c d(a,n), and continue the derivation: a x≡c(m o d n)a x=c+b n a d x=c d+b d n a x≡c(m o d n)a x=c+b n a d x=c d+b d n Now we want to switch b b and n n so n d n d could be seen more clearly as the modulo and continue it as following: a d x=c d+b n d a d x≡c d(m o d n d)a d x=c d+b n d a d x≡c d(m o d n d) Note that a d a d and n d n d from our derivation above is the p p and q q in the Lemma 2 respectively. Also note that as d d is g c d(a,n)g c d(a,n), so g c d(a d,n d)=1 g c d(a d,n d)=1. Hence by Lemma 2: a d x≡c d(m o d n d)a d x≡c d(m o d n d) is our solution to equation a x+b y=c a x+b y=c. As an example, let us solve 6 x−10 y=4⟺6 x=4+10 y⟺6 x≡4(m o d 10)6 x−10 y=4⟺6 x=4+10 y⟺6 x≡4(m o d 10). G c d(6,10)=2 G c d(6,10)=2, and 2\|4 2\|4, so by Theorem 1 that equation has a solution. From our derivation, the solution is 6 2 x=4 2(m o d 10 2)⟺3 x=2(m o d 5)6 2 x=4 2(m o d 10 2)⟺3 x=2(m o d 5). By Lemma 2, we have a solution m o d u l o 5 m o d u l o 5. What it means is if we write the solution in Z 5 Z 5, we would have: 3¯x¯=2¯x¯=3¯−1 2¯3¯x¯=2¯x¯=3¯−1 2¯ As in Z 5 Z 5, 3¯2¯=1¯=3¯3¯−1 3¯2¯=1¯=3¯3¯−1, so 3¯−1=2¯3¯−1=2¯ and we have: x¯=3¯−1 2¯x¯=2¯2¯=4¯x¯=3¯−1 2¯x¯=2¯2¯=4¯ In Z 5 Z 5, x¯=4¯⟺x≡4(m o d 5)⟺x=4+5 s⟺x=4,9,13,⋯x¯=4¯⟺x≡4(m o d 5)⟺x=4+5 s⟺x=4,9,13,⋯. You can check that indeed x≡4(m o d 5)x≡4(m o d 5) is the solution. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 13, 2020 at 17:27 answered Sep 13, 2020 at 14:43 Ari Royce HidayatAri Royce Hidayat 1,159 12 12 silver badges 12 12 bronze badges 2 Nice answer, +1. Could you please help me by extending your answer for all (i.e., non-integer too) solutions. I request that if not give solution, then kindly give reference to look for.jiten –jiten 2021-01-21 08:04:53 +00:00 Commented Jan 21, 2021 at 8:04 @jiten Diophantine equations necessarily require the solutions to be integers. The solutions for all real numbers is trivial; it is simply a line. In algebra, all solutions are in the form (x,c b−a b x)(x,c b−a b x), which can simply be derived from a x+b y=c a x+b y=c =>b y=c−a x b y=c−a x =>y=c b−a b x y=c b−a b x crxyz –crxyz 2023-05-20 02:48:02 +00:00 Commented May 20, 2023 at 2:48 Add a comment\| You must log in to answer this question. Protected question. 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Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 4Existence of Natural Solutions to Linear Diophantine Equation 0Use the Euclidean algorithm steps to find ALL the integer solutions of the equation 3Why do the modular multiplicative inverse of an integer a a always exists when it is coprime with the modulus? 1How to find multiplicative inverse of 17 17 in Z 26 Z 26? 2Diophantine Equation: how to find k such as solutions are natural numbers? 0Asking for help in problem 4 of section 2.5 of David Burton book 2How do you find all the answers to this Diophantus equation: 462 x+273 y=63 462 x+273 y=63? 0Do there exist whole number solutions to 27 y+23=32 x 27 y+23=32 x and 81 y+85=128 x 81 y+85=128 x? 0Solving 2444=960 x(mod 2618)2444=960 x(mod 2618) 0All possible values of k k such that a≤k≤b a≤k≤b and k k is a linear combination of a and b See more linked questions Related 11How to find an integer solution for general Diophantine equation a x+b y+c z+d t...=N a x+b y+c z+d t...=N 2Find Solutions To Some Diophantine Equations 3Conjecture about linear diophantine equations 1Solving Quadratic Diophantine Equation with initial solutions. 2General solutions for the Diophantine equation a x+b y+c x y=d a x+b y+c x y=d 1Particular solution to a diophantine in terms of n n Hot Network Questions Change default Firefox open file directory How do trees drop their leaves? 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2257	https://www.workybooks.com/resources/math-vocabulary/equivalent-ratios	Equivalent Ratios - Definition, Examples, Quiz, FAQ, Trivia Learn how ratios work and discover how different ratios can represent the same relationship Educational Standards Common Core: 6.RP.A.3, 6.RP.A.3.A, 6.RP.A.3.B Learning Sections What are Ratios? A ratio is a way to compare two or more quantities. It shows how much of one thing there is compared to another. We write ratios with a colon (:) between the numbers. For example, if you have 3 apples and 5 oranges, the ratio of apples to oranges is 3:5. This means for every 3 apples, there are 5 oranges. Ratios are everywhere in our daily lives: Key Concept Ratios compare quantities without units. They show the relationship between amounts. What are Equivalent Ratios? Equivalent ratios are different ratios that represent the same relationship between quantities. They are like different ways of expressing the same proportion. For example, the ratio 2:3 is equivalent to 4:6 because both represent the same proportional relationship. If you double both numbers in 2:3, you get 4:6. Think of equivalent ratios as fractions that have the same value. Just like 1/2 is equivalent to 2/4, ratios can be equivalent too! Visualizing Equivalent Ratios Remember Equivalent ratios can be found by multiplying or dividing both terms by the same number. How to Find Equivalent Ratios There are two main ways to find equivalent ratios: Multiplication Method Multiply both terms of the ratio by the same number. Division Method Divide both terms of the ratio by the same number. Important You can multiply or divide by any number except zero. The number must be the same for both parts of the ratio. Examples of Equivalent Ratios Let's look at some examples of equivalent ratios in different contexts: Example 1: Recipe Scaling A recipe calls for 2 cups of flour and 1 cup of sugar (ratio 2:1). If you want to make double the recipe, you'd use 4 cups of flour and 2 cups of sugar (ratio 4:2). The ratios 2:1 and 4:2 are equivalent. Example 2: Map Scale A map has a scale of 1 cm:10 km. This is equivalent to 2 cm:20 km and 5 cm:50 km. All represent the same proportional relationship between distance on the map and actual distance. Example 3: Mixing Paint To make purple paint, you mix 3 parts red with 2 parts blue (3:2). This is equivalent to 6 parts red and 4 parts blue (6:4) or 9 parts red and 6 parts blue (9:6). All these mixtures will produce the same shade of purple. Example 4: Classroom Ratios In a classroom, there are 4 girls for every 5 boys (4:5). If the class size doubles, there would be 8 girls for every 10 boys (8:10). The ratios 4:5 and 8:10 are equivalent. Practice Tip To check if two ratios are equivalent, divide the first number by the second number in each ratio. If you get the same decimal, they are equivalent! Equivalent Ratios Practice Quiz Test your understanding of equivalent ratios with this 5-question quiz. Choose the correct answer for each question. Frequently Asked Questions Here are answers to common questions about equivalent ratios: Math Trivia Discover interesting facts about ratios and proportions: Ancient Ratios The concept of ratios dates back to ancient Egypt and Babylon. Egyptian architects used ratios to design pyramids, maintaining consistent proportions regardless of size. Golden Ratio The golden ratio (approximately 1.618:1) appears throughout nature - in seashells, flower petals, and even human body proportions. Artists and architects have used it for centuries for its pleasing proportions. Scale Models Scale models use ratios to represent real objects. A 1:24 scale model car means that 1 unit on the model equals 24 units on the real car. All dimensions maintain this proportional relationship. Largest Ratio The largest ratio ever calculated in mathematics is Graham's number, which is so large that the observable universe isn't big enough to write it down. It appears in solutions to certain problems in Ramsey theory. Related Resources 1 Visual Model Equivalent Fractions Using visual models students will recognize and generate equivalent fractions 1 5th Grade Equivalent Fractions Strengthen your 5th graders' understanding of equivalent fractions with this engaging worksheet alig... 1 Identify Equivalent Fractions This comprehensive worksheet is designed to reinforce 5th-grade students' understanding of equivalen... 1 Equivalent Fractions Missing Numbers This worksheet is great for practicing and learning equivalent fractions! Students will need to make... 1 Tenths and Hundredths Equivalent Fractions Download now for practice expressing a fraction with denominator 10 as an equivalent fraction with d... 1 Equivalent Fractions This equivalent fractions worksheet is great for practicing this skill without the aid of visual rep... 1 Equivalent Fractions Visual Model This equivalent fractions worksheet is a great introduction to this skill. Students use visual model... 1 4th Grade Tenths and Hundredths Equivalent Fractions Download now for practice expressing a fraction with denominator 10 as an equivalent fraction with d... 1 How Many 10ths and 100ths Equivalent Fractions Download now for practice expressing a fraction with denominator 10 as an equivalent fraction with d... 1 Equivalent Fraction Resource This equivalent fractions worksheet is great for practicing this skill without the aid of visual rep... Related Resources Visual Model Equivalent Fractions Using visual models students will recognize and generate equivalent fractions 5th Grade Equivalent Fractions Strengthen your 5th graders' understanding of equivalent fractions with this engaging worksheet alig... Identify Equivalent Fractions This comprehensive worksheet is designed to reinforce 5th-grade students' understanding of equivalen... Equivalent Fractions Missing Numbers This worksheet is great for practicing and learning equivalent fractions! Students will need to make... Tenths and Hundredths Equivalent Fractions Download now for practice expressing a fraction with denominator 10 as an equivalent fraction with d... Equivalent Fractions This equivalent fractions worksheet is great for practicing this skill without the aid of visual rep... Equivalent Fractions Visual Model This equivalent fractions worksheet is a great introduction to this skill. Students use visual model... 4th Grade Tenths and Hundredths Equivalent Fractions Download now for practice expressing a fraction with denominator 10 as an equivalent fraction with d... How Many 10ths and 100ths Equivalent Fractions Download now for practice expressing a fraction with denominator 10 as an equivalent fraction with d... Equivalent Fraction Resource This equivalent fractions worksheet is great for practicing this skill without the aid of visual rep... Curriculum Resources by Grade Find learning resources for: Related Topics X and Y Axis Learn all about the x and y axis in coordinate geometry with clear definitions, visual examples, interactive quizzes, and interesting facts about coordinate systems. Feet to Centimeters (feet to cm) Learn how to convert feet to centimeters with easy explanations, conversion charts, examples, and interactive quizzes. Perfect for K-12 students learning measurement units. 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2258	https://www.dreambox.com/math/skills/multiplication/order-of-operations-pemdas	Skip to content Math Resources What is the Order of Operations in Math (PEMDAS)? Does the order in which we solve math problems matter? Yes! Let’s learn about the order of operations in math, aka PEMDAS, to solve multi-step expressions. Author Amber Watkins Published: Oct 2024 Key takeaways • The order of operations teaches the order in which to solve multi-operational expressions. • The order of operations can easily be remembered using the acronym PEMDAS. • As your child successfully masters the order of operations, they will be able to solve more advanced problems with ease. When reading a math problem, we usually start from left to right. So it is natural to want to solve problems in the same way. Does this method get us the right answer for every problem? Let’s solve the following problem in two ways to see how order makes a difference: 3 – 2 x 5 = 1. You can solve this problem bysubtractingfirst. 3 – 2 x 5 1 x 5 5 2. Or you can solve it bymultiplyingfirst. 3 – 2 x 5 3 – 10 -7 In both examples, we took the same actions: we multiplied and subtracted, but switched the order. As a result, we found two very different answers. How can we know which is the right answer? By following the math order of operations. What is the Order of Operations in Math? Operations are the actions we take in math. The most common math operations are adding, subtracting, multiplying, dividing, simplifying exponents, and finding square roots. Some math problems have more than one kind of operation in the problem. How can we solve those types of problems? We follow the order of operations! What is the order of operations? The order of operations tells us the order in which to perform operations when the problem contains more than one kind of operation. Order of Operations Rules Many times you will see the order of operations called PEMDAS. What is PEMDAS? PEMDAS is an acronym that helps us remember the order in which to solve multi-operational problems. What does PEMDAS stand for? PEMDAS is an acronym that stands for Parenthesis, Exponents, Multiplication or Division, Adding or Subtracting. Now that we know the Pemdas meaning, let’s discuss the Pemdas Rule. The most important PEMDAS Rule is when solving order of operation problems in math, you must follow the order of the acronym. First, you solve whatever operations are in the parenthesis.() Second, you calculate any exponents. x² Third, you multiply or divide. x or ÷ Fourth, youadd or subtract.+ or – BODMAS vs PEMDAS You may also hear the order of operations called BODMAS. What is BODMAS? BODMAS stands for Brackets, Orders (Powers or Roots), Division or Multiplication, Adding or Subtracting. Brackets are another name for parenthesis and orders are another name for exponents. Although the terms used for operations are different than Pemdas, both Bodmas and Pemdas follow the same rules of order of operations. What are those rules and how can we use them to solve PEMDAS problems? Table of contents What is it? Order of operations BODMAS vs PEMDAS How to solve Practice Get more practice with PEMDAS and math with DreamBox! Start Free Trial Turn math into playtime with DreamBox Math DREAMBOX MATH Get started for FREE today! Purchase DreamBox Math Try DreamBox for 14 Days How to Solve Order of Operations Problems Let’s solve this Order of Operations problem as we learn some PEMDAS rules.4² – ( 3 x 5 ) + 9 ፥ 3 x 2 Pemdas Rulefor ParenthesisWhen solving order of operation problems, first complete the operations that are found inside the parenthesis or brackets. In our example problem, we would multiply 3 x 5 first because it is in theparenthesis.PE M D A S 4² –( 3 x 5 )+ 9 ፥ 3 x 2 4² –15+ 9 ፥ 3 x 22.Pemdas Rulefor ExponentsExponents should always be simplified before moving to the next step in the order of operations. If you begin multiplying, dividing, adding, or subtracting while there is still an exponent in the problem, you may have missed this step. This is how we would simplify theexponents.PEM D A S 4²– 15 + 9 ፥ 3 x 216 – 15 + 9 ፥ 3 x 23. Pemdas Rulefor Multiplying or DividingWhen using the order of operations, you may have noticed the word “or” is used when multiplying or dividing. When following the order of operations in math, you can multiply or divide in any order. Simply complete the operation that comes first. We have both multiplication and division in this problem, so we will go from left to right. First is division, then is multiplication. P EM DA S 16 – 15 +9 ፥ 3x 2 16 – 15 +3 x 2 16 – 15 + 6 4.Pemdas Rulefor Adding or Subtraction When using the order of operations, you may have also noticed the word “or” is used when adding or subtracting. When following the order of operations in math, you can add or subtract in any order. Simply complete the operation that comes first. Finally, we would add or subtract depending on which operation comes first. PEMDAS 16 – 15+ 6 1+ 6 7 Now that we know some of the PEMDAS rules, let’s go over some PEMDAS examples together. PEDMAS Examples PEMDAS EXAMPLE 1: Solve 3 + ( 8 x 9 ) = Explanation: First, we see parenthesis comes first in the PEMDAS acronym. So we would multiply 8 x 9 because it is in the parenthesis. Therefore, 8 x 9 is equal to 72. Then we would add 3 and 72 to get 75 as our answer. Our Work for this PEMDAS example would look like this: 3 +( 8 x 9 )3 + 7275 PEMDAS EXAMPLE 2:Solve 12 ፥ 3 + 14 = First, we see that division comes before adding in PEMDAS. So we divide 12 by 3 which is equal to 4. Next, add 4 and 14 which equals 18.Our Work for this PEMDAS example would look like this:12 ፥ 3+ 144 + 1418 PEDMAS Practice Questions Click on the boxes below to see the answers! Use Pemdas to solve 2² + 12 – ( 3 x 2) = First, multiply 3 x 2 because it is in the parenthesis. Then, simplify the exponent 2² which equals 4. Next, add 4 + 12 which equals 16. Finally, subtract 6 which equals 10. 2² + 12 –(3 x 2)=2²+ 12 – 6 =4 + 12– 6 =16 – 6=10 Use Pemdas to Solve (3 x 2) + (12 ፥ 3 ) = First, multiply 3 x 2 because it is in the first parenthesis. Next, divide 12 by 3 because it is in the second parenthesis. Finally, add 6 + 4 which equals 10.(3 x 2)+ (12 ፥ 3 )6 +(12 ፥ 3 )6 + 410 Use Pemdas to Solve 100 ፥ 10 – (6 ፥ 2 ) = First, divide 6 by 2 because it is in the parenthesis. Next, divide 100 by 10. Finally, subtract 3 from 10 which equals 7. 100 ፥ 10 –(6 ፥ 2 )100 ፥ 10– 310 – 37 The math program that drives results Get started today! DreamBox adapts to your child’s level and learning needs, ensuring they are appropriately challenged and get confidence-building wins. Purchase DreamBox Math Try DreamBox for 14 Days FAQs About Order of Operations What does PEMDAS stand for? Pemdas stands for Parenthesis, Exponents, Multiplication or Division, Adding or Subtracting. Do you multiply or divide first in PEMDAS? In PEMDAS problems, multiplication or division can happen in any order. The Pemdas rule is to complete whichever operation comes first. Which math operation comes first? In Pemdas problems, always complete whichever operation is inside of the parenthesis first. What does BODMAS stand for? BODMAS stands for Brackets, Orders, Division or Multiplication, Adding or Subtracting. Are PEMDAS and BODMAS the same? Yes. PEMDAS and BODMAS are both acronyms to represent the order of operations, but they use different words to describe the same operation. For example, PEMDAS uses the terms parenthesis and exponents. BODMAS uses the terms brackets and powers. Take at home math practice to the next level Empowering parents and educators to make math practice more impactful. Plus, your kids will love it. Try DreamBox for Free! About the Author Amber Watkins Hello, my name is Miss Amber. I am an Education Specialist with a degree in Early Childhood Education. I have over 12 years of experience teaching and tutoring elementary through college level math. I also create both video and written math content that has helped educate thousands of students online. Knowing that my work in math education makes such an impact leaves me with an indescribable feeling of pride and joy!
2259	https://m.youtube.com/watch?v=_EmzkyT6ev8&pp=0gcJCdgAo7VqN5tD	Proof of the derivative of lnx: A Step-by-Step Proof and Explanation Math with Alex 1820 subscribers 311 likes Description 12812 views Posted: 23 Aug 2023 In this video, we will prove the derivative of ln(x) using the limit definition of the derivative, also known as the First Principle. I will explain it step by step to ensure that you can easily follow along and understand. If there's any part of the video that you find confusing or if you have any questions, feel free to leave a comment. If you're eager to explore more differentiation techniques for various functions, click the link below. Stay tuned for a variety of upcoming differentiation videos. Additionally, I recommend watching the easy integration videos using the Box Rule below, as they will complement your learning experience. Also, there are various other videos available on my channel and more will be uploaded continuously, so feel free to visit and check them out. derivatives #differentiation #mathwithalex 30 comments Transcript: Introduction In this video, we will prove why the derivative of natural logarithm of x, ln(x), is 1/x using the limit definition of the derivative, also known as the First Principle. To do this, we will utilize the limit definition of the derivative formula and the definition of Euler's number shown below the screen. Proof Let's first write down the statement we want to prove and the necessary formulas. Here, we have the limit definition of the derivative formula and Euler's number. Let's define f(x) as ln(x). Then, f(x + h) will be ln(x + h). By substituting these expressions into the limit definition of the derivative formula, the numerator becomes ln(x + h) - ln(x), and the denominator remains as h. Looking at the numerator, we can simplify it using the logarithmic property of subtracting logarithms as ln((x + h)/x). We can rearrange the denominator by expressing h as a fraction, allowing us to move 1/h to the front of the logarithm. And if we express the fraction inside the logarithm as two separate fractions, (x+h)/x can be rewritten as x/x + h/x. Since x/x is equal to 1, we can simplify the expression further to the limit of 1/h ln(1 + h/x). Now, let's move the 1/h that was multiplied to the natural logarithm to the exponent position inside the logarithm. This gives us the limit of ln(1 + h/x)^(1/h). It looks similar to Euler's number shown in the top right corner of the screen, but it requires further manipulation. Therefore, let's substitute h/x with n. As h approaches 0, n will also approach 0. Multiplying both sides by x gives us h = nx, and taking the reciprocal of the expression gives us 1/h = 1/nx. By substituting all occurrences of "h" with "n" in the above limit, we can express it in terms of "n" approaching 0. The expression "h/x" inside the natural logarithm becomes "n" and the exponent "1/h" becomes "1/(nx)". Rewriting the exponent 1/(nx) as 1/n times 1/x, the limit becomes n approaches 0 of ln((1+n)^(1/n))^(1/x). Taking the exponent 1/x outside the logarithm using the properties of logarithms, we get 1/x ln(1+n)^(1/n)", and the 1/x term is independent of the limit as "n" approaches 0, so we can move it back in front of the limit. Thus, the expression becomes 1/x limit of ln(1+n)^(1/n). Conclusion Since Euler's number "e" is defined as the limit of ln(1+n)^(1/n) when n approaches 0, we can replace the limit with "e", simplifying the expression to 1/x ln(e), which is equal to 1/x. Therefore, we conclude that the derivative of ln(x) is 1/x. Today, we have demonstrated why the derivative of the natural logarithm of x is 1/x using the limit definition of the derivative and the first principle. Now you can use the proven result to find the derivative of logarithmic functions. On my channel, you will find many videos covering the proofs of derivatives for various functions and quick methods to perform differentiations easily. I believe they will be helpful for your learning. I want to express my gratitude to all of you who watched today's video. Your subscriptions and likes bring great joy to me. Thank you once again, and I'll see you next time. This has been your Math Guardian Alex
2260	https://math.stackexchange.com/questions/3091733/chebyshevs-sum-inequality-proof	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Chebyshev's Sum Inequality Proof Ask Question Asked Modified 6 years, 8 months ago Viewed 676 times 1 $\begingroup$ So I was reading up on Chebyshev's Sum Inequality, and I was a little confused about the first proof presented on Wikipedia. Specifically, the line which reads "opening the brackets". What does this phrase mean exactly, and how does the author transition to the next line? Thank you! real-analysis proof-explanation rearrangement-inequality Share edited Jan 29, 2019 at 6:15 the_fox 5,96833 gold badges2525 silver badges4949 bronze badges asked Jan 29, 2019 at 4:10 D. J.D. J. 3155 bronze badges $\endgroup$ Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ \begin{align} \sum_j \sum_k (a_j - a_k)(b_j - b_k) &= \sum_j \sum_k (a_j b_j - a_k b_j - a_j b_k + a_k b_k) \ &= \sum_j n a_j b_j - \sum_j \sum_k a_k b_j - \sum_j \sum_k a_j b_k + \sum_k n a_k b_k \ &= \cdots \end{align} Share answered Jan 29, 2019 at 4:15 angryavianangryavian 93.8k77 gold badges7474 silver badges145145 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ The Rearrangement inequality says: Let $a_1\ge\dots\ge a_n$ and $b_1\ge\dots\ge b_n$. For all permutation $\sigma\in S_n$ prove that: $$\sum_{i=1}^na_ib_{n-i+1}\le\sum_{i=1}^na_ib_{\sigma(i)}\leq\sum_{i=1}^na_ib_i.$$ We see that $\sum\limits_{i=1}^na_ib_i$ is a biggest sum between all sums as $\sum\limits_{i=1}^na_ib_{\sigma(i)}$. Thus, after opening the brackets we obtain: $$n\sum\limits_{i=1}^na_ib_i\geq\sum_{i=1}^na_i\sum_{i=1}^nb_i.$$ We took here all only cyclic permutations of $(1,2,...,n).$ For example. Let $a\geq b\geq c$ and $x\geq y\geq z$. Thus, by Rearrangement $$3(ax+by+cz)\geq(a+b+c)(x+y+z)$$ because after opening the brackets we obtain: $$(a+b+c)(x+y+z)=(ax+by+cz)+(ay+bz+cx)+(az+bx+cy)\leq$$ $$\leq(ax+by+cz)+(ax+by+cz)+(ax+by+cz)=3(ax+by+cz).$$ Share edited Jan 29, 2019 at 5:37 answered Jan 29, 2019 at 5:20 Michael RozenbergMichael Rozenberg 208k3131 gold badges171171 silver badges294294 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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2261	https://www.nagwa.com/en/videos/516187068683/	Question Video: Understanding Unit Vectors in Relation To Scalar Multiplication \| Nagwa Question Video: Understanding Unit Vectors in Relation To Scalar Multiplication \| Nagwa Sign Up Sign In English English العربية English English العربية My Wallet Sign Up Sign In My Classes My Messages My Reports My Wallet My Classes My Messages My Reports Question Video: Understanding Unit Vectors in Relation To Scalar Multiplication Mathematics • Third Year of Secondary School Consider the vector 𝐀 = 5𝐢 − 2𝐣 − 4𝐤. Is the unit vector in the direction of 𝐀 the same as the unit vector in the direction of 3𝐀? Pause Play % buffered 00:00 00:00 0:00 Unmute Mute Disable captions Enable captions Settings Captions English Quality 0 Speed Normal Captions Go back to previous menu Disabled English EN Quality Go back to previous menu Speed Go back to previous menu 0.5×0.75×Normal 1.25×1.5×1.75×2× Exit fullscreen Enter fullscreen Play 04:08 Video Transcript Consider the vector 𝐀 five 𝐢 minus two 𝐣 minus four 𝐤. Is the unit vector in the direction of 𝐀 the same as the unit vector in the direction of three 𝐀? We know that vector 𝐀 can be rewritten in the form five, negative two, negative four. We are also asked to consider the vector three 𝐀. This involves multiplying vector 𝐀 by the scalar or constant three. This involves multiplying each component by the scalar three. Three multiplied by five is 15. Three multiplied by negative two is negative six. And three multiplied by negative four is negative 12. This means that three 𝐀 is equal to 15, negative six, negative 12. The unit vector 𝐕 hat is equal to one over the magnitude of vector 𝐕 multiplied by vector 𝐕. This is the same as dividing the vector by its magnitude. We know that to calculate the magnitude of any vector, we find the square root of the sum of the squares of each component. The magnitude of vector 𝐀 is equal to the square root of five squared plus negative two squared plus negative four squared. This is equal to the square root of 25 plus four plus 16. This simplifies to the square root of 45, which in turn is equal to three root five. The magnitude of vector 𝐀 is three root five. The unit vector in the direction of vector 𝐀 is therefore equal to one over three root five multiplied by five, negative two, negative four. We could rewrite this as the vector five over three root five, negative two over three root five, and negative four over three root five. Let’s now consider the unit vector in the direction of three 𝐀. The magnitude of vector three 𝐀 is equal to the square root of 15 squared plus negative six squared plus negative 12 squared. This is equal to the square root of 405, which in turn simplifies to nine root five. The magnitude of vector three 𝐀 is equal to nine root five. We might notice at this point that this is three times the magnitude of vector 𝐀. This leads us to a general rule. The magnitude of 𝑘𝐕 is equal to 𝑘 multiplied by the magnitude of vector 𝐕. This means that three multiplied by the magnitude of vector 𝐀 is equal to three multiplied by three root five. The unit vector in the direction of three 𝐀 is therefore equal to one over nine root five multiplied by the vector 15, negative six, negative 12. Factoring out a three from the vector gives us three over nine root five multiplied by the vector five, negative two, negative four. This is the same as one over three root five multiplied by the vector five, negative two, negative four. We can therefore conclude that the answer is yes, the unit vector in the direction of 𝐀 is the same as the unit vector in the direction of three 𝐀. This will be true of any vector multiplied by a positive scalar. As long as our value of 𝑘 is positive, then the unit vector in the direction of 𝐀 will be the same as the unit vector in the direction of 𝑘𝐀. Lesson Menu Lesson Lesson Plan Lesson Presentation Lesson Video Lesson Explainer Lesson Playlist Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Interactive Sessions Chat & Messaging Realistic Exam Questions Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company About Us Contact Us Privacy Policy Terms and Conditions Careers Tutors Content Lessons Lesson Plans Presentations Videos Explainers Playlists Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy Accept
2262	https://www.nidcd.nih.gov/health/vestibular-schwannoma-acoustic-neuroma-and-neurofibromatosis	Vestibular Schwannoma (Acoustic Neuroma) & Neurofibromatosis \| NIDCD Skip to main content An official website of the United States government Here's how you know Here's how you know U.S. Department of Health & Human Services \| National Institutes of Health Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock () or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Due to HHS and NIH restructuring, the information provided on nidcd.nih.gov is not being updated currently. Questions can be directed to Ask NIH. Menu NIDCD Employee Intranet \| A–Z Index \| En español Home Health Info Health Info Hearing, Ear Infections, and Deafness Balance Taste and Smell Voice, Speech, and Language Información en español Statistics Health Resources Clinical Studies Adult Hearing Health Care Research Research Extramural Research Intramural Research Scientific Workshop and Meeting Reports Clinical Studies Global Health Grants & Funding Grants & Funding Types of Grants and Funding How to Apply for a Grant All NIDCD Funding Opportunities The BRAIN Initiative and the NIDCD Training Training About NIDCD's Research Training Program Types of Research Training Funding Opportunities Research Training in NIDCD Laboratories (Intramural) More Training Resources News & Events News & Events NIDCD News Grantee News Multimedia Events About Us About Us Messages from the NIDCD Director Mission Strategic Plans NIDCD History and Milestones Organizational Structure and Staff Congressional Testimony and the NIDCD Budget Advisory Groups and Review Committees Contact Us Get Involved Breadcrumb Home Health Information Vestibular Schwannoma (Acoustic Neuroma) and Neurofibromatosis On this page: What is a vestibular schwannoma (acoustic neuroma)? How is a vestibular schwannoma diagnosed? How is a vestibular schwannoma treated? What is the difference between unilateral and bilateral vestibular schwannomas? What is being done about vestibular schwannoma? Where can I find more information about vestibular schwannomas? What is a vestibular schwannoma (acoustic neuroma)? Inner ear with vestibular schwannoma (tumor) A vestibular schwannoma (also known as acoustic neuroma, acoustic neurinoma, or acoustic neurilemoma) is a benign, usually slow-growing tumor that develops from the balance and hearing nerves supplying the inner ear. Source: NIH/NIDCD A vestibular schwannoma (also known as acoustic neuroma, acoustic neurinoma, or acoustic neurilemoma) is a benign, usually slow-growing tumor that develops from the balance and hearing nerves supplying the inner ear. The tumor comes from an overproduction of Schwann cells—the cells that normally wrap around nerve fibers like onion skin to help support and insulate nerves. As the vestibular schwannoma grows, it affects the hearing and balance nerves, usually causing unilateral (one-sided) or asymmetric hearing loss, tinnitus (ringing in the ear), and dizziness/loss of balance. As the tumor grows, it can interfere with the face sensation nerve (the trigeminal nerve), causing facial numbness. Vestibular schwannomas can also affect the facial nerve (for the muscles of the face) causing facial weakness or paralysis on the side of the tumor. If the tumor becomes large, it will eventually press against nearby brain structures (such as the brainstem and the cerebellum), becoming life-threatening. How is a vestibular schwannoma diagnosed? Unilateral/asymmetric hearing loss and/or tinnitus and loss of balance/dizziness are early signs of a vestibular schwannoma. Unfortunately, early detection of the tumor is sometimes difficult because the symptoms may be subtle and may not appear in the beginning stages of growth. Also, hearing loss, dizziness, and tinnitus are common symptoms of many middle and inner ear problems (the important point here is that unilateral or asymmetric symptoms are the worrisome ones). Once the symptoms appear, a thorough ear examination and hearing and balance testing (audiogram, electronystagmography, auditory brainstem responses) are essential for proper diagnosis. Magnetic resonance imaging (MRI) scans are critical in the early detection of a vestibular schwannoma and are helpful in determining the location and size of a tumor and in planning its microsurgical removal. How is a vestibular schwannoma treated? Early diagnosis of a vestibular schwannoma is key to preventing its serious consequences. There are three options for managing a vestibular schwannoma: (1) surgical removal, (2) radiation, and (3) observation. Sometimes, the tumor is surgically removed (excised). The exact type of operation done depends on the size of the tumor and the level of hearing in the affected ear. If the tumor is small, hearing may be saved and accompanying symptoms may improve by removing it to prevent its eventual effect on the hearing nerve. As the tumor grows larger, surgical removal is more complicated because the tumor may have damaged the nerves that control facial movement, hearing, and balance and may also have affected other nerves and structures of the brain. The removal of tumors affecting the hearing, balance, or facial nerves can sometimes make the patient’s symptoms worse because these nerves may be injured during tumor removal. As an alternative to conventional surgical techniques, radiosurgery (that is, radiation therapy—the “gamma knife” or LINAC) may be used to reduce the size or limit the growth of the tumor. Radiation therapy is sometimes the preferred option for elderly patients, patients in poor medical health, patients with bilateral vestibular schwannoma (tumor affecting both ears), or patients whose tumor is affecting their only hearing ear. When the tumor is small and not growing, it may be reasonable to “watch” the tumor for growth. MRI scans are used to carefully monitor the tumor for any growth. What is the difference between unilateral and bilateral vestibular schwannomas? Unilateral vestibular schwannomas affect only one ear. They account for approximately 8 percent of all tumors inside the skull; approximately one out of every 100,000 individuals per year develops a vestibular schwannoma. Symptoms may develop at any age but usually occur between the ages of 30 and 60 years. Most unilateral vestibular schwannomas are not hereditary and occur sporadically. Approximately one out of every 100,000 individuals per year develops a vestibular schwannoma. Bilateral vestibular schwannomas affect both hearing nerves and are usually associated with a genetic disorder called neurofibromatosis type 2 (NF2). Half of affected individuals have inherited the disorder from an affected parent and half seem to have a mutation for the first time in their family. Each child of an affected parent has a 50 percent chance of inheriting the disorder. Unlike those with a unilateral vestibular schwannoma, individuals with NF2 usually develop symptoms in their teens or early adulthood. In addition, patients with NF2 usually develop multiple brain and spinal cord related tumors. They also can develop tumors of the nerves important for swallowing, speech, eye and facial movement, and facial sensation. Determining the best management of the vestibular schwannomas as well as the additional nerve, brain, and spinal cord tumors is more complicated than deciding how to treat a unilateral vestibular schwannoma. Further research is needed to determine the best treatment for individuals with NF2. Scientists believe that both unilateral and bilateral vestibular schwannomas form following the loss of the function of a gene on chromosome 22. (A gene is a small section of DNA responsible for a particular characteristic like hair color or skin tone). Scientists believe that this particular gene on chromosome 22 produces a protein that controls the growth of Schwann cells. When this gene malfunctions, Schwann cell growth is uncontrolled, resulting in a tumor. Scientists also think that this gene may help control the growth of other types of tumors. In NF2 patients, the faulty gene on chromosome 22 is inherited. For individuals with unilateral vestibular schwannoma, however, some scientists hypothesize that this gene somehow loses its ability to function properly. What is being done about vestibular schwannoma? There are three options for managing a vestibular schwannoma: (1) surgical removal, (2) radiation, and (3) observation. Scientists continue studying the molecular pathways that control normal Schwann cell development to better identify gene mutations that result in vestibular schwannomas. Scientists are working to better understand how the gene works so they can begin to develop new therapies to control the overproduction of Schwann cells in individuals with vestibular schwannoma. Learning more about the way genes help control Schwann cell growth may help prevent other brain tumors. In addition, scientists are developing robotic technology to assist physicians with acoustic neuroma surgery. Where can I find more information about vestibular schwannomas? The NIDCD maintains a directory of organizations that provide information on the normal and disordered processes of hearing, balance, taste, smell, voice, speech, and language. 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2263	https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(LibreTexts)/07%3A_Rational_Expressions_and_Equations/7.04%3A_Complex_Rational_Expressions	7.4: Complex Rational Expressions - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 7: Rational Expressions and Equations Elementary Algebra (LibreTexts) { } { "7.01:_Simplifying_Rational_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_Multiplying_and_Dividing_Rational_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Adding_and_Subtracting_Rational_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Complex_Rational_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.05:_Solving_Rational_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.06:_Applications_of_Rational_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.07:_Variation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.08:_7.E_Review_Exercises_and_Sample_Exam" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Real_Numbers_and_Their_Operations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Linear_Equations_and_Inequalities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Graphing_Lines" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Solving_Linear_Systems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Polynomials_and_Their_Operations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Factoring_and_Solving_by_Factoring" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Rational_Expressions_and_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Radical_Expressions_and_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Solving_Quadratic_Equations_and_Graphing_Parabolas" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Appendix-_Geometric_Figures" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Tue, 19 Aug 2025 03:32:51 GMT 7.4: Complex Rational Expressions 18371 18371 Joshua Halpern { } Anonymous Anonymous 2 false false [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:anonymous", "licenseversion:30", "program:hidden", "cssprint:dense", "source@ ] [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:anonymous", "licenseversion:30", "program:hidden", "cssprint:dense", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Bookshelves 3. Algebra 4. Elementary Algebra (LibreTexts) 5. 7: Rational Expressions and Equations 6. 7.4: Complex Rational Expressions Expand/collapse global location 7.4: Complex Rational Expressions Last updated Aug 19, 2025 Save as PDF 7.3: Adding and Subtracting Rational Expressions 7.5: Solving Rational Equations Page ID 18371 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Definitions 3. Method 1: Simplify Using Division 1. Example 7.4.1 2. Example 7.4.2 3. Example 7.4.3 4. Example 7.4.4 5. Exercise 7.4.1 Method 2: Simplify Using the LCD Example 7.4.5 Note Example 7.4.6 Example 7.4.7 Exercise 7.4.2 Key Takeaways Exercise 7.4.3 Complex Rational Expressions Exercise 7.4.4 Discussion Board Topics Learning Objectives Simplify complex rational expressions by multiplying the numerator by the reciprocal of the divisor. Simplify complex rational expressions by multiplying numerator and denominator by the least common denominator (LCD). Definitions A complex fraction is a fraction where the numerator or denominator consists of one or more fractions. For example, 3 4 1 2 Simplifying such a fraction requires us to find an equivalent fraction with integer numerator and denominator. One way to do this is to divide. Recall that dividing fractions involves multiplying by the reciprocal of the divisor. 3 4 1 2=3 2 4⁢⋅2 1 1⁢=3 2 M⁢e⁢t⁢h⁡o⁢d 1:u⁢s⁢i⁢n⁢g⁡d⁢i⁢v⁢i⁢s⁢i⁢o⁢n An alternative method for simplifying this complex fraction involves multiplying both the numerator and denominator by the LCD of all the given fractions. In this case, the LCD = 4. 3 4⋅4 1 2⋅4=3 2 M⁢e⁢t⁢h⁡o⁢d 2:u⁢s⁢i⁢n⁢g⁡t⁢h⁡e L⁢C⁢D A complex rational expression is defined as a rational expression that contains one or more rational expressions in the numerator or denominator or both. For example, 1 2+1 x 1 4−1 x 2 We simplify a complex rational expression by finding an equivalent fraction where the numerator and denominator are polynomials. As illustrated above, there are two methods for simplifying complex rational expressions, and we will outline the steps for both methods. For the sake of clarity, assume that variable expressions used as denominators are nonzero. Method 1: Simplify Using Division We begin our discussion on simplifying complex rational expressions using division. Before we can multiply by the reciprocal of the divisor, we must simplify the numerator and denominator separately. The goal is to first obtain single algebraic fractions in the numerator and the denominator. The steps for simplifying a complex algebraic fraction are illustrated in the following example. Example 7.4.1 Simplify: 1 2+1 x 1 4−1 x 2 Solution: Step 1: Simplify the numerator and denominator. The goal is to obtain a single algebraic fraction divided by another single algebraic fraction. In this example, find equivalent terms with a common denominator in both the numerator and denominator before adding and subtracting. 1 2+1 x 1 4−1 x 2=1 2⋅x x+1 x⋅2 2 1 4⋅x 2 x 2−1 x 2⋅4 4 =x 2⁢x+2 2⁢x x 2 4⁢x 2−4 4⁢x 2 E⁢q⁢u⁢i⁢v⁢a⁢l⁢e⁢n⁢t⁢f⁡r⁢a⁢c⁢t⁢i⁢o⁢n⁢s w⁢i⁢t⁢h⁡c⁢o⁢m⁢m⁢o⁢n d⁢e⁢n⁢o⁢m⁢i⁢n⁢a⁢t⁢o⁢r⁢s =x+2 2⁢x x 2−4 4⁢x 2 Add the fractions in the numerator and denominator. At this point we have a single algebraic fraction divided by a single algebraic fraction. Step 2: Multiply the numerator by the reciprocal of the divisor. x+2 2⁢x x 2−4 4⁢x 2⁢=x+2 2⁢x⋅⁢4⁢x 2 x 2−4 Step 3: Factor all numerators and denominators completely. =x+2 2⁢x⋅4⁢x 2(x+2)⁢(x−2) Step 4: Cancel all common factors. =4⁢x 2⋅(x+2)2⁢x⁢(x+2)⁢(x−2)=4⁢x 2 2⁢x⋅(x+2)2⁢x⁢(x+2)⁢(x−2)=2⁢x(x−2) Answer: 2⁢x x−2 Example 7.4.2 Simplify:1 x−1 x−2 4 x 2−2⁢x Solution: 1 x−1 x−2 4 x 2−2⁢x=1 x⋅(x−2)(x−2)−1 x−2⋅x x 4 x 2−2⁢x=x−2 x⁢(x−2)−x x⁢(x−2)4 x 2−2⁢x=x−2−x x⁢(x−2)4 x 2−2⁢x=−2 x⁢(x−2)4 x 2−2⁢x=−2 x⁢(x−2)⋅x 2−2⁢x 4=−1 x̸2⋅x̸⁢(x−2)=−1 2 Answer: −1 2 Example 7.4.3 Simplify: 1−4 x−21 x 2 1−2 x−15 x 2 Solution: The LCD of the rational expressions in both the numerator and denominator is x 2. Multiply by the appropriate factors to obtain equivalent terms with this as the denominator and then subtract. 1−4 x−21 x 2 1−2 x−15 x 2=1 1⋅x 2 x 2⁢−4 x⋅⁢x x⁢−21 x 2 1 1⋅x 2 x 2⁢−2 x⋅⁢x x⁢−15 x 2=x 2 x 2−4⁢x x 2−21 x 2 x 2 x 2−2⁢x x 2−15 x 2=x 2−4⁢x−21 x 2 x 2−2⁢x−15 x 2 We now have a single rational expression divided by another single rational expression. Next, multiply the numerator by the reciprocal of the divisor and then factor and cancel. Answer: x−7 x−5 Example 7.4.4 Simplify: 1−1 x 2 1 x−1 Solution: 1−1 x 2 1 x−1=1 1⋅x 2 x 2⁢−1 x 2 1 x−1 1⋅x x=x 2−1 x 2 1−x x=x 2−1 x 2⋅x 1−x=(x+1)⁢(x−1)x x 2⁢⋅x−1⋅(x−1)=x+1−1⋅x=−x+1 x Answer: −x+1 x Exercise 7.4.1 Simplify: 1 81−1 x 2 1 9+1 x Answer x−9 9⁢x Method 2: Simplify Using the LCD An alternative method for simplifying complex rational expressions involves clearing the fractions by multiplying the expression by a special form of 1. In this method, multiply the numerator and denominator by the least common denominator (LCD) of all given fractions. Example 7.4.5 Simplify: 1 2+1 x 1 4−1 x 2 Solution: Step 1: Determine the LCD of all the fractions in the numerator and denominator. In this case, the denominators of the given fractions are 2,x,4, and x 2. Therefore, the LCD is 4⁢x 2. Step 2: Multiply the numerator and denominator by the LCD. This step should clear the fractions in both the numerator and denominator. 1 2+1 x 1 4−1 x 2=(1 2+1 x)⁢⋅4⁢x 2(1 4−1 x 2)⁢⋅4⁢x 2 D⁢i⁢s⁢t⁢r⁢i⁢b⁢u⁢t⁢e.=1 2⁢⋅4⁢x 2⁢+1 x⁢⋅4⁢x 2 1 4⁢⋅4⁢x 2⁢−1 x 2⁢⋅4⁢x 2 C⁢a⁢n⁢c⁢e⁢l.=2⁢x 2+4⁢x x 2−4 This leaves us with a single algebraic fraction with a polynomial in the numerator and in the denominator. Step 3: Factor the numerator and denominator completely. =2⁢x 2+4⁢x x 2−4=2⁢x⁢(x+2)(x+2)⁢(x−2) Step 4: Cancel all common factors. =2⁢x⁢(x+2)(x+2)⁢(x−2)=2⁢x x−2 Answer: 2⁢x x−2 Note This was the same problem that we began this section with, and the results here are the same. It is worth taking the time to compare the steps involved using both methods on the same problem. Example 7.4.6 Simplify: 1−2 x−15 x 2 3−14 x−5 x 2 Solution: Considering all of the denominators, we find that the LCD is x 2. Therefore, multiply the numerator and denominator by x 2: At this point, we have a rational expression that can be simplified by factoring and then canceling the common factors. =(x+3)⁢(x−5)(3⁢x+1)⁢(x−5)C⁢a⁢n⁢c⁢e⁢l.=x+3 3⁢x+1 Answer: x+3 3⁢x+1 It is important to point out that multiplying the numerator and denominator by the same nonzero factor is equivalent to multiplying by 1 and does not change the problem. Because x 2 x 2=1, we can multiply the numerator and denominator by x 2 in the previous example and obtain an equivalent expression. Example 7.4.7 Simplify: 1 x+1+3 x−3 2 x−3−1 x+1 Solution: The LCM of all the denominators is (x+1)⁢(x−3). Begin by multiplying the numerator and denominator by these factors. 1 x+1+3 x−3 2 x−3−1 x+1=(1 x+1+3 x−3)⁢⋅(x+1)⁢(x−3)(2 x−3−1 x+1)⁢⋅(x+1)⁢(x−3)D⁢i⁢s⁢t⁢r⁢i⁢b⁢u⁢t⁢e.=1⁢(x+1)⁢(x−3)x+1+3⁢(x+1)⁢(x−3)x−3 2⁢(x+1)⁢(x−3)x−3−1⁢(x+1)⁢(x−3)x+1 C⁢a⁢n⁢c⁢e⁢l.=(x−3)+3⁢(x+1)2⁢(x+1)−1⁢(x−3)S⁢i⁢m⁢p⁢l⁢i⁢f⁡y.=x−3+3⁢x+3 2⁢x+2−x+3=4⁢x x+5 Answer: 4⁢x x+5 Exercise 7.4.2 Simplify: 1 y−1 4 1 16−1 y 2 Answer −4⁢y y+4 Key Takeaways Complex rational expressions can be simplified into equivalent expressions with a polynomial numerator and polynomial denominator. One method of simplifying a complex rational expression requires us to first write the numerator and denominator as a single algebraic fraction. Then multiply the numerator by the reciprocal of the divisor and simplify the result. Another method for simplifying a complex rational expression requires that we multiply it by a special form of 1. Multiply the numerator and denominator by the LCM of all the denominators as a means to clear the fractions. After doing this, simplify the remaining rational expression. An algebraic fraction is reduced to lowest terms if the numerator and denominator are polynomials that share no common factors other than 1. Exercise 7.4.3 Complex Rational Expressions Simplify. (Assume all denominators are nonzero.) 1 2 5 4 7 8 5 4 10 3 20 9 −4 21 8 7 2 3 5 6 7 4 14 3 1−3 2 5 4−1 3 1 2−5 1 2+1 3 1+3 2 1−1 4 2−1 2 1+3 4 5⁢x 2 x+1 25⁢x x+1 7+x 7⁢x x+7 14⁢x 2 3⁢y x y 2 x−1 5⁢a 2 b−1 15⁢a 3(b−1)2 1+1 x 2−1 x 2 x+1 3−1 x 2 3⁢y−4 6−1 y 5 y−12 10−y y 2 1 5−1 x 1 25−1 x 2 1 x+1 5 1 25−1 x 2 1 x−1 3 1 9−1 x 2 1 4+1 x 1 x 2−1 16 16−1 x 2 1 x−4 2−1 y 1−1 4⁢y 2 1 x+1 y 1 y 2−1 x 2 1 2⁢x−4 3 1 4⁢x 2−16 9 2 25−1 2⁢x 2 1 5−1 2⁢x 4 25−1 4⁢x 2 1 5+1 4⁢x 1 y−1 x 4−2 x⁢y 1 a⁢b+2 1 a+1 b 1 y+1 x x⁢y 3 x 1 3−1 x 1−4 x−21 x 2 1−2 x−15 x 2 1−3 x−4 x 2 1−16 x 2 3−1 2⁢x−1 2⁢x 2 2−2 x+1 2⁢x 2 1 2−5 x+12 x 2 1 2−6 x+18 x 2 1 x−4 3⁢x 2 3−8 x+16 3⁢x 2 1+3 10⁢x−1 10⁢x 2 3 5−1 10⁢x−1 5⁢x 2 x−1 1+4 x−5 x 2 2−5 2⁢x−3⁢x 24⁢x+3 1 x−3+2 x 1 x−3 x−3 14 x−5+1 x 2 1 x 2+13 x−10 1 x+5+4 x−2 2 x−2−1 x+5 3 x−1−2 x+3 2 x+3+1 x−3 x x+1−2 x+3 x 3⁢x+4+1 x+1 x x−9+2 x+1 x 7⁢x−9−1 x+1 x 3⁢x+2−1 x+2 x x+2−2 x+2 x x−4+1 x+2 x 3⁢x+4+1 x+2 a 3−8⁢b 3 2 7 a−2 b 27⁢a 3+b 3 a b 3⁢a+b 1 b 3+1 a 3 1 b+1 a 1 b 3−1 a 3 1 a−1 b x 2+y 2 x y+2 x 2−y 2 2 x⁢y x y+4+4⁢y⁢x x⁢y+3+2⁢y⁢x 1+1 1+1 2 2−1 1+1 3 1 1+1 1+x x+1 x 1−1 x+1 1−1 x x−1 x 1 x−x x−1 x 2 Answer 1. 2 5 3 2 4 5 −6 11 10 3 x 5 3⁢(x−1)y⁢x x+1 2⁢x−1 −2 3 5⁢x x+5 −3⁢x x+3 −4⁢x+1 x x⁢y x−y 2⁢x+5 5⁢x x−y 4⁢x⁢y−2 x+y y 2⁢x 2 x−7 x−5 3⁢x+1 2⁢x−1 1 3⁢x−4 x⁢(x−1)x+4−5 x 2 3⁢x−6−2⁢x−3 5⁢x+18 x+12 (x−1)⁢(3⁢x+4)(x+2)⁢(x+3) x+1 3⁢x+2 7⁢b−2⁢a 2⁢b⁢(3−8⁢b 3) a 2−b⁢a+b 2 b 2⁢a 2 2⁢(x 2+y 2)x⁢(x+2⁢y)⁢(x 2−y 2) 5 3 x+1 x+2 1 x+1 Exercise 7.4.4 Discussion Board Topics Choose a problem from this exercise set and clearly work it out on paper, explaining each step in words. Scan your page and post it on the discussion board. Explain why we need to simplify the numerator and denominator to a single algebraic fraction before multiplying by the reciprocal of the divisor. Two methods for simplifying complex rational expressions have been presented in this section. Which of the two methods do you feel is more efficient, and why? Answer 1. Answers may vary Answers may vary This page titled 7.4: Complex Rational Expressions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform. Back to top 7.3: Adding and Subtracting Rational Expressions 7.5: Solving Rational Equations Was this article helpful? 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2264	https://www.sciencedirect.com/science/article/abs/pii/S0020025524013586	A novel linear discriminant analysis based on alternate ratio sum minimization - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (48) Cited by (2) Information Sciences Volume 689, January 2025, 121444 A novel linear discriminant analysis based on alternate ratio sum minimization☆ Author links open overlay panel Xiaojun Yang a b c d e, Chuanjie Cao a, Keyi Zhou a, Siyuan Peng a, Zheng Wang f, Liang Lin g, Feiping Nie f Show more Add to Mendeley Share Cite rights and content Abstract Linear discriminant analysis (LDA) and its variants are popular supervised dimension reduction methods, which have been widely used to handle high-dimensional data. Since most of previous LDA methods are developed based on the trace ratio (TR) criterion, they usually obtain the low-dimensional data feature with weak discriminative ability, due to the projections with small variance. The ratio sum (RS) criterion was developed to alleviate this drawback. However, the conventional ratio sum is formulated to maximize the arithmetic mean of items which suffers from the domination of the largest objectives and might deteriorate the recognition accuracy in practical applications. In this paper, a novel ratio sum minimization based linear discriminant analysis (RSM-LDA) method is proposed in this paper. Specifically, a new ratio sum minimization (RSM) criterion is developed, which is based on the properties of the harmonic mean and effectively avoids the dominance problem to discover more discriminative features of the data. However, obtaining a closed solution for the RSM-LDA problem is challenging. For this purpose, three optimization methods are used to solve the optimization problem of RSM-LDA, and the inherent relationships between the methods are discussed. Experimental results demonstrate that the proposed RSM-LDA method has better performance in classification tasks on several datasets, when compared with some comparison methods based dimensionality reduction methods. In addition, combined with other experimental results, the effectiveness of RSM-LDA was verified. Introduction With technology development, data dimensionality is increasing quickly in many fields, such as remote sensing image recognition , sensor networks , multiagent systems , and so on. High-dimensional data usually bring many challenges to the speed and performance of learning methods due to the curse of dimensionality . To solve this problem, many effective dimensionality reduction methods have been proposed in the past decades, which pay more attention to improving the discriminative ability of data mapped to a low-dimensional subspace . Generally speaking, dimensionality reduction methods can be classified into two groups, i.e., unsupervised and supervised methods. The most representative unsupervised and supervised dimensionality reduction methods are the principal component analysis (PCA) and linear discriminant analysis (LDA) respectively. Particularly, the target of PCA is to find a projection matrix that can project data from the original high-dimensional space into a low-dimensional subspace where the sample variance is maximized. Unlike PCA, LDA not only focuses on the total data but also takes advantage of label information of data for greater discriminative ability. The LDA aims to find the optimal low-dimensional representation by maximizing the between-class scatter matrix while minimizing the within-class scatter matrix to maximize the separability of the data after dimensionality reduction. Since the original LDA method can only be appropriate for coping with the binary classification problems , the generalized LDA method is proposed for the multi-classification problems . However, despite the demonstrated effectiveness of LDA and its variant algorithms in various applications, LDA research is primarily constrained by the following four issues: 1) To obtain an approximate closed-form solution, LDA often requires transformation into another form, introducing the possibility of errors between the approximate solution and the true value . 2) LDA relies on the non-singularity of the scatter matrix it constructs. When the number of samples is too small, it will lead to a small sample size problem (SSS) . 3) LDA needs to assume that the dataset satisfies Gaussian distribution and is not suitable for application to non-Gaussian distribution datasets. This is mainly because LDA focuses on the global structure and cannot effectively extract local information . 4) LDA tends to select features with smaller variance, retaining poorer discriminant dimensions . For the first problem, it is mainly due to the difficulty in obtaining closed-form global optimum solution for TR-based problems , only an approximate solution can be calculated by solving these problems. An approximate solution can be obtained by utilizing the generalized eigenvalue decomposition to solve the ratio trace (RT) problem transformed from the TR problem , . Since the solution of the RT criterion is a suboptimal solution , the projection matrix can fix the objective function value of the RT criterion by any non-singular matrix. Although these LDA methods based RT criterion can eliminate the impact as much as possible by incorporating orthogonal or uncorrelated constraints, the performance of them is still unstable in classification tasks. Hence, many new criteria have been proposed to obtain better solutions for the TR optimization problem. For instance, the maximum margin criterion (MMC) transformed the TR criterion into a trace difference (TD) criterion to obtain the projection matrix . Jia et al. proposed a TR model called the decomposed Newton's method (DNM) , which both transformed the TR problem into a TD problem and introduced eigenvalue perturbation theory to solve the TD problem. Wang et al. developed a novel formulation of trace ratio linear discriminant analysis (TRLDA), transforming the original trace ratio problem into a quadratic problem on the Stiefel manifold to find an optimal solution that minimizes the gap between the approximate solution and the true value. To address the second problem, the most straightforward solution to the non-invertibility issue of the divergence matrix is to employ pseudo-inversion for calculation. For example, Orthogonal LDA (OLDA), Uncorrelated LDA (ULDA), and Null space LDA (NLDA) all use pseudo-inverse calculation to solve the SSS problem. However, methods based on pseudo-inverse lack theoretical proof of achieving optimality. Therefore, their common flaw is unstable performance. Another common approach to address the SSS problem is to conduct PCA as a preprocessing step on the dataset, eliminating the null space, followed by LDA for dimensionality reduction . PCA is consistently employed to eliminate the null space, playing a pivotal role in the realm of feature extraction and dimensionality reduction. In contrast to the two-stage approach, Zhao et al. introduced a unified LDA and PCA framework. This framework eliminates the zero space in the covariance matrix through PCA, rendering LDA suitable for addressing small sample problems. Furthermore, kernel discriminant analysis (KDA) can perform discriminant analysis in the feature space implied by the kernel function, effectively handling the SSS problem, and due to its nonlinear property, it can improve performance and robustness . A revised TRLDA (RTRLDA) is proposed, which overcomes the SSS problem by revising the formation. To address the challenge of LDA's inapplicability to non-Gaussian distributed datasets, some researchers introduce graph embedding frameworks to capture local structural information to improve the discriminative performance to cope with complex real-world datasets. Inspired by locality preserving projections (LPP) , local fisher discriminant analysis (LFDA) ) achieves better performance on multi-modal data by maximizing between-class divergence and minimizing local within-class divergence. The regularized least square based discriminative projection (RLSDP) method achieved feature extraction by solving a model that minimizes the reconstructed residuals of test samples on the same class via ratio trace criterion. Moreover, certain researchers have recognized that the effectiveness of graph embedding-based methods is contingent on the graph constructed from the original data . This dependency makes the algorithm's discriminative performance susceptible to noise. They argue that the intrinsic geometry of the data is concealed in the low-dimensional space, emphasizing the importance of finding the neighborhood of each sample in the subspace rather than in the original space , . Hence, numerous researchers have developed graph-embedding LDA based on subspace learning. This approach aims to acquire lower-dimensional neighbor relationships and mitigate the impact of noise. For example, the locality adaptive discriminant analysis (LADA) method was proposed by continuous iterative learning to update the similarity matrix and the subspace simultaneously to ensure that the neighbor relationship learned is optimal. Wang et al. developed a model with dynamic maximum entropy graph (DMEG) method on similarity matrix to mine the optimum underlying local structure of data to overcome the above shortcoming. In addition, some researchers look at non-Gaussian problems from another angle, that is, when facing non-Gaussian data, it is not feasible to think that each point in each class contributes equally to the arithmetic mean center. Zhao et al. introduced a novel robust LDA (RLDA) that assigns weights to the center point of each class. This is achieved by constructing the center point of each class based on the principle that close points contribute more significantly to the center point construction, whereas distant points make a smaller contribution. Nie et al. proposed a novel and efficient local embedding learning via landmark-based dynamic connections (LDC), which uses multiple center points to represent different sub-clusters in the same class and establishes a relationship between each point and the center point. This method enhances the exploration of the local structure of the data with increased precision. In addition, since the graph embedding method takes a lot of time to construct the graph, some researchers have introduced the anchor point strategy to accelerate the algorithm. For example, fast and adaptive locality discriminant analysis (FALDA) introduces anchor points to save training time, and the anchor points implicitly divide the dataset into sub-blocks, making it suitable for non-Gaussian distribution datasets. The final issue primarily stems from the fact that most LDA variant algorithms are developed based on the trace ratio criterion. References , all pointed out that the trace ratio criterion tends to select features with small variance. To solve this problem, Liu et al. proposed a noise-insensitive tracking ratio criterion for feature extraction. Wang et al. proposed a maximization RS criterion, that is, after projection, it maximizes the sum of the ratios of between-class scatter and within-class scatter in each dimension. Although maximizing the RS criterion can in principle avoid the problem that the TR criterion tends to select the projection direction with small variance, the research is not enough. Because the form of maximizing the sum suffers from the domination of the largest objectives, the projection direction with a small variance may be selected. To tackle this issue, inspired by , using the harmonic mean to deal with the worst-case problem, we propose a novel ratio sum minimization linear discriminant analysis (RSM-LDA) in this paper. We extend the original RS framework of LDA by utilizing the harmonic mean. Compared with the original maximization RS criterion, this method effectively avoids the dominance problem of the maximum target by paying more attention to the projection direction with a small variance and ensures that the variance of all projection directions cannot be too small. However, minimizing RS brings difficulties to the optimization process. For this reason, we develop 3 optimization algorithms to solve this problem. The contributions of this paper can be summarized as follows: 1.A new ratio sum minimization (RSM) criterion is developed for LDA, which extends the current RS framework. The RSM criterion based on harmonic mean can pay more attention to the projection direction with small variance, so as to extract the projection directions with similar quality, make the variances between projection directions smoother, and ensure a larger global variance sum. This can effectively avoid domination problems. 2.Three optimization methods (i.e., gradient, greedy, and alternate) are adopted to solve the optimization problem of the proposed RSM-LDA method. The inherent relationship between them is discussed, and the computational complexity of RSM-LDA is analyzed. 3.Extensive experiments on multiple datasets show that RSM-LDA can retain features with strong discriminative ability, thereby obtaining high classification accuracy, and outperforming the compared methods in most cases. The rest of this article is structured as follows: In Section 2, the related work is briefly introduced. In Section 3, the motivation is analyzed, and the optimization problem of RSM-LDA is presented. In Section 4, three optimization problems are used to solve the optimization problem of RSM-LDA. The theoretical derivation of the three methods of the ratio sum model is presented. And we discuss the relationship and the computational complexity between these methods. In Section 5, extensive experiments are conducted to validate the effectiveness of RSM-LDA. Finally, the conclusion is proposed in Section 6. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Related work In this section, some important notations are introduced. LDA and RSLDA are briefly reviewed. Proposed method In this section, we proposed the ratio sum minimization criterion, discussed its difference from the maximization ratio sum criterion, and analyzed its principles for solving the domination problem. The domination problem is defined as follows: Domination problem The objective function is the sum of terms. The objective function value suffers from the domination of the term with largest objective value. First, let's discuss the shortcomings of the maximizing RS criterion. Reflecting on RSLDA, its approach to Optimization algorithms In this section, three popular optimization algorithms are used to optimize the objective function of RSM-LAD. The demonstration diagram of the optimization process of the three methods is shown in Fig. 1. Experiment In this section, we experimentally compare several state-of-the-art dimensionality reduction algorithms with the proposed algorithm to demonstrate the effectiveness of RSM-LAD. All experiments were performed in Matlab R2020a with Windows 10 operating system and Intel Core i5-9500@3.00Hz. Conclusion In this article, a novel linear discriminant analysis method called RSM-LDA is proposed for classification tasks, which uses the minimize RS criterion to replace the maximize RS criterion in RSLDA, thereby alleviating the influence of the domination problem that the maximize RS criterion may suffer and improving the overall quality of the projection matrix. Additionally, considering the optimization challenges posed by the minimizing RS criterion, three optimization methods have been developed CRediT authorship contribution statement Xiaojun Yang: Writing – review & editing, Writing – original draft, Project administration, Funding acquisition, Conceptualization. Chuanjie Cao: Writing – review & editing, Writing – original draft, Validation, Software, Resources, Methodology, Data curation. Keyi Zhou: Visualization, Software. Siyuan Peng: Supervision, Formal analysis. Zheng Wang: Methodology, Investigation. Liang Lin: Supervision, Investigation, Formal analysis. Feiping Nie: Writing – review & editing, Writing – original Declaration of Competing Interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. Recommended articles References (48) X. Yang et al. Limited-energy output formation for multiagent systems with intermittent interactions J. Franklin Inst. (2021) S. Wold et al. Principal component analysis Chemom. Intell. Lab. Syst. (1987) Y.-F. Guo et al. A generalized Foley-Sammon transform based on generalized Fisher discriminant criterion and its application to face recognition Pattern Recognit. Lett. (2003) L.-F. Chen et al. A new lda-based face recognition system which can solve the small sample size problem Pattern Recognit. (2000) Y. Liu et al. Efficient semi-supervised feature selection with noise insensitive trace ratio criterion Neurocomputing (2013) W. Yang et al. A regularized least square based discriminative projections for feature extraction Neurocomputing (2016) L. Zhang et al. Graph-optimized locality preserving projections Pattern Recognit. (2010) Z. Wang et al. Local structured feature learning with dynamic maximum entropy graph Pattern Recognit. (2021) F. Nie et al. Orthogonal vs. uncorrelated least squares discriminant analysis for feature extraction Pattern Recognit. Lett. (2012) X. Yang et al. Fast spectral embedded clustering based on structured graph learning for large-scale hyperspectral image IEEE Geosci. Remote Sens. Lett. (2020) C.A. Musluoglu et al. Distributed trace ratio optimization in fully-connected sensor networks Joshua B. Tenenbaum et al. A global geometric framework for nonlinear dimensionality reduction Science (2000) R. Wang et al. Fast and orthogonal locality preserving projections for dimensionality reduction IEEE Trans. Image Process. (2017) P.N. Belhumeur et al. Eigenfaces vs. fisherfaces: recognition using class specific linear projection IEEE Transactions on Pattern Analysis and Machine Intelligence (1997) R.A. Fisher The use of multiple measurements in taxonomic problems Ann. Eugen. (1936) C.R. Rao The utilization of multiple measurements in problems of biological classification J. R. Stat. Soc., Ser. B, Methodol. (1948) S. Balakrishnama et al. Linear discriminant analysis-a brief tutorial (1998) Y. Saad Numerical Methods for Large Eigenvalue Problems (2011) H. Wang et al. Trace ratio vs. ratio trace for dimensionality reduction H. Li et al. Efficient and robust feature extraction by maximum margin criterion Adv. Neural Inf. Process. Syst. (2004) Y. Jia et al. Trace ratio problem revisited IEEE Trans. Neural Netw. (2009) J. Wang et al. A novel formulation of trace ratio linear discriminant analysis IEEE Trans. Neural Netw. Learn. Syst. (2021) X. Zhao et al. Joint principal component and discriminant analysis for dimensionality reduction IEEE Trans. Neural Netw. Learn. Syst. (2019) J. Yang et al. Kpca plus lda: a complete kernel Fisher discriminant framework for feature extraction and recognition IEEE Trans. Pattern Anal. Mach. Intell. (2005) View more references Cited by (2) Fast self-supervised discrete graph clustering with ensemble local cluster constraints 2025, Neural Networks Show abstract Spectral clustering (SC) is a graph-based clustering algorithm that has been widely used in the field of data mining and image processing. However, most graph-based clustering methods ignore the utilization of additional prior information. This information can help clustering models further reduce the difference between their clustering results and ground-truth, but is difficult to obtain in unsupervised settings. Moreover, traditional graph-based clustering algorithms require additional hyperparameters and full graph construction to obtain good performance, increasing the tuning pressure and time cost. To address these issues, a simple fast self-supervised discrete graph clustering (FSDGC) is proposed. Specifically, the proposed method has the following features: (1) a novel self-supervised information, based on ensemble local cluster constraints, is used to constrain the sample indicator matrix; (2) the anchor graph technique is introduced for mining the structure between samples and anchors to handle large scale datasets. Meanwhile, a fast coordinate ascent (CA) optimization method, based on self-supervised constraints, is proposed to obtain discrete indicator matrices. Experimental clustering results demonstrate that FSDGC has efficient clustering performance. ### An Autoencoder-like Non-Negative Matrix Factorization with Structure Regularization Algorithm for Clustering 2025, Symmetry ☆ This work was supported in part by the National Natural Science Foundation under Grant 62306080 and 62373112, in part by the Guangzhou Science and Technology Plan Project under Grant 202206010104, and in part by Key Laboratory open FoundationWDZ20215250118. View full text © 2024 Elsevier Inc. All rights are reserved, including those for text and data mining, AI training, and similar technologies. 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2265	https://math.stackexchange.com/questions/210110/whats-the-name-of-the-approximation-1xn-approx-1-xn	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams What's the name of the approximation $(1+x)^n \approx 1 + xn$? Ask Question Asked Modified 4 years, 5 months ago Viewed 80k times 17 $\begingroup$ A good approximation of $(1+x)^n$ is $1+xn$ when $\|x\|n << 1$. Does this approximation have a name? Any leads on estimating the error of the approximation? approximation Share asked Oct 9, 2012 at 22:39 Martin C. MartinMartin C. Martin 29611 gold badge22 silver badges88 bronze badges $\endgroup$ 2 7 $\begingroup$ First order Taylor approximation or linearization. $\endgroup$ Cheerful Parsnip – Cheerful Parsnip 2012-10-09 22:45:43 +00:00 Commented Oct 9, 2012 at 22:45 $\begingroup$ Taylor approximation is exactly what I was looking for. Somehow, given that (1+x)^n has a finite expansion, I thought this was about finite series rather than infinite series, and didn't even think of the Taylor expansion. Your comment made me realize that the finite expansion IS the Taylor series, so I can use the integral form of the remainder to estimate how good of an approximation we have. Thanks! $\endgroup$ Martin C. Martin – Martin C. Martin 2012-10-10 18:40:23 +00:00 Commented Oct 10, 2012 at 18:40 Add a comment \| 3 Answers 3 Reset to default 10 $\begingroup$ I would just call it the first order truncation of the Binomial series. If you want more terms of the series, then it's given by $$(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \mathcal{O}(x^4)$$ for the full series, you can visit the link I provided. You may also be interested in Bernoulli's inequality Share answered Oct 9, 2012 at 22:44 EuYuEuYu 42.8k99 gold badges9696 silver badges142142 bronze badges $\endgroup$ 1 $\begingroup$ I wonder if the O(x^4) notation is a bit misleading, because the approximation depends on n and only hold for nx and x small. See en.wikipedia.org/wiki/Binomial_approximation $\endgroup$ Christian Fries – Christian Fries 2021-04-21 18:44:05 +00:00 Commented Apr 21, 2021 at 18:44 Add a comment \| 4 $\begingroup$ I would say it comes from the Bernoulli inequality. You can read it up on Wiki Share answered Oct 9, 2012 at 22:43 Jean-SébastienJean-Sébastien 4,67311 gold badge2121 silver badges4141 bronze badges $\endgroup$ Add a comment \| 3 $\begingroup$ Wikipedia calls it Binomial Approximation. Share answered Apr 21, 2021 at 18:41 Christian FriesChristian Fries 62644 silver badges55 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions approximation See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 0 How is this approximation derived? 0 Estimate (1+a)^n from above. Related How does this square root approximation work? 0 approximation of x=sin(x) error Hardy's approximation for the cosine Convergence of piecewise linear approximation of Bezier curve 1 Name and derivation of the approximation $\frac{1+x}{1+y} - 1 \approx x-y$? 1 Is $\approx$ an equivalence relation? 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2266	https://emergency.med.ufl.edu/files/2013/02/febrile-seizures-clinical-practice-guideline.pdf	DOI: 10.1542/peds.2008-0939 2008;121;1281-1286 Pediatrics Febrile Seizures Steering Committee on Quality Improvement and Management, Subcommittee on the Child With Simple Febrile Seizures Febrile Seizures: Clinical Practice Guideline for the Long-term Management of located on the World Wide Web at: The online version of this article, along with updated information and services, is rights reserved. Print ISSN: 0031-4005. Online ISSN: 1098-4275. Grove Village, Illinois, 60007. Copyright © 2008 by the American Academy of Pediatrics. All and trademarked by the American Academy of Pediatrics, 141 Northwest Point Boulevard, Elk publication, it has been published continuously since 1948. PEDIATRICS is owned, published, PEDIATRICS is the official journal of the American Academy of Pediatrics. A monthly at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from CLINICAL PRACTICE GUIDELINE Febrile Seizures: Clinical Practice Guideline for the Long-term Management of the Child With Simple Febrile Seizures Steering Committee on Quality Improvement and Management, Subcommittee on Febrile Seizures ABSTRACT Febrile seizures are the most common seizure disorder in childhood, affecting 2% to 5% of children between the ages of 6 and 60 months. Simple febrile seizures are deﬁned as brief (15-minute) generalized seizures that occur once during a 24-hour period in a febrile child who does not have an intracranial infection, metabolic disturbance, or history of afebrile seizures. This guideline (a revision of the 1999 American Academy of Pediatrics practice parameter [now termed clinical practice guideline] “The Long-term Treatment of the Child With Simple Febrile Seizures”) addresses the risks and beneﬁts of both continuous and intermittent anticonvulsant therapy as well as the use of antipyretics in children with simple febrile seizures. It is designed to assist pediatricians by providing an analytic framework for decisions regarding possible therapeutic interventions in this pa-tient population. It is not intended to replace clinical judgment or to establish a protocol for all patients with this disorder. Rarely will these guidelines be the only approach to this problem. Pediatrics 2008;121:1281–1286 The expected outcomes of this practice guideline include: 1. optimizing practitioner understanding of the scientiﬁc basis for using or avoid-ing various proposed treatments for children with simple febrile seizures; 2. improving the health of children with simple febrile seizures by avoiding therapies with high potential for adverse effects and no demonstrated ability to improve children’s long-term outcomes; 3. reducing costs by avoiding therapies that will not demonstrably improve children’s long-term outcomes; and 4. helping the practitioner educate caregivers about the low risks associated with simple febrile seizures. The committee determined that with the exception of a high rate of recurrence, no long-term effects of simple febrile seizures have been identiﬁed. The risk of developing epilepsy in these patients is extremely low, although slightly higher than that in the general population. No data, however, suggest that prophylactic treatment of children with simple febrile seizures would reduce the risk, because epilepsy likely is the result of genetic predisposition rather than structural damage to the brain caused by recurrent simple febrile seizures. Although antipyretics have been shown to be ineffective in preventing recurrent febrile seizures, there is evidence that continuous anticonvulsant therapy with phenobarbital, primidone, or valproic acid and intermittent therapy with diazepam are effective in reducing febrile-seizure recurrence. The potential toxicities associated with these agents, however, outweigh the relatively minor risks associated with simple febrile seizures. As such, the committee concluded that, on the basis of the risks and beneﬁts of the effective therapies, neither continuous nor intermittent anticonvulsant therapy is recommended for children with 1 or more simple febrile seizures. INTRODUCTION Febrile seizures are seizures that occur in febrile children between the ages of 6 and 60 months who do not have an intracranial infection, metabolic disturbance, or history of afebrile seizures. Febrile seizures are subdivided into 2 categories: simple and complex. Simple febrile seizures last for less than 15 minutes, are generalized (without a focal component), and occur once in a 24-hour period, whereas complex febrile seizures are prolonged (15 minutes), are focal, or occur more than once in 24 hours.1 Despite the frequency of febrile seizures (2%–5%), there is no unanimity of opinion about management options. This clinical practice guideline addresses potential therapeutic interventions in neurologically normal children with simple febrile seizures. It is not intended for patients with complex febrile seizures and does not pertain to children with previous neurologic insults, known central nervous system abnor-www.pediatrics.org/cgi/doi/10.1542/ peds.2008-0939 doi:10.1542/peds.2008-0939 All clinical reports from the American Academy of Pediatrics automatically expire 5 years after publication unless reafﬁrmed, revised, or retired at or before that time. The guidance in this report does not indicate an exclusive course of treatment or serve as a standard of medical care. Variations, taking into account individual circumstances, may be appropriate. Key Word fever Abbreviation AAP—American Academy of Pediatrics PEDIATRICS (ISSN Numbers: Print, 0031-4005; Online, 1098-4275). Copyright © 2008 by the American Academy of Pediatrics PEDIATRICS Volume 121, Number 6, June 2008 1281 at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from malities, or a history of afebrile seizures. This clinical practice guideline is a revision of a 1999 American Acad-emy of Pediatrics (AAP) clinical practice parameter, “The Long-term Treatment of the Child With Simple Febrile Seizures.”2 For a child who has experienced a simple febrile seizure, there are potentially 4 adverse outcomes that theoretically may be altered by an effective therapeutic agent: (1) decline in IQ; (2) increased risk of epilepsy; (3) risk of recurrent febrile seizures; and (4) death. Neither a decline in IQ, academic performance or neurocognitive inattention nor behavioral abnormalities have been shown to be a consequence of recurrent simple febrile seizures.3 Ellenberg and Nelson4 studied 431 children who experienced febrile seizures and observed no sig-niﬁcant difference in their learning compared with sib-ling controls. In a similar study by Verity et al,5 303 children with febrile seizures were compared with con-trol children. No difference in learning was identiﬁed, except in those children who had neurologic abnormal-ities before their ﬁrst seizure. The second concern, increased risk of epilepsy, is more complex. Children with simple febrile seizures have approximately the same risk of developing epilepsy by the age of 7 years as does the general population (ie, 1%).6 However, children who have had multiple simple febrile seizures, are younger than 12 months at the time of their ﬁrst febrile seizure, and have a family history of epilepsy are at higher risk, with generalized afebrile sei-zures developing by 25 years of age in 2.4%.7 Despite this fact, no study has demonstrated that successful treatment of simple febrile seizures can prevent this later development of epilepsy, and there currently is no evi-dence that simple febrile seizures cause structural dam-age to the brain. Indeed, it is most likely that the in-creased risk of epilepsy in this population is the result of genetic predisposition. In contrast to the slightly increased risk of developing epilepsy, children with simple febrile seizures have a high rate of recurrence. The risk varies with age. Chil-dren younger than 12 months at the time of their ﬁrst simple febrile seizure have an approximately 50% prob-ability of having recurrent febrile seizures. Children older than 12 months at the time of their ﬁrst event have an approximately 30% probability of a second febrile seizure; of those who do have a second febrile seizure, 50% have a chance of having at least 1 additional recur-rence.8 Finally, there is a theoretical risk of a child dying during a simple febrile seizure as a result of documented injury, aspiration, or cardiac arrhythmia, but to the com-mittee’s knowledge, it has never been reported. In summary, with the exception of a high rate of recurrence, no long-term adverse effects of simple febrile seizures have been identiﬁed. Because the risks associ-ated with simple febrile seizures, other than recurrence, are so low and because the number of children who have febrile seizures in the ﬁrst few years of life is so high, to be commensurate, a proposed therapy would need to be exceedingly low in risks and adverse effects, inexpensive, and highly effective. METHODS To update the clinical practice guideline on the treat-ment of children with simple febrile seizures, the AAP reconvened the Subcommittee on Febrile Seizures. The committee was chaired by a child neurologist and con-sisted of a neuroepidemiologist, 2 additional child neu-rologists, and a practicing pediatrician. All panel mem-bers reviewed and signed the AAP voluntary disclosure and conﬂict-of-interest form. The guideline was re-viewed by members of the AAP Steering Committee on Quality Improvement and Management; members of the AAP Sections on Neurology, Pediatric Emergency Med-icine, Developmental and Behavioral Pediatrics, and Epidemiology; members of the AAP Committees on Pe-diatric Emergency Medicine and Medical Liability and Risk Management; members of the AAP Councils on Children With Disabilities and Community Pediatrics; and members of outside organizations including the Child Neurology Society and the American Academy of Neurology. A comprehensive review of the evidence-based liter-ature published since 1998 was conducted with the aim of addressing possible therapeutic interventions in the management of children with simple febrile seizures. The review focused on both the efﬁcacy and potential adverse effects of the proposed treatments. Decisions were made on the basis of a systematic grading of the quality of evidence and strength of recommendations. The AAP established a partnership with the Univer-sity of Kentucky (Lexington, KY) to develop an evidence report, which served as a major source of information for these practice-guideline recommendations. The speciﬁc issues addressed were (1) effectiveness of continuous anticonvulsant therapy in preventing recurrent febrile seizures, (2) effectiveness of intermittent anticonvulsant therapy in preventing recurrent febrile seizures, (3) ef-fectiveness of antipyretics in preventing recurrent febrile seizures, and (4) adverse effects of either continuous or intermittent anticonvulsant therapy. In the original practice parameter, more than 300 medical journal articles reporting studies of the natural history of simple febrile seizures or the therapy of these seizures were reviewed and abstracted.2 An additional 65 articles were reviewed and abstracted for the update. Emphasis was placed on articles that differentiated sim-ple febrile seizures from other types of seizures, that carefully matched treatment and control groups, and that described adherence to the drug regimen. Tables were constructed from the 65 articles that best ﬁt these criteria. A more comprehensive review of the literature on which this report is based can be found in a forth-coming technical report (the initial technical report can be accessed at content/full/pediatrics;103/6/e86). The technical report also will contain dosing information. The evidence-based approach to guideline development requires that the evidence in support of a recommendation be identiﬁed, appraised, and summarized and that an ex-plicit link between evidence and recommendations be de-ﬁned. Evidence-based recommendations reﬂect the quality of evidence and the balance of beneﬁt and harm that is 1282 AMERICAN ACADEMY OF PEDIATRICS at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from anticipated when the recommendation is followed. The AAP policy statement “Classifying Recommendations for Clinical Practice Guidelines”9 was followed in designating levels of recommendations (see Fig 1 and Table 1). RECOMMENDATION On the basis of the risks and beneﬁts of the effective therapies, neither continuous nor intermittent anticon-vulsant therapy is recommended for children with 1 or more simple febrile seizures. ●Aggregate evidence quality: B (randomized, controlled trials and diagnostic studies with minor limitations). ●Beneﬁt: prevention of recurrent febrile seizures, which are not harmful and do not signiﬁcantly in-crease the risk for development of future epilepsy. ●Harm: adverse effects including rare fatal hepatotox-icity (especially in children younger than 2 years who are also at greatest risk of febrile seizures), thrombo-cytopenia, weight loss and gain, gastrointestinal dis-turbances, and pancreatitis with valproic acid and hy-peractivity, irritability, lethargy, sleep disturbances, and hypersensitivity reactions with phenobarbital; lethargy, drowsiness, and ataxia for intermittent diaz-epam as well as the risk of masking an evolving central nervous system infection. ●Beneﬁts/harms assessment: preponderance of harm over beneﬁt. ●Policy level: recommendation. BENEFITS AND RISKS OF CONTINUOUS ANTICONVULSANT THERAPY Phenobarbital Phenobarbital is effective in preventing the recurrence of simple febrile seizures.10 In a controlled double-blind study, daily therapy with phenobarbital reduced the rate of subsequent febrile seizures from 25 per 100 subjects per year to 5 per 100 subjects per year.11 For the agent to be effective, however, it must be given daily and main-tained in the therapeutic range. In a study by Farwell et al,12 for example, children whose phenobarbital levels were in the therapeutic range had a reduction in recur-rent seizures, but because noncompliance was so high, an overall beneﬁt with phenobarbital therapy was not identiﬁed. The adverse effects of phenobarbital include hyperac-tivity, irritability, lethargy, sleep disturbances, and hy-persensitivity reactions. The behavioral adverse effects FIGURE 1 Integrating evidence-quality appraisal with an assessment of the anticipated balance between beneﬁts and harms if a policy is conducted leads to designation of a policy as a strong recommendation, recommendation, option, or no recommendation. RCT indi-cates randomized, controlled trial. TABLE 1 Guideline Deﬁnitions for Evidence-Based Statements Statement Deﬁnition Implication Strong recommendation A strong recommendation in favor of a particular action is made when the anticipated beneﬁts of the recommended intervention clearly exceed the harms (as a strong recommendation against an action is made when the anticipated harms clearly exceed the beneﬁts) and the quality of the supporting evidence is excellent. In some clearly identiﬁed circumstances, strong recommendations may be made when high-quality evidence is impossible to obtain and the anticipated beneﬁts strongly outweigh the harms. Clinicians should follow a strong recommendation unless a clear and compelling rationale for an alternative approach is present. Recommendation A recommendation in favor of a particular action is made when the anticipated beneﬁts exceed the harms but the quality of evidence is not as strong. Again, in some clearly identiﬁed circumstances, recommendations may be made when high-quality evidence is impossible to obtain but the anticipated beneﬁts outweigh the harms. Clinicians would be prudent to follow a recommendation but should remain alert to new information and sensitive to patient preferences. Option Options deﬁne courses that may be taken when either the quality of evidence is suspect or carefully performed studies have shown little clear advantage to 1 approach over another. Clinicians should consider the option in their decision-making, and patient preference may have a substantial role. No recommendation No recommendation indicates that there is a lack of pertinent published evidence and that the anticipated balance of beneﬁts and harms is presently unclear. Clinicians should be alert to new published evidence that clariﬁes the balance of beneﬁt versus harm. PEDIATRICS Volume 121, Number 6, June 2008 1283 at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from may occur in up to 20% to 40% of patients and may be severe enough to necessitate discontinuation of the drug.13–16 Primidone Primidone, in doses of 15 to 20 mg/kg per day, has also been shown to reduce the recurrence rate of febrile seizures.17,18 It is of interest that the derived phenobar-bital level in a Minigawa and Miura study17 was below therapeutic (16 g/mL) in 29 of the 32 children, sug-gesting that primidone itself may be active in preventing seizure recurrence. As with phenobarbital, adverse ef-fects include behavioral disturbances, irritability, and sleep disturbances.18 Valproic Acid In randomized, controlled studies, only 4% of children taking valproic acid, as opposed to 35% of control sub-jects, had a subsequent febrile seizure. Therefore, val-proic acid seems to be at least as effective in preventing recurrent simple febrile seizures as phenobarbital and signiﬁcantly more effective than placebo.19–21 Drawbacks to therapy with valproic acid include its rare association with fatal hepatotoxicity (especially in children younger than 2 years, who are also at greatest risk of febrile seizures), thrombocytopenia, weight loss and gain, gastrointestinal disturbances, and pancreatitis. In studies in which children received valproic acid to prevent recurrence of febrile seizures, no cases of fatal hepatotoxicity were reported.15 Carbamazepine Carbamazepine has not been shown to be effective in preventing the recurrence of simple febrile seizures. Antony and Hawke13 compared children who had been treated with therapeutic levels of either phenobarbital or carbamazepine, and 47% of the children in the carbam-azepine-treated group had recurrent seizures compared with only 10% of those in the phenobarbital group. In another study, Camﬁeld et al22 treated children (whose conditions failed to improve with phenobarbital ther-apy) with carbamazepine. Despite good compliance, 13 of the 16 children treated with carbamazepine had a recurrent febrile seizure within 18 months. It is theoret-ically possible that these excessively high rates of recur-rences might have been attributable to adverse effects of carbamazepine. Phenytoin Phenytoin has not been shown to be effective in pre-venting the recurrence of simple febrile seizures, even when the agent is in the therapeutic range.23,24 Other anticonvulsants have not been studied for the continu-ous treatment of simple febrile seizures. BENEFITS AND RISKS OF INTERMITTENT ANTICONVULSANT THERAPY Diazepam A double-blind controlled study of patients with a his-tory of febrile seizures demonstrated that administration of oral diazepam (given at the time of fever) could re-duce the recurrence of febrile seizures. Children with a history of febrile seizures were given either oral diaze-pam (0.33 mg/kg, every 8 hours for 48 hours) or a placebo at the time of fever. The risk of febrile seizures per person-year was decreased 44% with diazepam.25 In a more recent study, children with a history of febrile seizures were given oral diazepam at the time of fever and then compared with children in an untreated con-trol group. In the oral diazepam group, there was an 11% recurrence rate compared with a 30% recurrence rate in the control group.26 It should be noted that all children for whom diazepam was considered a failure had been noncompliant with drug administration, in part because of adverse effects of the medication. There is also literature that demonstrates the feasibil-ity and safety of interrupting a simple febrile seizure lasting less than 5 minutes with rectal diazepam and with both intranasal and buccal midazolam.27,28 Al-though these agents are effective in terminating the seizure, it is questionable whether they have any long-term inﬂuence on outcome. In a study by Knudsen et al,29 children were given either rectal diazepam at the time of fever or only at the onset of seizure. Twelve-year follow-up found that the long-term prognosis of the children in the 2 groups did not differ regardless of whether treatment was aimed at preventing seizures or treating them. A potential drawback to intermittent medication is that a seizure could occur before a fever is noticed. Indeed, in several of these studies, recurrent seizures were likely attributable to failure of method rather than failure of the agent. Adverse effects of oral and rectal diazepam26 and both intranasal and buccal midazolam include lethargy, drowsiness, and ataxia. Respiratory depression is ex-tremely rare, even when given by the rectal route.28,30 Sedation caused by any of the benzodiazepines, whether administered by the oral, rectal, nasal, or buccal route, have the potential of masking an evolving central ner-vous system infection. If used, the child’s health care professional should be contacted. BENEFITS AND RISKS OF INTERMITTENT ANTIPYRETICS No studies have demonstrated that antipyretics, in the absence of anticonvulsants, reduce the recurrence risk of simple febrile seizures. Camﬁeld et al11 treated 79 chil-dren who had had a ﬁrst febrile seizure with either a placebo plus antipyretic instruction (either aspirin or acetaminophen) versus daily phenobarbital plus antipy-retic instruction (either aspirin or acetaminophen). Re-currence risk was signiﬁcantly lower in the phenobarbi-tal-treated group, suggesting that antipyretic instruction, including the use of antipyretics, is ineffective in pre-venting febrile-seizure recurrence. Whether antipyretics are given regularly (every 4 hours) or sporadically (contingent on a speciﬁc body-temperature elevation) does not inﬂuence outcome. Acetaminophen was either given every 4 hours or only for temperature elevations of more than 37.9°C in 104 children. The incidence of febrile episodes did not differ 1284 AMERICAN ACADEMY OF PEDIATRICS at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from signiﬁcantly between the 2 groups, nor did the early recurrence of febrile seizures. The authors determined that administering prophylactic acetaminophen during febrile episodes was ineffective in preventing or reducing fever and in preventing febrile-seizure recurrence.31 In a randomized double-blind placebo-controlled trial, acetaminophen was administered along with low-dose oral diazepam.32 Febrile-seizure recurrence was not reduced, compared with control groups. As with acet-aminophen, ibuprofen also has been shown to be inef-fective in preventing recurrence of febrile seizures.33–35 In general, acetaminophen and ibuprofen are consid-ered to be safe and effective antipyretics for children. However, hepatotoxicity (with acetaminophen) and re-spiratory failure, metabolic acidosis, renal failure, and coma (with ibuprofen) have been reported in children after overdose or in the presence of risk factors.36,37 CONCLUSIONS The subcommittee has determined that a simple febrile seizure is a benign and common event in children be-tween the ages of 6 and 60 months. Nearly all children have an excellent prognosis. The committee concluded that although there is evidence that both continuous antiepileptic therapy with phenobarbital, primidone, or valproic acid and intermittent therapy with oral diaze-pam are effective in reducing the risk of recurrence, the potential toxicities associated with antiepileptic drugs outweigh the relatively minor risks associated with sim-ple febrile seizures. As such, long-term therapy is not recommended. In situations in which parental anxiety associated with febrile seizures is severe, intermittent oral diazepam at the onset of febrile illness may be effective in preventing recurrence. Although antipyretics may improve the comfort of the child, they will not prevent febrile seizures. SUBCOMMITTEE ON FEBRILE SEIZURES, 2002–2008 Patricia K. Duffner, MD, Chairperson Robert J. Baumann, MD, Methodologist Peter Berman, MD John L. Green, MD Sanford Schneider, MD STEERING COMMITTEE ON QUALITY IMPROVEMENT AND MANAGEMENT, 2007–2008 Elizabeth S. Hodgson, MD, Chairperson Gordon B. Glade, MD Norman “Chip” Harbaugh, Jr, MD Thomas K. McInerny, MD Marlene R. Miller, MD, MSc Virginia A. Moyer, MD, MPH Xavier D. Sevilla, MD Lisa Simpson, MB, BCh, MPH Glenn S. Takata, MD LIAISONS Denise Dougherty, PhD Agency for Healthcare Research and Quality Daniel R. Neuspiel, MD Section on Epidemiology Ellen Schwalenstocker, MBA National Association of Children’s Hospitals and Related Institutions STAFF Caryn Davidson, MA REFERENCES 1. Nelson KB, Ellenberg JH. Prognosis in children with febrile seizures. Pediatrics. 1978;61(5):720–727 2. American Academy of Pediatrics, Committee on Quality Im-provement, Subcommittee on Febrile Seizures. The long-term treatment of the child with simple febrile seizures. Pediatrics. 1999;103(6 pt 1):1307–1309 3. Chang YC, Guo NW, Huang CC, Wang ST, Tsai JJ. Neurocog-nitive attention and behavior outcome of school age children with a history of febrile convulsions: a population study. Epi-lepsia. 2000;41(4):412–420 4. Ellenberg JH, Nelson KB. Febrile seizures and later intellectual performance. Arch Neurol. 1978;35(1):17–21 5. Verity CM, Butler NR, Golding J. Febrile convulsions in a national cohort followed up from birth. II: medical history and intellectual ability at 5 years of age. BMJ. 1985;290(6478): 1311–1315 6. Nelson KB, Ellenberg JH. Predictors of epilepsy in children who have experienced febrile seizures. N Engl J Med. 1976;295(19): 1029–1033 7. Annegers JF, Hauser WA, Shirts SB, Kurland LT. Factors prog-nostic of unprovoked seizures after febrile convulsions. N Engl J Med. 1987;316(9):493–498 8. Berg AT, Shinnar S, Darefsky AS, et al. Predictors of recurrent febrile seizures: a prospective cohort study. Arch Pediatr Adolesc Med. 1997;151(4):371–378 9. American Academy of Pediatrics, Steering Committee on Qual-ity Improvement and Management. Classifying recommenda-tions for clinical practice guidelines. Pediatrics. 2004;114(3): 874–877 10. Wolf SM, Carr A, Davis DC, Davidson S, et al. The value of phenobarbital in the child who has had a single febrile seizure: a controlled prospective study. Pediatrics. 1977;59(3):378–385 11. Camﬁeld PR, Camﬁeld CS, Shapiro SH, Cummings C. The ﬁrst febrile seizure: antipyretic instruction plus either phenobarbital or placebo to prevent recurrence. J Pediatr. 1980;97(1):16–21 12. Farwell JR, Lee JY, Hirtz DG, Sulzbacher SI, Ellenberg JH, Nelson KB. Phenobarbital for febrile seizures: effects on intel-ligence and on seizure recurrence [published correction ap-pears in N Engl J Med. 1992;326(2):144]. N Engl J Med. 1990; 322(6):364–369 13. Antony JH, Hawke SHB. Phenobarbital compared with car-bamazepine in prevention of recurrent febrile convulsions. Am J Dis Child. 1983;137(9):892–895 14. Knudsen Fu, Vestermark S. Prophylactic diazepam or pheno-barbitone in febrile convulsions: a prospective, controlled study. Arch Dis Child. 1978;53(8):660–663 15. Lee K, Melchior JC. Sodium valproate versus phenobarbital in the prophylactic treatment of febrile convulsions in childhood. Eur J Pediatr. 1981;137(2):151–153 16. Camﬁeld CS, Chaplin S, Doyle AB, Shapiro SH, Cummings C, Camﬁeld PR. Side effects of phenobarbital in toddlers: behav-ioral and cognitive aspects. J Pediatr. 1979;95(3):361–365 17. Minagawa K, Miura H. Phenobarbital, primidone and sodium valproate in the prophylaxis of febrile convulsions. Brain Dev. 1981;3(4):385–393 18. Herranz JL, Armijo JA, Arteaga R. Effectiveness and toxicity of phenobarbital, primidone, and sodium valproate in the pre-PEDIATRICS Volume 121, Number 6, June 2008 1285 at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from vention of febrile convulsions, controlled by plasma levels. Epilepsia. 1984;25(1):89–95 19. Wallace SJ, Smith JA. Successful prophylaxis against febrile convulsions with valproic acid or phenobarbitone. BMJ. 1980; 280(6211):353–354 20. Mamelle N, Mamelle JC, Plasse JC, Revol M, Gilly R. Preven-tion of recurrent febrile convulsions: a randomized therapeutic assay—sodium valproate, phenobarbitone and placebo. Neuro-pediatrics. 1984;15(1):37–42 21. Ngwane E, Bower B. Continuous sodium valproate or pheno-barbitone in the prevention of “simple” febrile convulsions. Arch Dis Child. 1980;55(3):171–174 22. Camﬁeld PR, Camﬁeld CS, Tibbles JA. Carbamazepine does not prevent febrile seizures in phenobarbital failures. Neurology. 1982;32(3):288–289 23. Bacon CJ, Hierons AM, Mucklow JC, Webb JK, Rawlins MD, Weightman D. Placebo-controlled study of phenobarbitone and phenytoin in the prophylaxis of febrile convulsions. Lancet. 1981;2(8247):600–604 24. Melchior JC, Buchthal F, Lennox Buchthal M. The ineffective-ness of diphenylhydantoin in preventing febrile convulsions in the age of greatest risk, under 3 years. Epilepsia. 1971;12(1): 55–62 25. Rosman NP, Colton T, Labazzo J, et al. A controlled trial of diazepam administered during febrile illnesses to prevent re-currence of febrile seizures. N Engl J Med. 1993;329(2):79–84 26. Verrotti A, Latini G, di Corcia G, et al. Intermittent oral diaz-epam prophylaxis in febrile convulsions: its effectiveness for febrile seizure recurrence. Eur J Pediatr Neurol. 2004;8(3): 131–134 27. Lahat E, Goldman M, Barr J, Bistritzer T, Berkovitch M. Com-parison of intranasal midazolam with intravenous diazepam for treating febrile seizures in children: prospective randomized study. BMJ. 2000;321(7253):83–86 28. McIntyre J, Robertson S, Norris E, et al. Safety and efﬁcacy of buccal midazolam versus rectal diazepam for emergency treat-ment of seizures in children: a randomized controlled trial. Lancet. 2005;366(9481):205–210 29. Knudsen FU, Paerregaard A, Andersen R, Andresen J. Long term outcome of prophylaxis for febrile convulsions. Arch Dis Child. 1996;74(1):13–18 30. Pellock JM, Shinnar S. Respiratory adverse events associated with diazepam rectal gel. Neurology. 2005;64(10):1768–1770 31. Schnaiderman D, Lahat E, Sheefer T, Aladjem M. Antipyretic effectiveness of acetaminophen in febrile seizures: ongoing prophylaxis versus sporadic usage. Eur J Pediatr. 1993;152(9): 747–749 32. Uhari M, Rantala H, Vainionpaa L, Kurttila R. Effect of acet-aminophen and of low dose intermittent doses of diazepam on prevention of recurrences of febrile seizures. J Pediatr. 1995; 126(6):991–995 33. van Stuijvenberg M, Derksen-Lubsen G, Steyerberg EW, Habbema JDF, Moll HA. Randomized, controlled trial of ibu-profen syrup administered during febrile illnesses to prevent febrile seizure recurrences. Pediatrics. 1998;102(5). Available at: www.pediatrics.org/cgi/content/full/102/5/e51 34. van Esch A, Van Steensel-Moll HA, Steyerberg EW, Offringa M, Habbema JDF, Derksen-Lubsen G. Antipyretic efﬁcacy of ibuprofen and acetaminophen in children with febrile seizures. Arch Pediatr Adolesc Med. 1995;149(6):632–637 35. van Esch A, Steyerberg EW, Moll HA, et al. A study of the efﬁcacy of antipyretic drugs in the prevention of febrile seizure recurrence. Ambul Child Health. 2000;6(1):19–26 36. Easley RB, Altemeier WA. Central nervous system manifesta-tions of an ibuprofen overdose reversed by naloxone. Pediatr Emerg Care. 2000;16(1):39–41 37. American Academy of Pediatrics, Committee on Drugs. Acet-aminophen toxicity in children. Pediatrics. 2001;108(4): 1020–1024 1286 AMERICAN ACADEMY OF PEDIATRICS at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from DOI: 10.1542/peds.2008-0939 2008;121;1281-1286 Pediatrics Febrile Seizures Steering Committee on Quality Improvement and Management, Subcommittee on the Child With Simple Febrile Seizures Febrile Seizures: Clinical Practice Guideline for the Long-term Management of & Services Updated Information including high-resolution figures, can be found at: References at: This article cites 36 articles, 21 of which you can access for free Citations les This article has been cited by 2 HighWire-hosted articles: Rs) 3 Peer Reviews (P Post-Publication R has been posted to this article: 3 One P Subspecialty Collections try Neurology & Psychiatry following collection(s): This article, along with others on similar topics, appears in the Permissions & Licensing tables) or in its entirety can be found online at: Information about reproducing this article in parts (figures, Reprints Information about ordering reprints can be found online: at Univ of Florida Health Science Center on July 1, 2009 www.pediatrics.org Downloaded from
2267	https://www.youtube.com/watch?v=JU8g_-9sRgA	For geometric sequence find the minimum value of n when with S_n is greater than 630, second term 5 Ms Shaws Math Class 50900 subscribers 15 likes Description 2137 views Posted: 3 Aug 2021 Transcript: hi everyone for a geometric sequence um and our sequence is a real number with our second term is five and our fifth term is 40 we're going to find the minimum value of a positive integer n such that our sum of n is greater than 630. all right so first we need to find n and what we're going to do is use the fact that our a sub 2 and that is a times r remember your first term for geometric series a the second term would be a times your common ratio the third term would be uh a times your common ratio two times so that would be squared so that equals five and our fifth term is going to be similarly uh your first term times common ratio four times so um r to the power of four and that equals 40. now this is equation 1 in equation 2. so what we want to do is solve for r and a so we can use this formula here all right so let's divide 2 divided by one all right so we take a r to the power of four equals forty uh divide it by um the second term so that's going to be a r and 5. so the a's cancel and you get r cubed equals 8 therefore r equals 2. all right so since r is 2 to find a let's just substitute it in here we need our first term so a times two equals five so our first term is five divided by two all right so since we have that we can use our sum it's going to be our first term here which is 5 halves and then our r is 2 to the power of n minus 1 all divided by 2 minus 1. this equals 5 halves times 2 to the power of n minus 1 and this has to be greater than 630. all right so let's multiply both sides through by two-fifths and that's going to give me uh 2n minus 1 has to be greater than 252. when you multiply those so you get 252 or and when you add 1 to both sides 2 to the power of n is greater than 253 all right now since we have and i think i'll just move on so i'll write this down again we have 2 to the power of n is greater than 253 all right now we have to find the minimum value so let's look and see what we have we know that 2 to the power of 7 is 128 that's not going to work all right so let's look at 2 to the power of 8 2 to the power of 8 is going to be 256. so this will work because this has to be greater than 253. this is greater than 253 and we have to have a positive integer so that means our n has to be greater than or equal to eight thank you have a nice day bye bye [Music] you
2268	https://helpingwithmath.com/scientific-notation/	Skip to content Helping with Math Home » Math Theory » Numbers » Scientific Notation Scientific Notation Math is used by scientists at all levels to explain their thoughts. A physicist must frequently employ extremely small or very big numbers when explaining natural parameters. Some figures are so small that they are difficult to express using standard notation. The mass of an electron, for example, is 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 005 Some numbers are so big that traditional units such as millions, billions, and trillions cannot be used to describe them. The mass of the Sun, for example, is believed to be 8,000,000,000,000,000,000,000,000,000,000 kg, or 8 followed by 30 zeros. It can be difficult to write such high numbers. Engineers and physical scientists implement scientific notation to describe these values efficiently and simply. Hence we can define scientific notation as follows Definition A number is expressed in scientific notation as a product of any integer between 1 and 10 to the 10th power. Because it shortens the notation, scientific notation is most utilized when dealing with huge quantities or numbers with numerous digits. General form of scientific notation Scientific notation has a general representation of a x 10b (where “b” is an integer and “a” is any real value between 1 and 10). Only include significant figures in the real number, “a,” when writing in scientific notation. The coefficient, base, and exponent are the three main components of scientific notation. Examples of scientific Notation 960,000,000 can be expressed as 9.6 x 108 in scientific notation. a = 9.6 b = 8 (according to general form) 0015000 can be written as 1.5 x 104 Here a = 1.5 b = 8 Some more examples of scientific notation 0.00000045 is written in scientific notation as 4.5 x 10-7 675,000,000,000 can be written as 6.75 x 1011 00583000 can be written as 5.83 x 105 0.000005000 can be written as 5x 106 OR 0.5 x 105 So, in these examples we analyzed that value of ‘’a’’ cannot be greater than 10 and b can be any integer without distinguish of positive or negative integer. Rules for scientific notations We must use the following rule to determine the power or exponent of ten: The starting point should always be ten. The exponent has to be a non-zero integer, which might be positive or negative. The coefficient’s magnitude is more than or equal to one, but it should be less than ten. Positive and negative numbers, as well as whole and decimal numbers, can be used as coefficients. The remainders of the number’s significant digits are carried by the mantissa. With the help of the representation below, we can see how many places we need to shift the decimal point after the single-digit value. If the given integer is a multiple of 10, the digits must be shifted to the left, and the power of ten will be positive. 480000000 = 4.8 × 10^8 4230000000 = 4.23 × 10^9 60500000 = 6.05 x 10^7 0.000000098 = 9.7 x 10^-8 0.0000212 = 2.12 x 10^-5 are some instances of scientific notation: 480000000 = 4.8×10^8 1230000000 = 1.23 × 10^9 50500000 = 5.05 x 10^7 0.000000097 = 9.7 x 10-8 0.0000312 = 3.12 x 10-5 How to write Scientific notation? The large or very small number written in ordinary form is called the standard form of the number. We can convert standard form to scientific notation and scientific notation to standard form by following simple steps. First we see the steps for conversion of standard form to scientific notation. Step 1: Shift the decimal point to the left till you have a value that is higher than 1 but less than 10. Step 2: Look at the number of decimal places to the left of the decimal point and use that number to calculate the positive power of ten. Step 3: Multiply the decimal (from Step 1) by a factor of ten (in Step 2). For a value of less than 1 Step 1: Shift the decimal point to the right until you can get a value that is more than or equal to Or less than 10 and equal to 1 Step 2: Check the amount of decimal places the decimal point has been shifted to Correct, then utilize that amount as a negative power of ten Step 3: Multiply the decimal (from Step 1) by a factor of ten (in Step 2). For example 0.00025 = 2.5 × 10^-4 A standard form number less than 1 has negative coefficient. Conversion of scientific notation to standard form Let’s see if we can return our new scientific notation numbers to their standard form. Step 1 Start the process with the number (not the power of ten part). Step 2 For each power of 10 you have, move the decimal point one place. Move the decimal point to the right if the powers of ten are positive. Move the decimal point one place to the left if they’re negative. Step 3 When we’re out of digits, fill in with 0s. Step 4 If your number is large, ensure sure the decimal places to the left of the decimal point are divided by commas into three groups of three. For example Scientific notation 4.45 × 10^2 Take the number and drop the ten and its exponent, as well as the multiplication sign (power) 4.45 2. Because the power of ten was positive, move the decimal one point to the right for each power of ten. Hence 445 3. Because there are only three digits, no commas are required. Some example for conversion of scientific to general 1.45 × 10^-5 = 0.0000145 3.56 ×10^7 = 35,600,000 1.8945 ×10^10 = 18,945,000,000 Arithmetic in Scientific notation Numbers expressed in scientific notation can be added, subtracted, multiplied, and divided while remaining in scientific notation. The processes for adding and subtracting two numbers in scientific notation are listed below. By shifting the decimal point of its decimal number, rewrite the number with the smaller exponent to have the same exponent as the number with the larger exponent. Decimals should be added or subtracted. The power of ten is unchangeable. If needed, convert your result to scientific notation. Here’s an illustration. (5.8 × 10^4) + (4.12 × 10^5) First, observe the numbers 4 and 5 in the exponents. You’ll need to rewrite 5.8 × 10^4 to make it have a 5 exponent. Because the exponent must be increased by one, the decimal point will be moved one place to the left. 5.8 ×10^5 replace 0. 57 × 10^4 Re – write this problem now. (0.58 × 10^5)+(4.12 ×10^5) Then, to use what we’ve learned about decimal addition, add the decimal numbers together the power of ten remains constant. => 0.58+4.12 = 4.70 is the coefficient part of the scientific notation 4.70 × 10^5 is the required answer. Similar for subtracting the numbers in scientific notation For example (1.23 × 10^6) – (1.20 × 10^6) =1.23-1.20 :: exponent is same so it is preserved 0.03 × 10 ^6 is the required answer after subtraction The processes for multiplication or division numbers in scientific notation are outlined below. Divide or multiply the decimal numbers. Add/subtract the exponents of the powers of ten to multiply/divide them. If necessary, convert your solution to scientific notation. Here’s an example of two decimal integers being multiplied. Multiply (3.4 × 10^2) by 6.2 × 10^6. First, use what you’ve learned about decimal multiplication to multiply the decimal numbers. 3.4×6.2=21.08 After that, add the exponents of the powers of ten. 10^2 × 10^6=10^(2+6) = 10^4 Combine the results now. (3.4 × 10^2).(6.2 × 10^6)==(3.4 × 6.2) × (10^2 × 10^6) = 21.08 × 104 Convert your solution to scientific notation at the end. You must rewrite 21.08 × 10^8 in such a way that the decimal value is at least 1 but not more than 10. Move the decimal point to the left one space. Increase the exponent on the 10 by one to keep the overall value the same. 21.08 × 10^8 becomes 2.108 × 10^9 Here’s an illustration of how to divide two decimal integers. Divide (8.4 × 10^5)(1.4 × 10^2) First, use what we have learned about 8.4 ÷ 1.4= 6 to divide the decimal values. Subtract the exponents from the powers of ten to divide them. It’s important to keep in mind that subtracting a negative number is just the same as addition the positive version. 10^5 ÷ 10^2 = 10^(5−2) = 10^3 Combine the results now. (8.4 × 10^5) ÷ (1.4 × 10^2) = (8.4 ÷ 1.4) × (10^5 ÷ 10^2) 6 × 10^7 Last but not least, double-check that your answer is written in scientific notation. Because 6 is less than 10 but greater than 1, your answer is in scientific notation. The solution is (8.4 × 10^ 5) ÷ (1.4 × 10^2) = 6.0 × 10^7. How to write scientific notations in calculator Take, for instance, the value 1.52 x 105 on a TI-30 calculator. To type this number in scientific notation, start by typing 1.52 into the calculator. After then, press the [2nd] key, followed by the x^-1 key, which has EE displayed above it. “x 10 to the power of” is what EE stands for. Type the exponent 5 after you’ve pressed the [EE] key. The display should look like this: 1.52 105 , with a 05 in smaller print in the upper right corner of the display. The [EE] key is not available on the TI-30XS Multi-View TM scientific calculators. Instead, they offer a shortcut key, [x10n], which can be used to enter scientific notation exponents. For a TI-83/84 calculator Click the mode button and select SCI on the top line to keep the number you input remains in scientific notation. When you press the enter key when in NORMAL mode, the number will be extended. Let’s utilize the same 1.52 x 10 ^5 example as the TI-30. 1.52 should be typed in. Then press the [2nd] key, followed by the [ ] “comma” key (it’s the key with the EE above it). After that, enter the exponent. It will be written as 1.52E05, which stands for 1.52 x 10 ^5 Why we use scientific notation When we’re interested in a career in math, engineering, or science (or already work in one of these professions). We’ll almost certainly need to use scientific notation. Computer scientists and astronomy, in particular, rely on scientific notation on a daily basis since they work with microscopic particles all the way up to gigantic celestial objects and require a system that can manage such a wide range of numbers. One of the benefits of scientific notation is that it helps you to be more precise with your numbers, which is important in those fields. Rather than rounding to a figure that is simpler to say or write, scientific notation allows you to be extremely precise with your numbers without making them unmanageable. Writing scientific papers In writing a scientific study article, you may need to utilize scientific notation because scientific investigations can contain very large or very small figures that must be precise. Consider the following scenario: If you’re dealing with the mass of particles or lengths in the universe, you don’t want to see pages full of numbers with digit after digit or numbers with seemingly endless zeroes! You also don’t want to round up or down too much, as this could skew your results and undermine your trustworthiness. Use of scientific notation in science and engineering Scientists and engineers frequently collaborate with both large and small groups of people. In this case, the standard practice of utilizing commas and leading zeroes proves to be extremely inconvenient. Scientific notation is a technique of representation that is more compact and less prone to errors. There are two parts to the number: an accuracy part (the mantissa) and a scale part (the significance) (the exponent, being a power of ten). 23,000, for example, may be written as 23 times 10 to the third power (that is, times one thousand). The exponent can be thought of as the lot of locations to the left of the decimal point. Because writing “times 10 to the X power” is inconvenient, a shorthand approach is utilized, in which the letter E replaces “times 10 to the X power” (which stands for exponent). As a result, 23,000 can be expressed as 23E3. 45E9 represents the value of 45,000,000,000. It’s worth noting that this figure might alternatively be written as 4.5E10 or perhaps even 0.45E11. The only distinction between scientific and engineering notation is that the exponent in engineering notation is always a multiple of three. As a result, 45E9 is correct engineering notation, however 4.5E10 is not. E is commonly denoted by a “EE” or “EXP” button on most scientific calculators. Depressing the keys 4 5 EE 9 would be the procedure of entering the value 45E9. The exponent is negative for fractional values and can be thought of as the number of places the decimal point must be pushed to the right. 0.00067 can so be written as 0.67E3, 6.7E4, or even 670E6. Only the first also last of these three can be used as engineering notation. Engineering notation takes a step further by replacing the multiples of three for the exponent with a set of prefixes. The prefixed are as follows: 10^6 = E6 = Mega (M) 10^3 = E3 = Kilo (k) 10^9 = E9 = Giga (G) 10^12 = E12 = Tera (T) 10^-6 = E-6 = Micro () and so forth Examples 23,000 volts, for example, can be expressed as 23E3 volts or simply 23 kilovolts. This writing is much simpler than the standard form for processing a large range of numbers, in addition to being more concise. Simply multiply the precise parts and add the exponents when multiplication. Divide the accuracy portions and eliminate the exponents when dividing. For example, multiplying 23,000 by 0.000003 may appear to be a difficult process. This is 23E3 times 3E6 in technical notation. 69E3 is the outcome (that is, 0.069). It will become second nature after enough experience that kilo (E3) multiplied by micro (E6) equals milli (E3). This will make lab calculations a lot easier. 42,000,000 divided by 0.002 equals 42E6 divided by 2E3, or 21E9 (the exponent is 6 minus a negative 3, or 9) Before adding different, make absolutely sure the exponents are equal (scaling if necessary) before adding or subtracting the precision sections. For instance, 2E3 plus 5E3 equals 7E3. 2E3 plus 5E6 is like 2E3 plus 5000E3, or 5002E3 by comparison (or 5.002E6). Conclusion Overall, scientific notation is a handy technique to write and work with extremely large or extremely small numbers. Scientific notation is useful for persons undertaking academic and professional work in math and science, even if it may seem difficult to conceive applying it in everyday life. A scientific notation calculator and converter make using this shorthand more easily for anyone studying or working in these subjects. 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2269	https://www.cuemath.com/current-formula/	Current Formula The current formula is derived from Ohm's law. Current is defined as the flow of electrons in an electric circuit. The flow of electrons occurs due to potential differences. The current is also known as the rate of change of charge with time. Current is represented by I and SI unit of current is Ampere. Let us learn the application of the electric current formula in the section below. What Is the Current Formula? Ohm's Law states that the voltage (V) across a conductor is equal to the product of the current (I) flowing through it and the resistance (R) of the conductor. According to Ohm's law, the current is the ratio of the potential difference (voltage) and the resistance. Thus, the electric current formula is given by: I = V/R where I represent current in Ampere (A), V is the potential difference in Volt (V) R is the resistance in Ohm (Ω). This current equation can be used to calculate the current in a circuit if the voltage and resistance are known, or to calculate the current or resistance if the other two values are known. Let us see the applications of the current formula in the following solved examples section. Want to find complex math solutions within seconds? Use our free online calculator to solve challenging questions. With Cuemath, find solutions in simple and easy steps. Book a Free Trial Class Examples Using Current Formula Example 1: In an electric circuit, the potential difference and the resistance are given as 20V and 4Ω respectively. Calculate current flowing in the circuit. Solution: To find: Current (I) flowing in the circuit. Given: V = 20 V, R = 4 Ω Using current equation, I = V/R I = 20/4 I = 5 Answer: Current flowing in the circuit is 5 Ampere. Example 2: The total current flowing in an electric circuit is 50 Amp whereas the resistance of the wires is 14 Ohm. Find the potential difference. Solution: To find the potential difference: Given: I = 50 A, R = 14 Ω Using electric current formula I = V/R 50 = V/14 V = 50 × 14 V = 700 Answer: Potential difference is 700 V. Example 3: In an electric circuit, the potential difference is 20 V and the value of current is 5 Amp respectively. Using the current formula, find the resistance of the circuit. Solution: To find the resistance (R) of the circuit: Given: V = 20 V, I = 5 Amp Using current formula R = V/I R = 20/5 R = 4 Ω Answer: The resistance of the circuit 4 Ω. FAQs on Current Formula How Do You Calculate Current Using Current Formula? If the voltage (V) and resistance (R) of any circuit is given we can use the electric current formula to calculate the current, i.e., I = V/R (amps). How Do You Calculate Voltage Using Current Formula? If the current (I) and resistance (R) of any circuit is given we can mold the current formula to calculate the voltage, i.e., V = IR (Volts). Can the Current Formula be Used for Both DC and AC Circuits? The current equation (I = V/R) is used for both direct current (DC) and alternating current (AC) circuits. However, it's important to note that for AC circuits, the resistance (R) is often replaced by impedance (Z) since AC circuits involve the effects of both reactance and resistance. How Do You Calculate Resistance Using the Current Formula? If the current (I) and potential difference (V) of any circuit are given we can mould the current formula to calculate the resistance, i.e., R = V/I (Ohms Ω). What Is the Definition of Electric Current Formula? Write Its SI Unit. The current is the ratio of the potential difference and the resistance. It is represented as (I). The current formula is given as I = V/R. The SI unit of current is Ampere (Amp). What Happens if the Resistance is Zero in the Current Formula? The current equation is I = V/R. If the resistance, R = 0, I = V/R would result in an undefined value for current (I/0). In practical terms, it means when an extremely high current can flow, potentially causing damage to the circuit or equipment, i.e., it would imply a short circuit. 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2270	https://www.youtube.com/watch?v=Giess9dYmGs	13. Regulation of Fatty Acid Oxidation \| Biochemistry \| USMLE Step 1 High-Yield Review MedSchool Simplified – High Yield Medical Lectures 74300 subscribers 3 likes Description 46 views Posted: 16 Aug 2025 𝐒𝐮𝐛𝐬𝐜𝐫𝐢𝐛𝐞 𝗙𝐨𝐫 𝗠𝐨𝐫𝐞 𝗜𝐧𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧 𝐨𝐧 𝗛𝐞𝐚𝐥𝐭𝐡 👩‍⚕‍ 𝐚𝐧𝐝 𝗠𝐞𝐝𝐢𝐜𝐢𝐧𝐞💉🩺💊 📌𝗜𝗻𝘀𝘁𝗮𝗴𝗿𝗮𝗺 : Regulation of Fatty Acid Oxidation \| Biochemistry \| USMLE Step 1 High-Yield Review This high-yield USMLE Step 1 lecture explains the Regulation of Fatty Acid Oxidation, a critical metabolic process that ensures the body switches efficiently between carbohydrate and fat utilization depending on energy demands. Fatty acid oxidation, primarily β-oxidation, occurs in the mitochondrial matrix and is the main source of ATP during fasting, prolonged exercise, and low-insulin states. Understanding its hormonal, enzymatic, and transport regulation is key for mastering integrated biochemistry, physiology, and pathology. 🧠⚡ Hormonal Regulation: – Insulin (fed state) inhibits fatty acid oxidation by promoting fatty acid synthesis and malonyl-CoA accumulation. – Glucagon and epinephrine (fasted state) stimulate oxidation by activating hormone-sensitive lipase (HSL) in adipose tissue → releases free fatty acids into the blood. Transport Regulation (Carnitine Shuttle Control): – Long-chain fatty acids require carnitine palmitoyltransferase I (CPT I) to enter mitochondria. – Malonyl-CoA inhibits CPT I, preventing simultaneous fatty acid synthesis and oxidation (avoids a futile cycle). – During fasting, AMP-activated protein kinase (AMPK) phosphorylates and inactivates acetyl-CoA carboxylase, lowering malonyl-CoA → relieving CPT I inhibition. Enzyme-Level Control: – Hormone-sensitive lipase mobilizes triglycerides in adipose tissue (activated by PKA in low-insulin states). – Acyl-CoA dehydrogenases catalyze β-oxidation steps inside mitochondria. – Deficiencies (e.g., MCAD deficiency) cause hypoketotic hypoglycemia — a high-yield Step 1 condition. Clinical Integration: – Primary carnitine deficiency → impaired transport of fatty acids into mitochondria → muscle weakness, hypoketotic hypoglycemia. – Secondary carnitine deficiency (e.g., chronic liver disease) → reduced availability of carnitine for oxidation. – MCAD deficiency → accumulation of medium-chain fatty acids, seizures, vomiting in fasting states. – Malonyl-CoA decarboxylase defects → impaired switch from fed to fasting metabolism. By the end of this lecture, you’ll be able to: – Map hormonal and transport regulation of β-oxidation – Recognize key enzyme checkpoints – Apply this knowledge to metabolic disease questions on Step 1 FattyAcidOxidation #BetaOxidation #CarnitineShuttle #USMLEStep1 #MetabolicPathways #BiochemistryReview #HormonalRegulation #MalonylCoA #CPT1 #MCADDeficiency #NBMEReview #Step1Prep #WhiteboardMedicine #DrGBhanuPrakash #USMLEBuzzwords #MedicalEducationUSA #USMLE2025 #EnergyMetabolism #AMPK #FastingMetabolism Transcript: [Music] [Music] Dear student, I am Dr. Prakash Mongli, professor of biochemistry. So in this fatty acid oxidation lecture I will be explaining you regulation of fatty acid oxidation. Let's get into the details. When it comes to regulation of fatty acid oxidation uh regulation of fatty acid oxidation is revolving around a single molecule that you must remember to remember the regulation of fatty acid oxidation and that is melonile coa. Make sure you remember this name of this molecule. Melanile COA. How do you get melanile COA? Melanile COA. It is basically coming from an acetile COA. Acetile CO is a twocarbon molecule as I'm showing here. Here it is. Acetile COA which is a twocarbon molecule will be converted to melan COA and this will be done by enzyme acetile COAS. Now acetile coercase is the one that will be converting acetile koi into melanile koi and this this reaction will be going on whenever person is in fed condition. So under fed condition fed state as you know there will be release of insulin and this insulin especially in the liver it is going to induce induce it induces this acetyl coaroxilace and also it is going to keep this enzyme under dephosphorilated state thereby this enzyme is active and when this enzyme is active it is going to convert more and more acetile coa into melan coa so that melanile coa is going into fatty acid synthesis. In fact, the purpose of making melanile coa is to make a fatty acid. Fatty acid synthesis. A fatty acid is synthesized using a melan coa. So that's what is the purpose of insulin. While this melan coa is elevated when the fatty acid is synthesized, fatty acid oxidation should not go on. Remember this because fatty acid synthesis and fatty acid oxidation that is beta oxidation these are antagonistic pathways always antagonistic pathways should not run simultaneously that's why whenever melan coa levels are high in the cytoplasm this melan coa it will have a negative effect on a CPT1 enzyme CPT1 The CPT1 enzyme it is there over outer mitochondrial membrane and it is involved in the transport of longchain fatty acids into the mitochondrial matrix by carnitine eststerification. So CPT1 over outer mitochondrial membrane what it does it is going to add carnitin to fatty coa to make it as fatty cornet. We have seen that in in our previous chapter where uh mechanism of transport of longchain fatty acids into the matrix of mitochondria. So whenever fatty acid synthesis is going on at the same time fatty acid oxidation should not go on remember that. So melanile coa when it is synthesized at high concentration in the cytoplasm. So the melan co what it does it is going and inhibiting a enzyme which is located over the outer mitochondrial membrane. Uh that name of the enzyme is CPT1. So the CPT1 enzyme let me write it down. Cpt1 enzyme is involved in the eststerification of fatty coa with the connotin to make fatty cornin and the longchain fatty acid. That's how it is transported inside the matrix of mitochondria. We have seen that in our fatty acid oxidation lecture in the beginning. Now whenever melan co is inhibiting this CPT1 so that means longchain fatty acids are not transported into the matrix of mitochondria. That means bet oxidation do not go on. So when fatty acid synthesis is going on under fed condition oxidation should not go on in the matrix of mitochondria. That's how it is regulated. So the melan co it is kind of controlling the beta oxidation by not allowing longchain fatty acids to go into the matrix of mitochondria. Okay. Now that happens under fed condition. So insulin will make sure melan co levels are high and that melan koa will have a negative effect on CPT1 and also under fasting condition what will happen? So whenever person is in fasting condition fasting state so fasting hormone is glucagon and this glucagon what it does it is going to have a negative effect on acetile coaaroxilase enzyme. Basically glucagon is going to phosphorilate acetile coercase and when acetile coaaroxilase is phosphorilated it is no longer converting acetyl coa into melan coa when acetile co is not converted to melanile coa under fasting condition so the levels of melan coa will decrease and when the levels of melan coa decreases so its negative effect on CPT1 is not there that means CPT1 is working because the negative effect is not there. So fatty acids longchain fatty acids are going through the inner mitochondrial membrane and undergo oxidation. So at the same time when the fatty acid oxidation is going on fatty acid synthesis will decrease because melan co levels are less melanile co is the one that has to go into fatty acid synthesis. So like this glucagon will increase bet oxidation of fatty acid by keeping melanile coil levels low and gluc insulin is going to keep fatty acid synthesis high bet oxidation low because melan co levels are elevated. So like this it is all around melan coa concentration that you must remember this is in fed condition and fasting condition. Now let's move on to see what is the what happens whenever a person is exercising during exercise like activity in the especially in the skeletal muscle. So in the muscle uh how this fatty acid oxidation is controlled. So in the skeletal muscle ams will increase whenever exercise whenever a person do exercise. So this AM is going to activate AM activated protein kynise and thereby it will have a negative effect on acetile coaroxilase. Let's look at that in the next slide. So here is a regulation of fatty acid oxidation in this skeletal muscle and cardiac tissue. This figure is showing you how a fatty acid oxidation is regulated in the muscle here. Now the AM whenever muscle contraction is going on so the AM level rises and this AM is going to activate a protein called AM activated protein kynise and this AM activated protein kynise has two roles. One of the role of AM activated protein kynise is to keep low melan coa levels by not allowing conversion of acetile coa into melan coa. You are seeing that in a number one reaction here. Right? So the number one is ACC. ACC2 is acetile coercaroxase is written as ACC2. Here is the acetile coarox ACC. Now acetile coarox is negatively modulated byk that is ampokin activated protein kynise. it is going to phosphorilate acetyl coarox and keep the enzyme activity low thereby acetyl co is not converted into melan co and that's one thing the second thing is same ampk that is activated protein kynise it is going to act positively on another enzyme called melan coicaroxilase and that melanile coicarox is written here as m co a dc M coc Mco COADC is melanile coa decarboxulate. So what this enzyme do? As the name says decarboxulation it is going to decaroxulate melan coa into acetyl coa thereby it decreases the levels of melan co. So like this increased AM during muscle contraction activate AM activated protein kynise that's going to inhibit acetile coaroxilates thereby decrease the conversion of estile koi to melan ka that will decrease melanile co levels on the other hand whatever melan ko that are already there they will be decarboxilated to acetile coa done by melan ka decarbox that is also activated byp PK because AMPK is going to keep melanile coated caroxillayase active thereby further decreasing melan co levels. So that means ampk is going to decrease the formation of melan ka by keeping acetile coarox inactive and also it is going to increase the degradation of melan ko into acetile co thereby by keeping melanile kicarox active. So overall amp PK it is going to totally decrease melan coil levels. When the melan coil levels are totally decreased which is shown in number two reaction here. So it's negative effect on CPT1 is not there. So the CPT1 here CPT1 carnitin palile transfer is one negative effect is not there. So fatty acidile fatty acids are easily flowing into the matrix of mitochondria and they will undergo oxidation to give energy needs of the skeletal muscle and cardiac tissue. So like this the regulation of beta oxidation of fatty acids it is all centered around melan coa concentration and its effect on CPT1. Anything that increases the concentration of melan coa will decrease vit oxidation which is done by insulin under fed condition and anything that decreases the melan coa concentration will increase bet oxidation because cpt1 will be free to work and that is done by epinephrine glucagon and also during exercise am which is increasing amp protein kynise am activated protein kynes Okay. So two enzymes that you need to remember here. One is acetile coercase. Other is melanile coa decarbox. When acetile coaroxilase is inhibited melanile co levels will decrease. When melanile coa decarbox levels are mean when melan co decaroxil is activated so again melan co levels will decrease because melan co is diverted into acetile co formation. So overall fatty acid oxidation will be regulated in this way. That's all about regulation of fatty acids. Thank you.
2271	https://www.clinicalkey.com/#!/content/book/3-s2.0-B9780323354790001410	ClinicalKey - Lead with Answers Skip to search Skip to navigation Skip to content Skip to help and feedback ClinicalKey Search Browse Tools Verifying Access… English CE CME/MOC CME/MOC Store Login Register Account options for Settings CME/MOC CME/MOC CME/MOC CE Change Language Store Saved Content Presentations Change Organization Remote Access Log Out Admin Portal Account options for Logged in via ClinicalKey Global Guest Users Activate Remote Access Log Out Help Call Us Email Us Help Search for diagnoses, conditions, drugs and more...Search for diagnoses, conditions, drugs and more... Main Navigation Close Verifying Access… Anonymous (LoginorRegister) Account options for Logged in via ClinicalKey Global Guest Users (Activate) Browse Books Journals Clinical Overviews Drug Monographs Guidelines Patient Education Multimedia Procedure Videos Tools Clinical Calculators IV Compatibility Clinical Comparison Adverse Reaction Drug Interaction Drug Identifier Do Not Crush More Change Role Profile Settings CME/MOC CME/MOC CME/MOC CE Change Language Store Saved Content Presentations Change Organization Remote Access Log Out Admin Portal Help & Feedback Call Us Email Us Help About ClinicalKey Learn more about other offerings and editions Search for conditions, treatments, drugs, books, journals, and more Choose search scope All Types All Types Books Journals Clinical Overviews Clinical Trials Drug Monographs Guidelines Patient Education Multimedia Procedure Videos Clinical Calculators Clinical Focus Search Inform your clinical decisions Clinical Overviews Comprehensive clinical topic summaries Drug Monographs Drug and dosing information to guide treatment Calculators Interactive tools to support clinical decisions Guidelines Latest clinical guidelines from leading organizations Patient Education Educate and empower your patients Deepen your specialty and medical knowledge Books Browse our collection of reference books Journals Browse our collection of journals Procedure Videos Watch videos that demonstrate best practice Multimedia Image and video collection organized by specialty Learn more about the World's First Clinical Insight Engine. Get news and updates Content Updates Content Updates: Summer 2025 View the latest content changes from Summer 2025. New Books Added in May View the latest content changes from May 2025. New Books Added in April Here’s a look at the content changes for April 2025. Other Updates ClinicalKey AI - Start your Free Trial Today ClinicalKey AI combines trusted, evidence-based clinical content with conversational search powered by generative AI to support clinicians in delivering high-quality patient care. Gen AI Academy for Health Elevate your AI expertise and transform your clinical practice. Sign up for free CME-Accredited Course now! Trending Topics How to use ClinicalKey ClinicalKey Quick App Sheet This instructional guide outlines how to download and access the ClinicalKey mobile app and utilize its key features. ClinicalKey Adoption Kit Institutional administrators can utilize this suite of assets designed to help you drive adoption and usage of ClinicalKey within your organization. Training Calendar Find upcoming ClinicalKey training sessions and events. Trending Topics Measles View search results for Measles. Clinical Rounds: a new section to The Lancet's clinical content The Lancet will now include cases that expand clinical reasoning and highlight diagnostic nuance or cover novel treatments or techniques. ; 246711 Google Play - Go to our Google Play page Apple App Store - Go to our Apple App Store page Facebook - Go to our Facebook page LinkedIn - Go to our LinkedIn page X - Go to our X page Elsevier (opens in a new window) Contact Us (opens in a new window) Resource Center (opens in a new window) Registered User Agreement (opens in a new window) Help (opens in a new window) Terms & Conditions (opens in a new window) Privacy Policy (opens in a new window) Accessibility (opens in a new window) We use cookies to help provide and enhance our service and tailor content. By continuing you agree to the Cookie Settings. All content on this site: Copyright © 2025 Elsevier Inc. or its licensors and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. (opens in a new window)
2272	https://www.sciencedirect.com/topics/neuroscience/translation-termination	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Fidelity and Quality Control in Gene Expression Abstract Translation termination in eukaryotes occurs in response to a stop codon in the ribosomal A-site and requires two release factors (RFs), eRF1 and eRF3, which bind to the A-site as an eRF1/eRF3/GTP complex with eRF1 responsible for codon recognition. After GTP hydrolysis by eRF3, eRF1 triggers hydrolysis of the polypeptidyl-tRNA, releasing the completed protein product. This leaves an 80S ribosome still bound to the mRNA, with deacylated tRNA in its P-site and at least eRF1 in its A-site, which needs to be disassembled and released from the mRNA to allow further rounds of translation. The first step in recycling is dissociation of the 60S ribosomal subunit, leaving a 40S/deacylated tRNA complex bound to the mRNA. This is mediated by ABCE1, which is a somewhat unusual member of the ATP-binding cassette family of proteins with no membrane-spanning domain but two essential iron–sulfur clusters. Two distinct pathways have been identified for subsequent ejection of the deacylated tRNA followed by dissociation of the 40S subunit from the mRNA, one executed by a subset of the canonical initiation factors (which therefore starts the process of preparing the 40S subunit for the next round of translation) and the other by Ligatin or homologous proteins. However, although this is the normal sequence of events, there are exceptions where the termination reaction is followed by reinitiation on the same mRNA (usually) at a site downstream of the stop codon. The overwhelming majority of such reinitiation events occur when the 5′-proximal open reading frame (ORF) is short and can result in significant regulation of translation of the protein-coding ORF, but there are also rare examples, mainly bicistronic viral RNAs, of reinitiation after a long ORF. Here, we review our current understanding of the mechanisms of termination, ribosome recycling, and reinitiation after translation of short and long ORFs. View chapterExplore book Read full chapter URL: Chapter Molecular Cell Biology 2016, Encyclopedia of Cell BiologyW. Hu Translation Termination and mRNA Stability Eukaryotic mRNA translation termination requires two release factors, eRF1 and eRF3. Translation termination process can influence mRNA half-life. Specifically, it was observed that the N-terminal domain of eRF3, which is not required for translation termination, can interact with Pab1, and this interaction is involved in modulating mRNA stability (Hosoda et al., 2003). Disruption of this interaction results in translation-dependent stabilization of mRNA caused by decreased deadenylation rate (Hosoda et al., 2003). Interestingly, it was further found that certain deadenylase complexes can also bind to the same site on Pab1 that is involved in the interaction with eRF3 (Funakoshi et al., 2007). Thus, it has been postulated that eRF3 can regulate mRNA deadenylation by competitively binding to the Pab1, which then modulates the recruitment and activation of deadenylase complexes (Funakoshi et al., 2007). In addition to the release factors, other proteins that can modulate translation termination can also influence mRNA stability. For example, a recent characterized protein named Tpa1 can interact with the two release factors and regulate the readthrough of stop codons (Keeling et al., 2006). Interestingly, although the detailed mechanisms still remain elusive, knocking out this protein can have decreased deadenylation rate and increased mRNA stability (Keeling et al., 2006). Collectively, these results suggest that mRNA translation termination can results in mRNP conformational changes that can influence mRNA stability, likely via modulating mRNA deadenylation. View chapterExplore book Read full chapter URL: Reference work2016, Encyclopedia of Cell BiologyW. Hu Chapter Genome and Gene Structure 2013, Emery and Rimoin's Principles and Practice of Medical Genetics (Sixth Edition)Daniel H. Cohen 5.4.2.3 3′-Untranslated Sequences and Transcriptional Termination The 3′ ends of primary transcripts are determined by transcriptional termination signals located downstream of the ends of each coding region. However, the 3′ ends of mature mRNA molecules are created by cleavage of each primary precursor RNA and the addition of a several hundred nucleotide polyadenylate (poly(A)) tails (see Figure 5-5). The cleavage site is marked by the sequence 5′-AAUAAA-3′ located 15–20 nucleotides upstream of the poly(A) site and by additional GU-rich sequences 10–30 nucleotides downstream. Histone mRNAs, which do not have poly(A) tails, have stem-loop structures instead with cleavage of the primary transcript mediated by a distinct protein complex that includes the U7 snRNP (20). Some complex transcriptional units contain several potential polyadenylation and/or transcription termination sites. It is often difficult to distinguish the latter from the former as the product available for analysis (mRNA) has lost the portion of the 3′ terminus originally transcribed by RNA polymerase. Alternative polyadenylation (or termination) sites can determine final protein structure if the longer precursor RNA contains an exon not found in the shorter precursor RNA. In a simple case, two proteins with different carboxyl termini are formed. But if alternative exon splice sites are made available in the longer precursor RNA, proteins with entirely different sequences can be produced. The region from the translation termination codon to the poly(A) addition site may contain up to several hundred nucleotides of a 3′-UTR, which includes signals that affect mRNA processing and stability. Many mRNAs that are known to have a very short half-life contain AU-rich elements, 50–150 bp sequences containing AUUUA motifs that regulate mRNA stability (21). Other, less well-characterized sequences can have similar effects. Removal or alteration of these sequences can prolong the half-life of mRNA, indicating that such elements represent a general regulatory feature of mRNAs whose level of expression can be rapidly altered. View chapterExplore book Read full chapter URL: Book2013, Emery and Rimoin's Principles and Practice of Medical Genetics (Sixth Edition)Daniel H. Cohen Chapter Disorders of Protein Synthesis 2022, Advances in Protein Chemistry and Structural BiologySilvia Lombardi, ... Alessio Branchini 3.1 Mechanism, programmed readthrough and natural vs premature stop codons The central role of protein synthesis has forced cells to develop accurate mechanisms to avoid translational errors. However, although effective, the process of translation termination, during which a competition occurs for the recognition of the stop codon (Fig. 4A), is not 100% efficient. At stop codons located at the mRNA 3′ end, release factors highly outcompete aa-tRNAs, thus driving polypeptide chain release. In rare cases, a stop codon can be decoded by an aa-tRNA, with addition of an amino acid to the polypeptide chain, thus allowing elongation to proceed until the next in-frame stop codon is reached (Fig. 4B) (Rodnina et al., 2019). This mechanism is called ribosome or translational readthrough and can be driven by a near-cognate aa-tRNA (i.e., a tRNA whose anticodon is complementary to two out of three positions; see Fig. 1), or by recoding events mediated by specialized tRNAs carrying the 21st selenocysteine or 22nd pyrrolysine amino acids (Cobucci-Ponzano, Rossi, & Moracci, 2012; Namy, Rousset, Napthine, & Brierley, 2004; Zhang, Baranov, Atkins, & Gladyshev, 2005). In particular, pyrrolysine is encoded by the UAG stop codon and is restricted to several microbes, whereas selenocysteine is encoded by the UGA codon and can be found in bacteria, archaea and eukaryotes. Incorporation of selenocysteine requires a specific insertion sequence (selenocysteine inserting sequence, SECIS) and is necessary for the synthesis of selenoproteins (Fig. 4C) (Allmang & Krol, 2006; Cobucci-Ponzano et al., 2012; Zhang et al., 2005). Approximately 100 selenoprotein families have been discovered, among which the human selenoproteome comprises 25 genes, with half of them encoding proteins with known functions (Hatfield, Tsuji, Carlson, & Gladyshev, 2014; Kryukov et al., 2003). These include peroxidase (glutathione-dependent), reductase (NADPH-dependent) and deionidase (metabolism of thyroid hormones) activities, and the specialized selenoprotein P, bearing 10 selenocysteine residues, produced by the liver and involved in the delivery of selenium to other organs (Hatfield et al., 2014; Labunskyy, Hatfield, & Gladyshev, 2014; Reeves & Hoffmann, 2009). Ribosome readthrough was first identified in viruses as a way to expand their genetic information (Beier & Grimm, 2001). It was observed that, in Escherichia coli infected by RNA phage Qβ, the tRNATrp stimulated readthrough over the UGA stop codon at the end of the coat protein cistron, resulting in a longer coat protein essential for the production of infective Qβ particles (Hirsh, 1971). Since then, stop codon readthrough has been found to play important biological roles in several organisms, providing a mechanism to regulate gene expression through the production of different proteins from the same gene (Beznosková, Wagner, Jansen, von der Haar, & Valášek, 2015; Eswarappa et al., 2014). This phenomenon has been called programmed readthrough and its prevalence varies among organisms, with a few examples found also in humans (Loughran et al., 2018; Rodnina et al., 2019; Stiebler et al., 2014). Nevertheless, unregulated readthrough of NTCs is generally a rare event and the termination signals at the end of an open reading frame evolved several ways to promote efficient termination, especially among highly expressed transcripts. A paradigmatic example is represented by tandem stop codons, which act as “back-up” termination signals in case of readthrough (Fleming & Cavalcanti, 2019; Liang, Cavalcanti, & Landweber, 2005). Moreover, the nonstop decay surveillance mechanism recognizes and degrades those transcripts with stalling ribosomes bound to the poly(A) tail, thus preventing the generation and accumulation of unwanted C-terminally extended proteins (Klauer & van Hoof, 2012). In addition, recent studies suggested that translation into 3′ UTRs might reduce C-terminally extended protein levels through 3′ UTR-encoded peptides that destabilize the attached protein, either co- or post-translationally (Arribere et al., 2016). This scenario changes in the presence of a PTC, inserted for instance by a nonsense mutation. In this case, the frequency of spontaneous readthrough is about 10-fold higher (0.01–1%) than at NTCs (0.001–0.1%) (Bonetti, Fu, Moon, & Bedwell, 1995; Cassan & Rousset, 2001; Keeling, Wang, Conard, & Bedwell, 2012; Manuvakhova, Keeling, & Bedwell, 2000). This difference is likely due to the presence (NTCs) or lack (PTCs) of interactions between the termination complex and factors bound to the 3′ UTR of the mRNA. Indeed, during translation the mRNA is maintained in a closed-loop conformation whose function is both to protect the ends of the transcript from exonucleolytic degradation (Chen & Shyu, 2011), and to enhance recycling of translational components, thus increasing the frequency of translation initiation. This closed conformation is maintained by the association of three actors, namely, the cap-binding protein eIF4E, which is bound to the 5′-cap structure of the mRNA, the poly(A)-binding protein (PABP), attached to the poly(A) tail, and eIF4G, which binds both eIF4E and PABP resulting in mRNA circularization. Along with its role in the formation of the closed-loop complex, PABP also promotes translation termination by interacting with eRF3 and stimulating polypeptide chain release (Cosson et al., 2002). In the presence of a PTC, which is usually distant from the NTC and thus from the poly(A) tail, the interaction between PABP and eRF3 will likely be less efficient, leading to prolonged ribosomal pausing and increased aa-tRNA sampling, which in turn makes the PTC more susceptible to readthrough (Amrani et al., 2004). View chapterExplore book Read full chapter URL: Book series2022, Advances in Protein Chemistry and Structural BiologySilvia Lombardi, ... Alessio Branchini Chapter Functional Cell Biology 2016, Encyclopedia of Cell BiologyS. Djuranovic, H.S. Zaher Substrate Recognition NMD As mentioned earlier, all NMD targets share the common feature of having a stop codon at a noncanonical position. Due to the nature of the stop codon translation termination is speculated to be quite distinct from normal termination (Amrani et al., 2004). The exact mechanism by which the ribosome distinguish between a normal stop codon and a PTC is not fully understood, but it involves three conserved proteins: Upf1, Upf2, and Upf3 in S. cerevisiae (Cui et al., 1995; Leeds et al., 1991, 1992). These proteins are thought to bridge a connection between downstream signals and the terminating ribosome (Chakrabarti et al., 2011; Chamieh et al., 2008). The connection between the Upf proteins and the ribosome is thought to be stimulatory for NMD, but inhibitory for termination and they compete with connections that are under normal conditions stimulatory for termination. How do Upf proteins communicate signals to the ribosome? Early studies on NMD suggested that mRNAs harboring a PTC >~50 nt upstream of exon–exon junction sites triggered a robust NMD response (Nagy and Maquat, 1998; Thermann et al., 1998; Zhang et al., 1998). Later studies showed that during splicing the region just upstream of the exon–exon junction sites is coated with a protein complex fittingly called the exon junction complex (EJC) (Le Hir et al., 2000). During the pioneering round of translation, the EJC complexes are stripped off the mRNA by the ribosome. Given that most of the stop codons reside in the last exon of a gene, the presence of an EJC complex downstream of a terminating ribosome provides a rational molecular signature for NMD. The core EJC complex is composed of three proteins, eIF4A3, MLN51, and a MAGOH/Y14 heterodimer, which interacts with the C-terminus of the NMD factor Upf3 (Andersen et al., 2006; Bono et al., 2006; Buchwald et al., 2010). Upf3 has orthologoues in all examined eukaryotes and consistent with its interaction with the EJC complex it shuttles between the nucleus and cytoplasm (Kim et al., 2001). Upf2 acts as a bridge between Upf3 and Upf1, which in turn interacts with the release factors eRF1 and eRF3 as well as the phosphatidylinositol 3-kinase-related kinase SMG1 (Kashima et al., 2006; Kunz et al., 2006). Among NMD factors, Upf1 is the most conserved factor and is central to the process (Culbertson and Leeds, 2003). Sequence analysis as well as biochemical characterization of the protein revealed that it belongs to the ATP-dependent helicase superfamily (Cheng et al., 2007; Czaplinski et al., 1995). It binds RNA and ATP and uses the energy from ATP hydrolysis to unwind RNA in a 5′ to 3′ direction. In addition, the protein is subject to cycles of phosphorylation (by SMG1) and dephosphorylation on its unstructured C-terminus and N-terminus region (Grimson et al., 2004; Okada-Katsuhata et al., 2012; Yamashita et al., 2001). On NMD substrates, Upf1–Upf2–Upf3 form a stable complex and it has been suggested that interaction between Upf2 and a cysteine/histidine (CH) rich domain on Upf1 activates the ATPase and helicase activity of Upf1 (Chamieh et al., 2008). In the absence of Upf2, the CH domain of Upf1 interacts with the factor’s helicase domain forming a closed conformation that inhibits the unwinding activity of Upf1. In addition to their interaction with the EJC complex, Upf proteins have been shown to interact with translation factors, namely release factors, for which they appear to modulate termination efficiency (Kashima et al., 2006). Stop codon readthrough is significantly more efficient on PTCs relative to normal stop codons (Wang et al., 2001). What is clear from the many decades of studies on NMD is that the process is intimately coupled to translation termination. Termination of protein synthesis occurs when one of three nearly universal stop codons (UAG, UGA, and UAA) enter the A site of the ribosome. In eukaryotes, termination requires the two release factors eRF1 and eRF3. eRF1 is a tRNA mimic, which binds the A site of the ribosome and recognizes the stop codon. eRF3 is a GTPase and similar to EF1A (Frolova et al., 1996), which binds aa-tRNAs to deliver them to the ribosome during elongation, is thought to form a ternary complex with eRF1 and GTP (Mitkevich et al., 2006). Once on a terminating ribosome, GTP is hydrolyzed, eRF1 undergoes a conformational change allowing the conserved GGQ domain to engage the active site of the ribosome to initiate peptide release. It has also been suggested that eRF3 interacts with polyA-binding protein (Pab1), and that this interaction stimulates peptide release (Kononenko et al., 2010). Following peptide release, the conserved recycling factor ABCE1 binds to eRF1 kicking eRF3 off; subunit dissociation ensues to complete the translation cycle (Pisarev et al., 2010). Based on this relatively short description of termination, it should be apparent that the process is amenable to modulation through differential protein–protein interaction network. Indeed on NMD substrates, as detailed earlier, termination is inhibited due to what has been suggested as direct competition between the Upf proteins and termination-stimulating factors (Ivanov et al., 2008). In particular, Upf1 has been shown to interact with eRF3, but whether this interaction on its own is sufficient to elicit NMD is debated. While the EJC model is elegant and appealing, it fails to explain all of the targets of NMD. An increasing number of NMD-targeted mRNAs do not have an intron downstream of a stop codon and as such are predicted not to harbor an EJC complex. Furthermore, NMD is robust in yeast, even though most of the pre-mRNAs are not spliced. It is worth noting that immunoprecipitation of eIF4AIII (a core component of the EJC) followed by deep sequencing suggests that the EJC can be deposited on mRNAs in a sequence-dependent and splicing-independent manner (Sauliere et al., 2012; Singh et al., 2012). So the possibility that the EJC is found on the UTR of mRNAs lacking an intron downstream of the stop codon cannot be ruled out. Furthermore, for all of these noncanonical targets, the mRNAs share the common feature of having a long UTR. The UTR model suggests that Upf1 binds to mRNAs nonspecifically and during translation the ribosome actively displaces the factor (Hogg and Goff, 2010). Since the ribosome does not traverse the 3′-UTR, mRNAs with long UTR are expected to have a larger number of Upf1 proteins and hence are subject to NMD. Alternatively, it has been suggested that the polyA tail of mRNAs with long 3′-UTRs is distant from the stop codon, precluding any stimulatory effect PAB1 might have on termination. In contrast, others have argued that in the absence of a downstream EJC complex Upf2 and Upf3 interact with a ribosome-bound Upf1, which on its own is able to sense the efficiency of termination (Stalder and Muehlemann, 2008). In summary, Upf1 appears to be the primary factor in determining what constitutes an NMD target whatever the initiating signal might be. Future work is needed to clarify the mechanism of the initial recognition of a PTC. NGD NGD is triggered by a stalled ribosome and requires the conserved protein Dom34 (or Pelota) and Hbs1. Interestingly, Dom34 and Hbs1 share structural features with eRF1 and eRF3, respectively (Atkinson et al., 2008; Chen et al., 2010). Consistent with these structural similarities, the factors interact with the A site of the ribosome (Becker et al., 2011) but because Dom34 lacks the GGQ domain of eRF1, the complex does not promote peptide release. Instead, in vitro biochemical studies showed that the factors promote subunit dissociation (Shoemaker et al., 2010). Moreover, the recycling factor ABCE1 (Rli1 in yeast) plays an important role during this reaction (Shoemaker and Green, 2011). How does Dom34–Hbs1 complex recognize a stalled ribosome? How does it compete effectively with the ternary complex of EF1–aa-tRNA–GTP that normally binds the A site during elongation? Important clues into this process emerged from recent biochemical characterization showing an inverse correlation between the efficiency of subunit dissociation activity and the length of the mRNA downstream of the P site (Pisareva et al., 2011; Shoemaker and Green, 2011). In particular, Dom34–Hbs1 appears to prefer ribosomal complexes with little sequence downstream of the P site. Later structural studies showed that this length dependency is bestowed by Hbs1. The factor binds the ribosome in the mRNA entry tunnel and hence can only bind when the mRNA is short (Becker et al., 2011). NSD As mentioned earlier, NSD is similar to NGD except that the former involves ribosomes stalling at the end of the mRNA. As a result, the processes proceed in a similar fashion utilizing the same factors. A notable exception is the requirement of Ski7 for NSD in some species of yeast (van Hoof, 2005). The factor is a translational GTPase related to eRF3 and Hbs1. However, in contrast to eRF3 and Hbs1, which have binding partners (eRF1 and Dom34, respectively), no binding partner for Ski7 has been identified. As a result, the mechanism of the recognition of NSD targets in yeast remains poorly understood. It is worth noting that Ski7 interacts with the exosome, thereby linking ribosome recognition of NSD targets to their ultimate degradation (van Hoof et al., 2002). View chapterExplore book Read full chapter URL: Reference work2016, Encyclopedia of Cell BiologyS. Djuranovic, H.S. Zaher Chapter Advances in Immunology 2023, Advances in ImmunologyXingxian Liu, ... Xiaoyu Hu 4 Translation regulation in macrophages Despite the intricate regulation of transcript isoform alterations and stability, the translation process emerges as the principal rate-limiting step in determining the final protein-coding gene expression profile. Remarkably, translation machinery accounts for approximately half of the total energy consumption in mammalian cells (Lindqvist, Tandoc, Topisirovic, & Furic, 2018). Such a substantial energy expenditure necessitates meticulous regulation, with cells requiring precise control to ensure a balanced translation of specific subsets of mRNA. This regulatory precision is essential for various biological processes (Bhat et al., 2015; Carpenter et al., 2014). In the context of immunity, inflammatory genes undergo translational regulation to coordinate a timely and accurate immune response. A notable instance of this occurs in mouse macrophages, where global translation upregulation was observed. In response to infection with Leishmania donovani, 27% of detectable transcripts exhibited altered translational efficiency (Chaparro et al., 2020). Moreover, the intricate regulation of specific transcripts has been reported under diverse inflammatory conditions. 4.1 Translation initiation and regulation The translation process encompasses three distinct stages: initiation, elongation, and termination. The intricate molecular composition of translation initiation machinery and its regulatory mechanisms have been elucidated biochemically. This process commences with the assembly of a functional 40S ribosomal subunit in conjunction with associated factors, forming the 43S pre-initiation complex (43S PIC), which initiates subsequent translation activation (Jackson, Hellen, & Pestova, 2010). At the heart of this initiation process lies the formation of the ternary complex (TC), a pivotal step characterized by the assembly of a trimeric complex involving eIF2 (comprising α, β, and γ subunits), initiator methionyl tRNA (tRNAi-Met), and GTP (Jackson et al., 2010). The phosphorylation of the α subunit of eIF2 by four eIF2α kinases—heme-regulated eIF2α kinase (HRI), protein kinase R (PKR), protein kinase R-like endoplasmic reticulum kinase (PERK), and general control nonderepressible 2 (GCN2)—obstructs the formation of the TC, marking a critical point of regulation (Costa-Mattioli & Walter, 2020). Despite their distinct abilities to sense various cellular stresses and participate in diverse functional pathways in response to stress, these kinases converge on the phosphorylation of eIF2α, constituting what is known as the integrated stress response (ISR) (Costa-Mattioli & Walter, 2020). The ISR equips cells with the capacity to reconfigure translation, enabling the synthesis of essential proteins and stress response factors. Notably, the activity of eIF2α kinases has been associated with the activation of several pattern recognition receptors (PRRs), either directly or indirectly (Carpenter et al., 2014). In instances of viral infections or intracellular pathogen exposure, such as Yersinia, L. monocytogenes, and C. trachomatis, macrophages exhibit increased eIF2α phosphorylation (Shrestha et al., 2012). Numerous reports underscore the role of ISR in regulating the innate immune response, utilizing genetic approaches to manipulate the presence of the four eIF2α kinases and their resulting pro- and anti-inflammatory phenotypes (Carpenter et al., 2014). Nevertheless, scant attention has been directed toward investigating translation reprogramming in myeloid cells lacking eIF2α kinases. Identifying the specific subset of ISR-regulated mRNAs that undergo differential translation and contribute to the inflammatory outcome remains an intriguing area of exploration. The recruitment of the 43S PIC to the mRNA template is facilitated by the eIF4F complex (Jackson et al., 2010), which includes the mRNA 5’-cap-binding subunit (eIF4E), a large scaffolding protein (eIF4G), and the DEAD box RNA helicase (eIF4A), ultimately leading to the assembly of the 48S PIC. A pivotal player in this process is eIF4E. While eIF4E is indispensable for cap-dependent translation, it exhibits greater sensitivity to mRNA with long and structured 5′ UTRs. The activity of eIF4E is primarily regulated through phosphorylation and its availability (Bhat et al., 2015). Phosphorylation of eIF4E at serine 209 is orchestrated by the Mitogen-Activated Protein Kinase (MAPK)-dependent Mitogen-Interacting Kinase (MNK). Activation of MNK is triggered by MAPKs, including p38 and ERK. In the context of innate immunity, viral infections contribute to the phosphorylation of eIF4E (Kleijn, Vrins, Voorma, & Thomas, 1996). Studies following this observation revealed that the S209A mutation of eIF4E in mice led to reduced susceptibility to VSV infection and increased type I interferon production. The diminished activity of S209A eIF4E was attributed to decreased translation of IκBα, the negative regulator of NFκB signaling (Herdy et al., 2012). Further investigations unveiled that MNK2–eIF4E signaling also negatively regulates TLR signaling by impacting IκBα translation (Bao et al., 2017). In the context of TLR ligand stimulation, the MNK–eIF4E axis becomes activated through IRAK2, which relies on Notch–RBP-J signaling. This activation process enhances the translation of IRF8, a crucial M1 macrophage transcription factor, thus promoting the production of pro-inflammatory mediators such as IL-12 and iNOS. Notably, this translation modulation occurs without affecting Irf8 mRNA levels. Additionally, the attenuation of Hes1 translation—a Notch signaling-dependent repressor of the inflammatory response—was observed upon MNK1/2 ablation. Conversely, IFN-γ inhibits the MNK–eIF4E axis, resulting in the suppression of translation events (Su et al., 2015). Recent findings demonstrated that IL-4 stimulation triggers stronger eIF4E phosphorylation compared to IFN-γ in macrophages. This distinctive phosphorylation pattern was associated with the selective translation of anti-inflammatory genes, mediated by MNK2 (Bartish et al., 2020) (Fig. 1). However, the deletion of MNK1 or MNK2, along with point mutations at the S209 phospho-site, led to reduced translation of specific subsets of mRNA without significantly affecting the global translation rate, as demonstrated by various studies (Bhat et al., 2015; Carpenter et al., 2014). eIF4E is a proto-oncogene that is highly expressed in cancer cells and contributes to tumor growth and malignancy (De Benedetti & Graff, 2004). Therefore, the availability of eIF4E is tightly controlled to regulate translation. The mammalian target of rapamycin complex 1 (mTORC1) acts as a central hub for regulating eIF4E availability by phosphorylating the eIF4E-binding protein (4E-BP) (Ma & Blenis, 2009). Pathogen infections or activation of TLR signaling lead to mTORC1 activation, subsequently phosphorylating two major mTOR substrates: ribosomal protein S6 kinase (S6K) and 4E-BP. Upon mTOR activation, 4E-BP undergoes hyper-phosphorylation, releasing eIF4E to initiate cap-dependent translation initiation. Genetic deletion of 4E-BPs in mice has revealed alterations in the translatome and inflammatory outcomes, underscoring the pivotal role of the mTOR–4E-BP pathway in immune regulation. Mice lacking both 4E-BP1 and 4E-BP2 were found to be resistant to vesicular stomatitis virus (VSV) infection. In this context, mouse embryonic fibroblasts (MEFs) and plasmacytoid dendritic cells (pDCs) from double knockout mice exhibited increased type I interferon production compared to wildtype controls. Polysome profiling of MEFs from these double knockout mice demonstrated an enhanced association of Irf7 mRNA with the polysome fraction, suggesting that 4E-BPs act as negative regulators, modulating the translation of inflammatory genes (Colina et al., 2008). The protozoan parasite Leishmania employs its protease GP63 to cleave mTOR, evading macrophage clearance. However, the knockout of 4E-BP1/2 rescued the anti-parasite immune response by enhancing the type I interferon response (Jaramillo et al., 2011). Recent investigations have highlighted the significance of the 4E-BP–eIF4E axis in regulating myeloid cell translation to bolster innate immunity. For instance, the translation of pro-inflammatory chemokines CCL5 and CXCL10 is negatively modulated by 4E-BP1/2 in LPS-stimulated macrophages in an mTOR-dependent manner (William et al., 2019). Furthermore, 4E-BP-deficient adipose tissue macrophages displayed heightened inflammatory gene expression upon exposure to a high-fat diet, attributed to enhanced Irf8 mRNA translation (Pearl et al., 2020). However, recent findings have introduced a paradox. While previous literature suggested an anti-inflammatory role of 4E-BPs, a recent study indicated that the translation of anti-inflammatory mediator IL-10 is also regulated by 4E-BP. Remarkably, 4E-BP1/2 double knockout macrophages exhibited reduced bactericidal activity, in contrast to the expected anti-inflammatory function of 4E-BPs (William et al., 2018) (Fig. 1). In summary, the mTOR–4E-BP–eIF4E pathway indeed plays a crucial role in regulating the translation of inflammatory genes in the context of innate immunity. In addition to the general regulation achieved through the activation of the translation initiation machinery, a multitude of genes undergo specific translation modulation. This layer of regulation hinges on the presence of distinct transcript motifs and their recognition by interacting RNA-binding proteins (RBPs), predominantly during the translation initiation process. A well-characterized instance of gene-specific regulation in innate immunity is the IFN-γ-activated inhibitor of translation (GAIT) complex (Mukhopadhyay, Jia, Arif, Ray, & Fox, 2009). The GAIT complex is a heterotetrameric assembly composed of glutamylprolyl tRNA synthetase (EPRS), heterogeneous nuclear ribonucleoprotein Q (hnRNPQ or NSAP1), glyceraldehyde-3-phosphate dehydrogenase (GAPDH), and ribosomal protein L13a (RPL13A). The formation of the GAIT complex unfolds in two phases. Within the initial 8 h following IFN-γ exposure, EPRS and HNRPQ initially assemble to generate the non-functional ’pre-GAIT complex’ (Sampath et al., 2004). The second phase takes place after 12–24 h of IFN-γ exposure, during which RPL13A is phosphorylated and binds to GAPDH and the ‘pre-GAIT complex,’ culminating in the formation of a functional GAIT complex (Kapasi et al., 2007; Mazumder et al., 2003). The GAIT complex binds to the stem-loop structure of specific transcripts, exerting translation repression through a direct interaction between RPL13A from the GAIT complex and the translation initiation factor eIF4G. This interaction impedes the association between eIF4G and eIF3, thereby attenuating translation initiation (Kapasi et al., 2007). The genetic removal of the GAIT complex component RPL13A results in an elevated inflammatory response (Poddar et al., 2013; Poddar, Kaur, Baldwin, & Mazumder, 2016). Polysome profiling has identified distinct chemokines and their receptor genes as being regulated by the GAIT complex in monocytes (Vyas et al., 2009). Notably, VEGFA stands as a well-validated target gene for the GAIT complex, with a finely tuned regulation mechanism under inflammatory and stress conditions (Ray et al., 2009; Yao et al., 2013) (Fig. 1). In response to IFN-γ stimulation of myeloid cells, the GAIT complex assembles on the 3 UTR of VEGFA mRNA, thereby repressing translation. Under hypoxic conditions, heterogeneous nuclear ribonucleoprotein L (HNRNP L), an RNA-binding protein, undergoes phosphorylation and forms a complex with DRBP76 and HNRNPA2/B1. This newly formed complex binds to the VEGFA transcript at a site adjacent to the GAIT complex, counteracting the inhibitory effects of the GAIT complex. This dynamic interplay promotes VEGFA translation and angiogenesis (Ray & Fox, 2007; Yao et al., 2013). Another example is the RNA-binding protein 2’-5’-oligoadenylate synthetase L1 (OASL1), which regulates IRF7 translation. OASL1 induction follows TLR ligand or viral stimulation and curtails IRF7 translation initiation by binding to IRF7’s 5’ UTR, impeding PIC scanning (Lee, Kim, Oh, & Kim, 2013). Deletion of OASL1 enhances type I interferon production and imparts virus resistance in mice (Lee et al., 2013). ARE-binding proteins also emerge as regulators of translation. For instance, TTP associates with the ARE in TNF mRNA to repress TNF translation. When phosphorylated under LPS stimulation, TTP is substituted by HuR, promoting efficient translation (Tiedje et al., 2012). Other ARE-binding proteins have been identified to regulate translation rather than their conventional role in modulating mRNA stability (Damgaard & Lykke-Andersen, 2011; Guillemin, Kumar, Wencker, & Ricci, 2022; Piecyk et al., 2000; Qi et al., 2012). 4.2 Translation elongation and its regulation Despite the extensive research into translation initiation, the translation elongation process and its regulation remain relatively understudied. Polypeptide synthesis heavily relies on ribosome elongation, a process that demands a considerable amount of energy. To ensure a balanced expenditure of energy with protein output, tight regulation of translation elongation is necessary. Elongation initiates promptly after translation initiation has occurred, with an 80S ribosome positioned at an AUG start codon and a methionyl-tRNAiMet residing in the P site (Dever, Dinman, & Green, 2018; Schuller & Green, 2018). Eukaryotic elongation factor 1A (eEF1A) in its GTP-bound state facilitates the transport of aminoacyl-tRNA to the A site of the ribosome. The pairing of the tRNA anticodon with the codon induces hydrolysis of eEF1A–GTP, releasing eEF1A–GDP from the A site. Following this, the ribosome triggers the formation of a peptide bond, subsequently followed by ribosome translocation. The latter is mediated by the hydrolysis of eEF2–GTP. This translocation event facilitates the entry of the next aminoacyl-tRNA into the A site. These elongation steps are iteratively carried out until an in-frame stop codon enters the ribosome’s A site, marking the translation termination point (Fabbri, Chakraborty, Robert, & Vagner, 2021). Current insights into the regulation of elongation predominantly focus on eEF2 activity. Eukaryotic elongation factor 2 kinase (eEF2K) is a protein kinase responsible for phosphorylating and inhibiting its sole known substrate, eEF2 (Fabbri et al., 2021; Liu & Proud, 2016). A major regulator upstream of eEF2K–eEF2 is mTORC1. Upon stimulation by growth factors and nutrients, mTORC1 activates the well-known substrate S6K. Subsequently, S6K phosphorylates eEF2K at Ser366, inactivating it and thereby activating eEF2 for the promotion of elongation (Wang et al., 2001). The MAPK signaling pathway also contributes to eEF2K phosphorylation at Ser366, largely through ERK-activated p90 ribosomal S6 kinase (p90 RSK) (Anjum & Blenis, 2008; Wang et al., 2001). The N-terminus kinase domain of RSKs shares considerable homology with S6K, implying potential shared substrates, including eEF2K phosphorylation at S366 (Roux & Topisirovic, 2018). Furthermore, eEF2K can be activated by the phosphorylation of AMP-activated protein kinase (AMPK) at a distinct site. This activation occurs as a response to energy scarcity, leading to eEF2 inhibition and subsequently suppressing translation elongation (Browne, Finn, & Proud, 2004). The translation regulation orchestrated by the eEF2K–eEF2 axis in immunity has been the subject of limited investigation, with only a handful of studies shedding light on its role. In the context of LPS-stimulated macrophages, MAP3K8, or MAPK kinase kinase 8, emerged as a pivotal player in mTORC1 activation, subsequently impacting mRNA translation, including pivotal genes such as Tnf, Il6, and Ccl2. Notably, the absence of MAP3K8 led to a reduction in inhibitory eEF2K phosphorylation, suggesting that MAP3K8-mediated MAPK signaling might influence myeloid cell translation by modulating eEF2K activity (Lopez-Pelaez et al., 2012). Subsequent investigations further unveiled the role of MAPK p38γ and p38δ in mediating elongation activation within the TLR signaling pathway (Gonzalez-Teran et al., 2013). Genetic deletion of p38γ and p38δ in myeloid cells of mice yielded a reduction in lethality and tissue damage in a liver sepsis model induced by LPS. Notably, double knockout (DKO) macrophages exhibited decreased inhibitory phosphorylation of eEF2K, consequently leading to increased phosphorylation of eEF2. Strikingly, the absence of p38γ and p38δ resulted in reduced TNF-α production, while Tnf mRNA levels remained unaltered. Intriguingly, biochemical experiments demonstrated diminished association between Tnf mRNA and eEF2 in DKO macrophages. Additionally, the knockdown of eEF2 led to decreased TNF-α production, although the same effect was not observed for IL-6 (Gonzalez-Teran et al., 2013). This study distinctly underscored the critical role of the eEF2K–eEF2 axis in orchestrating the elongation process, specifically influencing TNF-α expression at the translation level, thereby expanding beyond the realms of classical transcriptional regulation and mRNA stability (Fig. 2). Recently, our research group has uncovered a novel facet of cytokine release syndrome (CRS) by revealing the role of translation elongation in the regulation of IL-6. This finding emerged from the exploration of the distinctive action of the β1 adrenergic receptor blocker metoprolol, which demonstrated a substantial reduction in serum IL-6 level and a consequential amelioration of CRS in patients undergoing chimeric antigen receptor T cell (CAR-T) therapy (Yang et al., 2023). Strikingly, metoprolol consistently exhibited the capacity to diminish IL-6 protein abundance in primary human monocytes without altering IL6 mRNA level, suggesting the translation-mediated IL-6 regulation. Deeper mechanistic insight was acquired through ribosome profiling of human monocytes, revealing ribosome stalling on the 5’ UTR of the IL6 transcript. This stalling phenomenon was coupled with a decrease in IL-6 translation efficiency following metoprolol treatment, indicating the potential suppression of translation elongation by metoprolol. Further mechanistic exploration demonstrated a reduction in eEF2K phosphorylation and a concomitant increase in eEF2 phosphorylation in response to metoprolol treatment. Notably, inhibition of eEF2K activation led to heightened IL-6 production at the translation level. Complementing these findings, RNA sequencing data unveiled the downregulation of genes associated with Rap1 and cAMP signaling upon metoprolol treatment. Intriguingly, our investigation identified RAPGEF3 (also known as EPAC1) as an indispensable component of Rap1 signaling downstream of G-protein-coupled receptor (GPCR) activation. Knockdown or chemical inhibition of EPAC1 recapitulated the effects of metoprolol, effectively attenuating IL-6 translation via the activation of eEF2K (Yang et al., 2023) (Fig. 2). Our study offers a novel perspective, not only shedding light on the pathogenesis of CRS, but also revealing a unique connection between catecholamine-mediated GPCR signaling and the translation elongation of the pro-inflammatory cytokine IL-6. View chapterExplore book Read full chapter URL: Book series2023, Advances in ImmunologyXingxian Liu, ... Xiaoyu Hu Chapter RNA Turnover in Eukaryotes: Analysis of Specialized and Quality Control RNA Decay Pathways 2008, Methods in EnzymologyDaiki Matsuda, ... Lynne E. Maquat 3.6 Use of translational inhibitors to study NMD Nonsense-mediated mRNA decay provides a posttranscriptional process for regulating gene expression that requires ongoing translation. The finding that cycloheximide added to cultured cells at 100 μg/ml for 2 h results in an increase in the abundance of a PTC-containing mRNA relative to the corresponding PTC-free mRNA provides one indication that PTC-containing mRNA is targeted for NMD (Carter et al., 1995; Ishigaki et al., 2001). Additionally, the NMD of PTC-containing mRNA will be abrogated upon the transient expression of (i) poly(A)-interacting protein (Paip)2, which inhibits the interaction of PABPC1 and the poly(A) tail of mRNA, (ii) eIF2α S51D, which is a constitutively inactive phosphomimetic variant of eIF2α, or (iii) human immunodeficiency virus (HIV)-1 or -2 protease or poliovirus 2A protease, each of which cleaves eIF4G (Chiu et al., 2004; Lejeune et al., 2004). Notably, when using one of the proteases to inhibit translation, the time course of eIF4G cleavage should be compared to the time course of any increase in PTC-containing mRNA abundance, as the proteases can cleave cellular proteins in addition to eIF4G, although at different (generally slower) rates than they cleave eIF4G. For this reason, viral proteases should not be the sole test for dependence on ongoing translation. To test the effect of a translational inhibitor on the level of a putative NMD target, HeLa or Cos7 cells (1 × 106/100-mm dish) are transiently transfected with (i) an NMD test plasmid or, for comparative analysis, its nonsense-free counterpart, (ii) a reference plasmid that controls for variations in transfection efficiency and sample recovery, (iii) a reporter plasmid, such as pGL2 (Promega; Chiu et al., 2004) that produces firefly luciferase and can be used to measure the extent of translational inhibition (notably, our often-used reference plasmid, phCMV-MUP, can also be used for this purpose, as MUP is secreted from cells within 30 min of its synthesis), and (iv) an effector plasmid or transcript that produces either one of the aforementioned translational inhibitors or, as a negative control, no translational inhibitor. Effector plasmids include (i) pMT2-HA-eIF2α S51D or, as a negative control, HA-eIF2α WT (Srivastava et al., 1998; Chiu et al., 2004), (ii) a plasmid encoding poliovirus 2A protease (a gift from Richard Lloyd) or, as a negative control, pCI empty vector (Lejeune et al., 2004), and (iii) pcDNA3-HA-Paip2 or, as a negative control, pcDNA3 empty vector (Chiu et al., 2004; Khaleghpour et al., 2001). As an alternative to cotransfecting plasmid encoding a translational inhibitor, cells can be retransfected 18 h after the first transfection with in vitro-synthesized mRNA (1 μg) (Lejeune et al., 2004) generated from plasmids containing HIV-1 or HIV-2 protease cDNA using Lipofectamine 2000 (gifts from Théophile Ohlmann). As a control, in vitro-synthesized mRNA encoding a protein that does not inhibit translation can be assayed in parallel (Lejeune et al., 2004). Cells are harvested after an additional 16 to 20 h and assayed as described earlier. In transfections that involve a luciferase-producing reporter construct, translation inhibition is determined using a luminometer to measure luciferase activity. Importantly, activity is normalized to the level of luciferase mRNA, as quantitated using RT-PCR, to control for variations in transfection efficiency. In transfections that utilize a MUP-producing reporter construct, the level of MUP is quantitated by Western blotting. In parallel, the level of PTC-containing mRNA is compared to that of PTC-free mRNA after normalization to the level of reference mRNA to examine the effect of the translational inhibitor. In our hands, the extent to which Gl NMD or GPx1 NMD decreased ranged from two- to ninefold, depending on the inhibitor (Chiu et al., 2004; Lejeune et al., 2004). A test for dependence on translation can also be accomplished by transiently expressing a suppressor tRNA. For example, an NMD target that harbors a nonsense mutation that generates a UAG codon and that does not harbor a UAG normal termination codon will increase in abundance or half-life upon the transient expression of an amber suppressor tRNA that directs the incorporation of serine at UAG codons. In this test, mouse L-M(TK−) cells (≈4 × 106/100-mm dish) are transfected with (i) an NMD test plasmid or, for comparative analysis, its nonsense-free counterpart, (ii) a reference plasmid, and (iii) p53ts Su− or p53ts Su+ (up to 7.5 μg), which consist of a polyoma vector that replicates in L cells so as to overexpress wild-type serine tRNA or a UAG-suppressing derivative (Belgrader et al., 1993), respectively. The amount of each plasmid for transfection should be determined empirically to achieve the most efficient nonsense codon suppression, which can be determined by, for example, Western blot analysis using antibody against the protein encoded by the test mRNA, if it is available. In parallel, the level of the prospective NMD target mRNA should be compared to the level of its nonsense-free counterpart after normalization to the reference mRNA level. This analysis will enable a test for a suppressor tRNA-mediated partial inhibition of NMD. In our hands, an approximate twofold inhibition of NMD was observed for TPI 23Ter mRNA (Belgrader et al., 1993). In contrast to the aforementioned inhibitors, which affect the translation of both CBP80/20-bound mRNA and its remodeled eIF4E-bound mRNA product, transient expression of 4E-BP1 inhibits only eIF4E-bound mRNA and not CBP80/20-bound mRNA. Therefore, the abundance or half-life or an NMD target should not be affected by 4E-BP1 expression. View chapterExplore book Read full chapter URL: Book series2008, Methods in EnzymologyDaiki Matsuda, ... Lynne E. Maquat Chapter Advances in Immunology 2023, Advances in ImmunologyXingxian Liu, ... Xiaoyu Hu 4.2 Translation elongation and its regulation Despite the extensive research into translation initiation, the translation elongation process and its regulation remain relatively understudied. Polypeptide synthesis heavily relies on ribosome elongation, a process that demands a considerable amount of energy. To ensure a balanced expenditure of energy with protein output, tight regulation of translation elongation is necessary. Elongation initiates promptly after translation initiation has occurred, with an 80S ribosome positioned at an AUG start codon and a methionyl-tRNAiMet residing in the P site (Dever, Dinman, & Green, 2018; Schuller & Green, 2018). Eukaryotic elongation factor 1A (eEF1A) in its GTP-bound state facilitates the transport of aminoacyl-tRNA to the A site of the ribosome. The pairing of the tRNA anticodon with the codon induces hydrolysis of eEF1A–GTP, releasing eEF1A–GDP from the A site. Following this, the ribosome triggers the formation of a peptide bond, subsequently followed by ribosome translocation. The latter is mediated by the hydrolysis of eEF2–GTP. This translocation event facilitates the entry of the next aminoacyl-tRNA into the A site. These elongation steps are iteratively carried out until an in-frame stop codon enters the ribosome’s A site, marking the translation termination point (Fabbri, Chakraborty, Robert, & Vagner, 2021). Current insights into the regulation of elongation predominantly focus on eEF2 activity. Eukaryotic elongation factor 2 kinase (eEF2K) is a protein kinase responsible for phosphorylating and inhibiting its sole known substrate, eEF2 (Fabbri et al., 2021; Liu & Proud, 2016). A major regulator upstream of eEF2K–eEF2 is mTORC1. Upon stimulation by growth factors and nutrients, mTORC1 activates the well-known substrate S6K. Subsequently, S6K phosphorylates eEF2K at Ser366, inactivating it and thereby activating eEF2 for the promotion of elongation (Wang et al., 2001). The MAPK signaling pathway also contributes to eEF2K phosphorylation at Ser366, largely through ERK-activated p90 ribosomal S6 kinase (p90 RSK) (Anjum & Blenis, 2008; Wang et al., 2001). The N-terminus kinase domain of RSKs shares considerable homology with S6K, implying potential shared substrates, including eEF2K phosphorylation at S366 (Roux & Topisirovic, 2018). Furthermore, eEF2K can be activated by the phosphorylation of AMP-activated protein kinase (AMPK) at a distinct site. This activation occurs as a response to energy scarcity, leading to eEF2 inhibition and subsequently suppressing translation elongation (Browne, Finn, & Proud, 2004). The translation regulation orchestrated by the eEF2K–eEF2 axis in immunity has been the subject of limited investigation, with only a handful of studies shedding light on its role. In the context of LPS-stimulated macrophages, MAP3K8, or MAPK kinase kinase 8, emerged as a pivotal player in mTORC1 activation, subsequently impacting mRNA translation, including pivotal genes such as Tnf, Il6, and Ccl2. Notably, the absence of MAP3K8 led to a reduction in inhibitory eEF2K phosphorylation, suggesting that MAP3K8-mediated MAPK signaling might influence myeloid cell translation by modulating eEF2K activity (Lopez-Pelaez et al., 2012). Subsequent investigations further unveiled the role of MAPK p38γ and p38δ in mediating elongation activation within the TLR signaling pathway (Gonzalez-Teran et al., 2013). Genetic deletion of p38γ and p38δ in myeloid cells of mice yielded a reduction in lethality and tissue damage in a liver sepsis model induced by LPS. Notably, double knockout (DKO) macrophages exhibited decreased inhibitory phosphorylation of eEF2K, consequently leading to increased phosphorylation of eEF2. Strikingly, the absence of p38γ and p38δ resulted in reduced TNF-α production, while Tnf mRNA levels remained unaltered. Intriguingly, biochemical experiments demonstrated diminished association between Tnf mRNA and eEF2 in DKO macrophages. Additionally, the knockdown of eEF2 led to decreased TNF-α production, although the same effect was not observed for IL-6 (Gonzalez-Teran et al., 2013). This study distinctly underscored the critical role of the eEF2K–eEF2 axis in orchestrating the elongation process, specifically influencing TNF-α expression at the translation level, thereby expanding beyond the realms of classical transcriptional regulation and mRNA stability (Fig. 2). Recently, our research group has uncovered a novel facet of cytokine release syndrome (CRS) by revealing the role of translation elongation in the regulation of IL-6. This finding emerged from the exploration of the distinctive action of the β1 adrenergic receptor blocker metoprolol, which demonstrated a substantial reduction in serum IL-6 level and a consequential amelioration of CRS in patients undergoing chimeric antigen receptor T cell (CAR-T) therapy (Yang et al., 2023). Strikingly, metoprolol consistently exhibited the capacity to diminish IL-6 protein abundance in primary human monocytes without altering IL6 mRNA level, suggesting the translation-mediated IL-6 regulation. Deeper mechanistic insight was acquired through ribosome profiling of human monocytes, revealing ribosome stalling on the 5’ UTR of the IL6 transcript. This stalling phenomenon was coupled with a decrease in IL-6 translation efficiency following metoprolol treatment, indicating the potential suppression of translation elongation by metoprolol. Further mechanistic exploration demonstrated a reduction in eEF2K phosphorylation and a concomitant increase in eEF2 phosphorylation in response to metoprolol treatment. Notably, inhibition of eEF2K activation led to heightened IL-6 production at the translation level. Complementing these findings, RNA sequencing data unveiled the downregulation of genes associated with Rap1 and cAMP signaling upon metoprolol treatment. Intriguingly, our investigation identified RAPGEF3 (also known as EPAC1) as an indispensable component of Rap1 signaling downstream of G-protein-coupled receptor (GPCR) activation. Knockdown or chemical inhibition of EPAC1 recapitulated the effects of metoprolol, effectively attenuating IL-6 translation via the activation of eEF2K (Yang et al., 2023) (Fig. 2). Our study offers a novel perspective, not only shedding light on the pathogenesis of CRS, but also revealing a unique connection between catecholamine-mediated GPCR signaling and the translation elongation of the pro-inflammatory cytokine IL-6. View chapterExplore book Read full chapter URL: Book series2023, Advances in ImmunologyXingxian Liu, ... Xiaoyu Hu Chapter The Cell 2015, Systems Biology in Toxicology and Environmental HealthPaul D. Ray, Rebecca C. Fry Decoding the Genome (II): Translation The process of deciphering the DNA code into proteins, which will carry out the myriad functions of the cell, is a multistep process. Transcription “reads” the DNA, producing an RNA “blueprint.” This RNA blueprint is then used to assemble a polypeptide chain. This chain of amino acids will then undergo folding and further processing, producing an active protein, whether catalytically active enzyme, or a structural component like actin. Compared with transcription, the process of translation is peculiar in that the RNA is processed by a molecular machine called the ribosome, which is composed of specialized ribosomal RNAs. Other nonprotein coding RNAs, aptly termed transfer RNA, are used to transport the individual amino acids to the ribosome where they are attached to the growing polypeptide chain. RNA If an apt metaphor for DNA is a set of “master plans” which describe how cellular functions are to be executed through the generation of proteins, then RNA is the actual “blueprint” which directs the detailed assembly of proteins. However, RNA is not simply a schematic for protein assembly, but serves as effectors of cellular function as well. 1. : Messenger RNA (mRNA). Contains the amino acid sequence for the assembly of proteins. 2. : Noncoding RNA (ncRNA). RNA that does not code for protein assembly. a. : Ribosomal RNA (rRNA). The structural and catalytic RNA component of ribosomes which direct protein synthesis. b. : Transfer RNA (tRNA). Structural RNA that transports amino acids to the ribosome for incorporation into the peptide chain. c. : Regulatory RNAs. RNA transcribed from DNA that is not utilized for protein synthesis but instead regulates gene expression, and protein translation. – : Micro RNA (miRNA) – : Small interfering RNA (siRNA) – : Enhancer RNA Processing of mRNA In eukaryotes, after the pre-mRNA has been transcribed, extensive modifications take place. The 5′ end undergoes capping, the 3′ end undergoes polyadenylation, and the coding region undergoes splicing. Furthermore, tRNA and rRNA undergo posttranscriptional modifications; however, for the scope of this chapter only mRNA processing will be described. Capping The 5′ end is capped with a guanylate residue methylated at the N-7 position. This residue is attached to the adjacent residue by a 5′ to 5′ triphosphate linkage, and this adjacent residue may be methylated as well. Capping prevents exonuclease degradation. Polyadenylation Polyadenylation of the 3′ end occurs before the mRNA leaves the nucleus. This polyadenylate tail, around 100–200 nucleotides long, protects the mRNA from the degradatory action of phosphatases and nucleases. While prokaryote genes are continuous, that is, each base pair in the DNA sequence is conserved in the mRNA, eukaryotic mRNA is smaller than the template DNA sequence due to the splicing out of introns and the fusing together of exons. The 5′ capping and 3′ polyadenylation are not translated. Export of mRNA from the nucleus into the cytosol relies on polyadenylation of the 3′ end of the strand. Splicing Splicing occurs in the nucleus. During transcription, the RNA associates with various proteins, forming ribonucleoprotein particles. Splicing cleaves the 5′ and 3′ ends of the intron sequences at splice sites, which are GU (5′) and AG (3′). Another site involved in intron cleavage is known as the branch point, usually located 20–40 bp upstream from the 3′ end. The branch sequence is variable, PyNPyPuAPy, except for the A. The process of intron removal first involves the looping of the 5′ G looping downstream into contact with the invariable A of the branch site. The 2′ hydroxyl residue of the A performs a nucleophilic attack on the phosphodiester backbone of the 5′ G site. This causes the release of the upstream exon from the 5′ end of the intron. The resultant AG site of the released 3′ end of the exon then nucleophilically attacks the phosphodiester backbone of the 3′ G site of the intron, thus fusing the two exons and releasing the looped intron. This process is mediated by small nuclear ribonucleoproteins (snRNPs), composed of 100–200 nucleotide RNAs and ≥10 proteins, that bind to the pre-mRNA through complementary sequences at the splice and branch sites. Splicing is carried out by 50–60 S particle called the spliceosome. Gene expression is regulated at several levels, transcriptionally, translationally, and even at the splicing stage. Alternative splicing of some genes gives rise to several different mRNA isoforms, which are translated into proteins of differing size and function. Protein Synthesis After processing, the mRNA is exported out of the nucleus to ribosomes, the primary site of protein synthesis where it is “read” and translated into protein. The mRNA is bound by the ribosome, and amino acids attached to tRNAs bind to the ribosome and mRNA, based on complementary pairing of the tRNA codons with the mRNA codons. The amino acids are linked together by peptide bonds, thus forming a growing polypeptide chain. Amino Acids and Peptide Bonding Amino acids are composed of a central α-carbon, which is bonded to a hydrogen atom, an amino group (+NH3), a carboxyl group (COO−), and a distinct side chain (R) that confers specificity to each amino acid. The side chain groups are biologically important, and are classified by whether they are polar or nonpolar, and whether they contain basic or acidic residues. Amino acids are linked by covalent bonding between the α-carbon of an amino acid and the amino group of another. Codons The information necessary for protein assembly are encoded in mRNA through codons, a three base sequence. Specifically, a codon is a triplet nucleotide sequence. There are 64 possible codons, 61 of them representing the 20 amino acids, and the remaining three serving as stop codons, which are translation termination sequences. While an amino acid may be represented by more than one codon, no single codon encodes more than one amino acid. Methionine and tryptophan are the only amino acids represented by a single codon. Some amino acids are represented by up to six codons. The first and second bases of the codon are usually conserved, with the third base variable, known as the “wobble” base. This variability prevents deleterious phenotypes caused by mutations of the third base of the codon. Codons base pair with the anticodon of tRNA. In addition to coding for methionine, AUG is the initiating codon. Transfer RNA (tRNA) Transfer RNAs (tRNAs) contain an anticodon sequence which pairs with complementary mRNA codons. The conserved, three-dimensional shape of tRNAs arises in part from the interaction between bases in different regions of the RNA sequence. These interactions give rise to the secondary “cloverleaf” structure of tRNA; this base pairing results in three stem-loops and an open-ended stem formed by the pairing of the 5′ and 3′ ends. The latter stem is known as the acceptor stem, which binds an amino acid. Of the three stem loops, the anticodon loop contains the three nucleotide base sequence which pairs with the mRNA codon during translation. There are more possible codons (64) than individual tRNAs. As mentioned previously, the third base of the codon (5′ to 3′) is the wobble base which allows for an amino acid to be coded by several different codons. The first base of the anticodon (5′ to 3′) is a wobble base; if the base is G, U, or I (inosine), there exists variations in hydrogen bonding which allows the anticodon to base pair with more than one codon. While several amino acid types may bind to a tRNA, only one amino acid at a time may bind. Amino acids are attached to tRNA at the acceptor stem by the enzyme aminoacyl–tRNA synthetase in two steps. This enzyme catalyzes the hydrolysis of ATP, covalently bonding an amino acid to AMP, forming aminoacyl–AMP which remains bound to the enzyme. This aminoacyl residue is subsequently transferred and bound to the 3′ end of the acceptor stem through an ester linkage, forming aminoacyl–tRNA. Specificity is achieved by there being a synthetase for each type of amino acid which recognizes the tRNAs specific for that amino acid. However, a synthetase may recognize a structurally similar amino acid; this possible error is corrected at the second step. If the incorrect amino acid is attached to a tRNA, the synthetase detects this error through an editing site and hydrolyzes the incorrect amino acid. Another level of specificity is achieved by the synthetase being able to recognize specific tRNAs. Ribosomal RNA (rRNA) and Protein Synthesis The ribosome is the site of protein synthesis. The ribosome is composed primarily of two ribosomal RNA (rRNA) subunits, also known as ribozymes, catalytically active RNA. The smaller subunit reads the mRNA transcript while the larger subunit facilitates the synthesis of the amino acid chain. There are three binding sites in the ribosome; the aminoacyl (A) site, the peptidyl (P) site, and the exit (E) site. These sites are oriented in the context of the mRNA strand 5′ to 3′, E–P–A, due to the movement of the ribosome toward the 3′ end of mRNA. The ribosome reads the mRNA by codon, that is, the ribosome moves along the mRNA three bases at a time. Initiation of translation in eukaryotes is mediated by eukaryotic initiation factors (eIF). The first step is the assembly of a 43 S preinitiation complex. Methionine (Met) is the first amino acid of the polypeptide chain, due to the initiation codon being AUG, methionine is transferred by tRNAi, an initiator tRNA. Met-tRNAi, in a complex with eIF2 and GTP, is transferred to the 40 S ribosomal subunit, which is bound by eIF1A and eIF3. The second step in initiation is the recruitment of mRNA to the ribosome. The 5′ cap of the mRNA orients the ribosome to an AUG codon. Binding of eIF4E to the 5′ cap brings the cap into proximity of the poly A tail by binding to the poly A binding protein (Pab1p). This bridge between the poly A tail and the 5′ cap is also mediated by the association of several other eIFs with the eIF4E–Pab1p complex. The 40 S–eIF complex, upstream of the start codon, moves 5′ to 3′ along the mRNA strand until it encounters the correct start codon, which is identified by two determining factors: the presence of the Kozak sequence, a consensus sequence surrounding a start codon (ACCAUGG), and the lack of a hairpin loop downstream of an AUG codon. In step three, the 80 S initiation complex is formed by association of the 60 S ribosome with the 48 S preinitiation complex, which consists of the mRNA–eIF complex. The GTP recruited by Met-tRNAi in step one is hydrolyzed, releasing the initiation factors (Figure 13). Translation occurs in three steps similar to transcription. During the initiation phase of translation, the mRNA is the focus of a ribosome assembly, and the first tRNA is attached to the start codon. The second phase of elongation, the tRNA transfers an amino acid to the tRNA of the next codon. The ribosome then proceeds to the next codon of the mRNA, assembling the amino acid chain. The termination phase sees the ribosome releasing the polypeptide upon encountering the stop codon. Translation in eukaryotes occurs mainly on the rough ER. The aminoacyl tRNA synthetase is a multidomain enzyme which catalyzes the ATP-dependent bonding of an amino acid to the corresponding tRNA for transport to the ribosome, called an aminoacyl–tRNA. Once at the ribosome, the aminoacyl–tRNA is complementary base paired to the corresponding mRNA codon. The aminoacyl site binds the aminoacyl–tRNA with the complementary mRNA codon (Figure 13, 2). The peptidyl site stabilizes the aminoacyl–tRNA with the enlarging polypeptide chain. A peptide bond is made between the amino acid of the tRNA at the aminoacyl site and the amino acid at the peptidyl site (Figure 13, 3). The exit site holds the tRNA after transfer of the amino acid. The tRNA at the peptidyl site, now without an amino acid, is transferred to the exit site. The tRNA at the aminoacyl site is transferred to the peptidyl site (Figure 13, 4). The tRNA at the exit site is ejected (Figure 13, 5) and another aminoacyl–tRNA enters the aminoacyl site (Figure 13, 1) to repeat the process. Protein Structure and Folding The function of a protein is dependent upon its three-dimensional structure. The polypeptide chain synthesized by the ribosome undergoes folding to achieve its final conformation. Misfolded proteins may be functionally inert, or dysfunctional, and susceptible to degradation; several pathological states arise from protein misfolding . Protein folding at essence is the attainment of a thermodynamically favorable arrangement of the amino acid residues. The structure of a protein can be categorized into four different levels. The primary structure is the specific amino acid sequence of the polypeptide chain. The secondary structure is the spatial arrangement of the peptide backbone, stabilized by hydrogen bonding between amide and carbonyl groups of the peptide chain. The α-helix and β-pleated sheets are common secondary structures. Tertiary structure is the three-dimensional conformation, taking into account the amino acid side chains. The arrangement of peptide subunits to form a large multisubunit protein is known as the quaternary structure . Protein folding may occur as the peptide chain is being synthesized at the ribosome, or after transport to the cytoplasm or in the ER . View chapterExplore book Read full chapter URL: Book2015, Systems Biology in Toxicology and Environmental HealthPaul D. Ray, Rebecca C. Fry Chapter Disorders of Protein Synthesis 2022, Advances in Protein Chemistry and Structural BiologySilvia Lombardi, ... Alessio Branchini 3 Ribosome readthrough 3.1 Mechanism, programmed readthrough and natural vs premature stop codons The central role of protein synthesis has forced cells to develop accurate mechanisms to avoid translational errors. However, although effective, the process of translation termination, during which a competition occurs for the recognition of the stop codon (Fig. 4A), is not 100% efficient. At stop codons located at the mRNA 3′ end, release factors highly outcompete aa-tRNAs, thus driving polypeptide chain release. In rare cases, a stop codon can be decoded by an aa-tRNA, with addition of an amino acid to the polypeptide chain, thus allowing elongation to proceed until the next in-frame stop codon is reached (Fig. 4B) (Rodnina et al., 2019). This mechanism is called ribosome or translational readthrough and can be driven by a near-cognate aa-tRNA (i.e., a tRNA whose anticodon is complementary to two out of three positions; see Fig. 1), or by recoding events mediated by specialized tRNAs carrying the 21st selenocysteine or 22nd pyrrolysine amino acids (Cobucci-Ponzano, Rossi, & Moracci, 2012; Namy, Rousset, Napthine, & Brierley, 2004; Zhang, Baranov, Atkins, & Gladyshev, 2005). In particular, pyrrolysine is encoded by the UAG stop codon and is restricted to several microbes, whereas selenocysteine is encoded by the UGA codon and can be found in bacteria, archaea and eukaryotes. Incorporation of selenocysteine requires a specific insertion sequence (selenocysteine inserting sequence, SECIS) and is necessary for the synthesis of selenoproteins (Fig. 4C) (Allmang & Krol, 2006; Cobucci-Ponzano et al., 2012; Zhang et al., 2005). Approximately 100 selenoprotein families have been discovered, among which the human selenoproteome comprises 25 genes, with half of them encoding proteins with known functions (Hatfield, Tsuji, Carlson, & Gladyshev, 2014; Kryukov et al., 2003). These include peroxidase (glutathione-dependent), reductase (NADPH-dependent) and deionidase (metabolism of thyroid hormones) activities, and the specialized selenoprotein P, bearing 10 selenocysteine residues, produced by the liver and involved in the delivery of selenium to other organs (Hatfield et al., 2014; Labunskyy, Hatfield, & Gladyshev, 2014; Reeves & Hoffmann, 2009). Ribosome readthrough was first identified in viruses as a way to expand their genetic information (Beier & Grimm, 2001). It was observed that, in Escherichia coli infected by RNA phage Qβ, the tRNATrp stimulated readthrough over the UGA stop codon at the end of the coat protein cistron, resulting in a longer coat protein essential for the production of infective Qβ particles (Hirsh, 1971). Since then, stop codon readthrough has been found to play important biological roles in several organisms, providing a mechanism to regulate gene expression through the production of different proteins from the same gene (Beznosková, Wagner, Jansen, von der Haar, & Valášek, 2015; Eswarappa et al., 2014). This phenomenon has been called programmed readthrough and its prevalence varies among organisms, with a few examples found also in humans (Loughran et al., 2018; Rodnina et al., 2019; Stiebler et al., 2014). Nevertheless, unregulated readthrough of NTCs is generally a rare event and the termination signals at the end of an open reading frame evolved several ways to promote efficient termination, especially among highly expressed transcripts. A paradigmatic example is represented by tandem stop codons, which act as “back-up” termination signals in case of readthrough (Fleming & Cavalcanti, 2019; Liang, Cavalcanti, & Landweber, 2005). Moreover, the nonstop decay surveillance mechanism recognizes and degrades those transcripts with stalling ribosomes bound to the poly(A) tail, thus preventing the generation and accumulation of unwanted C-terminally extended proteins (Klauer & van Hoof, 2012). In addition, recent studies suggested that translation into 3′ UTRs might reduce C-terminally extended protein levels through 3′ UTR-encoded peptides that destabilize the attached protein, either co- or post-translationally (Arribere et al., 2016). This scenario changes in the presence of a PTC, inserted for instance by a nonsense mutation. In this case, the frequency of spontaneous readthrough is about 10-fold higher (0.01–1%) than at NTCs (0.001–0.1%) (Bonetti, Fu, Moon, & Bedwell, 1995; Cassan & Rousset, 2001; Keeling, Wang, Conard, & Bedwell, 2012; Manuvakhova, Keeling, & Bedwell, 2000). This difference is likely due to the presence (NTCs) or lack (PTCs) of interactions between the termination complex and factors bound to the 3′ UTR of the mRNA. Indeed, during translation the mRNA is maintained in a closed-loop conformation whose function is both to protect the ends of the transcript from exonucleolytic degradation (Chen & Shyu, 2011), and to enhance recycling of translational components, thus increasing the frequency of translation initiation. This closed conformation is maintained by the association of three actors, namely, the cap-binding protein eIF4E, which is bound to the 5′-cap structure of the mRNA, the poly(A)-binding protein (PABP), attached to the poly(A) tail, and eIF4G, which binds both eIF4E and PABP resulting in mRNA circularization. Along with its role in the formation of the closed-loop complex, PABP also promotes translation termination by interacting with eRF3 and stimulating polypeptide chain release (Cosson et al., 2002). In the presence of a PTC, which is usually distant from the NTC and thus from the poly(A) tail, the interaction between PABP and eRF3 will likely be less efficient, leading to prolonged ribosomal pausing and increased aa-tRNA sampling, which in turn makes the PTC more susceptible to readthrough (Amrani et al., 2004). 3.2 Readthrough-inducing compounds Readthrough-mediated suppression of PTCs is promoted certain small molecular weight compounds (Bidou, Allamand, Rousset, & Namy, 2012; Lee & Dougherty, 2012; Linde & Kerem, 2008), thus highlighting their potential application to treat genetic disorders caused by nonsense mutations (Fig. 4B). The first and most studied readthrough-inducing molecules are aminoglycosides, a class of structurally related antibiotics characterized by a common 2-deoxystreptamine core (ring II) linked to a glucopyranosyl (ring I) at position 4. Aminoglycosides exert their antibiotic activity by binding to the bacterial ribosome and inhibiting protein synthesis (Fan-Minogue & Bedwell, 2007). In particular, they bind to specific nucleotides in the decoding center inducing A1492 and A1493 to flip out from the internal loop of helix 44, inducing a conformational change that partially resembles that caused by cognate aa-tRNA binding (Fig. 5), thus resulting in extensive amino acid misincorporation and translation inhibition (Ogle, 2001). In humans, aminoglycosides can be safely used as antibiotics because their binding to the eukaryotic ribosome is much less efficient. Extensive mutational analyses revealed that the major determinants of the different sensitivity to aminoglycoside between prokaryotes and eukaryotes are the yeast G1645 and A1754 residues, corresponding to E. coli A1408 and G1491 residues. These divergent nucleotides cause the aminoglycoside binding pocket within helix 44 to be shallower in eukaryotes, thus preventing stable insertion and binding of the drug (Fan-Minogue & Bedwell, 2007). Nevertheless, in the presence of a PTC, the impact of aminoglycosides on the eukaryotic ribosome is sufficient to reduce discrimination between near-cognate aa-tRNAs and release factors, thus enhancing amino acid (mis)incorporation at PTCs and thus readthrough (Keeling & Bedwell, 2002; Manuvakhova et al., 2000). The molecular basis of aminoglycosides effects is still unclear, but recent structural analysis of yeast ribosomes in complex with different aminoglycosides revealed that these compounds interact with the eukaryotic ribosome at multiple sites and that the mechanism by which each aminoglycoside induces PTC readthrough is specific (Prokhorova et al., 2017). In particular, aminoglycosides containing 6′-OH substituent in ring I (e.g., G418) bind to the pocket within helix 44, thus facilitating near-cognate aa-tRNA accommodation and competition with the release factor. On the other hand, aminoglycosides containing a 6′-NH2 substituent (e.g., gentamicin) employ an alternative mechanism based on inter-subunit rotation effects that are likely to hamper the interaction between release factors and the ribosome (Prokhorova et al., 2017). However, in the context of a therapeutic approach based on suppression of nonsense mutations, the lifelong or even prolonged administration of aminoglycosides is hardly feasible due to persistent and serious ototoxicity and nephrotoxicity as severe side effects (Guthrie, 2008; Lopez-Novoa, Quiros, Vicente, Morales, & Lopez-Hernandez, 2011). The mechanism underlying this toxicity is currently unclear, but generation of reactive oxygen species is thought to be the main cause of cell death (Xie, Talaska, & Schacht, 2011). Several approaches have been tempted to overcome aminoglycosides toxic effects, such as liposome encapsulation (Schiffelers, 2001) and co-administration with antioxidants (Campbell et al., 2007), as well as other compounds able to reduce the interaction with cellular components (Du, Keeling, Fan, Liu, & Bedwell, 2009) or to increase the readthrough-inducing activity of aminoglycosides (Baradaran-Heravi et al., 2016). In addition, a different approach stemmed from the idea that distinct structures of aminoglycosides are responsible for nonsense suppression properties and toxicity. Therefore, aminoglycosides have been chemically modified to reduce toxicity and increase their therapeutic potential. In particular, the paromomycin derivatives NB30 and NB54 showed up to 15-fold reduced toxicity (Nudelman et al., 2009), whereas the G418 derivatives NB74 and NB84 showed reduced toxicity, increased activity and the ability to cross the blood–brain barrier in mice (Gunn et al., 2014). Moreover, NB124 was shown to reduce NMD-mediated degradation of the target transcript (Bidou, Bugaud, Belakhov, Baasov, & Namy, 2017), and the neomycin derivative pyranmycin (TC007) was able to increase the lifespan of spinal muscular atrophy mice (Mattis, Tom Chang, & Lorson, 2012). A different approach based on high-throughput screening of small molecular weight compound libraries led to the identification of non-aminoglycoside molecules with readthrough induction activity (Borgatti, Altamura, Salvatori, D'Aversa, & Altamura, 2020; Du et al., 2009; Welch et al., 2007). One paradigmatic example is PTC124 (also known as ataluren or Translarna®), an orally bioavailable oxadiazole compound with PTC suppression activity that has been shown to be safe, with minimal off-target side-effects and no antibacterial activity (Welch et al., 2007). Recent studies demonstrated that ataluren acts by inhibiting the release factor complex activity (Huang et al., 2022; Ng, Li, Ghelfi, Goldman, & Cooperman, 2021) and, although its mechanism of action has yet to be completely clarified, encouraging results have been obtained. Different diseases are currently being evaluated in clinical trials (aniridia, NCT02647359; colorectal and endometrium cancers, NCT04014530; epilepsy, NCT0275826), but results are unfortunately conflicting. Ataluren has been approved by the European Medicines Agency (EMA) for Duchenne muscular dystrophy patients with nonsense mutations (Haas et al., 2015; McDonald et al., 2017), whereas the lack of conclusive results obtained in pre-clinical and clinical trials led to discontinuation of the development of ataluren for cystic fibrosis (Kerem et al., 2014). Other promising readthrough-inducing compounds identified include RTC13 and RTC14 (Du, Damoiseaux, et al., 2009), amlexanox (Gonzalez-Hilarion et al., 2012), PTC414 (Moosajee et al., 2016), 2,6-diaminopurine (Trzaska et al., 2020), SRI-41315 (Sharma et al., 2021), TCP-1109 (Otani et al., 2022), NV848 (Bezzerri et al., 2022), and AC1903 (Baradaran-Heravi et al., 2022). In addition, the aminoglycoside analog ELX-02 (previously referred to as NB124) is under clinical investigation (Crawford, Alroy, Sharpe, Goddeeris, & Williams, 2020; Kerem, 2020; Leubitz et al., 2019). 3.3 The determinants of readthrough The establishment of therapeutic approaches based on nonsense suppression has been hampered to date by their extremely variable outcome. Indeed, the efficacy of readthrough induction depends on several factors, including the amount of the target transcript(s) (related to the occurrence of NMD), the efficiency of PTC suppression, and the features of the full-length proteins arising from readthrough events. Moreover, a recent study by Karki and colleagues explored the tissue-specificity of translational readthrough, which seems to be regulated by the expression levels and alternative splicing isoforms of termination factors (e.g., eRF1), and by the relative abundance of readthrough-prone suppressor tRNAs (Karki et al., 2021). Here, we will focus on the molecular determinants (nucleotide/protein contexts) of readthrough (Fig. 6). 3.3.1 The sequence context The PTC sequence context, namely, the stop codon and the surrounding sequences, is a major determinant of readthrough efficiency (Tate, Cridge, & Brown, 2018). It is widely established that the three stop codons are decoded with different accuracy by the ribosome, and are thus differently susceptible (leakiness) to readthrough with the rank order of UGA > UAG > UAA (Fig. 6A) (Bidou et al., 2004; Du et al., 2002; Loughran et al., 2014; Manuvakhova et al., 2000). Early statistical studies of natural termination sequences had suggested that the actual termination signal is a tetranucleotide, thus highlighting the influence of the nucleotide downstream of the stop codon (+ 4 position, with the first nucleotide of the termination codon marked as + 1) (McCaughan, Brown, Dalphin, Berry, & Tate, 1995). Later studies confirmed that the + 4 position is strongly involved in termination fidelity and readthrough efficiency, with an up to sixfold difference conferred solely by the 4th nucleotide (Cridge, Crowe-McAuliffe, Mathew, & Tate, 2018; Manuvakhova et al., 2000). Structural studies of the eukaryotic ribosomal complex containing eRF1 revealed that, when a stop codon is placed in the ribosomal A site, the base at position + 4 is pulled into the A site by mRNA compaction, where it is stabilized by stacking against G626 of 18S rRNA (Ben-Shem et al., 2011; Brown, Shao, Murray, Hegde, & Ramakrishnan, 2015; Matheisl, Berninghausen, Becker, & Beckmann, 2015; McCaughan et al., 1995). Stacking of G626 is less stable for pyrimidines, which help explaining the readthrough-favorable role of cytosine and uracil at the + 4 position. Extensive studies in yeast showed that the UGAC and UAGU signals are the most efficient in mediating readthrough, followed by UGAN and UAGN, whereas the UAAN signals, and in particular the UAAA sequence, mediate the most accurate termination (Fig. 6A) (Manuvakhova et al., 2000). Besides the tetranucleotide termination signal, both 5′ and 3′ surrounding sequences (from − 2 to + 9 positions) showed a non-random distribution of nucleotides in natural contexts (Bonetti et al., 1995). Concerning the 5′ sequence, the presence of two adenines in positions − 2 and − 1 has been associated with the highest levels of readthrough (Harrell, 2002; Tork, 2004), whereas the 3′ sequence context modulates termination fidelity, with a significant but complex role of the + 5, + 6 and + 8 nucleotides (Cridge et al., 2018). The structural basis for these effects is unclear, although the P site codon (including positions − 2 and − 1) appears to directly contact the top of helix 44 of 18S rRNA (Tork, 2004), and the + 4 and + 5 nucleotides to stack respectively with G626 and C1698 bases of 18S rRNA (Brown et al., 2015). This highlights the presence of a complex network of interactions between the ribosome and the sequences surrounding the stop codon, although the effects of 5′ and 3′ sequences are subtle and subordinated to the stop codon present. Overall, readthrough efficiency is hardly predictable based strictly on the sequence context, but most authors agree on the high susceptibility of the UGA stop codon followed by UAG and UAA, the latter being the less responsive. Even though the levels of aminoglycoside-induced readthrough were believed to be simply superimposed on the basal level allowed by each sequence context (Manuvakhova et al., 2000), it has been recently suggested that the sequence context also affects the optimal readthrough-inducing drug type as well as the optimal drug concentration (Schilff, Sargsyan, Hofhuis, & Thoms, 2021). This should be considered to identify susceptible nonsense mutations and optimal treatment in the context of a personalized PTC suppression approach, whose choice should be driven by experimental evidence whenever possible. 3.3.2 The reinserted amino acid Another critical but largely underestimated factor affecting the readthrough outcome is the amino acid reinserted at the PTC position. Readthrough occurs when a stop codon is misrecognized by the ribosome, which incorporates an aa-tRNA instead of terminating translation. Extensive studies in yeast and mammalian expression systems have shown that defined subsets of aa-tRNAs, and thus amino acids, are actually introduced into the growing polypeptide by the ribosome during readthrough, with UAA and UAG codons that can be suppressed by either glutamine, tyrosine or lysine, whereas UGA codons can be decoded by tryptophan, arginine or cysteine (Fig. 6B) (Blanchet et al., 2014, 2018; Roy, Leszyk, Mangus, & Jacobson, 2015). Comparison of several studies showed that, even though the set of incorporated amino acids is the same, different experimental systems and/or readthrough-inducing compounds can affect the relative frequencies of amino acid insertion at a specific PTC (Blanchet et al., 2014; Pranke et al., 2018; Roy et al., 2015). In particular, a comprehensive study performed in yeast showed that different readthrough-inducing conditions such as loss of Upf factors, defective release factors, or the presence of aminoglycosides, were all able to change the proportion of amino acids reinserted during readthrough (Roy et al., 2015). The structural basis for aa-tRNA selection has been partly unveiled. Recent studies showed that the ribosomal decoding center tolerates nonstandard Watson-Crick base pairs in the A site, with the shape of the base pair being crucial, rather than the number of hydrogen bonds formed (Westhof, Yusupov, & Yusupova, 2014). In particular, aa-tRNA insertion at a PTC occurs by mispairing at either position 1 or 3 of the stop codon. The UAG codon showed a preference for mismatches at position 1, whereas UGA favored mispairings at position 3, and UAA showed equal distribution at both positions (Roy et al., 2015). Moreover, identification of the amino acids reinserted during readthrough suggested that, at the third codon position, the A-C mispairing is favored over A-G and G-G, whereas in the first position the U-G mispairing is predominant, likely due to its geometrical mimicry of a standard base pairing (Demeshkina, Jenner, Westhof, Yusupov, & Yusupova, 2013; Roy et al., 2015; Westhof et al., 2014). The amino acids reinserted during PTC readthrough dictate the features of the resulting full-length proteins produced, and thus represent a crucial determinant of the outcome in the view of a nonsense suppression approach. Indeed, the result of readthrough is an assortment of full-length proteins with (potential) different missense changes at the site of the PTC, which can have major impact on protein stability and/or function. Reinsertion of the wild-type residue or a tolerant non-conserved position would favor the production of a functional protein. On the other hand, if the PTC affects an essential position (e.g., a catalytic residue), the protein arising from readthrough would be full-length but probably dysfunctional or non-functional. Therefore, careful evaluation of each PTC should be performed to drive the selection of nonsense mutations potentially amenable for suppression. Overall, the specific features of each PTC (sequence context and role of the residue in the protein) should be carefully considered to help predicting and interpreting the results of a nonsense suppression approach based on readthrough. View chapterExplore book Read full chapter URL: Book series2022, Advances in Protein Chemistry and Structural BiologySilvia Lombardi, ... Alessio Branchini Related terms: Prion Exon Stop Codon Amino Acid Nonsense Mediated mRNA Decay Sex Hormone Binding Globulin Synapsin I Ribosome Codon Open Reading Frame View all Topics
2273	https://math.libretexts.org/Courses/Hope_College/Math_125%3A_Hope_College/05%3A_Trigonometry_Essentials/5.05%3A_Graphs_of_the_Sine_and_Cosine_Functions	Published Time: 2022-06-28T20:32:11Z 5.5: Graphs of the Sine and Cosine Functions - Mathematics LibreTexts Processing math: 100% Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 5: Trigonometry Essentials Math 125: Hope College { } { "5.01:_Angles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.02:_Right_Triangle_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.03:_Unit_Circle" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.04:_The_Other_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.05:_Graphs_of_the_Sine_and_Cosine_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Algebra_Essentials" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Function_Essentials" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Derivative_Essentials" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Applications_of_the_Derivative" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Trigonometry_Essentials" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Differentiation_Rules_and_Applications" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Fri, 01 Nov 2024 19:12:08 GMT 5.5: Graphs of the Sine and Cosine Functions 106365 106365 Katharine Vance { } Anonymous Anonymous 2 false false [ "article:topic", "Cosine Function", "period", "amplitude", "authorname:openstax", "horizontal shift", "periodic functions", "sinusoidal function", "midline", "license:ccby", "showtoc:no", "transcluded:yes", "source-math-1519", "program:openstax", "licenseversion:40", "source@ ] [ "article:topic", "Cosine Function", "period", "amplitude", "authorname:openstax", "horizontal shift", "periodic functions", "sinusoidal function", "midline", "license:ccby", "showtoc:no", "transcluded:yes", "source-math-1519", "program:openstax", "licenseversion:40", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Hope College 4. Math 125: Hope College 5. 5: Trigonometry Essentials 6. 5.5: Graphs of the Sine and Cosine Functions Expand/collapse global location 5.5: Graphs of the Sine and Cosine Functions Last updated Nov 1, 2024 Save as PDF 5.4: The Other Trigonometric Functions 6: Differentiation Rules and Applications Page ID 106365 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Preview Video section 5.5 1. Learning Objectives Graphing Sine and Cosine Functions CHARACTERISTICS OF SINE AND COSINE FUNCTIONS Investigating Sinusoidal Functions Determining the Period of Sinusoidal Functions PERIOD OF SINUSOIDAL FUNCTIONS Example 5.5.1: Identifying the Period of a Sine or Cosine Function Solution Exercise 5.5.1 Determining Amplitude AMPLITUDE OF SINUSOIDAL FUNCTIONS Example 5.5.2: Identifying the Amplitude of a Sine or Cosine Function Solution Exercise 5.5.2 Analyzing Graphs of Variations of y=sin x and y=cos x VARIATIONS OF SINE AND COSINE FUNCTIONS Example 5.5.3: Identifying the Phase Shift of a Function Solution Exercise 5.5.3 Example 5.5.4: Identifying the Vertical Shift of a Function Solution Exercise 5.5.4 How to: Given a sinusoidal function in the form f(x)=A sin(B x−C)+D,identify the midline, amplitude, period, and phase shift Example 5.5.5: Identifying the Variations of a Sinusoidal Function from an Equation Solution Exercise 5.5.5 Example 5.5.6: Identifying the Equation for a Sinusoidal Function from a Graph Solution Exercise 5.5.6 Example 5.5.7: Identifying the Equation for a Sinusoidal Function from a Graph Solution Exercise 5.5.7 Graphing Variations of y=sin x and y=cos x Given the function y=A sin(B x), sketch its graph. Example 5.5.8: Graphing a Function and Identifying the Amplitude and Period Solution Exercise 5.5.8 How to: Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph Example 5.5.9: Graphing a Transformed Sinusoid Solution Exercise 5.5.9 Example 5.5.10: Identifying the Properties of a Sinusoidal Function Solution Using Transformations of Sine and Cosine Functions Example 5.5.11: Finding the Vertical Component of Circular Motion Solution Exercise 5.5.10 Example 5.5.12: Finding the Vertical Component of Circular Motion Solution Exercise 5.5.11 Example 5.5.13: Determining a Rider’s Height on a Ferris Wheel Solution Media Key Equations Key Concepts Homework Exercises 5.5 WeBWorK Problems: Written Problems: Verbal Graphical Algebraic Technology Real-World Applications Preview Video section 5.5 Please fill in your notes as you view the video and answer the questions. Login with LibreOne to view this question NOTE: If you typically access ADAPT assignments through an LMS like Canvas, you should open this page there. Login Learning Objectives Graph variations of y=sin(x) and y=cos(x). Use phase shifts of sine and cosine curves. White light, such as the light from the sun, is not actually white at all. Instead, it is a composition of all the colors of the rainbow in the form of waves. The individual colors can be seen only when white light passes through an optical prism that separates the waves according to their wavelengths to form a rainbow. Figure 5.5.1: Light can be separated into colors because of its wavelike properties. (credit: "wonderferret"/ Flickr) Light waves can be represented graphically by the sine function. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions. Graphing Sine and Cosine Functions Recall that the sine and cosine functions relate real number values to the x- and y-coordinates of a point on the unit circle. So what do they look like on a graph on a coordinate plane? Let’s start with the sine function. We can create a table of values and use them to sketch a graph. Table 5.5.1 lists some of the values for the sine function on a unit circle. Table 5.5.1\| x \| 0 \| π 6 \| π 4 \| π 3 \| π 2 \| 2 π 3 \| 3 π 4 \| 5 π 6 \| π \| \| sin(x) \| 0 \| 1 2 \| √2 2 \| √3 2 \| 1 \| √3 2 \| √2 2 \| 1 2 \| 0 \| Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. See Figure 5.5.2. Figure 5.5.2:The sine function Notice how the sine values are positive between 0 and π, which correspond to the values of the sine function in quadrants I and II on the unit circle, and the sine values are negative between π and 2 π, which correspond to the values of the sine function in quadrants III and IV on the unit circle. See Figure 5.5.3. Figure 5.5.3:Plotting values of the sine function Now let’s take a similar look at the cosine function. Again, we can create a table of values and use them to sketch a graph. Table 5.5.2 lists some of the values for the cosine function on a unit circle. Table 5.5.2\| x \| 0 \| π 6 \| π 4 \| π 3 \| π 2 \| 2 π 3 \| 3 π 4 \| 5 π 6 \| π \| \| cos(x) \| 1 \| √3 2 \| √2 2 \| 1 2 \| 0 \| −1 2 \| −√2 2 \| −√3 2 \| −1 \| As with the sine function, we can plots points to create a graph of the cosine function as in Figure 5.5.4. Figure 5.5.4:The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [−1,1]. In both graphs, the shape of the graph repeats after 2 π, which means the functions are periodic with a period of 2 π. A periodic function is a function for which a specific horizontal shift, P, results in a function equal to the original function: f(x+P)=f(x) for all values of x in the domain of f. When this occurs, we call the smallest such horizontal shift with P>0 the period of the function. Figure 5.5.5 shows several periods of the sine and cosine functions. Figure 5.5.5 Looking again at the sine and cosine functions on a domain centered at the y-axis helps reveal symmetries. As we can see in Figure 5.5.6, the sine function is symmetric about the origin. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because sin(−x)=−sin x. Now we can clearly see this property from the graph. Figure 5.5.6:Odd symmetry of the sine function Figure 5.5.7 shows that the cosine function is symmetric about the y-axis. Again, we determined that the cosine function is an even function. Now we can see from the graph that cos(−x)=cos x. Figure 5.5.7:Even symmetry of the cosine function CHARACTERISTICS OF SINE AND COSINE FUNCTIONS The sine and cosine functions have several distinct characteristics: They are periodic functions with a period of 2 π. The domain of each function is (−∞,∞) and the range is [−1,1]. The graph of y=sin x is symmetric about the origin, because it is an odd function. The graph of y=cos x is symmetric about they- y-axis, because it is an even function. Investigating Sinusoidal Functions As we can see, sine and cosine functions have a regular period and range. If we watch ocean waves or ripples on a pond, we will see that they resemble the sine or cosine functions. However, they are not necessarily identical. Some are taller or longer than others. A function that has the same general shape as a sine or cosine function is known as a sinusoidal function. The general forms of sinusoidal functions are y=A sin(B x−C)+D and y=A cos(B x−C)+D Determining the Period of Sinusoidal Functions Looking at the forms of sinusoidal functions, we can see that they are transformations of the sine and cosine functions. We can use what we know about transformations to determine the period. In the general formula, B is related to the period by P=2 π\|B\|. If \|B\|>1, then the period is less than 2 π and the function undergoes a horizontal compression, whereas if \|B\|<1, then the period is greater than 2 π and the function undergoes a horizontal stretch. For example, f(x)=sin(x), B=1, so the period is 2 π,which we knew. If f(x)=sin(2 x), then B=2, so the period is π and the graph is compressed. If f(x)=sin(x 2), then B=1 2, so the period is 4 π and the graph is stretched. Notice in Figure 5.5.8 how the period is indirectly related to \|B\|. Figure 5.5.8 PERIOD OF SINUSOIDAL FUNCTIONS If we let C=0 and D=0 in the general form equations of the sine and cosine functions, we obtain the forms y=A sin(B x) y=A cos(B x) The period is 2 π\|B\|. Example 5.5.1: Identifying the Period of a Sine or Cosine Function Determine the period of the function f(x)=sin(π 6 x). Solution Let’s begin by comparing the equation to the general form y=A sin(B x). In the given equation, B=π 6, so the period will be P=2 π\|B\|=2 π π 6=2 π⋅6 π=12 Exercise 5.5.1 Determine the period of the function g(x)=cos(x 3). Answer 6 π Determining Amplitude Returning to the general formula for a sinusoidal function, we have analyzed how the variable B relates to the period. Now let’s turn to the variable A so we can analyze how it is related to the amplitude, or greatest distance from rest. A represents the vertical stretch factor, and its absolute value \|A\| is the amplitude. The local maxima will be a distance \|A\| above the vertical midline of the graph, which is the line x=D; because D=0 in this case, the midline is the x-axis. The local minima will be the same distance below the midline. If \|A\|>1, the function is stretched. For example, the amplitude of f(x)=4 s i n x is twice the amplitude of f(x)=2 sin x If \|A\|<1, the function is compressed. Figure 5.5.9 compares several sine functions with different amplitudes. Figure 5.5.9 AMPLITUDE OF SINUSOIDAL FUNCTIONS If we let C=0 and D=0 in the general form equations of the sine and cosine functions, we obtain the forms y=A sin(B x)and y=A cos(B x) The amplitude is A, and the vertical height from the midline is \|A\|. In addition, notice in the example that \|A\|=a m p l i t u d e=1 2∣m a x i m u m−m i n i m u m\| Example 5.5.2: Identifying the Amplitude of a Sine or Cosine Function What is the amplitude of the sinusoidal function f(x)=−4 sin(x)? Is the function stretched or compressed vertically? Solution Let’s begin by comparing the function to the simplified form y=A sin(B x). In the given function, A=−4, so the amplitude is \|A\|=\|−4\|=4. The function is stretched. Analysis The negative value of A results in a reflection across the x-axis of the sine function, as shown in Figure 5.5.10. Figure 5.5.10 Exercise 5.5.2 What is the amplitude of the sinusoidal function f(x)=1 2 sin(x)? Is the function stretched or compressed vertically? Answer 1 2 compressed Analyzing Graphs of Variations of y=sin x and y=cos x Now that we understand how A and B relate to the general form equation for the sine and cosine functions, we will explore the variables C and D. Recall the general form: y=A sin(B x−C)+D and y=A cos(B x−C)+D or y=A sin(B(x−C B))+D and y=A cos(B(x−C B))+D The value C B for a sinusoidal function is called the phase shift, or the horizontal displacement of the basic sine or cosine function. If C>0, the graph shifts to the right. If C<0, the graph shifts to the left. The greater the value of \|C\|, the more the graph is shifted. Figure 5.5.11 shows that the graph of f(x)=sin(x−π) shifts to the right by π units, which is more than we see in the graph of f(x)=sin(x−π 4), which shifts to the right by π 4 units. Figure 5.5.11 While C relates to the horizontal shift, D indicates the vertical shift from the midline in the general formula for a sinusoidal function. See Figure 5.5.12. The function y=cos(x)+D has its midline at y=D. Figure 5.5.12 Any value of D other than zero shifts the graph up or down. Figure 5.5.13 compares f(x)=sin x with f(x)=sin x+2, which is shifted 2 units up on a graph. Figure 5.5.13 VARIATIONS OF SINE AND COSINE FUNCTIONS Given an equation in the form f(x)=A sin(B x−C)+D or f(x)=A cos(B x−C)+D, C B is the phase shift and D is thevertical shift. Example 5.5.3: Identifying the Phase Shift of a Function Determine the direction and magnitude of the phase shift for f(x)=sin(x+π 6)−2. Solution Let’s begin by comparing the equation to the general form y=A sin(B x−C)+D. In the given equation, notice that B=1 and C=−π 6. So the phase shift is C B=−π 6 1=−π 6 or π 6 units to the left. Analysis We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation shows a minus sign before C. Therefore f(x)=sin(x+π 6)−2 can be rewritten as f(x)=sin(x−(−π 6))−2. If the value of C is negative, the shift is to the left. Exercise 5.5.3 Determine the direction and magnitude of the phase shift for f(x)=3 cos(x−π 2). Answer π 2; right Example 5.5.4: Identifying the Vertical Shift of a Function Determine the direction and magnitude of the vertical shift for f(x)=cos(x)−3. Solution Let’s begin by comparing the equation to the general form y=A cos(B x−C)+D. In the given equation, D=−3 so the shift is 3 units downward. Exercise 5.5.4 Determine the direction and magnitude of the vertical shift for f(x)=3 sin(x)+2. Answer 2 units up How to: Given a sinusoidal function in the form f(x)=A sin(B x−C)+D,identify the midline, amplitude, period, and phase shift Determine the amplitude as \|A\|. Determine the period as P=2 π\|B\|. Determine the phase shift as C B. Determine the midline as y=D. Example 5.5.5: Identifying the Variations of a Sinusoidal Function from an Equation Determine the midline, amplitude, period, and phase shift of the function y=3 sin(2 x)+1. Solution Let’s begin by comparing the equation to the general form y=A sin(B x−C)+D. A=3, so the amplitude is \|A\|=3. Next, B=2, so the period is P=2 π\|B\|=2 π 2=π. There is no added constant inside the parentheses, so C=0 and the phase shift is C B=0 2=0. Finally, D=1, so the midline is y=1. Analysis Inspecting the graph, we can determine that the period is π, the midline is y=1, and the amplitude is 3. See Figure 5.5.14. Figure 5.5.14 Exercise 5.5.5 Determine the midline, amplitude, period, and phase shift of the function y=1 2 cos(x 3−π 3). Answer midline: y=0; amplitude: \|A\|=1 2; period: P=2 π\|B\|=6 π; phase shift: C B=π Example 5.5.6: Identifying the Equation for a Sinusoidal Function from a Graph Determine the formula for the cosine function in Figure 5.5.15. Figure 5.5.15 Solution To determine the equation, we need to identify each value in the general form of a sinusoidal function. y=A sin(B x−C)+D y=A cos(B x−C)+D The graph could represent either a sine or a cosine function that is shifted and/or reflected. When x=0, the graph has an extreme point, (0,0). Since the cosine function has an extreme point for x=0, let us write our equation in terms of a cosine function. Let’s start with the midline. We can see that the graph rises and falls an equal distance above and below y=0.5. This value, which is the midline, is D in the equation, so D=0.5. The greatest distance above and below the midline is the amplitude. The maxima are 0.5 units above the midline and the minima are 0.5 units below the midline. So \|A\|=0.5. Another way we could have determined the amplitude is by recognizing that the difference between the height of local maxima and minima is 1, so \|A\|=1 2=0.5. Also, the graph is reflected about the x-axis so that A=−0.5. The graph is not horizontally stretched or compressed, so B=1; and the graph is not shifted horizontally, so C=0. Putting this all together, g(x)=−0.5 cos(x)+0.5 Exercise 5.5.6 Determine the formula for the sine function in Figure 5.5.16. Figure 5.5.16 Answer f(x)=sin(x)+2 Example 5.5.7: Identifying the Equation for a Sinusoidal Function from a Graph Determine the equation for the sinusoidal function in Figure 5.5.17. Figure 5.5.17 Solution With the highest value at 1 and the lowest value at −5, the midline will be halfway between at −2. So D=−2. The distance from the midline to the highest or lowest value gives an amplitude of \|A\|=3. The period of the graph is 6, which can be measured from the peak at x=1 to the next peak at x=7,or from the distance between the lowest points. Therefore, P=2 π\|B\|=6. Using the positive value for B,we find that B=2 π P=2 π 6=π 3 So far, our equation is either y=3 sin(π 3 x−C)−2 or y=3 cos(π 3 x−C)−2.For the shape and shift, we have more than one option. We could write this as any one of the following: a cosine shifted to the right a negative cosine shifted to the left a sine shifted to the left a negative sine shifted to the right While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes y=3 cos(π 3 x−π 3)−2 or y=−3 cos(π 3 x+2 π 3)−2 Again, these functions are equivalent, so both yield the same graph. Exercise 5.5.7 Write a formula for the function graphed in Figure 5.5.18. Figure 5.5.18 Answer two possibilities: y=4 sin(π 5 x−π 5)+4 or y=−4 sin(π 5 x+4 π 5)+4 Graphing Variations of y=sin x and y=cos x Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations. Instead of focusing on the general form equations y=A sin(B x−C)+D and y=A cos(B x−C)+D we will let C=0 and D=0 and work with a simplified form of the equations in the following examples. Given the function y=A sin(B x), sketch its graph. Identify the amplitude, \|A\|. Identify the period, P=2 π\|B\|. Start at the origin, with the function increasing to the right if A is positive or decreasing if A is negative. At x=π 2\|B\| there is a local maximum for A>0 or a minimum for A<0, with y=A. The curve returns to the x-axis at x=π\|B\|. There is a local minimum for A>0 (maximum for A<0) at x=3 π 2\|B\| with y=–A. The curve returns again to the x-axis at x=π 2\|B\|. Example 5.5.8: Graphing a Function and Identifying the Amplitude and Period Sketch a graph of f(x)=−2 sin(π x 2). Solution Let’s begin by comparing the equation to the form y=A sin(B x). Step 1. We can see from the equation that A=−2, so the amplitude is 2. \|A\|=2 Step 2. The equation shows that B=π 2, so the period is P=2 π π 2=2 π⋅2 π=4 Step 3. Because A is negative, the graph descends as we move to the right of the origin. Step 4. The x-intercepts are at the beginning of one period, x=0, the horizontal midpoints are at x=2 and at the end of one period at x=4. The quarter points include the minimum at x=1 and the maximum at x=3. A local minimum will occur 2 units below the midline, at x=1, and a local maximum will occur at 2 units above the midline, at x=3. Figure 5.5.19 shows the graph of the function. Figure 5.5.19 Exercise 5.5.8 Sketch a graph of g(x)=−0.8 cos(2 x). Determine the midline, amplitude, period, and phase shift. Answer Figure 5.5.20 midline: y=0; amplitude: \|A\|=0.8; period: P=2 π\|B\|=π; phase shift: C B=0 or none How to: Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph Express the function in the general form y=A sin(B x−C)+D or y=A cos(B x−C)+D. Identify the amplitude, \|A\|. Identify the period, P=2 π\|B\|. Identify the phase shift, C B. Draw the graph of f(x)=A sin(B x) shifted to the right or left by C B and up or down by D. Example 5.5.9: Graphing a Transformed Sinusoid Sketch a graph of f(x)=3 sin(π 4 x−π 4). Solution Figure 5.5.21: A horizontally compressed, vertically stretched, and horizontally shifted sinusoid Step 1. The function is already written in general form: f(x)=3 sin(π 4 x−π 4).This graph will have the shape of a sine function, starting at the midline and increasing to the right. Step 2.\|A\|=\|3\|=3. The amplitude is 3. Step 3. Since \|B\|=\|π 4\|=π 4, we determine the period as follows. P=2 π\|B\|=2 π π 4=2 π⋅4 π=8 The period is 8. Step 4. Since C=π 4, the phase shift is C B=π 4 π 4=1. The phase shift is 1 unit. Step 5. Figure 5.5.21 shows the graph of the function. Exercise 5.5.9 Draw a graph of g(x)=−2 cos(π 3 x+π 6). Determine the midline, amplitude, period, and phase shift. Answer Figure 5.5.22 midline: y=0; amplitude: \|A\|=2; period: P=2 π\|B\|=6; phase shift: C B=−1 2 Example 5.5.10: Identifying the Properties of a Sinusoidal Function Given y=−2 c o s(π 2 x+π)+3, determine the amplitude, period, phase shift, and horizontal shift. Then graph the function. Solution Begin by comparing the equation to the general form. y=A cos(B x−C)+D Step 1. The function is already written in general form. Step 2. Since A=−2, the amplitude is \|A\|=2. Step 3.\|B\|=π 2, so the period is P=2 π\|B\|=2 π p i 2=2 π⋅2 π=4. The period is 4. Step 4. C=−π,so we calculate the phase shift as C B=−π π 2=−π⋅2 π=−2. The phase shift is −2. Step 5. D=3,so the midline is y=3, and the vertical shift is up 3. Since A is negative, the graph of the cosine function has been reflected about the x-axis. Figure 5.5.23 shows one cycle of the graph of the function. Figure 5.5.23 Using Transformations of Sine and Cosine Functions We can use the transformations of sine and cosine functions in numerous applications. As mentioned at the beginning of the chapter,circular motioncan be modeled using either the sine or cosine function. Example 5.5.11: Finding the Vertical Component of Circular Motion A point rotates around a circle of radius 3 centered at the origin. Sketch a graph of the y-coordinate of the point as a function of the angle of rotation. Solution Recall that, for a point on a circle of radius r, the y-coordinate of the point is y=r sin(x), so in this case, we get the equation y(x)=3 sin(x). The constant 3 causes a vertical stretch of the y-values of the function by a factor of 3, which we can see in the graph in Figure 5.5.24. Figure 5.5.24 Analysis Notice that the period of the function is still 2 π; as we travel around the circle, we return to the point (3,0) for x=2 π,4 π,6 π,....Because the outputs of the graph will now oscillate between –3 and 3, the amplitude of the sine wave is 3. Exercise 5.5.10 What is the amplitude of the function f(x)=7 cos(x)? Sketch a graph of this function. Answer Figure 5.5.25 Example 5.5.12: Finding the Vertical Component of Circular Motion A circle with radius 3 ft is mounted with its center 4 ft off the ground. The point closest to the ground is labeled P, as shown in Figure 5.5.26. Sketch a graph of the height above the ground of the point P as the circle is rotated; then find a function that gives the height in terms of the angle of rotation. Figure 5.5.26 Solution Sketching the height, we note that it will start 1 ft above the ground, then increase up to 7 ft above the ground, and continue to oscillate 3 ft above and below the center value of 4 ft, as shown in Figure 5.5.27. Figure 5.5.27 Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection. Second, we see that the graph oscillates 3 above and below the center, while a basic cosine has an amplitude of 1, so this graph has been vertically stretched by 3, as in the last example. Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, we find that y=−3 cos(x)+4 Exercise 5.5.11 A weight is attached to a spring that is then hung from a board, as shown in Figure 5.5.28. As the spring oscillates up and down, the position y of the weight relative to the board ranges from –1 in. (at time x=0) to –7 in. (at time x=π) below the board. Assume the position of y is given as a sinusoidal function of x. Sketch a graph of the function, and then find a cosine function that gives the position y in terms of x. Figure 5.5.28 Answer y=3 cos(x)−4 Figure 5.5.29 Example 5.5.13: Determining a Rider’s Height on a Ferris Wheel The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. Solution With a diameter of 135 m, the wheel has a radius of 67.5 m. The height will oscillate with amplitude 67.5 m above and below the center. Passengers board 2 m above ground level, so the center of the wheel must be located 67.5+2=69.5 m above ground level. The midline of the oscillation will be at 69.5 m. The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes. Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve. Amplitude: 67.5, so A=67.5 Midline: 69.5, so D=69.5 Period:30, so B=2 π 30=π 15 Shape: −cos(t) An equation for the rider’s height would be y=−67.5 cos(π 15 t)+69.5 where t is in minutes and y is measured in meters. Media Access these online resources for additional instruction and practice with graphs of sine and cosine functions. Amplitude and Period of Sine and Cosine Translations of Sine and Cosine Graphing Sine and Cosine Transformations Graphing the Sine Function Key Equations Sinusoidal functions f(x)=A sin(B x−C)+D f(x)=A cos(B x−C)+D Key Concepts Periodic functions repeat after a given value. The smallest such value is the period. The basic sine and cosine functions have a period of 2 π. The function sin x is odd, so its graph is symmetric about the origin. The function cos x is even, so its graph is symmetric about the y-axis. The graph of a sinusoidal function has the same general shape as a sine or cosine function. In the general formula for a sinusoidal function, the period is P=2 π\|B\|. See Example 5.5.1. In the general formula for a sinusoidal function, \|A\| represents amplitude. If \|A\|>1, the function is stretched, whereas if \|A\|<1, the function is compressed. See Example 5.5.2. The value C B in the general formula for a sinusoidal function indicates the phase shift. See Example 5.5.3. The value D in the general formula for a sinusoidal function indicates the vertical shift from the midline. See Example 5.5.4. Combinations of variations of sinusoidal functions can be detected from an equation. See Example 5.5.5. The equation for a sinusoidal function can be determined from a graph. See Example 5.5.6 and Example 5.5.7. A function can be graphed by identifying its amplitude and period. See Example 5.5.8 and Example 5.5.9. A function can also be graphed by identifying its amplitude, period, phase shift, and horizontal shift. See Example 5.5.10. Sinusoidal functions can be used to solve real-world problems. See Example 5.5.11, Example 5.5.12, and Example 5.5.13. Homework Exercises 5.5 WeBWorK Problems: none Written Problems: Please do #28 instead of #30 from the assignment. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions Verbal 1) Why are the sine and cosine functions called periodic functions? Answer The sine and cosine functions have the property that f(x+P)=f(x) for a certain P. This means that the function values repeat for every P units on the x-axis. 2) How does the graph of y=sin x compare with the graph of y=cos x? Explain how you could horizontally translate the graph of y=sin x to obtain y=cos x. 3) For the equation A cos(B x+C)+D, what constants affect the range of the function and how do they affect the range? Answer The absolute value of the constant A (amplitude) increases the total range and the constant D (vertical shift) shifts the graph vertically. 4) How does the range of a translated sine function relate to the equation y=A sin(B x+C)+D? 5) How can the unit circle be used to construct the graph of f(t)=sin t? Answer At the point where the terminal side of t intersects the unit circle, you can determine that the sin t equals the y-coordinate of the point. Graphical For the following exercises, graph two full periods of each function and state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x>0. Round answers to two decimal places if necessary. 6) f(x)=2 sin x 7) f(x)=2 3 cos x Answer amplitude: 2 3;period: 2 π;midline: y=0;maximum: y=2 3 occurs at x=0;minimum: y=−2 3 occurs at x=π;for one period, the graph starts at 0 and ends at 2 π. 8) f(x)=−3 sin x 9) f(x)=4 sin x Answer amplitude: 4; period: 2 π;midline: y=0;maximum y=4 occurs at x=π 2;minimum: y=−4 occurs at x=3 π 2;one full period occurs from x=0 to x=2 π 10) f(x)=2 cos x 11) f(x)=cos(2 x) Answer amplitude: 1; period: π;midline: y=0;maximum: y=1 occurs at x=π;minimum: y=−1 occurs at x=π 2;one full period is graphed from x=0 to x=π 12) f(x)=2 sin(1 2 x) 13) f(x)=4 cos(π x) Answer amplitude: 4; period: 2; midline: y=0;maximum: y=4 occurs at x=0;minimum: y=−4 occurs at x=1 14) f(x)=3 cos(6 5 x) 15) y=3 sin(8(x+4))+5 Answer amplitude: 3; period: π 4; midline: y=5; maximum: y=8 occurs at x=−4+21 π 16≈0.123; minimum: y=2 occurs at x=−4+23 π 16≈0.516; horizontal shift: −4; vertical translation 5; one period occurs from x=−4+22 π 16≈0.320 to x=−4+26 π 16≈1.105 16) y=2 sin(3 x−21)+4 17) y=5 sin(5 x+20)−2 Answer amplitude: 5; period:2 π 5; midline: y=−2; maximum: y=3 occurs at x=−4+13 π 10≈0.084; minimum: y=−7 occurs at x=−4+15 π 10≈0.712; phase shift: −4; vertical translation: −2; one full period can be graphed on x=−4+7 π 5≈0.398 to x=−4+9 π 5≈1.655 For the following exercises, graph one full period of each function, starting at x=0. For each function, state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x>0. State the phase shift and vertical translation, if applicable. Round answers to two decimal places if necessary. 18) f(t)=2 sin(t−5 π 6) 19) f(t)=−cos(t+π 3)+1 Answer amplitude: 1; period: 2 π; midline: y=1; maximum: y=2 occurs at t=2 π 3≈2.094; minimum: y=0 occurs at t=2 π 3≈5.24; phase shift: −π 3; vertical translation: 1; one full period is from t=2 π 3≈2.094 to t=8 π 3≈8.378 20) f(t)=4 cos(2(t+π 4))−3 21) f(t)=−sin(1 2 t+5 π 3) Answer amplitude: 1; period: 4 π; midline: y=0; maximum: y=1 occurs at t=11 π 3≈11.52; minimum: y=−1 occurs at t=5 π 3≈5.24; phase shift: −10 π 3; vertical shift: 0; one full period is from t=2 π 3≈2.094 to t=14 π 3≈14.661 22) f(x)=4 sin(π 2(x−3))+7 23) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below. Figure 5.5.23 Answer 23. amplitude: 2; midline: y=−3 period: 4; equation: f(x)=2 sin(π 2 x)−3 24) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below. Figure 5.5.24 25) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below. Figure 5.5.25 Answer 25. amplitude: 2; period: 5; midline: y=3 equation: f(x)=−2 cos(2 π 5 x)+3 26) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below. Figure 5.5.26 27) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below. Figure 5.5.27 Answer 27. amplitude: 4; period: 2; midline: y=0 ; equation: f(x)=−4 cos(π(x−π 2)) 28) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below. Figure 5.5.28 29) Determine the amplitude, midline, period, and an equation involving the cosine function for the graph shown in Figure below. Figure 5.5.29 Answer 29. amplitude: 2; period: 2; midline y=1 equation: f(x)=2 cos(π x)+1 30) Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure below. Figure 5.5.30 Algebraic For the following exercises, let f(x)=sin x. 31) On [0,2 π), solve f(x)=0. 32) On [0,2 π), solve f(x)=1 2. Answer π 6, 5 π 6 33) Evaluate f(π 2). 34) On [0,2 π), f(x)=√2 2. Find all values of x. Answer π 4, 3 π 4 35) On [0,2 π), the maximum value(s) of the function occur(s) at what x-value(s)? 36) On [0,2 π), the minimum value(s) of the function occur(s) at what x-value(s)? Answer 3 π 2 37) Show that f(−x)=−f(x). This means that f(x)=sin x is an odd function and possesses symmetry with respect to ____. For the following exercises, let f(x)=cos x 38) On [0,2 π), solve the equation f(x)=cos x=0 Answer π 2, 3 π 2 39) On [0,2 π), solve f(x)=1 2. 40) On [0,2 π), find the x-intercepts of f(x)=cos x. Answer π 2, 3 π 2 41) On [0,2 π), find the x-values at which the function has a maximum or minimum value. 42) On [0,2 π), solve the equation f(x)=√3 2. Answer π 6, 11 π 6 Technology 43) Graph h(x)=x+sin x on [0,2 π]. Explain why the graph appears as it does. 44) Graph h(x)=x+sin x on [−100,100]. Did the graph appear as predicted in the previous exercise? Answer The graph appears linear. The linear functions dominate the shape of the graph for large values of x. 45) Graph f(x)=x sin x on [0,2 π] and verbalize how the graph varies from the graph of f(x)=x sin x. 46) Graph f(x)=x sin x on the window [−10,10] and explain what the graph shows. Answer The graph is symmetric with respect to the y-axis and there is no amplitude because the function is not periodic. 47) Graph f(x)=sin x x on the window [−5 π,5 π] and explain what the graph shows. Real-World Applications 48) A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function h(t) gives a person’s height in meters above the ground t minutes after the wheel begins to turn Find the amplitude, midline, and period of h(t). Find a formula for the height function h(t). How high off the ground is a person after 5 minutes? Answer Amplitude: 12.5; period: 10; midline: y=13.5 h(t)=12.5 sin(π 5(t−2.5))+13.5 26 ft This page titled 5.5: Graphs of the Sine and Cosine Functions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 5.4: The Other Trigonometric Functions 6: Differentiation Rules and Applications Was this article helpful? Yes No Recommended articles 8.1: Graphs of the Sine and Cosine FunctionsIn the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and creat... 2.1: Graphs of the Sine and Cosine FunctionsIn this section, we will interpret and create graphs of sine and cosine functions 8.2: Graphs of the Sine and Cosine FunctionsIn the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and creat... 6.1: Graphs of the Sine and Cosine FunctionsIn the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and creat... 7.2: Graphs of the Sine and Cosine FunctionsIn the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and creat... Article typeSection or PageAuthorOpenStaxLicenseCC BYLicense Version4.0OER program or PublisherOpenStaxShow Page TOCnoTranscludedyes Tags amplitude Cosine Function horizontal shift midline period periodic functions sinusoidal function source@ source-math-1519 © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? 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2274	https://www.archtoolbox.com/calculating-volume/	Calculating Volume - Archtoolbox Search Archtoolbox for... Menu Design & Documentation▼ Architectural Concepts Construction Documentation Measurement / Scale / Geometry Sustainability & Health Materials & Systems▼ Concrete Masonry Metals Wood / Plastic / Composites Thermal & Moisture Protection Doors / Windows / Openings Finishes Specialties Equipment Furnishings Vertical Circulation Fire Suppression Plumbing HVAC Electrical Site & Landscape Professional Practice▼ Architectural References Career Development Project Delivery Store Calculating Volume Design & Documentation>Measurement, Scale, and Geometry Here is a series of diagrams that demonstrate how to calculate the volume of three-dimensional shapes. Volume of a Block, Prism, or Cylinder The volume of a block, prism, or cylinder is easily calculated by multiplying the area of the base by the height/altitude of the object. When the bases are not parallel, the height or altitude of the object is from the center point of one base to the center point of the other base. Our article,Calculating Area, can help you determine how to calculate the area of a shape. Volume = (area of base x height) Volume of a Block, Prism, or Cylinder Volume of a Cone or Pyramid The volume of cones and pyramids is calculated by multiplying the base by the height and then dividing by 3. Our article,Calculating Area, can help you determine how to calculate the area of a shape. Volume = (area of base x height) / 3 Volume of a Cone or Pyramid Article Updated: May 6, 2021 Help make Archtoolbox better for everyone. If you found an error or out of date information in this article (even if it is just a minor typo), please let us know. Subscribe to the Archtoolbox Newsletter Receive a curated email with industry news focusing on practice, leadership, technology, and career growth. Subscribe Helpful Tools for Architects and Building Designers ### BBToolsets Bluebeam Tools and Templates for Architects - Customized by Architects for Architects. ### Fee Development Tools The Archtoolbox Guide and Templates for Developing a Fee for Architectural Services. ### Negotiate a Raise The Archtoolbox Guide to Negotiating a Salary Increase Related Articles Calculating Slope and Common Slopes in Architecture Calculating Area Newest Articles Becoming a Licensed Architect in the United States Civil Engineering Plan Symbols Ethics in Architectural Practice Civil Engineering Abbreviations Steep Sloped Roofing Systems About Archtoolbox About Archtoolbox Contact Archtoolbox Terms of Service and Privacy Policy Follow Archtoolbox YouTube Twitter LinkedIn This website uses cookies. Display and/or use of the information contained within this site constitutes acceptance of the Terms of Service and Privacy Policy. It is important that all local codes and standards are adhered to. Please consult a professional architect, engineer, consultant, or agency for advice about specific projects, buildings, conditions, codes, and/or regulations. As an Amazon Associate, Archtoolbox earns from qualifying purchases. We may also earn commissions from other products we recommend. Copyright © 2025 Archtoolbox. All rights reserved. Archtoolbox is a Trademark of Aggregate Digital LLC.
2275	https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_13?srsltid=AfmBOoqRYUQEEudFfsQR2zcAuFng4nVi5cbxeyyCBXyFuJbiE1cQSvdf	Art of Problem Solving 1984 AIME Problems/Problem 13 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1984 AIME Problems/Problem 13 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1984 AIME Problems/Problem 13 Contents 1 Problem 2 Solutions 2.1 Solution 1 2.2 Solution 2 2.3 Solution 3 2.4 Solution 4 3 See also Problem Find the value of Solutions Solution 1 We know that so we can repeatedly apply the addition formula, . Let , , , and . We have , so and , so . Thus our answer is . Solution 2 Apply the formula repeatedly. Using it twice on the inside, the desired sum becomes . This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning. Solution 3 On the coordinate plane, let , , , , , , , , , and . We see that , , , and . The sum of these four angles forms the angle of triangle , which has a cotangent of , which must mean that . So the answer is Solution 4 Recall that and that . Then letting and , we are left with Expanding , we are left with See also 1984 AIME (Problems • Answer Key • Resources) Preceded by Problem 12Followed by Problem 14 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions Retrieved from " Category: Intermediate Trigonometry Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2276	https://www.youtube.com/watch?v=kxKAjhAUwPo	Residue calculus integration example with simple poles, Complex Analysis Dr. Bevin Maultsby 6440 subscribers 20 likes Description 614 views Posted: 14 Jun 2024 We evaluate the improper integral of 1/(x^4+1) from −∞ to ∞ using residue calculus following these steps: identify singularities of the extended complex function, construct a contour (the upper semicircle and real axis) that encloses certain singularities, calculate residues at these singularities, and then apply Cauchy's residue theorem to find the integral's value. (Fun fact: you can compute the residues of f a bit quicker using this technique: In greater detail, I start by extending the function into the complex plane as f(z)=1/(z^4+1), note that the degree of the denominator is at least two greater than the numerator. Next, I identify the singularities of the function, which are the roots of z^4=-1, finding four distinct solutions that lie on the circle \|z\|=1 in the complex plane. To evaluate the integral, I construct a contour in the upper half of the complex plane that encloses two of the four singularities so that we can use Cauchy's residue theorem to compute the integral around this contour. I then compute the residues at the enclosed singularities, simplifying the complex expressions involved. Finally, I demonstrate how the integral over the real line can be obtained by considering the limit of the contour integral as the radius of the semicircular part goes to infinity, subtracting the contribution from the semicircular path, which tends to zero. The final result of the integral is derived by summing the computed residues and applying Cauchy's residue theorem. complexanalysis #mathematics #Contourintegration #integration #CauchysResidueTheorem #residuecalculus #mathtutorial ##ResidueComputation 8 comments Transcript: In this video, we want to evaluate the improper integral of 1/(x^4 + 1) from negative infinity to infinity using residue calculus. We will use techniques from complex analysis to evaluate this integral by extending it into the complex plane. I've divided this up into several steps. The first thing we'll do is look at this integrand function, so we'll turn it into a function of the complex variable z and find its singularities. Then we will design a suitable contour in the complex plane which will give us a way to evaluate this integral on the real line. Then we will need to compute some residues, and we'll put it all together at the very end. Okay, so first, let's analyze the function f(z) = 1/(z^4 + 1). Let me just point out here for a moment that the degree of the denominator is 4, which is greater than or equal to 0 + 2, which is the degree of the numerator plus 2. When we have this kind of rational function with respect to z, that's what we want to see when we're doing this kind of contour computation that we're going to do. So I'm going to set this aside for now but then we'll mention it again at the end. Okay, the next step now that we've identified f(z) is to find the singularities of this function. So those are the values of z in the complex plane that make the denominator zero, or equivalently, z^4 = -1. There are different ways that you could approach this. If you're watching this, you might instantly know what the solutions to this equation are, but let me take you through the process. What I would do is take -1 and turn it into a complex exponential using Euler's formula. So -1 being one unit down the real axis to the left, that's going to be like e^(iπ). So with Euler's formula, that's cos(π) + i sin(π), so -1 + 0i. But we could also get to that number if we did e^(iπ) plus rotations by 2π radians. So let me indicate that with 2πik, where k indicates the number of rotations. We can solve for z now by taking the fourth root of each side, so z is going to be e^(iπ/4 + πik/2). Now, notice the second term is indicating rotations by factors of π/2 or 90 degrees. Rotating by π/2 four times returns you back to where you started. So the solutions to this equation that we're looking to solve, z^4 = -1, are going to correspond to this index k being 0, 1, 2, or 3. So that will give us four distinct solutions. Let me graph these in the complex plane. Each lives on the circle of modulus one. When k = 0, we have rotated around the circle by π/4, that puts us about here. I'm going to name the singularity z₀, and to write it concisely, that is (1 + i)/√2. Rotating by π/2 in the complex plane lands us here, that's our next singularity. That's the complex number (-1 + i)/√2, but I'm actually going to realize that it's z₁. And that's because if you don't see that we rotate by π/4 to get to z₀, then π/4 again to get to i, and then π/4 again, so doing the rotation that gives us z₁ lands us on the second pole, so that's z₁. And if you prefer, you could write that as (-1 + i)/√2. Another rotation by π/2 lands us here. I'm going to call that -z₀, so it's (-1 - i)/√2. And then one more rotation puts us here, and that is -z₁, so (1 - i)/√2. Let me say one more thing about how we can write f now that we've identified these four singularities. I'm going to give myself a little bit more room. We could take f now and write the denominator as a product of four factors. So I can say f(z) is 1/((z - z₀)(z + z₀)(z - z₁)(z + z₁)). And notice here each singularity has order one. Now that we've worked out the singularities of f, let's figure out what contour we should use in order to evaluate our starting integral. To determine a suitable contour, what we want to realize is that we're dealing with an integral over the entire real line, so we need to select a contour that includes this interval and is also closed up. So we'll close it through the upper half of the complex plane using the upper semicircle in the first and second quadrants. I'm going to name this contour I just drew γ_ρ for the radius, and then this upper semicircle part I'm going to call that C_ρ, and then it's the upper semicircle, so let me throw a + on there. Okay, within this contour, we have enclosed two of the four singularities for our function f(z). So for suitably large values of the radius ρ, the singularities z₀ and z₁ are enclosed within the semicircle. Since we're actually doing an integral from negative infinity to infinity, you want to imagine that ρ is going to infinity, so there's no worry about actually enclosing these. So our next step so that we can use Cauchy's residue theorem is to compute the residue of f at each of these singularities. We'll do them one at a time. Let's start with the residue of f at z₀, that's the pole in the first quadrant. I've put the entire expression here, how you compute residue, but this will simplify in just a minute. So it's going to be the limit as z approaches z₀ of this expression, which is written in terms of this letter m or this index m. m is the order of the pole, that's how many times the pole appears in the full factorization of our denominator. Our poles were all simple, which means that m = 1 for each. So looking back at this limit that we need to take, the first expression 1/(n - 1)! turns into 1/0! 0! is 1, so this goes away, or basically becomes the number one. This differentiation operation tells us how many times to differentiate a product. When m = 1, we're saying differentiate zero times, so we will not take any derivatives. And then what would we have differentiated? It's the product of z minus the pole to the n power times the original function f(z), but here m is just one, so we can get rid of that. So when your pole is simple, the residue is the limit as z approaches z₀ of (z - z₀) f(z), that's all we need to do. Let's just simplify our expression. So I've removed those extraneous terms that we don't need when m = 1. In this product here, I've also taken f(z) and written it out in terms of its factorization, because now we can say that the term z - z₀ cancels out with its appearance in the factorization of f in that denominator, leaving us just with the product of the three other factors after that cancellation. We can go ahead and evaluate this limit. So when z goes to z₀, z + z₀ is going to go to 2z₀. So we start to write this out. We will have 1 all over 2z₀(z - z₁)(z + z₁). This simplifies, in fact, it probably would have simplified more easily if I had taken these latter two terms here and left them as z^2 + i. What we have is 1/(2z₀(z^2 + i)), which you could also get by fully expanding this. Alright, one last thing to do here: z^2, that was rotation by π/4, and then rotation by π/4 again, this puts us at i. So that z^2 + i is 2i, so we can write this as 1/(4iz₀). You could, of course, simplify this further, but I'm going to leave it there for now because I think it might make adding the two residues easier at the end. Now that we've done this, what I would like for you to do is pause and work out the residue of f at the pole in the second quadrant, and if you would like, you can leave it in a form similar to this one. Okay, let's see if this is what you got. The first thing I did was cancel out z - z₁ with the factor as it appears in the denominator. You could then recognize that this factoring I've done here as z^2 - i might be more advantageous to work that way. Evaluating this limit, we'll get 1/(z₁^2) so that's z to the 6th power minus i times 2z₁. Algebraically, you might want to think of this as (z^2)^3 instead, so that's i cubed. Or you could imagine rotating six times by π/4 in the complex plane. Regardless, both of them will land you on -i for that first term, so that this residue, left in a form similar to the first computation that we did, is -4iz₁. Now let's put it all together to finish the integral. Using residue calculus, we can say that this improper integral from negative infinity to infinity of 1/(x^4 + 1) is like going around our semicircular contour, taking away the upper semicircle, and then letting the radius go to infinity, so that just actually puts us down on the real line. So I've written that idea as the limit as the radius ρ goes to infinity of the contour integral, so the closed loop, minus the limit as the radius goes to infinity of the integral of f around the upper semicircle. This latter expression, though, is actually going to be zero because of that observation I made at the very beginning, which was that the degree of the denominator was at least two larger than the degree of the numerator. Now, to evaluate the integral of f around our closed contour, we can use Cauchy's residue theorem. So this will be 2πi times the sum of the residues that we computed. So that's going to be 2πi times, let me pull out the 1/(4i) that each of the residues had, so that's 1/(4i), and then that leaves us with adding together 1/z₀ - 1/z₁. The second one was negative. Alright, this leading product is going to be π/2, and then 1/z₀ is √2/(1 + i), and 1/z₁ is going to be √2/(1 - i). √2/2 is 1/√2, so let me pull the √2 out and combine that with π/2 to write π/√2. And then we'll add the two remaining fractions together. That's going to be like finding a common denominator. So we'll have (1 + i - (1 - i)) in the numerator. And then for the denominator, we'll have 1 1 and then the 1 i - 1 i, those will cancel, and then we'll have +2i. So overall, that's going to be -2i. I'll go ahead and write it as (1 + i)^2 first. Alright, in the top, the imaginary numbers cancel out, leaving us with 1 - 1 all over -2i. So actually, that whole fraction is just 1. So the final result is that this integral is π/√2. Alright, and now we're all done. I hope you enjoy this exercise in residue calculus.
2277	https://www.quora.com/Factorial-function-How-many-zeroes-are-there-in-100	Something went wrong. Wait a moment and try again. Trailing Zero Factorial (function) Functions in Mathematics Combinatorial Math Mathematics Number Theory Trailing Zeroes in Factor... 5 Factorial (function): How many zeroes are there in 100!? · To find how many trailing zeros are in 100!, we can use the formula that counts the number of times 10 is a factor in the numbers from 1 to 100. Since 10 is made up of the factors 2 and 5, and there are usually more factors of 2 than 5, we only need to count the number of times 5 is a factor in the numbers from 1 to 100. The formula to calculate the number of trailing zeros in n! is: Number of trailing zeros=⌊n5⌋+⌊n52⌋+⌊n53⌋+… For n=100: Calculate \left\lfloor \frac{100}{5} To find how many trailing zeros are in 100!, we can use the formula that counts the number of times 10 is a factor in the numbers from 1 to 100. Since 10 is made up of the factors 2 and 5, and there are usually more factors of 2 than 5, we only need to count the number of times 5 is a factor in the numbers from 1 to 100. The formula to calculate the number of trailing zeros in n! is: Number of trailing zeros=⌊n5⌋+⌊n52⌋+⌊n53⌋+… For n=100: Calculate ⌊1005⌋: ⌊1005⌋=⌊20⌋=20 Calculate ⌊10025⌋: ⌊10025⌋=⌊4⌋=4 Calculate ⌊100125⌋: ⌊100125⌋=⌊0.8⌋=0 Now, add these values together: 20+4+0=24 Thus, the number of trailing zeros in 100! is 24. Shobhit Varshney B.Tech at SRM University, Kattankulathur (2015–present) · Upvoted by Dwight House , Ph. D. Mathematics, Duke University (1972) · 7y Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. It can be solved in two ways - Let’s look at how trailing zeros are formed in the first place. A trailing zero is formed when a multiple of 5 is multiplied with a multiple of 2. Now all we have to do is count the number of 5’s and 2’s in the multiplication. Each pair of 2 and 5 will cause a trailing zero. Since we have only 24 5’s, we can only make 24 pairs of 2’s and 5’s thus the number of trailing zeros in 100 factorial is 24. The question can also be answered using the simple Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. It can be solved in two ways - Let’s look at how trailing zeros are formed in the first place. A trailing zero is formed when a multiple of 5 is multiplied with a multiple of 2. Now all we have to do is count the number of 5’s and 2’s in the multiplication. Each pair of 2 and 5 will cause a trailing zero. Since we have only 24 5’s, we can only make 24 pairs of 2’s and 5’s thus the number of trailing zeros in 100 factorial is 24. The question can also be answered using the simple formula given below: The above formula gives us the exact number of 5s in n! because it will take care of all multiples of 5 which are less than n. Not only that it will take care of all multiples of 25, 125, etc. (higher powers of 5). Tip: Instead of dividing by 25, 125, etc. (higher powers of 5); it would be much faster if you divided by 5 recursively. Let us use this to solve a few examples: Q) What is the number of trailing zeroes in 100! ? [100/5] = 20 Now we can either divided 100 by 25 or the result in the above step i.e. 20 by 5. [20/5 ]= 4. It is less than 5, so we stop here. The answer is - 20+ 4 = 24 (direct answer in just few secs) Q) What is the number of trailing zeroes in 200! ? [200/5] = 40 Now we can either divided 200 by 25 or the result in the above step i.e. 40 by 5. [40/5 ]= 8 [8/5]=1. It is less than 5, so we stop here. The answer is - 40 + 8 + 1= 49 Q) What is the number of trailing zeroes in 1123!? [1123/5] = 224 [224/5] = 44 [44/5] = 8 [8/5]=1. It is less than 5, so we stop here. The answer is - 224 + 44 + 8 + 1 =277 If you have any questions, please feel free to ask in the comment section. Sponsored by Amazon Web Services (AWS) Want to level up your AWS game? Register for re:Invent 2025. 5 days of advanced architecture workshops, serverless deep dives, and coding labs. Las Vegas, Dec 1-5. Usha Keswani Studied Arts Degrees in Economics with Honors · Author has 94 answers and 47.5K answer views · 5y Originally Answered: How do you find the total number of zeroes at the end of 100!, where the symbol '!' represents factorial? · 100!=100×99×98×………………………………….×3×2×1 SOLUTION; (1) see what nos from 1 to 100 when multiplied end in zero (2) number ending with zero multiplied with another number will end in zero (3)A number ending in 5 multiplied by even number will end in zero (4)nos.25 ,75 when multiplied by 4 generate an extra zero (5)no.50 generate an extra zero ,will have at least two zeros at the end when multiplied by even number Now, Make a table 100———2 zeros 95–———1 zero 90———1 85———1 80——-1 75——-2 70——1 65——-1 60——-1 55——-1 50——-2 45——1 40———1 35——1 30———1 25——-2 20——1 15——1 10——-1 5——-1 Total 24 Zeros Answer 24 Zeros in 100! Related questions How many consecutive zeros are there at the end of 100 factorials? What is the number of zeros 100! (100 factorial) ends with? What is the factorial of 1 to 100? What is the value of 100 factorial? How many number of zeros present in 152 factorial ? Pradeep Rangarajan (Prad) 7y To get underway, one must know concept of zeros in factorial. Zeros start from 5! (Five factorial); 5! = 5×4×3×2×1 = 120. So from 5! onwards any factorial would fetch you zeros at the end. (Basically unit digit is zero). 6! =6×5! 7!= 7×6×5! So, 100! = 100×99×…..×5! It is obvious that 100 has 2 zeros in addition to the one zero in 5!. Now to the point, To find the number of zeros at the end of A! Is Mod(A/5) +mod(A/25) + mod(125) + mod(A/625)… Basically, denominators are powers of 5. Now the question is for 100! \|100/5\| +\|100/25\| 20+ 4 = 24. There are 24 zeroes in 100!. Hope it helped. Cheers :) Faheel Ahmad Studied at Jamia Millia Islamia · 10y Originally Answered: How many zeroes are in 100 factorial? · There are 30 zeroes in 100! 100! = 9332621544394415268169923e266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000 There are 24 trailing and 6 non-trailing zeroes. Don't forget those 6. Sponsored by Mutual of Omaha Retiring soon and need Medicare advice? Be prepared for retirement with a recommendation from our Medicare Advice Center. Jorge Sawyer B.S. in Mathematics & Physics, Lafayette College (Graduated 2010) · Author has 432 answers and 1.4M answer views · Updated 10y Originally Answered: How many zeroes are in 100 factorial? · The number of zeros in n! is equal the of 5 in n!. This is because there will always be more factors of 2 than 5 in n! and 10=2×5. To compute how many factors of 5 are in n! you can use: ∑∞i=1⌊n5i⌋=⌊n5⌋+⌊n25⌋+⌊n125⌋+⋯ In the case of n=100 you get ⌊1005⌋+⌊10025⌋=20+4=24. Interestingly you can approximate the number of trailing zeros in n! as \frac{n}{ The number of zeros in n! is equal the of 5 in n!. This is because there will always be more factors of 2 than 5 in n! and 10=2×5. To compute how many factors of 5 are in n! you can use: ∑∞i=1⌊n5i⌋=⌊n5⌋+⌊n25⌋+⌊n125⌋+⋯ In the case of n=100 you get ⌊1005⌋+⌊10025⌋=20+4=24. Interestingly you can approximate the number of trailing zeros in n! as n4. In general you can approximate the multiplicity of any prme p as np−1. Related questions How many zeros are in factorial 1000 to the power factorial 1000? How many zeroes are in Factorial 1000? Why does n ∑ i = 0 i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 ? How large is 100 factorial? What is the value of zero factorial (0!)? Kurt Mager Enjoys solving math problems · Author has 17.6K answers and 7.3M answer views · 5y Originally Answered: How many trailing zeros are in 100 factorial (i.e. 100!)? · There are 24. First identify the factors that are multiples of 5. 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100 Of those, 10 of them are divisible by 10. That gives 10 zeroes so far. One hundred has two trailing zeroes, so that adds one more for a total of 11 so far. Now we are just interested in factors of 5. 5 has a factor of 5 15 has a factor of 5. Two so far 25 has two factors of 5. Four so far. 35 has a factor of 5. Five so far. 45 has a factor of 5. That's six. 50 has one. That's seven 55 has one. That's eight. 65 has one. That's nine. 75 has two. That's eleven. 85 has There are 24. First identify the factors that are multiples of 5. 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100 Of those, 10 of them are divisible by 10. That gives 10 zeroes so far. One hundred has two trailing zeroes, so that adds one more for a total of 11 so far. Now we are just interested in factors of 5. 5 has a factor of 5 15 has a factor of 5. Two so far 25 has two factors of 5. Four so far. 35 has a factor of 5. Five so far. 45 has a factor of 5. That's six. 50 has one. That's seven 55 has one. That's eight. 65 has one. That's nine. 75 has two. That's eleven. 85 has one. That's twelve. 95 has one. That's thirteen. Of the remaining factors in 100! you have more than enough factors of two to make each of these fives into tens, so all told there will be 24 trailing zeroes. Sponsored by CDW Corporation Looking for secure, flexible solutions to support constituents? Gemini for Google Workspace and CDW have tools that are secure and compatible with existing tech. Lindsay Stevens 90% off flights w Mighty Travels Premium · Author has 345 answers and 2.5M answer views · 2y The factorial function is a mathematical function that calculates the product of all integers from 1 to a given number. The factorial function is written as "!" and pronounced "factorial." CMA K S Narayanan Cost Accountant having 24 years Telecom Experience · Author has 7K answers and 13.4M answer views · 5y Originally Answered: How do you find the total number of zeroes at the end of 100!, where the symbol '!' represents factorial? · Distinct Prime factors of 100 = 2 & 5. We need to determine how many factors comprising 2 & 5 are present in 100! No of factors comprising 2 as factor is (1 + 3 + 6 + 12 + 25 + 50 = 97 ) No of factors comprising 5 as factor is (4 + 20 = 24 ). 2 x 5 produces a zero in final product. => min(24,97) number of zeroes are present in 100!. Ans : 24 zeroes :-) Distinct Prime factors of 100 = 2 & 5. We need to determine how many factors comprising 2 & 5 are present in 100! No of factors comprising 2 as factor is (1 + 3 + 6 + 12 + 25 + 50 = 97 ) No of factors comprising 5 as factor is (4 + 20 = 24 ). 2 x 5 produces a zero in final product. => min(24,97) number of zeroes are present in 100!. Ans : 24 zeroes :-) Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Mohammad Afzaal Butt B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.8M answer views · 6y Originally Answered: How many 0s are in 100!? · We need to find number of zeros at the end of 100!. This is equivalent to finding the largest power of 10 that divides 100!. The prime factors of 10 are 2 and 5. If a and b are the largest powers of 2 and 5 respectively so that2a and5bdivides 100!, then a>b.The largest power of 5 dividing 100! is ⌊1005⌋+⌊10052⌋+⌊10053⌋+⋯where⌊⌋is the floor function. We need to find number of zeros at the end of 100!. This is equivalent to finding the largest power of 10 that divides 100!. The prime factors of 10 are 2 and 5. If a and b are the largest powers of 2 and 5 respectively so that2a and5bdivides 100!, then a>b.The largest power of 5 dividing 100! is ⌊1005⌋+⌊10052⌋+⌊10053⌋+⋯where⌊⌋is the floor function. =20+4+0=24 ∵the largest exponent of 2 is at least 24. It follows that 1024 is the largest power of 10 that divides 100!. Thus 100! ends in 24 zeros Ramneek Singh Student · 9y Originally Answered: 100 factorial has how many zeroes? · There is a direct formulae to do this. If you need to find the power of a number in factorials. Say if you want to find the power of "a" in x ! All you need to find is- [ x / a ] + [ x / a 2 ] + [ x / a 3 ] . . . , where [a] is the closest integer less than or equal to a. For eg. [ 3.6 ] = 3 . You will need to carry on the series till you get the denominator greater tgan the numerator(because the closest integer smaller than it would be 0). If you understand it , such problems are easy to do. I HAVE PROVIDED AN EXPLANATION IN THE END , FOR THE FORMUALE . BACK to your question- To find the number of zeroes There is a direct formulae to do this. If you need to find the power of a number in factorials. Say if you want to find the power of "a" in x ! All you need to find is- [ x / a ] + [ x / a 2 ] + [ x / a 3 ] . . . , where [a] is the closest integer less than or equal to a. For eg. [ 3.6 ] = 3 . You will need to carry on the series till you get the denominator greater tgan the numerator(because the closest integer smaller than it would be 0). If you understand it , such problems are easy to do. I HAVE PROVIDED AN EXPLANATION IN THE END , FOR THE FORMUALE . BACK to your question- To find the number of zeroes in the end we need to find the power of 10 in 100! 10 = 2 ∗ 5 So wee need to power of 2 and 5 in 100! . However it is easy to see that power of 2! would be more than that of 5!. So we just need to find power of 5 as there would be sufficient 2s for them to make 10. By our formulae- Power of 5 equals [math] [100/5] + [100/25] + [100/125] [/math] we don't need to do it furthur as it is equal to zero. So the power of 5 in 100! is 24. Therefore there would be 24 zeros at the end. NOW THE EXPLANATION OF THE FORMULAE- So let us see if we want to find the power of 2 in 100!. 100 ! = 1 ∗ 2 ∗ 3 ∗ 4...100 How many multiples of 2s are there between 0 and 100? 50, right? But wait what about 4 we will get one extra 2 from it. Same with multiples of 4. So the number of extra twos are equal to the number od multiples of 4 till 100. That is equal to 25. Now consider 8 and its multiples We will get still antoher extra 2 from them. So we need to add number of multiples of 8 to our answer i.e. 12 . But then there are 16, 32 ,64 and their multiples. We would get more extra twos from them. We just need to count the number of multiples of them. And same goes with our formuale, we are just counting the multiples of 2 , 2 2 , 2 3 , 2 4 , 2 5 , 2 6 by it. [ 100 / 2 ] + [ 100 / 4 ] + [ 100 / 8 ] + [ 100 / 16 ] + [ 100 / 32 ] + [ 100 / 64 ] = 50 + 25 + 12 + 6 + 3 + 1 = 97 Hope it helps! Harshal Chaudhari PGDBM from Sydenham Institute of Management Studies, Research and Entrepreneurship Education (SIMSREE) (Graduated 2019) · 9y Originally Answered: How many zeroes are in 100 factorial? · A zero can be obtained by (2 into 5) which is 10 Therefore we have to see how many 2 and 5 exist in 100! As we can see only 24 powers of 5 are available , so (2 and 5) can form 24 pairs together. Hence 24 zeros exist in 100! A zero can be obtained by (2 into 5) which is 10 Therefore we have to see how many 2 and 5 exist in 100! As we can see only 24 powers of 5 are available , so (2 and 5) can form 24 pairs together. Hence 24 zeros exist in 100! Gyan Ranjan Masters in Maths from IIT Madras · Author has 55 answers and 213.1K answer views · 9y I would take this question as 'How many zeroes does 100! ends with?' To answer this,let's think when will a 0 be created.When a product of 5 and 2 happens, a zero is created. Clearly in 100!,powers of 5 arise lesser than powers of 2. So,all that we need is the no. of times 5 or power of 5 occurs,which can be got by the simple formula: Power of 5 in 100! = [100/5]+[100/25]+[100/125]+... So,the no.of zeroes would be 24 :) Related questions How many consecutive zeros are there at the end of 100 factorials? What is the number of zeros 100! (100 factorial) ends with? What is the factorial of 1 to 100? What is the value of 100 factorial? How many number of zeros present in 152 factorial ? How many zeros are in factorial 1000 to the power factorial 1000? How many zeroes are in Factorial 1000? Why does n ∑ i = 0 i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 ? How large is 100 factorial? What is the value of zero factorial (0!)? How many zeroes are in 100 trillion? How many zeroes are in n factorial? How many zeros are between 1 to 100? If 2 + 3 = 13 3 + 4 = 25 4 +5 = 41 then 5 + 6 =? What is the number with 100 zeros? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2278	https://www.electronics-tutorials.ws/filter/filter_3.html	Register to download premium content! Tutorials AC Circuits Amplifiers Attenuators Binary Numbers Boolean Algebra Capacitors Combinational Logic Counters DC Circuits Diodes Electromagnetism Filters Inductors Input/Output Devices Logic Gates Miscellaneous Circuits Operational Amplifiers Oscillator Power Electronics Power Supplies RC Networks Resistors Sequential Logic Systems Transformers Transistors Waveform Generators eBooks Further Education Sitemap Page Contact Us English English Deutsch Polski Login Page Register Subscribe Register to download premium content! X Deutsch Polski Subscribe Register Login Page AC Circuits Amplifiers Attenuators Binary Numbers Boolean Algebra Capacitors Combinational Logic Connectivity Counters DC Circuits Diodes Electromagnetism Filters Inductors Input and Output Devices Logic Gates Miscellaneous Circuits Operational Amplifiers Oscillator Power Electronics Power Supplies Premium RC Networks Resistors Resources Sequential Logic Systems Tools Transformers Transistors Waveform Generators Premium Content Further Education Sitemap Page Contact Us Home / Filters / Passive High Pass Filter Passive High Pass Filter A High Pass Filter is the exact opposite to the low pass filter circuit as the two components have been interchanged with the filters output signal now being taken from across the resistor Where as the low pass filter only allowed signals to pass below its cut-off frequency point, ƒc, the passive high pass filter circuit as its name implies, only passes signals above the selected cut-off point, ƒc eliminating any low frequency signals from the waveform. Consider the circuit below. The High Pass Filter Circuit In this circuit arrangement, the reactance of the capacitor is very high at low frequencies so the capacitor acts like an open circuit and blocks any input signals at VIN until the cut-off frequency point ( ƒC ) is reached. Above this cut-off frequency point the reactance of the capacitor has reduced sufficiently as to now act more like a short circuit allowing all of the input signal to pass directly to the output as shown below in the filters response curve. Frequency Response of a 1st Order High Pass Filter The Bode Plot or Frequency Response Curve above for a passive high pass filter is the exact opposite to that of a low pass filter. Here the signal is attenuated or damped at low frequencies with the output increasing at +20dB/Decade (6dB/Octave) until the frequency reaches the cut-off point ( ƒc ) where again R = Xc. It has a response curve that extends down from infinity to the cut-off frequency, where the output voltage amplitude is 1/√2 = 70.7% of the input signal value or -3dB (20 log (Vout/Vin)) of the input value. Also we can see that the phase angle ( Φ ) of the output signal LEADS that of the input and is equal to +45o at frequency ƒc. The frequency response curve for this filter implies that the filter can pass all signals out to infinity. However in practice, the filter response does not extend to infinity but is limited by the electrical characteristics of the components used. The cut-off frequency point for a first order high pass filter can be found using the same equation as that of the low pass filter, but the equation for the phase shift is modified slightly to account for the positive phase angle as shown below. Cut-off Frequency and Phase Shift The circuit gain, Av which is given as Vout/Vin (magnitude) and is calculated as: High Pass Filter Example No1 Calculate the cut-off or “breakpoint” frequency ( ƒc ) for a simple passive high pass filter consisting of an 82pF capacitor connected in series with a 240kΩ resistor. Second-order High Pass Filter Again as with low pass filters, high pass filter stages can be cascaded together to form a second order (two-pole) filter as shown. Second-order High Pass Filter The above circuit uses two first-order filters connected or cascaded together to form a second-order or two-pole high pass network. Then a first-order filter stage can be converted into a second-order type by simply using an additional RC network, the same as for the 2nd-order low pass filter. The resulting second-order high pass filter circuit will have a slope of 40dB/decade (12dB/octave). As with the low pass filter, the cut-off frequency, ƒc is determined by both the resistors and capacitors as follows. In practice, cascading passive filters together to produce larger-order filters is difficult to implement accurately as the dynamic impedance of each filter order affects its neighbouring network. However, to reduce the loading effect we can make the impedance of each following stage 10x the previous stage, so R2 = 10R1 and C2 = 1/10th of C1. High Pass Filter Summary We have seen that the Passive High Pass Filter is the exact opposite to the low pass filter. This filter has no output voltage from DC (0Hz), up to a specified cut-off frequency ( ƒc ) point. This lower cut-off frequency point is 70.7% or -3dB (dB = -20log VOUT/VIN) of the voltage gain allowed to pass. The frequency range “below” this cut-off point ƒc is generally known as the Stop Band while the frequency range “above” this cut-off point is generally known as the Pass Band. The cut-off frequency, corner frequency or -3dB point of a high pass filter can be found using the standard formula of: ƒc = 1/(2πRC). The phase angle of the resulting output signal at ƒc is +45o. Generally, the high pass filter is less distorting than its equivalent low pass filter due to the higher operating frequencies. A very common application of this type of passive filter, is in audio amplifiers as a coupling capacitor between two audio amplifier stages and in speaker systems to direct the higher frequency signals to the smaller “tweeter” type speakers while blocking the lower bass signals or are also used as filters to reduce any low frequency noise or “rumble” type distortion. When used like this in audio applications the high pass filter is sometimes called a “low-cut”, or “bass cut” filter. The output voltage Vout depends upon the time constant and the frequency of the input signal as seen previously. With an AC sinusoidal signal applied to the circuit it behaves as a simple 1st Order high pass filter. But if we change the input signal to that of a “square wave” shaped signal that has an almost vertical step input, the response of the circuit changes dramatically and produces a circuit known commonly as an Differentiator. The RC Differentiator Up until now the input waveform to the filter has been assumed to be sinusoidal or that of a sine wave consisting of a fundamental signal and some harmonics operating in the frequency domain giving us a frequency domain response for the filter. However, if we feed the High Pass Filter with a Square Wave signal operating in the time domain giving an impulse or step response input, the output waveform will consist of short duration pulse or spikes as shown. The RC Differentiator Circuit Each cycle of the square wave input waveform produces two spikes at the output, one positive and one negative and whose amplitude is equal to that of the input. The rate of decay of the spikes depends upon the time constant, ( RC ) value of both components, ( t = R x C ) and the value of the input frequency. The output pulses resemble more and more the shape of the input signal as the frequency increases. PreviousPassive Low Pass Filter NextPassive Band Pass Filter Read more Tutorials inFilters 1. Capacitive Reactance 2. Passive Low Pass Filter 3. Passive High Pass Filter 4. Passive Band Pass Filter 5. Active Low Pass Filter 6. Active High Pass Filter 7. Active Band Pass Filter 8. Butterworth Filter Design 9. Second Order Filters 10. State Variable Filter 11. Band Stop Filter 12. Sallen and Key Filter 13. Decibels 82 Comments Join the conversationCancel reply Error! Please fill all fields. HALİT ÖZER Alçak geçiren filtresini üç voltluk devre üzerinde şematik olarak devrenin neresine bağlanır ikincisi üç voltluk devre için parazit azaltıcı veye kesici olarak RC filtrenin bileşenlerin değerleri kaç olmalıdır rica etsem verirmisiniz şimdiden teşekkür ediyorum lütfen en kısa zamanda bekliyorum. Posted on September 11th 2025 \| 3:53 am Reply Vidya The circuit of low pass filter and high pass filter is same as that of integrator and differentiator Posted on August 13th 2025 \| 9:05 am Reply Wayne Storr Yes, the basic circuit is exactly the same. The difference is the input signal, step response or sinusoidal. Then clearly you have missed the point of the tutorial. Posted on August 13th 2025 \| 2:54 pm Reply Erllinda amoscoo Bshw Posted on July 27th 2025 \| 10:58 am Reply Nandana Naishad In RC high pass differentiator the capacitive reactance should be more than the resistance of their resistor? Posted on April 30th 2025 \| 3:42 pm Reply Wayne Storr A capacitors reactance, XC changes with input frequency Posted on May 01st 2025 \| 10:32 am Reply elms What’s up, І would like to subscribe for this website to take neԝest updates, sⲟ where can i do it please assist. Posted on December 11th 2023 \| 11:58 am Reply Eduardo Dumalag hello, i interested all of this..thanks! Posted on October 24th 2023 \| 8:21 am Reply Dabeeggu Kabassima Very Interesting presentation! God bless you! Posted on August 02nd 2023 \| 7:32 pm Reply Daniel Dresser Hello, How does one build a passive high pass filter for the shortwave frequencies? Thanks Posted on July 05th 2023 \| 6:33 am Reply Mahi suthar What is circut use in audio mixer low cut circutAnd how its freq. (Cutt off. Variable Posted on May 21st 2022 \| 5:03 am Reply Ganpudi Jain 1.Design low pass constant-k type T-section and π-section filter with fc = 8KHz and R0 = 600 ohm. Compute α and β for the filters for f = 10 KHz and 20 KHz. Also determine the frequency at which the attenuation is 20 dbs2.For the given low pass constant k-type filter determine the nominal characteristic impedance and the cut-off frequency. Also draw π-section low pass filter. 3.A high pass constant-k filter with fc = 30 KHz is when used with terminating resistance of 500 ohm. Design a suitable T-section and π-section filter.Also determine(i)Design a suitable T-section and π-section filter(ii)Determine the frequency required to produce a maximum attenuation at 20 KHz.(iii)Determine the attenuation constant for the frequency at 20 KHz and 40 KHz.(iv)Determine the phase constant for the frequency at 20 KHz and 40 KHz.(v)Calculate the Zot and Zoπ at 10 KHz.4.Suppose you were installing a high-power stereo system in your car, and you wanted to build a simple filter for the” tweeter” (high-frequency) speakers so that no bass (low-frequency) power is wasted in these speakers. Modify the schematic diagram below with a filter circuit of your choice:“Tweeter”“Tweeter” “Woofer” “Woofer5.Design a suitable filter to produce the characteristic shown on the right: The following capacitors are available: 10 μF, 22 nF, and 0.47 pF.Draw the circuit diagram of the completed filter.Part II6.The following circuit shows one form of filter. (a) What is the name of this type of filter?(b) Calculate the reactance of the capacitor at 100 Hz.(c) What is the impedance of the circuit at 100 Hz..(d) Calculate the break frequency for this filter.(e) Calculate the output voltage at 100 Hz if VIN= 5 V.(f) Sketch the characteristic of this filter, labelling all critical values7.Design a constant k-type band pass T-section and π-section filter having cut –off frequencies of 2000 Hz and 5000 Hz with the characteristic impedance of 500 ohm. Find the resonant frequency.8.Design a constant k-type band stop T-section and π-section filter having cut –off frequencies of 3000 Hz and 6000 Hz with the characteristic impedance of 500 ohm. Find the resonant frequency Posted on April 23rd 2022 \| 10:57 am Reply Abhi Could you please send the solutions for the questions that you have sent here ?? Posted on April 29th 2022 \| 6:30 am Reply Nebojsa Kolaric Again great tutorail thank you Posted on April 16th 2022 \| 2:32 pm Reply Nitin Thank you very much Posted on February 13th 2022 \| 6:49 pm Reply Ayad Fyad Thank you very much>>>><<<<< Posted on December 04th 2021 \| 9:01 pm Reply SANJIT MANDAL requesting you send me Posted on December 05th 2021 \| 6:36 am Reply Junell Baruiz Need to learn about this topic.. Posted on November 24th 2021 \| 1:39 pm Reply Bob How to build a passive crossover for a two channel stereo amp to connect between the amp and speakers that will roll off the low frequency between 80Hz and 100Hz to protect the speakers from damage and the active subwoofer has a low pass filter to roll off the top end between 80Hz and 100Hz. The amp is approximately 100 wpc. Thank you very much, Bob Posted on September 27th 2021 \| 6:01 pm Reply Sanjit Mandal Kindly sent me circuit Posted on September 15th 2021 \| 11:24 am Reply Debasish Mahanta Good Posted on August 24th 2021 \| 6:15 am Reply Sanjit Mandal Very good circuit diagram . Requesting you sent me Posted on August 26th 2021 \| 8:23 am Reply Mohammad Waseem Akram` May I please know the application of Low Pass, High Pass, Band Pass and Band Stop Filter in communication circuits in detail?? Posted on June 01st 2021 \| 10:47 am Reply Musa Sen Thanks a lot, really clear and simple explanation Posted on January 01st 2021 \| 6:35 pm Reply Excellent and presicse text , that comes with age , think so , should be wrong that young minds Reach that stage early. Posted on November 08th 2020 \| 6:56 am Reply Read more Tutorials inFilters 1. Capacitive Reactance 2. Passive Low Pass Filter 3. Passive High Pass Filter 4. Passive Band Pass Filter 5. Active Low Pass Filter 6. Active High Pass Filter 7. Active Band Pass Filter 8. Butterworth Filter Design 9. Second Order Filters 10. State Variable Filter 11. Band Stop Filter 12. Sallen and Key Filter 13. 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2279	https://umem.org/educational_pearls/?author=265	UMEM Educational Pearls - University of Maryland School of Medicine, Department of Emergency Medicine SOM Home Quick Links Quick Links Departments Departments Anatomy & Neurobiology Neurosurgery Pharmacology Physical Therapy and Rehabilitation Science Anesthesiology Biochemistry Molecular Biology Dermatology Diagnostic Radiology Emergency Medicine Epidemiology & Public Health Family Medicine Medical Research Technology Medicine Microbiology & Immunology Neurology OB/GYN Ophthalmology Orthopaedics Otorhinolaryngology Pathology Pediatrics Physiology Psychiatry Radiation Oncology Surgery Institutes Institutes Institute for Genome Sciences Institute of Human Virology Institute for Global Health Centers & Programs Centers & Programs Comparative Medicine Personalized and Genomic Medicine Minority Health & Health Disparities Neuroscience Oncology Trauma Graduate Programs Graduate Programs GPILS-Graduate Program in Life Sciences Masters Programs Combined Degrees CIBR Cores Offices of the Dean Offices of the Dean Academic Administration Admissions Development Faculty Affairs & Professional Development Information Services Medical Education Policy & Planning Postdoctoral Scholars Public Affairs Research Student Affairs Student Research Email MyUMB Directory Map Calendar Tools Tools Email MyUMB Directory Map Calendar Apply Here Quick Links Quick Links Departments Departments Anatomy & Neurobiology Neurosurgery Pharmacology Physical Therapy and Rehabilitation Science Anesthesiology Biochemistry Molecular Biology Dermatology Diagnostic Radiology Emergency Medicine Epidemiology & Public Health Family Medicine Medical Research Technology Medicine Microbiology & Immunology Neurology OB/GYN Ophthalmology Orthopaedics Otorhinolaryngology Pathology Pediatrics Physiology Psychiatry Radiation Oncology Surgery Institutes Institutes Institute for Genome Sciences Institute of Human Virology Institute for Global Health Centers & Programs Centers & Programs Comparative Medicine Personalized and Genomic Medicine Minority Health & Health Disparities Neuroscience Oncology Trauma Graduate Programs Graduate Programs GPILS-Graduate Program in Life Sciences Masters Programs Combined Degrees CIBR Cores Offices of the Dean Offices of the Dean Academic Administration Admissions Development Faculty Affairs & Professional Development Information Services Medical Education Policy & Planning Postdoctoral Scholars Public Affairs Research Student Affairs Student Research Tools Tools Email MyUMB Directory Map Calendar Department Department Chairman's Welcome Department History Our Faculty Our Faculty Faculty Profiles Fellows Profiles Our Staff Our Staff UMEMA Staff Profiles APT Contact Information Department Calendar Prehospital and Disaster Medicine Prehospital and Disaster Medicine Rotations/Education Rotations/Education Anne Arundel County Fire Department Baltimore City Fire Dept Baltimore County Fire Dept Maryland State Police Express Care STAT MedEvac EMS Didactics EMS Education EMS Resident Rotation Medical Student Elective Pharmacy Resident Rotation Wilderness EMS Disaster Medicine Disaster Medicine FEMA Incident Command Courses Triage Disaster Didactics Contact Activities Calendar Links Research International International International Home Publications UMEM Blog FAQ Department Links Education Education Vice Chairman's Welcome Educational Hub Medical Student Training Medical Student Training Med Student Opportunities Residency Fellowship Programs Fellowship Programs Fellowships Research Ultrasound Health Policy Simulation Emergency Cardiovascular Medicine (EM/CVD) Contact Links Apply Now Health Policy Health Policy Health Policy Calendar Health Policy Fellowship Health Policy Elective Curriculum Health Policy Advocacy Health Policy Research Health Policy Education Health Policy Faculty Video Training Video Training Procedures Oral Boards Educational Pearls CME Courses and Conferences Weekly Resident Conference Schedule SchoolED Newsletter Archives Hospitals Hospitals Hospitals Home Academic Sites Academic Sites Our Physicians University of Maryland Medical Center Mercy Medical Center Baltimore VA Medical Center University of Maryland Medical Center - 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Its prognostic usefulness (i.e. in cases wherein treatment is not initiated) has been validated and should be applied in emergent settings to determine optimal patient candidates for tPA treatment. For example,NIHSS> 20 in patients over 75 years old = 45% mortality; NIHSS>17 in patients with atrial fibrillation = positive predictive value for poor outcome of 96%; NIHSS of 6 or less = good spontaneous recovery. An abbreviated version of the NIHSS has been validated and assesses those components which are the best indicators of prognosis. Therefore,when unable to perform a full NIHSS, one should strongly consider using this toolrather than not performing a stroke scale assessment at all. This abbreviated version consists of only 5 categories which assess ability to see (1. best gaze; 2. best visual), walk (3. motor function of lef t leg; 4. motor functio n of right leg), and talk (5. best language).Can patient "see, walk, and talk?" This scale is scored from 0 to 16, with 16 representing the worst prognosis. (see attached abbreviated NIH SS). Show References The National Institute of Neurological Disorders and Stroke rt-PA Stroke Study Group. Tissue plasminogen activator for acute ischemic stroke. N Engl J Med 333:1581-1587, 1995. Tirschwell, et al. "Shortening the NIH Stroke Scale for Use in the Prehospital Setting."Stroke 2002; 33: 2801-2806. Attachments 1112281605_Abbreviated_NIH_Stroke_Scale.docx (98 Kb) ShareFacebookLinkedInXCopy LinkPrintEmail Title:Management of Increased Intracranial Pressure Category:Neurology Keywords: increased intracranial pressure, opening pressure (PubMed Search) Posted: 12/21/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD When performing a lumbar puncture,anopening intracranial pressure (ICP)greater than 20 to 25 mm of H2O is elevated. If it is thought that a patient's headache is due to elevated pressure,cerebrospinal fluid (CSF) can be therapeutically removed. It is typically recommended that the pressure not be lowered by more than 50%of the amount above which it is normal. The source of elevated ICP should be determined and addressed.Common causes of increased intracranial pressure include: ---Venous drainage obstruction (i.e.cerebral venous sinus thrombosis). --- Endocrine (i.e. obesity, hypothyroidism,Cushing's disease,Addison's disease). --- Medications (i.e.vitamin A, cyclosporine,lithium,lupron, oral contraceptives, amiodorone,and antiobiotics such as tetracyclines and sulfonamides). ---Other conditions (i.e. pregnancy,steroid withdrawal,acromegaly,polycystic ovary syndrome,systemic lupus erythematosus,sleep apnea,HIV). ShareFacebookLinkedInXCopy LinkPrintEmail Title:Botulism Category:Neurology Keywords: botulism, descending paralysis, clostridium botulinum, weakness (PubMed Search) Posted: 12/14/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD Whilebotulism is a rare condition (about 145 reported cases annually), it should still be considered in cases of descending neuromuscular weakness, as it can cause rapid loss of respiratory function and death (mortality < 8%). Check patient's vital capacity. Botulismresults from ingesting (onset of symptoms 6 to 48 hours) or having contamination of a wound(onset 4-14 days; associated with intravenous drug use) with Clostridium b otulinum, an anaerobic, spore-forming bacteria; it has been used as abio-terrorist agentas well. Patients typically present with anticholinergic symptoms and the four "D's" - (1) dry moth, (2) dysarthria, (3) diplopia, and (4) dysphagia. The definitive diagnosis is made by isolating the toxin in serum and/or stool. Treatment is supportive and might include use of equine trivalent anti-toxin and human botulismimmunoglobulin. Antibiotic and anti-cholinergic therapy has not been shown to be particularly effective. Show References Fernandez-Frackelton M. Bacteria. In: Marx J, Hockerberger R, Walls R, ed. Rosen's Emergency Medicine: Concepts and Clinical Practice. Phila: Mosby; 2009:1686-9. Shearer p, Jagoda A. Neuromuscular Disorders. In: Marx J, Hockerberger R, Walls R, ed. Rosen's Emergency Medicine: Concepts and Clinical Practice. Phila: Mosby; 2009:1415. ShareFacebookLinkedInXCopy LinkPrintEmail Title:Micturation Syncope Category:Neurology Keywords: mictuation syncope, syncope, vagus nerve, vasovagal syncope (PubMed Search) Posted: 12/7/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD Micturation syncope is a relatively rare phenomenon (2.4 to 8.4% of fainting episodes) which most commonly affects males, and can often be diagnosed by simply taking a thorough history. Straining to urinatetriggers the vagus nerve which results inhypotension and bradycardia; in turn, cardiac output and brain perfusion is decreased, often resulting in diaphoresis, pallor, and weakness, followed by syncope or fainting. This process istransient and vital signs as well as consciousness typically return to normal rapidly. When evaluating a patient for syncope, pay close attention for the presence of the following factors in order to make the diagnosis: -- occurs during or immediately following urination, often when bladder is full. -- occurs at night or after standing from the recumbent position of a deep sleep to urinate. -- risk factors: enlarged prostate, alpha blocker therapy, dehydration, alcohol, fatigue. Sometimes defecation, coughing, or severe vomiting can also result in syncope. Show References Faintness and syncope. In: Ropper AH, et al. Adams & Victor's Principles of Neurology. 9th ed. New York, N.Y.: The McGraw-Hill Companies; 2009. Seizures and syncope. In: Aminoff MJ, et al. Clinical Neurology. 7th ed. New York, N.Y.: The McGraw-Hill Companies; 2009. ShareFacebookLinkedInXCopy LinkPrintEmail Title:First-time Adult Seizure Management Category:Neurology Keywords: seizure, epilepsy (PubMed Search) Posted: 11/30/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD Seizures occur commonly and it is estimated that 1 of 26 people will develop epilepsy at some point in their life. A first seizure provoked by an acute brain insult is less likely to recur (3-10%) than a first-time unprovoked seizure (30-50% over the next 2 years). As an emergency provider managing an adult who presents with their first-ever seizure, there are four primary questions that require answering: Was it in fact a true seizure? (often associated with tongue biting, urinary/bowel incontinence, preceding aura, post-ictal phase; examples of seizure mimics include syncope (i.e. cardiogenic, neurogenic, vasovagal), vertigo, myoclonic jerking, psychogenic convulsions, movement disorders.) Does the patient have epilepsy? (defined a having at least 2 unprovoked epileptic seizures by any immediately identifiable cause.) What type of epilepsy?(cryptogenic (i.e. of unknown etiology) or symptomatic (i.e. caused by prior central nervous system insult such as brain injury.) What is the cause? (metabolic panels to assess for uremia, electrolyte and glucose abnormalities, and drug intoxications should be performed, as well brain imaging to determine the presence of focal intracranial lesions.) Many patients do not require anticonvulsant medication following a single, first time seizure; A general consensus is that such therapy should be strongly considered for initiation after a second episode of seizure activity. Show References Engel J Jr. Report of the ILAE classification core group. Epilepsia. Sep 2006;47(9):1558-68. Hesdorffer DC, Logroscino G, Benn EK, Katri N, Cascino G, Hauser WA. Estimating risk for developing epilepsy: a population-based study in Rochester, Minnesota. Neurology. Jan 4 2011;76(1):23-7. Pohlmann-Eden B, Beghi E, Camfield C, Camfield P. The first seizure and its management in adults and children. BMJ. Feb 11 2006;332(7537):339-42. ShareFacebookLinkedInXCopy LinkPrintEmail Title:What to tell Bell's palsy patients about their prognosis? Category:Neurology Keywords: bell palsy, bell's palsy (PubMed Search) Posted: 11/23/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD Studies have shown that the natural history of Bell's Palsy without treatmentis such that 85% show signs of recovery within 3 weeksof symptom onset, and71%experience complete recovery. Of the remaining individuals who do not completely recover, 13% experience persistent mild sequelae and 16% have residual weakness, synkinesis, and/or contracture. Those with incomplete lesions (i.e. incomplete paralysis) are more likely to return to normal function (94%), while only 60% of those with clinically complete lesions return to normal function. Herpes zoster is associated with more severe paresis and a worse prognosis. When little to no recovery is seen within the first 21 days following symptom onset, the prognosis is less favorable. Show References Jabor MA, Gianoli G. Management of Bell's Palsy.Journal of LA State Med Soc. 1996; 148(7): 279. Peitersen E. Bell's palsy: the spontaneous course of 2,500 peripheral facial nerve palsies of different etiologies. Acta Otolaryngol Suppl. 2002. ShareFacebookLinkedInXCopy LinkPrintEmail Title:Recognizing and Managing Myasthenia Graves Category:Neurology Keywords: Myasthenia Graves, MG, edrophonium, Tensilon (PubMed Search) Posted: 11/16/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD Myasthenia Graves (MG) is an autoimmune disorder wherein antibodies attack acetylcholine nicotinic postsynaptic receptors at the myoneural junction, resulting in muscle fatigue (commonly bulbar) that worsens with use and improves with rest. MG flares are most commonly due to infection or inadequate treatmentwith cholinesterase inhibitors. The Tensilon (edrophonium) challenge testcan be used to help distinguish an MG crisis from a cholingergic crisis. Once the airway and ventilation are secure, escalating doses of edrophonium (i.e. 1 mg, then 3 mg, then 5 mg, up to a maximum of 10 mg total) can be administered with the goal of relieving the muscle weakness. If a true MG crisis is present, patients usually respond with dramatic improvement within 1 minute. Patients having a cholinergic crisis, on the other hand, typically respond with increased salivation, bronchopulmonary secretions, diaphoresis, and gastric motility. Monitor closely as edrophoniumcan cause significant bradycardia, heart block, and asystole (only 0.16% risk by reports, but have atropine nearby). Once the edrophonium wears off, patients having an MG crisis may develop increased secretions and respiratory distress as their muscle weakness returns, so manage expectantly and with caution. Show References Grob D, Brunner N, Namba T, Pagala M. Lifetime course of myasthenia gravis. Muscle Nerve. Feb 2008;37(2):141-9. Meriggioli MN, Sanders DB. Advances in the diagnosis of neuromuscular junction disorders. Am J Phys Med Rehabil. Aug 2005;84(8):627-38. Pascuzzi RM. Pearls and pitfalls in the diagnosis and management of neuromuscular junction disorders. Semin Neurol. Dec 2001;21(4):425-40. ShareFacebookLinkedInXCopy LinkPrintEmail Title:Treating Lithium Toxicity - To Dialyze or Not? Category:Neurology Keywords: lithium toxicity, hemodialysis, whole bowel irrigation (PubMed Search) Posted: 11/9/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD Remember that lithium overdoses should not be treated with oral activated charcoal, as these charged particles are not adequately absorbed by this method. Instead, whole bowel irrigationusing 500 mL to 2 liters of polyethylene glycol should be administered within the first 2-3 hours of presumed large ingestions(ie. at least 10 to 15 pills), with a goal of having the patient pass stool to the point of clear rectal effluent. Hemodialysis(HD) should be reserved to treat severe lithium toxicity, which is somewhat loosely defined as a serum level greater than 3.5 to 4 meq/L (mmol/L). For levels> 4 meq/L, most experts agree that HD should be performed regardless of whether associated symptoms are present. For levels> 2.5 meq/L with associated clinical signs/symptoms(i.e. tremulousness, dizziness, lethargy, seizure), conditions that would limit lithium excretion (i.e. renal insufficiency), or conditions that would limit ability to aggressively hydrate (i.e. CHF), HD should be performed. Show References Perrone J, Chatterjee P. "Lithium Poisoning." UpToDate. May 2011. Retrieved from: ShareFacebookLinkedInXCopy LinkPrintEmail Title:iPhone Use May Optimize the Care of Acute Stroke Patients Category:Neurology Keywords: stroke, iPhone, NIH Stroke Scale (PubMed Search) Posted: 11/2/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD A hugelimitation to effectively managing acute ischemic stroke in rural areas is the frequent lack of access to local experts in vascular neurology. While most guidelines encourage the use of telemedicine to overcome such barriers, the start up costs of such programs are sometimes prohibitive, particularly for small, rural practices. A recent, small study showed that providers may be able to use the iPhone as a primary or adjunctive tool with telemedicine, to properly diagnose and manage acute stroke. The study compared a face-to-face provider's NIH Stroke Scale (NIHSS) interpretation to that of a remote provider using an iPhone with FaceTime software that allows real-time streaming of audio and video. Agreement between providers was excellent(intraclass correlation coefficient 0.98); the NIHSS score of the providers did not differ by more than 1 point in 17 of the 20 cases; in only one category - ataxia - was agreement poor. TAKE HOME POINT: Streaming real-time video technology may offer an effective and economically feasible alternative to suboptimal acute stroke care in rural areas or an alternative/adjunct to pure telemedicine programs. (This is not an advertisement or endorsement for the iPhone.) Show References Anderson E, et al. Remote Assessment of Stroke Using the iPhone 4. Journal of Stroke and Cerebrovascular Diseases 2011. Neale T. iPhone Puts Stroke Patients, Docs Face to Face. MedPage Today. November 2011. Retrieved from: ShareFacebookLinkedInXCopy LinkPrintEmail Title:Clinical Findings Associated with Myasthenia Graves Category:Neurology Keywords: myasthenia graves, MG (PubMed Search) Posted: 10/26/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD Myasthenia Graves (MG) is often associated with several,distinct clinical findings which patients may have during their crisis in the emergency department. These findings may include the following: - Mask-like face - Eyelid weakness -- leads to ptosis -- exacerbated by sustained upward gaze -- improved by closing the eyes for a short while -Extraocular motion abnormality -- usually affects more than one extraocular muscle -- may be assymetrical -- may result in mild proptosis - Weak palatal muscles -- nasal-sounding voice -- nasal regurgitation of food - Weak jaw muscles - Absent gag reflex - Pupils normal ShareFacebookLinkedInXCopy LinkPrintEmail Title:Differentiating Central Retinal Artery vs. Vein Occlusion Fundoscopically Category:Neurology Keywords: fundoscopic examination, central retinal vein occlusion, central retinal artery occlusion (PubMed Search) Posted: 10/19/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD Differentiating Central Retinal Artery vs. Vein Occlusion Fundoscopically While there are several historical and clinical features that differentiatecentral retinal artery (CRA) occlusion from central retinal vein (CRV) occlusion, thefundoscopic examination can also be used to distinguish between the two. In CRA occlusion, the retina appears grossly swollen and pale, with a prominent fovea that would otherwise be obscured by a normal, pinkish-red background (see attached - Image 1). In CRV occlusion, the disc is massively swollen with splotches of hemorrhage and cotton wool spots diffusely (see attached - Image 2). Attachments 1110191840_Fundoscopic_Images_of_Central_Retinal_Artery_and_Vein_Occlusion.pdf (103 Kb) ShareFacebookLinkedInXCopy LinkPrintEmail Title:Performing Straight Leg Raise Test for Sciatica Category:Neurology Keywords: sciatica, straight leg raise test (PubMed Search) Posted: 10/12/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD The Straight Leg Raise (SLR) test can be used to determine if patient has true sciatica. The patient lies supine with one leg either straight or flexed at the knee with the sole of the foot flat on the stretcher. The other(affected) leg is kept straight and raised up by the examiner. The test is positive when raising the leg between 30 to 70 degrees causes pain to occur and radiate down the leg to at least below the knee, and often all the way down to the great toe (sensitivity 91%, specificity 26%). Sensitivity may improve with dorsi-flexion of the foot while the leg is elevated. The following do NOT indicate a positive test:pain of lower back only, without radiation to below knee; overtly excessive pain behavior; patient contraction of antagonist muscles that limit examiner's testing; tightness of buttock and hamstring muscles; nonspecific complaints. The SLR testcan also be performed with the patient in a sitting position, by stretching the sciatic nerve by extending the knee; the test is positive if pain radiates to below the knee. ShareFacebookLinkedInXCopy LinkPrintEmail Title:Using Corneal Reflex to Help Diagnose Pontine Injury: Clarification Category:Neurology Keywords: pontine stroke, pontine hemorrhage, corneal reflex, miosis, opiate abuse, opiate overdose (PubMed Search) Posted: 10/5/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD A normal corneal reflex is usually ABSENT in the setting of pontine injury, and typically PRESENT in patients presenting with an opiate overdose. ShareFacebookLinkedInXCopy LinkPrintEmail Title:Using Corneal Reflex to Help Diagnose Pontine Injury Category:Neurology Keywords: pontine stroke, pontine hemorrhage, corneal reflex, miosis, opiate abuse, opiate overdose (PubMed Search) Posted: 10/5/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD In patients presenting withbilateral miosis (i.e. pinpoint pupils) of unknown etiology, the astute clinician may consider acute pontine injury, opiate overdose, or medication-related causes as the source. In such cases, one should consider performing the simplecorneal reflex test to evaluate mid and lower pontine function. This test consists oflightly touching the cornea with the cotton swab of a Q-tip and observing blink responses in both eyes. It assesses afferent fifth nerve (sensory) and efferent seventh nerve (motor) function. A normal response is simultaneous (i.e. consensual) eye blinking. An abnormal response may be manifest by midline deviation, followed by relaxation, of the lower eyelids. TAKE HOME POINT: Corneal reflex testing is an easy way to help distinguish pontine injury from an opiate overdose in patients presenting with pinpoint pupils. Confirmatory studies by way of brain imaging should follow. ShareFacebookLinkedInXCopy LinkPrintEmail Title:What is a Marcus Gunn Pupil? Category:Neurology Keywords: marcus gunn pupil, afferent pupillary defect, swinging flashlight test (PubMed Search) Posted: 9/28/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD Marcus Gunn Pupil is a highly objective and significantclinical manifestation of a relative afferent pupillary defect (RAPD), indicating decreased pupillary response. The"swinging flashlight test" whereby a light is shone alternately back and forth between both eyes, can be used to elicit this finding. A normal response results in constriction of both pupils, which indicates intact direct and consensual pupillary light reflex. An abnormal responsedue to RAPD, however, results inminimal constriction of both pupils when the light in shone in the affected eye, causing one to perceive the presence of pupillary dilation. See the attached imagewhich contrasts a normal response (top) to an abnormal response (bottom). Marcus Gunn Pupil is most commonlyassociated with lesions at the level of the optic nerve (proximal to the optic chiasm) or severe retinal disease. Associated conditions include severe glaucoma, optic nerve tumors, and ocular trauma. Attachments 1109281911_Marcus_Gunn_Pupil.jpg (35 Kb) ShareFacebookLinkedInXCopy LinkPrintEmail Title:Using the Glascow Coma Scale (GCS) Category:Neurology Keywords: glascow coma scale, GCS (PubMed Search) Posted: 9/21/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD The Glascow Coma Scale (GCS) was originally derived from data from adult head injuries; its use is thereforenot always applicable or reliable in non-traumatic cases (particularly those which are complex) or children under a certain age. ScoringRange = 3 to 15. Severe is less than 9. Scores of 8 or less should prompt strong consideration for airway management via intubation. Themotor score is the most predictive and clinically useful component. See GCS Score below: EYE OPENING 4 = spontaneous 3 = to voice 2 = to pain 1 = none VERBAL RESPONSE 5 = orientated 4 = confused 3 = inappropriate 2 = incomprehensible 1 = none MOTOR RESPONSE 6 = obeys command 5 = localizes pain 4 = withdraw to pain 3 = decorticate 2 = decerebrate 1 = none ShareFacebookLinkedInXCopy LinkPrintEmail Title:Bell Palsy Category:Neurology Keywords: bell palsy, bell's palsy, cranial nerve seven palsy, facial paralysis (PubMed Search) Posted: 9/14/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD Bell Palsy is a condition associated with the acute onset of facial paralysis due to palsy of the seventh lower motor neuron (cranial nerve seven). The presence ofmastoid pain might be a diagnostic clue, as this symptom often precedes the onset of actual facial paresis. Other associated findings include:typically unilateralfacial muscle paralysisaffecting both the upper and lower parts of the face; tear overflow anddry eyes; altered taste; hyperacusis or sound sensitivity; sensation spared; no other cranial nerves involved. It is often associated with viruses such as HIV, Epstein-Barr, and Hepatitis B, but most commonly herpes simplex. If facial paralysis is bilateral, consider Lyme diseaseas a possible etiology. ShareFacebookLinkedInXCopy LinkPrintEmail Title:Encephalitis Category:Neurology Keywords: encephalitis, meningitis, acyclovir, headache, fever (PubMed Search) Posted: 9/7/2011 by Aisha Liferidge, MD Click here to contact Aisha Liferidge, MD Encephalitis, inflammation of the brain, is associated with the following signs and symptoms: fever, headache, altered mental status, neurologic deficit, hallucinations, behavioral changes, photophobia, seizures, neck stiffness (when associated with meningitis), preceding viral prodrome, recent mosquito/tick/animal bites, and/or immunocompromised state/use of immunosuppressant medications. The presence of focal neurologic deficit and/or altered mental status is more predictive of encephalitis than meningitis. The emergent management goal is to rule out and/or empirically treat bacterial meningitis and other treatable infectious sources such as Herpes Simplex Virus (HSV), Varicella Zoster Virus (VZV), and Cytomegalovirus (CMV); these carry significant mortality and morbidity risks. Remember to have patient's cerebrospinal fluid (CSF) specifically analyzed for etiologies such as these (i.e. via PCR). Treat presumed encephalitis aggressively by adding acyclovir to the antibiotic/steroid regimen administered, particularly when there is altered mental status and/or focal neurologic deficit. ShareFacebookLinkedInXCopy LinkPrintEmail Title:Recognizing Delirium Category:Neurology Keywords: delirium (PubMed Search) Posted: 8/31/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD Delirium is a symptom, not a diagnosis; the astute clinician must seek to discover and treat the underlying source of delirious states. It is a transient cognitive condition associated with decreased attention span and waxing and waning symptoms. Three types: (1) Hyperactive, (2) Hypoactive, (3) Mixed (daytime somnolence, nighttime agitation). In young patients, the cause is commonly due to toxins or trauma, while that for the elderly is typically infection or medication related. Five critical causes of delirium that must be recognized and treated immediately: Hypoxia Hypoglycemia Central nervous System infections Hypertensive encephalopathy Increased intracranial pressure Show References Smith J, Seirafi J. Delirium and Dementia. In: Marx J, Hockerberger R, Walls R, ed. Rosen's Emergency Medicine: Concepts and Clinical Practice. Phila: Mosby-Elsevier; 2006:1645-63. ShareFacebookLinkedInXCopy LinkPrintEmail Title:ROSIER Scale for Emergently Recognizing Stroke Category:Neurology Keywords: ROSIER scale, ischemic stroke (PubMed Search) Posted: 8/24/2011 by Aisha Liferidge, MD (Updated: 9/28/2025) Click here to contact Aisha Liferidge, MD While validated diagnostic tools such as the NIH Stroke Scale are often very helpful, particularly in terms of communicating with Neurologists, there are tools such asthe ROSIER (Recognition of Stroke in the Emergency Room) Scale which is a brief score designed to facilitate expedited diagnostic testing and treatment of stroke in the emergency department. The ROSIER Scale has been found to recognize stroke with 93% sensitivity, 83% specificity, 90% positive predictive value, and 88% negative predictive value. If the totalscore is > 0 (i.e. 1-6), then stroke is likely. If the total score is < or equal to 0, then stroke is unlikely, but can not be completely excluded. See attached ROSIER Scale for details. Show References Nor AM, Davis J, Sen B, et al. (November 2005). The Recognition of Stroke in the Emergency Room (ROSIER) Scale: Development and Validation of a Stroke Recognition Instrument. Lancet Neurology 4(11): 727-34. Attachments 1108241811_ROSIER_Scale_for_Stroke.doc (61 Kb) ShareFacebookLinkedInXCopy LinkPrintEmail Previous 1 2 3 4 5 6 ... 12 Next Download the UMEM Pearls App! Subscribe to daily UMEM Pearls (FREE!) via: Email Mailing List Browse a Pearl Category: Browse a Pearl Author: Search Pearls Contact Us Webmaster Site Index UMB Hotline Our Partners: University of Maryland, Baltimore University of Maryland Medical Center © University of Maryland School of Medicine, 655 W. Baltimore Street, Baltimore MD 21201 ✓ Thanks for sharing! AddToAny More…
2280	https://www.wordreference.com/conj/enverbs.aspx?v=founder	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| regular model: work verbs ending in -e: like work - model verbVariants of the regular models: 1. pass -s, -sh, -x, -o: +e 2. try -y>ie 3. omit -X>-XX 4. die -ie: -ie>y 5. agree -ee: +d Irregular past tense models: 1. cost invar. 2. feed vowel: long>short 3. find i>ou 4. know [o,a]>e 5. mean +t 6. panic -k- 7. pay -ay>aid 8. send -d>-t 9. sing i>a, u 10. show >ed, -n 11. stick a>u, i>u 12. sleep -ee_>-e_t 13. speak >o_e, o_en 14. spell >ed, -t 15. take -ake>-ook, -aken 16. think >-ought 17. wear -ear>-ore, -orn 18. write -i_e>o_e, i_en \| Firefox and Chrome users: install a shortcut (Firefox or Chrome) then type "conj founder" in your address bar for the fastest conjugations. founderIt is conjugated like: work --- \| \| \| \| \| \| \| --- --- --- \| \| infinitive: present participle: past participle: \| (to) founder foundering foundered \| \| \| definition \| in Spanish in French in Italian \| Open All Desktop View Indicative \| presentⓘAlso known as: present simple or simple present \| \| --- \| \| I \| founder \| \| you \| founder \| \| he, she, it \| founders \| \| we \| founder \| \| you \| founder \| \| they \| founder \| \| simple pastⓘAlso known as: past simple or preterit \| \| --- \| \| I \| foundered \| \| you \| foundered \| \| he, she, it \| foundered \| \| we \| foundered \| \| you \| foundered \| \| they \| foundered \| \| future \| \| --- \| \| I \| will founder \| \| you \| will founder \| \| he, she, it \| will founder \| \| we \| will founder \| \| you \| will founder \| \| they \| will founder \| Perfect tenses \| present perfect \| \| --- \| \| I \| have foundered \| \| you \| have foundered \| \| he, she, it \| has foundered \| \| we \| have foundered \| \| you \| have foundered \| \| they \| have foundered \| \| past perfectⓘAlso known as: pluperfect \| \| --- \| \| I \| had foundered \| \| you \| had foundered \| \| he, she, it \| had foundered \| \| we \| had foundered \| \| you \| had foundered \| \| they \| had foundered \| \| future perfect \| \| --- \| \| I \| will have foundered \| \| you \| will have foundered \| \| he, she, it \| will have foundered \| \| we \| will have foundered \| \| you \| will have foundered \| \| they \| will have foundered \| Continuous (progressive) and emphatic tenses \| present continuous \| \| --- \| \| I \| am foundering \| \| you \| are foundering \| \| he, she, it \| is foundering \| \| we \| are foundering \| \| you \| are foundering \| \| they \| are foundering \| \| past continuous \| \| --- \| \| I \| was foundering \| \| you \| were foundering \| \| he, she, it \| was foundering \| \| we \| were foundering \| \| you \| were foundering \| \| they \| were foundering \| \| present emphatic \| \| --- \| \| I \| do founder \| \| you \| do founder \| \| he, she, it \| does founder \| \| we \| do founder \| \| you \| do founder \| \| they \| do founder \| \| past emphatic \| \| --- \| \| I \| did founder \| \| you \| did founder \| \| he, she, it \| did founder \| \| we \| did founder \| \| you \| did founder \| \| they \| did founder \| Compound continuous (progressive) tenses \| present perfect \| \| --- \| \| I \| have been foundering \| \| you \| have been foundering \| \| he, she, it \| has been foundering \| \| we \| have been foundering \| \| you \| have been foundering \| \| they \| have been foundering \| \| past perfect \| \| --- \| \| I \| had been foundering \| \| you \| had been foundering \| \| he, she, it \| had been foundering \| \| we \| had been foundering \| \| you \| had been foundering \| \| they \| had been foundering \| \| future \| \| --- \| \| I \| will be foundering \| \| you \| will be foundering \| \| he, she, it \| will be foundering \| \| we \| will be foundering \| \| you \| will be foundering \| \| they \| will be foundering \| \| future perfect \| \| --- \| \| I \| will have been foundering \| \| you \| will have been foundering \| \| he, she, it \| will have been foundering \| \| we \| will have been foundering \| \| you \| will have been foundering \| \| they \| will have been foundering \| Conditional \| present \| \| --- \| \| I \| would founder \| \| you \| would founder \| \| he, she, it \| would founder \| \| we \| would founder \| \| you \| would founder \| \| they \| would founder \| \| perfectⓘAlso known as: past conditional \| \| --- \| \| I \| would have foundered \| \| you \| would have foundered \| \| he, she, it \| would have foundered \| \| we \| would have foundered \| \| you \| would have foundered \| \| they \| would have foundered \| \| present continuous \| \| --- \| \| I \| would be foundering \| \| you \| would be foundering \| \| he, she, it \| would be foundering \| \| we \| would be foundering \| \| you \| would be foundering \| \| they \| would be foundering \| \| perfect continuous \| \| --- \| \| I \| would have been foundering \| \| you \| would have been foundering \| \| he, she, it \| would have been foundering \| \| we \| would have been foundering \| \| you \| would have been foundering \| \| they \| would have been foundering \| Imperative \| present \| \| --- \| \| \| – \| \| (you) \| founder! \| \| \| – \| \| (we) \| let's founder! \| \| (you) \| founder! \| \| \| – \| Subjunctive \| present \| \| --- \| \| I \| founder \| \| you \| founder \| \| he, she, it \| founder \| \| we \| founder \| \| you \| founder \| \| they \| founder \| \| past \| \| --- \| \| I \| foundered \| \| you \| foundered \| \| he, she, it \| foundered \| \| we \| foundered \| \| you \| foundered \| \| they \| foundered \| \| past perfectⓘAlso known as: pluperfect subjunctive \| \| --- \| \| I \| had foundered \| \| you \| had foundered \| \| he, she, it \| had foundered \| \| we \| had foundered \| \| you \| had foundered \| \| they \| had foundered \| \| future \| \| --- \| \| I \| should founder \| \| you \| should founder \| \| he, she, it \| should founder \| \| we \| should founder \| \| you \| should founder \| \| they \| should founder \| Blue letters in conjugations are irregular forms. (example) Red letters in conjugations are exceptions to the model. (example) Report a problem. \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| regular model: work verbs ending in -e: like work - model verbVariants of the regular models: 1. pass -s, -sh, -x, -o: +e 2. try -y>ie 3. omit -X>-XX 4. die -ie: -ie>y 5. agree -ee: +d Irregular past tense models: 1. cost invar. 2. feed vowel: long>short 3. find i>ou 4. know [o,a]>e 5. mean +t 6. panic -k- 7. pay -ay>aid 8. send -d>-t 9. sing i>a, u 10. show >ed, -n 11. stick a>u, i>u 12. sleep -ee_>-e_t 13. speak >o_e, o_en 14. spell >ed, -t 15. take -ake>-ook, -aken 16. think >-ought 17. wear -ear>-ore, -orn 18. write -i_e>o_e, i_en \| Firefox and Chrome users: install a shortcut (Firefox or Chrome) then type "conj founder" in your address bar for the fastest conjugations. founderIt is conjugated like: work --- \| \| \| \| \| \| \| --- --- --- \| \| infinitive: present participle: past participle: \| (to) founder foundering foundered \| \| \| definition \| in Spanish in French in Italian \| Open All Desktop View Indicative \| presentⓘAlso known as: present simple or simple present \| \| --- \| \| I \| founder \| \| you \| founder \| \| he, she, it \| founders \| \| we \| founder \| \| you \| founder \| \| they \| founder \| \| simple pastⓘAlso known as: past simple or preterit \| \| --- \| \| I \| foundered \| \| you \| foundered \| \| he, she, it \| foundered \| \| we \| foundered \| \| you \| foundered \| \| they \| foundered \| \| future \| \| --- \| \| I \| will founder \| \| you \| will founder \| \| he, she, it \| will founder \| \| we \| will founder \| \| you \| will founder \| \| they \| will founder \| Perfect tenses \| present perfect \| \| --- \| \| I \| have foundered \| \| you \| have foundered \| \| he, she, it \| has foundered \| \| we \| have foundered \| \| you \| have foundered \| \| they \| have foundered \| \| past perfectⓘAlso known as: pluperfect \| \| --- \| \| I \| had foundered \| \| you \| had foundered \| \| he, she, it \| had foundered \| \| we \| had foundered \| \| you \| had foundered \| \| they \| had foundered \| \| future perfect \| \| --- \| \| I \| will have foundered \| \| you \| will have foundered \| \| he, she, it \| will have foundered \| \| we \| will have foundered \| \| you \| will have foundered \| \| they \| will have foundered \| Continuous (progressive) and emphatic tenses \| present continuous \| \| --- \| \| I \| am foundering \| \| you \| are foundering \| \| he, she, it \| is foundering \| \| we \| are foundering \| \| you \| are foundering \| \| they \| are foundering \| \| past continuous \| \| --- \| \| I \| was foundering \| \| you \| were foundering \| \| he, she, it \| was foundering \| \| we \| were foundering \| \| you \| were foundering \| \| they \| were foundering \| \| present emphatic \| \| --- \| \| I \| do founder \| \| you \| do founder \| \| he, she, it \| does founder \| \| we \| do founder \| \| you \| do founder \| \| they \| do founder \| \| past emphatic \| \| --- \| \| I \| did founder \| \| you \| did founder \| \| he, she, it \| did founder \| \| we \| did founder \| \| you \| did founder \| \| they \| did founder \| Compound continuous (progressive) tenses \| present perfect \| \| --- \| \| I \| have been foundering \| \| you \| have been foundering \| \| he, she, it \| has been foundering \| \| we \| have been foundering \| \| you \| have been foundering \| \| they \| have been foundering \| \| past perfect \| \| --- \| \| I \| had been foundering \| \| you \| had been foundering \| \| he, she, it \| had been foundering \| \| we \| had been foundering \| \| you \| had been foundering \| \| they \| had been foundering \| \| future \| \| --- \| \| I \| will be foundering \| \| you \| will be foundering \| \| he, she, it \| will be foundering \| \| we \| will be foundering \| \| you \| will be foundering \| \| they \| will be foundering \| \| future perfect \| \| --- \| \| I \| will have been foundering \| \| you \| will have been foundering \| \| he, she, it \| will have been foundering \| \| we \| will have been foundering \| \| you \| will have been foundering \| \| they \| will have been foundering \| Conditional \| present \| \| --- \| \| I \| would founder \| \| you \| would founder \| \| he, she, it \| would founder \| \| we \| would founder \| \| you \| would founder \| \| they \| would founder \| \| perfectⓘAlso known as: past conditional \| \| --- \| \| I \| would have foundered \| \| you \| would have foundered \| \| he, she, it \| would have foundered \| \| we \| would have foundered \| \| you \| would have foundered \| \| they \| would have foundered \| \| present continuous \| \| --- \| \| I \| would be foundering \| \| you \| would be foundering \| \| he, she, it \| would be foundering \| \| we \| would be foundering \| \| you \| would be foundering \| \| they \| would be foundering \| \| perfect continuous \| \| --- \| \| I \| would have been foundering \| \| you \| would have been foundering \| \| he, she, it \| would have been foundering \| \| we \| would have been foundering \| \| you \| would have been foundering \| \| they \| would have been foundering \| Imperative \| present \| \| --- \| \| \| – \| \| (you) \| founder! \| \| \| – \| \| (we) \| let's founder! \| \| (you) \| founder! \| \| \| – \| Subjunctive \| present \| \| --- \| \| I \| founder \| \| you \| founder \| \| he, she, it \| founder \| \| we \| founder \| \| you \| founder \| \| they \| founder \| \| past \| \| --- \| \| I \| foundered \| \| you \| foundered \| \| he, she, it \| foundered \| \| we \| foundered \| \| you \| foundered \| \| they \| foundered \| \| past perfectⓘAlso known as: pluperfect subjunctive \| \| --- \| \| I \| had foundered \| \| you \| had foundered \| \| he, she, it \| had foundered \| \| we \| had foundered \| \| you \| had foundered \| \| they \| had foundered \| \| future \| \| --- \| \| I \| should founder \| \| you \| should founder \| \| he, she, it \| should founder \| \| we \| should founder \| \| you \| should founder \| \| they \| should founder \| Blue letters in conjugations are irregular forms. (example) Red letters in conjugations are exceptions to the model. (example) Report a problem. \| regular model: work verbs ending in -e: like Variants of the regular models: Irregular past tense models: founder \| \| \| \| \| \| \| --- --- --- \| \| infinitive: present participle: past participle: \| (to) founder foundering foundered \| \| \| definition \| in Spanish in French in Italian \| Indicative \| presentⓘAlso known as: present simple or simple present \| \| --- \| \| I \| founder \| \| you \| founder \| \| he, she, it \| founders \| \| we \| founder \| \| you \| founder \| \| they \| founder \| \| simple pastⓘAlso known as: past simple or preterit \| \| --- \| \| I \| foundered \| \| you \| foundered \| \| he, she, it \| foundered \| \| we \| foundered \| \| you \| foundered \| \| they \| foundered \| \| future \| \| --- \| \| I \| will founder \| \| you \| will founder \| \| he, she, it \| will founder \| \| we \| will founder \| \| you \| will founder \| \| they \| will founder \| Perfect tenses \| present perfect \| \| --- \| \| I \| have foundered \| \| you \| have foundered \| \| he, she, it \| has foundered \| \| we \| have foundered \| \| you \| have foundered \| \| they \| have foundered \| \| past perfectⓘAlso known as: pluperfect \| \| --- \| \| I \| had foundered \| \| you \| had foundered \| \| he, she, it \| had foundered \| \| we \| had foundered \| \| you \| had foundered \| \| they \| had foundered \| \| future perfect \| \| --- \| \| I \| will have foundered \| \| you \| will have foundered \| \| he, she, it \| will have foundered \| \| we \| will have foundered \| \| you \| will have foundered \| \| they \| will have foundered \| Continuous (progressive) and emphatic tenses \| present continuous \| \| --- \| \| I \| am foundering \| \| you \| are foundering \| \| he, she, it \| is foundering \| \| we \| are foundering \| \| you \| are foundering \| \| they \| are foundering \| \| past continuous \| \| --- \| \| I \| was foundering \| \| you \| were foundering \| \| he, she, it \| was foundering \| \| we \| were foundering \| \| you \| were foundering \| \| they \| were foundering \| \| present emphatic \| \| --- \| \| I \| do founder \| \| you \| do founder \| \| he, she, it \| does founder \| \| we \| do founder \| \| you \| do founder \| \| they \| do founder \| \| past emphatic \| \| --- \| \| I \| did founder \| \| you \| did founder \| \| he, she, it \| did founder \| \| we \| did founder \| \| you \| did founder \| \| they \| did founder \| Compound continuous (progressive) tenses \| present perfect \| \| --- \| \| I \| have been foundering \| \| you \| have been foundering \| \| he, she, it \| has been foundering \| \| we \| have been foundering \| \| you \| have been foundering \| \| they \| have been foundering \| \| past perfect \| \| --- \| \| I \| had been foundering \| \| you \| had been foundering \| \| he, she, it \| had been foundering \| \| we \| had been foundering \| \| you \| had been foundering \| \| they \| had been foundering \| \| future \| \| --- \| \| I \| will be foundering \| \| you \| will be foundering \| \| he, she, it \| will be foundering \| \| we \| will be foundering \| \| you \| will be foundering \| \| they \| will be foundering \| \| future perfect \| \| --- \| \| I \| will have been foundering \| \| you \| will have been foundering \| \| he, she, it \| will have been foundering \| \| we \| will have been foundering \| \| you \| will have been foundering \| \| they \| will have been foundering \| Conditional \| present \| \| --- \| \| I \| would founder \| \| you \| would founder \| \| he, she, it \| would founder \| \| we \| would founder \| \| you \| would founder \| \| they \| would founder \| \| perfectⓘAlso known as: past conditional \| \| --- \| \| I \| would have foundered \| \| you \| would have foundered \| \| he, she, it \| would have foundered \| \| we \| would have foundered \| \| you \| would have foundered \| \| they \| would have foundered \| \| present continuous \| \| --- \| \| I \| would be foundering \| \| you \| would be foundering \| \| he, she, it \| would be foundering \| \| we \| would be foundering \| \| you \| would be foundering \| \| they \| would be foundering \| \| perfect continuous \| \| --- \| \| I \| would have been foundering \| \| you \| would have been foundering \| \| he, she, it \| would have been foundering \| \| we \| would have been foundering \| \| you \| would have been foundering \| \| they \| would have been foundering \| Imperative \| present \| \| --- \| \| \| – \| \| (you) \| founder! \| \| \| – \| \| (we) \| let's founder! \| \| (you) \| founder! \| \| \| – \| Subjunctive \| present \| \| --- \| \| I \| founder \| \| you \| founder \| \| he, she, it \| founder \| \| we \| founder \| \| you \| founder \| \| they \| founder \| \| past \| \| --- \| \| I \| foundered \| \| you \| foundered \| \| he, she, it \| foundered \| \| we \| foundered \| \| you \| foundered \| \| they \| foundered \| \| past perfectⓘAlso known as: pluperfect subjunctive \| \| --- \| \| I \| had foundered \| \| you \| had foundered \| \| he, she, it \| had foundered \| \| we \| had foundered \| \| you \| had foundered \| \| they \| had foundered \| \| future \| \| --- \| \| I \| should founder \| \| you \| should founder \| \| he, she, it \| should founder \| \| we \| should founder \| \| you \| should founder \| \| they \| should founder \| Blue letters in conjugations are irregular forms. (example) Red letters in conjugations are exceptions to the model. (example) Report a problem.
2281	https://faculty.bard.edu/cullinan/disccomp.pdf	THE DISCRIMINANT OF A COMPOSITION JOHN CULLINAN This lemma may be useful in the following contexts: a) to compute the discriminant of an iterated rational function, or b) to extract the discriminant of f or g from that of f ◦g. Let k be a ﬁeld and f, g ∈k[x]. Suppose deg(f) = ϵ and deg(g) = δ. Suppose further that neither f nor g has zero discriminant and that Res(f, g) ̸= 0, i.e. f and g have no common roots. Lemma. With all notation as above, the discriminant of the composition f ◦g is given by the formula disc f ◦g = (−1)ϵδ(3ϵδ−2δ−1)/2ℓ(f)δ−1ℓ(g)ϵ(ϵδ−δ−1)(disc f)δ Res(f ◦g, g′), where ℓdenotes the leading term of the polynomial. Proof. The deﬁnition of discriminant in terms of the resultant is gives us disc f ◦g = (−1)( ϵδ 2 )ℓ(f ◦g)−1 Res(f ◦g, (f ′ ◦g)g′). Applying the identity Res(P, QR) = Res(P, Q) Res(P, R), we focus on the resultant Res(f ◦g, (f ′ ◦g)g′): Res(f ◦g, (f ′ ◦g)g′) = (−1)ϵδ((ϵ−1)δ)ℓ(f ′ ◦g)ϵδ Y {θ : f ′(g(θ))=0} f(g(θ)) = (−1)ϵδ((ϵ−1)δ)ℓ(f ′ ◦g)ϵδ   Y {ρ : f ′(ρ)=0} f(ρ)   δ = (−1)ϵδ((ϵ−1)δ)ℓ(f ′ ◦g)ϵδ (ϵℓ(f)−ϵ Res(f, f ′) δ = (−1)ϵδ((ϵ−1)δ)ℓ(f ′ ◦g)ϵδ(ϵℓ(f)−ϵδℓ(f)δ(disc f)δ. Putting this resultant back into the discriminant formula above and working out the leading terms yields the desired formula. □ Department of Mathematics, Bard College, Annandale-On-Hudson, NY 12504 E-mail address: cullinan@bard.edu
2282	https://www.doubtnut.com/qna/642509047	Find the eqution of the curve passing through the point (1,1), if the tangent drawn at any point P(x,y) on the curve meets the coordinate axes at A and B such that P is the mid point of AB. More from this Exercise To find the equation of the curve passing through the point (1, 1) where the tangent at any point P(x,y) meets the coordinate axes at points A and B such that P is the midpoint of AB, we can follow these steps: Step 1: Understand the Geometry Let P(x,y) be a point on the curve. The tangent at this point intersects the x-axis at point A and the y-axis at point B. Since P is the midpoint of AB, we can denote the coordinates of A and B. Step 2: Determine Coordinates of Points A and B Assume the coordinates of point A are (2x,0) and the coordinates of point B are (0,2y). This is because the midpoint P is given by: P=(0+2x2,2y+02)=(x,y) Step 3: Find the Slope of Line AB The slope of the line AB can be calculated using the coordinates of points A and B: slope of AB=0−2y2x−0=−yx Step 4: Relate the Slope to the Derivative Since the slope of the tangent at point P is also given by dydx, we can set: dydx=−yx Step 5: Separate Variables We can rearrange the equation to separate the variables: dyy=−dxx Step 6: Integrate Both Sides Now, we integrate both sides: ∫dyy=−∫dxx This gives us: ln\|y\|=−ln\|x\|+C where C is the constant of integration. Step 7: Simplify the Equation Using properties of logarithms, we can rewrite the equation: ln\|y\|+ln\|x\|=C which simplifies to: ln\|xy\|=C Exponentiating both sides, we get: \|xy\|=eC Let k=eC, thus: xy=k Step 8: Use the Initial Condition We know the curve passes through the point (1,1): 1⋅1=k⟹k=1 Thus, the equation becomes: xy=1 Step 9: Final Form of the Equation Rearranging gives us the final equation of the curve: xy=1 To find the equation of the curve passing through the point (1, 1) where the tangent at any point P(x,y) meets the coordinate axes at points A and B such that P is the midpoint of AB, we can follow these steps: Step 1: Understand the Geometry Let P(x,y) be a point on the curve. The tangent at this point intersects the x-axis at point A and the y-axis at point B. Since P is the midpoint of AB, we can denote the coordinates of A and B. Step 2: Determine Coordinates of Points A and B Assume the coordinates of point A are (2x,0) and the coordinates of point B are (0,2y). This is because the midpoint P is given by: P=(0+2x2,2y+02)=(x,y) Step 3: Find the Slope of Line AB The slope of the line AB can be calculated using the coordinates of points A and B: slope of AB=0−2y2x−0=−yx Step 4: Relate the Slope to the Derivative Since the slope of the tangent at point P is also given by dydx, we can set: dydx=−yx Step 5: Separate Variables We can rearrange the equation to separate the variables: dyy=−dxx Step 6: Integrate Both Sides Now, we integrate both sides: ∫dyy=−∫dxx This gives us: ln\|y\|=−ln\|x\|+C where C is the constant of integration. Step 7: Simplify the Equation Using properties of logarithms, we can rewrite the equation: ln\|y\|+ln\|x\|=C which simplifies to: ln\|xy\|=C Exponentiating both sides, we get: \|xy\|=eC Let k=eC, thus: xy=k Step 8: Use the Initial Condition We know the curve passes through the point (1,1): 1⋅1=k⟹k=1 Thus, the equation becomes: xy=1 Step 9: Final Form of the Equation Rearranging gives us the final equation of the curve: xy=1 Similar Questions The tangent at any point P on the circle x2+y2=4 meets the coordinate axes at AandB . Then find the locus of the midpoint of AB. Find the equation of the curve passing through the point (0, -2) given that at any point (x,y) on the curve the product of the slope of its tangent and y coordinate of the point is equal to the x-coordinate of the point. Find the equation of a curve passing through (1,1) and whose slope of tangent at a point (x, y) is −xy. Find the equation of a curve passing through the point (0,1) if the slope of the tangent to the curve at any point (x,y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point. Find the equation of the curve passing through the point (1,0) if the slope of the tangent to the curve at any point (x,y)isy−1x2+x. Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point. Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point. Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point. Find the equation of the curve through the point (1,0), if the slope of the tangent to the curve at any point (x,y) is y−1x2+x Find the equation of the curve passing through the point, (5,4) if the sum of reciprocal of the intercepts of the normal drawn at any point P(x,y) on it is 1. Recommended Questions Find the eqution of the curve passing through the point (1,1), if the ... A curve y=f(x) passes through point P(1,1) . The normal to the cu... A curve C has the property that if the tangent drawn at any point P... Tangent is drawn at any point P of a curve which passes through (1,1) ... A curve y = f(x) passes through point P(1,1). The normal to curve at p... बिंदु (1, 1) से गुजरने वाले उस वक्र का समीकरण ज्ञात कीजिए जिसके किसी ब... The tangent at any point P to a curve C intersects the coordinate axes... A curve y=f(x) passes through the point P(1,1). The normal to the curv... एक वक्र के किसी बिन्दु P पर एक स्पर्शी खींची जाती है, जोकि (1,1) से ह... Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. 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2283	https://chemspacenk.quora.com/P5-What-is-the-Q-value-of-a-nuclear-reaction	P5 What is the Q value of a nuclear reaction? - ChemSpace@NareshKumar - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In ChemSpace@NareshKumar Chemistry; Biochemistry; Science News/History; Higher Education Follow · 999 107 Avinash Singh Assistant Professor at SRM University Delhi - NCR Sonepat (2020–present) · Updated 2y · P5 What is the Q value of a nuclear reaction? The Q value of a nuclear reaction refers to the energy released or absorbed during that reaction. It is calculated by taking the difference between the total mass of the reactants and the total mass of the products and multiplying it by the square of the speed of light (c) according to Einstein's mass-energy equivalence principle (E=mc^2). Mathematically, the Q value of a nuclear reaction can be expressed as: Q = (Σm[reactants] - Σm[products])c^2 In this equation, Σm[reactants] represents the total mass of the reactants, and Σm[products] represents the total mass of the products. The Q value provides insight into the energetics of a nuclear reaction. If the Q value is positive, it indicates that the reaction is exothermic, meaning it releases energy. Conversely, a negative Q value suggests an endothermic reaction, where energy is absorbed during the reaction. The Q value affects the reaction rate. The larger the positive Q value of a reaction, the faster the reaction proceeds. It can also be calculated by the formula: Q = (mr - mp)931 MeV where mr and mp are the masses of reactants and products in atomic mass unit. The Q value is an essential parameter in nuclear physics. It plays a significant role in various applications, including nuclear power generation, nuclear weapons, and understanding the stability and decay of atomic nuclei. It is analogous to the heat of a reaction or the enthalpy change (ΔH) of a reaction. The difference is for exothermic reaction, ΔH is negative and for endothermic reaction, ΔH is positive. 344 views · View upvotes · View 1 share · Upvote · 9 6 9 1 About the Author Avinash Singh Asst. Professor, Chemistry, PhD Assistant Professor at SRM University Delhi - NCR Sonepat 2020–present PhD from Bhabha Atomic Research Centre (BARC)Graduated 2018 Lives in Mumbai, Maharashtra, India 2013–present 136.6K content views 394 this month Active in 3 Spaces Joined August 2014 View more in ChemSpace@NareshKumar About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2284	https://pmc.ncbi.nlm.nih.gov/articles/PMC12009114/	Disorders of bone and mineral metabolism in pregnancy and lactation: A case based clinical review - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Osteoporos Sarcopenia . 2025 Mar 19;11(1):1–8. doi: 10.1016/j.afos.2025.03.002 Search in PMC Search in PubMed View in NLM Catalog Add to search Disorders of bone and mineral metabolism in pregnancy and lactation: A case based clinical review Manju Chandran Manju Chandran a Osteoporosis and Bone Metabolism Unit, Department of Endocrinology, Singapore General Hospital, Singapore Find articles by Manju Chandran a,⁎, Sarah Ying Tse Tan Sarah Ying Tse Tan b Obesity and Metabolic Unit, Department of Endocrinology, Singapore General Hospital, Singapore Find articles by Sarah Ying Tse Tan b Author information Article notes Copyright and License information a Osteoporosis and Bone Metabolism Unit, Department of Endocrinology, Singapore General Hospital, Singapore b Obesity and Metabolic Unit, Department of Endocrinology, Singapore General Hospital, Singapore ⁎ Corresponding author. Osteoporosis and Bone Metabolism Unit, Department of Endocrinology, Singapore General Hospital, 20 College Road, Academia Singapore 169856, Singapore. manju.chandran@singhealth.com.sg Received 2024 Sep 27; Revised 2025 Mar 1; Accepted 2025 Mar 8; Issue date 2025 Mar. © 2025 The Korean Society of Osteoporosis. Publishing services by Elsevier B.V. This is an open access article under the CC BY-NC-ND license ( PMC Copyright notice PMCID: PMC12009114 PMID: 40256359 Abstract Bone and mineral metabolism in the human body undergoes significant adaptations during pregnancy and lactation to meet the physiological demands of both the mother and fetus. The growing fetus requires approximately 30 g of calcium, with 80% of this transferred from the mother during the third trimester. These adaptations involve complex hormonal changes, such as increased parathyroid hormone-related peptide (PTHrP) and 1,25-dihydroxyvitamin D, ensuring the mother maintains calcium balance despite fetal demands. However, these changes can also exacerbate pre-existing metabolic bone disorders, presenting unique challenges during pregnancy. This narrative review, framed around illustrative case examples, focuses on the management of metabolic bone disorders in pregnancy. Relevant case studies of hypercalcemia, hypocalcemia, hypophosphatemia, and osteoporosis and chronic kidney disease mineral bone disorder are reviewed to illustrate the biochemical changes, clinical implications, and therapeutic strategies available during pregnancy and lactation. We analyze literature from case reports and existing guidelines to provide practical clinical recommendations. The review highlights critical pregnancy-related metabolic adaptations, such as increased intestinal calcium absorption and skeletal resorption. Disorders like primary hyperparathyroidism (PHPT) and familial hypocalciuric hypercalcemia present significant maternal and fetal risks, including miscarriage, growth restriction, and neonatal complications. Early identification and tailored treatment, including hydration, parathyroidectomy, and vitamin D supplementation, mitigate these risks, with surgical interventions in PHPT improving pregnancy outcomes compared to conservative management. Management of metabolic bone disorders during pregnancy and lactation requires a nuanced approach to meet the dual needs of the mother and fetus. Keywords: Pregnancy, Osteoporosis, Primary hyperparathyroidism, Hypocalcemia, Hypophosphatemia 1. Introduction Bone mineral homeostasis that is maintained in the non-pregnant state via the intricate relationship between the various components of the calcium-parathyroid hormone-vitamin D axis, undergoes significant changes to adapt to the physiological needs of pregnancy. The adaptations that occur may also impact on preexisting disorders of bone and mineral metabolism that the woman may have. During pregnancy, the growing fetus depends on maternal calcium and minerals for skeletal development. Throughout the course of pregnancy, the fetus accretes approximately 30 g calcium, 20 g phosphorus and 0.8 g magnesium . Eighty percent of this accretion occurs during the third trimester . This requirement is met through placental transfer from the mother. Several mechanisms allow maternal adaptation to meet this demand. Many hormones including parathyroid hormone-related peptide (PTHrP), estradiol, prolactin, and placental lactogen rise during pregnancy. Increased 1α-hydroxylation of 25(OH)D in the kidneys, results in an elevation of 1,25(OH)D levels. The rise in maternal 1,25(OH)D levels subsequently stimulates increased intestinal absorption of calcium reaching almost 400 mg/day by the third trimester. This is two-fold that of a non-pregnant woman. Secondly, there is also an increase in maternal skeletal resorption during pregnancy. These adaptations allow maternal calcium levels to remain stable despite the increased fetal demands. Though total serum calcium level decreases, due to a fall in serum albumin secondary to hemodilution, the physiologically important fraction, namely ionised calcium, remains stable. Serum phosphate and magnesium concentrations remain normal throughout pregnancy. Serum parathyroid hormone (PTH) concentrations are low particularly in the first trimester, and subsequently increase to the mid-normal range by term. On the other hand, PTHrP, which is produced by breast and placental tissue, rises during pregnancy and lactation. Serum 1,25(OH)D concentrations rise 2 to 5-fold in early pregnancy and remain elevated till delivery. Urinary calcium levels increase secondary to absorptive hypercalciuria, and the 24-h urinary calcium creatinine ratio (UCCR) often exceeds the normal range. This has implications on the use of fractional excretion of calcium for the diagnosis of certain conditions such as familial hypocalciuric hypercalcemia (FHH) in pregnancy. Table 1 summarises the physiologic changes in calcium and mineral metabolism that occur during pregnancy and lactation. Table 1. Physiologic maternal changes in calcium and mineral metabolism in pregnancy and lactation. \| \| Pregnancy \| Lactation \| --- \| Total serum calcium \| ↓ \| ↔ \| \| Ionised calcium \| ↔ \| ↔ \| \| Parathyroid hormone (PTH) \| First trimester: ↓ Third trimester: ↔ \| ↓ \| \| Parathyroid hormone-related peptide (PTHrP) \| ↑ \| ↑ \| \| 1,25(OH)D \| ↑ \| ↔ \| \| Serum Phosphate \| ↔ \| ↔ \| \| Serum Magnesium \| ↔ \| ↔ \| Open in a new tab ↑: increase; ↔: remain the same; ↓: decrease. Fetal calcium homeostasis relies heavily on maternal calcium supply. Fetal parathyroid development begins at the 6th week of gestation, but fetal PTH concentrations remain low. Instead, maternal PTHrP drives an active influx of maternal calcium via the placenta. Fetal calcium concentrations rise throughout pregnancy, reaching a peak of 2.5–2.75 mmol/L in the third trimester. Post-delivery, neonatal calcium levels fall, reaching a physiologic nadir at day two. This is accompanied by a rise in neonatal PTH, which then maintains calcium homeostasis in the newborn. The management of mineral and bone metabolic problems in pregnancy poses a challenge to physicians. In this review, we aim to provide guidance on the management of these disorders in pregnancy. The cases presented in this review are not derived from actual patient records but are constructed based on the authors' clinical experience managing similar presentations. They are designed to illustrate key diagnostic and therapeutic considerations in a manner that aligns with real-world clinical practice. 2. Hypercalcemia in pregnancy 2.1. Case example A 36-year-old lady (G3P1) was referred for hypercalcemia (serum calcium 3.01 mmol/L, reference 2.06–2.46 mmol/L) at 14 weeks’ gestation. An ultrasound performed as part of a health screening package three years ago had shown medullary nephrocalcinosis. Work-up then had revealed a PTH-dependent hypercalcemia with serum adjusted for albumin calcium 2.61 mmol/L (Normal Reference (NR):2.15–2.58 mmol/L) and intact PTH (iPTH) 9.9 pmol/L (NR: 0.8–6.8 pmol/L). Urine calcium creatinine clearance ratio was 0.0203. A diagnosis of PHPT had been made. A parathyroid sestamibi scan done at that time had reported subtle delayed tracer washout over the inferior poles of the bilateral thyroid beds and a concurrent ultrasound had reported hypoechoic extrathyroidal lesions posterior to the inferior poles of both thyroid lobes. She however had declined surgery at that point as she was asymptomatic and calcium levels were stable. At current consultation, she reported constipation but was otherwise asymptomatic. She revealed that she had suffered a miscarriage 2 years ago when she was 10 weeks pregnant. There was no family history of calcium disorders or any endocrinopathy. Physical examination was unremarkable. Serum ionised calcium was 1.62 mmol/L (NR: 1.18–1.37 mmol/L) and iPTH 8.6 pmol/L (NR: 0.9–6.2 pmol/L). A repeat neck ultrasound was not able to visualize the bilateral inferior parathyroid adenomas reported on the earlier scan. Genetic testing was recommended, but the patient declined. 2.2. Clinical questions 1. What are the maternal and fetal consequences of hypercalcemia during pregnancy? 2. How should hypercalcemia in pregnancy be evaluated? 3. What are the management options for primary hyperparathyroidism in pregnancy, including pharmacological options, and what do we know about the safety and optimum timing of parathyroidectomy in pregnancy? 2.3. Discussion 2.3.1. Primary hyperparathyroidism The maternal adaptations in calcium metabolism, such as increased intestinal calcium absorption and increased skeletal resorption, may lead to worsening hypercalcemia in pregnancy. Factors such as physical inactivity and bedrest in late pregnancy also contribute to increased skeletal resorption, further aggravating hypercalcemia. A large primary hyperparathyroidism (PHPT) database reported by Pal et al. described that out of 386 women with PHPT, 8 had gestational PHPT (2.1%). Other smaller available series describe an incidence of < 1% [3,4]. However, the true incidence of hypercalcemia in pregnancy is likely to be underestimated, for several reasons. Firstly, calcium levels are not measured in routine or even complicated pregnancies. Moreover, hypercalcemia may be missed if only total serum calcium levels are checked, due to the fall in serum albumin concentrations. Symptoms of hypercalcemia such as nausea, constipation, polyuria and fatigue are non-specific and difficult to distinguish from those of normal pregnancy. In fact, some studies suggest that the majority of cases of primary hyperparathyroidism in pregnancy may go undiagnosed, with miscarriage occurring even before the diagnosis is made . Gestational hypercalcemia is associated with significant maternal and fetal morbidity and mortality. Pregnancy can exacerbate existing hypercalcemia, resulting in severe hypercalcemia or even precipitate a hypercalcemic crisis. Mothers are also at higher risk of complications such as hyperemesis, nephrolithiasis, pancreatitis, and fractures. The risk of pregnancy-related complications such as pre-eclampsia and eclampsia are also increased. In the fetus, hypercalcemia is associated with an increased risk of miscarriage, growth restriction and premature birth . Post-delivery, neonatal hypocalcemia can occur due to failure of neonatal parathyroid glands to upregulate following prolonged suppression in-utero. This can be associated with severe neonatal complications such as arrhythmias, laryngospasm, seizures, and even death . While most cases of neonatal hypocalcemia are transient and resolve by 3–5 months, cases of permanent hypoparathyroidism have also been reported [7,8]. Primary hyperparathyroidism is the most common cause of hypercalcemia in pregnancy . As in non-pregnant individuals, this is most commonly due to a single parathyroid adenoma. However, given the younger age of presentation, familial causes of primary hyperparathyroidism or parathyroid hyperplasia should be considered, including multiple endocrine neoplasia (MEN) 1, MEN-2A, and hyperparathyroidism jaw tumour syndrome . Clinical evaluation for associated symptoms and significant family history should be undertaken. Conventional imaging options for localisation of primary hyperparathyroidism in pregnancy are limited. 99m Technetium-Sestamibi scintigraphy is not recommended, due to the potential radiation risk. Likewise, computed tomography and positron emission tomography (PET) scans should be avoided . Thus, parathyroid ultrasound remains the first-line imaging modality for pregnant women with primary hyperparathyroidism. While ultrasound is easily accessible and poses no radiation risk, its sensitivity is limited in cases of small adenomas, parathyroid hyperplasia, ectopic parathyroid glands, or concomitant thyroid disease . It is also operator-dependent. Reported sensitivities range from 16% (for double adenomas) to 79% (for single adenomas) . Another option that is feasible in pregnancy is needle aspiration for PTH, which has been described in case reports . Lastly, genetic testing for the conditions previously described should also be considered, particularly in the following scenarios: young age, multi-gland disease, parathyroid cancer, clinical features of MEN syndrome or jaw-tumour syndrome, and if there is a positive family history . Treatment of PHPT in pregnancy often presents a clinical conundrum. Most of the medications used for non-pregnant individuals with PHPT have safety concerns in pregnancy. Subcutaneous calcitonin is a category B medication in pregnancy: it does not cross the placenta and can be used safely to suppress bone resorption and promote urinary calcium excretion . Cinacalcet, a calcimimetic, is a category C medication. While its use has been described anecdotally in case reports, a potential risk of neonatal hypocalcemia due to suppression of the fetal parathyroids exists [16,17]. Similarly, bisphosphonates are also category C medications. Since bisphosphonates are retained in the skeleton for a long time, the fetus may be exposed even if the use is stopped before pregnancy. A recent systematic review of 13 studies including 108 pregnancies of women exposed to bisphosphonates before or during pregnancy described cases of "spontaneous abortion (N=6), congenital malformations (N=4), hypocalcemia (N=4), preterm birth (N=3), and low birth weight (N=3) . However, due to the data being largely retrospective and the lack of control groups, a direct causal relationship between bisphosphonate use and adverse fetal outcomes cannot be ascertained from this review. There is also a lack of studies detailing the long-term consequences of in-utero bisphosphonate exposure on future skeletal health of offspring . Denosumab is a category D medication and should not be used in pregnancy, as animal studies have described an increased risk of miscarriage and an osteopetrotic-like disorder in offspring . Thus, hydration remains the only long-term therapeutic option for control of hypercalcemia in pregnancy. Parathyroidectomy in pregnancy also has several challenges and considerations. Miscarriage risk with general anesthesia is highest during the first trimester . Thus, the optimal timing for parathyroidectomy is during the early second trimester, with close control of hypercalcemia using hydration prior to surgery. Perioperatively, a multi-disciplinary team management for close maternal and fetal monitoring is required. In terms of surgical considerations, given the fact that options for pre-operative localisation imaging are limited, patients may require a bilateral neck exploration if a single adenoma was not confidently identified on imaging. In a case series of patients with PHPT in pregnancy, 13 of 15 patients who underwent parathyroidectomy in pregnancy required a bilateral neck exploration . The two who underwent targeted surgery had sestamibi scans done prior to pregnancy. Intraoperative iPTH measurement is also important to ensure that the offending gland or glands have been removed . Two large case series have reviewed the impact of surgical versus conservative treatment on pregnancy outcomes for gestational PHPT. The first, conducted by Norman et al. in a specialist parathyroid clinic, looked at 32 women with PHPT in pregnancy . This included a total of 77 pregnancies (mean maternal calcium level 2.84 mmol/L), of which 62 were conservatively managed. Almost half of the pregnancies that were conservatively treated resulted in miscarriage, with a rate 3.5-fold higher than that of the general population. The risk of miscarriage correlated with the extent of maternal hypercalcemia. In patients with calcium levels ≥ 2.85 mmol/L, fetal death occurred more commonly than live birth. Pregnancy loss occurred between 7 and 23 weeks’ gestation, with most miscarriages occurring from week 8–13. On the other hand, outcomes were better amongst the 15 patients who underwent parathyroidectomy in the second trimester. PHPT was cured, and all 15 had uneventful deliveries of healthy term infants. Thus, this study showed that amongst women with PHPT in pregnancy, higher maternal calcium levels were associated with greater risk of miscarriage, and that parathyroidectomy resulted in better outcomes compared to conservative treatment. The second study, conducted by Hirsch et al. , took place in a community setting. It reviewed data of women 20–40 years old with screening calcium levels done. This included 74 patients with PHPT in pregnancy, with 124 pregnancies during the study period. Outcomes were compared to a control group of pregnant women with normocalcemia. The extent of hypercalcemia in this cohort was mild, with a mean calcium level of 2.7 mmol/L, and only 19% with levels above 2.74 mmol/L. The majority of these patients were conservatively treated. There was no increased rate of pregnancy complications or miscarriage compared to controls. Thus, this study illustrated that there is a role for conservative treatment of PHPT in pregnancy in women with mild hypercalcemia. Putting the findings of these studies together, it would be prudent to recommend that surgical management of PHPT be considered in women whose calcium levels are ≥ 2.85 mmol/L, or if they are ≥ 2.75 mmol/L with prior unexplained pregnancy loss. Parathyroidectomy should ideally be performed in the early second trimester, as most miscarriages occur at 10–15 weeks’ gestation. On the other hand, there is a role for conservative treatment in women with mild hypercalcemia ie, ≤ 2.7 mmol/L . Ultimately, treatment should be individualized based on the extent of hypercalcemia, overall pregnancy risk, and patient preference. The challenges of management of PHPT in pregnancy emphasize the importance of pre-conception counselling in women of childbearing age with pre-existing PHPT. In Norman's study, 72% of the women had at least one pregnancy loss before entering the study . Many in fact had known hypercalcemia, but it was not pursued until they became pregnant again. Thus, women with known PHPT who are planning for pregnancy should undergo pre-conception counselling, and definitive treatment should be considered prior to pregnancy. 2.3.2. Familial hypocalciuric hypercalcemia Familial hypocalciuric hypercalcemia (FHH) is another important cause of hypercalcemia in pregnancy. FHH represents a group of autosomal dominant, heterozygous disorders, caused by dysfunction of the calcium sensing receptor and signalling proteins . As in non-pregnant individuals, FHH in pregnancy is typically chronic and asymptomatic. However, a challenge may arise when it comes to differentiating FHH from PHPT in pregnancy. The absorptive hypercalciuria of pregnancy can lead to an increase in 24h urinary calcium creatinine ratios (UCCR) in patients with FHH, resulting in indeterminate levels ranging from 0.01 to 0.02 [26,27]. Thus, genetic testing should be considered early if UCCR is indeterminate and makes it difficult biochemically to differentiate FHH from PHPT. No maternal adverse outcomes have been reported in women with FHH in pregnancy. However, fetal outcomes would depend on whether the fetus is genetically affected or not. If the fetus is genetically concordant (heterozygous), no adverse outcomes will be seen . However, if the fetus is unaffected, the elevated maternal calcium levels may lead to suppression of fetal parathyroid glands in-utero, giving rise to neonatal hypocalcemia. The final rare scenario is if the fetus has a homozygous mutation. This would result in neonatal severe hyperparathyroidism, which can be fatal . Thus, post-delivery, neonates require close monitoring of their calcium levels. Prenatal testing is not routinely required as FHH is a benign condition but should be considered if both parents have FHH in view of the risk of neonatal severe hyperparathyroidism described above. 3. Hypocalcemia in pregnancy 3.1. Case example A 28-year-old primigravida presented at 28 weeks' gestation with contractions and impending pre-term delivery. She also reported intermittent tingling, cramps, and numbness in her fingers and peri-oral region. These symptoms were present prior to pregnancy but were mild and had worsened during her pregnancy. She had a history of a previous total thyroidectomy for Graves’ disease three years ago, for which she was on levothyroxine 100 μg daily. Laboratory investigations showed hypocalcemia with a serum albumin adjusted calcium of 1.69 mmol/L (NR: 2.06–2.46 mmol/L), hyperphosphatemia (serum phosphate: 2 mmol/L, NR: 0.9–1.5 mmol/L), and hypoparathyroidism (iPTH: 0.5 pmol/L, NR: 1.0–6.3 pmol/L). Her electrocardiogram showed a prolonged QTc of 560 ms. 3.2. Clinical questions 1. What medications are safe in the treatment of hypocalcemia in pregnancy? 2. What is the target calcium level in pregnancy? 3. How should maternal and fetal calcium levels be monitored post-delivery? 3.3. Discussion Adequate calcium during pregnancy is essential for the development of the endochondral skeleton . PTHrP plays an important role in calcium metabolism during pregnancy. It stimulates the preferential transfer of calcium across the placenta, even in the presence of inadequate maternal calcium . However, when maternal calcium levels are severely low, calcium levels will be insufficient to meet fetal requirements. This then leads to stimulation of the fetal parathyroid glands (secondary hyperparathyroidism), which causes demineralisation of the fetal skeleton, fractures, and low birth weight [31,32]. Hypocalcemia also poses a risk for miscarriage as it increases uterine irritability. Women with known hypoparathyroidism have varying courses during pregnancy. This is postulated to be due to the variation in PTHrP levels and oral calcium intake. Some women may see a substantial improvement in hypoparathyroidism, due to the effects of PTHrP and increased intestinal absorption. On the other hand, some may develop worsening hypocalcemia due to the increased fetal demand. Lastly, some may have no change as the increased fetal demand is balanced by the increase in PTHrP. In view of the varying response, close monitoring is recommended in pregnancy. Ionised or albumin adjusted calcium levels should be monitored every 2–4 weeks . Calcium in the low-normal range should be targeted, with careful avoidance of hypercalcemia . Calcium, calcitriol, and vitamin D supplements are all safe in pregnancy. However, thiazides and recombinant PTH are category C medications during pregnancy . In the post-partum period, close monitoring of maternal calcium is imperative, especially in the first few days after delivery. In the first two days post-partum, patients may have transient hypocalcemia due to sudden loss of placental PTHrP. Calcium levels then rise during lactation due to PTHrP production from lactating breasts; the extent of the rise in calcium levels depends on the intensity and exclusivity of breastfeeding . During this period, it is important to monitor for hypercalcemia, and supplements may need to be reduced or stopped. Finally, during weaning, calcium levels may fall again, requiring further adjustment of medications. If there are no plans to nurse post-delivery, medications can be decreased to the pre-pregnancy doses . 4. Hypophosphatemia in pregnancy 4.1. Case example A 30-year-old lady with X-linked hypophosphatemic rickets (XLH) consults regarding her desire to conceive. She was diagnosed at the age of 14 months and has been stable on supplemental phosphate (1 g of elemental phosphate in 3 divided doses), calcitriol 0.5 μg twice daily, and cholecalciferol 1000 unit daily. She has no other medical history. She enquires about the safety of her current medications in pregnancy, as well as the risk of her offspring inheriting XLH. 4.2. Clinical questions 1. How should hypophosphatemia in pregnancy be managed? 2. Which pharmacological options are safe in pregnancy? 4.3. Discussion Phosphate levels typically remain stable in pregnancy and the causes of hypophosphatemia in pregnancy are similar to that in a non-pregnant individual. However, there are several pertinent conditions that call for attention amongst pregnant women. In the first trimester, hypophosphatemia may be seen in women with hyperemesis gravidarum who are managed with parenteral supplementation and who develop refeeding syndrome, resulting in an intracellular shift of phosphate . Cases of severe hypophosphatemia complicated by rhabdomyolysis and hemolysis have been reported in women with hyperemesis gravidarum [36,37]. Thus, close monitoring of phosphate levels is imperative in patients recovering from hyperemesis gravidarum. An important cause of hypophosphatemia in pregnancy is XLH. As the most common congenital cause of hypophosphatemia, it is not infrequently encountered in women of childbearing age. It is caused by mutations in the phosphate regulating endopeptidase X-linked (PHEX) gene, which encodes a cell-surface-bound protein-cleavage enzyme predominantly expressed in osteoblasts, osteocytes, and teeth . The loss of PHEX function results in enhanced secretion of the phosphaturic hormone fibroblast growth factor (FGF)-23 and a reduction in calcitriol levels, leading to renal phosphate wasting and hypophosphatemia . Given the X-linked dominant inheritance pattern and X chromosome inactivation (lyonization) where one X chromosome is randomly turned off, females with XLH generally have been thought to exhibit a milder phenotype compared to males, with less pronounced skeletal complications though this does not seem to have been confirmed in recent literature [40,41]. Nonetheless, close biochemical monitoring is required for women with XLH in pregnancy. Phosphate and calcitriol supplementation are safe in pregnancy, and doses may need to be increased . Subcutaneous burosumab, a monoclonal antibody to FGF-23, has limited data in pregnancy. Reproductive toxicity has been demonstrated in animal studies, and thus its use is not recommended. For offspring who inherit XLH, serum phosphate levels may be normal within the first 3–4 months after birth, but close monitoring is required, to allow for intervention before the development of inevitable rachitic skeletal abnormalities . Offspring should also undergo PHEX mutation genetic testing since an affected female would pass the pathogenic variant to 50% of her offspring. An important acquired cause of hypophosphatemia in pregnancy is that secondary to the use of Ferric carboxymaltose (FCM) injections. FCM increases circulating concentrations of bioactive FGF-23, leading to increased urinary excretion of phosphate, and a fall in serum phosphate levels. Two randomised controlled trials looked at the effects of FCM compared to iron isomaltoside on phosphate levels in iron deficiency anemia. Both studies demonstrated that FCM was associated with a much higher incidence of hypophosphatemia, occurring in up to 75% in the first 35 days, with a mean nadir phosphate level of 0.6 mmol/L 14 days after receiving the injection . In contrast, the incidence of hypophosphatemia with iron isomaltoside was much lower, at 7%–8%. Although these studies excluded pregnant women, FCM has been used in pregnant women with iron deficiency anemia . It is therefore important to be cognizant about the high incidence of hypophosphatemia with FCM, and to use it with caution in pregnant women. 5. Osteoporosis in pregnancy and lactation 5.1. Case example A 35-year-old lady presented with persistent lower back pain that had occurred spontaneously during the third trimester of her first pregnancy. It had been treated conservatively at the time. Her pregnancy and delivery had been uneventful, and she breastfed her child post-delivery. However, the lower back pain persisted for more than a year after delivery. There had been no preceding trauma or falls, and the pain was not associated with lower limb weakness or numbness. She had a medical history significant for mild asthma that was managed with beclomethasone (steroid) inhaler. She also had lactose intolerance and had a low dietary calcium intake of 229 mg per day. She was a non-smoker. There was no family history of osteoporosis or metabolic bone disorders. Biochemical investigations including calcium, phosphate, iPTH, 25(OH)D, kidney function, blood count and thyroid function test were normal. However, a lumbar spine x-ray revealed an L1 compression fracture. She was also found to have low bone mineral density with a Z-score of −3.2 at the lumbar spine, −2.9 at the neck of femur, and −2.3 at the total hip. 5.2. Clinical questions 1. What are the reasons for osteoporotic fractures in pregnancy? 2. What form of treatment is recommended and safe in pregnancy? 3. What are the risks of further osteoporotic fractures during subsequent pregnancies? 5.3. Discussion There are several factors that may lead to a decrease in bone mineral density (BMD) during pregnancy and lactation. Firstly, there are increased fetal and neonatal calcium demands during pregnancy (with an average of 30 g of calcium accreted by the fetus during pregnancy) and lactation (daily loss of 280–400 mg of calcium) . With these increasing demands, there may be insufficient dietary calcium or vitamin D intake in women who avoid dairy or are lactose intolerant. Secondly, the rise in PTHrP concentrations in pregnancy and lactation may be excessive in some women, thus worsening skeletal resorption . Lastly, mechanical factors such as increased weight-bearing of pregnancy, lordotic posture, and carrying the child post-partum also place stress on the spine. Other important pre-existing risk factors include low body mass index, conditions causing low estrogen levels (such as anorexia nervosa), endocrinological conditions (such as hyperthyroidism), malabsorptive gastrointestinal disorders, bone or connective tissue disorders, smoking, medications that may induce bone loss (such as glucocorticoids or anti-epileptics), and a family history of osteoporosis and the use of medications that may induce bone loss such as anti-epileptics and glucocorticoids [45,46]. A 27% increase in the relative risk of osteoporotic fractures has been found in patients using even inhalational corticosteroids long-term . Women with these underlying risk factors therefore require closer monitoring and tailored management before, during, and after pregnancy to reduce the risk of further bone loss during pregnancy and lactation. In view of radiation exposure concerns, very few studies have been conducted to estimate BMD changes during normal pregnancy. Most existing studies report a decline in BMD of about 5% during pregnancy, particularly in the lumbar spine . During lactation, studies describe a fall in bone mineral content at trabecular sites by 3%–10% after 2–6 months of lactation, with smaller losses at cortical sites . This rate of bone loss exceeds even that occurring right after menopause (1%–3%) . These bone density losses of lactation are usually substantially reversed within 6 months of cessation of lactation. This is irrespective of how much bone density was lost initially, and parity and lactation do not appear to increase the long-term risk of osteoporosis or fractures . Fragility fractures during pregnancy and lactation are rare, but if they do occur, vertebral fractures are the ones that most commonly do. Their incidence may be under-recognised and under-reported, given the frequency of back pain in pregnancy. Vertebral fractures typically occur during a first pregnancy, during the third trimester, or first few months in the post-partum period . Most of these women do not have a previous history of bone or mineral metabolism disorders. At diagnosis, BMD is low, but this improves spontaneously 6–12 months after cessation of lactation. Previous case series described a low risk of recurrence with future pregnancies . Another entity distinct from osteoporosis in pregnancy is that of the likely misnamed transient osteoporosis of the hip. This is a rare condition causing skeletal fragility localised to one or both hips but is not associated with systemic skeletal resorption . Hip BMD is usually low and out of keeping with the lumbar spine measurement. Transient osteoporosis of the hip in pregnancy typically occurs in the third trimester and is postulated to be associated with risk factors such as femoral venous stasis, fetal pressure on the obturator nerve, reduced activity and bedrest. Women may present with hip pain, limping, or even hip fractures. Magnetic resonance imaging (MRI) of the hip reveals bone marrow edema of the femoral head and neck, commonly associated with a joint effusion . The MRI and BMD changes typically resolve spontaneously within 3–12 months post-partum. The overall treatment strategy for osteoporosis in pregnancy and lactation is that of watchful waiting, as spontaneous recovery of BMD does occur even in women who have suffered fractures . However, several supportive measures are important. Nutrition, calcium, and vitamin D intake should be optimised, with a recommend a dietary intake of 1200 mg/day of calcium. While the role of vitamin D in calcium metabolism and bone health is well-documented, the association between this micronutrient and pregnancy outcomes is still debated and the desirable 25(OH)D levels that should be maintained during pregnancy are still unclear. An intake of 600 IU/day of vitamin D as recommended by the WHO appears to be prudent [59,60]. What impact osteoporosis during pregnancy has on fetal, neonatal and childhood bone health has not been studied. A meta-analysis of three trials suggests moderate-to high-dose vitamin D supplementation in pregnancy might increase offspring BMD in early childhood, but further trials are required to confirm this finding . Pharmacologic therapy of pregnancy and lactation associated osteoporosis should be reserved for severe or recalcitrant cases and should be delayed for 12–18 months until the extent of spontaneous recovery has been determined. Case series have described the successful use of bisphosphonate treatment, with improvements of BMD at the spine (by 10.2%–23.0%) and femoral neck (by 2.6%–7.5%) at 1–3 years [55,62]. However, the concern about bisphosphonates is its accumulation in bone, possibly leading to fetal exposure in subsequent pregnancies . The optimal duration of treatment is also unknown. Denosumab therapy for osteoporosis of pregnancy and lactation is limited to case reports. Its use during pregnancy is not recommended due to limited data in humans. Reproductive toxicity including increased mortality and growth impairment have been described in animal studies . Available case reports describe its use in the early post-partum period, with an improvement of BMD at the spine (16.5%–32%) at 12–18 months . However, duration of therapy and an appropriate exit strategy need to be carefully planned given the risk of rebound vertebral fractures upon cessation of therapy . Teriparatide therapy for patients with pregnancy and lactation-associated osteoporosis has been described in retrospective cohort studies and case reports. A multi-center retrospective cohort study comparing women treated with teriparatide with calcium and vitamin D (N=19) versus those with calcium and vitamin D only (N=8) reported that those in the teriparatide treatment arm achieved a greater increase in lumbar spine BMD at 24 months (32.9±13.4% vs 12.2±4.2%) . These findings were similarly reflected in other cohort studies and case reports . Importantly, another study described that the use of sequential therapy with anti-resorptives after Teriparatide treatment for women with pregnancy and lactation-associated osteoporosis did not affect BMD gains . Thus, Teriparatide is a promising option for the treatment of pregnancy and lactation-associated osteoporosis. However, further randomised controlled data would be required to determine its efficacy and safety in this population. Table 2 summarises the current evidence, benefits, and risks of osteoporosis pharmacotherapy for osteoporosis in pregnancy and lactation. Table 2. Current evidence and risks of osteoporosis pharmacotherapy in pregnancy and lactation. \| Medication \| Evidence \| Benefits \| Risks \| --- --- \| \| Bisphosphonates \| Case series \| • Improvement of BMD at the spine (10.2%–23.0%) and femoral neck (2.6%–7.5%) at 1–3 years. \| • Fetal exposure in subsequent pregnancies due to accumulation in bone. • Reports of spontaneous abortion, congenital malformations, neonatal hypocalcemia; although no clear association described. • Long-term consequences on growth and skeletal development unknown. \| \| Denosumab \| Case reports \| • Improvement of BMD at spine (16.5%–32%) at 12–18 months. \| • Maternal: drop in spine BMD and risk of rebound vertebral fractures following cessation of therapy. • Fetal: unknown effects; increased mortality and impaired growth reported in animal studies. \| \| Teriparatide \| Retrospective cohort studies and case reports \| • Improvement of BMD at the spine (7.4%–36%) at 18 months. \| • None reported; but lack of long-term and randomised data. \| Open in a new tab In addition to the above strategies, the duration of breastfeeding post-partum should be balanced against the risk of progressive BMD loss and fracture risk. 6. Chronic kidney disease-mineral bone disorders (CKD-MBD) in pregnancy 6.1. Case example A 32-year-old woman (G2P1) with stage 3a chronic kidney disease (CKD) due to IgA nephropathy presents at 10 weeks' gestation for prenatal care. She was diagnosed 5 years ago and has been followed by a nephrologist for mild proteinuria (800 mg/day) and stable kidney function (eGFR ∼50 mL/min/1.73 m 2 pre-pregnancy). She has a history of secondary hyperparathyroidism (SHPT) with chronically elevated PTH and mild hyperphosphatemia. She was on calcitriol 0.5 μg daily and calcium carbonate 1g three times a day before pregnancy to maintain calcium homeostasis. Due to persistent hyperphosphatemia, sevelamer 800 mg three times a day had been started six months prior but was discontinued after confirmation of pregnancy on her nephrologist's advice. She had also been on a low-phosphate diet. At her first prenatal visit, she reports fatigue and occasional muscle cramps. She has no history of fractures. She is concerned about the effects of her kidney disease on her pregnancy and her baby’s skeletal development. Laboratory Investigations at 10 weeks’ gestation showed serum calcium of 2.06 mmol/L (NR: 2.06–2.46 mmol/L), serum phosphate 1.89 mmol/L (NR: 0.9–1.51 mmol/L), PTH 24.8 pmol/L (NR: 1.0–6.3 pmol/L), 25(OH)D 22 ng/mL, Serum creatinine 123 mmol/L (eGFR: 49 mL/min/1.73 m 2, and urinary protein/creatinine ratio 850 mg/g. She is currently taking prenatal vitamins with 800 IU vitamin D3 daily, along with calcium carbonate and calcitriol under nephrology guidance. 6.2. Clinical questions 1. What are the maternal and fetal risks associated with CKD-MBD in pregnancy? 2. How should secondary hyperparathyroidism and mineral imbalances be managed in pregnant women with CKD? 3. Which phosphate binders and vitamin D analogs are safe during pregnancy? 4. How should maternal and fetal bone health be monitored in CKD-MBD? 6.3. Discussion Women with pre-existing CKD face increased maternal and fetal risks in pregnancy, with the risk increasing with increasing degrees of kidney dysfunction . Pregnancy results in several alterations to renal physiology, such as changes in glomerular and tubular function. This can lead to declining renal function during pregnancy, especially in women with moderate to severe CKD, or concomitant hypertension and/or diabetes . These changes in renal physiology during pregnancy may negatively impact maternal calcium and phosphate balance. Maternal risks include worsening kidney function due to increased glomerular filtration during pregnancy, hyperphosphatemia-related vascular calcification and cardiovascular risks, exacerbation of secondary hyperparathyroidism (SHPT), potentially leading to bone resorption and increased fracture risks, hypertension, and preeclampsia. Fetal risks include impaired skeletal mineralization due to maternal hyperphosphatemia and elevated PTH, increased risk of intrauterine growth restriction (IUGR), neonatal hypocalcemia due to prolonged fetal PTH suppression in utero and preterm delivery. Poorly controlled CKD-MBD leading to neonatal complications, including transient neonatal hyperparathyroidism or neonatal fractures, has also been reported. There is scanty data on the appropriate management of women with CKD-MBD in pregnancy and these patients require a collaborative multi-disciplinary approach involving the nephrologist, obstetrician, and endocrinologist to manage the bone mineral abnormalities that occur in them. Pre-conception counselling and planning is essential to ensure that kidney function, blood pressure, and other comorbidities are optimised before pregnancy. During pregnancy, the Nephrologist should ensure intensive hypertension control, suppression of proteinuria, and adjustment of dialysis/immunosuppression in women with end-stage kidney disease (ESKD) or kidney transplant respectively. Close fetal and placental surveillance should be performed by the obstetrician. Management of mineral metabolism during pregnancy depends largely on the severity of CKD-MBD and whether the woman has pre-existing secondary or tertiary hyperparathyroidism. Calcium, phosphate, and iPTH levels should be assessed regularly, targeting calcium and phosphate levels within the normal range. Calcium-based binders are safe in pregnancy and lactation, but non-calcium-based binders such as sevelamer and lanthanum should be stopped pre-conception as animal studies have reported reduced or irregular ossification . Calcimimetics such as cinacalcet are category C in pregnancy with animal data suggesting low risk and can be continued in women with poorly controlled hypercalcemia. Women with secondary hyperparathyroidism should also be continued on 1,25(OH)D supplementation. Women on dialysis should also have the dialysate composition of calcium and phosphate modified based on individual needs . 7. Conclusions Management of metabolic bone disorders during pregnancy and lactation requires careful balance due to the dual needs of the mother and fetus. Early diagnosis, close monitoring, and individualized treatment, particularly in cases of hyper and hypocalcemia, hypophosphatemia and osteoporosis are essential to prevent adverse maternal and fetal outcomes. Optimal timing of interventions, such as parathyroidectomy, and the cautious use of pharmacological treatments, can significantly improve both maternal health and fetal development outcomes. 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2285	https://www.americanimmigrationcouncil.org/story/third-generation-apple-farmer-barney-hodges-cant-find-enough-americans-to-harvest-his-200-acre-farm/	Third-Generation Apple Farmer Barney Hodges Can’t Find Enough Americans to Harvest His 200-Acre Farm - American Immigration Council Skip to content The American Immigration Council is a non-profit, non-partisan organization. Sign up for our emails. 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Storytelling Belonging Begins With Us Get Involved Get Involved Volunteer Corporate Partnerships Institutional Donors Events Donate Today Contact Us Donate Now The American Immigration Council is a non-profit, non-partisan organization. Sign up for our emails. Home / Stories / Third-Generation Apple Farmer Barney Hodges Can’t Find Enough Americans to Harvest His 200-Acre Farm Third-Generation Apple Farmer Barney Hodges Can’t Find Enough Americans to Harvest His 200-Acre Farm Published: September 22, 2016 AgricultureLabor-Intensive IndustriesStories Share: Barney Hodges III is a third-generation apple farmer and the second generation to run his family’s farm in Vermont. Like his father and his grandfather before him, Hodges depends on migrant labor to keep the family business alive—a farm that pumps $3 million into the local economy each year. These days the family farm is a 200-acre apple orchard in Cornwall, Vermont, in the agriculturally rich Champlain Valley. Formed by the Lake Champlain basin, the valley’s glacial-fed soil and moderate temperatures have attracted farmers since the colonial period. Its broad tapestry of farms, known today as the “breadbasket of Vermont,” produces beef, pork, poultry and eggs, honey, flowers and vegetables, much of New England’s dairy, and a wide variety of apples, peaches and berries. This work requires a skilled workforce. My Jamaican workers are the best workers I could get for this job. They know all the elements of my farm. They know the quality standards of my farm and how to achieve that. What this fertile valley can’t produce, however, is the labor needed to harvest the crops. At the end of each August, the apples at Hodges’ Sunrise Orchards reach maturity, and it is then that he requires a good 50 to 60 skilled pickers willing to work 10 hours a day, six to seven days a week, for six to eight weeks, through October. Hodges has never been able to fill this need with American workers, who can secure year-round, or less physically grueling, jobs elsewhere. “If we were willing to pay $20 to $25 an hour for that labor, maybe we would be able to get American workers to do that work,” Hodges says. Except, he says, “We wouldn’t sell any apples. . . . Would you pay $12 for a half peck [5 pounds] of apples?” “If I was to go and be a good Samaritan and say, I’m going to pay that $20 to $25 an hour, then there might be some good Samaritans who say, That’s great, I want to support that, I’m going to buy his apples,” he says. “But I need an entire region to support that, from Florida to Boston to Texas. . . . It would require a monumental shift in the understanding of food quality, food production, and food economics.” By using the H-2A visa, the only legal avenue to hire foreign agricultural labor, Hodges can reliably produce apples at market prices, selling them to supermarkets for $0.50 to $0.75 per pound wholesale. He grows upwards of 150,000 bushels a year, of Paula Red, McIntosh, Empire, Courtland, Honey Crisp, Red Delicious, Gala, and several varieties for hard cider. It is enough to support an extended family of eight and to hire 15 year-round local, American workers. All told, Sunrise Orchards funnels more than $3 million into the local economy, through the purchase of farm services, farm supplies, and personal living expenses. A sizable portion of that economic boost comes directly from Hodges’ temporary migrant workers; as many as 12 non-farm jobs exist for every farm job, including those undertaken by H-2A workers. And if the H-2A visa program didn’t exist? Hodges says he would stop farming. “That program represents a legal method for businesses in our country to get temporary highly skilled, trained agricultural workers,” he says. “The program is not perfect, but it’s a very helpful, necessary program for my sector.” Like many American farmers, Hodges’ family has been relying on migrant labor since the 1940s, when the U.S. brought skilled farm workers from Mexico and the West Indies to help fill labor shortages created by World War II. Many of Hodges’ workers return every year. He has known many since he was a child scampering through the fields. The H-2A program is not cheap. Hodges says it costs him about $90,000 to hire 60 H-2A workers, or $1,500 per worker, an amount that includes visa fees and roundtrip transportation but not housing, which employers must provide. He estimates the two bunkhouses he built equate to an additional $700 per worker per season over a 20-year period. All these costs are in addition to wages, which the federal government sets annually for each state. In Vermont, the 2016 H-2A wage rate is $11.74 per hour. In full, the extra cost of bringing in a migrant worker through the H-2A program adds about $4 to $6 to each worker’s hourly wage. But farmers are grateful for the high quality of work they receive. “This work requires a skilled workforce,” says Hodges. “My Jamaican workers are the best workers I could get for this job. They know all the elements of my farm. They know the quality standards of my farm and how to achieve that.” Furthermore, says Hodges, the program beats the alternative used by some farms: the hiring of undocumented workers, without government oversight or mandated safety and health standards. Such a “don’t ask, don’t tell” policy “is a huge violation of human rights and a crazy approach,” he says. That said, he believes the H-2A program remains in dire need of improvement if farmers are to continue to stay in business. Under the current program, for example, farmers must apply at least 60 days before harvest for the exact number of workers they will need. For Hodges, that’s at least 60 days before he knows how many apples his trees are going to produce and how many pickers he’s going to need. The requirement makes sense in theory; the U.S. government wants to ensure each job is advertised in time to give U.S. workers the opportunity to apply. In practice, however, the provision serves little purpose since U.S. workers rarely do apply for the jobs. Hodges recruits across the East Coast and receives just one or two applications a season from American workers. Those that do show up to work invariably quit within three days. One year he under-estimated the crop, ended up with too few H-2A workers, and “had to overwork people, and leave crops on the ground,” he says. “I’m never going to make that mistake again.” Now he requests more workers than he needs, a system that prevents crop loss but makes for an inefficient payroll. “We’re really kind of cornered into knowing numbers of workers and guaranteeing work for a commodity that’s controlled by Mother Nature,” Hodges says. “What I would like is to have more flexibility to bring workers up if you need them. It only takes a couple days. 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2286	https://math.stackexchange.com/questions/2837627/locus-of-points-a-fixed-distance-from-a-hyperplane	euclidean geometry - Locus of points a fixed distance from a hyperplane - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Locus of points a fixed distance from a hyperplane Ask Question Asked 7 years, 3 months ago Modified7 years, 3 months ago Viewed 173 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I want to specify a subspace of a high dimension Euclidean space. The subspace consists of those points whose closest proximity to a specified hyperplane is a fixed value r. Here are some examples. Example 1. Consider a 3D space, and a line (the hyperplane) running though it. The subspace is the surface of cylinder of radius r whose axis is the line. Example 2. Consider a 3D space, and a plane (the hyperplane) running though it. The subspace consists of the two planes parallel to the original one and a distance r from it. I want this subspace specified in the form of a locus of points. In other words, a set of parameters uniquely specifies any point in the subspace. For instance in example 1, any point in the subspace could be specified by 3 parameters, e.g. one parameter would specify a point on the line r would specify a circle radius r, centred on the point and orthogonal to the line theta (an angle) would specify a point on that circle. (You would have to define where theta=zero was and the direction of rotation of theta but I don't think you would need more parameters for that, just the adoption of some convention.) I've been able to specify a b dimensional hyperplane in a b + c dimensional euclidean space (b<b+c), but am now stuck. I appreciate any help you can give me. What about this. One could start with the polar coordinates for a point on a b dimensional sphere. Covert from polar to Cartesian, to express these b parameters as a b-vector in b dimensional Euclidean space. Place the b dimensional sphere in a b + c dimensional Euclidean space. Place it at position zero in each of the c new dimensions. Now add c parameters which are the positions of the b-sphere in the c new dimensions. You have now created a parameterisation for one example of the desired subspace. The c dimensions are the hyperpane. You can then rotate and translate your c+b parameterisation in c+b dimensional space to make it coincide with any specific c dimensional hyperplane. You set the radius of the b-sphere to a specific value. Your final parameterisation has b+c-1 parameters. euclidean-geometry locus Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jul 2, 2018 at 12:20 Desmond CampbellDesmond Campbell asked Jul 1, 2018 at 14:20 Desmond CampbellDesmond Campbell 21 2 2 bronze badges 8 For example 1 would need to specify where the zero for the angle θ θ is, depending on the point on the line.coffeemath –coffeemath 2018-07-01 14:31:22 +00:00 Commented Jul 1, 2018 at 14:31 Both of your examples can be represented implicitly by second-degree Cartesian equations.amd –amd 2018-07-01 17:50:18 +00:00 Commented Jul 1, 2018 at 17:50 What, specifically, is your question?amd –amd 2018-07-01 17:50:52 +00:00 Commented Jul 1, 2018 at 17:50 1 @coffeemath - I have edited the question to reflect your point.Desmond Campbell –Desmond Campbell 2018-07-01 21:42:56 +00:00 Commented Jul 1, 2018 at 21:42 @amd - I want a parameter space such that each point in the parameter space maps to a single point in the subspace (defined above). I guess the inverse is not necessary, a point in the subspace could map to multiple points in the parameter space, e.g.for an angle parameter values 360 degrees apart might map to the same subspace point. However I'd probably restrict the parameter space so as to make the mapping one-to -one in both directions, e.g. 0 <= theta < 360 degrees.Desmond Campbell –Desmond Campbell 2018-07-02 11:01:44 +00:00 Commented Jul 2, 2018 at 11:01 \|Show 3 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. You could think about stereographic projection: If you have a linear subspace U U of dimension m m of a Euclidean vector space V V of dimension n n, you could pick a vector in U⊥U⊥, call it w w, and consider (orthonormal) co-ordinates (x 1,…,x n−1)(x 1,…,x n−1) on the space w⊥w⊥ orthogonal to w w, chosen such that (x 1,…,x m)(x 1,…,x m) give co-ordinates on U U. Now apply the inverse stereographic projection map (you can look up the formula I guess) to the co-ordinates (x m+1,…,x n−1)(x m+1,…,x n−1), and leave the variables x 1,…,x m x 1,…,x m fixed. That gives you a map from R n−1 R n−1 to your locus. Unfortunately this is not quite surjective, and misses out a translate of U U in the direction w w. Still, the formulas are nicer than finding spherical co-ordinates I think. One small point: 'hyperplane' generally refers to a subspace of dimension one less than the ambient vector space. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jul 2, 2018 at 13:29 tprincetprince 116 3 3 bronze badges 2 Would I end up with a point expressed in the Cartesian coordinate system? Let's call the Cartesian axes C_1, C_2, ... C_n. A point would be an n-vector. Each element of the vector would be function of (probably all) the n-1 parameters. Yes?Desmond Campbell –Desmond Campbell 2018-07-04 22:30:30 +00:00 Commented Jul 4, 2018 at 22:30 Sorry for the delay - but yes, that's right, inverse stereographic projection will give a vector each element of which is a rational function (a fraction with polynomials in numerator and denominator) in the n-1 variables. The first m coordinates are just the coordinates on U, so the entry is just x_i, but the rest of the co-ordinate entries use everything from x_{m+1} to x_{n-1}.tprince –tprince 2018-07-14 12:56:17 +00:00 Commented Jul 14, 2018 at 12:56 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions euclidean-geometry locus See similar questions with these tags. 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2287	https://wecanfigurethisout.org/ENERGY/Web_notes/Bigger_Picture/Where_do_we_go_Supporting_Files/Efficiency%20in%20Electricity%20Generation%20-%20EURELECTRIC.pdf	EFFICIENCY IN Report drafted by: EURELECTRIC “Preservation of Resources” Working Group’s “Upstream” Sub-Group in collaboration with VGB July 2003 ELECTRICITY GENERATION %2Fwww.eurelectric.org%2FDownload%2FDownload.aspx%3FDocumentID%3D13549&usg=AFQjCNH_41MwVJ57EVzrpWHxpDmc1WkZ-w&sig2=wxMKAcHEdOyGSTRLO6MODg&cad=rja 2 Efficiency in Electricity Generation July 2003 This report has been drafted by: EURELECTRIC “Preservation of Resources” Working Group’s “Upstream” Sub-Group in collaboration with VGB Members of the Drafting Team: EURELECTRIC Upstream Sub-Group: Livio HONORIO, Chairman (PT) Jean-Guy BARTAIRE (FR), Rolf BAUERSCHMIDT (DE), Tapio OHMAN (FIN), Zoltan TIHANYI (HU), Hans ZEINHOFER (AT), John F. SCOWCROFT (EURELECTRIC), Vasco DE JANEIRO (EURELECTRIC) VGB: Hartmut KRUGER (DE), Hans-Joachim MEIER (DE), Daniel OFFERMANN (DE), Ulrich LANGNICKEL (DE) The Union of the Electricity Industry - EURELECTRIC, formed as a result of a merger in December 1999 of the twin electricity industry associations, UNIPEDE and EURELECTRIC, is the sector association representing the common interests of the European electricity industry and its worldwide affiliates and associates. Its mission is to contribute to the development and competitiveness of the electricity industry and to promote the role of electricity in the advancement of society. Union of the Electricity Industry - EURELECTRIC Boulevard de l’Impératrice, 66 – B-1000 Brussels Tel: +32 2 515 1000 - Fax: +32 2 515 1010 E-mail: eurelectric@eurelectric.org Website: www.eurelectric.org VGB PowerTech was founded in Leuna (East Germany) in 1920. It is a voluntary association of power as well as heat generating utilities. Its main objective is joint support and improvement of operational safety, availability, efficiency and environmental compatibility of power plants (fossil-fired, nuclear, renewables) both in operation or under construction. Further, VGB is involved in standardization as well as in elaboration of technical guidelines and regulations in the field of thermal power plants. Within the framework of its legal possibilities, VGB supports the work on the mandatory pressure vessel- and steam boiler regulations and nuclear regulations in Germany and also in the European Union. VGB PowerTech e.V. Postfach 10 39 32, D-45039 Essen Klinkestraße 27 - 31, D-45136 Essen Tel: +49 201 81280 - Fax: +49 201 81283 E-Mail: info@vgb.org Website: www.vgb.org Copyright © Union of the Electricity Industry – EURELECTRIC, VGB July 2003 – Ref. 2003 – 030 - 0548 All rights reserved 3 TABLE OF CONTENTS I. INTRODUCTION 4 II. ELECTRICITY GENERATION: main purpose 4 III. EFFICIENCY IN ELECTRICITY GENERATION BASED ON THERMAL PROCESSES 4 3.1 Introduction 4 3.2 Power plant efficiency 7 3.2.1 Converting oil into electricity 7 3.2.2 Converting coal into electricity 7 3.2.3 Converting natural gas into electricity 8 3.2.4 Converting biomass and biogas into electricity 8 3.2.5 Converting nuclear energy into electricity 9 3.2.6 Converting geothermal energy into electricity 10 3.2.7 Converting thermal energy into electricity and useful heat in CHP units 10 IV. EFFICIENCY IN NON-THERMAL ELECTRICITY GENERATION BASED ON RENEWABLE RESOURCES 11 4.1 Introduction 11 4.2 Converting solar energy into electricity 11 4.3 Converting wind energy into electricity 12 4.4 Converting hydro energy into electricity 12 4.5 Converting energy from the ocean into electricity 12 V. EFFICIENCY IN DECENTRALISED GENERATION TECHNOLOGIES 12 VI. IMPACTS ON THE ENVIRONMENT 14 6.1 Fossil and nuclear fuels 14 6.2 Renewable energy sources 15 6.3 Comparison of specific CO2 emissions 15 VII. POWER PLANT SCALES AND GENERATION COSTS 16 VIII. ELECTRICITY GENERATION INDICATORS IN THE EU 18 8.1 EU-15 electricity generation mix 20 8.2 Candidate countries electricity generation mix 20 IX. THE BENEFITS OF IMPROVING EFFICIENCY 20 9.1 Effects of slight efficiency improvement 22 X. FINAL REMARKS AND NEXT STEPS 23 XI. GLOSSARY 24 4 I. INTRODUCTION This report gathers state-of-the-art knowledge on energy efficiency in electricity generation based on thermal processes, on renewable energy sources and on distributed generation technologies. The aim of the report is to support policy analysis made not only by electricity companies’ analysts but also by policymakers, consultants and other stakeholders. The efficiencies presented in this report are related to the most recent generation technologies on the market or expected to be available in the near future. The report is the result of collaborative work between the Union of the Electricity Industry-EURELECTRIC and VGB PowerTech’s experts. The report is divided into two distinct parts: one on energy efficiency values in electricity generation (Chapter III, IV and V), which is the core of the report, and the other part on related issues such as impacts on the environment (Chapter VI), power plant scales and generation costs (Chapter VII), indicators (Chapter VIII) and benefits of improving energy efficiency (Chapter IX). II. ELECTRICITY GENERATION: main purpose Electric energy generation is the conversion of other kinds of energy, mainly primary energy, into electrical energy. Examples: Hydro power plant The conversion of the water “mechanical” energy into electrical energy Nuclear power plant The conversion of the nuclear energy released by nuclear fuel into electrical energy Fossil-fuel power plant The conversion of the chemical energy of fossil fuel into electrical energy Fuel Cell The conversion of the chemical energy from an oxygenation controlled reaction directly into electrical energy Generally, the process of generating electricity goes through several transformations, as there is little primary energy directly convertible into electricity. For instance, in a thermal power station the primary energy is converted to high temperature steam, as an intermediate heat source, then into mechanical energy in the turbines physically connected with the generators where the electric energy is produced. Direct energy conversion might represent greater efficiency since it means that electricity could be generated without intermediate equipment. III. EFFICIENCY IN ELECTRICITY GENERATION BASED ON THERMAL PROCESSES 3.1 INTRODUCTION In thermal power plants the steam is generated by burning fuels or from the heat released by nuclear fission or is extracted from underground geothermal reservoirs. The different energy resources used may be grouped as follows: • Fossil fuels such as coal, oil, natural gas; • Fuels artificially prepared, such as hydrogen, alcohol and acetylene; • Converted fuels, such as methane and biogas; • Nuclear fuels • Geothermal steam 5 Fuels converting into electricity may also be grouped into solid, liquid and gaseous fuels, as follows: Solid fuels: − Fuel wood 1 − Forest products 1 − Coal: anthracite; bituminous coal; sub-bituminous coal; lignite (brown coal) − Peat: peat is considered as a substance somehow between forest product and coal − Carbon wastes Liquid fuels: These fuels result from refining crude oil: − The lighter products first to distillate are liquefied petroleum gases (LPG) − The following distillate products will give gasoline, petrol and gas-oil − The residue, which is not distillate, is fuel-oil. There are also on the market some mixtures of gas-oil and thick fuel-oil which result in: − Diesel-oil − Burner-oil − Thin fuel-oil Other liquid fuels: − Alcohol (especially ethanol) Gaseous fuels − Natural gas is a mixture of hydrocarbons, chiefly methane (CH4) − Liquefied petroleum gases (LPG), butane; propane − Manufactured gas: derived from the industrial petrochemical process − Other fuel gases: hydrogen, acetylene, et al. Fossil fuels Fuel is an organic substance used for its energy content. The energy content of a fossil fuel, before any treatment or conversion, corresponds to primary energy. A fuel is characterised, giving the common feature inherent to its heat energy generation, by the calorific value. Calorific value (GCV-Gross Calorific Value or NCV-Net Calorific Value) is the quantity of heat released by the complete combustion of a unit quantity of a fuel in a well-determined condition. Its calculation may, or not, take into account the vapour condensation of the water, determining the GCV or the NCV. Generation Efficiency The electric power plant efficiency η is defined as the ratio between useful electricity output from the generating unit, in a specific time unit, and the energy value of the energy source supplied to the unit, within the same time.2 1 Considered as renewable 2 The type of energy converted in a fuel-burning installation is variable. The output of the conversion process may either be electricity (power), heat or a mixture of both, which makes it difficult to define efficiency of the process (it is even more complex in a three-product system of electricity, heat and a high-quality syn-gas product, i.e. produced in gasification plants). Different energy conversion processes have different thermodynamic limitations. Therefore, the term “efficiency” should only be used for one process with one energy source and one energy product, specifically referring to the output, i.e. “electrical efficiency”. 6 In physics theory, η of a thermal electricity generation process is limited by the Carnot efficiency. Carnot efficiency = (T source – T sink) / T source Example of the Carnot efficiency: A heat engine supplied with steam at 543ºC (T source = 816K) And the choice of a sink in a river at 23ºC (T sink = 296K) In theory, the Carnot efficiency η = 64% In practice, the process efficiency is less than that ideal maximum, about 40%3. Example: Energy value of fuel supplied in a time unit (available energy) 1 toe (tonne of oil equivalent) <> 11,628 kWh in NCV (implicit “heat equivalent of 1 kWh” = 86 grams of oil equivalent) Useful electricity output from the power station in a time unit (electricity supplied) 1 toe <> 4,505 kWh (implicit average heat consumption per 1 kWh = 222 grams of oil equivalent) Thermoelectric power plant efficiency η = 4,505/11,628 = 38.7% (in NCV) (Or η = 86/222 = 38.7%) Energy value of fuel – the denominator – represents its heat content, which is the product of the burned mass times the NCV or GCV. Electricity output – the numerator – represents the net power output of the power station in the same period. The difference between these two terms represents the losses. Transformation sequence: Example At the fuel CV - Calorific value (= 100%) The solid, liquid and gaseous fuels used in a thermal power plant are mainly hard coal, lignite, fuel-oils, gas-oil and natural gas. Their values of GCV and NCV are on an average the following: GCV NCV Heavy fuel-oil 42.6 MJ/kg (= 10,175 kcal/kg) 40.57 MJ/kg (= 9,690 kcal/kg) Light fuel-oil 43.3 MJ/kg (= 10,342 kcal/kg) 41.2 MJ/kg (= 9,840 kcal/kg) Burner-oil 44.1 MJ/kg (= 10,533 kcal/kg) 42.16 MJ/kg (= 10,070 kcal/kg) Gas-oil 45.7 MJ/kg (= 10,915 kcal/kg) 43.75 MJ/kg (= 10,450 kcal/kg) Natural gas 42.0 MJ/m3 (=10,032 kcal/ m3) 37.9 MJ/m3 (= 9,052 kcal/m3) Hard coal 35.4 MJ/kg (=8,448 kcal/kg) 34.1 MJ/kg (=8,145 kcal/kg) Lignite 24.0 MJ/kg (=5,732 kcal/kg) 23.0 MJ/kg (=5,493 kcal/kg) 1 calorie = 4.1868 Joules Processes like CHP (see page 10) have two different efficiencies. When added, these data together represent the utilisation of the input energy fuel. This is called “fuel utilisation” or “overall efficiency”. In electricity statistics, the share of fuels used for electricity generation is estimated based on conventional/design values of the two different efficiencies mentioned above. 3 Efficient turbine technologies are currently close to 60% 7 Note: All figures are calculated for water and ash free fuels. For hard coal and lignite containing water and ashes the values are lower (e.g. lignite: water content 40-60% and ash content of 5%, NCV of 9.2 MJ/kg instead of 23 MJ/kg). Generally in Europe, for calculating efficiency the NCV is applied Ref: VDI 4660 conversion factors for specific emissions from energy conversion systems The efficiency of a fossil fuelled power station with once-through water-cooling depends directly on the “Steam generator” efficiency and the “Turbine plant” efficiency. The losses occur in the steam generator (e.g. head losses, mass losses and steam cooling) when it passes from the steam generator to the turbine; these represent about 10%. At the turbine input CPST - Calorific power transmitted in the input steam to the turbine (= 90) The “Turbine” efficiency is the most influential factor of a power plant’s efficiency, and takes into account the fundamental heat rejected in the condenser (cold source). This efficiency is the difference between the ideal efficiency of the turbine and the sum of the losses – internal and external – of the turbine. At the generator input MPG - Mechanical power transmitted to the generator (= 40) The efficiency of a large electrical generator is typically 99%. At the generator output GEP - Gross electric power at the generator terminals (= 39) From the gross output without electrical power consumed by the station auxiliaries and the losses in “generator transformers” it is possible to obtain the net value. At the transmission grid input NEP - Net electric power supplied to the grid (= 36) η - Power station efficiency η = NEP/CV (= 36%) Losses CV-NEP (= 64%) 3.2 POWER PLANT EFFICIENCY This section indicates efficiency values for converting fuels into electricity in today’s average thermal power plant. It is important to note that these values do not represent fuel or power plant availability, but efficiency as defined in part 3.1. 3.2.1 CONVERTING OIL INTO ELECTRICITY Steam turbine fuel-oil power plant 38 to 44% Ref: Figures agreed through peer review between EURELECTRIC and VGB experts 3.2.2 CONVERTING COAL INTO ELECTRICITY An appropriate coal to be used in a thermal power station is “steam coal”. Conventional power plants with pulverized coal firing have efficiencies as follows: 8 Steam turbine coal-fired power plant 39 to 47% Ref: VGB 3.2.2.1 CONVERTING COAL IN NEW COAL COMBUSTION TECHNOLOGIES INTO ELECTRICITY Coal and other non-gaseous fossil fuels can also be converted into electricity (and heat in CHP power plants) in combined gas-steam-cycle if the fuel is gasified in advance. Such IGCC (Integrated Gasification Combined Cycle) power plants offer large potential for higher efficiencies. On the other hand, these plants are very complex and difficult to operate, which reduces flexibility and availability. Other advanced techniques concentrate on special firing systems like fluidised bed combustion (FBC; attractive for medium scale and low-quality coal) and increased steam parameters (600°C, 270 bar and more; affords new materials). Currently, there are four main available “Clean Coal Combustion Technologies” in various sizes: Pulverised coal boilers with ultra-critical steam parameters Up to 47% Ref.: Nordjyllandsvaerket power plant 47% with sea water cooling (Denmark) Atmospheric Circulating Fluidised Bed Combustion (CFBC) > 40% Ref: Gardanne power plant (France) Pressurised Fluidised Bed Combustion (PFBC) > 40% Ref: Cottbus: 74 MWe. 220 th. (Germany) Coal fired IGCC > 43% Ref: Buggenum: 43% (Netherlands); Puertollano (330 MWe); 45% (Spain) 3.2.3 CONVERTING NATURAL GAS INTO ELECTRICITY In the case of gas turbines: Large gas turbine (MW range) up to 39% Ref: Gas turbine World Handbook, 2000/2001 3.2.3.1 CONVERTING NATURAL GAS INTO ELECTRICITY IN COMBINED CYCLE In the case of CCGT (Combined Cycle Gas Turbine processes) power is generated more efficiently than in a simple gas turbine cycle: the hot exhaust gases of the gas turbine are used to produce steam that generates electricity in a steam turbine cycle. Large gas fired CCGT power plant up to 58% Ref.: Mainz-Wiesbaden “GuD” plant (“GuD” – Gas and Steam / Siemens) 3.2.4 CONVERTING BIOMASS AND BIOGAS INTO ELECTRICITY Biomass results from the joint combustion of organic materials of vegetal or animal origin, and also including materials resulting from their transformation. Biogas is a mixture resulting from the anaerobic fermentation of organic materials. Biomass and biogas 30-40% Biomass gasification combined cycle power plant 40% Ref: Energy from Biomass, Principles and Applications 9 We may consider as main sources of biomass the following: forest; waste materials from forestry and sewage; skin and residues from agro-industrial activities; residues from agricultural plantation; sewage from animal wastes; urbane waste; energy farm. Waste-to-energy Utilisation of waste for power generation should be treated as “Renewable” because it prevents the use of exhaustible fuels. Moreover, it reduces the need for waste landfills and related methane emissions. The incineration of biomass and organic waste is CO2 neutral, because the carbon dioxide that is released into the atmosphere practically offsets the CO2 absorbed by biomass during its growth. Waste-to-electricity power plant 22 to 28% Ref: VGB 3.2.5 CONVERTING NUCLEAR ENERGY INTO ELECTRICITY As far as nuclear energy is concerned, the fact that the fission of one gram of U235 releases approximately 24 MWh or 1 MWday (MWd) of thermal energy makes it convenient to use the concept of combustion rate, also known as “burn-up”, which is expressed in MWdays per tonne of heavy metal4 (MWd/tHM). Over the past 30 years, burn-up has steadily increased: for light water reactors, the most common type in the western world, it has moved from 33,000 to around 65,000 MWd/tHM and is expected to increase further. The total thermal energy released by nuclear fuel is proportional to the burn-up it reaches at the end of its reactor life5. One fuel assembly containing typically 460 kg of uranium and reaching a burn-up of 65,000 MWd/tU would therefore release 65,000 MWd/tU x 0.46 tU x 24h/d = 717,600 MWh of thermal energy over its reactor life. The thermal efficiency of a nuclear power station is defined in exactly the same way as for any other thermal plant: it is the efficiency of the thermodynamic cycle by which the heat generated by the fuel is converted into steam through steam generators. The thermal efficiency of a conventional nuclear power station is around 33%. Nuclear power plant 33% to 36% Ref: Figures agreed through peer review between EURELECTRIC and VGB experts Therefore, to generate 1,000 MW of electrical power (MWe) in a nuclear plant it is needed around 3,000 MW of thermal power from the fission reaction. One day generating and fuel consumption: For a generating capacity of 1,000 MWe, the energy output in one day is: (1 x 109 J/s) x (86,400 s) = 0,864 x 1014 J Each day fuel consumption: 1,000 MWe coal-fired power plant burns about 8,000 tons of coal; 1,000 MWe nuclear power plant has to undergo fission of about 3.2 kg mass of U235. 4 Heavy metal referring to the fissionable material composing the fuel 5 The relation between the burn-up achievable and the initial heavy metal content of the fuel assembly is more complex 10 3.2.6 CONVERTING GEOTHERMAL ENERGY INTO ELECTRICITY Geothermal energy comes from the thermal earth inner activity, mainly where there is volcano activity. The deposits of heat may be exploited with almost constant power supply. Once steam reaches earth surface through wells, it is used to produce electricity, in some cases used for non-electric purposes (e.g. building heating), or saving energy otherwise produced through conventional methods. Inside geothermal plants steam supplies power to move the turbines producing electricity. Waste water derived from steam is then injected in deep wells in order to keep a constant pressure level and to avoid steam pollution. In some areas of the world, including Europe, geothermal energy plays a leading role. The type of use – heating or power generation – depends on the quantity and quality (level of temperature) of the geothermal source. In some regions, it has been produced commercially in the range of hundreds of MW for many decades [EU Blue Book on Geothermal Resources]. Geothermal power plant up to 15% for 190°C Ref: EGEC / Geothernet The efficiency of existing organic Rankine Cycle plants generally range from 10% to 15.5% for resources at 100°C to 160°C and is slightly higher (17%) for temperatures up to 190°C with a two-phase geothermal fluid [quote from EGEC / Geothernet]. Advanced cycles like the Kalina Cycle offer large potential but are not commercially available. Regarding the high density and the constant availability of the energy source – that is, for a renewable technology, only comparable with hydro – the focus is not on increasing efficiencies but at reducing costs. Just for heating purposes the use of heat pumps is very attractive, especially if the temperature of the geothermal source is not very high (low quality). Heat pumps require external energy input like electricity but are able to generate much more heat (at medium quality) than the quantity included in the fuel for generating this electricity. For domestic heating, even the upper ground or ambient air suffices as geothermal source. 3.2.7 CONVERTING THERMAL ENERGY INTO ELECTRICITY AND USEFUL HEAT IN CHP UNITS In the case of Combined Heat and Power (CHP), or co-generation, part of the converted thermal energy is used for generating useful heat: either by utilising the low-temperature steam at the steam-turbine exit for district heating or branching off a certain amount of steam directly from the steam turbine i.e. for process heat. This reduces the electrical efficiency slightly (~14 % of extracted heat for district heating), but the input fuel energy is better used in total. The loss of electrical output results from the pressure difference by condensing steam at back pressure instead of vacuum conditions. For high temperature steam extraction, the loss is higher. For example, a 112 MW (electric) plant operating in a mode without heat extraction has an electrical efficiency of 36.3%. By producing 152 MW additional heat the overall efficiency increases to 84.9%. Example: Power of gas turbine 69,100 kW Power of back-pressure steam turbine 44,700 kW Auxiliary power consumption 1,400 kW Net power output of plant 112,400 kW Heat input gas turbine 230,000 kW Heat input supplementary firing 79,600 kW Process steam output 152,000 kW Electrical efficiency 36.3 % Thermal efficiency of heat production (only) 48.6 % Overall efficiency 84.9 % The “overall efficiency” is higher than the electrical efficiency and results from adding the efficiency of the generated heat (= useful heat / energy of fuel supplied). The overall efficiency is therefore defined as: 11 Overall efficiency = (Electrical Power Output + Useful Heat Output) / Total Fuel Input. Comparing separated heat and power supply to CHP or two different CHP solutions on the basis of overall efficiencies is possible with the same amount of electricity and heat at uniform temperature levels6. CHP applications provide potential for better fuel utilisation especially if the volume of heat demand is high and relatively constant (in the summer period too), as in industry or in some northern regions of Europe. Examples for CHP power stations in Finland show highest figures for heat output and overall efficiency compared to others and in contrast to other countries, without any subsidies being provided. IV. EFFICIENCY IN NON-THERMAL ELECTRICITY GENERATION BASED ON RENEWABLE RESOURCES 4.1 INTRODUCTION Renewable energies are sources of energy that renew themselves constantly through natural processes and, seen on a human-time scale, will never run out. Renewable energies come from three primary sources: solar radiation; heat from inner earth; tidal power. These three sources can be used either directly or indirectly, in particular the form of biomass, wind, wave energy and ambient heat. Renewable energy sources (RES) can be converted into electricity, heat and also fuel. 4.2 CONVERTING SOLAR ENERGY INTO ELECTRICITY Solar systems for electricity generation purposes are based on the concentration of sunlight. There are three different concentration solar power systems: parabolic trough systems; solar power tower; parabolic dish technology using a stirling motor Their efficiency values are the following: Parabolic trough 14 – 18% Power tower 14 – 19% Dish stirling 18 – 23% Ref: Figures agreed through peer review between EURELECTRIC and VGB experts Solar energy may also be used directly to produce electricity (photovoltaic effect) that involves photovoltaic cells and, sometimes, grouped on photovoltaic panels. Although it is difficult to generate a high output solar energy compared with fossil fuel or nuclear energy, solar energy is of major importance because it is a non-polluting and renewable energy source. The efficiency value of photovoltaic cells is the ratio of the electrical energy produced by the cells to the incident solar radiant energy. Photovoltaic cells Up to 15% Ref: VGB 6 Directly comparing different CHP applications, or comparing CHP with separate heat and power generation could be misleading, if the quality of the energy converted is not taken into account (2 nd law of thermodynamics). For pure electricity production, the thermodynamic limits have to be considered and for pure heat production this comparison does not show a benefit for CHP, since simple heating boilers, i.e. for domestic heating, have a comparable or even higher quality in “overall efficiency” than CHP units. 12 4.3 CONVERTING WIND ENERGY INTO ELECTRICITY Energy derived from the wind results from the solar energy on the different stratum of the atmosphere. A typical turbine with 40 meters blade diameter and an 8 m/s wind speed, extract about 400 kW from the air of which about 35% can be converted into electric power. Wind turbine Up to 35% Ref: VGB - this figure is a maximum because it does not increase with the total performance. Even 4 MW off-shore converters will not exceed 35%. Example: German coast on-shore: up to 2,500 hours per year; off-shore: up to 4,500 hours per year. 4.4 CONVERTING HYDRO ENERGY INTO ELECTRICITY The electrical efficiency of a hydroelectric power station depends mainly on the type of water turbine. The electricity generated by moving water comes from large hydroelectric power plants and also from smaller ones, such as: mini-power and micro-power plants. It is worth mentioning that more than 90% of total hydro power generated in the EU comes from large hydro. The installed capacity of a small hydroelectric power plant is generally a few MW (<5 MW with an efficiency between 80 and 85 %). Large hydro power plant Up to 95% Small hydro power plant Up to 90% Ref: VGB 4.5 CONVERTING ENERGY FROM THE OCEAN INTO ELECTRICITY Tidal energy results from submarine turbines moving from the rise and fall of sea level due to the gravitational forces of the moon and sun. A dam is used to store the water and a turbine to enable useful energy production. Tidal power plant Up to 90% Ref: Electricité de France (EDF) V. EFFICIENCY IN DECENTRALISED GENERATION TECHNOLOGIES Short-term energy-storage technologies such as mechanical flywheel, chemical batteries and fuel cells, magnetic superconducting, electric ultra-capacitor, can be incorporate in a multi-energy system7. Applications with high electrical demands and lower heat demands will be suited to fuel cells. Electrical efficiencies are as follows: Protons Exchange Membrane Fuel Cell (PEMFC) 40 % Phosphoric Acid Fuel Cell (PAFC) 40 % Solid Oxide Fuel Cell (SOFC) 46 % (aimed > 60%, pressurized) Melted Carbonates Fuel Cell (MCFC) 52 % (aimed: 65 %) Ref: VGB 7 A study considered the following storage technology: batteries (lead-acid and advanced), flywheels (low speed and high speed), ultracapacitors, compressed air energy storage, superconducting magnetic energy storage, pumped hydro electric storage, and hydrogen storage. 13 Fuel cells produce electricity, heat and water. The waste heat recovered from the fuel cells may be used (e.g. for building heating) with effects in the increase of overall efficiency of the system. The overall efficiency is at around 85 to 95%. Microturbines Small and micro-turbines (up to 100 kW) 17 to 22% Ref: Capstone As micro turbines, diesel/gas motors are often used for decentralised CHP applications: Diesel engine as decentr. CHP unit, electrical share 20% to 40% and above 45% at large scale Ref: Figures agreed through peer review between EURELECTRIC and VGB experts The graph below summarises the efficiency in various generation technologies: Efficiency in Electricity Generation 0 10 20 30 40 50 60 70 80 90 100 Large hydro power plant Small hydro power plant Tidal power plant Large gas fired CCGT power plant Melted carbonates fuel cell (MCFC) Pulverised coal boilers with ultra-critical steam parameters Solid oxide fuel cell (SOFC) Coal fired IGCC Atmospheric Circulating Fluidised Bed Combustion (CFBC) Pressurised Fluidised Bed Combustion (PFBC) Biomass gasification combined cycle power plant Protons exchange membrane fuel cell (PEMFC) Phosporic acid fuel cell (PAFC) Large gas turbine (MW range) Steam turbine coal-fired power plant Steam turbine fuel-oil power plant Wind turbine Nuclear power plant Biomass and biogas Waste-to-electricity power plant Diesel engine as decentralised CHP unit (electrical share) Solar dish stirling Small and micro turbines (up to 100 kW) Photovoltaic cells Geothermal power plant Solar parabolic trough Solar power tower Efficiency (%) Efficiency (%) 14 VI. IMPACTS ON THE ENVIRONMENT 6.1 FOSSIL AND NUCLEAR FUELS The impacts on the environment due to the use of fossil fuels in thermal power plants are the following: emissions of gases causing acidification (e.g. sulphur dioxides and nitrogen oxides), greenhouse gas emissions (e.g. carbon dioxide, methane, sulphur hexafluoride, etc), ashes and dust emissions to air. Example: 1,000 MW of generation capacity, 6,600 full load hours per annum = 6.6 TWh electricity production per annum result in the following emissions (the diversity of fuel mixes in each country gives different average values): Ref: VGB The table below shows relative power plant emissions per unit of electricity generated in the UK : Fuel Source SO2 NOx CO2 Coal (average UK) 1.00 1.00 1.00 Coal (typically imported) 0.55 1.00 1.00 Coal (incorporating fuel emission control) 0.10 0.60 1.05 Heavy Fuel-Oil 1.20 0.75 0.85 Natural Gas 0.00 0.25 0.50 For example, considering the CO2 intensity8, the values may range from 32 gCO2/kWh – in a system with Nuclear + Hydro + Biomass + Wind – to 1,000 gCO2/kWh – in a system based exclusively on coal. The table below contains the total SO2, NOX and CO2 emissions in the European Union. Gas (kt) 1980 1990 2000 2005 2010 2020 SO2 12,214 8,424 3,115 2,259 1,736 1,059 NOx 3,214 2,672 1,653 1,400 1,268 1,053 CO2 894,342 888,562 862,676 832,352 847,068 891,369 Ref: Eurprog, EURELECTRIC 8 Pulverised coal plant with desulphurisation – assuming CO2 emissions of 961 gCO2/kWh; a co-generation installation with CO2 emissions of 399-434 gCO2/kWh Hard Coal Lignite Oil Gas-CC Nuclear El. Efficiency % 42.0% 40.0% 44.0% 57.0% 34.0% Fuel Consumpt. t/a 2,000,000 7,600,000 1,289,768 920,000 20 Oxygen Cons. t/a 3,800,000 4,800,000 3,270,047 1,600,000 0 CO2 emiss. t/a 5,200,000 6,600,000 4,496,314 2,200,000 0 SO2 emiss t/a 3,800 4,300 3,134 1,200 0 NOx emiss t/a 3,800 4,300 3,134 3,500 0 Dust emiss t/a 600 640 470 200 0 Radioactivity kBq/a 80 90 0 0 52,800 Ash t/a 150,000 950,000 2,000 0 0 Gypsum t/a 75,000 110,000 220,000 0 0 15 Lignite Hard Coal Natural Gas Nucl. Power Solar Power Wind Power IGCC Fuel Oil Oil Power Plant Hydro Power Photovoltaic Plant Site: Germ. Equa-torial Areas Site: Coastal Area 1.02 0.91 0.83 0.79 0.76 0.58 0.47 0.36 0.02 0.006 0.19 0.14 0.02 Emissions in kg CO2/kWh Operation Fuel Supply Construction Steam Power Plant Gas-Turbine Cycle Carbon Dioxyde Emissions per kWh IGCC Steam Cycle Steam-Turbine Cycle Gas+ Steam Comb. Cycle PWR 1300 MW Run-of-River Plant according to Siemens / Voss / VDI-GET 1999 6.2 RENEWABLE ENERGY SOURCES To illustrate the environmental benefits associated with a non-polluting source an example is provided. Considering a 10 MW wind farm with an average production of 23.5 GWh/year, compared with an alternative thermal production, the avoided emissions of SO2, NOx, CO2 and ashes, are as follows: Wind Farm 10 MW; 6,5-7 m/s; 23,5 GWh/year - avoided emissions: Considering an alternative thermal production: SO2 NOx CO2 Ashes and particles Based on fuel-oil 127 t 60 t 19,000 t 6.5 t Based on Coal 183 t 122 t 25,000 t 1,400 t Ref: Jorge A Gil Saraiva, 1996 Power plants using renewable energy sources do not emit greenhouse gases and other emissions during operation. On the other hand, except for hydro and geothermal power, the net output of such a plant is comparably small, because the energy input (e.g. solar radiation, wind) is not constant and its density is low. Therefore, the expenditure of energy and materials for plant construction per electricity generated is high. 6.3 COMPARISON OF SPECIFIC CO2 EMISSIONS For an overall comparison of specific CO2 emissions (i.e. kg CO2/kWh) a full “Life-Cycle Balance”, including site erection and fuel supply, is necessary. The graph below shows that the specific emissions cannot be neglected for solar power; on the other hand, nuclear power is very competitive in this sense. Finally, carbon dioxide emissions are not the only criterion for climate change issues. Some other greenhouse gases like methane have higher Global Warming Potential (GWP) than CO2. Therefore, gas pipeline leakages for example could have a considerable impact on the Life-Cycle Balance of gas-based power supply, but they are difficult to assess (2% to 8%). 16 Note that the values presented in the graph below differ from the values “Intensity CO2 emissions from Thermal Electricity Generation” in the table in page 18, because they refer to specific CO2 emissions not only per thermal electricity generation but per total electricity generation. CO2 emissions per electricity generation in the European Union 0 200 400 600 800 1000 1200 1400 Austria Belgium Germany Denmark Spain Finland France Great Britain Greece Ireland Italy Luxembourg Netherlands Portugal Sweden Switzerland Norway Czech Rep. Poland Slovakia ktCO2/TWh 1990 2000 Ref: Eurprog, EURELECTRIC [Luxembourg, 1999 = 7,197 ktCO2/TWh, data from 2000 not available] VII. POWER PLANT SCALES AND GENERATION COSTS The following table gives an idea of the typical power plant scale. Type Scale (kW) Nuclear Plant 1,300,000 Coal Plant 500,000 Gas Turbine, Combined-cycle 250,000 Gas Turbine, Single-cycle 100,000 Industrial Co-generation Plant 50,000 Wind Turbine 1,000 Micro-turbine 50 Residential Fuel Cell 7 Household Solar Panel 3 Stirling Engine 1 Ref: Figures agreed through peer review between EURELECTRIC and VGB experts Total costs of electricity generation per kWh generated comprise the specific costs for capital investment, operation & maintenance and fuel. Regarding full life-cycle balance decommissioning and external effects of emissions (e.g. health, climate) have to be included, although external costs are difficult to assess: • The total costs of fossil based power generation vary from 3 to 4 eurocent/kWh at current fuel prices, even for natural gas in combined cycle generation. Although investment costs are very low for gas based plants, gas prices vary considerably from time to time in comparison to hard coal and lignite (in some countries fuel tax has to be added). External costs: significant emission reductions are being achieved, but there still remains a negative impact, especially CO2, that studies show to be 20%, and in some pessimistic view up to 60% of the total costs. 17 • In general, nuclear systems have in total, competitive costs when compared with fossil fuel generation systems. The investment costs are higher, but today most of these plants only produce for fuel costs and therefore are attractive. The latest design studies like the European Pressurised Water Reactor [EPR] have been completed under the prerequisite to be competitive to coal and even gas at total costs. The external costs are low due to near zero emissions; however, the nuclear waste issue should also be taken into account. • Hydro and geothermal power are – if at large scale – the only renewable sources providing sufficient energy density and availability to generate power at attractive costs. If investments are paid, large hydro power plants have the lowest total costs, i.e. less than 2 eurocent/kWh. External costs are difficult to estimate (i.e. displacing CO2-emitting plants) but comparatively low. • Wind turbines, biomass and solar thermal power plants have total costs of at least 5 eurocent/kWh under optimum conditions. At normal conditions, 10 eurocent/kWh can easily be exceeded. This counts as well for small hydro plants (< 5 MW). • Total costs from photovoltaic power generation are high: 35 eurocent/kWh can be exceeded even under sunny conditions. To give an example: in Germany photovoltaic power is subsidised around 50 eurocent/kWh, which covers approximately only half of the total costs for photovoltaic application in this country. External costs are relatively small compared to this dimension, but comparable to those of gas-fired power plants. A key factor for a sustainable – that means ecologically, economically and socially compatible – power supply is the cost of saved kgCO2/kWh. Several studies show [Wagner, Munich, 1997] that this can be done through measures that are relatively cheap to realise – i.e. retrofitting advanced turbine blades – when related to increased power output. In comparison, erecting wind converters could cost up to 10 times more per tCO2 reduced and photovoltaic systems over 100 times more (see graph below). Consequently, the most cost-effective CO2 abating technique is improving conventional power supply. Of. 12.01.02 1 0 50 100 150 200 Retrofit new GTCC Nucl. PP (EPR) Hydro Adv. Coal Techn. CO2-Sequestr. wind photovolt. € / t avoided CO2 Source: medium values of figures based on • own calculations including external costs after Voss, 2001, and other publications, see VGB PowerTech, and • collected data following Wagner, 1998, Pruschek, Göttlicher (Coal-PP with CO2-Sequestration, study), 1999/2001 Costs of CO2-Emissions avoided by specific measures > 1000 18 VIII. ELECTRICITY GENERATION INDICATORS IN THE EU The following table presents indicators that characterise some electricity generation aspects in the EU Member States. Two series of values are presented in the table below: one related to 1985 and the other to 1999. Some indicators 1985 > 1999 Energy intensity (Gross Inland Consumption/ GDP) toe/1990 MEUR Import dependency (%) Electricity Generated/ Capita kWh/inhabitant Intensity CO2 emissions from Thermal Electricity Generation tCO2 emissions from power generation/ GWh Non-Nuclear thermal electricity generated AUSTRIA 1985>1999 220.5 186.5 65.3 66.1 5,913.9 7,458.8 541 619 BELGIUM 1985>1999 329.6 313.3 69.3 76.5 5,813.5 8,263.8 838 590 DENMARK 1985>1999 213.8 160.2 77.6 -13.6 5,679.6 7,305.9 928 748 FINLAND 1985>1999 297.2 261.2 59.2 51.7 10,139.7 13,439.3 699 587 FRANCE 1985>1999 251.5 230.9 54.1 51.9 6,226.7 8,856.8 846 740 GERMANY 1985>1999 315.0 228.1 42.1 59.2 6,706.5 6,768.4 944 809 GREECE 1985>1999 308.4 341.9 60.7 66.1 2,791.8 4,725.6 1,009 852 IRELAND 1985>1999 322.4 202.7 60.1 83.1 3,414.1 5,888.5 757 733 ITALY 1985>1999 182.9 180.3 82.0 80.9 3,281.4 4,607.6 672 580 LUXEMBOURG 1985>1999 448.1 269.6 99.0 97.3 2,560.2 2,374.1 1,205 385 NETHERLANDS 1985>1999 320.9 260.6 5.8 29.7 4,342.1 5,480.9 599 600 PORTUGAL 1985>1999 297.2 351.3 75.2 89.9 1,908.3 4,331.7 697 655 SPAIN 1985>1999 235.9 238.9 60.6 76.6 3,314.4 5,302.4 907 761 SWEDEN 1985>1999 290.6 244.8 42.2 35.1 16,421.4 17,535.3 1,105 455 UNITED KINGDOM 1985>1999 312.9 252.6 -15.4 -20.3 5,257.7 6,163.5 886 618 Ref: 2001-Annual Energy Review, January 2002 – European Commission Data from 1999 EU-15 USA Japan Average Thermal Efficiency 39.9% 33.3% 44.9% Consumption/GDP [toe/MEUR] 231.3 396.2 198.2 CO2 emissions/capita [tCO2/inhabitant] 8.2 20.7 9.1 Import dependency 47.6% [686.6 Mtoe] 24.9% [565.2 Mtoe] 79.4% [409.2 Mtoe] Ref: 2001-Annual Energy Review, January 2002 – European Commission 19 Reference Year: 1999 Countries with electricity generated/capita >7500 kWh/inhabitant Import Dependency (%) 0 1 0 20 30 40 50 60 70 80 90 BEL FIN FR SWE Energy Intensity [toe/1990 MEUR] 0 50 1 00 1 50 200 250 300 BEL FIN FR SWE Intensity CO2 emissions from Thermal Electricity Generation (tCO2/GWh) 0 1 00 200 300 400 500 600 700 800 BEL FIN FR SWE Countries with electricity generated/capita 5500-7500 kWh/inhabitant Import Dependency (%) -40 -20 0 20 40 60 80 1 00 AT DK GER IRL UK Energy Intensity [toe/1990 MEUR] 0 50 1 00 1 50 200 250 AT DK GER IRL UK Intensity CO2 emissions from Thermal Electricity Generation (tCO2/GWh) 0 1 00 200 300 400 500 600 700 800 AT DK GER IRL UK Countries with electricity generated/capita < 5500 kWh/inhabitant Import Dependency (%) 0 1 0 20 30 40 50 60 70 80 90 1 00 GR IT LUX NL PT ES Energy Intensity [toe/1990 MEUR] 0 50 1 00 1 50 200 250 300 350 GR IT LUX NL PT ES Intensity CO2 emissions from Thermal Electricity Generation (tCO2/GWh) 0 1 00 200 300 400 500 600 700 800 GR IT LUX NL PT ES Ref: 2001-Annual Energy Review, January 2002 – European Commission 20 8.1 EU-15 ELECTRICITY GENERATION MIX The thermal power plants based on non-renewable resources represent 85.2% of the total electricity generation in the European Union. Electricity Generation in the EU-15 Year 2000 Conventional thermal 51.8% Nuclear 33.4% Hydro & other renewables 14.8% EU-15 Electricity Generation in the year 2000 = 2,448.6 TWh Ref: Eurostat 8.2 ACCESSION COUNTRIES ELECTRICITY GENERATION MIX The generation mix of the future EU Members States (e.g. Estonia, Latvia, Lithuania, Malta, Cyprus, Czech Republic, Slovakia, Hungary, Poland, Slovenia) which will enter the European Union in 2004, is the following: Electricity Generation in future 10 EU Member States Year 2000 Conventional thermal 75.9% Nuclear 17.9% Hydro & other renewables 6.2% Future 10 EU Member States Electricity Generation in the year 2000 = 302,1 TWh Ref: Eurprog (data from Malta and Estonia was not available) IX. THE BENEFITS OF IMPROVING EFFICIENCY Considering the scarcity of non-renewable energy resources such as: gas-oil, light fuel-oil, heavy fuel-oil, anthracite, coal, butane, natural gas - and the environmental impacts associated with the use of such resources, such as acidification and climate change, there is a need to consider the analysis of the following resources: − Renewable energy sources: Solar energy, hydro power, tidal power, geothermal power, biomass, wind energy − Direct energy conversion: Fuel cells, magneto hydrodynamic generators, thermionic converters, semiconductor thermoelectric converters − CHP supply and combined cycle: Combined Heat and Power, Combined Cycle Power Station − Multi-energy systems: combining traditional grid power with new technologies (fuel cells, others)9 − Advanced combustion technologies and efficient process: Integrated Gasification Combined Cycle, Fluidised bed combustion, Flue Gas desulphurisation process − Bio-fuel: Methane, Biogas, Bio-diesel − Nuclear For these resources it is necessary to identify their availability, their potential for technical improvement and the way to promote their implementation. 9 One example of a multi-energy system could be a combination of a grid power, fuel cells, flywheel energy storage, and diesel engine generator for achieving the quality, availability, and system efficiency desired by the customer from its energy source 21 It is also important to measure the progress for each country on a regular basis, i.e. every four-five years, with performance indicators, concerning the efficiency process, renewable and combined heat and power contributions, greenhouse gas reductions, such as: − Overall efficiency of power plants (%) − Electricity RES and CHP as share of total electricity production (%) − Overall CO2 emissions Main Issues on Preservation of Resources on the Supply-Side Renewables Direct energy conversion Multi-energy systems CHP supply and combined cycle Better combustion technologies and efficient processes Bio-fuels 4-5 year reporting performance indicators Nuclear POWER GENERATION 22 9.1 EFFECTS OF SLIGHT EFFICIENCY IMPROVEMENT Increasing efficiency even at a very small scale has a remarkable impact on emissions reductions and on fuel consumption. The following table gives an overview on the main figures of fuel consumption and emissions for state-of-the-art power plants and different types of fuel including figures for increased efficiency of +0.1%. Emissions of SO2, NOx follow the latest EU Directive on Large Combustion Plants. Hard Coal + 0.1% efficiency Lignite + 0.1% efficiency Oil + 0.1% efficiency Gas-CC + 0.1% efficiency Nuclear + 0.1% efficiency Output MWel 800 900 500 300 1,000 Load factor h-peak/a 6,000 7,500 1,000 4,500 7,500 Energy output MWhel/a 4,800,000 6,750,000 500,000 1,350,000 7,500,000 El. Efficiency % 42.0% 42.1% 40.0% 40.1% 44.0% 44.1% 57.0% 57.1% 34.0% 34.1% Fuel Consupt t/a 1,400,000 -3,382 7,790,000 -19,400 98,000 -200 189,000 -331 22 -0.067 Oxygen cons t/a 2,700,000 -6,500 4,900,000 -12,220 248,000 -600 328,000 -574 0 CO2 emiss. t/a 3,800,000 -8,938 6,750,000 -16,800 341,000 -800 451,000 -790 0 SO2 emiss t/a 2,700 -7 4,400 -11 240 -1 250 0 0 NOx emiss t/a 2,700 -7 4,400 -11 240 -1 720 -1 0 Dust emiss t/a 400 -1 700 -2 40 0 40 0 0 Ash t/a 107,000 -254 970,000 -2,400 150 0 0 0 Ref: Calculations made by VGB following the Emission Limit Values (ELV) required by the Directive on Large Combustion Plant The table below includes two specific examples, gas fired CHP and a coal fired power plant located in Finland. Here, the efficiency improvement of +0.1% result in 1,000 to 1,500 tons of reduced CO2 emissions per year. Reference year 2000 480 MW Natural Gas-fired CHP station 160 MW Coal-fired CHP station Vuosaari B Salmisaari B Capacity MWel 480 160 District heat MW 420 270 Maximum fuel effect MW 974 510 Electricity generation GWh 3,137 710 District heat generation GWh 2,877 1,299 Fuel input GWh 6,635 2,469 Load Factor (h as peak) () 6,812 4,841 Improved global efficiency From 90.6% to 90.7% From 88.0% to 88.1% Reduce fuel burn 540t of natural gas 367t of coal Reduce environmental emissions tCO2 1,487 892 tSO2 - 1.21 tNO2 0.93 1.15 tDust - 0.1 tFly ash - 41 tBottom ash - 8 tLime - 5 tFlue gas end product - 10 Ref: Helsinki Energy () Fuel input (MWh) / Maximum fuel effect (MW) There are many different technologies to meet future energy demand inter alia: Improving efficiencies of established processes, Developing and commissioning of advanced and new techniques, some in more decentralised supply structures, 23 Fuel switching and replacing old power plants with stations state-of-the-art. To estimate the impacts on the environment, not only fuel consumption and emissions at the operating period of a power plant need to be calculated, but also construction of the site, fuel supply and decommissioning have to be taken into account. A full Life-Cycle Balance may show in detail that not every new or advanced technique is of advantage. X. FINAL REMARKS AND NEXT STEPS • Coal and other fossil fuels dominate world-wide power supply due to their cost-effective usage in large central units – both industrialised countries and developing countries. Though a high standard of fossil fuel conversion techniques has been achieved up to now, further improvement of classical processes is possible, i.e. through advanced steam parameters with new materials. Emission abatement through efficiency improvement at coal fuelled power plant is comparably cheap and therefore has great effects on fuel consumption and environmental impacts. • Nuclear power avoids large-scale greenhouse gas emissions. Consequently, expanding its use may compensate for fossil fuel power emissions, especially in industrialised countries, which are typically able to raise high capital costs. • Renewable energy and advanced conversion techniques such as fuel cells will be the future pillar of the fuel mix for power supply – with considerable potential for resource preservation. Much effort will have to be made to reduce costs and to optimise the full life cycle. • CHP applications have great potential for better fuel utilisation. However, extending CHP has to be calculated carefully for every individual case in comparison to efficient but separate power and heat generation, especially if there is no constant heat demand for the heat load generated. • Distributed power plants of small scale, especially fuel cells, will contribute more and more to power supply and might contribute to fuel preservation due to their specific high efficiency. Large units will be necessary for economically feasible long term base power supply. Therefore, the two power supply concepts are not competitors but rather complementary. • For some fuels such as coal, large power plants represent the only way for ecological and economically efficient utilisation. In contrast to gas and oil, coal resources are distributed over many regions of the world and there are vast reserves which could last for hundreds of years. 24 XI. GLOSSARY Battery A DC voltage source containing two or more cells that directly convert chemical energy into electrical energy. Co-generation A process that generates two different types of usable energy. However, co-generation is in most cases used to describe a process in which waste heat from power generation (such as gas turbine, micro-turbine and fuel cells) is captured and used for other purposes. For example, exhaust heat from a gas turbine can be transferred directly through a heat exchanger and used to heat water for residential or commercial use. Combined heat and power (CHP) Another term for co-generation. See above. Distributed generation (DG) Any power generation capability at the point of use rather than a central power generation facility. Types of DG include conventional combustion generators as well as micro-turbines, fuel cells, photovoltaics and wind generators. Efficiency Useable energy output divided by energy input. Emission Discharge of substances into the atmosphere, water or soil. The point or area from which the discharge takes place is called the “source”. The term is used to describe the discharge, the amount of discharge, and the rate of discharge (per m3 or kWh), which can be indicated as concentration or specific emission. The term can also be applied to noise, heat, etc. Energy indicator Indicator used either to chart trends in the energy or economic situation of a given geographical or economical unit over a period of time, or to compare the energy situations of different units. Energy indicators may also be used to chart macroeconomic trends or changes in the standard of living, given the importance of energy to a country’s economy on the one hand and to household expenditure on the other. Note: Energy consumption per capita, frequently used as a standard of living indicator, must be used for this purpose with great care, as a high level of consumption may be the result of bad management (and vice-versa), and differences in accounting systems and procedures may conceal major differences. Energy intensity Ratio between gross energy domestic consumption (see Gross Inland Consumption) or final energy consumption (see Final Energy Consumption) and gross domestic product (see Gross Domestic Product). Note: The indicator is very important for charting trends in the energy content of an economic system or of a country’s energy efficiency. Energy sources All sources (primary or derived) from which useful energy can be recovered directly or by transformation. The terms “energy sources”, “forms of energy”, “energy agents”, “energy” and “energy vectors” are interchangeable in many contexts. Note: It is recommended that each source of energy be called by its specific name, as the generic terms can lead to confusion. For 25 example, “new energy” can be applied to energy sources that have been exploited more systematically or with the help of more sophisticated techniques. Likewise, “conventional energy” which often refers to fossil fuels and partly to electrical energy is a very relative concept, and one which changes with time. As for “renewable energy sources”, they may be continually renewable (permanent flux), renewable in short cycles (annual, for example), over a period of one or more generations; or may be partly or completely renewable. The names “energy flux” and “energy stock” are sometimes used to distinguish between “renewable energy sources” and “non-renewable energy sources. Furthermore, some terms such as “soft energy sources” or “hard energy sources” which do not correspond to any physical reality have a sociological rather than technical or economic meaning. All these generic terms can thus be used validly with a very indicative and qualitative meaning. Final energy consumption Energy consumed by the final user for all energy purposes (1) Fuel cell A device that generates a relatively small amount of electricity from fuels through an electro-chemical process rather than from combustion. In the most common case, a catalyst strips electrons from hydrogen proton and oxygen in the air to produce water and heat as a by-product. Hydrogen fuel cells are characterised as having high electrical efficiencies with zero harmful emissions. Gas turbine An engine (based upon the jet engines used on aircraft) used to power electrical generators in major power plants. Gas turbines run on natural gas and are characterised by having low investment costs, short construction time, no need for thermal cycle cooling, low environmental impact and very low maintenance requirements. Power output can vary from a couple of megawatts to several hundred megawatts. Gross domestic product Total production of goods and services by the subjects of a country and foreigners within national borders (2) Gross inland energy consumption Gross consumption minus bunkers: final energy consumption plus energy sector’s own consumption and losses. Higher heating value (Gross calorific value) Quantity of heat liberated by the complete combustion of a unit volume or weight of a fuel in the determination of which the water produced is assumed completely condensed and the heat recovered. Level of energy dependency Quotient of net energy imported and the total amount of energy consumed within a given geographical or economic unit, for a given time period, e.g. one year. It is also possible to calculate this ratio for a specific energy source. It is also possible to work out the level of energy independence, which is the quotient of primary energy production divided by total energy consumption within a geographic or economic unit. This gives a rough indication of the coverage of needs by primary energy production. a) These two levels are not complementary in that changes in stocks may mean that the two percentages, when added together, do not equal 100%. b) When a country is a net exporter of energy, the level of energy dependency may be negative. Lower heating value (net calorific Quantity of heat liberated by the complete combustion of a unit 26 value) volume or weight of a fuel in the determination of which the water produced is assumed to remain as a vapour and the heat not recovered. Micro-grid A group of power generators, connected with intelligent switchgear and remote controls to supply the electricity demands of local users. A typical micro-grid might have an output of about 10 MW. Micro-turbine A small turbine-engine powered electric generation plant. Like the larger gas turbine used in major power generating facilities, micro-turbines also have one moving part: the turbine shaft. Micro-turbines are very low maintenance, efficient, fairly quiet and come in various outputs ranging from 30kW to 75 kW. Micro-turbines can run on numerous fuels and can also be linked together to provide outputs of several megawatts. Net energy of an energy-producing installation The gain in energy obtained from an energy-production installation during an assumed lifespan; in other words, the amount of energy produced during that period, all the energy required for the construction, operation and subsequent dismantling of the installation. Performance of consumer equipment The ratio of the useful energy produced by the consumer equipment to the energy supplied to it. Note: A distinction is made between theoretical performance in set conditions and actual performance in real conditions. The second level of performance is generally lower than the first. Photovoltaic(s) (PVs) Pertains to the direct conversion of light into electricity. Photovoltaic cell Component commonly called a solar cell that can convert light energy into electrical energy. Cells can be combined to form an array to provide greater overall output. Pollutant Any physical or chemical characteristic or material present in environmental media (air, water, soil), emitted either by human activities or by natural processes, and adversely affecting man or the environment. Photovoltaic module A number of photovoltaic cells electrically interconnected in either series or parallel and mounted together, usually in a sealed unit of convenient size for shipping, handling and assembling into panels and/or arrays. The term “module” is often used interchangeably with the term “panel”. Uninterruptible power supply (UPS) A device that supplies back-up power to electronic equipment. When an electrical surge, sag or outage is sensed, can instantly switch from grid-supplied power to a back-up power supply such as a battery. The end result is a constant, clean supply of power to the end-using equipment. UPS’s come in sizes ranging from units that can supply a couple of hundred watts to a laptop, to units that supply megawatts of power to entire data centres. NOTES: 1. Own consumption by energy industries: The consumption of self-produced and of purchased energy by energy producers and transformers in operating their installations, (e.g. heating, light). 27 Note: For hydraulic pumping, the balance resulting from pumping (difference between electricity produced and electricity consumed for pumping) is generally ascribed to the consumption by the electricity sector; the consumption by auxiliaries is included in this entry. 2. Gross National Product (GNP): Total productions of goods and services by the subject of a country at home and abroad. In national income accounting, it is a measure of the performance of nation’s economy, within a specific accounting period (usually a year). 2.1 The concepts and measuring relevance are increasingly being brought into question because, as a monetary measure, many welfare transactions are not taken into account (such as the shadow economy, exchange of goods or services, household work). 2.2 As a rule the GNP is simultaneously drawn up and presented in three aspects: its formation, distribution, and consumption. 2.3 The nominal GNP (market price) is fundamentally distinguished from the real GNP (prices of the base year), with the sole purpose of recording the quantitative changes. Equivalences Typical reference values 1 kWh = 0,860 Mcal 1 cal = 4,1868 Joules 1 tonne of Oil = 10 000 Mcal 1 tonne of Coal = 7 000 Mcal 1 tonne of Natural Gas = 10 790 Mcal Heat equivalent of 1 kWh 1 kWh = 86 gram Oil equivalent 1 kWh = 123 gram Coal equivalent 1 kWh = 79.7 gram Natural Gas equivalent Electricity equivalent of 1 tonne of Fuel 1 tonne of Oil <> 11 628 kWh 1 tonne of Coal <> 8 140 kWh 1 tonne of Natural Gas <> 12 547 kWh Note: The unit “toe” (tonne of oil equivalent) is commonly used to compare the energy content of different fuels. 28 References World Energy Council: Energy Dictionary – 1992 Sears and Zemansky’s – University Physics with Modern Physics. Young & Freedman –Tenth edition - 2000 Waste to Energy in the Great Lisbon. Armando Manuel de Jesus Branco. The International UIE/EDP Conference “Electricity for a sustainable urban development” – Lisbon 1/4November 2000 The trend and the Benefits of Multi-Energy Services for a New Generation of Industrial and Commercial Customers. Ben Banerjee, Arshad Mansoor, Art Mannion. The International UIE/EDP Conference “Electricity for a sustainable urban development” – Lisbon 1/4November 2000 National Strategies for Management of Fluctuations in Wind Power and CHP .The case of Denmark. Henrik Lund. The International UIE/EDP Conference “Electricity for a sustainable urban development” – Lisbon 1/4November 2000 Roof-and facade integration of PV systems in a laboratory building, renovation of the ECN building 31 with PV. Tjerk Reijenga and Henk Kaan. The International UIE/EDP Conference “Electricity for a sustainable urban development” – Lisbon 1/4November 2000 Hybrid distributed generation systems using renewable. Energy sources to enhance fuel cell technologies. Annette Von Jouanne, Alan Wallace, Alexandre Yokochi. The International UIE/EDP Conference “Electricity for a sustainable urban development” – Lisbon 1/4November 2000 The value of micro-generation in liberalised electricity markets. Andrew Wright. The International UIE/EDP Conference “Electricity for a sustainable urban development” – Lisbon 1/4November 2000 2001-Annual Energy Review, January 2002 – European Commission Parques Eólicos em Portugal: da ideia à realização . Jorge A Gil Saraiva – Revista Energia Solar Jan/Jun96 Energia Nuclear. Mitos e Realidades. Jaime Oliveira e Eduardo Martinho - Edição de O MIRANTE 2000 Photovoltaic Technologies and Their Future Potential (A Thermie Programme Action – For the Commission of the European Communities. Directorate-General XVII for Energy 1993) Les photopiles en quête de punch. Sylvain Le Gall. – Sciency&Vie –Mars 2001 Le Gaz naturel remonte le courant. Jean Bernard. - Sciency&Vie –Mars 2001 Energy from Biomass, Principles and Applications (A Thermie Programme Action – European Commission. Directorate-General for Energy DGXVII) L’hydrogène tombera-t-il pile? Riadh Elloumi et Nicolas Forget – Sciency&Vie –Mars 2001 Renewable Energy The federal Minister for Research and Technology (BMFT) - Bonn, Dec 1992 29 New Renewable Energy Norwegian Developments, July 1998 Dicionário de Terminologia Energética APE (Associação Portuguesa de Energia ) 3ª edição , 2001 ”A new unit for Finland” by Risto Tarjanne and Sauli Rissanen – Nuclear Enginnering International – January 2001 ”Boiling Point” by James Luckey –Energy Markets –November 2000 ”Photovoltaics: technology overview” – M.ª Green – Energy Policy – November 2000 (Special Issue “The viability of Solar Photovoltaics”) ”Keynote Address on Technology Innovation in a Competitive Environment”, Richard E. Balzhiser – Electric Power Research Institute (Conference IEA/EDF/UNIPEDE “New Electricity 21: Designing a Sustainable Electric System for the Twenty-First Century” – Paris 22/24 May 1995) ”Rapporteur’s Report on Workshop 3: Power Producers and Global Climate Change Issues” Jeffrey Skeer –International Energy Agency (Conference IEA/EDF/UNIPEDE “New Electricity 21: Designing a Sustainable Electric System for the Twenty-First Century” – Paris 22/24 May 1995) Global Warming Reduction Through Generation Efficiency Improvements, A.F.Armor-Fossil Power Plants and G.T.Preston-Electric Power Research Institute (Conference IEA/EDF/UNIPEDE “New Electricity 21: Designing a Sustainable Electric System for the Twenty-First Century” – Paris 22/24 May 1995) “The contribution of Energy efficiency and Renewables to the reduction of CO2 Emissions” –José António Paredes and Ignacio Mateo –Union Eléctrica Fenosa (Conference IEA/EDF/UNIPEDE “New Electricity 21: Designing a Sustainable Electric System for the Twenty-First Century” – Paris 22/24 May 1995) “Biomass for Electricity” – P.Barlucci, G.Neri and G.Trebbi – ENEL (Conference IEA/EDF/UNIPEDE “New Electricity 21: Designing a Sustainable Electric System for the Twenty-First Century” – Paris 22/24 May 1995) “Overview of current and future Clean Coal Technologies: A utility point of view” –Alain Darthenay -EDF (Conference IEA/EDF/UNIPEDE “New Electricity 21: Designing a Sustainable Electric System for the Twenty-First Century” – Paris 22/24 May 1995) “Competitive Energy Markets: The Effective Route to improving the Environment” – Dr D.J. Swinden – UK (Conference IEA/EDF/UNIPEDE “New Electricity 21: Designing a Sustainable Electric System for the Twenty-First Century” – Paris 22/24 May 1995) Schroder Salomon Smith Barney “The new Electric Economy. Powering the future”- May 2001 Arthur D. Little White Paper “Reliability and Distributed Generation” – 2000 www.hexis.ch W.A.Benesch “Planning New Coal-fired Power Plants” – VGB PowerTech, June 2001 30 K.Huttenhofer , A.Lezuo “ Cogeneration Power Plant Concepts Using Advanced Gas Turbines” –VGB PowerTech, June 2001 Susan M Schoenung “MPS Review Energy Storage. The long and short of it” – May2001 Modern Power Systems Gas turbine World Handbook, 2000/2001 Mainz-Wiesbaden “GUD” plant (“GUD” – Gas and Steam/Siemens) Nordjyllandsvaerket Puertollano
2288	https://forum.puzzlebaron.com/forum/puzzle-baron/logic-puzzles/29419-any-set-of-well-known-techniques-tips-for-a-new-logic-puzzles-player	Any set of well-known techniques/ tips for a new Logic Puzzles player? - Puzzle Baron Login or Sign Up Logging in... - [x] Remember me Log in Or Sign Up Forgot password or user name? Log in with [x] Search in titles only - [x] Search in Logic Puzzles only Search Advanced Search Forums Online Puzzles Printable Puzzles Galleries Latest Posts Today's Posts Member List Calendar Home Forum Puzzle Baron Logic Puzzles If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. Any set of well-known techniques/ tips for a new Logic Puzzles player? Collapse X Collapse Posts Latest Activity Photos Page of 2 Filter Time All Time Today Last Week Last Month Show All Discussions only Photos only Videos only Links only Polls only Events only Filtered by: Clear All new posts Previous12templateNext Paul671919 Member Join Date: Apr 2022 Posts: 3 Share Image 28 Tweet #1 Any set of well-known techniques/ tips for a new Logic Puzzles player? 04-10-2022, 03:22 PM I'm a new Logic Puzzles player and struggling to get up to speed - I seem to keep making avoidable mistakes, and end up solving a very low percentage. Is there some guidebook or available list of general techniques, tips that people have found useful and apply to these puzzles? I'm not trying to ace every puzzle, it's just discouraging to work 30-40 minutes on something (where the average solve time is 12-15 minutes!) only to find that I've flubbed something. Probably over a long time of doing these I'd figure out my own techniques but it would be nice to jump-start my learning so that I can get to a competent level faster. Any advice/help would be much appreciated. Tags:None admin Administrator Join Date: Apr 2007 Posts: 732 Share Image 30 Tweet #2 04-10-2022, 03:25 PM Try this: Logic Puzzles \| How to Solve a Logic Puzzle (puzzlebaron.com) If you enjoy our puzzles, please consider upgrading to a premium account to remove all ads and help support us financially. Thanks for your support! Comment Post Cancel kittyboo Member Join Date: Feb 2019 Posts: 3 Share Image 33 Tweet #3 04-10-2022, 05:44 PM That tutorial looks very useful, I probably should have gone through it when I started on the site. Oops. One of the great advantages of this site over paper puzzles is the "Clear Errors" button, and I'd advise using it a lot. After every move, even. It helps you catch mistakes early before they propagate through the whole puzzle. The hint button can be useful as well. When you get stuck, instead of guessing, get a hint and pay attention to the reasoning it gives. Sometimes this is more useful than others, of course. Apart from that, you don't say what level or size you're working on, but I do think it's worth starting small and working up, just like it's worth starting with the easy ones. I still don't do the largest ones because they're just too much information for me to hold in my head. Good luck! Comment Post Cancel Paul671919 Member Join Date: Apr 2022 Posts: 3 Share Image 35 Tweet #4 04-11-2022, 06:37 AM Thank you very much! The tutorial looks really good, and the advice to try using the site rather than paper sounds really good too. thanks for taking the time to write back to me. Comment Post Cancel boxeeboxeebox Member Join Date: Feb 2019 Posts: 29 Share Image 37 Tweet #5 04-11-2022, 03:42 PM If you're playing on the site, a big help for me was changing the way the puzzle is displayed (or rather, the icons used for the checks and crosses). If you go to User Settings - Puzzle Baron User Profile Settings and scroll down a bit, you'll see a section called "Custom Grid Icons" and you can pick the set you prefer there. I like V3 since it offers much greater contrast between the true and false squares and helps me identify what the tutorial calls "pseudo-true pairs" and when there's only one possibility left in a column/row. Comment Post Cancel JedMedGrey Member Join Date: Dec 2020 Posts: 129 Share Image 40 Tweet #6 04-11-2022, 11:26 PM Welcome, Paul - #1 - don't rush. #2 - stick with it! #3 - start with the smallest and easiest puzzles. #4 - Use the tutorial. #5 - Read through all of the clues before you start marking anything. Then mark the really clear ones first, especially those that are "True", as doing so will automatically fill in a bunch of squares. #6 - figure out a reward for yourself as a treat when you do well on a puzzle. #7 - don't give up, but don't get mad if you seem to be slow. You will get faster, the puzzles will start to seem easier, but it might take a few weeks. JMG Comment Post Cancel KarmaQ Member Join Date: Sep 2020 Posts: 1 Share Image 42 Tweet #7 04-13-2022, 11:49 AM Hi Paul, I'll admit I have never used the tutorial on this site, but I've been doing these puzzles since I was a kid. My dad used to buy me Dell Logic Puzzle books to take on road trips to keep me entertained. So this site is a bit of escape for me and I often use it to clear my head when I'm stuck on a problem at work. But to your question: Everyone has their own methods I imagine so I'll just tell you what I do. (Also, kudos for the tip on contrasting icons. I've never been in the settings field so had no idea that was an option. ) I don't read through all the clues before starting. For me, that's too much of an info dump all at once. But I'm also a gal who has like 50 browser windows open in my mind and on my screen. I take the clues one at a time in the order they are given, marking what I know to be true from each one. Here are 3 common types of clues: used in these puzzles: a. A clue that is definitive - meaning there is one distinctive true piece of information. Ex. 1:Adrienne's film received 8 nominations. So you may not know what film title received 8 nominations yet, but you definitely know whose film got that many. So at the Intersection of Adrienne and 8 nominations, you'd mark true. Ex. 2 Sea of Dreams was Not directed by Maddie.In this case, you know definitively which film isNOTMaddie's. At the intersection of Maddie and Sea of Dreams, you'd mark false. The great thing about working these puzzles on the site is that it will highlight the row and color where you're about to mark something so watch those side highlights to avoid accidentally marking the wrong box. This can happen if you're going too fast. b. Either/or clue. Clue: The action film is either the film that received 2 nominations OR the film directed by Paul. Although the clue starts off talking about the action film, you can't mark anything true or false about the action film yet (on your first pass through the clues). But the clue does give you definitive information about both Paul and the # of nominations. If the action film is Either A or B - A cannot = B. So Paul's film did NOT get 2 nominations. You mark that intersection as false. Later, as other clues give you more details about the action film, you will be able to determine whether it is A or B. C. A is more/less than B clue. Clue: Sea of Dreams received fewer nominations than Paul's film. This clue gives you 3 pieces of information. Sea of Dreams is NOT Paul's film. Mark that intersection as false. Sea of Dreams does NOT have the MOST nominations. At the intersection of Sea of Dreams and the biggest number of nominations, mark as False. Paul's film does NOT have the LEAST nominations. At the intersection of Paul and the smallest number of nominations, mark as false. As you read through more clues, you will get information that builds from these and helps you to eliminate possibilities. Below is a section of a larger puzzle grid to illustrate the clues above: (x = False mean the same. I'm using words to illustrate what came from each clue more clearly and the Xs to illustrate what was inferred. Ex. If Adrienne has 8 noms, no other director can have that so everyone but Adrienne gets an X for the row with 8 noms. # of NominationsAdrienne Maddie Paul ZoeFilm StyleABC Murders Comedy Capers Sea of Dreams Annie Get your Gun 2 xFalse 4 x 6 x x x 8Truex x x xFalse Film Title ABC Murders Comedy Capers x x Sea of Dreams xFalseFalse Annie Get your Gun Here are some things you would probably determine upon a second read through of the above clues: If Adrienne has the most nominations: (from the first sample clue) and Sea of Dreams does NOT have the most..... Adrienne's film is NOT Sea of Dreams. Mark that intersection as False. Based on the last clue, we know Paul doesn't have the least amount of nominations. But we also know Adrienne has the most. So the most nominations Paul can have is 6. Also, based on the last clue, Sea of Dreams has fewer nominations than Paul. So Sea of Dreams can NOT have 6 nominations. Mark the intersection of Sea of Dreams and 6as False. I go through the clues one at a time, and mark what facts each clue offers on the first pass. Then I go through the clues a second time, using them to build on each other. Then it's just rinse and repeat as you continue to whittle down choices. So take the clue: The action film is either the film that received 2 nominations OR the film directed by Paul. Eventually, other clues would eliminate one of those possibilities. A clue might say: "The action film has more nominations than Comedy Capers." That would tell you that A. The action film is NOT Comedy Capers - and it has more than 2 nominations. B. Since the action is A or B - and we've just eliminated A (2 nominations), the action film is Paul's film. C. That also means Comedy Capers is NOT Paul's film. D. Comedy Capers also does NOT have the most nominations.... so it's also NOT Adrienne's film. E. Since Comedy Capers has fewer nominations than both Paul and Adrienne's films, it can AT MOST have 4 nominations. So we mark it as false for having either 6 or 8. See how that one clue (having built on information from previous clues) helped us eliminate so many possibilities? Hope it helps. Happy puzzling! Comment Post Cancel kgf Member Join Date: Aug 2015 Posts: 15 Share Image 46 Tweet #8 04-14-2022, 12:33 PM I guess going through all the clues first may be helpful, but it just seems like a time waster to me. However, I do scan the ends of all the clues for these words: "are all different", then I start with that clue. Sometimes a puzzle will have one clue that has multiple pieces: "Scott, the person who bought the bike, the person who paid $375, the person who went second, and Nellie are all different people". That's a lot of information right there, so use this to eliminate a bunch of possibilities before moving on to other clues. There's never more than one of these large clues. Some puzzles have smaller versions with only three pieces, and sometimes there are two clues like that. I always start with these clues. On some puzzles though, they throw a wrench in the works by phrasing it differently: " The seven people are.." Same kind of clue, just worded differently. Also, use the grid itself. (This is really hard to put into words) For example, say Scott could have bought seven different items at an auction, but you've eliminated five of the possibilities. So Scott has all those boxes filled in as negatives except for the two remaining possibilities. Scan across the grid for other columns that have those 2 items blocked off as negatives - now you know that since those columns can't be either of the two items that Scott may have won, those columns can't be associated with Scott. Use the grid as a visual aid, not just the text clues. Comment Post Cancel brilliantbelle Member Join Date: Dec 2021 Posts: 2 Share Image 48 Tweet #9 04-15-2022, 03:52 PM Start with what you know is true. Then go to the clues that are like "Table 1, Bryan, and the person who ordered fish are all different diners" because then you know that Bryan isn't at table 1 AND he didn't order fish. Go from there. Comment Post Cancel contrary Premium Member Join Date: Feb 2019 Posts: 16 Share Image 51 Tweet #10 04-17-2022, 07:36 PM If your goal is to learn logic puzzles, go slower, read the clues, fill in everything you can based on the clue and if you get stuck use a hint. When you use a hint don't just mark the thing the hint tells you, stop and figure out what the hint actually means and how the clues interact with one another. Each hint should be a learning opportunity to learn to see the clues better. Once you know how to solve logic puzzles and want to get faster, start trying to solve things in the top row before you go back through and use the rest of the grid. Comment Post Cancel CALLIE10 Member Join Date: Oct 2019 Posts: 1 Share Image 53 Tweet #11 05-02-2022, 12:28 PM I think I have solved maybe 2 puzzles in the hard section (7) without having to do trial and error or hints and having no errors. I usually go through the clues 3-4 times and then if I don't have enough clues (which I almost NEVER have enough clues or the right kind of clues) I just try putting an X somewhere and then checking to see if it's right. 50% accuracy on that. After I get a correct answer in top left corner, I always x off the boxes to the bottom or right that correlate. Make sure when you get a "Sarah had 3 more than Kansas" that you go down and x off that Sarah can't come from Kansas. If it's "Of Sarah and Kansas, one is right-handed and the other made 120 points" x off the connection of Sarah and Kansas. Then I usually shove that clue to the used section and come back to it later to see if I can answer it. Comment Post Cancel AKLong Member Join Date: Jun 2019 Posts: 1 Share Image 55 Tweet #12 05-03-2022, 06:07 PM Easiest method for me is to mark what is true 1st (ex: Mark wore blue shirt) then go back & determine the other answers. Start with small puzzles & work your way up to the bigger ones. Comment Post Cancel act_now Premium Member Join Date: Feb 2019 Posts: 2 Share Image 58 Tweet #13 05-04-2022, 10:51 PM These are good suggestions. One thing I tried that really cut down my time was starting with the last clue and working my way up. Comment Post Cancel JedMedGrey Member Join Date: Dec 2020 Posts: 129 Share Image 60 Tweet #14 05-05-2022, 09:38 AM per CALLIE10 "Make sure when you get a "Sarah had 3 more than Kansas" that you go down and x off that Sarah can't come from Kansas. If it's "Of Sarah and Kansas, one is right-handed and the other made 120 points" x off the connection of Sarah and Kansas. "Also x out the right-handed and 120 pts connection! act_now - I too often find that clues further down on the list are more straightforward and allow one to quickly discard earlier clues with little analysis. I posted earlier about reading all the clues first, and got some negative comments on that. I did/do not mean that you have to understand all of the details in each clue, but by quickly going through the list, you can mark the obviously true or false relationships, e.g., "Tony paid $6.75" or "Brenda did not buy ham and cheese." The TRUE relationships are particularly helpful and time-saving as so many boxes get filled in automatically. Sometimes there will be another clue that you can then automatically discard, such as "Tony did not pay $9.75." Or a clue will contain within it a relationship you can then easily see is false, such as "Of Tony and the person who ordered club soda, one paid $8.75 and the other had turkey on rye." Well, you already know Tony paid $6.75, so he must have bought turkey on rye, and whoever paid $8.75 had club soda. By entering the TRUE Tony = $6.75 , you don't have to spend as much time thinking about the Tony v club soda/$8.75 v turkey on rye clue. I then do the clues that say such things as "Tony, ham and cheese, and $9.75 represent 3 different people" or "The five contestants are: blah, blah, blah". Some books/guides say to do these types of clues first, as it can be easier to make sure all of the relationships are entered before the graph gets too cluttered. I have tried both ways, but not really analyzed which method works better for me. Then look at the tougher clues for those that share common items. This can really help you narrow the choices for different items to paired with. By quickly reading all of the clues first, it can stick in your mind that Brenda (or the price, or the drink, or the sandwich) is mentioned in more than one clue , and you can compare those clues before tackling the more obscure ones. As you practice, it will become easier to see the relationships and understand the clues. Above all, celebrate your achievements and have fun! Comment Post Cancel margjack Member Join Date: Jul 2020 Posts: 2 Share Image 62 Tweet #15 05-09-2022, 10:02 AM Starting out, my suggestion is to not worry about your time at first and don't guess - if you get stuck, use a hint. this will help you learn how to work the puzzles better. Comment Post Cancel Previous12templateNext Help Contact Us Privacy Go to top Powered by vBulletin® Version 6.1.2 Copyright © 2025 MH Sub I, LLC dba vBulletin. All rights reserved. All times are GMT-5. This page was generated at 04:47 PM. Working... Yes No OK OK Cancel 😀 😂 🥰 😘 🤢 😎 😞 😡 👍 👎 ☕
2289	https://simple.wikipedia.org/wiki/Lysosome	Lysosome - Simple English Wikipedia, the free encyclopedia Jump to content [x] Main menu Main menu move to sidebar hide Getting around Main page Simple start Simple talk New changes Show any page Help Contact us About Wikipedia Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Give to Wikipedia Create account Log in [x] Personal tools Give to Wikipedia Create account Log in [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 References Lysosome [x] 75 languages Afrikaans العربية Azərbaycanca বাংলা Беларуская Български Bosanski Català Čeština Dansk Deutsch Eesti Ελληνικά English Español Esperanto Euskara فارسی Français Gaeilge Gaelg Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Íslenska Italiano עברית Jawa ქართული Қазақша Kurdî Latina Latviešu Lëtzebuergesch Lietuvių Lingua Franca Nova Magyar Македонски മലയാളം Bahasa Melayu Nederlands 日本語 Norsk bokmål Occitan Oʻzbekcha / ўзбекча Plattdüütsch Polski Português Română Русский Shqip සිංහල Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Sunda Suomi Svenska Tagalog தமிழ் Татарча / tatarça తెలుగు ไทย Тоҷикӣ Türkçe Українська Tiếng Việt 吴语粵語中文 Change links Page Talk [x] English Read Change Change source View history [x] Tools Tools move to sidebar hide Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sandbox Edit interlanguage links Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikidata item From Simple English Wikipedia, the free encyclopedia Duration: 10 seconds.0:10 Lysosome A lysosome is a cellorganelle. They are like spheres and they have hydrolyticenzymes which can break down almost all kinds of biomolecules, including proteins, nucleic acids, carbohydrates, lipids, and cellular debris. They contain more than 50 different enzymes. By convention, lysosome is the term used for animal celis. In plant cells, vacuoles do similar functions. With a wider definition, lysosomes are found in the cytoplasm of plant and protists as well as animalcell. Lysosome diagram Lysosomes work like the digestive system to break down, or digest, proteins, acids, carbohydrates, dead organelles, and other unwanted materials. They break up larger molecules into smaller molecules. Those smaller molecules can then be used again as building blocks for other large molecules.Sometimes, when the cell itself is dying or is dead the lysosomes will eat up the cell. This is why they are also known as 'suicide bags' of cells. References [change \| change source] ↑De Duve C. 2005. The lysosome turns fifty. Nature Cell Biology7 (9): 847–9. ↑Lysosome. Yale University. Archived 2015-10-30 at the Wayback Machine ↑ Jump up to: 3.03.1Fullick, Ann (2008). Edexcel AS Biology Students' Book. pp.142-143. ISBN978-1-4058-9632-0. This short article about biology can be made longer. You can help Wikipedia by adding to it. Retrieved from " Category: Organelles Hidden categories: Webarchive template wayback links Biology stubs This page was last changed on 9 December 2024, at 02:55. Text is available under the Creative Commons Attribution-ShareAlike License and the GFDL; additional terms may apply. See Terms of Use for details. Privacy policy About Wikipedia Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Lysosome 75 languagesAdd topic
2290	https://www.postgresql.org/docs/9.3/datatype-numeric.html	August 14, 2025: PostgreSQL 17.6, 16.10, 15.14, 14.19, 13.22, and 18 Beta 3 Released! Documentation → PostgreSQL 9.3 Supported Versions: Current (17) / 16 / 15 / 14 / 13 Development Versions: 18 / devel Unsupported versions: 12 / 11 / 10 / 9.6 / 9.5 / 9.4 / 9.3 / 9.2 / 9.1 / 9.0 / 8.4 / 8.3 / 8.2 This documentation is for an unsupported version of PostgreSQL. You may want to view the same page for the current version, or one of the other supported versions listed above instead. \| PostgreSQL 9.3.25 Documentation \| \| \| \| \| --- --- \| Prev \| Up \| Chapter 8. Data Types \| Next \| 8.1. Numeric Types Numeric types consist of two-, four-, and eight-byte integers, four- and eight-byte floating-point numbers, and selectable-precision decimals. Table 8-2 lists the available types. Table 8-2. Numeric Types \| Name \| Storage Size \| Description \| Range \| \| smallint \| 2 bytes \| small-range integer \| -32768 to +32767 \| \| integer \| 4 bytes \| typical choice for integer \| -2147483648 to +2147483647 \| \| bigint \| 8 bytes \| large-range integer \| -9223372036854775808 to +9223372036854775807 \| \| decimal \| variable \| user-specified precision, exact \| up to 131072 digits before the decimal point; up to 16383 digits after the decimal point \| \| numeric \| variable \| user-specified precision, exact \| up to 131072 digits before the decimal point; up to 16383 digits after the decimal point \| \| real \| 4 bytes \| variable-precision, inexact \| 6 decimal digits precision \| \| double precision \| 8 bytes \| variable-precision, inexact \| 15 decimal digits precision \| \| smallserial \| 2 bytes \| small autoincrementing integer \| 1 to 32767 \| \| serial \| 4 bytes \| autoincrementing integer \| 1 to 2147483647 \| \| bigserial \| 8 bytes \| large autoincrementing integer \| 1 to 9223372036854775807 \| The syntax of constants for the numeric types is described in Section 4.1.2. The numeric types have a full set of corresponding arithmetic operators and functions. Refer to Chapter 9 for more information. The following sections describe the types in detail. 8.1.1. Integer Types The types smallint, integer, and bigint store whole numbers, that is, numbers without fractional components, of various ranges. Attempts to store values outside of the allowed range will result in an error. The type integer is the common choice, as it offers the best balance between range, storage size, and performance. The smallint type is generally only used if disk space is at a premium. The bigint type is designed to be used when the range of the integer type is insufficient. SQL only specifies the integer types integer (or int), smallint, and bigint. The type names int2, int4, and int8 are extensions, which are also used by some other SQL database systems. 8.1.2. Arbitrary Precision Numbers The type numeric can store numbers with a very large number of digits and perform calculations exactly. It is especially recommended for storing monetary amounts and other quantities where exactness is required. However, arithmetic on numeric values is very slow compared to the integer types, or to the floating-point types described in the next section. We use the following terms below: the precision of a numeric is the total count of significant digits in the whole number, that is, the number of digits to both sides of the decimal point. The scale of a numeric is the count of decimal digits in the fractional part, to the right of the decimal point. So the number 23.5141 has a precision of 6 and a scale of 4. Integers can be considered to have a scale of zero. Both the maximum precision and the maximum scale of a numeric column can be configured. To declare a column of type numeric use the syntax: NUMERIC(precision, scale) The precision must be positive, the scale zero or positive. Alternatively: NUMERIC(precision) selects a scale of 0. Specifying: NUMERIC without any precision or scale creates a column in which numeric values of any precision and scale can be stored, up to the implementation limit on precision. A column of this kind will not coerce input values to any particular scale, whereas numeric columns with a declared scale will coerce input values to that scale. (The SQL standard requires a default scale of 0, i.e., coercion to integer precision. We find this a bit useless. If you're concerned about portability, always specify the precision and scale explicitly.) Note: The maximum allowed precision when explicitly specified in the type declaration is 1000; NUMERIC without a specified precision is subject to the limits described in Table 8-2. If the scale of a value to be stored is greater than the declared scale of the column, the system will round the value to the specified number of fractional digits. Then, if the number of digits to the left of the decimal point exceeds the declared precision minus the declared scale, an error is raised. Numeric values are physically stored without any extra leading or trailing zeroes. Thus, the declared precision and scale of a column are maximums, not fixed allocations. (In this sense the numeric type is more akin to varchar(n) than to char(n).) The actual storage requirement is two bytes for each group of four decimal digits, plus three to eight bytes overhead. In addition to ordinary numeric values, the numeric type allows the special value NaN, meaning "not-a-number". Any operation on NaN yields another NaN. When writing this value as a constant in an SQL command, you must put quotes around it, for example UPDATE table SET x = 'NaN'. On input, the string NaN is recognized in a case-insensitive manner. Note: In most implementations of the "not-a-number" concept, NaN is not considered equal to any other numeric value (including NaN). In order to allow numeric values to be sorted and used in tree-based indexes, PostgreSQL treats NaN values as equal, and greater than all non-NaN values. The types decimal and numeric are equivalent. Both types are part of the SQL standard. 8.1.3. Floating-Point Types The data types real and double precision are inexact, variable-precision numeric types. In practice, these types are usually implementations of IEEE Standard 754 for Binary Floating-Point Arithmetic (single and double precision, respectively), to the extent that the underlying processor, operating system, and compiler support it. Inexact means that some values cannot be converted exactly to the internal format and are stored as approximations, so that storing and retrieving a value might show slight discrepancies. Managing these errors and how they propagate through calculations is the subject of an entire branch of mathematics and computer science and will not be discussed here, except for the following points: If you require exact storage and calculations (such as for monetary amounts), use the numeric type instead. If you want to do complicated calculations with these types for anything important, especially if you rely on certain behavior in boundary cases (infinity, underflow), you should evaluate the implementation carefully. Comparing two floating-point values for equality might not always work as expected. On most platforms, the real type has a range of at least 1E-37 to 1E+37 with a precision of at least 6 decimal digits. The double precision type typically has a range of around 1E-307 to 1E+308 with a precision of at least 15 digits. Values that are too large or too small will cause an error. Rounding might take place if the precision of an input number is too high. Numbers too close to zero that are not representable as distinct from zero will cause an underflow error. Note: The extra_float_digits setting controls the number of extra significant digits included when a floating point value is converted to text for output. With the default value of 0, the output is the same on every platform supported by PostgreSQL. Increasing it will produce output that more accurately represents the stored value, but may be unportable. In addition to ordinary numeric values, the floating-point types have several special values: Infinity -Infinity NaN These represent the IEEE 754 special values "infinity", "negative infinity", and "not-a-number", respectively. (On a machine whose floating-point arithmetic does not follow IEEE 754, these values will probably not work as expected.) When writing these values as constants in an SQL command, you must put quotes around them, for example UPDATE table SET x = 'Infinity'. On input, these strings are recognized in a case-insensitive manner. Note: IEEE754 specifies that NaN should not compare equal to any other floating-point value (including NaN). In order to allow floating-point values to be sorted and used in tree-based indexes, PostgreSQL treats NaN values as equal, and greater than all non-NaN values. PostgreSQL also supports the SQL-standard notations float and float(p) for specifying inexact numeric types. Here, p specifies the minimum acceptable precision in binary digits. PostgreSQL accepts float(1) to float(24) as selecting the real type, while float(25) to float(53) select double precision. Values of p outside the allowed range draw an error. float with no precision specified is taken to mean double precision. Note: Prior to PostgreSQL 7.4, the precision in float(p) was taken to mean so many decimal digits. This has been corrected to match the SQL standard, which specifies that the precision is measured in binary digits. The assumption that real and double precision have exactly 24 and 53 bits in the mantissa respectively is correct for IEEE-standard floating point implementations. On non-IEEE platforms it might be off a little, but for simplicity the same ranges of p are used on all platforms. 8.1.4. Serial Types The data types smallserial, serial and bigserial are not true types, but merely a notational convenience for creating unique identifier columns (similar to the AUTO_INCREMENT property supported by some other databases). In the current implementation, specifying: CREATE TABLE tablename ( colname SERIAL ); is equivalent to specifying: CREATE SEQUENCE tablename_colname_seq; CREATE TABLE tablename ( colname integer NOT NULL DEFAULT nextval('tablename_colname_seq') ); ALTER SEQUENCE tablename_colname_seq OWNED BY tablename.colname; Thus, we have created an integer column and arranged for its default values to be assigned from a sequence generator. A NOT NULL constraint is applied to ensure that a null value cannot be inserted. (In most cases you would also want to attach a UNIQUE or PRIMARY KEY constraint to prevent duplicate values from being inserted by accident, but this is not automatic.) Lastly, the sequence is marked as "owned by" the column, so that it will be dropped if the column or table is dropped. Note: Because smallserial, serial and bigserial are implemented using sequences, there may be "holes" or gaps in the sequence of values which appears in the column, even if no rows are ever deleted. A value allocated from the sequence is still "used up" even if a row containing that value is never successfully inserted into the table column. This may happen, for example, if the inserting transaction rolls back. See nextval() in Section 9.16 for details. Note: Prior to PostgreSQL 7.3, serial implied UNIQUE. This is no longer automatic. If you wish a serial column to have a unique constraint or be a primary key, it must now be specified, just like any other data type. To insert the next value of the sequence into the serial column, specify that the serial column should be assigned its default value. This can be done either by excluding the column from the list of columns in the INSERT statement, or through the use of the DEFAULT key word. The type names serial and serial4 are equivalent: both create integer columns. The type names bigserial and serial8 work the same way, except that they create a bigint column. bigserial should be used if you anticipate the use of more than 231 identifiers over the lifetime of the table. The type names smallserial and serial2 also work the same way, except that they create a smallint column. The sequence created for a serial column is automatically dropped when the owning column is dropped. You can drop the sequence without dropping the column, but this will force removal of the column default expression. \| \| \| \| --- \| Prev \| Home \| Next \| \| Data Types \| Up \| Monetary Types \|
2291	https://english.stackexchange.com/questions/42864/origin-of-world-weary	meaning - Origin of "world-weary" - English Language & Usage Stack Exchange Join English Language & Usage By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community English Language & Usage helpchat English Language & Usage Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Origin of "world-weary" Ask Question Asked 14 years ago Modified14 years ago Viewed 844 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. As a non native speaker, I thought that the words world weary referred to someone who carries the world on his/her shoulder or someone who wears/puts on the world. Then I noticed that the word weary has a different meaning. Is there any relationship between wear and weary in this context? Is there anyone who is able to explain the etymology of this phrase? meaning etymology Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Sep 22, 2011 at 7:43 Bogdan 2,277 6 6 gold badges 29 29 silver badges 39 39 bronze badges asked Sep 22, 2011 at 6:49 mirazmiraz 587 2 2 gold badges 8 8 silver badges 15 15 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. I always thought the usage world-weary was derived from the story of Atlas. Atlas is described as Atlas Telamon (Atlas the Enduring) and also as "The world-weary Atlas" when Hercules encounters him during his quest for The Apples of the Hesperides.(I couldn't find the Homeric greek for "the world-weary Atlas.") Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 22, 2011 at 7:29 AutoresponderAutoresponder 4,439 21 21 silver badges 25 25 bronze badges 4 It's a nice visualization, and may indeed be the source of the phrase, but I doubt the Greek etymology for the words "to wear" and "weary" is any closer than the English. The query seems to imply the words themselves are related.user13141 –user13141 2011-09-22 07:36:49 +00:00 Commented Sep 22, 2011 at 7:36 You're right, the two words are distinct. Maybe I was taking the question too literally- OP asks if there is a relation between wear and weary "in this context.Autoresponder –Autoresponder 2011-09-22 07:41:39 +00:00 Commented Sep 22, 2011 at 7:41 @onomatomaniak : I'll be happy to post my reply as a comment if that will be better.Autoresponder –Autoresponder 2011-09-22 07:49:00 +00:00 Commented Sep 22, 2011 at 7:49 1 This is exactly what I meant, but I could not remember the example of Atlas. I am glad that you mentioned about it. So in this context, there is a relationship :)miraz –miraz 2011-09-22 19:09:41 +00:00 Commented Sep 22, 2011 at 19:09 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Though the words look similar, they are not related in the way you'd thought. Their pronunciations are different, too. To wear is pronounced [wair] and is derived from O.E. werian, meaning "to clothe, put on". Weary is pronounced [weer-ee] and comes from O.E. werig, meaning "tired". World-weary has nothing to do, literally, with the act of wearing or carrying anything; as a mnemonic device, though, that may be a helpful way to think of it. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 22, 2011 at 7:39 answered Sep 22, 2011 at 7:23 user13141 user13141 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. In the words of the Oxford English Dictionary (OED), ‘world-weary’ means ‘Weary of the world; feeling or indicating feelings of weariness, boredom, or cynicism as a result of long experience of life.’ Its first recorded use is in 1750. The ‘weary’ element takes its sense from the OED’s second main definition of the adjective: ‘Discontented at the continuance or continued recurrence of something, and desiring its cessation; having one's patience, tolerance, zeal, or energy exhausted; “sick and tired” of something.’ Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 22, 2011 at 6:56 Barrie EnglandBarrie England 141k 10 10 gold badges 247 247 silver badges 408 408 bronze badges 2 I am afraid something is missing in my question. I know the definition of the phrase, however I am looking for a relationship between wear and weary. It seemed to me that a world weary person could be someone who carried the world, so that could be reason why she is unhappy about the life.miraz –miraz 2011-09-22 07:03:17 +00:00 Commented Sep 22, 2011 at 7:03 No, I don’t think that’s it. As I said, it derives from ‘weary’ meaning ‘discontented’, etc. Nothing to do with ‘wear’ as in ‘wearing clothes’. That has a quite different etymology.Barrie England –Barrie England 2011-09-22 07:22:15 +00:00 Commented Sep 22, 2011 at 7:22 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions meaning etymology See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 4Referring to wild animals as "game" 2Origin and evolution of the term 'amen corner' 2What is the origin of the expression, "I'm with stupid?" 0"to be a small person" - doeas it have second meanning? 0what's the origin of the phrase "little help"? 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2292	https://www.inchcalculator.com/convert/kilogram-to-gram/	Kilograms to Grams Converter Enter the weight in kilograms below to convert it to grams. Have a Question or Feedback? Result in Grams: Do you want to convert grams to kilograms? On this page: Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications. Cite This Page How to Convert Kilograms to Grams To convert a measurement in kilograms to a measurement in grams, multiply the weight by the following conversion ratio: 1,000 grams/kilogram. Since one kilogram is equal to 1,000 grams, you can use this simple formula to convert: grams = kilograms × 1,000 The weight in grams is equal to the weight in kilograms multiplied by 1,000. For example, here's how to convert 5 kilograms to grams using the formula above. grams = (5 kg × 1,000) = 5,000 g How Many Grams Are in a Kilogram? There are 1,000 grams in a kilogram, which is why we use this value in the formula above. 1 kg = 1,000 g Kilograms and grams are both units used to measure weight. Keep reading to learn more about each unit of measure. What Is a Kilogram? One kilogram is equal to 1,000 grams, 2.204623 pounds, or 1/1,000 of a metric ton. The formal definition of the kilogram changed in 2019. One kilogram was previously equal to the mass of the platinum-iridium bar, known as the International Prototype of the Kilogram, which was stored in Sèvres, France. The 2019 SI brochure now defines the kilogram using the Planck constant, and it is defined using the meter and second. It is equal to the mass of 1,000 cubic centimeters, or milliliters, of water. The kilogram, or kilogramme, is the SI base unit for weight and is also a multiple of the gram. In the metric system, "kilo" is the prefix for thousands, or 103. Kilograms can be abbreviated as kg; for example, 1 kilogram can be written as 1 kg. Learn more about kilograms. What Is a Gram? A gram is a unit of mass equal to 1/1,000 of a kilogram or 0.035274 ounces, and is equivalent to the mass of one cubic centimeter, or one milliliter, of water. The gram, or gramme, is an SI unit of weight in the metric system. Grams can be abbreviated as g; for example, 1 gram can be written as 1 g. Learn more about grams. Kilogram to Gram Conversion Table Table showing various kilogram measurements converted to grams. \| Kilograms \| Grams \| --- \| \| 0.001 kg \| 1 g \| \| 0.002 kg \| 2 g \| \| 0.003 kg \| 3 g \| \| 0.004 kg \| 4 g \| \| 0.005 kg \| 5 g \| \| 0.006 kg \| 6 g \| \| 0.007 kg \| 7 g \| \| 0.008 kg \| 8 g \| \| 0.009 kg \| 9 g \| \| 0.01 kg \| 10 g \| \| 0.02 kg \| 20 g \| \| 0.03 kg \| 30 g \| \| 0.04 kg \| 40 g \| \| 0.05 kg \| 50 g \| \| 0.06 kg \| 60 g \| \| 0.07 kg \| 70 g \| \| 0.08 kg \| 80 g \| \| 0.09 kg \| 90 g \| \| 0.1 kg \| 100 g \| \| 0.2 kg \| 200 g \| \| 0.3 kg \| 300 g \| \| 0.4 kg \| 400 g \| \| 0.5 kg \| 500 g \| \| 0.6 kg \| 600 g \| \| 0.7 kg \| 700 g \| \| 0.8 kg \| 800 g \| \| 0.9 kg \| 900 g \| \| 1 kg \| 1,000 g \| References More Kilogram & Gram Conversions Weight IconConvert from Kilograms Weight IconConvert to Grams Have Feedback or a Suggestion? About
2293	https://chem.libretexts.org/Courses/Sewanee%3A_The_University_of_the_South/Organic_Chemistry_Lab_Textbook/02%3A_Important_Calculations/2.02%3A_Important_Calculations_(Theoretical_Yield_and_Percent_Yield)	Skip to main content 2.2: Important Calculations (Theoretical Yield and Percent Yield) Last updated : Aug 13, 2024 Save as PDF 2.1: Important Calculations (Limiting Reagent) 3: General Techniques for Lab Page ID : 489211 ( \newcommand{\kernel}{\mathrm{null}\,}) The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control. Percent Yield Chemical reactions in the real world don't always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield. To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry.This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage. Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense. Typically, percent yields are understandably less than because of the reasons indicated earlier. However, percent yields greater than are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction. Example Potassium chlorate decomposes upon slight heating in the presence of a catalyst according to the reaction below: In a certain experiment, is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed and the oxygen gas is collected and its mass is found to be . What is the percent yield for the reaction? Solution First, we will calculate the theoretical yield based on the stoichiometry. Step 1: List the known quantities and plan the problem. Known Given: Mass of Molar mass Molar mass Unknown Theoretical yield Apply stoichiometry to convert from the mass of a reactant to the mass of a product: Step 2: Solve. The theoretical yield of is . Step 3: Think about your result. The mass of oxygen gas must be less than the of potassium chlorate that was decomposed. Now we will use the actual yield and the theoretical yield to calculate the percent yield. Step 1: List the known quantities and plan the problem. Known Actual yield Theoretical yield Unknown Percent yield Use the percent yield equation above. Step 2: Solve. Step 3: Think about your result. Since the actual yield is slightly less than the theoretical yield, the percent yield is just under . Summary Theoretical yield is calculated based on the stoichiometry of the chemical equation. The actual yield is experimentally determined. The percent yield is determined by calculating the ratio of actual yield/theoretical yield. 2.1: Important Calculations (Limiting Reagent) 3: General Techniques for Lab
2294	https://www.vedantu.com/jee-main/physics-work-done-by-torque	Work Done by Torque: Definition, Formula, Derivation & Solved Examples Sign In Courses Explore all Vedantu courses by class or target exam, starting at ₹9000 Find courses by class Find courses by target Long Term Courses Guaranteed improvement in marks or get your fees back One-to-one LIVE classes Learn one-to-one with a teacher for a personalised experience Topic specific courses Master any topic at just ₹1 Courses for Kids Courses for Kids Confidence-building & personalised learning courses for Class LKG-8 students English Superstar Age 4 - 8 Level based holistic English program Summer Camp For Lkg - Grade 10 Limited-time summer learning experience Spoken English Class 3 - 5 See your child speak fluently Learn Maths Class 1 - 5 Turn your child into a Math wizard Coding Classes Class 1 - 8 Learn to build apps and games, be future ready Free study material Get class-wise, author-wise, & board-wise free study material for exam preparation NCERT SolutionsCBSEJEE MainJEE AdvancedNEETQuestion and AnswersPopular Book Solutions Subject wise Concepts ICSE & State Boards Kids Concept Online TuitionCompetative Exams and Others Offline Centres Online Tuition Get class-wise, subject-wise, & location-wise online tuition for exam preparation Online Tuition By Class Online Tuition By Subject Online Tuition By Location More Know about our results, initiatives, resources, events, and much more Our results A celebration of all our success stories Child safety Creating a safe learning environment for every child Help India Learn Helps in learning for Children affected by the Pandemic WAVE Highly-interactive classroom that makes learning fun Vedantu Improvement Promise (VIP) We guarantee improvement in school and competitive exams Master talks Heartfelt and insightful conversations with super achievers Our initiatives Resources About us Know more about our passion to revolutionise online education Careers Check out the roles we're currently hiring for Our Culture Dive into Vedantu's Essence - Living by Values, Guided by Principles Become a teacher Apply now to join the team of passionate teachers Contact us Got questions? Please get in touch with us Vedantu Store JEE Main Physics Work Done By Torque Work Done by Torque – Formula, Derivation, and Applications Download PDF Study Material Important Questions Chapter Pages Revision Notes Formulas Difference Between Preparation Tips Exam Info Important Dates Eligibility Criteria Application Form Registration Correction Window Exam Centres Admit Card Reservation Criteria Slot Booking Weightage Exam Pattern Cut-off College Predictor Answer Key Result Colleges News Videos FAQs Syllabus Physics Syllabus Mathematics Syllabus Chemistry Syllabus Courses Class 11 JEE Course (2023-25) Class 12 JEE Course (2023-24) JEE Repeater Course (2023-24) Class 8 JEE Foundation Course Class 9 JEE Foundation Course Class 10 JEE Foundation Course JEE Main Coaching Previous Year Question Paper Subject wise question Paper JEE Main Yearwise Question Paper Practice Materials Practice Papers Sample Papers Mock Test Maths Mock Test Physics Mock Test Chemistry Mock Test Question Answers How do you derive the formula for work done by torque in rotational motion? Understanding the work done by torque is essential in rotational mechanics for JEE Main. This idea links applied force in a circle to changes in rotational energy. Many real-world mechanisms, such as wheels and engines, use this principle. Mastery of its definitions, formulas, and typical mistakes will strengthen your command of this classic topic. In rotational motion, when a force causes an object to rotate about an axis, it creates torque. The work done by torque measures how much energy is transferred as the object undergoes angular displacement. It relates directly to mechanical concepts like rotational work-energy theorem and is foundational for solving JEE Main physics problems. Definition and Physical Concept of Work Done by Torque The work done by torque is the energy imparted to or by a rotating body due to torque acting over an angular displacement. When constant torque is applied, the object rotates by a certain angle, and energy changes occur, typically seen as changes in rotational kinetic energy. For linear systems, force and displacement matter; for rotational ones, it’s torque and angular displacement. In JEE-level problems, you’ll often be asked to find work done when a torque acts through a given angle. This scenario is common in machines, flywheels, electric motors, and magnetic dipoles in a field. The distinction between torque’s role in rotation and force’s role in translation is a classic comparison point for JEE questions. Formula for Work Done by Torque and Key Variables The fundamental formula for work done by torque is: \| Physical Quantity \| Symbol \| SI Unit \| --- \| Torque \| τ \| N·m \| \| Angular Displacement \| θ \| radian \| \| Work Done by Torque \| W \| joule (J) \| W = τ × θ, if torque is constant and θ is in radians. If torque varies, use W = ∫ τ · dθ. Here, τ (torque) is the turning effect of force, θ is angular displacement, and W is work done by torque. These definitions often appear in quick-reference tables and problems throughout rotational motion of a rigid body and torque formula pages. Step-by-Step JEE Derivation of Work Done by Torque Recall the work done by a force: W = F · s, where s is linear displacement. Analogously, work in rotation is dW = τ dθ for a small angle dθ. Integrate both sides for finite rotation: W = ∫ τ · dθ. If τ is constant, this becomes W = τ θ. By the rotational work-energy theorem, W = ΔK.E. Here, ΔK.E. = ½ I (ω 2 - ω 0 2), where I is moment of inertia and ω is angular speed. This derivation is a staple in practice questions and appears in nearly every rotational mechanics chapter in work, energy, and power and rotational motion resources on Vedantu. Worked Example: Applying Work Done by Torque in a JEE Problem A constant torque of 12 N·m acts on a flywheel, causing an angular displacement of π/4 radians. Calculate the work done by torque on the flywheel. Given: τ = 12 N·m; θ = π/4 radians Formula: W = τ × θ Substitute: W = 12 × (π/4) = 3π J Final Answer: 3.14 × 3 = 9.42 J In a JEE Main context, answers should always use proper SI units and report the final value: work done is 9.42 J. Torque in magnetic field questions may use W = pE(cosθ i - cosθ f) for dipoles. Moment of inertia problems often relate torque-work to changes in rotational kinetic energy. Numerical problems sometimes require angular displacement to be found from angular acceleration data, connecting this principle to moment of inertia. Conceptual traps include mixing up force (linear work) and torque (rotational work). Exam Tips, Mistakes, and Applications of Work Done by Torque The formula W = τ θ applies only for constant torque scenarios. Always use radians (not degrees) for θ in JEE calculations. Work done by torque is zero if there’s no angular displacement. Negative work indicates torque opposes motion, e.g., braking systems. Applications include DC motors, magnetic dipoles, and rotational pulleys. Compare carefully: force–displacement (linear work) vs torque–angular displacement (rotational work). Errors often arise from using degrees or from neglecting direction of torque/displacement. Refer to difference between work and power to clarify these distinctions. Summary Table: Core Points on Work Done by Torque \| Aspect \| Rotational Motion \| Linear Motion \| --- \| Quantity Doing Work \| Torque (τ) \| Force (F) \| \| Displacement Type \| Angular Displacement (θ) \| Linear (s) \| \| Basic Formula \| W = τ θ \| W = F s \| \| SI Unit \| Joule (J) \| Joule (J) \| For last-minute revision, focus on the formula, conditions for its use, and units. Practising questions from rotational motion practice paper and work, energy, and power important questions can enhance grasp and speed. JEE aspirants often make errors in the direction (sign) or mismatching units in these problems. Review of solved examples in work, energy, and power revision notes is strongly recommended. The Vedantu Physics team regularly updates these with exam-style problems and error-proofed solutions. Competitive Exams after 12th Science CBSE JEE Main JEE Advanced NEET Olympiad CUET Latest Vedantu courses for you Grade 10 \| MAHARASHTRABOARD \| SCHOOL \| English Vedantu 10 Maharashtra Pro Lite (2025-26) Academic year 2025-26 ENGLISH Unlimited access till final school exam School Full course for MAHARASHTRABOARD students Physics Biology Chemistry Maths ~~₹27,500~~ (9% Off) ₹25,000 per year EMI starts from ₹2,083.34 per month Select and buy FAQs on Work Done by Torque – Formula, Derivation, and Applications How do you calculate work done by torque? Work done by torque is calculated by multiplying the constant torque (τ) applied to a body by the angular displacement (θ) it produces (in radians). The formula is: Work (W) = Torque (τ) × Angular displacement (θ) τ: Torque applied (in Nm) θ: Angular displacement (in radians) For a variable torque, use the integral: W = ∫ τ dθ. How is work done and torque related? Work done and torque are directly related in rotational motion, similar to how force and displacement are related in linear motion. Specifically: Work done (W) is the product of torque (τ) and angular displacement (θ). If a varying torque is applied, work is found by integrating torque with respect to angle. This relationship forms the basis of the rotational work-energy theorem. What is the formula for work done by torque? The formula for work done by torque is:W = τ × θ, where W is work, τ is torque in newton-metre (Nm), and θ is angular displacement in radians. For non-constant torque: Use W = ∫ τ dθ. This equation is essential for rotational motion calculations in physics and class 12 syllabi. What is the rotational work-energy theorem? The rotational work-energy theorem states that the work done by torque on a rigid body equals the change in its rotational kinetic energy. W = ΔK = (1/2)I(ω_2^2 - ω_1^2) I is moment of inertia ω is angular velocity This connects rotational work and energy principles in physics. What are the SI units of rotational work and torque? The SI unit of rotational work is the joule (J), and the SI unit of torque is newton-metre (Nm). Both rotational and linear work use joules Torque has the same dimensional formula as work, but represents rotation What is the work done by a constant torque acting on a body rotating through an angle? When a constant torque (τ) acts on a body and it rotates through an angle (θ) in radians, work done is: W = τ × θ This is a direct product, easy to apply in numerical problems. Derive the equation for work done by torque in rotational motion. The work done by torque is derived using the definition of work in rotational systems. For a small angular displacement dθ, small work done, dW = τ dθ. Integrating over total angle gives: W = ∫ τ dθ = τ θ for constant torque. This derivation is often asked in board exams. How is the formula for torque in terms of work written? Torque (τ) can be expressed in terms of work (W) and angular displacement (θ): τ = W / θ Here, W is work done by torque and θ is angular displacement in radians. This formula helps relate linear and rotational systems. Give an example problem: How much work is done by a torque of 5 Nm rotating a shaft through 90 degrees? First, convert angular displacement to radians: 90° = (π/2) radians. Work done, W = τ × θ = 5 Nm × (π/2) = (5π/2) Joules ≈ 7.85 J. This is a standard Class 11–12 numerical type question. What is the work done by torque in a magnetic field on a magnetic dipole? The work done by torque on a magnetic dipole of moment M in a magnetic field B: W = MB (cos θ_1 – cos θ_2) Where: M is magnetic dipole moment B is magnetic field strength θ_1 and θ_2 are initial and final angles This formula is directly used in Class 12 Physics for solving magnetic dipole problems. What is the dimensional formula of work done by torque? The dimensional formula for work done by torque is [ML 2 T-2], the same as energy and linear work. M = mass L = length T = time It helps in dimensional analysis and is vital for competitive exams. 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2295	https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_18?srsltid=AfmBOopRXyfwXN1iFQdcAeyB0CYjvc4E3Ks67RfezW9qLgs7etQxeT7n	Art of Problem Solving 2024 AMC 12A Problems/Problem 18 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2024 AMC 12A Problems/Problem 18 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2024 AMC 12A Problems/Problem 18 Contents 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 (Inscribing in Circle) 5 Solution 4(In case you have no time and that's what I did) 6 Solution 4.5 7 Solution 5 (the simplest solution ever) 8 Solution 6 9 Solution 7 (No Trig needed) 10 Solution 8 (Similar to Solution 1) 11 Video Solution, Fast, Quick, Easy! 12 See also Problem On top of a rectangular card with sides of length and , an identical card is placed so that two of their diagonals line up, as shown (, in this case). Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled in the figure? Solution 1 Let the midpoint of be . We see that no matter how many moves we do, stays where it is. Now we can find the angle of rotation () per move with the following steps: Since Vertex is the closest one and Vertex C will land on Vertex B when cards are placed. (someone insert diagram maybe) ~lptoggled, minor Latex edits by eevee9406 Solution 2 Let AC intersect BD at O, We want to find Since , So each time we rotate BD to AC for , and we need to rotate times to overlap a point with B Therefore, the answer is Note: If you don't remember ~luckuso ~minor edit by happyfroggy Solution 3 (Inscribing in Circle) By the Pythagorean Theorem, . So we have We note that Therefore, Call the rotated rectangle given in the problem's diagram Since we have So the angle formed between and is . In general, when the rectangle is rotated, the angle formed between the left vertex of the original rectangle and the right vertex of the new rectangle is So we can inscribe the rectangles in a circle with equally spaced points, and the rotated rectangles will have vertices on these points. We get the equation where is the number of rectangles used and is the number of points used on the circle for those rectangles (for example, 1 rectangle uses points, rectangles use , etc). The vertex will land on after we have used all points on the circle plus the additional points we get from creating a new rotated rectangle, so Then . ~grogg007 Solution 4(In case you have no time and that's what I did) and it eliminates all options except and . After one rotation it has turned , so to satisfy the problem, divide and get . Solution 4.5 Memorize that in a 15-75-90 right triangle, the ratios of the side lengths are , , and . So, we have that the diagonal of the rectangle forms a 15 degree angle. Drawing it out we see the answer is , and this makes sense because 15 times 6 is 90, thus rotating a vertex back to B. Solution 5 (the simplest solution ever) Look at the picture and draw the next one and continue draw down the line and then when you first hit B, count how many rectangles you’ve drawn (excluding the first which hasn’t been rotated), and you should get as the answer. ~EaZ_Shadow Solution 6 Process is the rotation around the center of the card point at the angle By applying the Law of Cosines, we get vladimir.shelomovskii@gmail.com, vvsss Solution 7 (No Trig needed) Let E and P be the intersection points, by symmetry axis of (since two rectangles are identical), we get: Now assume there exists a point on , where Thus: Connecting , we see that: So F must collide with P, and triangle is a -- triangle, so: and external angle would be ~D3rrr Solution 8 (Similar to Solution 1) Let the vertex of the second rectangular card closest to be and the unnamed vertex of the first card be . First, notice how the second rectangular card is a reflection of the first rectangular card about rotated about . This means that the diagonal from the second rectangular card would have been rotated an angle clockwise. This also means that side would have also been rotated an angle clockwise, so the angle external to would equal . When this process is repeated, a regular polygon is formed with half the number of sides of the polygon being the answer. To find , we can reflect triangle about to form an isosceles triangle. Then By the law of cosine: Simplifying: The resulting polygon from repeating the process would have sides and cards would have to have been placed. ~SandCanyon Video Solution, Fast, Quick, Easy! See also 2024 AMC 12A (Problems • Answer Key • Resources) Preceded by Problem 17Followed by Problem 19 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2296	https://www.mayoclinic.org/diseases-conditions/obsessive-compulsive-disorder/symptoms-causes/syc-20354432	Obsessive-compulsive disorder (OCD) - Symptoms and causes - Mayo Clinic This content does not have an English version. This content does not have an Arabic version. 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Type 3 or more letters to display suggested search options. When results are available, use up and down arrow keys to navigate. Press space key to fill the input with the suggestion or enter key to search with the suggestion.Search by keyword or phrase Search Diseases & Conditions Obsessive-compulsive disorder (OCD) Request an Appointment Symptoms & causes Diagnosis & treatment Doctors & departments Care at Mayo Clinic Print Overview Obsessive-compulsive disorder (OCD) features a pattern of unwanted thoughts and fears known as obsessions. These obsessions lead you to do repetitive behaviors, also called compulsions. These obsessions and compulsions get in the way of daily activities and cause a lot of distress. Ultimately, you feel driven to do compulsive acts to ease your stress. Even if you try to ignore or get rid of bothersome thoughts or urges, they keep coming back. This leads you to act based on ritual. This is the vicious cycle of OCD. OCD often centers around certain themes, such as being overly fearful of getting contaminated by germs. To ease contamination fears, you may wash your hands over and over again until they're sore and chapped. If you have OCD, you may be ashamed, embarrassed and frustrated about the condition. But treatment can be effective. Symptoms Obsessive-compulsive disorder usually includes both obsessions and compulsions. But it's also possible to have only obsession symptoms or only compulsion symptoms. You may or may not know that your obsessions and compulsions are beyond reason. But they take up a great deal of time, reduce your quality of life, and get in the way of your daily routines and responsibilities. Obsession symptoms OCD obsessions are lasting and unwanted thoughts that keeping coming back or urges or images that are intrusive and cause distress or anxiety. You might try to ignore them or get rid of them by acting based on ritual. These obsessions usually intrude when you're trying to think of or do other things. Obsessions often have themes, such as: Fear of contamination or dirt. Doubting and having a hard time dealing with uncertainty. Needing things to be orderly and balanced. Aggressive or horrific thoughts about losing control and harming yourself or others. Unwanted thoughts, including aggression, or sexual or religious subjects. Examples of obsession symptoms include: Fear of being contaminated by touching objects others have touched. Doubts that you've locked the door or turned off the stove. Intense stress when objects aren't orderly or facing a certain way. Images of driving your car into a crowd of people. Thoughts about shouting obscenities or not acting the right way in public. Unpleasant sexual images. Staying away from situations that can cause obsessions, such as shaking hands. Compulsion symptoms OCD compulsions are repetitive behaviors that you feel driven to do. These repetitive behaviors or mental acts are meant to reduce anxiety related to your obsessions or prevent something bad from happening. But taking part in the compulsions brings no pleasure and may offer only limited relief from anxiety. You may make up rules or rituals to follow that help control your anxiety when you're having obsessive thoughts. These compulsions are beyond reason and often don't relate to the issue they're intended to fix. As with obsessions, compulsions usually have themes, such as: Washing and cleaning. Checking. Counting. Ordering. Following a strict routine. Demanding reassurance. Examples of compulsion symptoms include: Hand-washing until your skin becomes raw. Checking doors over and over again to make sure they're locked. Checking the stove over and over again to make sure it's off. Counting in certain patterns. Silently repeating a prayer, word or phrase. Trying to replace a bad thought with a good thought. Arranging your canned goods to face the same way. Severity varies OCD usually begins in the teen or young adult years, but it can start in childhood. Symptoms usually begin over time and tend to vary in how serious they are throughout life. The types of obsessions and compulsions you have also can change over time. Symptoms generally get worse when you are under greater stress, including times of transition and change. OCD, usually thought to be a lifelong disorder, can have mild to moderate symptoms or be so severe and time-consuming that it becomes disabling. When to see a doctor There's a difference between being a perfectionist — someone who needs flawless results or performance — and having OCD. OCD thoughts aren't simply excessive worries about real issues in your life or liking to have things clean or arranged in a specific way. If your obsessions and compulsions affect your quality of life, see your doctor or mental health professional. Request an appointment There is a problem with information submitted for this request. Review/update the information highlighted below and resubmit the form. From Mayo Clinic to your inbox Sign up for free and stay up to date on research advancements, health tips, current health topics, and expertise on managing health. Click here for an email preview. Email Address 1 Error Email field is required Error Include a valid email address Learn more about Mayo Clinic’s use of data. We use the data you provide to deliver you the content you requested. To provide you with the most relevant and helpful information, we may combine your email and website data with other information we have about you. If you are a Mayo Clinic patient, we will only use your protected health information as outlined in our Notice of Privacy Practices. You may opt out of email communications at any time by clicking on the unsubscribe link in the email. Subscribe! Thank you for subscribing! You'll soon start receiving the latest Mayo Clinic health information you requested in your inbox. Sorry something went wrong with your subscription Please, try again in a couple of minutes Retry Causes The cause of obsessive-compulsive disorder isn't fully understood. Main theories include: Biology.OCD may be due to changes in your body's natural chemistry or brain functions. Genetics.OCD may have a genetic component, but specific genes have yet to be found. Learning. Obsessive fears and compulsive behaviors can be learned from watching family members or learning them over time. Risk factors Factors that may raise the risk of causing obsessive-compulsive disorder include: Family history. Having parents or other family members with the disorder can raise your risk of getting OCD. Stressful life events. If you've gone through traumatic or stressful events, your risk may increase. This reaction may cause the intrusive thoughts, rituals and emotional distress seen in OCD. Other mental health disorders.OCD may be related to other mental health disorders, such as anxiety disorders, depression, substance abuse or tic disorders. Complications Issues due to obsessive-compulsive disorder include: Excessive time spent taking part in ritualistic behaviors. Health issues, such as contact dermatitis from frequent hand-washing. Having a hard time going to work or school or taking part in social activities. Troubled relationships. Poor quality of life. Thoughts about suicide and behavior related to suicide. Prevention There's no sure way to prevent obsessive-compulsive disorder. However, getting treated as soon as possible may help keep OCD from getting worse and disrupting activities and your daily routine. By Mayo Clinic Staff Obsessive-compulsive disorder (OCD) care at Mayo Clinic Request an appointment Diagnosis & treatment Dec. 21, 2023 Print Show references Kara PJ, et al. Deep brain stimulation for obsessive compulsive disorder: Evolution of surgical stimulation target parallels changing model of dysfunctional brain circuits. Frontiers in Neuroscience. 2019; doi:10.3389/fnins.2018.00998. Allscripts EPSi. Mayo Clinic. 2019. Obsessive-compulsive and related disorders. In: Diagnostic and Statistical Manual of Mental Disorders DSM-5-TR. 5th ed. American Psychiatric Association; 2022. Accessed May 17, 2023. Blair Simpson H. Pharmacotherapy for obsessive-compulsive disorder in adults. Accessed May 17, 2023. Blair Simpson H. Obsessive-compulsive disorder in adults: Epidemiology, pathogenesis, clinical manifestations, course, and diagnosis. Accessed May 17, 2023. Abromowitz J. Psychotherapy for obsessive-compulsive disorder in adults. Accessed May 17, 2023. Obsessive-compulsive disorder. National Alliance on Mental Illness. Accessed May 17, 2023. Obsessive-compulsive disorder in children and adolescents. American Academy of Child & Adolescent Psychiatry. Accessed May 17, 2023. Obsessive-compulsive disorder: When unwanted thoughts or repetitive behaviors take over. National Institute of Mental Health. Accessed May 17, 2023. Mental health medications. National Institute of Mental Health. Accessed May 17, 2023. AskMayoExpert. Obsessive-compulsive disorder (OCD). Mayo Clinic; 2023. Depression. National Institute of Mental Health. Accessed May 17, 2023. Obsessive-compulsive disorder (OCD). Merck Manual Professional Version. Accessed May 17, 2023. Woody EZ, et al. Obsessive compulsive disorder (OCD): Current treatments and a framework for neurotherapeutic research. Advances in Pharmacology. 2019; doi:10.1016/bs.apha.2019.04.003. Cervin M. Obsessive-compulsive disorder: Diagnosis, clinical features, nosology and epidemiology. Psychiatric Clinics of North America. Accessed May 17, 2023. Medical review (expert opinion). Mayo Clinic. June 23, 2023. Related Associated Procedures Cognitive behavioral therapy Deep brain stimulation Electroconvulsive therapy (ECT) Psychotherapy Transcranial magnetic stimulation Show more associated procedures Obsessive-compulsive disorder (OCD) Symptoms & causes Diagnosis & treatment Doctors & departments Care at Mayo Clinic Advertisement Mayo Clinic does not endorse companies or products. Advertising revenue supports our not-for-profit mission. Advertising & Sponsorship Policy Opportunities Ad Choices Mayo Clinic Press Check out these best-sellers and special offers on books and newsletters from Mayo Clinic Press. NEW: Mayo Clinic Guide to Better Sleep Listen to Health Matters Podcast Mayo Clinic on Incontinence The Essential Diabetes Book FREE Mayo Clinic Diet Assessment Mayo Clinic Health Letter - FREE book CON-20199571 Diseases & Conditions Obsessive-compulsive disorder (OCD) Final days of 5X Challenge! 5X My Gift! 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2297	https://en.wikiversity.org/wiki/Parabola	Jump to content Search [dismiss] Why create a Wikiversity account? Contents 1 The General Parabola 2 An Example 3 Reverse-Engineering the Parabola 3.1 Method 1. By analytical geometry 3.2 Method 2. By algebra 4 Slope of the Parabola 4.1 Point at given slope 4.1.1 Implementation 4.1.1.1 Examples 5 Parabola and any chord 6 Parabola and two tangents 7 Area enclosed between parabola and chord 7.1 Introduction 7.2 Proof 7.3 A worked example 7.3.1 Method 1. By chord and parallel tangent 7.3.2 Method 2. By identifying the basic parabola. Parabola Add links Resource Discuss Read Edit Edit source View history Tools Actions Read Edit Edit source View history General What links here Related changes Permanent link Page information Cite this page Get shortened URL Download QR code Wikimedia Projects Commons Wikibooks Wikidata Wikinews Wikipedia Wikiquote Wikisource Wikispecies Wikivoyage Wiktionary Meta-Wiki Outreach MediaWiki Wikimania Print/export Create a book Download as PDF Printable version In other projects Appearance From Wikiversity In Cartesian geometry in two dimensions the is the locus of a point that moves so that it is always equidistant from a fixed point and a fixed line. The fixed point is called the and the fixed line is called the . Distance from to is non-zero. See Figure 1. The focus is point and the directrix is line The , point , is half-way between focus and directrix. A is the segment of a line joining any two distinct points of the parabola. The line segment is a chord. Because chord passes through the focus , it is called a The focal chord parallel to the directrix is called the The line through the focus and perpendicular to the directrix is the , sometimes called . Let an arbitrary point on the curve be . By definition, . This expression expanded gives: : . If the equation of the curve is expressed as: , then where the has coordinates and is the distance form vertex to focus. If the directrix is parallel to the axis, then the parabola is the same as the familiar quadratic function. The general parabola allows for a directrix anywhere with any orientation. The General Parabola [edit \| edit source] Let the directrix be where at least one of is non-zero. Let the focus be . Let be any point on the curve. Distance from point to focus = . Distance from point to directrix () = where . By definition these two lengths are equal: . . Square both sides: . . Expand and the result is: . has the form of the equation of the conic section where because this curve is a parabola. An Example [edit \| edit source] \| \| \| --- \| \| See Figure 2. The equation of the parabola is derived as follows: The equation of the parabola in Figure 2 is: Equation of directrix in normal form: Distance from to Distance from vertex to focus . Therefore, curve has shape of where . \| \| \| Caution: An interesting situation occurs if the focus is on the directrix. Consider the directrix: and the focus which is on the directrix. In this case the "parabola" has equation: . This seems to be the equation of a parabola because , but look closely. . The result is a line through the focus and normal to the directrix. If you solve for using the algebraic solutions, you will produce the values as above. However, the distance between focus and directrix = where distance \| \| Reverse-Engineering the Parabola [edit \| edit source] \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Given a parabola in form the aim is to produce the directrix and the focus. We will solve the example shown in Figure 3: , where: . Method 1. By analytical geometry[edit \| edit source] \| \| \| \| \| \| --- --- \| Find two tangents that intersect at a right angle. The simplest to find are those that are parallel to the axes. \| \| \| Put the equation of the parabola in the form of a quadratic in . At the tangent there is exactly one value of . Therefore the discriminant must be . In the general parabola therefore . In this example . Point has coordinates . The line is tangent to the curve at and has equation: . \| \| \| \| Put the equation of the parabola in the form of a quadratic in : . By using calculations similar to the above, and . Point has coordinates . The line is tangent to the curve at and has equation: . \| \| \| \| Point at the intersection of the two tangents has coordinates , and point is on the directrix, the equation of which is: . Put known values into the equation of the directrix: . Therefore and the equation of the directrix is: . \| \| \| \| The coordinates of points are known. Therefore chord is defined as: . Draw the line perpendicular to . The line is defined as: . Point at the intersection of lines is the focus with coordinates . \| \| Method 2. By algebra[edit \| edit source] \| \| \| \| --- \| After rearranging the above values, there are three equations to be solved for three unknowns: : The solutions are: \| \| \| If is the parabola becomes the quadratic: and: The directrix has equation: . If is , then: The directrix has equation: . \| \| \| \| If is the parabola becomes the quadratic: and: The directrix has equation: . If is , then: The directrix has equation: . These values agree with the corresponding values in the graph of to the right. \| \| \| Slope of the Parabola [edit \| edit source] Consider parabola and line Let point be any point on the line. Therefore Let the line intersect the parabola. Substitute the above value of into the equation of the parabola and expand: We want the line to be tangent to the curve. Therefore must have exactly one value and the discriminant is Discriminant = The above discriminant is a quadratic in : where: If the point is on the curve, then the line touches the curve at and: because for a parabola. When slope is displayed in this format, we see that slope is vertical if The line is tangent to the curve. \| \| \| Let the equation of a line be: in which case By using calculations similar to the above it can be shown that: . therefore: When slope is displayed in this format, we see that slope is zero if . The line is tangent to the curve. \| \| \| \| This examination of the parabola has produced two expressions for slope of the parabola: . where the point is any arbitrary point on the curve. Therefore . This formula for contains both tangents parallel to the axes and is derived without calculus. \| \| \| \| The formula from calculus below is simpler and unambiguous concerning sign. \| \| \| \| The slope of the parabola is where or: The slope of the parabola is vertical where or: \| \| \| \| Caution: If the curve is , the slope can never be vertical. If the curve is , the slope can never be . \| \| \| \| If and , the equation of the parabola becomes: and: \| Point at given slope [edit \| edit source] \| \| \| --- \| \| Given parabola defined by and slope where at least one of is non-zero, calculate point at which the slope is Let Then Let Then Substitute in the equation of the parabola and As shown below, with a little manipulation of the data, the same formula can be used to calculate \| \| \| Equation above is the equation of a straight line with slope Substitute for and the slope of becomes Equation is that of a line parallel to the axis of symmetry of the parabola. It is possible for both both to equal in which case the calculation of above fails as an attempt to divide by See caution under "Slope of the Parabola" above. \| \| Implementation [edit \| edit source] \| \| \| # python code def pointAtGivenSlope(parabola, tangent): s, t = tangent if s == t == 0: print('pointAtGivenSlope(): both s,t can not be 0.') return None def calculate_y(parabola, tangent): A, B, C, D, E, F = parabola s, t = tangent G = B s + 2 A t; H = 2 C s + B t; I = E s + D t return -(A I I - D G I + F G G)/(2 A H I - B G I - D G H + E G G) y = calculate_y(parabola, tangent) A, B, C, D, E, F = parabola x = calculate_y((C, B, A, E, D, F),(t, s)) return x, y \| Examples [edit \| edit source] \| \| \| \| --- \| A parabola is defined as \| \| \| Calculate coordinates of vertex. At vertex tangent has same slope as directrix. # python code parabola = A, B, C, D, E, F = 9, - 24, 16, 70, - 260, 1025 a = C.5 b = - B/(2 a) tangent = - a, b result = pointAtGivenSlope(parabola, tangent) print(result) (0.2,6.4) \| \| \| \| Calculate point at which tangent is vertical. # python code parabola = A, B, C, D, E, F = 9, - 24, 16, 70, - 260, 1025 tangent = 1, 0 result = pointAtGivenSlope(parabola, tangent) print(result) (-0.25,7.9375) \| \| Parabola and any chord [edit \| edit source] Refer to Figure 4. The curve has equation: The chord has equation: . Point has coordinates Line is tangent to the curve at Line is tangent to the curve at This section shows that point has coordinates where Point has coordinates where: , , and slope of tangent Point has coordinates where: , , and slope of tangent Points have coordinates Equation of tangent Equation of tangent At point of intersection Review the coordinates of points . The line with equation bisects the line segment and also the chord at point . Any chord parallel to has two tangents that intersect on the line . Any chord parallel to is bisected by the line . The coordinate of point Any chord that passes through the point has two tangents that intersect on the line . Angle \| \| \| Using: If , point is above the is positive and ° °. is acute and, as increases, increases, approaching °. If , point is on the ° and the line is the . If , point is below the is negative and ° °. is obtuse and, as increases, decreases, approaching °. \| Parabola and two tangents [edit \| edit source] Refer to Figure 5. The curve has equation: Point with coordinates is any point on the line Line is tangent to the curve at point . Line is tangent to the curve at point . This section shows that the chord passes through the point Equation of any line through point Let this line intersect the curve: We want this line to be a tangent to the curve, therefore has exactly one value and the discriminant is : where slope of tangent . slope of tangent . Slope of chord Intercept of chord on the axis Angle If ° In the basic parabola where the has coordinates . or If then and are the same point, the chord passes through the and the line is the . Area enclosed between parabola and chord [edit \| edit source] Introduction [edit \| edit source] \| \| \| See Figure 6. The curve is: . Integral is: . Area under curve Area under curve area area Area between chord and curve . Consider the chord . Call this the with value . The tangent through the origin is parallel to , and the perpendicular distance between is . Call this distance the with value . In this case the area enclosed between chord and curve the same as that calculated earlier. Consider the chord and curve By inspection the area between chord and curve Chord has equation in normal form. The line is parallel to and touches the curve at Distance from to chord Length of Area between chord and curve the same as that calculated earlier. These observations suggest that the area enclosed between chord and curve \| Proof [edit \| edit source] \| \| \| We prove this identity for the general case. See Figure 7. Slope of chord Find equation of chord Equation of chord Find equation of tangent We choose a value of that gives exactly one value. Therefore discriminant Equation of tangent in normal form: Equation of chord in normal form: Therefore distance between chord and tangent where Length of chord Area under chord Area under curve Area between chord and curve The aim is to prove that: or or or or where: Therefore and area enclosed between curve and chord = where is the length of the chord, and is the perpendicular distance between chord and tangent parallel to chord. \| A worked example [edit \| edit source] Consider parabola: and chord: The aim is to calculate area between chord and curve. Calculate the points at which chord and parabola intersect: Method 1. By chord and parallel tangent [edit \| edit source] \| \| \| Length of chord Equation of chord in normal form: Equation of parallel tangent in normal form: where and and Equation of chord in normal form: Equation of parallel tangent in normal form: Distance between chord and parallel tangent Area enclosed between chord and curve \| Method 2. By identifying the basic parabola. [edit \| edit source] \| \| \| See Figure 9. Calculate directrix, focus and axis. Focus is distance from directrix. Curve has the shape of where . Axis of symmetry has equation: Distance from point to axis of symmetry Distance from point to axis of symmetry Area between chord and curve where Calculate as above and area enclosed between chord and curve . \| Retrieved from " Add topic
2298	https://www.webelements.com/iodine/thermochemistry.html	\| Se \| Br \| Kr \| --- \| Te \| I \| Xe \| \| Po \| At \| Rn \| Actinium ☢ Aluminium Aluminum Americium ☢ Antimony Argon Arsenic Astatine ☢ Barium Berkelium ☢ Beryllium Bismuth Bohrium ☢ Boron Bromine Cadmium Caesium Calcium Californium ☢ Carbon Cerium Cesium Chlorine Chromium Cobalt Copernicium ☢ Copper Curium ☢ Darmstadtium ☢ Dubnium ☢ Dysprosium Einsteinium ☢ Erbium Europium Fermium ☢ Flerovium ☢ Fluorine Francium Gadolinium Gallium Germanium Gold Hafnium Hassium ☢ Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium ☢ Lead Lithium Livermorium ☢ Lutetium Magnesium Manganese Meitnerium ☢ Mendelevium ☢ Mercury Molybdenum Moscovium ☢ Neodymium Neon Neptunium Nickel Nihonium ☢ Niobium Nitrogen Nobelium Oganesson ☢ Osmium Oxygen Palladium Phosphorus Platinum Plutonium ☢ Polonium Potassium Praseodymium Promethium ☢ Protactinium ☢ Radium ☢ Radon ☢ Rhenium Rhodium Roentgenium ☢ Rubidium Ruthenium Rutherfordium ☢ Samarium Scandium Seaborgium ☢ Selenium Silicon Silver Sodium Strontium Sulfur Sulphur Tantalum Technetium Tellurium Tennessine ☢ Terbium Thallium Thorium ☢ Thulium Tin Titanium Tungsten Uranium ☢ Vanadium Xenon Ytterbium Yttrium Zinc Zirconium Iodine - 53I ▸▸ 🇬🇧 Iodine 🇺🇦 Йод 🇨🇳 碘 🇳🇱 Jood 🇫🇷 Iode 🇩🇪 Iod 🇮🇱 יוד 🇮🇹 Iodio 🇯🇵 ヨウ素 🇵🇹 Iodo 🇪🇸 Yodo 🇸🇪 Jod 🇷🇺 Иод Iodine - 53I: thermochemistry and thermodynamics ▸▸ I Essentials Physical properties Electron shell data Atom sizes Electronegativity Isotopes and NMR Crystal structure Thermochemistry History Uses Geology Biology Binary compounds Compound properties Element reactions List all I properties Temperatures Melting point: 386.85 [113.7 °C (236.66 °F)] K Boiling point: 457.4 [184.3 °C (363.7 °F)] K Liquid range: 70.55 K Critical temperature: 819 [546 °C (1015 °F)] K Superconduction temperature: (no data) K Enthalpies Enthalpy of fusion: 7.76 (per mol I atoms) kJ mol-1 Enthalpy of vaporisation: 20.9 (per mole I atoms) kJ mol-1 Enthalpy of atomisation: 107 kJ mol-1 Thermodynamic data This table gives a few thermodynamic data for iodine. Most values are those given in the NBS technical notes (reference 1) after conversion from the units used within those notes. Values labelled with an asterisk () are Committee on Data for Science and Technology (CODATA) agreed values for the thermodynamic properties of key chemical substances (reference 2). These values are published in a number of places including the WWW (reference 3). \| State \| ΔfH° \| ΔfG° \| S° \| CpH \| H°298.15‑H°0 \| --- --- --- \| \| Units \| kJ mol‑1 \| kJ mol‑1 \| J K‑1 mol‑1 \| J K‑1 mol‑1 \| kJ mol‑1 \| \| Solid \| 0 \| 0 \| 116.14 ± 0.30 \| 54.44 \| 13.196 ± 0.040 \| \| Gas (I2) \| 62.42 ± 0.08 \| 19.36 \| 260.687 ± 0.005 \| 36.9 \| 10.116 ± 0.001 \| \| Gas (atoms) \| 106.76 ± 0.04 \| 70.28 \| 180.787 ± 0.004 \| 20.79 \| 6.197 ± 0.001 \| Notes This tables gives a few thermodynamic data. Most values are those given in the NBS technical notes (reference 1) after conversion from the units used within those notes. Values labelled with an asterisk () are Committee on Data for Science and Technology (CODATA) agreed values for the thermodynamic properties of key chemical substances (reference 2). These values are published in a number of places including the WWW (reference 3). References R.H. Schumm, D.D. Wagman, S. Bailey, W.H. Evans, and V.B. Parker in National Bureau of Standards (USA) Technical Notes 270-1 to 270-8, 1973. J.D. Cox, DD., Wagman, and V.A. Medvedev, CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, USA, 1989.
2299	https://stackoverflow.com/questions/19676109/how-to-generate-all-the-permutations-of-a-multiset	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog How to generate all the permutations of a multiset? Ask Question Asked Modified 2 years, 11 months ago Viewed 13k times This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. A multi-set is a set in which all the elements may not be unique.How to enumerate all the possible permutations among the set elements? string algorithm combinations combinatorics backtracking Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications asked Oct 30, 2013 at 7:19 piyukrpiyukr 64133 gold badges1212 silver badges1818 bronze badges 1 1 I'm assuming that you know how to generate the permutations for a unique set. For a multi-set, sort the set, and just keep the same logic with one change, discard the permutation for next number in set if it is same as the current number. – Abhishek Bansal Commented Oct 30, 2013 at 7:42 Add a comment \| 6 Answers 6 Reset to default This answer is useful 13 Save this answer. Show activity on this post. Generating all the possible permutations and then discarding the repeated ones is highly inefficient. Various algorithms exist to directly generate the permutations of a multiset in lexicographical order or other kind of ordering. Takaoka's algorithm is a good example, but probably that of Aaron Williams is better moreover, it has been implemented in the R package ''multicool''. Btw, if you just want the total number of distinct permutations, the answer is the Multinomial coefficient: e.g., if you have, say, n_a elements 'a', n_b elements 'b', n_c elements 'c', the total number of distinct permutations is (n_a+n_b+n_c)!/(n_a!n_b!n_c!) Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Jul 1, 2014 at 9:46 ecatmur 158k2828 gold badges311311 silver badges386386 bronze badges answered Jan 22, 2014 at 13:51 ruggeroruggero 44033 silver badges77 bronze badges 1 2 The link is now broken. The algorithm with implementations in multiple languages is now at this link. – Varun Vejalla Commented Oct 20, 2019 at 15:16 Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. This is my translation of the Takaoka multiset permutations algorithm into Python (available here and at repl.it): `` def msp(items): '''Yield the permutations ofitemswhere items is either a list of integers representing the actual items or a list of hashable items. The output are the unique permutations of the items given as a list of integers 0, ..., n-1 that represent the n unique elements initems`. Examples for i in msp('xoxox'): ... print(i) [1, 1, 1, 0, 0] [0, 1, 1, 1, 0] [1, 0, 1, 1, 0] [1, 1, 0, 1, 0] [0, 1, 1, 0, 1] [1, 0, 1, 0, 1] [0, 1, 0, 1, 1] [0, 0, 1, 1, 1] [1, 0, 0, 1, 1] [1, 1, 0, 0, 1] Reference: "An O(1) Time Algorithm for Generating Multiset Permutations", Tadao Takaoka ''' def visit(head): (rv, j) = ([], head) for i in range(N): (dat, j) = E[j] rv.append(dat) return rv u = list(set(items)) E = list(reversed(sorted([u.index(i) for i in items]))) N = len(E) # put E into linked-list format (val, nxt) = (0, 1) for i in range(N): E[i] = [E[i], i + 1] E[-1][nxt] = None head = 0 afteri = N - 1 i = afteri - 1 yield visit(head) while E[afteri][nxt] is not None or E[afteri][val] < E[head][val]: j = E[afteri][nxt] # added to algorithm for clarity if j is not None and E[i][val] >= E[j][val]: beforek = afteri else: beforek = i k = E[beforek][nxt] E[beforek][nxt] = E[k][nxt] E[k][nxt] = head if E[k][val] < E[head][val]: i = k afteri = E[i][nxt] head = k yield visit(head) ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered May 19, 2017 at 2:31 smichrsmichr 19.5k33 gold badges3131 silver badges4343 bronze badges Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. sympy provides multiset_permutations. from the doc: ``` from sympy.utilities.iterables import multiset_permutations from sympy import factorial [''.join(i) for i in multiset_permutations('aab')] ['aab', 'aba', 'baa'] factorial(len('banana')) 720 sum(1 for _ in multiset_permutations('banana')) 60 ``` Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Sep 5, 2021 at 5:15 answered Oct 13, 2020 at 14:15 hiro protagonisthiro protagonist 47.2k1717 gold badges9393 silver badges119119 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Optimisation of smichr's answer, I unzipped the nxts to make the visit function more efficient with an accumulate() (the map() is faster than a list comprehension and it seemed shallow and pedantic to have to nest it in a second one with a constant index) ``` from itertools import accumulate def msp(items): def visit(head): '''(rv, j) = ([], head) for i in range(N): (dat, j) = E[j] rv.append(dat) return(rv)''' #print(reduce(lambda e,dontCare: (e+[E[e]],nxts[e]),range(N),([],head))) #print(list(map(E.getitem,accumulate(range(N-1),lambda e,N: nxts[e],initial=head)))) return(list(map(E.getitem,accumulate(range(N-1),lambda e,N: nxts[e],initial=head)))) u=list(set(items)) E=list(sorted(map(u.index,items))) N=len(E) nxts=list(range(1,N))+[None] head=0 i,ai,aai=N-3,N-2,N-1 yield(visit(head)) while aai!=None or E[ai]>E[head]: beforek=(i if aai==None or E[i]>E[aai] else ai) k=nxts[beforek] if E[k]>E[head]: i=k nxts[beforek],nxts[k],head = nxts[k],head,k ai=nxts[i] aai=nxts[ai] yield(visit(head)) ``` Here are the test results (the second has (13!/2!/3!/3!/4!)/10! = 143/144 times as many permutations but takes longer due to being more of a multiset, I suppose), mine seems 9% and 7% faster respectively: ``` cProfile.run("list(msp(list(range(10))))") cProfile.run("list(msp([0,1,1,2,2,2,3,3,3,3,4,4,4]))") original: 43545617 function calls in 28.452 seconds 54054020 function calls in 32.469 seconds modification: 39916806 function calls in 26.067 seconds 50450406 function calls in 30.384 seconds ``` I have insufficient reputation to comment upon answers, but for an items input list, Martin Böschen's answer has time complexity the product of the factorials of the number of instances of each element value times greater, or ``` reduce(int.mul,map(lambda n: reduce(int.mul,range(1,n+1)),map(items.count,set(items)))) ``` This can grow large quickly when computing large multisets with many occurrences. For instance, it will take 1728 times longer per permutation for my second example than my first. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Aug 24, 2022 at 1:09 DroneBetterDroneBetter 12122 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. There are O(1) (per permutation) algorithms for multiset permutation generation, for example, from Takaoka (with implementation) Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Oct 30, 2013 at 9:03 MBoMBo 80.6k55 gold badges6161 silver badges9898 bronze badges Add a comment \| This answer is useful -1 Save this answer. Show activity on this post. You can reduce your problem to enumerate all permutations of a list. The typcial permutation generation algorithm takes a list and don't check if elements are equal. So you only need to generate a list out of your multiset, and feed it to your permutation generating algorithm. For example, you have the multiset {1,2,2}. You transform it to the list [1,2,2]. And generate all permutations, for example in python: ``` import itertools as it for i in it.permutations([1,2,2]): print i ``` And you will get the output ``` (1, 2, 2) (1, 2, 2) (2, 1, 2) (2, 2, 1) (2, 1, 2) (2, 2, 1) ``` The problem is, that you get some permutations repeatedly. A simple solution would be just to filter them out: ``` import itertools as it permset=set([i for i in it.permutations([1,2,2])]) for x in permset: print x ``` Output: ``` (1, 2, 2) (2, 2, 1) (2, 1, 2) ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Oct 30, 2013 at 7:50 answered Oct 30, 2013 at 7:37 Martin BöschenMartin Böschen 1,7891616 silver badges2323 bronze badges 1 Here permutations are getting repeated. – Abhishek Bansal Commented Oct 30, 2013 at 7:39 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions string algorithm combinations combinatorics backtracking See similar questions with these tags. The Overflow Blog Documents: The architect’s programming language Research roadmap update, August 2025 Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... 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